url
string | fetch_time
int64 | content_mime_type
string | warc_filename
string | warc_record_offset
int32 | warc_record_length
int32 | text
string | token_count
int32 | char_count
int32 | metadata
string | score
float64 | int_score
int64 | crawl
string | snapshot_type
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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
http://mathhelpforum.com/algebra/10255-distance-rate-time-help.html | 1,529,682,022,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864546.30/warc/CC-MAIN-20180622143142-20180622163142-00227.warc.gz | 197,066,578 | 10,114 | # Thread: Distance, Rate, and Time... help?
1. ## Distance, Rate, and Time... help?
"A canoeist paddled 2 hours with a 3 km/h current to a fishing site. The return trip against the same current took 6 hours. Find the speed of the canoe in still water."
I'm supposed to solve this problem with two variables and write two equations, but I can't seem to figure out how to do that. If anyone could lend me a hand, it would be much appreciated.
2. Originally Posted by firefly_senshi6
"A canoeist paddled 2 hours with a 3 km/h current to a fishing site. The return trip against the same current took 6 hours. Find the speed of the canoe in still water."
I'm supposed to solve this problem with two variables and write two equations, but I can't seem to figure out how to do that. If anyone could lend me a hand, it would be much appreciated.
Assume the canoe travels at a speed v in still water.
The trip downstream was with the current, so according to an observer on the shore the boat moved with a speed of v + 3 km/h and travelled a distance of x km in 2 hours.
$\displaystyle v + 3 = \frac{x}{2}$
The trip upstream was against the current, so according to an observer on the short the boat moved with a speed of v - 3 km/h and travelled a distance (again) of x km in 6 hours.
$\displaystyle v - 3 = \frac{x}{6}$
We want v, so solve the first equation for x:
$\displaystyle x = 2(v + 3)$
Insert this value of x into the second equation:
$\displaystyle v - 3 = \frac{2(v+3)}{6}$
Now solve for v. I get v = 6 km/h.
-Dan
3. Ahh, thank you so much!! It makes sense now. | 420 | 1,577 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-26 | latest | en | 0.967148 |
http://blueshrapnel.com/cms/recursion/ | 1,695,804,798,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510284.49/warc/CC-MAIN-20230927071345-20230927101345-00111.warc.gz | 6,862,859 | 19,867 | # Recursion
## Steps in drawing out the recursive call.
How to write a method that calls itself. https://www.youtube.com/watch?v=MyzFdthuUcA – Dave Feinberg’s video series.
1. Write “if”.
There must be at least 2 cases:, a recursive case (where the method calls itself) and a base case (where the method does not).
2. Handle the simplest case(s). “Base Case”
Simplest = no recursive call needed (no further loop)
3. Write the recursive call(s).
On the next simpler input/state, it may help to store the recursive call
4. Assume the recursive call works .
Ask yourself: What does it do?
Ask yourself: How does it help?
## Factorial
### Base Case
If n! is defined as the product of all positive integers from 1 to n, then:
1! = 1*1 = 1
2! = 1*2 = 2
3! = 1*2*3 = 6
4! = 1*2*3*4 = 24
n! = 1*2*3*…*(n-2)*(n-1)*n
and so on.
Logically, n! can also be expressed n*(n-1)! .
Therefore, at n=1, using n! = n*(n-1)!
1! = 1*0!
which simplifies to 1 = 0!
### Java Recursive Factorial Method
```// precondition: n >= 0
public static in fact(int n) {
if (n=0) return 1;
else return fact(n-1) * n;
}
``` | 360 | 1,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-40 | latest | en | 0.784548 |
https://www.exceldemy.com/excel-divide-without-remainder/ | 1,709,517,905,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476409.38/warc/CC-MAIN-20240304002142-20240304032142-00554.warc.gz | 747,163,510 | 74,627 | # How to Divide Without Remainder in Excel (3 Effective Methods)
Get FREE Advanced Excel Exercises with Solutions!
Division is a very important mathematical operation. When a number (dividend) can’t be divided exactly by another number (divisor) then there remains a remainder. In that case, the result becomes a number with a fraction. We can easily remove the fractional part and get the quotient in Excel. In this article, I will show 3 effective methods to divide without remainder in Excel.
## How to Divide Without Remainder in Excel: 3 Effective Methods
We can divide a number without remainder in Excel following different methods. Here, I will show 3 effective methods to divide a number and get the quotient.
### 1. Use QUOTIENT Function
In this section, I will use the QUOTIENT function of Excel to divide without remainder. For demonstration, I have used a dataset containing Dividend, Divisor, and Quotient.
#### 1.1 Apply Numeric Value to Divide Without Remainder in Excel
We can use dividend and divisor numbers directly as the arguments of the QUOTIENT function. Let’s walk through the procedures.
• First, write the following formula in Cell D5 and press Enter.
`=QUOTIENT(112,5)`
In the formula, I have given 122 (dividend) and 5 (divisor) as the argument of the QUOTIENT function.
• After that, drag the Fill Handle down to cell D10.
• Finally, you will get results like the picture below.
#### 1.2 Use Cell Reference to Perform Division Without Remainder
Instead of directly including the numbers as the arguments of the QUOTIENT function, we can use cell reference also. I am showing the stepwise procedure to do that.
• First, write the following formula in cell D5 and press Enter.
`=QUOTIENT(B5,C5)`
• Further, use the AutoFill option of Excel to copy the formula in the cells below.
• Consecutively, we will see the result without the remainder in each cell.
### 2. Apply TRUNC Function
Another way to divide without remainder is to round the fractional result by using the TRUNC function. I am showing the stepwise procedures to do that in the following section.
• First, write the following formula in cell D5 and hit Enter.
`=TRUNC(B5/C5)`
• Afterward, use the Fill Handle to copy the formula in the following cells.
• Finally, we will see the division result without the remainder.
In the formula, the TRUNC function removes the fractional part which is the remainder of the result of division (B5/C5).
### 3. Insert ROUNDDOWN Function
We can also use the ROUNDDOWN function of Excel to remove the fractional part of the division result and get the quotient value. Follow the stepwise procedures given below.
• First, write the following formula in cell D5.
`=ROUNDDOWN(B5/C5,0)`
• Then, hit Enter.
• Further, use the Fill Handle to copy the formula in the following cells.
• Finally, we will see the results without the fractional part in each cell.
In the formula, the ROUNDDOWN function removes the fractional part from the division value (B5/C5) as we included 0 as the argument.
## How to Find Remainder in Excel
When a dividend number isn’t completely divisible by the divisor, there remains a remainder. In this section, I will show the method to find the remainder in Excel. Follow the stepwise procedures given below.
• First, write the following formula in cell D5.
`=MOD(B5,C5)`
• Then, press Enter.
• Further, use the Fill Handle to copy the formula in the cells below.
• Finally, we will see the remainder for each division.
In the formula, the MOD function takes the dividend value (B5) and divisor value (C5) as arguments and gives the remainder directly.
Read More: [Fixed!] Excel MOD Function Not Working
You can download the practice workbook from here.
## Conclusion
Sometimes we want to get a division result without the fractional part i.e. division without remainder. Here, I have shown 3 effective methods to divide without remainder in Excel. If you have faced any problems following the procedures, leave a comment.
## Related Articles
<< Go Back to Excel MOD Function | Excel Functions | Learn Excel
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Advanced Excel Exercises with Solutions PDF | 1,106 | 5,052 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-10 | latest | en | 0.843665 |
http://mathhelpforum.com/trigonometry/23422-verify-solution-equation.html | 1,553,575,723,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204790.78/warc/CC-MAIN-20190326034712-20190326060712-00248.warc.gz | 127,350,120 | 10,014 | 1. Verify solution of equation
HI!
Directions: verify that the x-values are solutions of the equation.
2cosx-1=0
(a.) x= pi/3
(b.) x= 5pi/3
Thanks again!
2. Originally Posted by overduex
HI!
Directions: verify that the x-values are solutions of the equation.
2cosx-1=0
(a.) x= pi/3
(b.) x= 5pi/3
Thanks again!
First, convert the radians to degrees
Then, use your calculator to find the decimal values for cosx
For example, in question a)
cos(pi/3) or cos60 = 0.5
Then sub in the value of 0.5 in the equation "2cosx - 1 = 0"
And solve:
2cos(0.5) - 1 = 0
3. Originally Posted by Macleef
Then, use your calculator to find the decimal values for cosx
I dearly hope that anyone that would be expected to be able to solve such an equation would already have memorized
$\displaystyle cos(60^o)= \frac{1}{2}$
and not need to rely on a calculator to give them that answer.
-Dan
,
,
verify that x value
Click on a term to search for related topics. | 286 | 955 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2019-13 | latest | en | 0.905271 |
http://www.loopandbreak.com/digital-circuits/ | 1,619,091,453,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039603582.93/warc/CC-MAIN-20210422100106-20210422130106-00104.warc.gz | 143,130,226 | 16,122 | ### Digital Circuits
All binary digital devices and systems employ high-speed electronic switches that perform Boolean operations. These switches are called logic gates. By combining logic gates, sophisticated digital systems can be built up. Even the most advanced computers are, at the basic level, comprised of logic gates.
#### Positive and Negative Logic
Usually, the binary digit 1 stands for “true” and is represented by a voltage of about +5 V. The binary digit 0 stands for “false” and is represented by about 0 V. This is positive logic. There are other logic forms, the most common of which is negative logic. In a common form of negative logic, the digit 1 (the logic high state) is represented by about 0 V, and the digit 0 (the logic low state) is represented by about +5 V. The remainder of these topics deals with positive logic.
#### Basic Gates
An inverter or NOT gate has one input and one output. It reverses the state of the input. An OR gate can have two or more inputs. If both, or all, of the inputs are 0 (low), then the output is 0. If any of the inputs is 1 (high), then the output is 1. An AND gate can have two or more inputs. If both, or all, of the inputs are 1, then the output is 1. Otherwise the output is 0.
#### Other Gates
An inverter or NOT gate (A), an AND gate (B), an OR gate (C), an XOR gate (D), a NAND gate (E), and a NOR gate (F).
Gate type Number of inputs Remarks
NOT
OR
1
2 or more
Changes state of input.
Output high if any inputs are high.
AND 2 or more Output low if any inputs are low.
Output high if all inputs are high.
NOR 2 or more Output low if any inputs are high.
Output high if all inputs are low.
NAND 2 or more Output high if any inputs are low.
Output low if all inputs are high.
XOR 2 Output high if inputs differ.
Output low if inputs are the same.
Sometimes an inverter and an OR gate are combined. This produces a NOR gate. If an inverter and an AND gate are combined, the result is a NAND gate. An exclusive OR gate, also called an XOR gate, has two inputs and one output. If the two inputs are the same (either both 1 or both 0), then the output is 0. If the two inputs are different, then the output is 1. The functions of logic gates are summarized in above table. Their schematic symbols are shown in above figure
#### Black Boxes
Logic gates can be combined to form circuits with many inputs and outputs. When two or more logic gates are combined, the outputs are always specific logical functions of the inputs. A complex combination of logic gates is sometimes called a black box.
The functions of a black box can always be determined using Boolean algebra, if the gates inside, and the way they are interconnected, is known. Conversely, if a certain complex logical function is needed for an application, a black box can be designed to perform that function by using
Boolean algebra to break the function down into components of NOT, OR, AND, NOR, NAND, and XOR.
#### Forms of Binary Data
In communications, binary (two-level) data is less susceptible to noise and other interference than analog or multilevel digital data. There are several forms.
• Morse code is the oldest binary means of sending and receiving messages. It is a binary code because it has only two possible states: on (key-down) and off (key-up). It is used mainly by amateur radio operators in their hobby activities. A “human ear/brain machine,” scrutinizing
a Morse code signal, is an amazingly effective digital communications receiver.
• Baudot, also called the Murray code, is a five-unit digital code not widely used by today’s digital equipment, except in some radioteletype communications.
• ASCII (American National Standard Code for Information Interchange) is a seven-unit code for the transmission of text and some programs. Letters, numerals, symbols, and control operations are represented. ASCII is designed for computers. There are 27, or 128, possible representations. Both upper- and lowercase letters can be represented, along with numerals and certain symbols.
#### Flip-flops
A flip-flop is also known as a sequential logic gate. In a sequential gate, the output state depends on both the inputs and the outputs. A flip-flop has two states, called set and reset. Usually, the set state is logic 1 (high), and the reset state is logic 0 (low). Here are some common types.• R-S flip-flop inputs are labeled R (reset) and S (set). The outputs are Q and −Q. (Often, rather than −Q, you will see Q′, or perhaps Q with a line over it.) The outputs are always in logically opposite states. The symbol for an R-S flip-flop, also known as an asynchronous flipflop, is shown in above figure A. The truth table (a specialized form of table denoting logic functions) for an R-S flip-flop is at below Table A.
• Synchronous flip-flop states change when triggered by the signal from a clock. In static triggering, the outputs change state only when the clock signal is either high or low. This type of circuit is sometimes called a gated flip-flop. In positive-edge triggering, the outputs change state at the instant the clock pulse is positive-going. The term edge triggering derives from the fact that the abrupt rise or fall of a pulse looks like the edge of a cliff (Fig. 26-3B). In negative-edge triggering, the outputs change state at the instant the clock pulse is negative going.
• Master/slave (M/S) flip-flop inputs are stored before the outputs are allowed to change state. This device essentially consists of two R-S flip-flops in series. The first flip-flop is called the master, and the second is called the slave. The master flip-flop functions when the clock output is high, and the slave acts during the next low portion of the clock output. This time delay prevents confusion between the input and output.
• J-K flip-flop operation is similar to that of an R-S flip-flop, except that the J-K has a predictable output when the inputs are both 1. Table 26-3B shows the input and output states for this type of flip-flop. The output changes only when a triggering pulse is received. The symbol for a J-K flip-flop is shown in Fig. 26-3C.
• R-S-T flip-flop operation is similar to that of an R-S flip-flop, except that a high pulse at the T input causes the circuit to change state.
• The T flip-flop has only one input. Each time a high pulse appears at the T input, the output state is reversed.
A:
R-S Flip-flop R S Q -Q
0 0 Q -Q
0 1 1 0
1 0 0 1
1 1 ? ?
B:
J-K Flip-flop R K Q -Q
0 0 Q -Q
0 1 1 0
1 0 0 1
1 1 -Q Q
#### Clocks
At A, the symbol for an R-S flip-flop. At B, pulse edges are either negative-going or positive-going. At C, the symbol for a J-K flip-flop.
In electronics, the term clock refers to a circuit that generates pulses at high speed and at precise intervals. It sets the tempo for the operation of digital devices. In a computer, the clock acts like a metronome for the microprocessor. Clock speeds are measured and expressed in hertz (Hz), kilohertz (kHz), megahertz (MHz), or gigahertz (GHz).
#### Counters
A counter consists of a set of flip-flops or equivalent circuits. Each time a pulse is received, the binary number stored by the counter increases by 1. A frequency counter measures the frequency of a wave by tallying the cycles in a given interval of time. The circuit consists of a gate, which begins and ends each counting cycle at defined intervals. The accuracy is a function of the length of the gate time; the longer the time base, the better the accuracy. The readout is in base-10 digital numerals. | 1,786 | 7,538 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2021-17 | latest | en | 0.900816 |
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# Sentance Correction Question 18Oct
Author Message
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18 Oct 2006, 13:57
Please check attached GIF file. Can any one explain me the answer
Attachments
Q1_B.GIF [ 6.62 KiB | Viewed 762 times ]
If you have any questions
New!
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18 Oct 2006, 14:02
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18 Oct 2006, 14:09
A is not answer since "those" donot match with "spinal coloumb" which is singular.
Can any one tell me why not E
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18 Oct 2006, 14:09
A or E. hmmm Pick A.
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18 Oct 2006, 14:10
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18 Oct 2006, 14:32
I think It is B
A a spinal column is compared with plural
B Correct
C Chimpanzees should be the subject ..otherwise ‘so were’ doesn’t match
D Chimpanzees should be the subject
E Chimpanzees should be the subject
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18 Oct 2006, 14:39
I am thinking D
Chimpanzee's spinal column resembled ape's (spinal column)
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18 Oct 2006, 14:42
B
Not E because , the spinal columns (blah blah) and so were unable to walk (spinal columns don't walk )
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18 Oct 2006, 14:43
D would also suggest the spinal columns are walking, in my opinion.
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18 Oct 2006, 14:53
londonluddite wrote:
D would also suggest the spinal columns are walking, in my opinion.
Good catch I totally missed that reading the next part.
In that case B is the best one, thanks london
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18 Oct 2006, 14:57
londonluddite wrote:
D would also suggest the spinal columns are walking, in my opinion.
agreed
CDE are all out b/c they state that the spinal columns are walking.
I like B better than A b/c it's shorter although PR says gmat likes using those/that of for apples and oranges.
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18 Oct 2006, 15:00
on second thought, i'm liking A better now. it seems clearer than B in comapring the spinal columns.
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18 Oct 2006, 15:09
Yea i'm torn between A&B, would def go for A if spinal columns was plural.
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18 Oct 2006, 16:33
B.
It should be either "a spinal column that resembled an ape's" or "spinal columns that resembled those of apes' ". Only the former is correctly expressed in B.
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18 Oct 2006, 17:02
Should be B
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18 Oct 2006, 17:18
MBA2ran wrote:
A is not answer since "those" donot match with "spinal coloumb" which is singular.
Can any one tell me why not E
I dont think THOSE was referring 'spinal column' I thought it was connecting to apes'. I think A & E has missing apostrophe though I dont know what the right answer is. I throw in my towels. whats a OA and OE?
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Re: Sentance Correction Question 18Oct [#permalink]
### Show Tags
18 Oct 2006, 19:14
I really think that all of them are wrong. But in the rush, I'd pick B, after all, the others try to say that the poor monkeys had a few columns each. Quite heavy to accept!
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18 Oct 2006, 19:49
that resemebed...that may explain
My choice is A.
_________________
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the way is difficult, but perfect trust makes it easy.
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19 Oct 2006, 00:48
Am up for (A) too.
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19 Oct 2006, 06:53
Modifier SC, leaves only A and B.
I'll take (B)
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Display posts from previous: Sort by | 2,050 | 6,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2017-22 | latest | en | 0.844749 |
https://cm-to.com/330-cm-to-inches/ | 1,716,193,875,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058254.21/warc/CC-MAIN-20240520080523-20240520110523-00873.warc.gz | 149,752,408 | 8,564 | # 330 Centimeter to Inch
Convert 330 (three hundred thirty zero) Centimeters to Inches (cm to in) with our conversion calculator.
330 Centimeters to Inches equals 129.9213 in.
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Convert 330 Centimeters to Inches (cm to in) with our conversion calculator.
330 Centimeters to Inches equals 129.9213 in.
To convert centimeters to inches, the given length in centimeters needs to be multiplied by the conversion factor from centimeters to inches. The conversion factor is 0.393701, because one inch is equal to 2.54 centimeters. Therefore, to convert 330 centimeters to inches, you multiply 330 by 0.393701. The calculation is as follows: 330 cm * 0.393701 = 129.92133 inches. This result indicates that 330 centimeters is equal to approximately 129.92 inches once the conversion has been done using the specified conversion factor.
The process of conversion is straight-forward and is a crucial aspect when dealing with measurements in different units, especially between metric and imperial systems. Understanding and applying the conversion factor correctly is imperative to ensure accurate results. In various fields such as engineering, manufacturing, and even daily activities, converting measurements is common. For instance, if someone is familiar with the metric system but is working with specifications given in the imperial system, converting centimeters to inches or vice versa becomes essential. This calculation demonstrates how one can easily switch between units to understand dimensions in their preferred system, thereby making planning and executing tasks more accessibly.
After converting 330 cm to inches, here are seven items that are approximately 330 cm (129.92 inches) in length:
1. Small Car: A very compact car can have a total length close to this measurement.
• Length of vehicle from front bumper to rear bumper.
2. Kayak: Certain models of kayaks are designed around this length for stability and speed.
• Length from the tip of the bow to the stern.
3. Sofa: A large sectional sofa could measure approximately 330 cm in total length.
• Total length from one end to the other, possibly including chaise.
4. Dining Table: Some extended dining tables can reach this length, seating up to 12 people.
• Length of tabletop surface.
5. King Size Bed: The width of a standard King Size bed is much less but combined with a custom headboard or end-of-bed bench, the total can reach 330 cm.
• Combined length of bed and additional furniture piece.
6. Row of Bar Stools: About 4 to 5 bar stools side by side could total 330 cm in length, depending on the model.
• Combined width from one end stool to the last.
7. Garden Bench: Some long garden benches designed for multiple seating can be around 330 cm.
• Total length of the bench from one end to the other.
These items show how the measurement of 330 cm can apply to a variety of objects, many of which we might encounter in everyday life or specific activities. | 702 | 3,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2024-22 | latest | en | 0.872039 |
http://sciencedocbox.com/Physics/71267007-Chapter-4-newton-s-laws.html | 1,558,407,754,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256215.47/warc/CC-MAIN-20190521022141-20190521044141-00222.warc.gz | 187,948,368 | 27,309 | # Chapter 4 Newton s Laws
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1 Chapter 4 Newton s Laws
2 Isaac Newton Some inventions and discoveries: 3 laws of motion Universal law of gravity Calculus Ideas on: Sound Light Thermodynamics Reflecting telescope In this chapter, we study his 3 laws of motion which have to do with how forces affect an object s motion.
3 What is a force? In everyday language it is a push or a pull. Some forces we will use in physics 1114: 1) Gravity (Weight) 2) Friction 3) Normal 4) Tension A force is always the interaction between two objects! For every force, you should notice what causes the force and what object the force acts on. The SI unit of force is a Newton (N) and the British unit of force is a pound (lb). More on this later
4 Newton s First Law An object remains at rest, or at a constant speed in a straight line, unless it is acted upon by a net external force. If an object is either stationary, or moving at a constant speed in a straight line, there is no net force acting on the object. There may be many forces acting on the object, but there is no net force acting on the object. Force is a vector, and the net force is the sum of all the forces acting on an object
5 Pushing a car You and your friend each push a car at rest with the same force Car has forces acting on it, but total force is 0 so it does not move Car now has a net force and will move
6 Interactive Question When the rocket engines of a starship are suddenly turned off, while traveling in empty space far from any stars or planets, the starship will A) stop immediately B) slowly slow down, and then stop C) go faster and faster D) move with a constant velocity
7 Interactive Question A rocket ship in space has its engines firing and is following path 1. At point 2, the engines shut off. Which path does the rocket ship follow? 1 2 A) B) C) D) E)
8 A) The forces acting to make the car go in the forward direction must be greater than the forces acting to make the car go in the backward direction, or the car would not go forward. B) The forces acting to make the car go in the forward direction must be equal to the forces acting to make the car go in the backward direction. C) There are no forces acting on the car at a constant velocity D) There is not enough information to say anything. Interactive Question You are driving your car down a straight road at a constant velocity of 65 mph. What can you conclude about the forces acting on your car?
9 Interactive Question If a single non-zero force is acting on an object, what kind of motion is not allowed? A) The object could be speeding up. B) The object could be slowing down. C) The object could be at moving at a constant velocity. D) The object could be turning. E) None of the above. They are all allowed
10 Inertia The force required to change an object s state of motion is a measure of the inertia of the body. This measure of inertia is called mass. Objects with large inertia, or mass, require a larger force to achieve the same change in the state of motion. A change in the state of motion is described as an acceleration. SI unit of mass is kilograms (kg) British unit of mass is slugs.
11 Newton s Second Law The acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass. The direction of the acceleration is in the direction of the net force acting on the object. F net = ma This is a vector equation. So it really means: F net,(horizontal) = ma (horizontal) F net,(vertical) = ma (vertical)
12 F net = ma The left hand side is the net force acting on the object with the mass m. The right hand side is the effect of the net force acting on that object. The SI unit of force is a Newton (N) = kg m/s 2 The British unit of force is a pound (lb) Newton s 1 st law is a special case of the 2 nd law when a=0. The acceleration in Newton s second law can be related to the acceleration in the equations of motion from Chapters 2 and 3.
13 Newton s Second Law of Motion F net = ma The same net force on an object with more mass will produce less acceleration.
14 Interactive Question A box has two forces acting on it as shown by the arrows which have the same length and point in opposite directions. What can you say about the motion of this box? A) It is definitely not moving B) It may be moving at a constant velocity or stationary C) It may be accelerating D) Not enough information is given
15 Interactive Question A box has two forces acting on it as shown by the arrows which have the different lengths and point in opposite directions. What can you say about the motion of this box? A) It is definitely not moving B) It is definitely moving to the right C) It may be moving at a constant velocity or stationary D) It may be accelerating but not necessarily E) It is definitely accelerating
16 Interactive Question A box has two forces acting on it as shown by the arrows which have the different lengths and point in opposite directions. What can you say about the motion of this box? A) It is definitely moving to the right B) It is definitely accelerating to the right C) Both of the above D) None of the above
17 Interactive Question A constant force F acts on block 1 with mass m and block 2 with mass 2m. Which graph correctly represents the accelerations of the blocks? (A) (B) (C) (D) 2 1
18 Interactive Question A net force F is required to give a mass m an acceleration a. If a net force of 6F is applied to a mass 2m, what acceleration results. A) a B) 2a C) 3a D) 4a E) 6a
19 Problem: You push horizontally on a box with a 40 N force. A 30 N frictional force opposes the motion. The box accelerates at a rate of 2.2 m/s 2. What is the mass of the box?
20 Problem: A 12,500 lb truck (which has a mass of 5680 kg) is traveling at 33 mi/hr when it applies its brakes and comes to a stop in 4.5 seconds. What was the average net force stopping the truck (in Newtons)?
21 Interactive Question A constant force is acting on an object. Which of these graphs best represents the velocity of the object? (A) (B) (C)
22 Interactive Question Two equal forces act on an object in the directions shown. If these are the only forces involved, what can you say is definitely true about the motion of the object? A) It is moving at a constant velocity. B) It is speeding up C) It is slowing down D) It is accelerating E) Nothing, not enough information
23 The Force of Gravity: Weight Consider an object that has only the force of gravity acting on it. We ll call this force W. (You ll see why later.) From Newton s second law F net = ma W = ma W But we also know, that the acceleration of this object will be a constant value we call g. So the strength of the force of gravity, W = mg This force of gravity pulling on an object is called weight.
24 Mass, Weight, and Inertia Mass is a measure of an object s inertia. The SI units of mass are kilograms (kg). Inertia is an object s resistance to a change in its motion so mass is used in Newton s 2 nd law, F net = ma. A larger force is required to produce the same acceleration for a larger mass. The mass of an object doesn t change as you move the object to a different planet or location. Weight is the gravitational force acting on the object. The SI units of weight are newtons (N). The weight of an object depends on the where the location of the object. In the absence of gravity the object has no weight, but still has mass.
25 We say that 1.0 kg is 2.2 lbs, but this is not an accurate statement because pounds is a unit of weight and kilograms is a unit of mass. Only on earth does a 1.0 kg mass have a weight of 2.2 lbs. In general, F = mg = (1.0 kg) (10 m/s 2 ) = 10 N = 2.2 lb because a mass of 1 kg feels a force of 9.8 N on earth. So 2.2 lb = 10 N, or 1 lb = 4.5 N
26 Problem: A car has a weight of 3500 lbs. What is the mass of the car in British units?
27 Interactive Question Object A is more massive than object B. Which of the following statements is true? A) Both object A and object B experience the same gravitational force. B) Both object A and object B experience the same gravitational acceleration. C) Object A experiences a greater gravitational force and greater gravitational acceleration than object B. D) More than one of the above is true
28 Interactive Question A ball is thrown through the air and follows the path shown. It is traveling from the left to the right. When the ball is at the position shown, which of the following diagrams correctly shows the forces acting on the ball, neglecting air resistance. (A) (B) (C) (D) (E)
29 Interactive Question A ball is thrown through the air and follows the path shown. It is traveling from the left to the right. When the ball is at the position shown, which of the following diagrams correctly shows the forces acting on the ball, not neglecting air resistance. (A) (B) (C) (D) (E)
30 When an object moves fast enough through the air we cannot ignore air resistance. Air resistance always acts opposite the direction of motion. For instance as a sky diver falls, her weight W stays the same but the force of air resistance R increases as her speed increases. When the force of air resistance has increased to equal the magnitude of her weight, the net force is zero so the acceleration is zero (F net = ma) The velocity is then at its maximum value, the terminal velocity
31 The Normal Force Consider an stationary object on a surface. According to Newton s second law, the object has no net force acting on it. The force of gravity is pulling the object toward the center of the earth. What other force is balancing out the force of gravity? We call the force of the table pushing up on the book the Normal force which means perpendicular. Whenever one object pushes on another it produces this normal force. N W
32 Problem: A book with a mass of 0.80 kg is stationary on a table. What is the normal force acting on the book?
33 The normal force is not always equal to the weight, as this problem illustrates. Problem: A book with a mass of 0.80 kg is stationary on a table. You tie a string to the book and pull up with a force of 2.5 N? What is the normal force acting on the book?
34 Problem: A book with a mass of 0.80 kg is stationary on a table. You have a weight of 640 N and sit on the book. What is the normal force acting on the book?
35 Interactive Question Consider a person standing in an elevator that is moving upward at a constant velocity. The upward normal force N exerted by the elevator floor on the person is A) larger than B) identical to C) smaller than the downward weight W of the person
36 Interactive Question Consider a person standing in an elevator that is accelerating upward. The upward normal force N exerted by the elevator floor on the person is A) larger than B) identical to C) smaller than the downward weight W of the person
37 Newton s Third Law Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first. This action-reaction pair of forces always acts on different objects, and thus never add to produce zero net force on a single object.
38 Let s look at a book sitting on a table in light of Newton s 2 nd and 3 rd laws. The two forces acting on the book are the weight of the book, W, (the force of the earth pulling down on the book), and the normal force, N, (the force of the table pushing up on the book.). Because of Newton s 2 nd law, F net = ma = 0, W = N. By Newton s 3 rd law, we also know that the book pushes down on the table with the same force as the table pushes up on the book, N, and that the book pulls up on the earth with the same force that the earth pulls down on the book, W. N N W W
39 Interactive Question A ping-pong ball collides with a bowling ball. Which of experiences a greater force from the other one due to the collision? A) The ping-pong ball B) The bowling ball C) They experience the same force D) The force on each depends on its velocity E) The force on each depends on its mass
40 Interactive Question A book is resting on the surface of a table. Consider the following four forces that arise in this situation. (1) the force of the earth pulling on the book (2) the force of the table pushing on the book (3) the force of the book pushing on the table (4) the force of the book pulling on the earth Which two forces form an action-reaction pair which obey Newton s third law? A) 1 and 2 C) 1 and 4 E) 3 and 4 B) 1 and 3 D) 2 and 4
41 Newton s 3 rd law, and Motion What makes a car accelerate forward? F F The car s wheels push against the road (a force on the road), and by Newton s 3 rd law, the road pushes back against the car (a force on the car). It is this force on the car which makes the car accelerate forward.
42 Friction, Newton s 3 rd Law, and Motion F WG is the force of the wheel on the ground in the horizontal direction. a F GW F WG
43 Problem for discussion: Suppose you throw a baseball. Your body exerts a force on the baseball and the baseball exerts an equal and opposite force back on your body. Why is the ball accelerated forward, but you are not accelerated backwards?
44 Interactive Question A horse pulls a cart along a flat level road. Consider the following four forces that arise in this situation. (1) The force of the horse pulling on the cart. (2) The force of the cart pulling on the horse. (3) The force of the horse pushing on the road. (4) The force of the road pushing on the horse. Which two forces form an action-reaction pair which obey Newton s third law. A) (1) and (4) C) (2) and (4) E) (2) and (3) B) (1) and (3) D) (3) and (4)
45 Third-Law Action/Reaction Pair The cart pulls back on the horse with the same magnitude of force as the horse pulls forward on the cart. So how does this system ever move? To analyze why the horse moves we P must look only at N C W C P the forces acting on the horse and to N H W H analyze why the cart moves we F C F H must look only at the forces acting on the cart.
46 Interactive Question An astronaut who is walking in space pushes on a spaceship with a force of 36 N. The astronaut has a mass of 92 kg and the spaceship has a mass of kg. Which statement is true? A) The astronaut will accelerate, but not the spaceship. B) No net force will be exerted on the astronaut or on the spaceship. C) A force will be exerted on the astronaut but not on the spaceship. D) The astronaut and the spaceship will have the same magnitude of acceleration. E) None of the above.
47 Problem: An astronaut who is walking in space pushes on a spaceship with a force of 36 N. The astronaut has a mass of 92 kg and the spaceship has a mass of kg. What happens?
48 Systems, Internal Forces, and External Forces You can always solve problems with more than one object by working with each object separately as a system. Sometimes it is more convenient to chose a number of objects as a system. Internal forces are forces that act only between objects in the system. External forces act between an object outside of the system and an object inside the system. In Newton s second law, only external forces need to be considered.
49 Interactive Question A train is accelerating to the right. There is no friction opposing the motion. How does the tension between the engine and the 2 nd car compare with the tension between the 2 nd car and the third car? A) It is less. B) It is the same. C) It is greater. D) More information is needed.
50 Problem: A train is accelerating at a rate of 5.0 m/s 2. There is no friction opposing the motion. What is the tension between each of the cars? 1500 kg 1000 kg 2500 kg
51 Problem: A 4 kg block and a 2 kg block can move on a horizontal surface. The blocks are pushed by a 24 N force in the positive x direction as shown. A frictional force of 8 N acts on the 4 kg block and a frictional force of 4 N acts on the 2 kg block. A) What is the net force acting on the two blocks? B) Determine the acceleration of the blocks. C) What is the force of the 2 kg block pushing on the 4 kg block? 24 N 4 kg 2 kg +x
### 4. As you increase your push, will friction on the crate increase also? Ans. Yes it will.
Ch. 4 Newton s Second Law of Motion p.65 Review Questions 3. How great is the force of friction compared with your push on a crate that doesn t move on a level floor? Ans. They are equal in magnitude and
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### Isaac Newton ( )
Isaac Newton (1642-1727) In the beginning of 1665 I found the rule for reducing any degree of binomial to a series. The same year in May I found the method of tangents and in November the method of fluxions | 10,456 | 43,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2019-22 | longest | en | 0.923416 |
https://assessingpsyche.wordpress.com/2014/07/10/two-visualizations-for-explaining-variance-explained/ | 1,670,146,725,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710968.29/warc/CC-MAIN-20221204072040-20221204102040-00775.warc.gz | 146,381,314 | 33,103 | Statistics
# Two visualizations for explaining “variance explained”
In my introductory statistics class, I feel uneasy when I have explain what variance explained means. The term has two things I don’t like. First, I don’t like variance very much. I feel much more comfortable with standard deviations. I understand that at a deep level variance is a more fundamental concept than the standard deviation. However, variance is a poor descriptive statistic because there is no direct visual analog for variance in a probability distribution plot. In contrast, the standard deviation illustrates very clearly how much scores typically deviate from the mean. So, variance explained is hard to grasp in part because variance is hard to visualize.
The second thing I don’t like about variance explained is the whole “explained” business. As I mentioned in my last post, variance explained does not actually mean that we have explained anything, at least in a causal sense. That is, it does not imply that we know what is going on. It simply means that we can use one or more variables to predict things more accurately than before.
In many models, if X is correlated with Y, X can be said to “explain” variance in Y even though X does not really cause Y. However, in some situations the term variance explained is accurate in every sense:
X causes Y
In the model above, the arrow means that X really is a partial cause of Y. Why does Y vary? Because of variability in X, at least in part. In this example, 80% of Y’s variance is due to X, with the remaining variance due to something else (somewhat misleadingly termed error). It is not an “error” in that something is wrong or that someone is making a mistake. It is merely that which causes our predictions of Y to be off. Prediction error is probably not a single variable. It it likely to be the sum total of many influences.
Because X and error are uncorrelated z-scores in this example, the path coefficients are equal to the correlations with Y. Squaring the correlation coefficients yields the variance explained. The coefficients for X and error are actually the square roots of .8 and .2, respectively. Squaring the coefficients tells us that X explains 80% of the variance in Y and error explains the rest.
# Visualizing Variance Explained
Okay, if X predicts Y, then the variance explained is equal to the correlation coefficient squared. Unfortunately, this is merely a formula. It does not help us understand what it means. Perhaps this visualization will help:
Variance Explained
If you need to guess every value of Y but you know nothing about Y except that it has a mean of zero, then you should guess zero every time. You’ll be wrong most of the time, but pursuing other strategies will result in even larger errors. The variance of your prediction errors will be equal to the variance of Y. In the picture above, this corresponds to a regression line that passes through the mean of Y and has a slope of zero. No matter what X is, you guess that Y is zero. The squared vertical distance from Y to the line is represented by the translucent squares. The average area of the squares is the variance of Y.
If you happen to know the value of X each time you need to guess what Y will be, then you can use a regression equation to make a better guess. Your prediction of Y is called Y-hat (Ŷ):
$\hat{Y}=b_0+b_1X=0+\sqrt{0.80}X\approx 0.89X$
When X and Y have the same variance, the slope of the regression line is equal to the correlation coefficient, 0.89. The distance from Ŷ (the predicted value of Y) to the actual value of Y is the prediction error. In the picture above, the variance of the prediction errors (0.2) is the average size of the squares when the slope is equal to the correlation coefficient.
Thus, when X is not used to predict Y, our prediction errors have a variance of 1. When we do use X to predict Y, the average size of the prediction errors shrinks from 1 to 0.2, an 80% reduction. This is what is meant when we say that “X explains 80% of the variance in Y.” It is the proportion by which the variance of the prediction errors shrinks.
# An alternate visualization
Suppose that we flip 50 coins and record how many heads there are. We do this over and over. The values we record constitute the variable Y. The number of heads we get each time we flip a coin happens to have a binomial distribution. The mean of a binomial distribution is determined by the probability p of an event occurring on a single trial (i.e., getting a head on a single toss) and the number of events k (i.e., the number of coins thrown). As k increases, the binomial distribution begins to resemble the normal distribution. The probability p of getting a head on any one coin toss is 0.5 and the number of coins k is 50. The mean number of heads over the long run is:
$\mu = pk=0.5*50=25$
The variance of the binomial distribution:
$\sigma^2 = p(1-p)k=0.5*(1-0.5)*50=12.5$
Before we toss the coins, we should guess that we will toss an average number of heads, 25. We will be wrong much of the time but our prediction errors will be as small as they can be, over the long run. The variance of our prediction errors is equal to the variance of Y, 12.5.
Now suppose that after tossing 80% of our coins (i.e., 40 coins), we count the number of heads. This value is recorded as variable X. The remaining 20% of the coins (10 coins) are then tossed and the total number of heads is counted from all 50 coins. We can use a regression equation to predict Y from X. The intercept will be the mean number of heads from the remaining 10 coins:
$\hat{Y} = b_0+b_1X=5+X$
In the diagram below, each peg represents a coin toss. If the outcome is heads, the dot moves right. If the outcome is tails, the dot moves left. The purple line represents the probability distribution of Y before any coin has been tossed.
X explains 80% of the variance in Y.
When the dot gets to the red line (after 40 tosses or 80% of the total), we can make a new guess as to what Y is going to be. This conditional distribution is represented by a blue line. The variance of the conditional distribution has a mean equal to Ŷ, with a variance of 2.5 (the variance of the 10 remaining coins).
The variability in Y is caused by the outcomes of 50 coin tosses. If 80% of those coins are the variable X, then X explains 80% of the variance in Y. The remaining 10 coins represent the variability of Y that is not determined by X (i.e., the error term). They determine 20% of the variance in Y.
If X represented only the first 20 of 50 coins, then X would explain 40% of the variance in Y.
X explains 40% of the variance in Y.
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## 4 thoughts on “Two visualizations for explaining “variance explained””
1. Ewa says:
Great visual explanations – thank you. Glad to have stumbled upon your blog!
2. Dawson says:
Incredible! Thank you so much for sharing this.
3. Joe says:
I really like the second visualization, thank you for sharing this! | 1,618 | 7,014 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2022-49 | latest | en | 0.959009 |
http://spotidoc.com/doc/153979/finding-the-percent-of-a-number-4.4-2-activity- | 1,603,727,227,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107891428.74/warc/CC-MAIN-20201026145305-20201026175305-00148.warc.gz | 98,951,209 | 10,351 | # Finding the Percent of a Number 4.4 2 ACTIVITY:
```English
4.4
Spanish
Finding the Percent of a Number
How can you use mental math to find the
S
STATE
STANDARDS
percent of a number?
MA.6.A.5.1
“I have a secret way for
finding 21% of 80.”
1
“10% is 8 and 1% is 0.8.”
“So 21% is 8 + 8 + 0.8 = 16.8.”
EXAMPLE: Finding 10% of a Number
a. How did Newton know that 10% of 80 is 8?
Write 10% as a fraction.
10
10
100
10% = —
per
1
10
=—
cent
Method 1: Using a Model
0%
0
10%
8
20%
30%
40%
50%
60%
70%
80%
90%
16
24
32
40
48
56
64
72
Method 2: Using Multiplication
1
10
1
10
80
10
10% of 80 = — of 80 = — × 80 = — = 8
b. How do you move the decimal point to find 10% of a number?
Move the decimal point one place to the left.
2
10% of 80. = 8.0
ACTIVITY: Finding 1% of a Number
Work with a partner.
a. How did Newton know that 1% of 80 is 0.8?
b. How do you move the decimal point to find 1% of a number?
168
Chapter 4
Fractions, Decimals, and Percents
100%
80
English
Spanish
3
EXAMPLE: Using Mental Math
Use mental math to find each percent of a number.
a. 12% of 40
b. 19% of 50
Think: 12% = 10% + 1% + 1%
10% of 40 = 4
Think: 19% = 10% + 10% − 1%
1% of 40 = 0.4
10% of 50 = 5
4 + 0.4 + 0.4 = 4.8
4
1% of 50 = 0.5
5 + 5 − 0.5 = 9.5
ACTIVITY: Using Mental Math
Work with a partner. Use mental math to find each percent of a number.
a. 20% tip for a \$30 meal
b. 18% tip for a \$30 meal
c. 6% sales tax on a \$20 shirt
d. 9% sales tax on a \$20 shirt
e. 6% commission on selling a \$200,000 house
f. 2% property tax on a \$200,000 house
g. 21% income tax on an income of \$40,000
h. 38% income tax on an income of \$80,000
5. IN YOUR OWN WORDS How can you use mental math to find the percent
of a number?
6. Describe two real-life examples of finding a percent of a number.
Use what you learned about finding the percent of a number to
complete Exercises 3 –10 on page 172.
Section 4.4
Finding the Percent of a Number
169
English
Spanish
Lesson
4.4
Lesson Tutorials
Finding the Percent of a Number
Words
Write the percent as a fraction or decimal. Then multiply.
Numbers
20% of 60 is 12.
1
5
Model
0%
20% 40%
60% 80% 100%
0
12
36
— × 60 = 12
0.2 × 60 = 12
EXAMPLE
1
24
48
60
Finding the Percent of a Number
Use a fraction to find the percent of the number.
a. Find 25% of 40.
b. Find 60% of 150.
3
5
1
4
60% of 150 = — × 150
25% of 40 = — × 40
30
10
1 × 40
=—
4
3 × 150
=—
5
= 10
= 90
1
1
So, 25% of 40 is 10.
0%
0
So, 60% of 150 is 90.
25%
50%
75%
100%
10
20
30
40
0% 20%
0
30
40%
60%
60
90
80% 100%
120
150
Use a fraction to find the percent of the number.
1. 90% of 20
EXAMPLE
Summer Vacation
Did you go on a
vacation this
past summer?
Yes
No
48%
52%
Note: 200 students surveyed
2
2. 75% of 32
3. 10% of 110
4. 30% of 75
Standardized Test Practice
How many students went on vacation?
A 48
○
B 96
○
C 100
○
D 104
○
From the survey, you can see that 48% out of 200 students said yes.
48% of 200 = 0.48 × 200
= 96
Write 48% as a decimal.
Multiply.
So, 96 students went on vacation. The correct answer is ○
B.
170
Chapter 4
Fractions, Decimals, and Percents
English
Spanish
Use a decimal to find the percent of the number.
5. 15% of 40
Exercises 3–18
EXAMPLE
3
6. 78% of 150
7. 35% of 16
8. 4% of 70
Using Mental Math
Your friend is bidding online for concert tickets. The current bid is
shown. The winning bid is 150% of the current bid. How much is the
winning bid?
View larger picture
Current bid:
US \$120.00
Time remaining:
1 hour 45 min
Method 1: Write 150% as a
decimal and multiply.
Method 2: Using mental math,
think 150% = 100% + 50%.
150% of 120 = 1.5 × 120
100% of 120 = 1 × 120 = 120
1
2
50% of 120 = — × 120 = 60
= 180
Add: 120 + 60 = 180
So, the winning bid is \$180.
9. WHAT IF? In Example 3, the winning bid is 225% of the
current bid. How much is the winning bid?
Exercises 23–30
EXAMPLE
4
Real-Life Application
The width of a rectangular room is 80% of its length. What is the area
of the room?
Find 80% of 15 feet.
4
5
80% of 15 = — × 15
3
Tim
e
4 × 15
=—
5
0%
0
20%
3
40%
60%
80%
100%
6
9
12
15
1
15
1
5 ftt
= 12
The width is 12 feet.
Use the formula for the area A of a rectangle.
U
A = 15 × 12 = 180
So, the area of the room is 180 square feet.
10. The width of a rectangular stage is 55% of its length. The stage
is 120 feet long. What is the area?
Section 4.4
Finding the Percent of a Number
171
English
Spanish
Exercises
4.4
Help with Homework
1. DIFFERENT WORDS, SAME QUESTION Which is different? Find “both” answers.
What is twenty percent of 30?
What is one-fifth of 30?
What is 20 multiplied by 30?
What is 0.2 times 30?
2. REASONING If 52 is 130% of a number, is the number greater or less than 52?
Explain.
6)=3
9+(- 3)=
3+(- 9)=
4+(- =
1)
9+(-
Find the percent of the number.
1
2
3. 20% of 60
4. 10% of 40
5. 18% of 70
7. 8% of 90
8. 14% of 20
9. 26% of 50
10. 3% of 60
11. 30% of 70
12. 75% of 48
13. 45% of 45
14. 92% of 19
15. 40% of 60
16. 38% of 22
17. 70% of 20
18. 87% of 55
19. ERROR ANALYSIS Describe and correct the
error in finding 40% of 75.
✗
6. 32% of 30
40% of 75 = 40% × 75 = 3000
20. MANGROVES Lake Worth, near West Palm Beach, had about 2120 acres of
mangrove trees 40 years ago. Only about 13% of the mangrove trees remain.
How many acres of mangrove trees remain?
21. SPIDER MONKEY The tail of the
spider monkey is 64% of its length.
What is the length of its tail?
55
22. CABLE A family pays \$45 each month
for cable television. The cost increases 7%.
in.
a. How many dollars is the monthly increase?
b. What is the new monthly cost?
Find the percent of the number.
172
3 23. 140% of 60
24. 120% of 33
25. 175% of 54
26. 250% of 146
27. 4.5% of 50
28. 0.7% of 40
29. 2.8% of 150
30. 7.2% of 235
Chapter 4
Fractions, Decimals, and Percents
English
Spanish
Copy and complete the statement using <, >, or =.
31. 80% of 60
60% of 80
32. 20% of 30
30% of 40
33. 120% of 5
0.8% of 250
34. 85% of 40
25% of 136
35. TIME How many minutes is 40% of 2 hours?
36. LENGTH How many inches is 78% of 3 feet?
37. GEOMETRY The width of the rectangle is 75% of its length.
a. What is the area of the rectangle?
b. The length of the rectangle is doubled. What percent
of the length is the width now? Explain your reasoning.
24 in.
back to between 68% and 75% of the starting height. A new ball is
dropped from 6 feet and bounces back 4 feet 1 inch. Does the ball
pass inspection? Explain.
39. REASONING You know that 15% of a number n is
12. How can you use this to find 30% of n? 45% of
n? Explain.
40. SURFBOARD You have a coupon for 10% off the
sale price of a surfboard.
a. What is the sale price of the surfboard?
b. What is the price of the surfboard after using the coupon?
c. Is taking 40% off the regular price the same as taking
30% off the regular price and then 10% off the sale
41.
On three geography tests, you earned grades of 88%, 94%,
and 90%. Each test was worth 150 points.
a. The final exam is worth 250 points. How many points do you need
on the final exam to earn 93% of the total points on tests?
b. What percent do you need on the final?
Multiply. Write the answer in simplest form.
2
3
42. — × 4
3
8
43. — × 4
SECTION 2.2
3
5
44. 6 × —
46. MULTIPLE CHOICE What is the quotient of 7.5 and 2.4?
A 0.0032
○
B 0.03125
○
C 0.32
○
Section 4.4
5
6
45. 12 × —
SECTION 3.5
D 3.125
○
Finding the Percent of a Number
173
``` | 2,774 | 7,270 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2020-45 | latest | en | 0.860672 |
http://schoolphysics.org/age16-19/medical%20physics/text/Back_damage/index.html | 1,642,360,991,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300010.26/warc/CC-MAIN-20220116180715-20220116210715-00000.warc.gz | 63,603,007 | 2,691 | # Back damage
In the section on vectors we looked at the tension in your back muscles when you bend over. One assumption that was made was that the reaction at the pelvis acted long the spine. If this is not so then we need to use the principle of moments to work out the tension in the back muscles.
Let's consider somebody bending over at an angle A to put a heavy case into the back of a car, probably one of the worst lifting situations.
Let the mass of the upper body of the person (about 2/3 total body mass) be W, the reaction at the pelvis be R acting along a line q to the spine (Figure 1) and the load that the person lifts be mg.
Taking moments about the base of the spine (P) gives:
TL sin 10 = (W + mg) L cos A
And so:
Tsin 10o =(W + mg) cos A
Notice that since the line of action of the reaction at the pelvis (R) runs through the point P here is no moment of S about P.
Example problem
A gardener of mass 80 kg lifts a mower of mass 50 kg into the back of their truck. In doing this they lean over at an angle of 60o to the horizontal. (g = 9.8 ms-2)
Calculate the tension in their back muscles.
Tsin 10o = (W + mg) cos A T sin 10 = 130xgxcos 60
T = 637/sin 10 = 3668 N
## Bending over and damage to your back – an alternative approach
You probably know that the correct way to lift something is by bending your knees and keeping your back straight. The following calculation shows very clearly why this is!
This can be demonstrated very well by using a mop to represent the spine and a string to represent the back muscles!
Consider a person bending over so that their spine makes an angle of A with the horizontal (see Figure 2).
Their back muscles make an angle of 10o with the spine and have a tension T. We will assume that the reaction (R) at the pelvis acts along the spine. It has been calculated that the weight of the upper body (W) is about 2/3 of the total body weight.
Resolving at right angles to the spine we have:
W cosA = T sin (10)
Example problem
If a person with a mass of 50 kg bends over at 60o then
T = 0.67 x 50gsin60/sin 10 = 1630 N = 3.3 times body weight | 552 | 2,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2022-05 | latest | en | 0.938143 |
https://cs.stackexchange.com/questions/47711/what-is-the-regular-expression-describing-this-language/64987 | 1,719,053,264,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862310.25/warc/CC-MAIN-20240622081408-20240622111408-00643.warc.gz | 160,531,836 | 40,984 | # What is the regular expression describing this language?
The language is $\{w \mid \text{$w$is any string except$11$and$111$}\}$ where the alphabet is $\{0,1\}$.
Drawing the DFA recognizing $\{v \mid \text{$v$is either$11$or$111$}\}$, then switching the accept and non-accept states, and finally examining this resulting structure, I thought of it as being:
$(ϵ∪0Σ^*∪1∪10Σ^*∪110Σ^*∪111Σ^+)$
My reasoning was that it has to accept the empty string, or a 0 followed by whatever or just a $1$ or, if it begins with a $1$, a $10$ followed by whatever or, if it begins with $11$, then a $110$ followed by whatever or a $111$ followed by at least a $0$ or a $1$
Does this appear to be correct? If I'm wrong, could you please tell me why?
Thank you,
• You could have written it more compactly, e.g. $\epsilon + 1 + (0 + 1(\epsilon + 1 + 11)0)(1 + 0)^*$, but the idea is basically the same. Commented Sep 29, 2015 at 23:25
• We discourage "please check whether my answer is correct" questions, as only "yes/no" answers are possible, which won't help you or future visitors. See here and here. Can you edit your post to ask about a specific conceptual issue you're uncertain about? As a rule of thumb, a good conceptual question should be useful even to someone who isn't looking at the problem you happen to be working on. If you just need someone to check your work, you might seek out a friend, classmate, or teacher.
– D.W.
Commented Sep 29, 2015 at 23:28
Basically here two things are happening.
1. Regular languages are closed under compliment operation
2. If L1 is accepted by DFA1 and DFA2 is obtained by interchanging the final states to non-final states and non-final states to final states of DFA1 then the language accepted by DFA2 is compliment of L1. That is, if
• DFA1 is a quintuple (Q, Σ, δ , q0, F)
And L1 is language accepted by DFA1
• DFA2 is (Q, Σ, δ , q0, (Q - F))
And L2 is language accepted by DFA2
Then L2 is compliment of L1.
In this case the languages
L1 = {w∣w is any string except 11 and 111} and L2 = {v∣v is either 11 or 111} are compliment o each other . | 604 | 2,090 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-26 | latest | en | 0.937586 |
http://gmatclub.com/forum/if-x-y-and-z-are-positive-integers-x-is-a-factor-of-2y-141201.html?fl=similar | 1,484,735,062,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280266.9/warc/CC-MAIN-20170116095120-00526-ip-10-171-10-70.ec2.internal.warc.gz | 123,528,575 | 59,075 | If x, y, and z are positive integers, x is a factor of 2y : GMAT Problem Solving (PS)
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# If x, y, and z are positive integers, x is a factor of 2y
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If x, y, and z are positive integers, x is a factor of 2y, and 3y is a factor of z, which of the following must also be an integer?
A) $$\frac{y}{x}$$
B) $$\frac{2y}{6}$$
C) $$\frac{xy}{3z}$$
D) $$\frac{zx}{3y}$$
E) $$\frac{zy}{3x}$$
[Reveal] Spoiler: OA
Last edited by Bunuel on 24 Oct 2012, 04:18, edited 2 times in total.
Renamed the topic and edited the question.
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LM wrote:
If x, y, and z are positive integers, x is a factor of 2y, and 3y is a factor of z, which of the following must also be an integer?
A) $$\frac{y}{x}$$
B) $$\frac{2y}{6}$$
C) $$\frac{xy}{3z}$$
D) $$\frac{zx}{3y}$$
E) $$\frac{zy}{3x}$$
$$x$$ is a factor of $$2y$$, means that $$\frac{2y}{x}=integer$$.
Similarly, $$3y$$ is a factor of $$z$$, means that $$\frac{z}{3y}=integer$$. Multiply both sides of this equation by integer $$x$$: $$\frac{z}{3y}*x=integer*x$$ --> $$\frac{zx}{3y}=x*integer=integer*integer=integer$$.
Answer: D.
Hope it's clear.
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Re: If x, y, and z are positive integers, x is a factor of 2y [#permalink]
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31 Oct 2012, 05:12
Bunuel wrote:
LM wrote:
If x, y, and z are positive integers, x is a factor of 2y, and 3y is a factor of z, which of the following must also be an integer?
A) $$\frac{y}{x}$$
B) $$\frac{2y}{6}$$
C) $$\frac{xy}{3z}$$
D) $$\frac{zx}{3y}$$
E) $$\frac{zy}{3x}$$
$$x$$ is a factor of $$2y$$, means that $$\frac{2y}{x}=integer$$.
Similarly, $$3y$$ is a factor of $$z$$, means that $$\frac{z}{3y}=integer$$. Multiply both sides of this equation by integer $$x$$: $$\frac{z}{3y}*x=integer*x$$ --> $$\frac{zx}{3y}=x*integer=integer*integer=integer$$.
Answer: D.
Hope it's clear.
Hello Bunuel,
Can we simplify the options and then attempt the Questions
For Option 5, ZY/3X can be simplified as 3Y*Y/3X --> Y*Y/X which may or may not be true.
For example looking at option 4, ZX/3Y, if simplify it as ---> Z=3y and then the eqn becomes 3y*y/ 3y which gives x only
Please confirm
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Re: If x, y, and z are positive integers, x is a factor of 2y [#permalink]
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01 Nov 2012, 06:13
mridulparashar1 wrote:
Bunuel wrote:
LM wrote:
If x, y, and z are positive integers, x is a factor of 2y, and 3y is a factor of z, which of the following must also be an integer?
A) $$\frac{y}{x}$$
B) $$\frac{2y}{6}$$
C) $$\frac{xy}{3z}$$
D) $$\frac{zx}{3y}$$
E) $$\frac{zy}{3x}$$
$$x$$ is a factor of $$2y$$, means that $$\frac{2y}{x}=integer$$.
Similarly, $$3y$$ is a factor of $$z$$, means that $$\frac{z}{3y}=integer$$. Multiply both sides of this equation by integer $$x$$: $$\frac{z}{3y}*x=integer*x$$ --> $$\frac{zx}{3y}=x*integer=integer*integer=integer$$.
Answer: D.
Hope it's clear.
Hello Bunuel,
Can we simplify the options and then attempt the Questions
For Option 5, ZY/3X can be simplified as 3Y*Y/3X --> Y*Y/X which may or may not be true.
For example looking at option 4, ZX/3Y, if simplify it as ---> Z=3y and then the eqn becomes 3y*y/ 3y which gives x only
Please confirm
We are not given that z=3y, we are given that z=3y*integer (3y is a factor of z). Now, if we substitute z in $$\frac{zx}{3y}$$, we'l get: $$\frac{zx}{3y}=\frac{(3y*integer)*x}{3y}=integer*x=integer$$.
Hope it helps.
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Re: If x, y, and z are positive integers, x is a factor of 2y [#permalink]
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19 Sep 2015, 01:35
if we take x=4, y=2 and z=12...
x(=2) is factor of 2y(=4)
and 3y(=6) is a factor of z(=12)......
D it is.......
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Re: If x, y, and z are positive integers, x is a factor of 2y [#permalink] 19 Sep 2015, 01:35
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# If x, y, and z are positive integers, x is a factor of 2y
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2,218 | 6,571 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2017-04 | latest | en | 0.832383 |
https://www.codespeedy.com/extract-matrix-elements-in-the-spiral-form-in-python3/ | 1,675,369,112,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500041.18/warc/CC-MAIN-20230202200542-20230202230542-00651.warc.gz | 708,259,800 | 16,075 | # How to Extract Matrix Elements in the Spiral Form in Python3?
## CLOCKWISE SPIRAL FORM EXTRACTION OF MATRIX ELEMENTS
Clockwise Spiral Form can be best described as:
• Let’s consider a matrix of m x n dimension, where m is the number of rows and n is the number of columns.
• Let’s take a point, the point starts from the first element (1,1).
• Then the point will move in the right direction until the end (1,n).
• From there, the point will move downwards until the last element of the matrix (m,n).
• Then, the point move towards the second column (2,n), it will not hit the first column.
• From there it will rise up to the second row (2,2), it will not touch the first row.
• This process continues until it covers all the elements.
Read more here: Spiral array model
The Clockwise Spiral Form is shown below in the image.
Spiral Matrix Python
Now, let’s take a look at the code snippet
### PROGRAM to Extract Matrix Elements in the Spiral Form in Python3 :
```# Python3 program to print the given matrix in spiral form
def spiralPrint(m,n,a):
k=0;l=0
''' k - starting row index
m - ending row index
l - starting column index
n - ending column index
i - iterator '''
while(k<m and l<n):
# Print the first row from
# the remaining rows
for i in range(l,n) :
print(a[k][i],end=" ")
k += 1
# Print the last column from
# the remaining columns
for i in range(k,m) :
print(a[i][n-1],end=" ")
n -= 1
# Print the last row from
# the remaining rows
if(k<m):
for i in range(n-1,(l-1),-1) :
print(a[m-1][i], end = " ")
m-=1
# Print the first column from
# the remaining columns
if(l<n):
for i in range(m - 1, k - 1, -1) :
print(a[i][l],end=" ")
l+=1
# Driver Code
R=int(input("Enter the number of rows:"))
C=int(input("Enter the number of columns:"))
a=[]
print("Enter the elements of the matrix:")
for i in range(R):
l=list(map(int,input().split(" ")))
a.append(l)
print("The elements are taken out from the matrix in the clockwise spiral direction.")
spiralPrint(R,C,a)```
OUTPUT 1:
```Enter the number of rows:3
Enter the number of columns:3
Enter the elements of the matrix:
1 2 3
4 5 6
7 8 9
The elements are taken out from the matrix in the clockwise spiral direction.
1 2 3 6 9 8 7 4 5```
OUTPUT 2:
```Enter the number of rows:4
Enter the number of columns:4
Enter the elements of the matrix:
1 2 3 4
4 5 6 7
7 8 9 1
2 3 6 4
The elements are taken out from the matrix in the clockwise spiral direction.
1 2 3 4 7 1 4 6 3 2 7 4 5 6 9 8```
So, hope this tutorial helped you to clear your doubts. | 743 | 2,511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2023-06 | latest | en | 0.641491 |
https://draftlessig.org/what-is-law-of-inertia-and-examples/ | 1,679,845,981,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945473.69/warc/CC-MAIN-20230326142035-20230326172035-00445.warc.gz | 257,157,777 | 14,687 | # What is law of inertia and examples?
## What is law of inertia and examples?
Law of Inertia Objects want to stay in rest or motion unless an outside force causes a change. For example, if you roll a ball, it will continue rolling unless friction or something else stops it by force. You can also think about the way that your body keeps moving forward when you hit the brake on your bike.
What is inertia class 9th?
Inertia. Defintion: Inertia is a property or tendency of every object to resist any change in its state of rest or of uniform Force and Laws of Motion. It is measured by the mass of an object.
### What is law of inertia for Class 8?
Summary. Inertia is the tendency of an object to resist a change in its motion. Because of inertia, a resting object will remain at rest, and a moving object will keep moving. Objects with greater mass have greater inertia.
What is called inertia?
1a : a property of matter by which it remains at rest or in uniform motion in the same straight line unless acted upon by some external force. b : an analogous property of other physical quantities (such as electricity) 2 : indisposition to motion, exertion, or change : inertness.
#### What are three types of inertia?
It is of Three Types: Inertia of rest: Tendency of a body to remain in the state of rest. Inertia of direction: Tendency of a body to remain in a particular direction. Inertia of motion: Tendency of a body to remain in a state of uniform motion.
What is inertia class 9 with example?
Ans: Inertia is the natural tendency of a body to resist any change in its state. For example, while sleeping, you are in a free state.
## Is the law of inertia?
Law of inertia, also called Newton’s first law, postulate in physics that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.
What is the first law of inertia?
### Who postulated the law of inertia?
Galileo Galilei
The law of inertia was first formulated by Galileo Galilei for horizontal motion on Earth and was later generalized by René Descartes. | 486 | 2,162 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2023-14 | latest | en | 0.944254 |
https://www.meritnation.com/ask-answer/question/solve-this-q-19-abcd-is-a-parallelogram-bisectors-of-angles/rectilinear-figures/11924893 | 1,591,273,277,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347439928.61/warc/CC-MAIN-20200604094848-20200604124848-00231.warc.gz | 773,854,395 | 9,449 | # Solve this: Q.19. ABCD is a parallelogram, bisectors of angles A and B meet at E which lies on DC. Prove that AB = 2 AD.
nhgnghnytu
• 0
ABCD is a parallelogram , SO we know
AB = CD and BC = DA
And
AB | | CD and BC | | DA
Here ∠ DAE = ∠ EAB ( As given AE is angle bisector of ∠ BAD )
And
∠ CBE = ∠ EBA ( As given BE is angle bisector of ∠ ABC )
Now we take AE as transversal line and we know AB | | CD
So
∠ EAB = ∠ DEA ( As they are alternate interior angles )
So,
∠ DEA = ∠ DAE ( As we know ∠ DAE = ∠ EAB )
So,
AD = DE( As we know base angle theorem , if base angles of any triangle is equal than opposite sides are also equal )
And
Now we take AE as transversal line and we know AB | | CD
So
∠ EBA = ∠ CEB ( As they are alternate interior angles )
So,
∠ CBE = ∠ CEB ( As we know ∠ CBE = ∠ EBA )
So,
BC = EC( As we know base angle theorem , if base angles of any triangle is equal than opposite sides are also equal )
Hence
AD = DE = BC = EC ---- ( 1 )
We know
AB = CD
And
AB = DE + EC ( As CD = DE + EC ) | 440 | 1,071 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2020-24 | latest | en | 0.653021 |
https://www.coursehero.com/file/6822579/Additional-Questions-Chp-7/ | 1,493,475,604,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123491.79/warc/CC-MAIN-20170423031203-00456-ip-10-145-167-34.ec2.internal.warc.gz | 888,231,517 | 59,007 | # Additional Questions Chp 7 - Yield-to-maturity in 2 years =...
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1. The yield-to-maturity on a bond is the interest rate you earn on your investment if interest rates do not change. If you actually sell the bond before it matures, your realized return is known as the holding period yield. Suppose that today, you buy a 12 percent annual coupon bond for \$1,000. The bond has 13 years to maturity. Two years from now, the yield-to-maturity has declined to 11 percent and you decide to sell. What is your holding period yield? The yield-to-maturity at the time of purchase must be 12 percent, which is the coupon rate, because the bond was purchased at par value.
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Unformatted text preview: Yield-to-maturity in 2 years = 12 percent - 1 percent = 11 percent This cannot be solved directly, so it's easiest to just use the calculator method to get an answer. You can then use the calculator answer as the rate in the formula just to verify that your answer is correct. 2. Bryceton, Inc. has bonds on the market with 13 years to maturity, a yield-to-maturity of 9.2 percent, and a current price of \$895.09. The bonds make semiannual payments. What is the coupon rate? Coupon rate = (\$39 × 2)/\$1,000 = 7.80 percent...
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## This note was uploaded on 02/28/2012 for the course FIN 101 taught by Professor Chen during the Spring '12 term at CSU San Bernardino.
### Page1 / 2
Additional Questions Chp 7 - Yield-to-maturity in 2 years =...
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Ask a homework question - tutors are online | 447 | 1,840 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-17 | longest | en | 0.931688 |
https://discuss.codechef.com/t/time-complexity/16252 | 1,601,029,603,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400223922.43/warc/CC-MAIN-20200925084428-20200925114428-00179.warc.gz | 354,047,441 | 4,762 | # Time complexity
What is the the time complexity for this particular piece of code?
for ( int i=1 ; i < n ; i++){
for(int j=i ; j < n ; j+=i ){
//do something
}
}
1 Like
How ?
• by doing dry run, you can see for i = 1 inner loop runs n/1 times, i = 2 loop runs (n-1)/2 similarly for other i’s also
• (n-1)/1+(n-2)/2+(n-3)/3+(n-4)/4 … …
• n(1+1/2+1/3+1/4+ … … ) - (1+1+1+1 … (n-1) times)
• approximately n*(1+1/2+1/3+1/4 … … )
• 1+1/2+1/3+1/4 …1/n = ln(n)+γ+1n−14n2+O(n−4) see here which can be roughly approximated to log(n)
• so time complexity is O(n*log(n))
Hope this helps!
correct me, if I missed something.
1 Like
Not sure, but @sandeep_007 's answer seems quite accurate.
Sieve of erastothenes code is similar, and it runs in O(N Log(LogN)) , so i think its same, i.e. O(N(log(logN)), but the difference between O(NlogN) and O(Nlog(logN)) is negligible
(except when calculating this complexity comes in your college exam. There the difference can be as much as between an A and B )
My answer is slightly different from that of @sandeep_007
Answer: O(k*log(k)) where k = n-1
• For i=1 inner loop goes from 1 to n-1 i.e inner loop runs k/1 times.
• For i=2 inner loop runs for approximately k/2 times.
• …and so on till i=k for which inner loop runs only 1 time.
• (k/1)+(k/2)+(k/3)+ . . . +(k/k) = k((1/1)+(1/2)+(1/3)+ . . . +(1/k))*
• (1/1)+(1/2)+(1/3)+ . . . +(1/k) is \sum_{i=1}^{k} \frac{1}{k} which is harmonic number H_k and H_k = O(log(k))
Hence more accurate answer would be O(k*log(k)) where k = n-1
Seems ok. A better approximation expression would have been
(\sum_{i=1}^n \frac{(n-i)}{i}) + O(n)
Obviously, O(n*(log(n))) would be bigger than O(n).
thanks really!! | 608 | 1,723 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2020-40 | latest | en | 0.863586 |
https://www.physicsforums.com/threads/help-derive-newtons-law-of-cooling.388156/ | 1,713,289,990,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817103.42/warc/CC-MAIN-20240416155952-20240416185952-00403.warc.gz | 850,944,745 | 15,028 | HELP Derive Newton's Law of cooling
• Doctor_Doom
In summary, Newton's Law of Cooling is a mathematical equation that describes the rate of heat loss of an object as it cools in a surrounding environment. It can be derived by conducting experiments and has many practical applications in various fields. The law is based on the Second Law of Thermodynamics and has some limitations, such as assuming a constant environment and no external factors.
Doctor_Doom
Hey guys was hoping someone could give me a hint for the derivation of Newton's Law of cooling stated in the following equation form:http://img263.imageshack.us/img263/3143/a6e5a27835e9e5738016606.png
http://img716.imageshack.us/img716/8728/75784887.jpg
Last edited by a moderator:
Bump!
Found plenty of derivations in the form of temperature or temperature change but nothing in the form of heat (energy) rate transfers.
1. What is Newton's Law of Cooling?
Newton's Law of Cooling is a mathematical equation that describes the rate of heat loss of an object as it cools in a surrounding environment. It states that the rate of change of temperature of an object is directly proportional to the difference between the object's temperature and the temperature of its surroundings.
2. How can Newton's Law of Cooling be derived?
Newton's Law of Cooling can be derived by conducting experiments and collecting data on the temperature of an object as it cools in a controlled environment. The data can then be plotted on a graph and analyzed to determine the rate of heat loss and the proportionality constant. This process can be repeated with different objects and environments to validate the law.
3. What is the significance of Newton's Law of Cooling?
Newton's Law of Cooling has many practical applications in various fields such as engineering, physics, and meteorology. It is used to predict the cooling rates of objects in different environments and to design efficient cooling systems. It also helps in understanding the thermal behavior of different materials.
4. How does Newton's Law of Cooling relate to the Second Law of Thermodynamics?
Newton's Law of Cooling is based on the Second Law of Thermodynamics, which states that heat always flows from a hotter object to a colder object. This law provides a mathematical framework to quantify the rate of heat transfer and the direction of heat flow in a cooling process.
5. Are there any limitations to Newton's Law of Cooling?
Yes, there are some limitations to Newton's Law of Cooling. It assumes that the surrounding environment is at a constant temperature and that there are no external factors affecting the cooling process. In real-life situations, this may not always be the case, and the law may not accurately predict the cooling rate.
• Introductory Physics Homework Help
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• Classical Physics
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0
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514 | 721 | 3,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-18 | latest | en | 0.948365 |
https://en.wikipedia.org/wiki/Hashiwokakero | 1,721,668,443,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517890.5/warc/CC-MAIN-20240722160043-20240722190043-00644.warc.gz | 198,437,062 | 18,820 | # Hashiwokakero
A Hashiwokakero puzzle (left) and one of its solutions. The number of bridges connected to each "island" must match the number written on that island.
Hashiwokakero (橋をかけろ Hashi o kakero; lit. "build bridges!") is a type of logic puzzle published by Nikoli.[1] It has also been published in English under the name Bridges or Chopsticks (based on a mistranslation: the hashi of the title, , means bridge; hashi written with another character, , means chopsticks). It has also appeared in The Times under the name Hashi. In France, Denmark, the Netherlands, and Belgium it is published under the name Ai-Ki-Ai.
## Rules
Hashiwokakero is played on a rectangular grid with no standard size, although the grid itself is not usually drawn. Some cells start out with (usually encircled) numbers from 1 to 8 inclusive; these are the "islands". The rest of the cells are empty.
The goal is to connect all of the islands by drawing a series of bridges between the islands. The bridges must follow certain criteria:[2]
• They must begin and end at distinct islands, travelling a straight line in between.
• They must not cross any other bridges or islands.
• They may only run orthogonally (i.e. they may not run diagonally).
• At most two bridges connect a pair of islands.
• The number of bridges connected to each island must match the number on that island.
• The bridges must connect the islands into a single connected group.
## Solution methods
Solving a Hashiwokakero puzzle is a matter of procedural force: having determined where a bridge must be placed, placing it there can eliminate other possible places for bridges, forcing the placement of another bridge, and so on.[3]
An island showing '3' in a corner, '5' along the outside edge, or '7' anywhere must have at least one bridge radiating from it in each valid direction, for if one direction did not have a bridge, even if all other directions sported two bridges, not enough will have been placed. A '4' in a corner, '6' along the border, or '8' anywhere must have two bridges in each direction. This can be generalized as added bridges obstruct routes: a '3' that can only be travelled from vertically must have at least one bridge each for up and down, for example.
It is common practice to cross off or fill in islands whose bridge quota has been reached.[2] In addition to reducing mistakes, this can also help locate potential "short circuits": keeping in mind that all islands must be connected by one network of bridges, a bridge that would create a closed network that no further bridges could be added to can only be permitted if it immediately yields the solution to the complete puzzle. The simplest example of this is two islands showing '1' aligned with each other; unless they are the only two islands in the puzzle, they cannot be connected by a bridge, as that would complete a network that cannot be added to, and would therefore force those two islands to be unreachable by any others.
Any bridge that would completely isolate a group of islands from another group would not be permitted, as one would then have two groups of islands that could not connect. This deduction, however, is not very commonly seen in Hashiwokakero puzzles.
Determining whether a Hashiwokakero puzzle has a solution is NP-complete, by a reduction from finding Hamiltonian cycles in integer-coordinate unit distance graphs.[4] There is a solution using integer linear programming in the MathProg examples included in GLPK.[5] A library of puzzles counting up to 400 islands as well as integer linear programming results are also reported.[6]
## History
Hashiwokakero first appeared in Puzzle Communication Nikoli in issue #31 (September 1990), although an earlier form of the puzzle appeared in issue #28 (December 1989).
3. ^ Malik, Reza Firsandaya; Efendi, Rusdi; Pratiwi, Eriska Amrina (March 2012), "Solving Hashiwokakero puzzle game with Hashi solving techniques and depth first search", Bulletin of Electrical Engineering and Informatics, 1 (1): 61–68, doi:10.11591/eei.v1i1.227 (inactive 31 January 2024)`{{citation}}`: CS1 maint: DOI inactive as of January 2024 (link) | 948 | 4,158 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-30 | latest | en | 0.949521 |
https://www.physicsforums.com/threads/kinematics-boat-question.712363/ | 1,586,542,717,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370511408.40/warc/CC-MAIN-20200410173109-20200410203609-00541.warc.gz | 1,057,681,881 | 15,317 | # Kinematics boat question
## Main Question or Discussion Point
a boat takes 3.1 hours to travel 15 km down a river, then takes 5.1 hours to travel the same distance upstream. how fast is the river traveling.
I found the boat is traveling at 4.84km/hr down the stream and 2.94 km/hr up it. not sure where to go next.
Any help would be appreciated.
## Answers and Replies
Related Classical Physics News on Phys.org
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d=rt
Let's say the river's speed is 1 kph and the boat's speed is 2 kph then the boat will travel downstream at 3 kph and upstream at 1 kph. So when you go downstream you add the speeds, but when going upstream, you subtract the river's speed from the boat's speed. Try working that backwards. | 194 | 760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-16 | longest | en | 0.916381 |
http://mathonline.wikidot.com/differentiation-and-uniformly-convergent-series-of-functions | 1,720,985,528,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514638.53/warc/CC-MAIN-20240714185510-20240714215510-00422.warc.gz | 24,362,905 | 5,112 | Differentiation and Uniformly Convergent Series of Functions
# Differentiation and Uniformly Convergent Series of Functions
On the Differentiation and Uniformly Convergent Sequences of Functions page we saw that if $(f_n(x))_{n=1}^{\infty}$ is a sequence of functions that are differentiable on $(a, b)$, $(f_n'(x))_{n=1}^{\infty}$ is uniformly convergent to a function $g(x)$ on $(a, b)$, and if there exists an $x_0 \in (a, b)$ such that the numerical sequence $(f_n(x_0))_{n=1}^{\infty}$ converges, then we can conclude that $(f_n(x))_{n=1}^{\infty}$ uniformly converges to some function $f(x)$ and that $f'(x) - g(x)$.
We will now state and prove an analogous theorem for series of differentiable functions.
Theorem 1: Let $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ be a series of differentiable functions on $(a, b)$. Suppose that $\displaystyle{\sum_{n=1}^{\infty} f_n'(x)}$ is uniformly convergent to $g(x)$ on $(a, b)$ and that there exists an $x_0 \in (a, b)$ such that the numerical series $\displaystyle{\sum_{n=1}^{\infty} f_n(x_0)}$ converges. Then $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ is uniformly convergent to some function $f(x)$ on $(a, b)$ and $f'(x) = g(x)$.
• Proof: For all $n \in \mathbb{N}$ we have that the sequence of partial sums, $(s_n(x))_{n=1}^{\infty}$ for $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ has general term:
(1)
\begin{align} \quad s_n(x) = \sum_{k=1}^{n} f_k(x) = f_1(x) + f_2(x) + ... + f_n(x) \end{align}
• Each $s_n$ is differentiable (as a finite sum of differentiable functions is differentiable) and taking the derivative of both sides gives us:
(2)
\begin{align} \quad s_n'(x) = \sum_{k=1}^{n} f_k'(x) = f_1'(x) + f_2'(x) + ... + f_n'(x) \end{align}
• Notice that this is simply the general term for the sequence of partial sums, $(s_n'(x))_{n=1}^{\infty}$ for the series $\displaystyle{\sum_{n=1}^{\infty} f_n'(x)}$, which is uniformly convergent to $g(x)$ on $(a, b)$.
• Now, since $\displaystyle{\sum_{n=1}^{\infty} f_n(x_0)}$ converges, the numerical sequence of partial sums $(s_n(x_0))_{n=1}^{\infty}$ converges.
• So $(s_n(x))_{n=1}^{\infty}$ is a sequence of functions that are differentiable, $(s_n'(x))_{n=1}^{\infty}$ uniformly converges to $g(x)$ on $(a, b)$, and there exists an $x_0 \in (a, b)$ such that the numerical sequence $(s_n(x_0))_{n=1}^{\infty}$ converges. So by the theorem presented on the Differentiation and Uniformly Convergent Sequences of Functions we have that $(s_n(x))_{n=1}^{\infty}$ is uniformly convergent on $(a, b)$ to some limit function $f(x)$, i.e., $\displaystyle{\sum_{n=1}^{\infty} f_n(x)}$ converges uniformly to $f(x)$ on $(a, b)$, and moreover, $f'(x) = g(x)$. $\blacksquare$ | 927 | 2,686 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-30 | latest | en | 0.731896 |
https://benidormclubdeportivo.org/what-are-vertices-of-a-cube/ | 1,652,807,923,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662519037.11/warc/CC-MAIN-20220517162558-20220517192558-00269.warc.gz | 185,443,368 | 7,021 | Vertices, faces and also edges come up a lot in major school geometry when youngsters are learning about the nature of 3d shapes. Below we define what every of this mean and how to job-related out the variety of vertices, faces and also edges for any kind of shape. We also include the variety of edges, faces and also vertices that the most typical shapes.
You are watching: What are vertices of a cube
This vocabulary is presented in the national curriculum at Year 2, and so the following information can be used with student throughout major school years. Also Year 1 pupils can start to engage with nature of shapes in this method if you desire to give them a head start!
Vertices, faces and also edges of typical 3d shapes
### What space vertices?
Vertices (or vertex in that singular form) space the suggest where two or an ext line segments or edge meet. These deserve to be used to explain 2d and 3d shapes. ‘Corners’ is a word supplied synonymous with vertices. Whereby possible, vertex and also vertices have to be offered instead. A cube has 8 vertices. 7 room visible here and also one is hidden.
### What are edges?
Edges room the heat segments that join one vertex to another and also are additionally where the shape’s deals with meet. These have the right to be provided to explain 2d and 3d shapes. Although numerous shapes have straight lines and straight edges, there are forms which have curved edges, such together a hemisphere. A cube will have 12 right edges as viewed below; 9 room visible and 3 room hidden.
FREE 3D shapes Independent Recap Worksheets
Help her Year 2 (and older) pupils review vertices, faces and edges v our complimentary Independent Recap worksheets.
### What room faces?
Faces room the level surface of a solid shape. Because that example, a cuboid has 6 faces. Once thinking about 2d and also 3d shapes, that is essential to understand that a 2d shape merely represents the challenge of a 3d shape.
It is likewise important to recognize that together our truth is created in 3 dimensions, that is impossible to physically handle 2d forms as we are surrounded by 3-dimensional shapes. Therefore, if in her classroom there is a drawer labelled ‘2d Shapes’ this have to be gotten rid of as it is teaching youngsters a misconception. Although an interactive ide for the classroom, 2d shapes can only exist as 2 dimensional drawings.
You deserve to have both level faces and also curved faces, but I find it useful to describe curved encounters as curved surfaces as it matches well through the visual of the shape.
There are three faces that space visible and three that are covert on the cube below.
### Vertices, faces and also edges of common 3d shapes
How many faces, edges and also vertices walk a cuboid have?
A cuboid has 8 vertices.
A cuboid has actually 12 edges.
A cuboid has actually 6 faces.
How countless faces, edges and also vertices walk a cylinder have?
A cylinder has 0 vertices.
A cylinder has actually 2 edges.
A cylinder has 2 faces and also 1 curved surface.
How countless faces, edges and vertices walk a hemisphere have?
A hemisphere has actually 0 vertices.
A hemisphere has 1 curved edge.
A hemisphere has actually 1 face and 1 curved surface.
How many faces, edges and vertices walk a cone have?
A cone has actually 1 vertex.
A cone has 1 edge.
A hemisphere has 1 face and 1 curved surface.
How numerous faces, edges and also vertices walk a tetrahedron have?
A tetrahedron has 4 vertices.
A tetrahedron has 6 edges.
A tetrahedron has 4 faces.
How numerous faces, edges and vertices does a ball have?
A sphere has actually 0 vertices.
A tetrahedron has actually 0 edges.
A tetrahedron has 1 bent surface.
How plenty of faces, edges and vertices walk a prism have?
A prism is a heavy object, geometric shape or polyhedron whereby the deals with of both ends room the same shape. Together such, students will certainly come throughout many species of prism throughout their schooling. Common ones encompass cubes, cuboids, triangular prisms, pentagonal prisms and hexagonal prisms.
ShapePictorialFacesEdgesVertices
Triangular Prism
596
Pentagonal Prism
71510
Hexagonal Prism
81812
Cube
6812
Cuboid6812
### When will certainly my boy learn about vertices, faces and also edges in main school?
Children should be formally presented to the vocabulary that vertices, faces and also edges in Year 2 once studying geometry. However, teachers may make the choice to present this vocabulary previously on.
Year 2 pupils must be maybe to:
identify and describe the nature of 3-D shapes, consisting of the variety of edges, vertices and also faces
The non-statutory guidance says that:
Pupils handle and also name a wide range of typical 2-d and also 3-d shapes including: quadrilaterals and polygons, and also cuboids, prisms and cones, and also identify the nature of each form (for example, variety of sides, variety of faces). Student identify, compare and sort shapes on the basis of their properties and also use vocabulary precisely, such as sides, edges, vertices and faces.
From this point on, the national curriculum walk not recommendation vertices, faces and edges clearly again, so teachers in various other year teams will have actually to proceed to usage this vocabulary once looking in ~ shape
A on slide from Third room Learning’s online maths intervention, using the relationship between 2d and 3d shapes to aid Year 3 pupils identify vertices, faces and edges.
### How do vertices, faces and also edges relate come other areas of maths?
Students will usage the expertise of vertices, faces and also edges once looking in ~ 2d shapes and 3d shapes. Learning what edge are and also identifying lock on compound forms is an important for detect the perimeter and area the 2d link shapes. That is critical foundation for later on years when dealing with different maths theorems, such together graph theory and parabolas.
### How carry out vertices, faces and edges called to genuine life?
Any thing in genuine life has vertices, faces and also edges. For example, a decision is one octahedron – it has eight faces, twelve edges and also six vertices. Understanding these properties for different three-dimensional shapes lays the structure for various markets such as architecture, interior design, engineering and more.
### Vertices, faces and edges instance questions
1. Describe what a peak is.
(Answer: A peak is where 2 lines meet)
2. How countless edges go a triangular prism have?
3. How plenty of vertices walk a cone have?
4. How numerous faces walk a cuboid have? What are the 2D forms of those faces?
(Answer: 6 faces. They have the right to have 2 square faces and 4 rectangular encounters or simply 6 rectangle-shaped faces.)
5. For every the common prisms (cubes, cuboids, triangle prisms, pentagonal prisms and hexagonal prisms) include the faces and also vertices together and also subtract the edges. What carry out you notification about the answers?
(Answer: The price is always 2. This is well-known as Euler’s formula (number of vertices – number of edges + variety of faces = 2)
Wondering about how to explain other an essential maths vocabulary to her children? inspect out our main Maths Dictionary, or try these:
You can uncover plenty the geometry class plans and also printable worksheets for primary school college student on the Third an are Learning Maths Hub.
See more: Question: What Percent Of 50 Is 3, 3 Is 50 Percent Of What Number
Online 1-to-1 maths lessons reliable by schools and also teachersEvery mainly Third an are Learning’s maths specialist tutors assistance thousands of major school children with weekly online 1-to-1 lessons and maths interventions. Since 2013 we’ve aided over 100,000 kids become more confident, maybe mathematicians. Learn much more or request a personalised quote come speak come us about your needs and how we can help.
Our virtual tuition for maths programme gives every son with their own experienced one come one maths tutor | 1,747 | 8,039 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2022-21 | latest | en | 0.963021 |
https://slideplayer.com/slide/6091640/ | 1,623,519,173,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487586239.2/warc/CC-MAIN-20210612162957-20210612192957-00231.warc.gz | 498,861,679 | 22,640 | # Choosing Statistical Procedures
## Presentation on theme: "Choosing Statistical Procedures"— Presentation transcript:
Choosing Statistical Procedures
Types of Inferential Statistics
Parametric Statistics: estimate the value of a population parameter from the characteristics of a sample Assumes the values in a sample are normally distributed Interval/Ratio level data required Nonparametric Statistics: No assumptions about the underlying distribution of the sample Used when the data do not meet the assumption for a nonparametric test (ordinal and nominal data)
What are We Testing Anyway?
Parametric Statistics: comparing the means of two or more groups relative to the variance within the groups Nonparametric Statistics: comparing the medians or ranks of two or more groups Testing the null hypothesis Statistical Significance: determination that the differences between groups are large enough to be unlikely to have occurred by chance
Steps in the Test of Significance
State the null hypothesis. State the alternative hypothesis. Set the level of significance associated with the null hypothesis (Type I Error). Select the appropriate test statistic. Compute the test statistic value. Determine the critical value needed for rejection of the null hypothesis for the particular statistic. Compare the obtained value to the critical value. Make a decision: Accept or reject the null hypothesis.
All These Tests! How do I know which one?
The type of statistical test you use depends on several factors: Number of independent variables Number of levels of the independent variable(s) Number of dependent variables Independent vs. dependent samples (between vs. within groups design) Scale of measurement of the dependent variable
Selecting Statistical Tests
What Can We Conclude? Intact Groups Design (Quasi-Experiments)
Subject Variables: conditions over which the experimenter has no direct control Cannot establish cause and effect Can only conclude that the groups are different from one another True Experiments Manipulated Variables: conditions that the experimenter controls directly and to which he randomly assigns participants Allows for cause and effect interpretations
The Power of a Test Power of a Test: the probability that one will reject H0 when it is a false statement. The power of a statistical analysis is represented as 1 – β. Reminder: β = the probability of making a Type II Error It is influenced by: μX - μ0 n σ α Type 2 error the probability of failing to reject a null hypothesis when it is false missed opportunity Power probability of NOT failing to reject the null when it is false the probability of accurately rejecting a false null hypothesis
Effect Size Statistics
Due to the role of N (sample size) in the formulae for parametric statistics, a large sample size can make a negligible difference between groups appear significant. Because of this, current APA guidelines recommend computing effect size statistics in addition to parametric comparisons.
Uses of the Effect Size Statistic
The effect size statistic can be used to: Estimate the true effect of the IV Compare the results of one research project to the results of other projects Estimate the power of the statistic Estimate the number of subjects one needs in a research project to maximize the chance of rejecting a false hypothesis (power) | 626 | 3,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2021-25 | latest | en | 0.839893 |
http://chestofbooks.com/architecture/Cyclopedia-Carpentry-Building-4-6/Conic-Sections.html | 1,519,274,560,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814002.69/warc/CC-MAIN-20180222041853-20180222061853-00665.warc.gz | 69,791,751 | 6,167 | If a plane intersects a cone the geometrical figures thus formed are called conic sections. A plane perpendicular to the base and passing through the vertex of a right circular cone forms an isosceles triangle. If the plane is parallel to the base the intersection of the plane and conical surfaces will be the circumference of a circle.
Fig.
Fig. 866.
Fig. 86c.
Fig. 86d illustrations of Method of Forming Conic Sections.
Ellipse. An ellipse is a curve formed by the intersection of a plane and a cone, Fig. 86a, or cylinder, Fig. 866, the plane being oblique to the axis but not cutting the base. An ellipse may be defined as a curve generated by a point moving in a plane in such a manner that the sum of the distances from the point to two fixed points shall always be constant.
The two fixed points are called foci, Fig. 87, and shall lie on the longest line that can be drawn in the ellipse which is called the major axis; the shortest line is called the minor axis, and is perpendicular to the major axis at its middle point, called the center.
Fig. 87. Ellipse.
Fig. 88. Parabola.
An ellipse may be constructed if the major and minor axes are given or if the foci and one axis are known.
Parabola. The parabola is a curve formed by the intersection of a cone and a plane parallel to an element of the cone, Fig. 86c. This curve is not a closed curve for the branches approach parallelism.
A parabola may be defined as a curve every point of which is equally distant from a line and a point.
The point is called the focus, Fig. 88, and the given line, the directrix. The line perpendicular to the directrix and passing through the focus is the axis. The intersection of the axis and the curve is the vertex.
Hyperbola. This curve is formed by the intersection of a plane and a cone, the plane being parallel to the axis of the cone, Fig. 86d.
Fig. 89. Hyperbola.
Like the parabola, the curve is not closed, the branches constantly diverging.
An hyperbola is defined as a plane curve such that the difference between the distances from any point in the curve to two fixed points is equal to a given distance.
The two fixed points are the foci and the line passing through them is the transverse axis, Fig. 89.
Rectangular Hyperbola. The form of hyperbola most used in Mechanical Engineering is called the rectangular hyperbola because it is drawn with reference to rectangular coordinates. This curve is constructed as follows: In Fig. 90, 0 X and 0 Y are the two coordinate axes drawn at right angles to each other. These lines are also called asymptotes. Assume A to be a known point on the curve. Draw A C parallel to 0 X and A D perpendicular to OX. Mark off any convenient points on A C such as E, F, G, and H, and through these points draw EE', FF', GG', and HH', perpendicular to 0 X. Connect E, F, G, H, and C with 0. Through the points of intersection of the oblique lines and the vertical line A D draw the horizontal lines LL', MM', NN', PP', and QQ'. The first point on the curve is the assumed point A, the second point is R, the intersection of LL' and EE'. The third is the intersection S of MM' and FF'. The other points are found in the same way.
In this curve the products of the coordinates of all points are equal. Thus LR X RE' = MS X SF' = NT X TG'. | 788 | 3,296 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-09 | latest | en | 0.939707 |
http://mathhelpforum.com/algebra/92624-bit-rusty-my-log-rearranging-print.html | 1,529,728,433,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864940.31/warc/CC-MAIN-20180623035301-20180623055301-00523.warc.gz | 200,220,082 | 3,626 | # Bit rusty on my log rearranging
• Jun 12th 2009, 06:18 AM
Joel
Bit rusty on my log rearranging
Can you help me with these
Rewrite these relations in index form (that is, without using logarithims):
(a) logα (x + y) = logα x + logα y
(b) log ₁₀ x = 3 + log ₁₀ y
(c) log₃ x = 4log₃y
(d) 2log₂x + 3log₂y – 4log₂z = 0
(e) xlogα 2 = log α y
(f) log α x - log α y = n log α z
(g) ˝log₂ x + = ⅓log₂ y – 1
(h) 2 log₃(2x + 1) = 3 log₃ (2x – 1)
• Jun 12th 2009, 06:44 AM
Amer
Quote:
Originally Posted by Joel
Can you help me with these
Rewrite these relations in index form (that is, without using logarithims):
(a) logα (x + y) = logα x + logα y
(b) log ₁₀ x = 3 + log ₁₀ y
(c) log₃ x = 4log₃y
(d) 2log₂x + 3log₂y – 4log₂z = 0
(e) xlogα 2 = log α y
(f) log α x - log α y = n log α z
(g) ˝log₂ x + = ⅓log₂ y – 1
(h) 2 log₃(2x + 1) = 3 log₃ (2x – 1)
$\displaystyle a)log_{a}(x+y) = log_{a} x + log_{a} y$
$\displaystyle log_{a} (x+y) = log_{a} xy$ I use $\displaystyle log A + log B = log (AB)$
$\displaystyle (x+y) = xy$
$\displaystyle b) log_{10} x =3 + log_{10} y$
$\displaystyle x=10^{3+log_{10} y }$
$\displaystyle x= 10^3 (10^{log_{10} y })$
$\displaystyle x= 1000(y)$ since $\displaystyle A^{log_{A} c} = c$
$\displaystyle e)xlog_{a} 2 = log_{a} y$
$\displaystyle log_{a}2^x= log_{a} y$ since $\displaystyle (a) log c = log c^a$
$\displaystyle 2^x = y$
(g) ˝log₂ x + = ⅓log₂ y – 1
$\displaystyle g)1/2log_{2}x = 1/3log_{2} y -1$
$\displaystyle log_{2} x^{1/2} - log_{2} y^{1/3} = -1$
$\displaystyle log_{a} y^{1/3} - log_{2} x^{1/2} = 1$ multiply with -1
$\displaystyle log_{a} \dfrac{y^{1/3}}{x^{1/2}} = 1$
$\displaystyle \dfrac{y^{1/3}}{x^{1/2}} = a$
• Jun 12th 2009, 06:58 AM
Amer
some log rules I attach them because in sometimes latex it is not clear if I use powers
Attachment 11865
• Jun 12th 2009, 07:43 AM
Soroban
Hello, Joel!
Some of your bases didn't show up . . . I'll use $\displaystyle b$.
Quote:
Rewrite in index form (that is, without using logarithims):
$\displaystyle (a)\;\;\log_b(x+y) \:=\:\log_b(x) + \log_b(y)$
We have: .$\displaystyle \log_b(x+y) \:=\:\log_b(xy)$
Then: .$\displaystyle x+y \:=\:xy$
Quote:
$\displaystyle (b)\;\;\log_b(x) \:= \:3 + \log_b(y)$
We have: .$\displaystyle \log_b(x) \;=\;\log_b(b^3) + \log_b(y) \;=\;\log_b(b^3y)$
Then: .$\displaystyle x \;=\;b^3y$
Quote:
$\displaystyle (c)\;\;\log_b(x) \:=\: 4\log_b(y)$
We have: .$\displaystyle \log_b(x) \;=\;\log_b(y^4)$
Then: .$\displaystyle x \;=\;y^4$
Quote:
$\displaystyle (d)\;\;2\log_b(x) + 3\log_b(y) - 4\log_b(z) \:=\:0$
We have: .$\displaystyle \log_b(x^2) + \log_b(y^3) - \log_b(z^4) \:=\:0 \quad\Rightarrow\quad \log_b\left(\frac{x^2y^3}{z^4}\right) \:=\:0$
Then: .$\displaystyle \frac{x^2y^3}{x^4} \:=\:1$
Quote:
$\displaystyle (e)\;\;x\log_b(2) \:=\: \log_b(y)$
We have: .$\displaystyle \log_b(2^x) \:=\:\log_b(y)$
Then: .$\displaystyle 2^x\:=\:y$
Quote:
$\displaystyle (f)\;\;\log_b(x) - \log_b(y) \:=\: n\log_b(z)$
We have: .$\displaystyle \log_b(x) - \log_b(y) - \log_b(z^n) \:=\:0 \quad\Rightarrow\quad \log_b\left(\frac{xy}{z^n}\right) \:=\:0$
Then: .$\displaystyle \frac{xy}{z^n} \;=\;1$
Quote:
$\displaystyle (g)\;\;\tfrac{1}{2}\log_b(x) \:=\: \tfrac{1}{2}\log_b(y) - 1$
We have: .$\displaystyle \log_b\left(x^{\frac{1}{2}}\right) \;=\;\log_b\left(y^{\frac{1}{2}}\right) - \log_b(b)$
. . $\displaystyle \log_b\left(\sqrt{x}\right) - \log_b\left(\sqrt{y}\right) + \log_b(b) \;=\;0 \quad\Rightarrow\quad \log_b\left(\frac{\sqrt{x}}{\sqrt{y}}\cdot b\right) \;=\;0$
Then: .$\displaystyle \frac{b\sqrt{x}}{\sqrt{y}} \:=\:1$
Quote:
$\displaystyle (h)\;\;2\log_b(2x + 1) \:=\: 3\log_b (2x – 1)$
We have: .$\displaystyle \log_b(2x+1)^2 \;=\;\log_b(2x-1)^3$
Then: .$\displaystyle (2x+1)^2 \;=\;(2x-1)^3$ | 1,644 | 3,755 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-26 | latest | en | 0.571479 |
https://brainly.ph/question/293883 | 1,485,272,071,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560284429.99/warc/CC-MAIN-20170116095124-00458-ip-10-171-10-70.ec2.internal.warc.gz | 793,333,441 | 9,898 | # A player hits a football which is initially at rest and moves 22 m/s. Find the: a. momentum of the football if it has a mass of 0.48 kg b. force exerted on it when the time of contact is 0.03s c. impulse in the football
1
by taravela
## Answers
2016-02-22T23:47:13+08:00
1. P= m x v
P= 0.48 kg x 22 m/s
P= 10.56 kg.m/s
2. F=m x a
F= 0.48 kg x 733 m/s^2
F= 352 N
3. I = f x t
I = 352 N x 0.03 s
I = 10.56 kg.m/s | 179 | 417 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2017-04 | latest | en | 0.783542 |
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# Is x^2 + y^2 > 100?
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Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2 [#permalink]
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28 Oct 2012, 09:44
Yes there is something dicey. In the solution he has added (1) and (2) and proved that X^2+y^2>100. Request some expert to reply!
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Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2 [#permalink]
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28 Oct 2012, 21:56
rajathpanta wrote:
Yes there is something dicey. In the solution he has added (1) and (2) and proved that X^2+y^2>100. Request some expert to reply!
I found this solution from Bunuel.
Statement 2. $$x^2 + y^2 + 2xy > 200$$
Since $$(x-y)^2$$ is cannot be lesser than 0, (square of a number is always positive or 0), the minimum value that 2xy can take is equal to $$x^2 + y^2$$.
So statement 2 can be changed to be, $$x^2 + y^2 + x^2 + y^2 > 200$$.
So, $$x^2 + y^2 > 100$$.
That is a convincing method to show that B is infact the right answer.
Kudos Please... If my post helped.
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Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2 [#permalink]
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29 Oct 2012, 04:41
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rajathpanta wrote:
Is x^2+y^2>100??
(1) 2xy<100
(2) (x+y)^2>200
To me its only B. because statement 2 boils down to x+y>$$\sqrt{200}$$
Can someone explain the OA
Merging similar topics. You are right, OA should be B, not C. See here: is-x-2-y-108343.html#p859197
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Re: Is x^2 + y^2 > 100? [#permalink]
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31 Oct 2012, 15:49
It's B for me.
$$(x + y)^2 = x^2 + 2xy + y^2$$
The largest possible value $$2xy$$ can reach is $$(x + y)^2/2$$, that only occurs when $$x = y$$.
When $$x = y$$, it turns out that $$x^2 + 2xy + y^2 > 200$$ is $$x^2 + 2xx + x^2 > 200$$, which can be rearranged in $$2x^2 + 2x^2 > 200$$, meaning that $$x^2 + y^2$$ is at least half of the stated value, while $$2xy$$ can be at most the other half.
Any value above 200 will require that $$x^2 + y^2 > 100$$, making 2) sufficient.
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Re: Is x^2 + y^2 > 100? [#permalink]
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18 Feb 2013, 19:58
Bunuel wrote:
(2) (x + y)^2 > 200 --> $$x^2+2xy+y^2>200$$. Now, as $$(x-y)^2\geq{0}$$ (square of any number is more than or equal to zero) then $$x^2+y^2\geq{2xy}$$
Hi Bunuel,
could you please explain why you followed with (x-y)^2 instead of (x+y)^2? Shouldn't (x-y)^2 be distributed as x^2-2xy+y^2 in which case it will be different from the original expression? Also, when you transfer 2xy to the other side from x^2+2xy+y^2, why do you keep it positive?
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Re: Is x^2 + y^2 > 100? [#permalink]
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19 Feb 2013, 05:43
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LinaNY wrote:
Bunuel wrote:
(2) (x + y)^2 > 200 --> $$x^2+2xy+y^2>200$$. Now, as $$(x-y)^2\geq{0}$$ (square of any number is more than or equal to zero) then $$x^2+y^2\geq{2xy}$$
Hi Bunuel,
could you please explain why you followed with (x-y)^2 instead of (x+y)^2? Shouldn't (x-y)^2 be distributed as x^2-2xy+y^2 in which case it will be different from the original expression? Also, when you transfer 2xy to the other side from x^2+2xy+y^2, why do you keep it positive?
We have $$(x + y)^2 > 200$$ which is the same as $$x^2+2xy+y^2>200$$.
Now, we need to find the relationship between x^2+y^2 and 2xy.
Next, we know that $$(x-y)^2\geq{0}$$ --> $$x^2-2xy+y^2\geq{0}$$ --> $$x^2+y^2\geq{2xy}$$.
So, we can safely substitute $$2xy$$ with $$x^2+y^2$$ in $$x^2+2xy+y^2>200$$ (as $$x^2+y^2$$ is at least as big as $$2xy$$ then the inequality will still hold true) --> $$x^2+(x^2+y^2)+y^2>200$$ --> $$2(x^2+y^2)>200$$ --> $$x^2+y^2>100$$. Sufficient.
Hope it's clear.
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Re: Is x^2 + y^2 > 100? [#permalink]
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19 Feb 2013, 21:14
Thanks, it's clear now!
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Re: Is x^2 + y^2 > 100? [#permalink]
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30 Mar 2013, 06:32
Bunuel wrote:
arvindg wrote:
Problem source: Veritas Practice Test
Is x^2 + y^2 > 100?
(1) 2xy < 100
(2) (x + y)^2 > 200
Is x^2 + y^2 > 100?
(1) 2xy < 100 --> clearly insufficient: if $$x=y=0$$ then the answer will be NO but if $$x=10$$ and $$y=-10$$ then the answer will be YES.
(2) (x + y)^2 > 200 --> $$x^2+2xy+y^2>200$$. Now, as $$(x-y)^2\geq{0}$$ (square of any number is more than or equal to zero) then $$x^2+y^2\geq{2xy}$$ so we can safely substitute $$2xy$$ with $$x^2+y^2$$ (as $$x^2+y^2$$ is at least as big as $$2xy$$ then the inequality will still hold true) --> $$x^2+(x^2+y^2)+y^2>200$$ --> $$2(x^2+y^2)>200$$ --> $$x^2+y^2>100$$. Sufficient.
Are you sure the OA is C?
Brunel I have one confusion,
As you said that x^2+y^2 is at least as big as 2xy and so
2(x^2 + y^2)>200 but if x^2 +y^2 is = 2xy then wont 2(x^2+y^2)<200 as 4xy<200. (2xy<100, First statement, two statements cannot contradice)
So till the time we are not sure how big is the difference between x^2+y^2 and 2xy can we really say if 2(x^2+y^2)>200
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Re: Is x^2 + y^2 > 100? [#permalink]
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31 Mar 2013, 08:46
anujtsingh wrote:
Bunuel wrote:
arvindg wrote:
Problem source: Veritas Practice Test
Is x^2 + y^2 > 100?
(1) 2xy < 100
(2) (x + y)^2 > 200
Is x^2 + y^2 > 100?
(1) 2xy < 100 --> clearly insufficient: if $$x=y=0$$ then the answer will be NO but if $$x=10$$ and $$y=-10$$ then the answer will be YES.
(2) (x + y)^2 > 200 --> $$x^2+2xy+y^2>200$$. Now, as $$(x-y)^2\geq{0}$$ (square of any number is more than or equal to zero) then $$x^2+y^2\geq{2xy}$$ so we can safely substitute $$2xy$$ with $$x^2+y^2$$ (as $$x^2+y^2$$ is at least as big as $$2xy$$ then the inequality will still hold true) --> $$x^2+(x^2+y^2)+y^2>200$$ --> $$2(x^2+y^2)>200$$ --> $$x^2+y^2>100$$. Sufficient.
Are you sure the OA is C?
Brunel I have one confusion,
As you said that x^2+y^2 is at least as big as 2xy and so
2(x^2 + y^2)>200 but if x^2 +y^2 is = 2xy then wont 2(x^2+y^2)<200 as 4xy<200. (2xy<100, First statement, two statements cannot contradice)
So till the time we are not sure how big is the difference between x^2+y^2 and 2xy can we really say if 2(x^2+y^2)>200
No. If $$x^2 +y^2 = 2xy$$, then we can simply substitute 2xy to get the same: $$x^2+(x^2+y^2)+y^2>200$$ --> $$2(x^2+y^2)>200$$.
How did you get 2(x^2+y^2)<200?
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Re: Is x^2 + y^2 > 100? [#permalink]
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31 Mar 2013, 12:34
If 2(x^2+y^2)>200 and (x^2+y^2)=2xy then 2(2xy)>200 and 2xy>100 but the statement one says that 2xy<100. So even though we are not considering 1st statement, the answer cannot contradict the first statement.
Pls correct me if I am wrong
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Re: Is x^2 + y^2 > 100? [#permalink]
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31 Mar 2013, 12:44
anujtsingh wrote:
If 2(x^2+y^2)>200 and (x^2+y^2)=2xy then 2(2xy)>200 and 2xy>100 but the statement one says that 2xy<100. So even though we are not considering 1st statement, the answer cannot contradict the first statement.
Pls correct me if I am wrong
Frankly I don't understand your point. I just wanted to show that we CAN substitute $$2xy$$ with $$x^2+y^2$$ if $$x^2+y^2\geq{2xy}$$.
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Re: Is x^2 + y^2 > 100? [#permalink]
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05 Jul 2013, 02:44
Bumping for review and further discussion*. Get a kudos point for an alternative solution!
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Re: Is x^2 + y^2 > 100? [#permalink]
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14 Jan 2014, 07:45
Bunuel wrote:
arvindg wrote:
Problem source: Veritas Practice Test
Is x^2 + y^2 > 100?
(1) 2xy < 100
(2) (x + y)^2 > 200
Is x^2 + y^2 > 100?
(1) 2xy < 100 --> clearly insufficient: if $$x=y=0$$ then the answer will be NO but if $$x=10$$ and $$y=-10$$ then the answer will be YES.
(2) (x + y)^2 > 200 --> $$x^2+2xy+y^2>200$$. Now, as $$(x-y)^2\geq{0}$$ (square of any number is more than or equal to zero) then $$x^2+y^2\geq{2xy}$$ so we can safely substitute $$2xy$$ with $$x^2+y^2$$ (as $$x^2+y^2$$ is at least as big as $$2xy$$ then the inequality will still hold true) --> $$x^2+(x^2+y^2)+y^2>200$$ --> $$2(x^2+y^2)>200$$ --> $$x^2+y^2>100$$. Sufficient.
Are you sure the OA is C?
Hello Bunuel,
I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition.
So I guess the OA is correct i.e. C ...
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Re: Is x^2 + y^2 > 100? [#permalink]
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14 Jan 2014, 08:03
joe26219 wrote:
Bunuel wrote:
arvindg wrote:
Problem source: Veritas Practice Test
Is x^2 + y^2 > 100?
(1) 2xy < 100
(2) (x + y)^2 > 200
Is x^2 + y^2 > 100?
(1) 2xy < 100 --> clearly insufficient: if $$x=y=0$$ then the answer will be NO but if $$x=10$$ and $$y=-10$$ then the answer will be YES.
(2) (x + y)^2 > 200 --> $$x^2+2xy+y^2>200$$. Now, as $$(x-y)^2\geq{0}$$ (square of any number is more than or equal to zero) then $$x^2+y^2\geq{2xy}$$ so we can safely substitute $$2xy$$ with $$x^2+y^2$$ (as $$x^2+y^2$$ is at least as big as $$2xy$$ then the inequality will still hold true) --> $$x^2+(x^2+y^2)+y^2>200$$ --> $$2(x^2+y^2)>200$$ --> $$x^2+y^2>100$$. Sufficient.
Are you sure the OA is C?
Hello Bunuel,
I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition.
So I guess the OA is correct i.e. C ...
There is no flaw in my reasoning. Statement (2) says: $$x^2+2xy+y^2>200$$. Next, we know that $$x^2+y^2\geq{2xy}$$ is true for any values of $$x$$ and $$y$$. So we can manipulate and substitute $$2xy$$ with $$x^2+y^2$$ in (2) (because $$x^2+y^2$$ is at least as large as $$2xy$$): $$x^2+(x^2+y^2)+y^2>200$$ --> $$x^2+y^2>100$$.
By the way the OA is B, not C. VeritasPrep corrected it: is-x-2-y-108343.html#p860573
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Re: Is x^2 + y^2 > 100? [#permalink]
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14 Jan 2014, 21:30
joe26219 wrote:
Bunuel wrote:
arvindg wrote:
Problem source: Veritas Practice Test
Is x^2 + y^2 > 100?
(1) 2xy < 100
(2) (x + y)^2 > 200
Is x^2 + y^2 > 100?
(1) 2xy < 100 --> clearly insufficient: if $$x=y=0$$ then the answer will be NO but if $$x=10$$ and $$y=-10$$ then the answer will be YES.
(2) (x + y)^2 > 200 --> $$x^2+2xy+y^2>200$$. Now, as $$(x-y)^2\geq{0}$$ (square of any number is more than or equal to zero) then $$x^2+y^2\geq{2xy}$$ so we can safely substitute $$2xy$$ with $$x^2+y^2$$ (as $$x^2+y^2$$ is at least as big as $$2xy$$ then the inequality will still hold true) --> $$x^2+(x^2+y^2)+y^2>200$$ --> $$2(x^2+y^2)>200$$ --> $$x^2+y^2>100$$. Sufficient.
Are you sure the OA is C?
Hello Bunuel,
I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition.
So I guess the OA is correct i.e. C ...
To put it in words, think of it this way:
Is x^2 + y^2 > 100?
(2) (x + y)^2 > 200
which means: x^2 + y^2 + 2xy > 200
Now you know that 2xy is less than or equal to x^2 + y^2.
If 2xy is equal to $$(x^2 + y^2)$$, $$(x^2 + y^2)$$ will be greater than 100 since the total sum is greater than 200.
If 2xy is less than $$(x^2 + y^2)$$, then anyway $$(x^2 + y^2)$$ will be greater than 100 (which is half of 200).
So statement 2 is sufficient alone.
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Re: Is x^2 + y^2 > 100? [#permalink]
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16 Jan 2014, 04:04
Bunuel wrote:
joe26219 wrote:
Bunuel wrote:
Is x^2 + y^2 > 100?
(1) 2xy < 100 --> clearly insufficient: if $$x=y=0$$ then the answer will be NO but if $$x=10$$ and $$y=-10$$ then the answer will be YES.
(2) (x + y)^2 > 200 --> $$x^2+2xy+y^2>200$$. Now, as $$(x-y)^2\geq{0}$$ (square of any number is more than or equal to zero) then $$x^2+y^2\geq{2xy}$$ so we can safely substitute $$2xy$$ with $$x^2+y^2$$ (as $$x^2+y^2$$ is at least as big as $$2xy$$ then the inequality will still hold true) --> $$x^2+(x^2+y^2)+y^2>200$$ --> $$2(x^2+y^2)>200$$ --> $$x^2+y^2>100$$. Sufficient.
Are you sure the OA is C?
Hello Bunuel,
I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition.
So I guess the OA is correct i.e. C ...
There is no flaw in my reasoning. Statement (2) says: $$x^2+2xy+y^2>200$$. Next, we know that $$x^2+y^2\geq{2xy}$$ is true for any values of $$x$$ and $$y$$. So we can manipulate and substitute $$2xy$$ with $$x^2+y^2$$ in (2) (because $$x^2+y^2$$ is at least as large as $$2xy$$): $$x^2+(x^2+y^2)+y^2>200$$ --> $$x^2+y^2>100$$.
By the way the OA is B, not C. VeritasPrep corrected it: is-x-2-y-108343.html#p860573
Thank you Bunuel.Got it.Your explanation was really helpful (the highlighted line esp.)
I guess any value of 2xy should be assumed to satisfy the given condition, as it is given.
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Re: Is x^2 + y^2 > 100? [#permalink]
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10 May 2014, 08:00
Am still confused. Consider, x^2 + y^2 = 200 for the minimum value. That means, x + y is minimum 200 ^ 0.5 which means for a minimum sum of x + y, x has to be (200^0.5)/2 which gives x = 7.07. Sum up x^2 + y^2 at x = y = 7.07 we get x^2+ y^2 = approx. 98.30. Can someone please explain where I am going wrong.
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Re: Is x^2 + y^2 > 100? [#permalink]
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10 May 2014, 08:29
mayank_sharma wrote:
Am still confused. Consider, x^2 + y^2 = 200 for the minimum value. That means, x + y is minimum 200 ^ 0.5 which means for a minimum sum of x + y, x has to be (200^0.5)/2 which gives x = 7.07. Sum up x^2 + y^2 at x = y = 7.07 we get x^2+ y^2 = approx. 98.30. Can someone please explain where I am going wrong.
Your question is not clear... For example, why are you considering x^2 + y^2 = 200?...
Anyway, x^2 + y^2 = 200 doe not mean that $$x+y=\sqrt{200}$$.
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Re: Is x^2 + y^2 > 100? [#permalink]
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Re: Is x^2 + y^2 > 100? [#permalink]
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17 Jun 2015, 06:11
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!
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Is X^2+Y^2>100 ..
----> is [(x+y)^2 + (x-y)^2]/2 >100
So we can change the question stem to.. is (x+y)^2+(x-y)^2>200?
Statement A can in invalidated easily. Use different values of x and y.
statement B says that (X+Y)^2>200
Substitute statement B in question stem -
is Something >200 [(X+Y)^2] + something >=0 [(X-Y)^2] >200 .... Ans is obviously yes.
So B is the correct Ans. I hope this makes sense!
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Re: Is x^2 + y^2 > 100? [#permalink] 17 Jun 2015, 06:11
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## Presentation on theme: "5.1 Monomials Monomial Standard Notation Scientific Notation."— Presentation transcript:
5.1 Monomials Monomial Standard Notation Scientific Notation
Definition of a Monomial Monomial are one term algebraic expression. A term is an coefficient times a variable to a power. 7x 3 The coefficient is 7 The variable is x The power is 3 Monomial powers are never fractions or negative. Monomial will never a variable as a denominator.
Negative exponents If you remember how to multiply variables, then learning about negative is easier. When you multiply like variables you add their exponents
Negative exponents When you divide like variables you subtract their exponents
Negative exponents When you divide like variables you subtract their exponents
Simplify with only positive exponents
Properties of Powers Power of Power Power of Product Power of Fractions
Simplify with only positive exponents
Scientific Notation A way to express really big or small number in terms of a power of ten. Scientific Notation needs to have one number in front of the decimal. 3,120,000 would become 3.12 X 10 6 0.00000451 would become 4.51 X 10 -6
Write Standard Notation in Scientific Notation 4,560,000 0.000092
You can multiply with scientific notation (3 X 10 4 )(5 X 10 5 ) = (3X5)(10 4 X10 5 ) = 15 x 10 9 = 1.5 x 10 x 10 9 = 1.5 x 10 10
You can multiply with scientific notation (1.8 x 10 -4 )(4 x 10 7 ) =
Division of scientific notation
Now something from the 70’s The power of TEN http://www.youtube.com/watch?v=0fKBhvDjuy0
Homework Page 226 – 228 # 19 – 59 odd 64,65, 80 – 84 even
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19.6 The Divide-And-Conquer Method
This presentation is a summary of the divide-and-conquer method. For more in-depth coverage of this algorithm, see Refs. [1, pp. 216-228], [19, pp. 359-363], and [26, pp. 229-232].
The recursive algorithm divides a symmetric tridiagonal matrix into submatrices and then applies the same algorithm to the submatrices. We will illustrate the method of splitting the problem into smaller submatrix problems using a 5 ×5 matrix
$T=\left[\begin{array}{ccccc}{a}_{1}& {b}_{1}& 0& 0& 0\\ {b}_{1}& {a}_{2}& {b}_{2}& 0& 0\\ 0& {b}_{2}& {a}_{3}& {b}_{3}& 0\\ 0& 0& {b}_{3}& {a}_{4}& {b}_{4}\\ 0& 0& 0& {b}_{4}& {a}_{5}\end{array}\right].$
Write A as a sum of two matrices as follows:
$T=\left[\begin{array}{ccccc}{a}_{1}& {b}_{1}& 0& 0& 0\\ {b}_{1}& {a}_{2}-{b}_{2}& 0& 0& 0\\ 0& 0& {a}_{3}-{b}_{2}& {b}_{3}& 0\\ 0& 0& {b}_{3}& {a}_{4}& {b}_{4}\\ 0& 0& 0& {b}_{4}& {a}_{5}\end{array}\right]+\left[\begin{array}{ccccc}0& 0& 0& 0& 0\\ 0& {b}_{2}& {b}_{2}& 0& 0\\ 0& {b}_{2}& {b}_{2}& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\end{array}\right].$
If we let
${T}_{1}=$
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# Bubble sort
Bubble sort is a simple sorting algorithm. It takes a lot of passes through the list to be sorted, comparing two items at a time, swapping these two items in the correct order if necessary. Bubble sort gets its name from the fact that the items that belong at the top of the list gradually "float" up there.
Bubble sort needs &Theta(n2) comparisons to sort n items and can sort in place. It is one of the simplest sorting algorithms to understand but is generally too inefficient for serious work sorting large numbers of elements.
It is essentially equivalent to insertion sort --- it compares and swaps the same pairs of elements, just in a different order. Naive implementations of bubble sort (like those below) usually perform badly on already-sorted lists (Θ(n2)), while insertion sort needs only Θ(n) operations in this case. Hence most modern algorithm textbooks strongly discourage use of the algorithm, or even avoid mentioning it, and prefer insertion sort instead. It is possible to reduce the best case complexity to Θ(n) if a flag is used to denote whether any swaps were necessary during the first run of the inner loop. In this case, no swaps would indicate an already sorted list. Also, reversing the order in which the list is traversed for each pass improves the efficiency somewhat.
The bubble sort algorithm is:
1. Compare adjacent elements. If the first is greater than the second, swap them.
2. Do this for each pair of adjacent elements, starting with the first two and ending with the last two. At this point the last element should be the greatest.
3. Repeat the steps for all elements except the last one.
4. Keep repeating for one fewer element each time, until you have no more pairs to compare.
Due to its simplicity, the bubble sort is often used to introduce the concept of an algorithm.
A simple implementation of bubble sort in Python:
```def bubblesort(array):
# Values for i from len(array)-1 down to 1 inclusive.
for i in range(len(array)-1, 0, -1):
# Values for j from 0 to i-1 inclusive.
for j in range(i):
if array[j] > array[j+1]:
array[j], array[j+1] = array[j+1], array[j]
```
Or in the C programming language:
```void bubbleSort(int *array, int length)
{
int i, j;
for(i = length - 1; i > 0; i--)
for(j = 0; j < i; j++)
if(array[j] > array[j+1]) /* compare neighboring elements */
{
int temp;
temp = array[j]; /* swap array[j] and array[j+1] */
array[j] = array[j+1];
array[j+1] = temp;
}
}
```
Note: You do not need the temp variable if using the xor swap algorithm. In C, the swapping can be done by:
``` array[j] = array[j] ^ array[j+1];
array[j+1] = array[j] ^ array[j+1];
array[j] = array[j] ^ array[j+1];
```
\n | 662 | 2,750 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2023-06 | latest | en | 0.900291 |
https://cs.stackexchange.com/questions/75160/higher-order-empirical-entropy-is-not-the-entropy-of-the-empirical-distribution | 1,718,264,741,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.74/warc/CC-MAIN-20240613060639-20240613090639-00877.warc.gz | 170,076,739 | 40,519 | # Higher order empirical entropy is not the entropy of the empirical distribution?
Basically, the problem is that I always thought that the (unnormalized) $k$th order empirical entropy $n\cdot H_k(x)$ (see "Backround" at the end of this post for more information) for a given string $x$ of length $n$ equals the entropy of the empirical distribution on the set of all words of length $n$ (the conditional probabilities are taken from the observed occurrences in $x$). Until today, I was totally sure that this is true, but it seems to be wrong for very easy examples (e.g. for $x=1011$).
To be more accurate, consider $\Sigma=\{0,1\}$, $x=1011$ and $k=1$ such that $$n\cdot H_k(x)=4\cdot H_1(x)=2.$$ On the other hand, the corresponding empirical distribution is a First-order Markov source described by
• $P(\text{The first letter is }0)=0$
• $P(\text{The first letter is }1)=1$
• $P(\text{Letter }0 \text{ immediately follows after letter }0)=0$
• $P(\text{Letter }1 \text{ immediately follows after letter }0)=1$
• $P(\text{Letter }0 \text{ immediately follows after letter }1)=0.5$
• $P(\text{Letter }1 \text{ immediately follows after letter }1)=0.5$
Now this probabilities yield a random variable $X$ on words of length $n=4$, where the Shannon entropy is $$H(X)=-\sum_{w\in\Sigma^4}P(w)\log P(w)=\frac{9}{4}.$$ (You only need to consider words of length $4$ which start with $1$ and do not contain $00$.)
But shouldn't be both measures intuitively give the same value? I did the calculations more than once, so I am pretty sure the mistake is in my understanding, not in the calculation. Or did I always thought something wrong? But it seems so clear that the empirical entropy is the entropy of the empirical distribution (also for higher order Markov sources).
Backround
For a given string $x\in\Sigma^n$ over an alphabet of size $k$, the ($0$-th order) empirical entropy is defined as $$H_0(x)=\frac{1}{n}\sum_{i=1}^{k}n_i\cdot\log \frac{n}{n_i},$$ where $n_i$ is the number of occurrences of the $i$th alphabet symbol. The $k$-order empirical entropy ($k\ge 1$) is then defined as $$H_k(x)=\frac{1}{n}\sum_{|w|=k}|S_w|H_0(S_w),$$ where $S_w$ is the string obtained by concatenating the characters immediately following occurrences of $w$ in $x$.
No, as your counterexample shows, it is not true. I don't understand why you previously thought it was true, either, so it's hard to say what might have gone wrong with your previous intuition.
We can say that
$$H(X) = H(X_1) + H(X_2 | X_1) + H(X_3 | X_1,X_2) + H(X_4 | X_1,X_2,X_3),$$
where $X_i$ denotes the $i$th character in $X$. that much is true.
In the example you gave, $H(X_1) = 0$, $H(X_2|X_1) = 1$, $H(X_3|X_1,X_2) = 0.5$, $H(X_4 | X_1,X_2,X_3) = 0.75$ (see footnote * for explanation). Thus $H(X) = 1 + 0.5 + 0.75 = 2.25$.
Do notice that $\Pr[X_i=1]$ and $H(X_{i+1}|X_1,\dots,X_i)$ both depend on $i$. This might be important.
Also, notice that $H(X_{i+1}|X_1,\dots,X_i)$ is unrelated to $H_1(x)$.
Perhaps you were expecting $\Pr[X_i=1]$ and $H(X_{i+1}|X_1,\dots,X_i)$ to be the same for all $i$? If that were true, there might be a relationship between $H_1(x)$ and $\Pr[X_i=1]$ and $H(X_{i+1}|X_1,\dots,X_i)$ (or something similar to them)... but it's not.
Footnote *: How did I calculate those? I used the formula
$$H(X_{i+1}|X_1,\dots,X_i) = \sum_v \Pr[(X_1,\dots,X_i) = v] \times H(X_{i+1}|(X_1,\dots,X_i) = v).$$
For $i=1,2$, we only need to consider two values of $v$; for $i=3$, we only need to consider three values of $v$.
• Thanks for your answer. I think my confusion is based on the fact that for the $0$th order empirical entropy, this is true, i.e. the empirical entropy is the Shannon entropy of the empirical distribution. So I thought it is true also for higher order Markov sources. Commented May 10, 2017 at 8:59
• Also, Paul Vitányi states in his paper "On empirical entropy" on side 5 that the Shannon entropy of the random variable $X$ "associated with the $k$th order empirical entropy computed from $x$" equals $n\cdot H_k(x)$. Do you have an idea which distribution on the set of words of length $n$ he could mean, i.e. which natural distribution on $\Sigma^n$ yields the Shannon entropy $n\cdot H_k(x)$ for $x\in\Sigma^n$ ? I always thought he refers to the empirical distribution as described in my question. Commented May 10, 2017 at 8:59 | 1,351 | 4,357 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-26 | latest | en | 0.869494 |
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# Which of the following inequalities is equal to |2x-|x||<6? A. -2<x<0
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Which of the following inequalities is equal to |2x-|x||<6? A. -2<x<0 [#permalink]
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04 Oct 2017, 00:45
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[GMAT math practice question]
Which of the following inequalities is equal to |2x-|x||<6?
A. -2<x<0
B. 0<x<6
C. -2<x<6
D. -2<x<2
E. -6<x<6
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Re: Which of the following inequalities is equal to |2x-|x||<6? A. -2<x<0 [#permalink]
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04 Oct 2017, 01:31
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MathRevolution wrote:
[GMAT math practice question]
Which of the following inequalities is equal to |2x-|x||<6?
A. -2<x<0
B. 0<x<6
C. -2<x<6
D. -2<x<2
E. -6<x<6
You can write the inequality as:
-6 < |2x-|x|| < 6
Case 1:
If x >= 0 ; |x| = x
|2x-|x|| becomes 2x - x = x
-6 < x < 6
But as x>=0 we only take the values as 0 < x < 6
Case 2:
If x < 0 ; |x| = -x
|2x-|x|| becomes 2x + x = 3x
-6 < 3x < 6
-2 < x < 2
But as x < 0 we only take the values as -2 < x < 0
Combining the results of Case 1 and 2
-2 < x < 6
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Re: Which of the following inequalities is equal to |2x-|x||<6? A. -2<x<0 [#permalink]
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04 Oct 2017, 00:47
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MathRevolution wrote:
[GMAT math practice question]
Which of the following inequalities is equal to |2x-|x||<6?
A. -2<x<0
B. 0<x<6
C. -2<x<6
D. -2<x<2
E. -6<x<6
Similar question: https://gmatclub.com/forum/which-of-the ... 47913.html
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Re: Which of the following inequalities is equal to |2x-|x||<6? A. -2<x<0 [#permalink]
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04 Oct 2017, 02:44
the maximum range provided in the answer is -6, -2, 2, 0 and 6
check for this range
-6, gives us that |-12-|-6||<6
|-12-6| <6
18<6 not true
Eliminate E
next value given in the answer series is -2
Holds good, so left value in the number line should be 2
Now check for right value by putting 6
Holds good so the range should be -2 to 6
So clearly answer should be C
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Re: Which of the following inequalities is equal to |2x-|x||<6? A. -2<x<0 [#permalink]
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06 Oct 2017, 01:42
=>
We should split real number of x into two cases x ≥ 0 and x < 0.
Case 1: x ≥ 0
|2x-|x||<6 ⬄ | 2x – x | < 6 ⬄ |x| < 6 ⬄ -6<x<6.
From the assumption of the case, x ≥ 0, we have 0 ≤ x < 6.
Case 2: x < 0
|2x-|x||<6 ⬄ | 2x + x | < 6 ⬄ |3x| < 6 ⬄ |x| < 2 ⬄ -2<x<2.
From the assumption of the case, x < 0, we have -2<x<0.
Thus, we have -2 < x < 6.
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Which of the following inequalities is equal to |2x-|x||<6? A. -2<x<0 [#permalink]
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09 Feb 2019, 03:24
Bunuel - Can you help me understand where did I go wrong?
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Which of the following inequalities is equal to |2x-|x||<6? A. -2<x<0 [#permalink] 09 Feb 2019, 03:24 | 1,705 | 4,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-24 | latest | en | 0.833845 |
https://dsp.stackexchange.com/questions/93496/how-do-you-properly-cut-out-negative-frequencies-from-fft-of-a-real-signal-if-it | 1,718,750,719,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861794.76/warc/CC-MAIN-20240618203026-20240618233026-00555.warc.gz | 195,475,104 | 41,140 | # How do you properly cut out negative frequencies from FFT of a real signal if it reduces sample size?
If a real time signal has $$N$$ samples, then the magnitude of its FFT will also have $$N$$ samples--half of which will have positive frequencies and half of which will have negative frequencies. The distribution of these magnitudes will be symmetric about frequency $$= 0$$. If you cut out the $$\frac{N}{2}$$ negative frequency samples because they're not physically meaningful, you've cut your sample by half. What's the proper way to handle this?
If it's at all helpful, I used Python's Scipy FFT to do the calculation.
This is a computational follow-up to this conceptual question: What is the physical significance of negative frequencies?
• What’s the proper way to handle what? If you want to reconstruct your signal, you’ll need the negative frequencies. Is that what you’re asking?
– Jdip
Commented Apr 4 at 18:49
• For my application, the negative frequencies don't make sense, so my thought is to simply ignore them and only use the positive frequencies. Would that make sense to do? Commented Apr 4 at 19:04
• What is your application? If it's just analysis and you don't need to modify/synthesize back to time-domain, then yes it makes sense.
– Jdip
Commented Apr 4 at 19:27
• uhm, keep in mind that your $N$ samples going into the FFT are real values for the FFT to display the Hermitian symmetry where the "negative frequencies" (those are the bins of the FFT between $X[\tfrac{N}2+1]$ and $X[N-1]$ inclusive) are a complex conjugate reflection of the positive frequencies (between $X[1]$ and $X[\tfrac{N}2-1]$ inclusive). The number of real parameters that define an $N$-point FFT is actually $2N$ because the imaginary parts of all of those samples of $x[n]$ going in are zero. Commented Apr 4 at 19:54
• All I want to see is the distribution of frequencies in the data. I don't need to do an inverse FFT to go back to time space, although if I did, then I would, of course, use the full set of positive and negative frequencies. Commented Apr 4 at 19:55
For a real input vector, the $$0^{th}$$ bin will be real. Then for bins $$0 < n < N/2$$, the values will be complex, so each will be good for two real numbers (I.e., $$a + jb$$ contains two real numbers: $$a$$ and $$b$$). Then, if N is even, the bin at $$n = N/2$$ will be real.
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## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
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Re: Star to Star Distances taken on a Second Hand Sextant
Date: 2019 Dec 6, 20:28 -0500
In my previous statement on this topic, I indicated that, for arc calibration, one should get at least 6 observations against the datum. That should require a bit more explanation, no need to just take my word for it.
We are concerned with the accuracy of the arc, but first we must know the repeatability of the observations. Arc calibration, once found, is very straight forward to use. The concept is that we simply add an offset to the displayed value, realizing the true value.
Accuracy can be considered to be the mean of your observations. Total them up and divide by the number of observations. Let's suppose I am measuring 10 degrees. The first observation is 5°, then 6°, 14° and 15°. Well, surprise, the mean is 10° and therefore my sextant is accurate, right?
How about if I am measuring 10°, and the observations are 7.9, 8.1, 8.0. That's terrible accuracy, but much better repeatability. Is it a piece of junk? Which sextant would you prefer?
[Note that I exaggerated these values for clarity, in either case, I'd be sure to find a different sextant]
Consequently, we care about repeatability of the instrument more so than the absolute accuracy. We can correct for inaccuracy, but only if we are repeatable.
So how is repeatability determined? By the examination of multiple observations! Repeat the measurement, find multiple errors. Apply some statistical rigor, via standard deviation, and viola!
Let us see how this would work out. Suppose, when the datum is 10°, I observe 5°,5°, 14° and 14°. The mean is 9.5°, so I should add 0.5° in this section of arc. So now my observations are 5.5°, 5.5°, 14.5° and 14.5°. That didn't help me much. Crappy repeatability means a crappy result.
Suppose I observed 9.1, 9.2, 9.0, 8.9. The mean is 9.05, an error of 0.95°. One can easily see the arc correction will have merit here.
So you read this far, and you think, "meh, so I'll take the readings". Think this through. You need 6 readings for each section of the arc you want an arc correction for. You should really get all of the arc sections done in the same night, so as to keep thermal variations to a minimum. That isn't a requirement, but for very tight results, you do want to minimize environmental factors. If you wanted every 10° of arc, for 90° of arc, that is a very strenuous 54 careful observations, one after the other until finished. I hope you are feeling peppy!
PS: I'm sure every sage navigator remembers the advice to only approach from one direction. But why? Repeatability. The root cause is the gear train and lost motion. Yet the result is poor repeatability of observation.
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Subject: Author: Start date: (yyyymm dd) End date: (yyyymm dd) | 892 | 3,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-14 | latest | en | 0.867807 |
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# Tips for drawing graphs of form y^2=... watch
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1. I'm having trouble drawing the graphs of equations in the form y^2=...
For example, I can easily draw y=|x²-9|-7 and I can easily transorm this to y²=|x²-9|-7 for all the obvious points, but its in between these points I can't really get a feel for the shape of the graph.
Anyone have any tips?
(I would attach an image but it's too big)
2. its just [|x^2 - 9| - 7] times itself
you need to remember that when a number is squared, it becomes bigger by a factor of itself, so when the number is less than one it will be smaller, when its considerably close to 0 it will be practically zero, and when tis considerably greater than 1, it will get huge.
practice drawing sin^2 x, cos^2 x, and other similar things. tan^2 x, (e^x)^2, and you should get the hang of it.. also remember for real functions the curve should be above the x-axis!
Im not sure there's an easier way than that.
3. The function is y²=|x²-9|-7 so it's y=(|x²-9|-7 )0.5
not y=(|x²-9|-7)2
4. (Original post by mik1w)
also remember for real functions the curve should be above the x-axis!
The standard hyperbola has a curvce above and below the x axis since
5. argh. thanks for correcting me, I dont know what Im typing half the time. I think some content is right but 1. I completely misread the question, 2. I misanswered the misread question.
sorry .. my bad. more often than not two wrongs don't make a right. (double negatives aren't not cool)
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Top tips from students who have already aced their exams | 613 | 2,298 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-51 | latest | en | 0.943439 |
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Unformatted text preview: = ix + iy + 2.5 ix + iy = 2.5 vab 20 ix = 1.25 A iy = 1.25 A 4 1/21/14 Example 1Ω 4Ω v1 7Ω 10 A 3Ω 6Ω v2 v1 = ? v2 = ? Find : P7 Ω = ? P1Ω = ? Hint: Use circuit reduction technique. Solution: 1Ω a a b 4Ω v1 3Ω / /6Ω = 7Ω 10 A c 3Ω d d 1Ω b 4Ω 7Ω 10 A v1 c 2Ω d v2 d d a a 6Ω ( 3) ( 6 ) = 2Ω ( 3) + ( 6 ) d v2 d 5 1/21/14 1Ω a a b 4Ω 7Ω 10 A v1 c v2 2Ω d d 1Ω + 4 Ω + 2Ω = 7Ω d a a 7Ω 10 A d 7Ω d d a a 7Ω 10 A 7Ω 7Ω / /7Ω = 3.5Ω d d d a a vad 10 A d 3.5Ω vad = (10 ) ( 3.5 ) = 35V d 6 1/21/14 1Ω a a 7Ω 10 A b i1 4 Ω i d i= Ohm’s Law: d d 35 7 i = 5A P7 Ω = ( i ) ( 7 ) i1 = 5 A P1Ω = ( i1 ) (1) i2 = i1 Current division: 3Ω 3Ω + 6Ω 2 i2 = v2 = ( i2 ) ( 6 ) Ohm’s Law: 2 v1 = 20V v1 = ( 5 ) ( 4 ) Ohm’s Law: 6Ω i2 10 = 5 + i1 KCL: v1 c 3Ω d vad = 35V 5 A 3 v2 = 10V Review Problem (1) a Head light(1) • a • + • − 12V b b • • a Head light(2) Given:head light draws 3A P12V = ? 7 1/21/14 SoluCon: a Head light(1) • a • + • − 12V b b • • a Head light(2) KCL node (a) a I = 3 + 3 = 6A 3A I P12V = − (12 ) ( 6 ) = −72W 12V P12V = 72W (delivered ) b• Problem (2) 40Ω Ix 12 mA Iy Ix = ? Iy = ? Iz = ? 3A 20Ω 30Ω 10Ω 2 mA 4 mA Iz P12 mA = ? P4 mA = ? 8 1/21/14 SoluCon Ix KCL node C: I x + 12 m = 4 m a Iy 12 mA b c I x = − 8 mA 4 mA I x + I y = 2m KCL node a: 2 mA −8 m + I y = 2 m d I y = 10 mA KCL node b: Iz d I z = I y − 12 m I y = I z + 12 m d I z = 10 m − 12 m I z = −2 mA Assume the voltages : vbc vcd Ix KVL a- c- b- a () − ( 40Ω ) ( I x ) + vbc + ( 20Ω ) I y = 0 a 40Ω Iy b + vbc 20Ω 30Ω − ( 40Ω ) ( −8 m )...
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## This document was uploaded on 03/12/2014 for the course EEL 3004C at University of Central Florida.
Ask a homework question - tutors are online | 963 | 1,901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2017-13 | longest | en | 0.516189 |
https://jeopardyapp.com/play/wam20-final-review | 1,604,094,771,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107911792.65/warc/CC-MAIN-20201030212708-20201031002708-00153.warc.gz | 295,757,780 | 6,584 | Slope, Rate of Change, and Trigonometry Graphing Surface Area, Volume and Capacity Financial Services and Personal Budgets Scale Representations
What is the formula for slope?
A ratio of rise to run which indicates how steeply something is slanted.
What is the number of degrees in a circle?
360
What is a prism?
A 3-D shape with ends that are the same shape and size and sides that are parallelograms.
What is the principal?
The original amount of money invested or borrowed.
What is a proportion?
A statement of equality between two rates or two ratios.
What is the sine ratio?
opposite/hypotenuse
What is interpolation?
Making inferences about points between given points on a graph.
What is a net?
A 2-D pattern that can be folded to make a 3-D shape.
What is simple interest?
Interest calculated as a percentage of the principal.
What is a ratio?
A comparison between two numbers with the same units.
What is the Pythagorean Theorem?
a² = c² - b²
What is extrapolation?
Making predictions about what the next points on a graph will look like.
What is volume?
The measure of the space a three-dimensional object occupies
What is compound interest?
Interest calculated
on the principal plus interest earned in prior compounding periods.
What is a rate?
A comparison between two numbers with different units.
The slope of a physical feature such as a road or hill, often expressed as a percentage.
What is a broken line graph?
A graph that uses points joined by line segments to display data.
What is capacity?
The amount a three-dimensional object can hold.
What is the rule of 72?
years to double = 72 ÷ (interest rate as a percent)
What is a scale statement?
A ratio that shows the relationship between the sizes of two objects.
What is zero slope?
A line on a graph is horizontal.
What is a histogram?
A graph that is used to represent continuous data so that the bars are touching. The width of each bar represents a range of numbers.
What is the general formula for volume of a prism?
V = area of the base X height
What is a conservative budget?
What is a scale factor?
A number by which all the dimensions of an original figure are multiplied to produce an enlargement or a reduction. | 480 | 2,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2020-45 | latest | en | 0.96061 |
https://en.wikipedia.org/wiki/One_half | 1,657,126,120,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104675818.94/warc/CC-MAIN-20220706151618-20220706181618-00212.warc.gz | 268,647,227 | 18,036 | # One half
One half prefixes hemi- (from Greek) semi-/demi- (from Latin) Binary 0.1 or 0.011111111111... Ternary 0.11111111111... Decimal 0.5 or 0.499999999999... Duodecimal 0.6 or 0.5BBBBBBBBBBBB... Hexadecimal 0.8 or 0.7FFFFFFFFFFF... Continued fraction [0; 1, 1] or [0; 2] Single-precision 3F00000016= 001111110000000000000000000000002
½
In UnicodeU+00BD
Different from
Different from¼, ¾
Related
One
One half is the irreducible fraction resulting from dividing one by two (2) or the fraction resulting from dividing any number by its double. Multiplication by one half is equivalent to division by two, or "halving"; conversely, division by one half is equivalent to multiplication by two, or "doubling". One half often appears in mathematical equations, recipes, measurements, etc. Half can also be said to be one part of something divided into two equal parts.
For instance, the area S of a triangle is computed.
S = base/2 × perpendicular height.
One half also figures in the formula for calculating figurate numbers, such as triangular numbers and pentagonal numbers:
${\displaystyle {\frac {n}{2}}{\left[\left(s-2\right)n-(4-s)\right]}}$
and in the formula for computing magic constants for magic squares
${\displaystyle M_{2}(n)={\frac {n}{2}}\left(n^{2}+1\right)}$
The Riemann hypothesis states that every nontrivial complex root of the Riemann zeta function has a real part equal to 1/2.
One half has two different decimal expansions, the familiar 0.5 and the recurring 0.49999999... It has a similar pair of expansions in any even base. It is common to believe these expressions represent distinct numbers: see the proof that 0.999... equals 1 for a detailed discussion of a related case. In odd bases, one half has no terminating representation, only a single representation with a repeating fractional component, such as 0.11111111... in ternary.
## Particularities in writing and language
Postal stamp, Ireland, 1940: one halfpenny postage due.
12 is also one of the few fractions to usually have a key of its own on typewriters (see fractions). It also has its own code point in some early extensions of ASCII at 171. In Unicode, it has its own code unit at U+00BD (decimal 189) in the C1 Controls and Latin-1 Supplement block and a cross-reference in the Number Forms block, rendering as "½." The HTML entity is ½.
One half is also one of the few fractions which are commonly expressed in natural languages by suppletion rather than regular derivation: in English, for example, compare "one half" against other regular formations like "one-sixth".
It is acceptable to write one half as a hyphenated word, "one-half" or non-hyphenated, "one half".
## Code points
In the Unicode standard, the symbol ½ has the code point U+00BD ½ Vulgar Fraction One Half (HTML #0189· inherited from Latin-1)[1]
### PC entry
• Alt+0189 (keeping Alt pressed until all 4 digits have been typed on the numeric keypad only) | 766 | 2,945 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2022-27 | latest | en | 0.886037 |
http://new-to-teaching.blogspot.fr/2017/01/ | 1,527,114,678,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865830.35/warc/CC-MAIN-20180523215608-20180523235608-00489.warc.gz | 213,758,435 | 20,847 | ## Tabs
### Probability Through Caine's Arcade
First day of our probability unit we watched this video.
Day 1: I chose this video, because most of my students are hispanic and I think the biggest thing in our school right now is empathy. We talked about having games of chance like in the video they just watched, what does chance mean? We did our first section of probability and told them they were going to create their own games, just like Caine did.
We did a 5 question check for understanding and told the students they needed to finish the bottom half of the checklist today.
Here are other great resources including the checklist: http://cainesarcade.com/schools/activitykits/
Students designing their cardboard games.
Day 2: We talked about conditional probability. Did another check for understanding on Kahoot. Then I got lots of cardboard boxes from our recycling bin and had to make a quick pit-stop at Dollar General for more cardboard boxes.
Day 3: I had a substitute teacher this day, but students started creating their boxes.
Day 4: We went over theoretical vs experimental probability. I gave students 7 minutes to finish their cardboard arcade games. Then we went over theoretical probability again. We talked about geometrical probability from Day 1. Students were given rulers and yardsticks and had to find the theoretical probability of successfully completing their arcade game.
Day 5: We finished the material for our probability unit. I gave students 5 minutes to make sure their game is playable and to finish anything on the checklist. Then we went over that I would give them 5 minutes to go play other games to get other groups experimental probability, then the partners would switch and the other partner would go play games.
I thought this unit was much better than the 3D dice activity from last year, this project was more hands-on and did a better job of combining the curriculum and the project together.
### Statistics Sampling through Articles
In our statistics unit for Algebra 2, we talk about measures of central tendency then we go over different types of sampling. The four we talk about random, convenience, systematic, and cluster. Students don't really know what these are, we talk about when people are surveyed there are different methods to survey those people. Then we go over survey biases. It is a pretty boring lesson, students already know mean, median, mode, and range. They don't really get why we go over types of sampling.
I trimmed down and this article from The Street:
To summarize the article it is about the new Mac Jr and Grand Mac, in the article it says, "McDonald's started testing the Grand Mac and Mac Jr. in more than 120 restaurants in the central Ohio and Dallas areas in April last year." This is the basis of what I wanted the students to pick up out of the article, but thought this might be a good chance to get them inferring reading in the math classroom.
Students read the article, I gave them 5 minutes to read and answer the following 5 questions:
1. What is the main point of the article?
2. What are two supporting details?
3. What type of sampling method was mentioned in the article?
4. Why do you think McDonald's chose that type of sampling method?
5. Do you think the authors view of McDonalds were positive or negative? Why?
I was surprised about the level of detail that students put into the article here were some sticky notes and students working on the article.
Next year I will try to put the article on ActivelyLearn, last year it was my go to place for articles and mathematics for Junior Standards Math. I will have to use more articles in math class, it was a good experience for me and my students.
### License Plate Combinations
When we come back from winter break we normally start our probability and statistics unit. I normally take a week for probability and a week for statistics which normally melts into three weeks. I've always thought nothing of changing it, but during winter break Dan Meyer posted "Plates Without States" Since we were going over permutations and combinations I thought this would be an excellent way to get students thinking about how many different combinations there are in license plates and why they make them like that.
To start the lesson I had students go through Dan Meyer State-Plate Game. Students were definitely engaged and loved playing against each other in their groups.
Next we talked about license plates and I separated it from combinations and permutations.
I gave all of the students a blank license plate and a card. The card had a name of a city or state and a population that students had to take in consideration.
Here are some of the license plates that students were working on.
When students were done with their license plates, they took a picture of their license plate and put it on SeeSaw. The last part they had to do was comment on three others the number of different combinations that they had with their license plate.
Here were a few students figuring out and commenting on other students post.
I like this activity much more and students realized how license plates play a role in local governments and how the population of an area can control the different license plates possible.
I posted almost all of them in the back of my room here are a bunch of different ones that are posted.
### Posting Teacher Goals
I was reading this blog post, I can't seem to find it now. It had 21 things teachers should try in 2017, number 19 was "Post Your Goals in Your Classroom."
I thought this would be an excellent way for students to see what I am working on in the classroom and maybe they can hold me more accountable.
There are lots of different things I want to do in 2017, I will post a list of what I want to do at the bottom.
Here are the three goals I posted in my classroom:
1. More Activities, Less Homework
I have been disappointed lately with our school's emphasis on homework and worksheets. I feel that some of our students are being pushed down and out with this emphasis. I want students to experiment with math and I want more formative assessments to understand my students knowledge.
Since we are 1-1 with iPads I see students trying Google the worksheet before attempting any of the problems. They know in other classrooms that they get their worksheets online and don't need to do the work and its easier.
Getting students using Desmos, WODB, and Estimation180 to challenge their thinking and their understanding of mathematics.
2. Students In Charge of their Own Learning
My students heavily rely on the teacher for their information. If they don't know an answer right away their hands go up. I want my students to be challenged, but also know that I am there to guide not to tell them the answer.
I want students to be able to go out and find the answer. If they don't know how to do something I want them to be able to go out and search for it, find a YouTube video.
3. Build Students Up with Growth Mindset
This last one is very similar to the second one. My last goal is for students to have a growth mindset, to start the year I normally have a BreakoutEDU box for students to do. I want students not to think of math as thing that "smart people" do.
### Things I have Planned or Want to Do in 2017
• Different types of seating
• Incorporating more VR
• Walking Classroom
• Incorporate more reading.
Here is a seating chart that I currently use and really like, plus my goals are posted!!
### Codename: Numbers
The 2016 game of the year was Codenames, but for Christmas my wife and I were given Codenames Pictures, which is an equally awesome game. The basic gameplay is that you have a partner or a group where one person gives a one word clue and a number which correlates to how many it is suppose to cover, the first one to get them all wins. There are neutral ones and one is an assassin which ends the game.
My thought: How cool would this be if you did this with numbers.
If I made board pieces that had a bunch of different numbers, students would use one word such as: even, odd, cubic, etc... this game would help build number sense. Since you could play it with groups of 4 it would be a great station activity.
More to come with actual student gameplay. | 1,731 | 8,309 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-22 | latest | en | 0.976935 |
http://applet-magic.com/nonlinear01.htm | 1,490,618,156,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189472.3/warc/CC-MAIN-20170322212949-00418-ip-10-233-31-227.ec2.internal.warc.gz | 21,757,148 | 3,707 | San José State University
Department of Economics
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Thayer Watkins
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USA
The Use of Forward Finite Differences in the Solution of Nonlinear Differential Equations
The analysis starts with the simplest case: dy/dt = -y2. The use of a forward difference approximation of dy/dt at t, i.e., (y(t+h)-y(t))/h; gives more accurately an approximation of dy/dt at t+h/2. Thus the finite difference scheme is really an attempt to find an approximate solution to:
#### (dy/dt)|t+h/2 = -y2(t) which is equivalent to (dy/dt) = -y2(t-h/2)
After dealing with this case an extention to the more general problem will be presented.
## The Solution of the Differential-Difference Equationdy/dt = - y2(t-h/2)
As stated above this equation arises in the use of a forward difference of size h as an approximation of the first derivative dy/dt.
Let y(t-h/2) be approximated by y(t)-(h/2)(dy/dt) so
#### dy/dt = -(y(t)-(h/2)(dy/dt))2= - y2 + hy(dy/dt) -(h2/4)(dy/dt)2
This differential equation can be written as
#### (h2/4)(dy/dt)2 + (1 - hy)(dy/dt) + y2 = 0
The solution for dy/dt, using the formula for the solution of a quadratic equation, is
#### dy/dt = ( -(1-hy) ± ((1-hy)2 - h2y2)1/2 )/(h2/2)
The Taylor series approximation of the square root term is
#### ((1-hy)2 - h2y2)1/2 = (1-hy) - h2y2/(2(1-hy)) <
Thus the solution for dy/dt using the plus form of the square root term is
#### dy/dt = - y2/(1-hy)
This differential equation has the solution
#### 1/y(t) - 1/y(0) + hln(y(t)/y(0)) = t
This transcendental equation does not have an explicit solution for y as a function of t, but having t as a function of y is just as good.
The solution for the backward differences is obtained by making h a negative value; i.e.,
## Numerical Solution to the Differential Equation
The numerical approach uses the scheme
#### (y(t+h)-y(t))/h = -y2which reduces to y(t+h) = y(t) - hy2(t)
An example with h=1 yields the results shown below.
As can be seen from the graph above, the numerical solution using forward differences tracks not the solution to the differential equation but instead it is an approximation to the solution of the corresponding differential-difference equation. Thus the numerical solution provides a good approximation to the solution of the differential equation only to the extent that the solution to the corresponding differential-difference equation appoximates the solution to the differential equation.
## The More General Case
The preceding analyis generalizes to the case
#### dy/dt = f(y) which has the solution F(y) - F(y0) = ∫y0y(dz/f(z)) = t
where y0 is the value of y at t=0.
For the forward difference the corresponding differential-difference equation is:
#### dy/dt = f(y(t-h/2)) which is approximately dy/dt = f(y(t)-(h/2)(dy/t)) which in turn is approximately equal to dy/dt = f(y(t)) - f'(y(t))(h/2)(dy/dt)
Putting this equation in differential form gives
#### (1+f'(y)(h/2))(dy/f(y)) = dy/f(y) + (h/2)(f'(y)dy/f(y)) = dt which, upon integration from 0 to t gives F(y) - F(y0) + (h/2)ln(f(y)/f(y0)) = t
The backward difference approximation leads to the same expression except for a negative sign for the logarithmic term.
Thus the effect of forward or backward differencing in the solution of a differential equation of the type dy/dt = f(y) is to shift the time at which a particular level of y is achieved by an amount proportional to the timestep and a logarithmic term; i.e., (h/2)ln(f(y)/f(y0)). The logarithmic term is a function of the ratio of dy/dt at time t to what it was at t=0. | 996 | 3,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-13 | longest | en | 0.881131 |
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Post subject: How To Solve Timur's BiaxePosted: Thu Sep 05, 2013 5:00 am
Joined: Sun Aug 26, 2012 10:01 am
Solving Timur's Biaxe
Because Timur and luckyguynt are producing the Biaxe, I thought a solving thread would be a good idea! You can play with the puzzle by downloading the software here. It’s the 3dcube32.rar file. You’ll have to email the email address given on the site to unlock the program.
This is my method in solving Timur's Biaxe Puzzle
1. Solve the corners very easily by rotating them into place; they only need to be orientated.
2. Solve the star points, (making sure I solve them to match the corners when using set up moves) using [R L' R' L R' L R L R' L' R L R 2L R' L' R]x4. This is the hardest part of the solve in my opinion because the set up moves can be VERY difficult to set up.
3. Solve the edges using this 3 cycle (of the edges on the front face, which makes any required set up moves easy (again, in my opinion)): [R' L R L']x7
4. Solve the centers using this 2-2 swap: [R' L R L']x21.
You might have noticed yourself, but the algorithm for the edges and the centers are both the same, just repeated a different amount of times. The [R' L R L']x7 which is used for solving the edges is a 3 cycle of edges AND a 2-2 swap of the centers which is fine however because we solve the centers afterwards because they're easier to set up. If we call ([R' L R L']x7) Y, then 3Y would cancel out the 3 cycles and leave a 2-2 swap.
They won't be required when solving because they're solved first, but I decided to find pure corner twisting algorithms:
Rotate both corners 90 degrees: [L R' L R L R L R' L' R' 2L]x4
Rotate the right corner 180 degrees: [L R L' R]x105
When using the software, the notation for inputting is as follows:
R=3
L=1
So if you want to type in an algorithm, you must replace all R’s with 3’s etc.
Also using ( )xY doesn’t work, you have to click the “run alg” button Y times.
Now for your convenience, here are the algorithms with the 3’s and 1’s:
Cycle 3 points: (31'3'13'1313'1'313113'1'3)x4
2-2 Swap centres: (3'131')x21
Cycle 3 edges: (3'131')x14
Rotate 2 corners 90 degrees: (13'131313'1'3'11)x3
Rotate single corner 180 degrees: (131'3)x105
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My YouTube, including a FF Siamese 2x2x2 Walkthrough
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Post subject: Re: How To Solve Timur's BiaxePosted: Wed Jan 22, 2014 7:13 pm
Joined: Fri Feb 18, 2011 5:49 pm
Location: New Jersey
I'm surprised this thread doesn't have any other replies yet! The biaxe was a pretty tough solve for such a small puzzle.
Here's my method:
1. Twist corners to fix orientation
2. Solve centers with simple 2-2 swaps: R' L R L'
3. Solve edges with 3-cycles: (R' L R L')x2
4. Solve the star-points with 3-cycles: [x,y]
x:
R2 L R2 L R2 L2 R2
y:
R' L2 R'
(L2 R' L2 R)x5
R L2 R
Although my 3-cycle algorithm is technically slightly shorter (61 moves versus 65 when you consider move-cancellations), I think Jack's is much easier to remember and use...
Jack, I'm surprised you don't solve the star-points last as well, since your algorithm is pure too. Solving them last makes the edge and center algorithms much much shorter. Also, I think having the other pieces already in place makes finding and undoing setup moves a lot easier for the star-points. Also, solving the edges and centers is so fast it's hardly a noteworthy loss of "investment" if you mess up the star points afterwards.
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Post subject: Re: How To Solve Timur's BiaxePosted: Wed Jan 22, 2014 7:43 pm
Joined: Thu Dec 31, 2009 8:54 pm
Location: Bay Area, California
Don't the small triangles come in two orbits?
My plan is 1) corner orientation, 2) edges, 3) centers, 4) the rest
The first three steps are short and easy but I don't have anything good for "the rest". I've been looking for something elegant without much luck yet.
Edit: I should mention that at some point I'm going to get around to programming it too. It has too many states for fully enumerate god's algorithm but with a pruning table for the other pieces it should be easy to search for pure cycles in the triangles. With only two axes I should be able to search to a depth of ~20 or so.
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Prior to using my real name I posted under the account named bmenrigh.
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Post subject: Re: How To Solve Timur's BiaxePosted: Thu Jan 23, 2014 1:28 am
Joined: Sun Aug 26, 2012 10:01 am
I haven't actually solved the biaxe, I just decided to see if I could find the algorithms necessary to do so due to the small production of then.
_________________
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Post subject: Re: How To Solve Timur's BiaxePosted: Thu Jan 23, 2014 5:50 pm
Joined: Fri Feb 18, 2011 5:49 pm
Location: New Jersey
I found an even shorter 3-cycle:
In full commutator/conjugate form it looks like it should be 52 moves:
((R2 L' : (L' , R2)) (L2 R2 L' : (L' , R2)) , (L2 R : (L2 R)))
But when you expand it, a lot of turns cancel out or combine and it's actually only 38 moves:
R2 L2 R2 L R2 L R2 L2 R2 L2 R2 L R2 L R' L2 R L R2 L' R2 L2 R2 L2 R2 L' R2 L' R2 L2 R2 L2 R2 L2 R' L2 R' L2
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Post subject: Re: How To Solve Timur's BiaxePosted: Sat Feb 22, 2014 8:43 pm
Joined: Thu Mar 28, 2013 7:14 pm
I just got mine! Quickly discovered the edge 3-cyles and center swaps as previously described. Kinda stuck on the star points but I'll figure it out. What a great little puzzle. Just two turning axes and it's a killer!
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Post subject: Re: How To Solve Timur's BiaxePosted: Thu Mar 27, 2014 4:45 pm
Joined: Sat Oct 29, 2011 9:28 am
Location: Sweden
Hi!
I also think this is a great little puzzle and today I finally find out how to 3 cycle these star points.
As said here before it's quite easy to solve the edges and centers. Here is my solution for the star points:
[x:y]
x: (R' L R L')x6 or (R L' R' L)x6
y: R' L' (2R L' 2R L)x5 L R
I think it's quite easy to remember and use backwards and mirror it.
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All times are UTC - 5 hours | 1,917 | 6,392 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2014-23 | latest | en | 0.916421 |
http://www.jiskha.com/display.cgi?id=1335676061 | 1,498,339,084,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320338.89/warc/CC-MAIN-20170624203022-20170624223022-00690.warc.gz | 533,410,797 | 3,799 | # Algebra 2
posted by .
If an object is thrown vertically upward with an initial velocity of v, from an original position of s, the height h at any time t is given by: h=-16t^2+vt+s(where h and s are in ft, t is in seconds and v is in ft/sec)
A package is thrown downward with an initial velocity of -5 ft/sec from a helicopter at 300 ft. A) what is the height of the package in 2 sec? B) how long does it take to hit the ground?
Round answer to 2 decimals Enter: blank ft,blank sec.
• Algebra 2 -
A. h = ho - (Vo*t + 16t^2).
h = 300 - (5*2 + 16*2^2 )= 226 Ft above
gnd.
B. h = 300 - (5t + 16t^2 = 0.
300 - 5t -16t^2 = 0.
-16t^2 - 5t + 300 = 0. | 228 | 650 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2017-26 | latest | en | 0.895908 |
https://www.varsitytutors.com/gre_subject_test_math-help/classifying-algebraic-functions/algebra/imaginary-numbers-complex-functions | 1,579,796,502,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250611127.53/warc/CC-MAIN-20200123160903-20200123185903-00232.warc.gz | 1,125,820,257 | 46,679 | # GRE Subject Test: Math : Imaginary Numbers & Complex Functions
## Example Questions
← Previous 1 3 4 5 6
### Example Question #241 : Gre Subject Test: Math
Evaluate:
Explanation:
We can set in the cube of a binomial pattern:
### Example Question #1 : Complex Imaginary Numbers
Simplify the following product:
Explanation:
Multiply these complex numbers out in the typical way:
and recall that by definition. Then, grouping like terms we get
### Example Question #1 : Irrational Numbers
Simplify:
Explanation:
Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.
Remember that , so .
Substitute in for
### Example Question #1 : Imaginary Numbers & Complex Functions
Simplify:
Explanation:
Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.
Remember that , so .
Substitute in for
### Example Question #11 : Number Theory
Simplify:
Explanation:
Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.
Remember that , so .
Substitute in for
### Example Question #1 : Equations With Complex Numbers
Solve for and
Explanation:
Remember that
So the powers of are cyclic. This means that when we try to figure out the value of an exponent of , we can ignore all the powers that are multiples of because they end up multiplying the end result by , and therefore do nothing.
This means that
Now, remembering the relationships of the exponents of , we can simplify this to:
Because the elements on the left and right have to correspond (no mixing and matching!), we get the relationships:
No matter how you solve it, you get the values .
### Example Question #2 : Imaginary Roots Of Negative Numbers
Simplify:
None of the Above
Explanation:
Step 1: Split the into .
Step 2: Recall that , so let's replace it.
We now have: .
Step 3: Simplify . To do this, we look at the number on the inside.
.
Step 4: Take the factorization of and take out any pairs of numbers. For any pair of numbers that we find, we only take of the numbers out.
We have a pair of , so a is outside the radical.
We have another pair of , so one more three is put outside the radical.
We need to multiply everything that we bring outside:
Step 5: The goes with the 9...
Step 6: The last after taking out pairs gets put back into a square root and is written right after the
It will look something like this:
### Example Question #1 : Imaginary Numbers & Complex Functions
Explanation:
There are two ways to simplify this problem:
Method 1:
Method 2:
Explanation: | 662 | 2,777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2020-05 | latest | en | 0.772228 |
https://arbital.greaterwrong.com/p/decimal_notation?l=4sl | 1,582,505,533,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145859.65/warc/CC-MAIN-20200223215635-20200224005635-00014.warc.gz | 281,817,571 | 7,361 | # Decimal notation
Seventeen is the number that represents as many things as there are x marks at the end of this sentence: xxxxxxxxxxxxxxxxx. Writing out numbers by saying “the number representing how many things there are in this pile:” gets unwieldy when the pile gets large. Thus, we represent numbers using the numerals 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Specifically, we write the number representing this many things: xxx as “3″, and the number representing this many things: xxxxxxxxxxx as 11, and the number seventeen as “17”. This is called “decimal notation,” because there are ten different symbols that we use. Numbers don’t have to be written down in decimal notation, it’s also possible to write them down in other notations such as binary notation. Some numbers can’t even be written out in decimal notation (in full); consider, for example, the number $$e$$ which, in decimal notation, starts out with the digits 2.71828… and just keeps going.
# How decimal notation works
How do you know that 17 is the number that represents the number of xs in this sequence: xxxxxxxxxxxxxxxxx? In practice, you know this because the rules of decimal notation were ingrained in you in a young child. But do you know those rules explicitly? Could you write out a series of rules for taking in some input symbols like ‘2’, ‘4’, and ‘6’ and using those to figure out how many pebbles to add to a pile?
The answer, of course, is this many:
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
But how do we perform that conversion in general?
In short, the number 246 represents $$(2 \cdot 100) + (4 \cdot 10) + (6 \cdot 1),$$ so as long as we know how to do addition and multiplication, and as long as we know what the basic numerals 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 mean, and as long as we know how to get to powers of 10 (1, 10, 100, 1000, …), then we can explicitly understand decimal notation.
(What do the basic numerals 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 mean? By convention, they represent as many things as are in the following ten sequences of xs: , x, xx, xxx, xxxx, xxxxx, xxxxxx, xxxxxxx, xxxxxxxx, and xxxxxxxxx, respectively.)
This explanation assumes that you’re already quite familiar with decimal notation. Explaining decimal notation from scratch to someone who doesn’t already know it (which was a task people actually had to do back when half the world was using Roman numerals, a much less convenient system for representing numbers) is a fun task; to see what that looks like, refer to Representing numbers from scratch.
# Other common notations
The above text made use of unary notation, which is a method of representing numbers by making a number of marks that correspond to the represented number. For example, in unary notation, 17 is written xxxxxxxxxxxxxxxxx (or ||||||||||||||||| or whatever, the actual marks don’t matter). This is perhaps somewhat easier to understand, but writing large numbers like 93846793284756 gets rather ungainly.
Historical notations include Roman numerals, which were a pretty bad way to represent numbers. (It took humanity quite some time to find good tools for representing numbers; the decimal notation that’s been ingrained in your head since early childhood is the result of many centuries worth of effort. It’s much harder to invent good representations of numbers when you don’t even have good tools for writing down and reasoning about numbers. Furthermore, the modern tools for representing numbers aren’t necessarily ideal!)
Common notations in modern times (aside from decimal notation) include binary notation (often used by computers), hexadecimal notation (which is a useful format for humans reading binary notation). Binary notation and hexadecimal notation are very similar to decimal notation, with the difference that binary uses only two distinct symbols (instead of ten), and hexadecimal uses sixteen.
Children:
• 0.999...=1
No, it’s not “infinitesimally far” from 1 or anything like that. 0.999… and 1 are literally the same number.
Parents:
• Mathematics
Mathematics is the study of numbers and other ideal objects that can be described by axioms.
• Maybe refer to the advantages of being compact over the unary system (all marks are worth one unit)? And have some redlinks to other notations, binary, hexadecimal, roman numerals?
• Maybe something other than æ would be better? like /////? | 1,630 | 4,903 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2020-10 | longest | en | 0.885017 |
https://atsa-es.github.io/atsa-labs/sec-boxjenkins-est-ARMA-params.html | 1,669,574,476,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710417.25/warc/CC-MAIN-20221127173917-20221127203917-00489.warc.gz | 143,204,762 | 14,937 | ## 5.7 Estimating ARMA parameters
### 5.7.1 AR(2) data
Simulate AR(2) data and add a mean level so that the data are not mean 0.
$$$\begin{gathered} x_t = 0.8 x_{t-1} + 0.1 x_{t-2} + e_t\\ y_t = x_t + m \end{gathered}$$$
m <- 1
ar2 <- arima.sim(n = 1000, model = list(ar = c(0.8, 0.1))) +
m
To see info on arima.sim(), type ?arima.sim.
### 5.7.2 Fit with Arima()
Fit an ARMA(2) with level to the data.
forecast::Arima(ar2, order = c(2, 0, 0), include.constant = TRUE)
Series: ar2
ARIMA(2,0,0) with non-zero mean
Coefficients:
ar1 ar2 mean
0.7684 0.1387 0.9561
s.e. 0.0314 0.0314 0.3332
sigma^2 estimated as 0.9832: log likelihood=-1409.77
AIC=2827.54 AICc=2827.58 BIC=2847.17
Note, the model being fit by Arima() is not this model
$$$y_t = m + 0.8 y_{t-1} + 0.1 y_{t-2} + e_t$$$
It is this model:
$$$(y_t - m) = 0.8 (y_{t-1}-m) + 0.1 (y_{t-2}-m)+ e_t$$$
or as written above: $$$\begin{gathered} x_t = 0.8 x_{t-1} + 0.1 x_{t-2} + e_t\\ y_t = x_t + m \end{gathered}$$$
We could also use arima() to fit to the data.
arima(ar2, order = c(2, 0, 0), include.mean = TRUE)
Warning in arima(ar2, order = c(2, 0, 0), include.mean = TRUE): possible
convergence problem: optim gave code = 1
Call:
arima(x = ar2, order = c(2, 0, 0), include.mean = TRUE)
Coefficients:
ar1 ar2 intercept
0.7684 0.1387 0.9561
s.e. 0.0314 0.0314 0.3332
sigma^2 estimated as 0.9802: log likelihood = -1409.77, aic = 2827.54
However we will not be using arima() directly because for if we have differenced data, it will not allow us to include and estimated mean level. Unless we have transformed our differenced data in a way that ensures it is mean zero, then we want to include a mean.
Try increasing the length of the simulated data (from 100 to 1000 say) and see how that affects your parameter estimates. Run the simulation a few times.
### 5.7.3 AR(1) simulated data
ar1 <- arima.sim(n = 100, model = list(ar = c(0.8))) + m
forecast::Arima(ar1, order = c(1, 0, 0), include.constant = TRUE)
Series: ar1
ARIMA(1,0,0) with non-zero mean
Coefficients:
ar1 mean
0.7091 0.4827
s.e. 0.0705 0.3847
sigma^2 estimated as 1.34: log likelihood=-155.85
AIC=317.7 AICc=317.95 BIC=325.51
### 5.7.4 ARMA(1,2) simulated data
Simulate ARMA(1,2) $x_t = 0.8 x_{t-1} + e_t + 0.8 e_{t-1} + 0.2 e_{t-2}$
arma12 = arima.sim(n = 100, model = list(ar = c(0.8), ma = c(0.8,
0.2))) + m
forecast::Arima(arma12, order = c(1, 0, 2), include.constant = TRUE)
Series: arma12
ARIMA(1,0,2) with non-zero mean
Coefficients:
ar1 ma1 ma2 mean
0.8138 0.8599 0.1861 0.3350
s.e. 0.0646 0.1099 0.1050 0.8145
sigma^2 estimated as 0.6264: log likelihood=-118.02
AIC=246.03 AICc=246.67 BIC=259.06
We will up the number of data points to 1000 because models with a MA component take a lot of data to estimate. Models with MA(>1) are not very practical for fisheries data for that reason.
### 5.7.5 These functions work for data with missing values
Create some AR(2) data and then add missing values (NA).
ar2miss <- arima.sim(n = 100, model = list(ar = c(0.8, 0.1)))
ar2miss[sample(100, 50)] <- NA
plot(ar2miss, type = "l")
title("many missing values")
Fit
fit <- forecast::Arima(ar2miss, order = c(2, 0, 0))
fit
Series: ar2miss
ARIMA(2,0,0) with non-zero mean
Coefficients:
ar1 ar2 mean
1.0625 -0.2203 -0.0586
s.e. 0.1555 0.1618 0.6061
sigma^2 estimated as 0.9679: log likelihood=-79.86
AIC=167.72 AICc=168.15 BIC=178.06
Note fitted() does not return the expected value at time $$t$$. It is the expected value of $$y_t$$ given the data up to time $$t-1$$.
plot(ar2miss, type = "l")
title("many missing values")
lines(fitted(fit), col = "blue")
It is easy enough to get the expected value of $$y_t$$ for all the missing values but we’ll learn to do that when we learn the MARSS package and can apply the Kalman Smoother in that package. | 1,520 | 3,894 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 4, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2022-49 | latest | en | 0.789649 |
https://kinmoremotor.com/blog/anti-electromagnetic-field-in-a-dc-motor/ | 1,606,372,146,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141186761.30/warc/CC-MAIN-20201126055652-20201126085652-00478.warc.gz | 358,470,274 | 33,633 | Anti-electromagnetic field in a DC motor
# Anti-electromagnetic field in a DC motor
The inverse electromagnetic field is caused by the coil rotation inside the stator magnetic field. Its unit is the same as the voltage, the volt. Let us briefly discuss the characteristics of the anti-electromagnetic fields in dc motors.
## How does a DC motor generate electromagnetic fields?
A DC motor has a magnetic field due to a permanent magnet or electromagnet. When supplied to the armature, the voltage generated across the armature windings is the opposite of the armature current. Because of the interacting magnetic field, the armature is forced to rotate. The voltage generated at both ends of the armature in a DC motor is called the anti-electromagnetic field or anti-electromagnetic field in a DC motor. The reverse electromagnetic field is always opposite to the supply voltage. To keep the armature of a DC motor rotating, the supply voltage must force the current through the armature winding to reverse the electromagnetic field.
Here is the mathematical expression of the inverse electromagnetic field:
Eb = Φ NZ / 60 * P/A
Among them, the magnetic flux/pole = Φ, armature speed = N, total number of armature conductors = Z, armature winding = A.
The figure below shows the DC motor, all parts labeled to understand how the emf generates, and the internal structure of the DC motor.
Therefore, the energy conversion in a DC motor is only due to the creation of this anti-electromagnetic field. When the DC motor runs without load, the voltage provided is almost equal to the anti-electromagnetic field generated by the DC motor. That is because dc motors require a small torque (Ta=KIa) to overcome friction and winding resistance.
It can be concluded from the above discussion that anti-electromagnetic fields play a dynamic role in the operation of DC motors. The anti-electromagnetic fields have another function, which is to make the motors adjust automatically. Then, how it makes the dc motor automatic regulation, let us through the following study.
## The counter electromagnetic field makes the DC motor adjust automatically.
Assuming that the motor operates under no-load conditions, at this stage, the DC motor only needs very little torque to adjust the friction and winding loss. As a result, the DC motor extracts less current. The inverse electromagnetic field and the current in a DC motor are mutually dependent, so the value of the inverse electromagnetic field will decrease with the decrease of the electric current. At this stage, the magnitude of the counter electromagnetic field is approximately equal to the voltage provided.
When a sudden load is applied on the shaft of the DC motor, the DC motor slows down. As the speed of the DC motor decreases, the amplitude of the anti-electromagnetic field will also decrease. A small inverse electromagnetic field will extract a large current from the power supply. The armature current induces a greater torque in the armature, which is needed in a DC motor. As a result, the DC motor runs continuously at a new speed.
The load on the DC motor is suddenly reduced. In this case, the drive torque on the DC motor is greater than the load torque. The driving torque increases the speed of the DC motor, which also increases the anti-electromagnetic field. The armature current is reduced by the high value of the inverse electromagnetic field. Thus, the armature current value produces a drive torque, which is equal to the load torque. As a result, the DC motor will rotate uniformly at the new speed.
Therefore, the generated counter emf operates the DC motor by adjusting the armature current to meet the load requirements. Here are some of the advantages of anti-emf.
## The advantage of the anti-electromagnetic field in DC motor
The inverse electromagnetic field enables the DC motor to be automatically regulated, which means that it operates at a constant speed from no-load to maximum load. A DC motor generates an armature current based on the load applied to it. We calculate the armature current by the following formula. The figure below also shows the armature current in the DC motor. I=(V-Eb)Ra
Abstract
The inverse electromagnetic field is vital for running a DC motor without generating fluctuations and effectively responding to the output load.
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# Show that W(5, sin2 t, cos 2t) = 0 for all t. Can you establish this result without
ISBN: 9780470458327 393
## Solution for problem 17 Chapter 4.1
Elementary Differential Equations | 10th Edition
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Problem 17
Show that W(5, sin2 t, cos 2t) = 0 for all t. Can you establish this result without directevaluation of the Wronskian?
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2.1 – Tangent and Velocity Problems Tangent line at (2,8) 1. Start from secant lines – (2.5, 2.5 ), (2.1, 2.1 ), (2.01, 2.01 ) 3 2. Find the slope using the slope formula – y -y / x 2x 1 2 1 a. 2.5 – 8 / 2.5 – 2 = 15.25 3 b. 2.1 – 8 / 2.1 – 2 = 12.61 c. 2.01 – 8 / 2.01 – 2 = 12.0601 3. The slopes...
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##### ISBN: 9780470458327
This full solution covers the following key subjects: . This expansive textbook survival guide covers 61 chapters, and 1655 solutions. Since the solution to 17 from 4.1 chapter was answered, more than 207 students have viewed the full step-by-step answer. Elementary Differential Equations was written by and is associated to the ISBN: 9780470458327. This textbook survival guide was created for the textbook: Elementary Differential Equations, edition: 10. The answer to “Show that W(5, sin2 t, cos 2t) = 0 for all t. Can you establish this result without directevaluation of the Wronskian?” is broken down into a number of easy to follow steps, and 22 words. The full step-by-step solution to problem: 17 from chapter: 4.1 was answered by , our top Math solution expert on 03/13/18, 08:19PM.
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Unlock Textbook Solution | 566 | 1,797 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2020-45 | latest | en | 0.856074 |
https://www.univerkov.com/what-amount-of-heat-will-need-to-be-transferred-to-the-mercury-thermometer-when-it-is-heated-from-room-temperature/ | 1,723,527,710,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641063659.79/warc/CC-MAIN-20240813043946-20240813073946-00661.warc.gz | 813,787,099 | 6,479 | # What amount of heat will need to be transferred to the mercury thermometer when it is heated from room temperature
What amount of heat will need to be transferred to the mercury thermometer when it is heated from room temperature t = 20 degrees to the body temperature of a healthy person (36.6)? Consider the mass of mercury equal to m = 2g. Will this amount of heat be equal to the amount of heat received by the thermometer during the temperature measurement?
t1 = 20 ° C.
t2 = 36.6 ° C.
m = 2 g = 2 * 10 ^ -3 kg.
C = 120 J / kg * ° C.
Q -?
The amount of thermal energy that goes into heating the mercury is determined by the formula: Q = C * m * (t2 – t1), where C is the specific heat of mercury, m is the mass of mercury, t2, t1 are the final and initial temperatures of mercury.
Q = 120 J / kg * ° C * 2 * 10 ^ -3 kg * (36.6 ° C – 20 ° C) = 3.984 J.
The thermometer received more heat from the body than was used to heat the mercury. Part of the heat was used to heat the thermometer itself.
Answer: it took Q = 3.984 J of thermal energy to heat the mercury.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. | 345 | 1,388 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-33 | latest | en | 0.944586 |
https://www.ribblu.com/cbse/class-10-maths-question-paper-post-term-exam-set-a-2017-18 | 1,721,416,015,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514917.3/warc/CC-MAIN-20240719170235-20240719200235-00365.warc.gz | 848,588,635 | 76,880 | ## CBSE Class 10 Maths Previous Year Question Papers
These Previous Year Question Papers Class 10 Maths are important for students to prepare for board exams. By solving the past years’ papers, students will get an idea of the Maths paper pattern and type of questions asked in the exam. So, to help students prepare better for the exam, here on ribbly students can freely download Past Year Question Papers for Class 10 Mathematics , Half Yearly Question Papers, Pre Board Examination Papers with solutions. All these papers are available in PDF format. They must solve these CBSE Previous Year Question Papers to score high marks in the Class 10 Maths exam. These previous year papers and board papers are in accordance with CBSE guidelines and prescribed syllabus and thus after solving them students get enough confidence and practice to face the forthcoming exams
Class 10 Maths Marks Distribution
Units Marks
Number Systems 06
Algebra 20
Coordinate Geometry 06
Geometry 15
Trigonometry 12
Mensuration 10
Statistics & Probability 11
Internal Assessment 20
Total 100
CBSE Class 10 Maths Syllabus
• Real Numbers
• Polynomials
• Pair of Linear Equations in Two Variables
• Quadratic Equations
• Arithmetic Progressions
• Triangles
• Coordinate Geometry
• Statistics
• Probability
• Introduction to Trigonometry
• Some Applications of Trigonometry
• Circles
• Constructions
• Area Related to Circles
• Surface Areas and Volumes
UNIT I: NUMBER SYSTEMS
1. REAL NUMBER
Fundamental Theorem of Arithmetic - statements after reviewing work done earlier and after illustrating and motivating through examples, Proofs of irrationality.
UNIT II: ALGEBRA
1. POLYNOMIALS Zeros of a polynomial. Relationship between zeros and coefficients of quadratic polynomials.
2. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES Pair of linear equations in two variables and graphical method of their solution, consistency/inconsistency. Algebraic conditions for number of solutions. Solution of a pair of linear equations in two variables algebraically - by substitution, by elimination. Simple situational problems.
3. QUADRATIC EQUATIONS Standard form of a quadratic equation ax2 + bx + c = 0, (a ≠ 0). Solutions of quadratic equations (only real roots) by factorization, and by using quadratic formula. Relationship between discriminant and nature of roots. Situational problems based on quadratic equations related to day to day activities to be incorporated.
4. ARITHMETIC PROGRESSIONS Motivation for studying Arithmetic Progression Derivation of the nth term and sum of the first n terms of A.P. and their application in solving daily life problems.
UNIT III: COORDINATE GEOMETRY
Coordinate Geometry
Review: Concepts of coordinate geometry, graphs of linear equations. Distance formula. Section formula (internal division).
UNIT IV: GEOMETRY
1. TRIANGLES Definitions, examples, counter examples of similar triangles.
1. (Prove) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
2. (Motivate) If a line divides two sides of a triangle in the same ratio, the line is parallel to the third side.
3.(Motivate) If in two triangles, the corresponding angles are equal, their corresponding sides are proportional and the triangles are similar.
4.(Motivate) If the corresponding sides of two triangles are proportional, their corresponding angles are equal and the two triangles are similar.
5.(Motivate) If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the two triangles are similar.
2. CIRCLES Tangent to a circle at, point of contact
1. (Prove) The tangent at any point of a circle is perpendicular to the radius through the point of contact.
2.(Prove) The lengths of tangents drawn from an external point to a circle are equal.
UNIT V: TRIGONOMETRY
1. INTRODUCTION TO TRIGONOMETRY Trigonometric ratios of an acute angle of a right-angled triangle. Proof of their existence (well defined); motivate the ratios whichever are defined at 0° and 90°. Values of the trigonometric ratios of 30°, 45°, and 60°. Relationships between the ratios.
2. TRIGONOMETRIC IDENTITIES Proof and applications of the identity sin2A + cos2A = 1. Only simple identities to be given.
3. HEIGHTS AND DISTANCES: Angle of elevation, Angle of Depression. (10)Periods Simple problems on heights and distances. Problems should not involve more than two right triangles. Angles of elevation / depression should be only 30°, 45°, and 60°.
UNIT VI: MENSURATION
1. AREAS RELATED TO CIRCLES Area of sectors and segments of a circle. Problems based on areas and perimeter / circumference of the above said plane figures. (In calculating area of segment of a circle, problems should be restricted to central angle of 60°, 90° and 120° only.
2. SURFACE AREAS AND VOLUMES Surface areas and volumes of combinations of any two of the following: cubes, cuboids, spheres, hemispheres and right circular cylinders/cones.
UNIT VII: STATISTICS AND PROBABILITY
1. STATISTICS Mean, median and mode of grouped data (bimodal situation to be avoided).
2. PROBABILITY (10) Periods Classical definition of probability. Simple problems on finding the probability of an event.
PRESCRIBED BOOKS:
1. Mathematics - Textbook for class IX - NCERT Publication
2. Mathematics - Textbook for class X - NCERT Publication
3. Guidelines for Mathematics Laboratory in Schools, class IX - CBSE Publication
4. Guidelines for Mathematics Laboratory in Schools, class X - CBSE Publication
5. Laboratory Manual - Mathematics, secondary stage - NCERT Publication
6. Mathematics exemplar problems for class IX, NCERT publication.
7. Mathematics exemplar problems for class X, NCERT publication.
Structure of CBSE Maths Sample Paper for Class 10 is
Type of Question Marks per Question Total No. of Questions Total Marks
Objective Type Questions 1 20 20
Short Answer Type Questions - I 2 6 12
Short Answer Type Questions - II 3 8 24
Long Answer Type Questions 4 6 24
Total 40 80
For Preparation of exams students can also check out other resource material
CBSE Class 10 Maths Sample Papers
CBSE Class 10 Maths Worksheets
CBSE Class 10 Maths Test Papers
CBSE Class 10 Maths Important Questions
CBSE Class 10 Maths Revision Notes
## Question Papers of Other Subjects of Class 10
CBSE Question Papers of Class 10 NTSE CBSE Question Papers of Class 10 Science CBSE Question Papers of Class 10 English CBSE Question Papers of Class 10 Social Science CBSE Question Papers of Class 10 Computer Science CBSE Question Papers of Class 10 Hindi CBSE Question Papers of Class 10 Sanskrit CBSE Question Papers of Class 10 German CBSE Question Papers of Class 10 French
In order to access the level of preparation done by any particular student he or she needs to solve Previous Year Question Papers. These papers act as perfect tools to practise for the final board exam. If one wants to get a clear look and feel of how final exam papers are framed in terms of level of difficulty, time and other aspects then , all students must make sure that they attempt these papers once their course revision is finished. Few benefits of solving Previous Question Papers are given below:
• Revising the subject is very good practice but until and unless one Solves the past question papers in the lookalike environment as in board exam or final class room exam, there is most likely that student may not be able to identify and check whether the understanding of all concepts of the subject are complete or not. It is only after students attempt the question paper in the same time frame he or she is able to judge the capability of solving the paper in the stipulated time frame. It highlights the weak areas if any and gives students ample amount of time to work on those areas and be better prepared before exams.
• Knowing everything is great but it is of no use unless the implementation and results are not matching with that. There is always a risk of the case in which in spite of knowing everything a student falls short of time to complete the entire question paper and thus loses marks. Generally CBSE Board papers and previous year questions are generally of 3 hour duration. So while practicing such papers it is imperative to create a final exam or board like environment at home and ensure that the Question paper is attempted only in 3 hours and then check whether it was possible to complete the paper in the desired amount of time. Often at first students take longer than expected, and thus they get early warning to practice more and increase the speed.
• Students with anxiety issues need previous year papers more than anyone for overcoming such issues. Since they do not know what questions will be asked in the CBSE board they create panic in their mind due to this fear of the unknown and get scared with the idea that they might not be able to do well in exams. Thus such students need to complete at-least 7-10 Question papers prior to the exams, to gain confidence and get into a better frame of mind. | 1,984 | 9,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-30 | latest | en | 0.864134 |
https://slopeinterceptform.net/slope-intercept-form-solver/ | 1,653,352,078,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562106.58/warc/CC-MAIN-20220523224456-20220524014456-00085.warc.gz | 597,712,587 | 14,589 | # Slope Intercept Form Solver
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
Slope Intercept Form Solver – One of the many forms employed to represent a linear equation, one of the most commonly found is the slope intercept form. It is possible to use the formula of the slope-intercept identify a line equation when that you have the straight line’s slope , and the y-intercept. It is the point’s y-coordinate at which the y-axis crosses the line. Find out more information about this particular linear equation form below.
## What Is The Slope Intercept Form?
There are three primary forms of linear equations: the standard slope, slope-intercept and point-slope. Although they may not yield identical results when utilized but you are able to extract the information line produced more quickly using an equation that uses the slope-intercept form. The name suggests that this form utilizes an inclined line, in which the “steepness” of the line reflects its value.
This formula is able to calculate the slope of a straight line, the y-intercept (also known as the x-intercept), where you can utilize a variety formulas available. The equation for this line in this particular formula is y = mx + b. The straight line’s slope is represented by “m”, while its y-intercept is signified with “b”. Every point on the straight line can be represented using an (x, y). Note that in the y = mx + b equation formula, the “x” and the “y” must remain as variables.
## An Example of Applied Slope Intercept Form in Problems
When it comes to the actual world in the real world, the slope-intercept form is commonly used to depict how an object or problem evolves over an elapsed time. The value of the vertical axis indicates how the equation handles the degree of change over what is represented through the horizontal axis (typically in the form of time).
A simple example of the use of this formula is to determine how many people live in a specific area as time passes. In the event that the area’s population increases yearly by a predetermined amount, the values of the horizontal axis will rise by a single point each year and the worth of the vertical scale will increase to reflect the increasing population by the fixed amount.
You may also notice the starting value of a particular problem. The starting value occurs at the y’s value within the y’intercept. The Y-intercept is the place at which x equals zero. By using the example of a problem above the beginning value will be at the time the population reading begins or when time tracking starts, as well as the changes that follow.
This is the place at which the population begins to be monitored to the researchers. Let’s say that the researcher starts with the calculation or measurement in the year 1995. This year will become considered to be the “base” year, and the x = 0 point would occur in the year 1995. So, it is possible to say that the population in 1995 corresponds to the y-intercept.
Linear equations that use straight-line formulas are almost always solved in this manner. The starting value is depicted by the y-intercept and the change rate is expressed as the slope. The principal issue with an interceptor slope form generally lies in the horizontal variable interpretation particularly when the variable is accorded to the specific year (or any other kind of unit). The first step to solve them is to make sure you comprehend the variables’ meanings in detail. | 716 | 3,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2022-21 | longest | en | 0.931499 |
https://math.stackexchange.com/questions/1079510/a-question-about-eliminating-square-roots | 1,560,767,678,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998473.44/warc/CC-MAIN-20190617103006-20190617125006-00405.warc.gz | 513,430,655 | 28,993 | # A question about eliminating square roots [duplicate]
If $\sqrt{x^2} = \pm x$, then why does $\sqrt{(x+2)^2} = x+2$ and not $\pm (x+2)$?
This is driving me crazy, so feel free to elucidate. Thanks!
---EDIT---
I'm not sure how the other questions' answers would help answer my own question, but it doesn't really matter now that I've figured it out. After thinking a bit about some of what @Daniel Hast typed on a similar question I asked, I realized that the reason why $\sqrt{(x+2)^2} = x+2$ was that more generally, given that $\sqrt{x} = r$ such that $r^2 = x$, squaring $r$ in $\sqrt{x} = r$ produced $(\sqrt{x})^2 = r^2 = x$ and subsequently $(\sqrt{x})^2 = x$.
## marked as duplicate by Peter Woolfitt, user147263, Bungo, coffeemath, peterwhyDec 24 '14 at 7:09
• It is not that $\sqrt{x^2}=\pm x$; if $x\ge0$, then $\sqrt{x^2}=x$, and if $x\le0$, then $\sqrt{x^2}=−x$. – peterwhy Dec 24 '14 at 7:11
It should be $|x+2|$. And you can find out if $x \geq -2$, then it equals $x+2$, and equals to $-x-2$ otherwise.
The usual definition, of a square root is as follows: if $x \geq 0$, then $y =\sqrt{x}$ is the unique positive number such that $y^2 = x$. So, we have $$\sqrt{x^2} = |x|$$ And similarly, $$\sqrt{(x+2)^2} = |x+2|$$ Along these lines, the solution for $y$ to $y^2 = x^2$ is $y = \pm \sqrt{x^2} = \pm x$, and $\sqrt{x^2}$ is by definition the positive number among these two solutions.
A common misconception in algebra is that $\displaystyle \sqrt{x^2}=\pm x$. However, square roots are actually ALWAYS positive or 0. The $\pm$ comes from solving the equation $x^2=n\implies x=\pm\sqrt n\implies|x|=\sqrt{n}$. Therefore your equation actually states that $|x+2|=x+2,$ only valid for $x\ge-2$ | 571 | 1,714 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2019-26 | latest | en | 0.885218 |
https://www.jaromirstetina.cz/kapan-armenia-egem/998cf2-integration-by-parts-formula | 1,680,142,702,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949093.14/warc/CC-MAIN-20230330004340-20230330034340-00382.warc.gz | 906,389,771 | 16,769 | # integration by parts formula
This handy formula can make your calculus homework much easier by helping you find antiderivatives that otherwise would be difficult and time consuming to work out. The 5 Strategies You Must Be Using to Improve 4+ ACT Points, How to Get a Perfect 36 ACT, by a Perfect Scorer. If you want to master the technique of integrations, I suggest, you use the integration by parts formula. Integration by parts twice Sometimes integration by parts can end up in an infinite loop. Skip to content. Let’s try it. Integration by parts is a special rule that is applicable to integrate products of two functions. ( Integration by Parts) Let $u=f (x)$ and $v=g (x)$ be differentiable functions. This is the currently selected item. Recall the formula for integration by parts. so that and . But there is only one function! The College Entrance Examination BoardTM does not endorse, nor is it affiliated in any way with the owner or any content of this site. Integration by parts is a technique for performing indefinite integration intudv or definite integration int_a^budv by expanding the differential of a product of functions d(uv) and expressing the original integral in terms of a known integral intvdu. So let's say that I start with some function that can be expressed as the product f of x, can be expressed as a product of two other functions, f of x times g of x. In general, your goal is for du to be simpler than u and for the antiderivative of dv to not be any more complicated than v. Basically, you want the right side of the equation to stay as simple as possible to make it easier for you to simplify and solve. Plug these new variables into the formula again: ∫ex sin(x) dx = sin(x) ex - (cos(x) ex −∫−sin(x) ex dx), ∫ex sin(x) dx = ex sin(x) - ex cos(x) −∫ ex sin(x)dx. u is the function u (x) In a similar manner by integrating "v" consecutively, we get v 1, v 2,.....etc. When using this formula to integrate, we say we are "integrating by parts". Integration by parts Calculator online with solution and steps. This is the integration by parts formula. We can also sometimes use integration by parts when we want to integrate a function that cannot be split into the product of two things. Integration is a very important computation of calculus mathematics. Alright, now I'm going to show you how it works on a few examples. A special rule, which is integration by parts, is available for integrating the products of two functions. LIPET. This gives a systematic list of what to try to set equal to u in the integration by parts formula. so that and . The key thing in integration by parts is to choose $$u$$ and $$dv$$ correctly. Sample Problem. It just got more complicated. The steps are: Wondering which math classes you should be taking? Struggling with the math section of the SAT or ACT? The key thing in integration by parts is to choose $$u$$ and $$dv$$ correctly. Choose u in this order LIPET. Many rules and formulas are used to get integration of some functions. a Quotient Rule Integration by Parts formula, apply the resulting integration formula to an example, and discuss reasons why this formula does not appear in calculus texts. With “x” as u, it’s easy to get du, so let’s start there. Integration by parts with limits. 3. Choose u based on which of these comes first: And here is one last (and tricky) example: Looks worse, but let us persist! Bernoulli’s formula is advantageously applied when u = x n ( n is a positive integer) For the following problems we have to apply the integration by parts two or more times to find the solution. The advantage of using the integration-by-parts formula is that we can use it to exchange one integral for another, possibly easier, integral. u = ln x. v' = 1. See how other students and parents are navigating high school, college, and the college admissions process. This is still a product, so we need to use integration by parts again. Ideally, your choice for the “u” function should be the one that’s easier to find the derivative for. Integration by parts - choosing u and dv How to use the LIATE mnemonic for choosing u and dv in integration by parts? Let and . Once you have your variables, all you have to do is simplify until you no longer have any antiderivatives, and you’ve got your answer! Things are still pretty messy, and the “∫cos(x) ex dx” part of the equation still has two functions multiplied together. In this case Bernoulli’s formula helps to find the solution easily. Solved exercises of Integration by parts. Integration by parts is an important technique of integration. You’ll have to have a solid … Theorem. If we chose u = 1 then u' would be zero, which doesn't seem like a good idea. Therefore, . In high school she scored in the 99th percentile on the SAT and was named a National Merit Finalist. logarithmic factor. In calculus, and more generally in mathematical analysis, integration by parts or partial integration is a process that finds the integral of a product of functions in terms of the integral of the product of their derivative and antiderivative. The moral of the story: Choose u and v carefully! ( f g) ′ = f ′ g + f g ′. In calculus, definite integrals are referred to as the integral with limits such as upper and lower limits. The main results are illustrated by SDEs driven by α-stable like processes. Factoring. so that and . Practice: Integration by parts: definite integrals. A helpful rule of thumb is I LATE. There are five steps to solving a problem using the integration by parts formula: #4: Plug these values into the integration by parts equation. ∫(fg)′dx = ∫f ′ g + fg ′ dx. It's also written this way, when you have a definite integral. Instead, integration by parts simply transforms our problem into another, hopefully easier one, which we then have to solve. In a way, it’s very similar to the product rule, which allowed you to find the derivative for two multiplied functions. Substituting into equation 1, we get . Integration by Parts Formulas Integration by parts is a special rule that is applicable to integrate products of two functions. Many rules and formulas are used to get integration of some functions. (2) Rearranging gives intudv=uv-intvdu. We choose = because its derivative of 1 is simpler than the derivative of , which is only itself. Integrate … The ilate rule of integration considers the left term as the first function and the second term as the second function. The application of integration by parts method is not just limited to the multiplication of functions but it can be used for various other purposes too. Scroll down the page for more examples and solutions. I'm going to write it one more time with the limits stuck in. A single integration by parts starts with d(uv)=udv+vdu, (1) and integrates both sides, intd(uv)=uv=intudv+intvdu. Integration by Parts with a definite integral Previously, we found $\displaystyle \int x \ln(x)\,dx=x\ln x - \tfrac 1 4 x^2+c$. In order to avoid applying the integration by parts two or more times to find the solution, we may us Bernoulli’s formula to find the solution easily. Recall the formula for integration by parts. Now that we have all the variables, let’s plug them into the integration by parts equation: All that’s left now is to simplify! Ask questions; get answers. Try the box technique with the 7 mnemonic. How do we choose u and v ? Learn which math classes high schoolers should take by reading our guide. Click HERE to return to the list of problems. The Integration by Parts formula is a product rule for integration. We call this method ilate rule of integration or ilate rule formula. Next lesson. The integrand is the product of the two functions. How to derive the rule for Integration by Parts from the Product Rule for differentiation? The integration-by-parts formula tells you to do the top part of the 7, namely . integration by parts formula is established for the semigroup associated to stochas-tic (partial) differential equations with noises containing a subordinate Brownian motion. This formula shows which part of the integrand to set equal to u, and which part to set equal to dv. As you can see, we start out by integrating all the terms throughout thereby keeping the equation in balance. Deriving the integration by parts standard formula is very simple, and if you had a suspicion that it was similar to the product rule used in differentiation, then you would have been correct because this is the rule you could use to derive it. Therefore, . To do this integral we will need to use integration by parts so let’s derive the integration by parts formula. A lot of times, a function is a product of other functions and therefore needs to be integrated. AMS subject Classification: 60J75, 47G20, 60G52. This is where integration by parts comes in! In this guide, we’ explain the formula, walk you through each step you need to take to integrate by parts, and solve example problems so you can become an integration by parts expert yourself. The acronym ILATE is good for picking $$u.$$ ILATE stands for. Formula : ∫udv = uv - ∫vdu. The rule can be thought of as an integral version of the product rule of differentiation. SAT® is a registered trademark of the College Entrance Examination BoardTM. So let's say that I start with some function that can be expressed as the product f of x, can be expressed as a product of two other functions, f of x times g of x. So, we are going to begin by recalling the product rule. This formula follows easily from the ordinary product rule and the method of u-substitution. Integrating by parts (with v = x and du/dx = e-x), we get:-xe-x - ∫-e-x dx (since ∫e-x dx = -e-x) = -xe-x - e-x + constant. First multiply everything out: Then take the antiderivative of ∫x2/3. 2 INTEGRATION BY PARTS 5 The second integral we can now do, but it also requires parts. Integration by Parts with a definite integral Previously, we found $\displaystyle \int x \ln(x)\,dx=x\ln x - \tfrac 1 4 x^2+c$. Integration by parts review. The 5 Strategies You Must Be Using to Improve 160+ SAT Points, How to Get a Perfect 1600, by a Perfect Scorer, Free Complete Official SAT Practice Tests. The first step is to select your u and dv. For steps 2 and 3, we’ll differentiate u and integrate dv to get du and v. The derivative of x is dx (easy!) This gives us: Next, work the right side of the equation out to simplify it. We use integration by parts a second time to evaluate . Recall the product rule: (where and are functions of ). The mathematical formula for the integration by parts can be derived in integral calculus by the concepts of differential calculus. and the antiderivative of sin(x) is -cos(x). Dave4Math » Calculus 2 » Integration by Parts (and Reduction Formulas) Here I motivate and … Integration by Parts Derivation. This topic will derive and illustrate this rule which is Integration by parts formula. 7 Example 3. The Product Rule states that if f and g are differentiable functions, then . We were able to find the antiderivative of that messy equation by working through the integration by parts formula twice. What I often do is to derive it from the Product Rule (for differentiation), but this isn't very efficient. Using the Integration by Parts formula . LIPET. Bernoulli’s formula is advantageously applied when u = x n ( n is a positive integer) For the following problems we have to apply the integration by parts two or more times to find the solution. The integration by parts formula can be a great way to find the antiderivative of the product of two functions you otherwise wouldn’t know how to take the antiderivative of. ∫ = − ∫ 3. En mathématiques, l'intégration par parties est une méthode qui permet de transformer l'intégrale d'un produit de fonctions en d'autres intégrales, dans un but de simplification du calcul. Integration by parts is a technique used in calculus to find the integral of a product of functions in terms of the integral of their derivative and antiderivative. The formula for integration by parts is: The left part of the formula gives you the labels (u and dv). ∫ ( f g) ′ d x = ∫ f ′ g + f g ′ d x. The trick we use in such circumstances is to multiply by 1 and take du/dx = 1. 7.1: Integration by Parts - … SOLUTION 2 : Integrate . 5 Example 1. The differentials are $du= f' (x) \, dx$ and $dv= g' (x) \, dx$ and the formula \begin {equation} \int u \, dv = u v -\int v\, du \end {equation} is called integration by parts. The goal when using this formula is to replace one integral (on the left) with another (on the right), which can be easier to evaluate. LIPET. Menu. The Integration by Parts formula may be stated as: $$\int uv' = uv - \int u'v.$$ I wonder if anyone has a clever mnemonic for the above formula. The main results are illustrated by SDEs driven by α-stable like processes. It is also possible to derive the formula of integration by parts with limits. Reduction Formula INTEGRATION BY PARTS Reduction Formula Example Example INTEGRATION BY PARTS Reduction Formula INTEGRATION BY PARTS Reduction Formula Example Example Reduction Formula INTEGRATION BY PARTS Reduction Formula Example Example Reduction Formula F132 F121 Sec 7.5 : STRATEGY FOR INTEGRATION Trig fns Partial fraction by parts Simplify integrand Power of … Learn which math classes high schoolers should take by reading our guide. Example. SOLUTIONS TO INTEGRATION BY PARTS SOLUTION 1 : Integrate . And from that, we're going to derive the formula for integration by parts, which could really be viewed as the inverse product rule, integration by parts. (fg)′ = f ′ g + fg ′. You’ll need to have a solid knowledge of derivatives and antiderivatives to be able to use it, but it’s a straightforward formula that can help you solve various math problems. Focusing just on the “∫cos(x) ex dx” part of the equation, choose another u and dv. AMS subject Classification: 60J75, 47G20, 60G52. Integration by Parts. The integration by parts formula can also be written more compactly, with u substituted for f(x), v substituted for g(x), dv substituted for g’(x) and du substituted for f’(x): You can use integration by parts when you have to find the antiderivative of a complicated function that is difficult to solve without breaking it down into two functions multiplied together. Choose a and , and find the resulting and . If there is a logarithmic function, try setting this equal to u, with the rest of the integrand equal to dv. Get the latest articles and test prep tips! 8 Example 4. That's really interesting. In English, to help you remember, ∫u v dx becomes: (u integral v) minus integral of (derivative u, integral v), Integrate v: ∫1/x2 dx = ∫x-2 dx = −x-1 = -1/x (by the power rule). Integration by Parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways. However, don’t stress too much over choosing your u and v. If your first choices don’t work, just switch them and integrate by parts with your new u and v to see if that works better. SOLUTIONS TO INTEGRATION BY PARTS SOLUTION 1 : Integrate . Using the Formula. This formula is very useful in the sense that it allows us to transfer the derivative from one function to another, at the cost of a minus sign and a boundary term. Transcription de la vidéo. In other words, this is a special integration method that is used to multiply two functions together. The formula for this method is: ∫ u dv = uv - ∫ v du. Click HERE to return to the list of problems. LIPET. We can use integration by parts again: Now we have the same integral on both sides (except one is subtracted) ... ... so bring the right hand one over to the left and we get: It is based on the Product Rule for Derivatives: Some people prefer that last form, but I like to integrate v' so the left side is simple. Add the constant, and you’re done; there are no more antiderivatives left in the equation: Find du and v (the derivative of sin(x) is cox(x) and the antiderivative of ex is still just ex. We illustrate where integration by parts comes from and how to use it. Sometimes integration by parts must be repeated to obtain an answer. What ACT target score should you be aiming for? ln x = (ln x)(1), we know. We'll then solve some examples also learn some tricks related to integration by parts. minus the integral of the diagonal part of the 7, (By the way, this method is much easier to do than to explain. … The following figures give the formula for Integration by Parts and how to choose u and dv. Welcome to advancedhighermaths.co.uk A sound understanding of Integration by Parts is essential to ensure exam success. Let u = x the du = dx. Deriving the integration by parts standard formula is very simple, and if you had a suspicion that it was similar to the product rule used in differentiation, then you would have been correct because this is the rule you could use to derive it. First choose which functions for u and v: So now it is in the format ∫u v dx we can proceed: Integrate v: ∫v dx = ∫cos(x) dx = sin(x) (see Integration Rules). It may seem complicated to integrate by parts, but using the formula is actually pretty straightforward. Choose a u that gets simpler when you differentiate it and a v that doesn't get any more complicated when you integrate it. Our new student and parent forum, at ExpertHub.PrepScholar.com, allow you to interact with your peers and the PrepScholar staff. We’ll start with the product rule. It is used for integrating the products of two functions. Well, that was a spectacular disaster! There are five steps to solving a problem using the integration by parts formula: #1: Choose your u and v #2: Differentiate u to Find du #3: Integrate v to find ∫v dx #4: Plug these values into the integration by parts equation #5: Simplify and solve It may seem complicated to integrate by parts, but using the formula is actually pretty straightforward. My Integrals course: https://www.kristakingmath.com/integrals-course Learn how to use integration by parts to prove a reduction formula. But it also requires parts v=g ( x ) ( 1 ), we are integrating parts! 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First multiply everything out: then take the antiderivative of that messy equation by working through the integration parts... Is simpler than the derivative for example, if we have to solve to using integration... | 5,409 | 23,882 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2023-14 | latest | en | 0.931457 |
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``` “Do noW” TemplaTe TAPE the note sheet you were
handed at the door into pg. of
The ancient Greeks concluded that it was
necessary to apply a continuous force to
keep an object in motion.
For example, if you set a book on a table and
give the book a quick push, the book slides
a short way then stops. To keep the book
moving, you need to keep pushing it.
The Greeks reasoned that the book stops
moving because you stop pushing it.
QUESTION
Based off of information you have learned, write a brief
paragraph letter to the Greeks, explaining what scientists
have discovered to help the Greeks understand why the book
stops moving.
A puck is moving on an
air-table. If the table is
so large that the puck
never hits an edge, what
will happen to the puck?
A It will never stop.
B It will stop in a short
distance.
C It will speed up.
Galileo’s Thought Experiment
• Galileo thought of a different way to explain
the observation of the motion of the book.
• He imagined a world without friction and
conducted an experiment.
I doubt the Greeks’
idea! Suppose…
Galileo’s Thought Experiment
The rail is
horizontal…
If ball-bearing is No force is needed to
released here… keep the ball-bearing
moving.
…it will never rise up,
and will move at
constant speed.
horizontal rail
Galileo’s Thought Experiment
– In the absence of friction, a moving object will
continue moving even if there is no force acting
on it.
– In other words it does not take a force to keep an
object moving, it takes a force – friction – to stop
an object that is already moving.
Newton’s 1st Law of Motion
Description
Objects at rest remain at
rest, and objects in motion
remain in motion with the
same velocity, unless acted
upon by an unbalanced
force.
Newton’s 1st Law of Motion
LAW
Newton’s 1st Law of Motion
Inertia: Tendency of an
object to resist changes in
motion.
mass inertia Mass of an object is
resistance to a measure of its
start to move from rest inertia.
change moving speed
Which Has More Inertia?
Which is harder to push, pull, turn, or stop?!?
Newton’s 1st Law of Motion
Everyday Examples
Unbalanced Force
Unbalanced Force
Object in Motion
Object at Rest
Activity: (Name of Activity Station)
Observations:
Newton’s 1st Law!)
Exit Ticket
A ball is at rest on the floor of a car moving at a
constant velocity. What will happen to the ball
if the car swerves suddenly to the left? | 598 | 2,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2013-48 | longest | en | 0.932234 |
http://shell.math.rutgers.edu/~greenfie/currentcourses/math291/old_diary.html | 1,537,727,916,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159570.46/warc/CC-MAIN-20180923173457-20180923193857-00491.warc.gz | 232,276,386 | 33,116 | ### Diary for Math 291, fall 2002
Now in reverse chronological order!
Date What happened
10/14/2002 Jason Sullivan kindly presented problem #5 of workshop #4. This is a problem which caused some difficulties.
We then went over some problems from chapter 14 in preparation for the exam. Ther instructor managed to use up lots of time by suggesting that students do a problem from the Chain Rule section, translating the Laplacian (sum of the "pure" second partials in x and y) from one coordinate system to another.
10/10/2002 Ah, well, the last real class before our first exam wasn't a total disaster but I didn't quite do what I wanted. Let me first tell what I thought I did, and then comment on what I might additionally have wished to do.
I summarized looking for extreme values for functions of 1 variable. Here goes:
• A point p in the real numbers is a local {maximum|minimum} for a function f if the domain of f includes an interval with p in the interior of the interval, and if for all x in that interval, f(x){<|>}=f(p). Now comments about this definition: the definition really is local. The interval doesn't have to be "big", just some interval. So if f(x)=x2-10-100x4 I bet that locally (near 0) f(x)'s values are positive (the effect of the x4 term is tiny) except that f(0)=0. So 0 is a local min of f. But certainly when |x| is large, the x4 term dominates, and so f does not have an absolute min. The word "strict" is sometimes applied as a modifier to "local" if the = sign is not needed. So the function f(x)=0 has all local maxes (and mins, actually!) but no strict local maxes or mins.
• If f is differentiable at a point p, and if f'(p) is not 0, then f does not have a local max or min at p. That's because from the definition of differentiability which we reviewed before, f(p+h)=f(p)+f'(p)h+higher order error. So when h is small, the error will be much less in absolute value than the f'(p)h term, and that will make values to the {right|left} bigger than f(p) if f'(p) is {posi|nega}tive, and values to the {left|right}less that f(p).
• p is a critical point of f if either f'(p) doesn't exist or f'(p) equals 0. There are lots of "rough" functions where f' doesn't exist, so that is a possibility which should not be discarded in practice, although in elementary courses it is frequently neglected. And finding p's where f'(p)=0 can be difficult with functions defined by even moderately complicated formulas.
• Local {max|min} must occur at critical points. Examples: |x|, +/-x2, +/-x3.
• How can one guarantee that a critical point is a local max or min? A simple "test" uses the second derivative, goes like this: suppose p has f'(p)=0 (so p is a critical point). Then:
Ifthen
f''(p)>0p is a local max.
f''(p)<0p is a local min.
As for the last line of the table, the examples x3 and +/-x4 show that the hypotheses can be fulfilled while the function has no local max or min, or that such a function can have a local max or min.
• This all really should be thought about in the context of Taylor's Theorem. This result is very important. If a function f has sufficiently many derivatives, then f(p+h)=f(p)+f'(p)h+(f''(p)/2)h2+...+(f(n)(p)/n!)hn+Error(f,p,h,n) where (the important thing!) the error term --> faster than hn. This last means precisely that the limit of Error(f,p,h,n)/hn is 0 as h-->0. Notice that this really looks like the definition of derivative for the case n=1. Taylor's Theorem yields "statements" like the 18th derivative test. (?) This could say something like this: if p is a critical point, and if f''(p)=f'''(p)=...f(17)(p)=0 and if f18(p) is not 0, then f has a local max or min at p depending on the sign of f(18)(p). The reason this is true is that Taylor's Theorem for n=18 is just (because of all the ludicrous hypotheses!) f(p)+(f(18)(p)/18!)h18+Error, and the Error term is negligible compared to the term immediately before it for |h| small. I don't know if anyone ever really states such a "test" because it would hardly ever be used.
How much of all this can be carried over to more than 1 variable? Much can but some surprises develop, most particularly in the local geometry of a critical point. Students in 1 variable calc usually don't like x3 which has an inflection point at 0: the tangent line crosses the graph. An analogous occurence (the tangent plane crossing the graph) will occur very often if n>1. So let's begin.
• A point p in Rn is a local max for a function f if there is some positive number R so that the domain of f includes all points at distance <R from p and if x is a point in Rn with ||x-p||<R then f(x)<=f(p). When n=2, "||x-p||<R" means points inside a circle of radius R centered at p. When n=3, "||x-p||<R" means points inside a sphere of radius R centered at p. For local min just change < to > in what was written. We had as examples in R4 functions like f(x)=x1300+5x2600+88x3900+22x46. Here f(0,0,0,0)=0. Because of the parity (even!) of the exponents and positivity of the coefficients, f of anything not (0,0,0,0) is positive. So (0,0,0,0) must be a local min of this f. No other "work" is necessary! We can of course get a local max by reversing all the signs. But notice that worse can happen. There are 16=24 choices of signs in this expression. Any of the other 14 choices of sign distribution result in the following behavior: values of f near (0,0,0,0) which are bigger than 0 and values which are less than 0. Since grad f at 0 is 0, the tangent plane is "horizontal" (parallel to the domain plane) and the graph of the function cuts through it, sort of an inflection behavior. This behavior is called a saddle point.
• If f is differentiable at p, and if the gradient of f at p is not 0, then: well, some partial derivative of f at p isn't 0, so in some "slice" with all varialbes but 1 fixed, we get a function with non-zero derivative at p, and by the 1 variable analysis above, in that slice, the functin can't have a local max or min at p. So if grad f is not 0 at p, p can;'t be a local max or min.
• p is a critical point of f if either f is not differentiable at p or grad f(p)=0.
• Local {max|min}'s must occur at critical points. We considered some examples in just two variables. f(x,y)=sqrt(x2+y2) can be differentiated from (0,0). grad f is not 0 away from (0,0). At (0,0) the gradient does not exist (the square root is in the denominator!) (0,0) is the only critical point, and certainly f(0,0)=0 and f(everywhere else)>0 (just look at the function, don't attempt to do anything very sophisticated!). Therefore for this function, (0,0) is a local min (actually an absolute min, in fact). The graph of this function is a right circular cone, with axis of symmetry the z-axis and vertex at (0,0,0), a "corner" (it is the graph of |x| revolved about the z-axis. Then f(x,y)=+/-x200+/-y300 provides 4 more examples. This function is differentiable at every point, and the only critical point is (0,0). For +/+ the c.p. is a local min, for -/- it is a local max, and for +/- or -/+ it is a saddle point.
Now we'll try to extend a second derivative test to several variables. This will be complicated. The idea is somehow to use "second order" information at a critical point to see if the critical point is a max, a min, or a saddle. Realize, though, that the test might well "fail" -- that is, it might be inapplicable just as the 1 variable second derivative test (last line in the table above, "no conclusion can be made") can also fail.
The simplest situation is 2 variables. Things already get complicated enough. We try to "bootstrap" from what we already know: that is we will try to use the second derivative test from one variable to allow us to get information for two variables.
Here are the starting assumptions: we have a function z=f(x,y) with a critical point at (a,b). So we know that grad f=0 at (a,b). If we now "slice" the graph in R3 of z=f(x,y) by planes perpendicular to the (x,y)-plane which go through the point (a,b), in each case we get a curve which has a critical point above (a,b). Let me try to describe these slices. Choose a two-dimensional direction: that is, a two-dimensional unit vector. We're lucky that such vectors have a simple description: u=cos(theta)i+sin(theta)j. A two-dimensional straight line through (a,b) in the direction of u is just (a+cos)theta)t,b+sin(theta)t). And f's values on that straight line are exactly f(a+cos(theta)t,b+sin(theta)t). What's the derivative of this function with respect to t? We will compute this, but first some discussion of it: this is the directional derivative of f in the u direction. It is also the derivative of the curve obtained by slicing the graph z=f(x,y) by the plane through (a,b,0), perpendicular to the (x,y)-plane, in the direction of u. So now:
```(d/dt)f(a+cos(theta)t,b+sin(theta)t)=
D1f(a+cos(theta)t,b+sin(theta)t)cos(theta)
+D2f(a+cos(theta)t,b+sin(theta)t)sin(theta).```
This computation uses the Chain Rule. It is, of course, exactly Duf, which was defined last time. Since (a,b) is a critical point, both D1f and D2f are 0 at (a,b). So the first directional derivative is 0. What we now need to do is take the t derivative again. We must be careful. Each of D1f and D2f are functions of 2 variables, and that each of the two entries in each of these two functions has some dependence on t: the Chain Rule needs to be used with care. So if we d/dt what's above (in this computation theta and a and b are constant), we will get:
```d/dt(D1f(a+cos(theta)t,b+sin(theta)t)cos(theta)
+D2f(a+cos(theta)t,b+sin(theta)t)sin(theta))=
D1D1f(a+cos(theta)t,b+sin(theta)t)(cos(theta))2
+D2D1f(a+cos(theta)t,b+sin(theta)t)(cos(theta)sin(theta))
+D1D2f(a+cos(theta)t,b+sin(theta)t)(sin(theta)cos(theta))
+D2D2f(a+cos(theta)t,b+sin(theta)t)(sin(theta))2```
and now "plug in" t=0. Remember that cross-partials are equal (fxy=fyx) and call A=fxx(a,b) and B=fxy(a,b) and C=fyy(a,b). Then what we have is
`A(cos(theta))2+2Bcos(theta)sin(theta)+C(sin(theta))2`
We could apply the second derivative test in one variable to this if we knew appropriate information about every slice. Thus we could conclude that the critical point (a,b) was a minimum if for every theta, the expression above was positive. If theta=0 this means that A=fxx(a,b) should be positive. But even more needs to be true. Suppose we divide the expression above by (cos(theta))2 and we call w=tan(theta) (this is weird!). Then what is written above is just A+2Bw+Cw2 and we would like to know what conditions on A and B and C would imply that this is always positive. We did something like this in the lecture 9/9. This positivity will occur exactly when the quadratic has no real reals. So (2B)2-4AC should be negative. (The parabola A+2Bw+Cw2 will then never intersect the w-axis and since A>0 it must always be positive.!)
This somewhat subtle reasoning leads to what is called the Second Derivative Test for functions of two variables. First, define the Hessian, H, to be
``` | fxx fxy |
H=det | |
| fxy fyy |```
and suppose that (a,b) is a critical point of f.
Ifand ifthen
H(a,b)>0fxx(a,b)>0p is a local max.
fxx(a,b)<0p is a local min.
This statement needs examples. But it does display the local "structure" of a critical point a bit. (I very briefly looked at f(x,y)=xy.) Please do some problems in section 14.7.
This should give you some idea of the complexity of critical point behavior in more than 1 variable. Although much is known already, this is actually an object of current research.
10/9/2002 I finally got around to something I should have discussed several lectures ago. First, a mean thing done to Maple: I typed z:=x^2*arctan(y*exp(y)); and then I asked Maple to compute diff(diff(z,y\$30),x\$3); which is the 30th derivative with respect to y of z followed by 3 derivatives with respect to x. The result was 0, and the computation time needed was 5.539 seconds on the PC in my office. Of course what Maple did was compute 30 derivatives of a complicated function, and these derivatives get very big in terms of data (this is called "expression swell"), and take time and storage space to manipulate. Then there were 3 x derivatives, and the result was 0. On the other hand, when I instructed Maple to do the x derivatives first and then the y derivatives, the program reported no elapsed time used (that means, less than a thousandth of a second).
What's going on? The result needed here is called in your text Clairaut's Theorem, and loosely states that the order of taking "mixed" partial derivatives (derivatives with respect to different variables) doesn't matter. More precisely, the result is: if f is a function of two variables, x and y, and if both fxy and fyx exist and are continuous, then they must be equal. I tried to motivate this with a brief allusion to spreadsheets (!) and then remarked that an actual proof is in appendix F of the text, and uses the Mean Value Theorem 4 times, in much the same manner as we've already seen (the lecture on 10/3). I didn't have the time or desire to give the proof.
Here, though, is an example to show that some hypotheses are needed to guarantee the equality of the mixed partials:
A function f with fxy NOT equal to fyx PDF Picture
The main topic of today's class was flying a rocket ship. Suppose we have a rocket ship flying through a nebula. The nebula is a cloud of matter, with qualities such as pressure and temperature. So I decided to consider the temperature. What contributes to the temperature of the nebula as perceived by the crew of the rocket ship? An analysis of this question mathematically using the tools we already have is possible if we describe everything algebraically. So suppose the flight path of the rocket ship is given as a parametric curve. That is, x=x(t), y=y(t), and z=z(t). This is the same as giving a position vector, R(t)=x(t)i+y(t)j+z(t)k. We know about this, and about the first and second derivatives of position (which are velocity and acceleration, respectively). The temperature depends on the location in the nebula, so that is a function T of (x,y,z), the coordinates of a point in the nebula: usually we write T(x,y,z).
The temperature of the point in the nebula outside the rocket ship at time t is T(x(t),y(t), z(t)). How does the temperature change? This is, of course, (d/dt)T=Tx(dx/dt)+Ty(dy/dt)+Tz(dz/dt). Some contemplation of this formula allows a certain "decoupling" to take place, that is, a separation of the influence of the rocket ship and of the temperature. We recognize that the velocity vector is V(t)=(dx/dt)i+(dy/dt)j+(dz/dt)k. And then we see that (d/dt)T is really a dot product of V(t) with another vector, namely Txi+Tyj+Tzk. This vector is called the gradient of T, written sometimes as (upside down triangle)T and sometimes called grad T. I'll call it grad T here because of typographical limitations.
There were some questions about grad T, so I invented a rather simple example, something like the following: if T(x,y,z)=3x2+4xy3+z4, then T(2,-1,1)=5. And grad T=6xi+12xy2+4z3. At p=(2,-1,2), grad T is 12i+24j+4k. grad T is a vector function of position, while T itself is a scalar function. Then (d/dt)T turns out to be the dot product of grad T with V(t).
Observation 1 We have separated the effects of temperature and the rocket. In fact, all that matters in the computation of (d/dt)T from the rocket is the velocity. If two rocket ships go through the same point with the same velocities, then (d/dt)T will be the same. All that matters is the velocity: the tangent direction to the flight path (a curve) and the speed (the lenght of V).
Observation 2 How can we make (d/dt)T big? If it isn't 0, we could always make the rocket ship move faster. If we increase the speed by a factor of M>0, then (d/dt)T will multiply by M. This is because (d/dt)T is ||V|| ||grad T|| cos(angle between V and grad T). Multiplication of V by M results in a multiplication of ||V|| by M, and nothing else changes. So (d/dt)T gets bigger.
Observation 3 But really the answer to the previous question was a bit silly. We should try to look at different directions. What direction will cause the derivative to be larger? So we defined the directional derivative of T with respect to a unit vector u: DuT. This was the rate of change of T with respect to a "rocket ship" moving with unit speed in the direction of u, and it is just u· grad T. Since this is ||u|| ||grad T|| cos(angle between them) and we've restricted ||u|| to be a unit vector, we see the only ingredient we've got to vary is the angle. Indeed, immediately we have:
This is because cosine's values are in [-1,1]. More is true, when grad T is not the 0 vector.
The unique unit vector maximizing the directional derivative is (grad T)/||grad T||.
The unique unit vector minimizing the directional derivative is -(grad T)/||grad T||.
This turns out to have many computational implications.
I went back to the example and found the directional derivative of that T in various directions, including the maximizing direction and the minimizing one.
Observation 4 Isothermals are collections of points where the temperature is constant. If T(2,-1,1) is 5, then (2,-1,1) is in the isothermal (the level set) associated to the temperature 5. If the rocket ship flies in an isothermal, then the rate of change of the temperature perceived by the rocket ship is 0. So (d/dt)T=0 and by the decoupling we've already seen, that means V(t)·grad T=0. So the velocity vector of such a flight is always perpendicular (normal) to the gradient vector. From this evidence and the knowledge that grad T at (2,-1,1) is 12i+24j+4k we were able to deduce that the plane tangent to the surface T(x,y,z)=9 is 12(x-2)+24(z+1)+4(z-1)=0. Directions normal to the gradient have zero first order change.
Here is a command to graph a chunk of the surface T(x,y,z)=5 and store the graph in the variable A:
A:=implicitplot3d(3*x^2+4*x*y^3+z^4=5,x=1..3,y=-2..0,z=0..2,
grid=[40,40,40],axes=normal,color=green):
Here is a command to graph and store as a variable B a piece of the candidate for the tangent plane which we computed:
B:=implicitplot3d(8*(x-2)+24*(y+1)+4*(z-1)=0,x=1..3,y=-2..0,z=0..2,
grid=[40,40,40],axes=normal,color=red):
And finally the command display3d(A,B) displays both graphs together. The result is shown below. Note that the tangent plane actually cuts through the surface (a saddle-more about this next time!).
Then I discussed how people might want to computationally maximize ("hill climbing") or minimize ("method of steepest descent") functions of many variables. To increase (respectively decrease) a function W of many variables, compute grad W, and move in the direction of grad W (respectively - grad W) "for a while". Then repeat (because the direction of grad W will likely change). There are many technicalities in all this: how big are the steps to move, and when should one terminate the procedure. Answers to these questions may depend on the specific nature of the functions and the situations. I looked at a specific function of 4 variables (I think it was p2exp(q-rs) and computed its (4-dimensional) gradient at (2,1,-2,1).
I sketched some level curves for the function 10-2x2-y2 and related the gradient of the function to the level curves: again the gradient is perpendicular to the level curves, and points in the direction of increasing function value.
I gave out and discussed the review problems.
10/7/2002 We discussed the chain rule using some examples of functions with given data points. What I wanted people to get out of this was a general philosophy that differentiation somehow measures first-order or linear effects. Therefore when computing the derivative or partial derivative, "kick" the designated variable and follow through the change using the various definitions of derviative and partial derivative. At the end, examine the compounded change and identify the first-order effect: that's the desired derivative.
Each of the questions on the data sheet involved what seemed to be a slightly different type of composition. They can all be analyzed in the same way, though.
For the club example, the chain rule is S'(t)=h'(k(t))k'(t). For diamond, the equation W'(t)=fx(h(t),k(t))h'(t)+fy(h(t),k(t))k'(t) applies. And heart is a consequence of Qx(x,y)=fx(h(x),g(x,y))h'(x)+fy((h(x),g(x,y))gx(x,y) and Qy(x,y)=fy((h(x),g(x,y))gy(x,y). There is the potential for great confusion in the varied appearances of x and y, and in where the derivatives are evaluated. Finally, spade follows from C'(t)=h'(f(t,2-3t))(fx(t,2-3t)+fy(t,2-3t)(-3)).
Then I tried to indicate how the chain rule can result in results that are notationally irritating when working with implicit functions. These examples are discussed in the text. If F(x,y)=Constant implicitly defines y as a function of x, then applying d/dx to this equation gives (Fx)+(Fy)(dy/dx)=0 or dy/dx=-Fx/Fy where certainly the idea of treating derivatives like fractions doesn't make sense! And in 3 variables the situation gets even worse. If F(x,y,z)=Constant implicitly defines z as a function of x and y, then the partial derivative of z as a function of x (with y held constant) is -Fx/Fz. Symmetrically working with each variable in turn we see that (zx)(xy)(yz)=-1.
Then I gave out workshop #5 and it was noted that the due date was wrong. This has since been corrected on the web.
10/3/2002 A function of one variable, f(x), is differentiable if f(x+h) can be written in the following way:
f(x+h)=f(x)+Ah+Error(f,x,h)h
where: 1) A is a number not depending on h but possibly depending on x and f (A is called f'(x) usually); 2) Error(f,x,h)-->0 as h-->0.
Parenthetical remark: I mentioned that if we knew a function of 1 variable was equal to its Taylor series, then the error term could be written as a sum of an infinite tail of the series, so that's what "Error" could be. Of course, many functions don't have a Taylor series and/or don't have a readily computable Taylor series, so the value of that observation is unclear!
I want to define something similar for functions of two variables. Preliminarily, f(x,y) will be differentiable if f(x+h,y+k) can be written in the following way:
f(x+h,y+k)=f(x,y)+Ah+Bk+Error1(f,x,y,h,k)h+Error2(f,x,y,h,k)k
where Error1 and Error2 both -->0 as (h,k)-->(0,0).
This is a complicated statement and needs some investigation. First, the equation is true for any selections of h and k. So we can look at any special values we care to. For example, if k=0 and h is not 0, we can rewrite the equation to become ((f(x+h,y)-f(x,y))/h)=A+Error1. But as h-->0, the right-hand side-->A (since the Error terms go to 0). That means the left-hand side has a limit. So if f is differentiable, then fx exists and equals A. Similarly (set h=0 and let k-->0) if f is differentiable, then fy exists and is B. Therefore,
If f is differentiable, the partial derivatives of f exist.
The converse of that statement is generally not true.
The converse of a simple implication reverses the hypothesis and the conclusion. For example, the converse of "If Fred is a frog, then Fred hops" is "If Fred hops, then Fred is a frog." Even this simple a statement shows that a converse need not be true.
The function we looked at last time, defined by f(x,y)=(xy)/sqrt(x2+y2) if (x,y) is not (0,0) and by 0 if x and y are both 0 has some interesting properties. First, f has partial derivatives at every point. Why? Away from (0,0), the partial derivatives can be computed by the standard algorithms of 1-variable calculus. At (0,0), we notice that f is 0 on both the x and y axes, so the partials both exist and are 0. But consider f(w,w) where w is a small positive number. Direct computation in the formula shows that f(w,w,) is (1/sqrt(2))w. But if f were differentiable at (0,0) the wonderfully complicated definition above would apply. The right-hand side of the formula is f(x,y)+fx(0,0)h+fy(0,0)k+Error1(f,x,y,h,k)h+Error2(f,x,y,h,k)k but f(0,0) and fx(0,0) and fy(0,0) are all 0. So f(w,w)=Error1(f,x,y,w,w)w+Error2(f,x,y,w,w)w. If we remember that f(w,w) is (1/sqrt(2))w and divide the equation by w, we get (1/sqrt(2))=Error1+Error2 and both of the errors-->0 as w-->0. So the right-hand side-->0 but the left-hand side does not. This rather intricate contradiction shows that this f is not differentiable at (0,0).
Several things still may not (probably are not!) clear to a student right now. What are the reasons for defining "differentiable" in this intricate manner? How could one check that specific functions defined by formulas are differentiable?
Here is one answer to the second question. The first will be answered several times during the rest of this course. Suppose we want to compare f(x+h, y+k) and f(x,y). Then consider:
f(x+h,y+k)-f(x,y)=f(x+h,y+k)+0-f(x,y)=f(x+h,y+k)-f(x,y+k)+f(x,y+k)-f(x,y)
. We have again written a 2-variable change as a succession of 1-variable changes.
The 1-variable Mean Value Theorem shows that f(x,y+k)-f(x,y)=fy(x,y+betak)k where |betak|<|k|. Also f(x+h,y+k)-f(x,y+k)=fx(x+alphah,y+k)k where |alphah|<|h|. Here's the big hypothesis coming: if the partial derivatives are continuous, then the difference between fy(x,y+betak) and fy(x,y) approaches 0 as k-->0. And the difference between fx(x+alphah,y+k) and fx(x,y) also goes to 0 as (h,k)-->(0,0). These differences both get put into the error terms, and so we see
If the partial derivatives are continuous, then the function is differentiable.
The contrast between differentiability and partial derivatives doesn't really have a good analogy in 1 variable: it seemed new to me when I first saw it and I needed time and effort to understand it. There are some further examples to illustrate the complicated logical relationships from a calculus course at MIT:
A function f so that fx and fy exist everywhere, but f is NOT differentiable PDF Picture A function f which is differentiable although fx and fy are NOT continuous PDF Picture
In this course, almost all the functions will be defined by formulas and the formulas can be differentiated, and, inside usually easily defined domains, the partial derivatives will be continuous, so the functions will be differentiable.
Then I went through problem #19 of section 14.4, a rather straightforward application of the linear approximation (that's the constant term plus the first degree terms) to an incremented value of the function. The numbers worked out nicely, and, in this example, the errors were fairly small.
I concluded by finding a tangent plane to a graph, I think the graph of z=5x2+2y4, a sort of a cup, at the point (2,1,22). I did this by examining the section of the graph where y=1. This gives a curve with equation z=5x2+2 whose derivative at (2,22) was 10x, or 20. Therefore 1i+0j+20k was tangent to the surface at (2,1,22). A similar analysis when x=2 shows that 0i+1j+8k was also tangent to the surface at that point. Therefore the cross product of these vectors would be perpendicular to the tangent plane. We computed the cross-product and it was -20i-8j+k (generally it will be -fxi-fyj+k) and the plane went through the point (2,1,22), so the desired tangent line is -20(x-2)-8(y-1)+(z-22)=0.
I remarked that on an exam I would rather not have such an answer "simplified" which brought up the question of
Exam conditions and schedule
• I'd like to give the first exam after we finish 14.5, 14.6, and 14.7, as planned on the schedule. I would also give out review material. So I tentatively would like to give the exam on Wednesday, October 16.
• I would like to eliminate time pressure on the exam, so I will try to get a room we can stay in for more than the standard period (more than 4:30--5:50). I will not use this as a reason to make an exam long or excessively difficult.
• I will try to schedule a review session on Tuesday evening, October 15.
• My feeling right now is that calculator use should be minimal on an exam in this course. I would like to restrict the use of calculators to the last 20 minutes of an exam.
• Some of the formulas may seem intricate to you (to me, too!). I would be happy to write a formula sheet which would be attached to the exam. Please let me know what formulas that you believe you will need.
10/2/2002 Matthew Gurkovich presented a solution to problem 3 in workshop 3. I thank him for this. I think the most difficult part of this problem is recognizing that the letter "t" serves two different purposes in the problem.
I analyzed the idea of differentiability for functions of 1 variable. What is the derivative? Some ideas are i) rate of change, ii) slope of the tangent line to the graph of the function, iii) a certain limit, and iv) velocity or acceleration. And certainly there are others.
As a limit, f'(x)=limh-->0(f(x+h)-f(x))/h. Of course this limit is paradoxical contrast to the basic "limit reason" for looking at continuous functions: those limits can be evaluated just by "plugging in", and here such an approach results in the forbidden 0/0.
Removal of the "limh-->" prefix to the defining equation above yields a statement which is generally false: there is an error involved. Also the division and minus signs are a complicating feature. So we transform the equation above into the following: f(x+h)=f(x)+f'(x)h+Error(f,x,h)h where Error(f,x,h) is something (possibly [probably!]) very complicated depending on f and x and h with the property that (with f and x held fixed) that Error(f,x,h)-->0 as h-->0. With perspective that further study of calculus gives, we know that if f is equal to its Taylor series, then the Error(f,x,h)h is just a sum of terms involving powers of h higher than 1, so at least one power of h can be factored out. The result is something that goes to 0 "faster" than h: order higher than first order. So a perturbation in the argument, x, to a function f(x), that is, x changes to x+h (h may be either positive or negative) yields f(x)+f'(x)h+Error(f,x,h)h. The first term is the old valuye, the second term is directly proportional to the change, h, with constant of proportionality f'(x), while the third term -->0 with more than first order in h.
In particular, this shows easily that if h-->0 then f(x+h)-->f(x), so differentiable functions (which are those having such a "decomposition") must be continuous. But more than continuity is involved. The simple example
``` 2x if x>0
f(x)= 0 if x=0
(-1/3)x if x<0```
shows that the same multiplier is needed on both sides of 0: 2 can't be deformed into -1/3 by concealing things in the error term. So this function is not differentiable.
The aim is to define a similar "good" decomposition for functions of more than 1 variable:
(old value)+(linear or first order change)+(error of higher order)
Now I looked at 2 variable functions. I defined fx, the partial derivative with respect to x, as the limit as h-->0 of (f(x+h,y)-f(x,y))/h, and fy, the partial derivative with respect to y, as the limit as k-->0 of (f(x,y+k)-f(x,y))/k, if either or both of these limits exist. The choice of letters (k and h) are conventional, and of course we could use anything.
We computed some partial derivatives for functions which looked like 5x2y5 and xexy2 and arctan(y/x). The routine differentiation algorithms (rules for derivatives) work here.
More suspicious examples were considered. The first is f(x,y)=0 if xy=0 and 1 otherwise. This function had the following properties: away from the coordinate axes (where xy=0) both fx and fy are 0. On the y-axis, fy is 0 and on the x-axis, fx is 0. At (0,0), both fx and fy exist and are 0. The limits for the other partial derivatives don't exist. But notice that f(0+h,0+k) compared to f(0,0)+fx(0,0)h+fy(0,0)k+ERROR where somehow ERROR should go to 0 faster than first order. But that means f(h,k)=(three zero terms)+ERROR. Since if both h and k are small positive numbers, we get 1=ERROR, apparently the decomposition is impossible! So (when the correct definition is given!) this f has partial derivatives at (0,0) but is not differentiable at (0,0)!
Then I looked at the function f(x,y)=r cos theta sin theta with x and y in polar coordinates. A more mysterious formula for f(x,y) results if we use x=r cos theta and y=r sin theta: f(x,y)=(xy)/sqrt(x2+y2). The graph of this function includes the x and y axes: f's values are zero there. But then the partial derivatives of f at (0,0) both must exist and are 0. But f(w,w)=.5w, so the change along the line y=x is certainly first order, but cannot be accounted for in the formula f(x+h,y+k)=f(x,y)+Ah+Bk+ERROR. If we take k=0, then a limit manipulation shows that A must be fx(x,y) and B must be fy(x,y). For this function, both partial derivatives exist at all points, and both are 0 at (0,0). Therefore f(0+h,0+k)=0+0+0+ERROR. But f(w,w) is .5w=ERROR, and the right-hand side is higher than first order as w-->0 and the left-hand side is not. More will follow about this tomorrow.
Here's the result of the Maple command
plot3d((x*y)/sqrt(x^2+y^2),x=-3..3,y=-3..3,grid=[30,30],axes=normal,color=pink);
and maybe the picture will help you understand the properties of the function. The colors seem fairly unreliable!
9/30/2002 I began by considering how continuity is defined. In one variable, a function f is continuous at x0 if limx-->x0 f(x) exists and equals f(x0). Combining this with the official definition of limit given last time, we see:
Definition of continuity A function f is continuous at x0 if, given any eps>0, there is a delta>0 so that if |x-x0|
The definition says that limits can be evaluated in the simplest possible fashion, just by "plugging in". It is an important definition which I wanted to work with.
First we looked at the function n(x,y), defined last time. I asked where n was continuous: that is, for which (x0,y0) does the limit of n(x,y) as (x,y)-->(x0,y0) exist and is it equal to n(x0,y0)? First we approached the question "emotionally": where is n continuous? After some discussion, it was decided that n would be continuous "off" the parabola, that is, for (x0,y0) where y0 is not equal to (x0)2. Away from the parabola, the graph of the function is quite flat (always 0). So if (x,y) is close enough, then n(x,y) is really 0 around (x0,y0). So it should be continuous. If (x0,y0) is on the parabola, though, the limit won't exist.
I then said that I wanted to work with the definition, and verify that: n(x,y) is continuous at (0,1) and n(x,y) is not continuous at (0,0).
How to verify that n(x,y) is continuous at (0,1): if we take delta to be, say, 1/2, then ||(a,b)-(0,1)||<1/2 means that (a,b) is not on the parabola (the 1/2 is actually chosen for this reason!), so that n(a,b)=0 and n(0,1)=0 also, and therefore |n(a,b)-n(0,1)|<eps for any positive eps.
The choice of delta here is rather easy and almost straightforward. In general, the choice of delta likely will depend on (x0,y0) and on eps.
Then we went on to try to show that n is not continuous at (0,0). Here we need to verify the negation of the continuity statement. Negations of complicated logical statements can be quite annoying to state. In this case, we need to do the following:
Negation of continuity There is at least one eps>0 so that for all delta>0 there is an (a,b) in R2 with ||(a,b)-(x0,y0)||eps.
Here (x0,y0) is (0,0) and n(0,0)=1. The values of n(a,b) are either 0 or 1 (n is actually a rather simple function!). So we guess that a useful eps to try will be 1/2. Then to get |n(a,b)-n(x0,y0)| at least 1/2 we'd better have n(a,b) equal to 0. That means (a,b) should be off the parabola. So we need (a,b) off the parabola and also within distance delta of (0,0). The suggestion was made that we take (a,b) to be (0,delta/2), and this does work. Note that the (a,b) varies with the delta.
What are the simplest functions usually considered to be continuous> Thank goodness the suggestion was made that polynomials are continuous because I had prepared an analysis of the continuity of the polynomial f(x,y,z)=x2y2-4yz. We expect that this polynomial (and, indeed, all polynomials) are actually continuous at every point. So the limits should be evaluated just by "plugging in". In fact, the limit of f(x,y,z) as (x,y,z)-->(3,1,2) should just be f(3,1,2) which is 1.
I verified the definition for this limit statement. The verification used entirely elementary methods, but was quite intricate is spite of that.
We looked at: |f(x,y,z)-1|=|x2y2-4yz-1|=|(x2y2-4yz)-(3212-4·1·2)|. Then the triangle inequality was used, so that the last term is less than or equal to |x2y2-3212|+|-4yz+4·1·2|. The second of these expressions seems a bit easier to handle, so: |-4yz+4·1·2|<|4|·|yz-1·2|. Here I tried to suggest that too much was changing. This is handled by making the difference equal to several one variable differences. So: |4|·|yz-1·2|=4|yz + 0 -1·2|= 4|yz-y2+y2-1·2|=4(|(yz-y2)+(y2-1·2)|). And again split up by the triangle inequality: 4|yz-y2|+4|y2-1·2|. The second term looks easiest to handle.
Get a "bound" on 4|y2-1·2|=8|y-1| by making |y-1| sufficiently small. Since we had split up the original difference into two parts and then split the part we were considering into two parts, I guessed that it would be good enough to make this less than eps/4. So we would need |y-1|<eps/32.
Now we considered 4|yz-y2|=4|y|·|z-2|. To get this less than eps/4 we would make |z-2| small. But to control the product we needed to control the size of |y|. Well, if (pulled out of the air!) |y-1|<1 then I knew that 0<y<2, so |y|<2. Therefore 4|y|·|z-2|<4·2·|z-2|. This will be less than eps/4 if |z-2|<eps/32. As I mentioned in class, the coincidence of the 32's made me uneasy.
Now we are half done. We still need to estimate the difference: |x2y2-3212|. Again we write it as a succession of differences of one variable: |x2y2-3212|=|x2y2-x212+x212-3212|. And again the triangle inequality leaves us with estimation of two pieces:|x2y2-x212| and |x212-3212|.
We do the second part first. |x212-3212|=|x2-32|=|x+3|·|x-3|. If, say, |x-3|<1 then x is between 2 and 4 so |x+3| is between 5 and 7, and therefore |x2-32|=|x+3|·|x-3|<7|x-3|. This will be less than eps/4 if we require |x-3|<eps/28. (Somehow the numbers came out differently in class!)
The final piece to handle is |x2y2-x212|=|x2|·|y-1|. Since we have already controlled |x| (it is less than 4) we know that |x2| is less than 16. Therefore |x2|·|y-1|<16|y-1|. This will be less than eps/4 if we require |y-1|<eps/64
If we collect all the restrictions on the variables, we see that the implication "if ||(x,y,z,)-(3,1,2)||<delta then |f(x,y,z)-f(3,1,2)|<eps" will be true when delta is chosen to be the minimum of all the blue restrictions. Therefore choose delta to be the minimum of 1, eps/32, eps/28, and eps/64. The technique I outlined is a bit painful, but it does work and it is "elementary".
A picture of what f(x,y,z) looks like doesn't seem to help very much. Here, for example, are three views of the output of the Maple command implicitplot3d(x^2*y^2-4*y*z=1,x=1..8,y=-1..3,z=-2..4,grid=[20,20,20],
color=green,axes=normal);
. The procedure implicitplot3d is loaded by with(plots); and plots implicitly defined surfaces, just as implicitplot itself plots implicitly defined curves. The option grid=[20,20,20] alters the sampling rate Maple uses. The default is [10,10,10], which makes quite a rough picture. On the other hand, one can ask for [50,50,50] which will take about 125 (53) as much time as the default. I just experimented at home. Sketching a sphere with the default grid took .070 seconds, and the [50,50,50] grid took 7.719 seconds. Indeed: in practical applications, the tradeoff between time and picture detail can be interesting.
Pictures of f(x,y,z)=1=f(3,1,2)
x-axis pointing "out" y-axis pointing "out" z-axis pointing "out"
I then discussed workload in the course.
• I expected that students would spend 10 to 12 hours per week outside of class on the course, doing workshop problems and the textbook homework problems.
• The workshop problems should be done neatly, with the pages fastened (stapled or with paperclips), with complete English sentences, with details of computations only indicated and not given.
• Students could hand in do-overs of one workshop problem per workshop. These writeups would need to be done individually. Other students could be consulted, but the writeups themselves would need to be done individually. These redone workshop problems would be due on Wednesday, October 2.
• I invited students to give oral presentations (5 points per problem bonus) of problems 3 and 4 and 5 of workshop #3, at most 5 minutes per problem, at the beginning of class on Wednesday. Please send me e-mail if you want to do this. I hope that by the end of the semester every student would have presented at least one problem to the class.
9/26/2002 The instructor rudely began by filling some boards with remarks about limits and continuity in 1 dimension. Consider f(x)=x2. A sketch was drawn. What happens to f(x) as x-->4? Clearly limx-->4x2=16. What does this mean? This is a limit statement, and the official definition of limit is as follows:
```limx-->af(x)=b means given any eps>0 there is a delta>0 so that
if 0<|x-a|<delta, then |f(x)-b|<eps.
(This is in bold because it is important in the history and theory of the subject!)```
Here due to the limitations of html, "eps" will be written in place of the usually used Greek letter "epsilon" and "delta", in place of the usually used Greek letter "delta".
Here this means: given eps>0, find delta>0 so that if |x-4|<delta, then |x2-4|<eps. In order to verify that this implication is correct, some connection between |x-4| and |x2-16|. But in fact |x2-16|=|x-4| |x+4|. In order to really be convinced that this is true, we need to show that when |x-4| is small, |x+4| is controlled. That is, it does no good to verify that a product of two factors is small by showing that one factor of a product is small, if the size of the other product can't be controlled.
Here is a small "computation": if |x-4|<1, then -1<x-4<1, so 3<x<5 so that 7<x+4<9 and consequently |x+4|<9. Now we can do the proof.
The official proof Suppose eps>0. Take delta to be the smaller of 1 and eps/9. Then if |x-4|
The relevance of the official definition of limit to real people in real life is maybe not too clear. First, revealing the official definition is an effort to encourage people not to interpret the limit statement as just "plugging in" a for x in a formula for f(x) (that's what we like to do, and in fact we do it for well-behaved functions -- exactly the continuous functions. The other observation is that the eps-delta connection is relevant in more detailed analyses of functions, where one tries to relate the "output tolerance" for an error (how close to f(b) are we?) to the input tolerance for error (how close to a need we be to produce at most an appropriate error in the output?).
Then I began analyzing functions in R2. We began by looking at f(x,y)=x2+y2. I drew a graph of this: the graph was a collection of points in R3. I also commented on the contour lines. I strongly recommended the Maple procedures plot3d and contourplot and contourplot3d. You need to type the command with(plots); before using these procedures.
Maple command followed by a picture of its output plot3d(x^2+y^2,x=-2..2,y=-2..2,axes=normal); contourplot(x^2+y^2,x=-2..2,y=-2..2,axes=normal,color=black,thickness=2); contourplot3d(x^2+y^2,x=-2..2,y=-2..2,axes=normal,color=black,thickness=2);
Functions defined by such simple formulas will be continuous, so lim(x,y)-->(x0,y0)x2+y2=(x0)2+(y0)2 "naturally". In fact a detailed verification is much like what I just did in one variable. I would like to concentrate on aspects of limit and continuity which are somewhat new because of more than 1 variable.
We considered the function g(x,y) defined piecewise by g(x,y)=1 if (x,y) is NOT (0,0) and which is 0 if (x,y)=(0,0). The graph is a plane parallel to the (x,y)-plane, 1 unit "up" in the z direction, except for the origin, which is back at (0,0,0). I asked when the limit as (x,y)-->(x0,y0) existed and what the value was. Some discussion followed, and the somewhat disconcerting truth was told: the limit always exists and it always is 1. This example can be done in one variable, though.
The piecewise function h(x,y) defined by
``` = 1 if x>0
h(x,y)=37 if x=0
= 2 if x<0```
has lim(x,y)-->(x0,y0)h(x,y) existing if x is NOT 0, and for x0>0 the limit is 1 while for x0<0 the limit is 2.
A much more subtle example is provided by m(x,y)=(x2-y2)/(x2+y2). This function is "fine" (continuous) away from (0,0). The limit along rays through the origin varies with the ray. Along the positive and negative x-axis the limit is 1, but along the positive and negative y-axis the limit is -1. Along y=Mx, the limit is (1-M2)/(1+M2). I tried to show this surface with a demonstration in class. It is interesting to view the surface using Maple. The procedure contourplot3d gave the "best" picture for me.
I finally looked at n(x,y). This is peculiar piecewise-defined function. Its value is 1 if y=x2 and 0 otherwise. It has the property that limits along any straight line through (0,0) exist, and all these limits are 0 BUT the limit as (x,y)-->(0,0) does NOT exist. I tried to explain this.
The text gives an example of a rational function: (xy2)/(x2+y4) (see p.890 in section 14.2) with similar properties, which maybe is harder to understand.
As a pop quiz, I asked students to create a function so that the limit as (x,y)-->(x0,y0) did not exist if x2+y2=1 but did exist for all other (x,y). I urged students to begin reading chapter 14.
9/25/2002 Hardly any "progress" was made. We did more and more problems from chapter 13. Attempts were made by valiant students to really get me to explain what CURVATURE and TORSION: Google reports only about 16,700 links with information about both of these.
Curvature, I tried to insist, referred to how much a curve bends. I gave another interpretation using the idea of the osculating circle. If a circle agrees "up to second order" (passes through a point, and first and second derivatives agree) with the graph of a function, then 1/(the radius of the circle) turns out to be the curvature. The circle is called the osculating circle. My online dictionary states:
```1. [Math.] (of a curve or surface) have contact of at least the second
order with; have two branches with a common tangent, with each branch
extending in both directions of the tangent.
2. v.intr. & tr. kiss.```
The osculating circle is a second order analog of a tangent line. The tangent line agrees with a curve up to first order (value and first derivative of the curve and tangent line should agree). The osculating circle does the same up to second order. So where the closest circle is small, the curve bends a lot.
Torsion is weirder. In a picture on this link an attempt is being made to show "high torsion when there is rapid departure from a plane."
In my Google search I found web pages dealing with the relationship of curvature and torsion to coronary arteries and blood flow, concrete, plasma flow, how birds and gnats and flies fly, "carbon nanotubes" (thin filaments), models of molecules, motion of robots and octopuses and cilia and flagella ... and so on. I found Maple routines for computation of curvature and torsion: lots of stuff, most of it quite technical in both its applications and its mathematics. Lots of stuff! I tried to argue that Problem #5 on Workshop #3 showed that curvature could be concealed easily. According to Einstein, "the Lord is subtle but not mean" (approximately) and that knowing that the structure of a curve means dealing with its curvature and torsion sometimes may make life easier. Problem #5 has exquisitely disguised simple curves: in a), a circle, and in b), a straight line.
The next few weeks would see an effort to analyze functions whose domain is in R2 or R3 or Rn and whose range was R. We will look at the concepts of limit, continuity, and derivative, and try to understand the real conceptual subtleties which occur with such functions.
9/23/2002 A valiant and not completely successful attempt to review all questions students had about textbook homework problems for most of the first two chapters. I'll need to spend time doing a few more problems on Wednesday.
I mentioned that one reason to consider an abstract version of vectors and inner products and lengths is that strong results involving other important examples can be learned. I suggested the following setup: a vector would correspond to a function on [0,1]. Vector addition and scalar multiplication would correspond to addition of functions and multiplication of functions by a constant. The dot product of two functions f and g would be defined by the integral from 0 to 1 of f(x) times g(x), so that the "length" of f would be the square root of the integral of f(x)^2 from 0 to 1. First, all the results we have proved about lengths and dot product remain correct. For example, the integral from 0 to 1 of exp(x)sin(x) will be bounded by the square root of the integral of exp(x)^2 multiplied by the square root of the integral of sin(x)^2. (Maple tells me that the first one is approximately .909 while the second is approximately .933.) So we have been "efficient" in learning how to organize our thoughts. Second, it turns out that this method of measuring the "size" of functions is essentially the same as the method of least squares, a widely used technique for estimating errors.
I just learned today that there is a web page with detailed solutions for many of the odd-numbered problems in the textbook. Sigh. You may want to look at www.hotmath.org
9/19/2002 We began with a problem for students: compute the curvature of the plane curve defined by
```x(t)=integral from 0 to t cos(w^2/2)dw
y(t)=integral from 0 to t sin(w^2/2)dw```
Most students were able to successfully see that this curve had curvature k=t, curvature which increased directly proportionately with travel along the curve. The integrals involved are called Fresnel integrals, and the curve resulting is called the Cornu spiral. The curve (and the integrals) arise in diffraction, and one link with a Java applet illustrating this is given. The spiral winds more and more tightly as the parameter increases.
Today is devoted to an investigation of space curves. The geometry of these curves, as seen from the point of view of calculus (called "differential geometry of space curves") is a subject which originated in the 1800's. The material presented here was stated in about 1850-1870. It has within the last few decades become very useful in a number of applications: robotics, material science (structure of fibers), and biochemistry (the geometry of big molecules such as DNA).
I'll carry along is a right circular helix as a basic example.
```x(t)=a cos(t)
y(t)=a sin(t)
z(t)=b t```
The quantities a and b are supposed to be positive real numbers. This helix has the z-axis as axis of symmetry. It lies "above" the circle with radius a and center (0,0) in the (x,y)-plane. The distance between two loops of the helix is 2Pi b.
If r(t)=x(t)i+y(t)j+z(t)k (the position vector), then r'(t)=x'(t)i+y'(t)j+z'(t)k=(ds/dt)T(t) is called the velocity vector. Here T(t) is called the unit tangent vector and is a unit vector in the direction of r'(t). ds/dt is the speed, and is sqrt(x'(t)2+y'(t)2+z'(t)2), the length of r'(t). We use ds/dt also to convert derivatives with respect to t to derivatives with respect to s, as last time (the Chain Rule).
Since T(t)·T(t)=1 differentiation together with commutativity of dot product gives 2T'(t)·T(t)=0, so T'(t) and T(t) are perpendicular. In fact, we are interested in dT/ds, which is the same as (1/(ds/dt))T'(t) (it is usually easier to compute T'(t) directly, however, and "compensate" by multiplying by the factor 1/(ds/dt)). Any non-zero(!) vector is equal to the product of its magnitude times a unit vector in its direction. For dT/ds, the magnitude is defined to be the curvature, and the unit vector is defined to be the unit normal N(t). This essentially coincides with what was done last time, when curvature was defined to be d(theta)/ds but, as a student remarked and I tried uncomfortably to acknowledge, there could be problems if dT/ds is zero or if d(theta)/ds was negative (example: look at how T and N change for y=x3 as x goes from less than 0 to greater than 0).
For the helix, we computed ds/dt (sqrt(a2+b2)) and T(t) (1/sqrt(a2+b2))(-a sint(t)i +a cos(t)j +bk) and also N(t) (-cos(t)i-sin(t)j, always pointing directly towards the axis of symmetry) and k, which was a/(a2+b2). I strongly suggested "checking" this computation by looking at what the formula "says" when a and b are large and small, and comparing this to the curves. We "complete" T and N to what is called a 3-dimensional frame by defining the binormal B(t) to be the cross-product of T(t) and N(t). Since T(t) and N(t) are orthogonal unit vectors, B(t) is a unit vector orthogonal to both of them. (This needs some thinking about, using properties of cross-product!). How does B(t) change? Since B(t)·B(t)=1, differentiation results in 2 B'(t)·B(t)=0, so B'(t) is orthogonal to B(t). But differentiation of B(t)=T(t)xN(T) results in B'(t)=T'(t)xN(t)+T(t)xN'(t). Since T'(t) is parallel to N(t), the first product is 0 (another property of cross-product!) so that B'(t) is a cross-product of T(t) with something. Therefore B'(t) is also perpendicular to T(t). Well: B'(t) is perpendicular to both T(t) and B(t), and therefore, since only one direction is left, B'(t) must be a scalar multiple of N(t). The final important definition here for space curves is: dB/ds is a product of a scalar and N(t). The scalar is - t. That is supposed to be the Greek letter tau, and the minus sign is put there so that examples (the most important is coming up!) will work out better. This quantity is called torsion, and is a measure of "twisting", how much a curve twists out of a plane. If a space curve does lie in a plane, and if everything is nice and continuous, then B will always point in one direction (there are only two choices for B, "up" and "down" relative to the plane, and by continuity only one will be used) so that the torsion is 0 since B doesn't change. The converse implication (not verified here!) is also true: if torsion is always 0, then the curve must lie in a plane!
For our example, we computed B(t) by directly computing the cross-product of T(t) and N(t). We got (I think!) (1/(1/sqrt(a2+b2))(b sin(t)i-a cos(t)j+a k) for B(t). This can be "checked" in several ways. First, that the candidate for B(t) has unit length, and then, that B(t) is orthogonal to both T(t) and N(t). This candidate passes those tests. Then we took d/dt of this B(t) and multiplied it by 1/(ds/dt)=(1/sqrt(a2+b2)). The result was (b/(a2+b2))(cos(t)i+sin(t)j). Checking all the minus signs (one in the definition of torsion and one in the result of N(t)) shows that here torsion is (b/(a2+b2)). Looking at the extreme values of a and b in this expression (a, b separately big and small) is not as revealing and/or as useful as with curvature, since a "feeling" for torsion isn't as immediate.
Then I looked at dN/ds, using the expression N=BxT. The result, after using the product rule carefully (remember that this product is not commutative!) is (dB/ds)xT+Bx(dT/ds) which, by the earlier equations, is -tNxT+kBxN which is tB-kT. So we have the following equations, called the Frenet-Serret equations (also called Darboux equations in mechanics):
```dT/ds= 0 + kN + 0
dN/ds=-kT+ 0 + tB
db/ds= 0 - tN + 0```
This is a collection of 3 3-dimensional vector equations, or a collection of 9 scalar differential equations. The remarkable fact is that if an initial point is specified for the curve, and an initial "frame" for the Frenet frame of T, N, and B, and if the curvature and torsion are specified, then the solutions to the differential equations above give exactly one curve. All the information about the curve is contained in the equations. So, for example, the motion of an airplane or a robot arm or the (geometric) structure of a long molecule are, in some sense, completely specified by k and t. Of course, this doesn't tell you really how to effectively control something so it moves or twists the way it is "supposed" to. The idea that the Frenet frame "evolves" in time, governed by the differential equations above, is useful.
Here are some pictures of various helices produced by Maple (the plural of "helix" is "helices").
The pictures below were produced using the command spacecurve([a*cos(t),a*sin(t),5*t],t=0..6*Pi,axes=normal,color=black,thickness=2,
scaling=constrained);
where a is 1 and 10 and 100 respectively. The procedure spacecurve is loaded as part of plots using the command with(plots);. I used the option scaling=constrained in order to "force" Maple to display the three curves with similar spacing on the axes. Otherwise the x and y variables would be much altered in each image. I hope that these pictures give some idea of what the curvature and torsion represent.
Some helices: x=a cos(t) & y=a sin(t) & z=bt
a=1 & b=5 k=.038 & t=.192 a=10 & b=5 k=.08 & t=.04 a=100 & b=5 k=.01 & t=.0005
9/18/2002 I wrote some simple vector differentiation "rules", dealing with how to differentiate A(t)+B(t) and f(t)A(t) and A(t)·B(t) and A(t)xB(t) if A(t) and B(t) are differentiable vector functions of t and f(t) is a differentiable scalar function of t. I miswrote one of these simple (!) rules, so was condemned to write out a proof until I found the error. I am sorry.
Then I tried to analyze the idea of how a curve bends. Curvature will be a measure of this bending.(Today a plane curve, in R2, and tomorrow a space curve, in R3.) I began an analysis that was apparently first done by Euler in about 1750. A curve is a parametric curve, where a point's position at "time" t is given by a pair of functions, (x(t),y(t)). Equivalently, we study a position vector, r(t)=x(t)i+y(t)j. Here x(t) and y(t) will be functions which I will feel free to differentiate as much as I want.
There are special test cases which I will want to keep in mind. A straight line does NOT bend, so it should have curvature 0. A circle should have constant curvature, since each little piece of a circle of radius R>0 is congruent to each other little piece, and, in fact, the curvature should get large with R gets small (R is positive), and should get small when R gets large (and looks more like a line locally). I also suggested that even y=x2 might be a good test to keep in mind, since there the curvature should be an even (symmetric with respect to the y-axis) function of x, and should be bell-shaped, with max at 0 and limits 0 as x goes to +/- infinity.
The problem is to somehow extract the geometric information from the parameterized curve. That is, if a particle moves faster, say, along a curve, it could seem like the same curve bends more. So what can we do?
We looked at theta, the angle that the velocity vector r'(t) makes with respect to the x-axis. How does theta change? After some discussion it was suggested that we look at the rate of change with respect to arclength along the curve: that is the same as asking for the rate of change with respect to travel along the curve at unit speed, and therefore somehow the kinetic information will not intrude on the geometry.
Arc length on a curve is computable with a definite integral: sqrt(x'2+y'2) integrated from t0 to t with dt is the arc length. This is rarely exactly computable with antidifferentiation using the usual family of functions. But ds/dt is just sqrt(x'2+y'2) by the Fundamental Theorem of Calculus. And the Chain Rule suggests that d*/ds(ds/dt)=d*/dt if * is some quantity of interest, such as theta.
By drawing a triangle we see that theta is arctan of y'/x'. Differentiation with respect to t shows that d(theta)/dt must be (y''x'-x''y')/(x'2+y2)2 (this uses the formula for the derivative of arctan, the Chain Rule, and the quotient rule. Then the previous results say that
d(theta)/dt=(y''x'-x'y'')/(x'2+y'2)3/2, a complicated formula.
Then we saw that this formula for a straight line gave 0, and this formula for a circle was 1/R, where R is the radius of the circle. We used y=mx+b for the line (so x=t and y=mt+b) and x=Rcos(t) and y=Rsin(t) for the circle. This fit well with the examples suggested earlier. And, in fact, on the curve y=x2, with the parameterization x=t and y=t2, the d(theta)/ds gave 4/(1+4x^2)3/2, also consistent with earlier considerations.
d(theta)/ds is curvature, usually called k (Greek letter kappa).
I defined the unit tangent vector, T, to be a unit vector in the direction of r'(t). Therefore r'(t)=(ds/dt)T, where ds/dt is the length of r'(t), and this is the speed. I differentiated the formula for r'(t) using one of the product rules we had stated earlier. Therefore I got r''(t)=(d2s/dt2) T+ (ds/dt)d/dt(T). But T is cos(theta)i+sin(theta)j, and differentiation with respect to t is the same as differentiation with respect to s multiplied by ds/dt. But differentiation with respect to s gives (-sin(theta)i+cos(theta)j) multiplied by the derivative of theta with respect to s, and this is k. All this put together is:
r''(t)= (d2s/dt2)T + k(ds/dt)2N
where N is (-sin(theta)i+cos(theta)j), a unit vector normal to T (check this by dot product!), where is called the unit normal.
We have decomposed acceleration into the normal and tangential directions.
I used this to show that notion in a straight line (k= 0) had no normal component, and therefore a particle moving in a straight line had no force needed transverse to its motion. On the other hand, in our circular situation, the curvature was a positive number, and as long as the particle was moving (ds/dt not equal to 0) a force was needed to keep it one the circle. This is because the curvature k was non-zero, and so were the other terms. This is not at all "intuitively clear" to me.
9/16/2002 I'll go to a lecture which will finish at about 7:30 PM, tomorrow, Tuesday. I will go to Hill 304 and I will be available for questions from my arrival until 9:00 PM. I reserve the right to go home, however, if no one wants to talk to me.
We continued with the problem from last time: p=(3,2,-1) and q=(2,0,1) and r=(1,1,2) are three points in space. Can I describe a simple way to tell if the point (x,y,z) is on the plane determined by these three points?
Here is an method. Suppose v is the vector from p to q (so v is -i-2j+2k) and w is the vector from p to r (so w is -2i-j+3k) Then vxw is -4i-j-3k, a vector normal (perpendicular, orthogonal) to the plane. If a=(x,y,z), then a is on the plane determined by p and q and r if the vector from p to a is orthogonal to vxw=-4i-j-3k. This means that (x-3)(-4)+(y-2)(-1)+(z--1)(-3)=0, which simplifies to -4x-y-3z+11=0. All the steps of this process are reversible, so (x,y,z) is on the plane exactly when that equation is satisfied.
More generally, the points whose coordinates (x,y,z) satisfy Ax+By+Cz+D=0 (with N=Ai+Bj+Ck NOT zero) form a plane, with normal vector N.
We easily checked by direct substitution that (1,2,3) is not on the plane.
Another parametric description of this plane is obtained by adding the vector from 0 to p to scalar multiples of the vectors v and w: the result must be on the plane. So if s and t are any numbers, then the vector 3i+2j-k (from 0 to p) +tv+sw is on the plane. This means (looking at components) if x and y and z satisfy:
```x=3+-t+-2s
y=2+-2t+-1s
z=-1+2t+3s```
for some real numbers s and t, they must be on the plane. I substituted this into the equation -4x-y-3z+11=0 and checked that everything canceled.
What is the distance of Fred=(1,2,3) to the plane described above? We found two ways to do this. First, take a point on the plane: we took p=(3,2,-1). Then I drew a picture to convince people that the distance would be the "projection" of the line segment from Fred to p on a vector normal to the plane. That is, we would need to multiply the distance from Fred to p by the cosine of the angle between the Fred-to-p vector and a normal. We have such a normal (it is inherent in the equation of the plane), and we had such a vector. Then distance can then be computed with a dot product multiplied by the distance from Fred to p. We computed this.
Here's an alternative way to get the distance: find the point (which we called Walter) on the plane which is closest to Fred. How could we find Walter? The vector from Fred to Walter is parallel to a normal to the plane, so the vector from Fred to Walter is a scalar multiple of any normal vector. We therefore got the equations (assuming now that Walter has coordinates (x,y,z)):
```x-1=-4t
y-2=-1t
z-3=-3t```
where t is the scalar. Then substituting x=-4t+1 and y=-t+2 and z=-3t+3 into the equation of the plane (-4x-y-3z+11=0) got one value of t, and this value of t gave the coordinates of Walter. And the distance from Fred to Walter is the distance from the plane to the point.
More generally now I started talking about vector functions of a scalar variable. Here R(t)=x(t)i+y(t)j+z(t)k. This describes the geometry (the path) and kinematics (movement) of a particle. I illustrated this by playing around with the equations x(t)=-4t+1 and y(t)=-1t+2 and z(t)=-3t+3. What is the geometric object described by:
x(t)=-4t+1 and y(t)=-1t+2 and z(t)=-3t+3. a line
x(t)=-8t+1 and y(t)=-2t+2 and z(t)=-6t+3. the same line
x(t)=4t+1 and y(t)=1t+2 and z(t)=3t+3. again the same line!
x(t)=-4t2+1 and y(t)=-1t2+2 and z(t)=-3t2+3. a closed half-line (a ray)
x(t)=-4(sin(t))+1 and y(t)=-1(sin(t))+2 and z(t)=-3(sin(t))+3. a closed line segment
The motion on the second line is in the direction of the first but twice as fast. The third line's motion is opposite the first. The t2 in the fourth gives an up and down effect from infinity to (1,2,3). The last example just oscillates back and forth on an interval. I recommended that students use the Maple procedure spacecurve to "see" what curves can look like.
So motion can be complicated.
I very briefly discussed what it means for a vector function to be differentiable: this works out to be the same as differentiability "in parallel" for each of the components. The same for integration. Then I began to discuss what R'(t), usually called the velocity vector, really means in terms of particle motion: the magnitude is the speed, and the direction is tangent to the curve. I needed to give some mechanical illustration of what a tangent vector to a curve might be.
We are currently skipping 12.6 and are jumping right into 13.1 and 13.2.
9/12/2002 The Maple field trip. Students worked through several pages of problems designed to give them some familiarity with Maple.
9/11/2002 We returned to considering the 3-dimensional vectors v=3i+j-k and w=4i+2j+3k. Writing a vector as a sum of perpendicular and parallel parts (compared to another vector) was vaguely (!) motivated by a picture of a block sliding on an inclined plane. We were able to write v as a sum: vperp + vparallel, where vperp was perpendicular to w and vparallel was parallel to w. We did this by finding vparallel first: its direction was the direction of w, so a unit vector in w's direction was created by writing (1/|w|)w. The magnitude of vparallel was obtained by looking at a triangle in the plane of v and w: the magnitude was |v|cos theta, where theta was the angle between v and w. Luckily we know cos theta from previous work with the dot product. So the magnitude is v · w /|w|. We computed all this and got vparallel. vperp was obtained by writing vperp = v - vparallel. A simple check was suggested: w· vperp was to be 0, since the vectors are supposed to be perpendicular. Indeed (thank goodness!) this dot product was 0. My personal success rate with hand computation of this kind is not high. A new definition: vectors are orthogonal if their dot product is 0.
I introduced a new product, called the cross product or the vector product. There are dot products in every dimension. Dot products, however, give scalars as the result. The cross product is more-or-less unique to 3 dimensions and involves making a choice of "handedness". Some people feel this is rather important to physical reality. Related to this is the concept of chirality, important in chemistry and physics as well as mathematics. But enough diversions! What's vxw? Here is what the text calls the physics definition:
vxw is a vector. The magnitude of vxw: in the plane determined by v and w, draw the parallelogram determined by v and w. The magnitude is the area of that parallelogram (an easy picture shows that the magnitude will be |v| |w| sin theta, where theta is the angle between v and w). The direction of vxw: curl the fingers of your right hand from v to w. The thumb will "naturally" point perpendicular to the plane determined by v and w. That direction is the direction of vxw.
I "computed" a simple multiplication table:
``` x i j k -------------------- i 0 k -j -------------------- j -k 0 i -------------------- k j -i 0 ```
This table already has some distressing or surprising information. Cross product has these properties:
• Squares are 0: vxv=0 always (the area of a one-dimensional parallelogram is 0).
• x is anticommutative: vxw=-wxv (the thumb points the other way!)
• x is not even necessarily associative: (ixj)xj=-i but ix(jxj)=0.
Therefore computationally one must be careful in both ordering and grouping factors! This can lead to errors.
I stated further properties of x:
• (v1+v2)xw= (v1xw)+(v2xw) for any vectors v1, v2, and w.
• (cv)xw=c(vxw) for any scalar c and any vectors v and w.
The second one is almost believable from the geometric definition (stretch one side of a parallelogram by a factor of c and then the area gets stretched by c). The first is not so clear, and I didn't "prove" it. Similar results ("linearity") are also true in the second factor of x.
I applied those results to compute vxw, where v and w were the vectors we used earlier in the lecture. We distributed addition across x and also let the scalars "float" to the front. The multiplication table written above was used, and we finally got a result.
More generally, a convenient algebraic method of computing vxw was stated using determinants. The determinant of a 2-by-2 array (a matrix)
```| a b |
| c d |
```
is ad-bc, while the determinant of a 3-by-3 array
```| a b c |
| c d e |
| f g h |
```
is a(det I) - b (det II) +c (det III) where
``` I II III
|| || ||
| d e | | c e | | c d |
| g h | | f h | | f g |
```
(these are called the minors of the larger matrix). There are many minus signs involved and ample opportunity for error. A course in linear algebra (Math 250 here) will explain why these formulas are interesting, but right now all I want are the definitions.
If v=ai+bj+ck and w=di+ej+fk, then vxw= the determinant of
```| i j k |
| a b c |
| d e f |
```
and this follows from the linearity in each factor and the entries of the multiplication table for x. We checked that this works for the specific v and w we started with.
I began a geometric application of · and x which I'll finish next time. Students should read 12.3, 12.4, and begin 12.5.
Class tomorrow is a Maple field trip, to ARC 118.
9/9/2002
• I tried to prevent my dog from getting to a lamppost by keeping the leash short enough. This led to the question of over- and under-estimating the quantity |v+w| where v and w are vectors. An overestimate is gotten from the triangle inequality: |v+w|=<|v|+|w|. An underestimate is obtained by a slightly more circuitous route: |v|=|v+0|=|v+(w-w)|=|v+w+(-w)|=<|v+w|+|-w|=|v+w|+|w| so that if we subtract |w| we get |v|-|w|=<|v+w| . This can give useful information if good choices of v and w are made (that is, with |v|>|w|).
• I then recited the Law of Cosines for triangles in the plane. Most people seemed to know this, more or less. It specializes to the Pythagorean Theorem when the angle is a right angle. I used the Law of Cosines with vectors for the sides of the specified angle to deduce that the cosine of the angle between two vectors v and w was equal to a ratio: the bottom of the ratio was the product of the lengths of v and w, and the top of the ratio was a product ac+bd if v=ai+bj and w=ci+dj. This quantity is called the dot product or scalar product or inner product.
• We generalized to Rn. Here vectors are the sum of scalar multiples of n unit vectors pointed along the coordinate axes: v=sumn=1najej. If w is the same sort of sum (with bj's as coordinates) then the dot product of v and w is v·w=sumj=1najbj. This product has some noteworthy properties.
1. If v and w are vectors, then v·w is a scalar (here, a real number).
2. v·w=w·v (commutativity).
3. If c is a real number, then c(v·w)=(cv)·w.
4. (v1+v2)·w=(v1·w)+(v2·w). Note Because of commutativity, the properties affecting the first factor work just as well with the second factor.
5. |v|=sqrt(v·v).
We "proved" a few of these, which really just involve not being afraid of the use of summation signs. Then the definition of the angle between two vectors is motivated by the 2-dimensional computation, so that: the angle between the vectors v and w =arccos( (v·w)/(|v| |w|) ).
• Why should the argument to the function arccos be in the domain of arccos? Examination of this question shows that |v·w| is "supposed" to be less than the product of |v| and |w|, which is certainly not immediately clear to me. That is, if one actually wrote out several 23-dimensional vectors and computed the quantities v·w and |v| and |w| the desired inequality becomes less obvious. It is indeed true, and called the Cauchy-Schwarz inequality, and we went through a mysterious rapid proof of it. If Q(t)=|v+tw|2, then Q(t) is a real-valued non-negative function of the real number t. Since |v+tw|2=(v+tw)·(v+tw), we can use the properties above to "expand this expression, and get Q(t)=(v·v)+(2v·w)t+(w·w)t2=C+Bt+At2. This is just a quadratic function, whose graph is a parabola. Since A is non-negative, the parabola opens "up". Since Q(t) is itself always non-negative valued, the quadratic cannot have two real roots, so the discriminant B2-4AC must be nonpositive. But substitute the values of A and B and C to see that (2v·w)2-4|v|2|w|2 is nonpositive. Add, divide by 4, take square roots: this is the Cauchy-Schwarz inequality.
• I then applied this to find the angle between two specific vectors in R3 (not known to me at this time), an example I'll continue with next time.
9/5/2002 The instructor asked students to work on problem 40 of section 12.1, in groups or individually. The work was collected.
We introduced vectors: directed line segments, "arrows" going from a tail to a head, quantities with magnitude and direction. We gave some simple physical interpretations (force, velocity, etc.). Two vector "representations" will be the same if the arrows are parallel and have the same length.
Vector addition is motivated by simple physical experiments. It is defined by looking at the heads and tails of the summands in the correct order. The vector sum of two vectors is another vector. Then vector addition is commutative and associative, and these "facts" can be verified (at least in R3) geometrically. The zero vector, 0, is special, with head=tail. It is an additive identity: any vector+0=the vector. And vectors have additive inverses, defined by reversing the assignment of head and tail. The sum of a vector and its additive inverse is the zero vector.
"Scalar" multiplication was discussed. The word scalar is a somewhat antique usage, and here will mean a real number. If v is a vector and c is a scalar, then cv is another vector. If c=0, cv is the 0 vector. If c>0, then cv is a vector in the direction of v, whose length is v's multiplied by c. And if c<0, then cv is a vector whose direction is opposite v's direction, with length is v's multiplied by -c. Scalar multiplication satisfies various properties (all in section 12.2 of the text).
We discussed the norm or length or magnitude of a vector, v: |v|. It is the distance from the head to the tail. It is non-negative and satisfies various properties. The most subtle is |v+w|=<|v|+|w|, the vector version of the triangle inequality. And |cv|=|c| |v|.
Then we began to use this to give over- and under-estimates of various vector sums. This will be continued next time.
Some time was devoted to discussing ideas about the first problem of the workshop set, and just a very small amount of time devoted to the second problem.
9/4/2002 The instructor discussed the course and collected student information. Brief explanation of why the study of calculus in Rn rather than just R2 or R3 might be interesting was given.
Presentation of rectangular coordinate systems (how to locate points) in R1 (the real line), R2 (the plane), and R3 (space), followed by generalization to Rn.
Distance introduced: this is a nonnegative real number. On the line, the distance between p and q, if these have coordinates a and b, say, is |a-b| or sqrt((a-b)2). "Simple" properties of this distance were given:
• The distance from p to p is 0.
• The distance from p to q equals the distance from q to p.
• The distance from p to r is less than or equal to the distance from p to q + the distance from q to r.
The first two properties were "clear" and the last needed some discussion. An example was given to show that equality is not necessarily correct.
A distance formula in R2 was suggested using the Pythagorean Theorem: the square root of the sum of the squares of the differences in coordinates of the points. Verification of the first two properties of distance for the plane suggested by those of the line was immediate. A rather lengthy algebraic verification of the third property was given. A sequence of reversible algebraic steps was applied to the inequality suggested until a statement about squares being nonnegative was obtained.
Generalization of the formula to Rn was given. A name was given to the third property of distance: the triangle inequality, and an appropriate picture was drawn: geometrically, the length of one side of a triangle is "clearly" less than or equal to the sum of the lengths of the other two sides. | 20,905 | 78,350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-39 | latest | en | 0.925844 |
https://brainacademy.pro/question/333452 | 1,708,823,871,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474573.20/warc/CC-MAIN-20240225003942-20240225033942-00149.warc.gz | 144,057,892 | 15,070 | Pat needs to determine the height of a tree before cutting it down to be sure that it will not fall on a nearby fence. the angle of elevation of the tree from one position on a flat path from the tree is and from a second position farther along this path it is what is the height of the tree?
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The height of the tree, which Pat needs to determine, before cutting it down to be sure that it will not fall on a nearby fence is 111 ft.
### What is right angle triangle property?
In a right angle triangle, the ratio of the opposite side to the base side is equal to the tangent angle between them.
Here, (b) is the opposite side, (a) is the base side.
Pat needs to determine the height of a tree before cutting it down to be sure that it will not fall on a nearby fence.
The angle of elevation of the tree from one position on a flat path from the tree is
From a second position,
Farther along this path it is,
Let the distance between tree and first position is x. Thus, from the trigonometry,
....1
The distance between tree and second position is (x+60) ft. Thus, again from the trigonometry,
......2
Compare the equation 1 and 2,
Put this value in equation 1 as,
Thus, the height of the tree, which Pat needs to determine, before cutting it down to be sure that it will not fall on a nearby fence is 111 ft. | 325 | 1,362 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-10 | latest | en | 0.953148 |
https://calculatoruniverse.com/percentage-change-word-problems/ | 1,721,119,995,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514742.26/warc/CC-MAIN-20240716080920-20240716110920-00531.warc.gz | 127,815,035 | 33,680 | # Percentage Change Word Problems
Instructions:
• Enter the initial and final values you want to calculate the percentage change for.
• Click "Calculate Percentage Change" to see the result along with the formula used.
• Your calculation history will be displayed below the results.
• Click "Clear Results" to clear the current calculation and history.
• Click "Copy Results" to copy the current calculation result to the clipboard.
Calculation History
## Percentage Change
1. Definition and Basic Concept:
• Percentage change is a way to express the relative change between two numbers as a percentage. It’s a measure of how much a quantity has increased or decreased in comparison to its original value.
• The formula for calculating percentage change is: Percentage Change=((New Value−Original Value)/Original Value)×100%
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1. Functionality:
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## Mathematical Principles of Percentage Change
1. Calculation Process:
• The key step in percentage change problems is identifying the “original value” and the “new value” from the context of the problem.
• The percentage change formula is then applied to these values to find the required solution.
## Applications of Percentage Change Calculations
1. In Educational Settings:
• Vital for teaching students how to interpret and solve real-world problems involving percentage changes.
2. In Finance and Economics:
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3. In Everyday Life:
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## Interesting Facts and Further Exploration
1. Historical Development of Percentages:
• The concept of percentages has been used for centuries in commerce and trade, evolving into a key mathematical tool in modern times.
2. Cross-Disciplinary Importance:
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## Conclusion
The Percentage Change Word Problems Calculator is an invaluable tool for anyone looking to enhance their understanding and proficiency in solving percentage change problems. By providing practical scenarios, this tool not only aids in reinforcing mathematical concepts but also sharpens problem-solving skills applicable in various real-world situations.
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Last Updated : 27 February, 2024
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### 10 thoughts on “Percentage Change Word Problems”
1. I appreciate the emphasis on the applications of percentage change in various contexts. It’s important for students and professionals to understand the real-world implications of this mathematical concept.
2. The formula for calculating percentage change is clearly explained. It’s a fundamental concept in mathematics and this article does a great job of breaking it down.
3. The conclusion wraps up the article nicely, emphasizing the practical and educational significance of percentage change calculations. Well-written and informative. | 815 | 4,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-30 | latest | en | 0.839221 |
https://studylib.net/doc/10419723/algebra-qualifying-examination-january-7--2008-instructions- | 1,606,893,045,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141692985.63/warc/CC-MAIN-20201202052413-20201202082413-00143.warc.gz | 483,472,802 | 12,097 | # Algebra Qualifying Examination January 7, 2008 Instructions:
```Algebra Qualifying Examination
January 7, 2008
Instructions:
• There are eight Algebra questions worth a total of 100 points. Individual point
values are listed next to each problem.
as well as the clarity and main steps of your reasoning. “Rough working” will
not be accepted: answers must be written in a structured and understandable
manner.
• You may use a calculator to check your computations (but you will not earn
points for using it as a step in your reasoning).
Notation: Throughout, Z denotes the integers, Q denotes the rational numbers, R
denotes the real numbers, and C denotes the complex numbers.
1. (15 points) Let n be a positive integer. Denote by Sn and An the symmetric
and alternating groups of degree n, respectively.
(a) For each positive integer n, give a composition series of Sn .
(b) Find the centralizer in S5 of the 3-cycle (345).
(c) Find the centralizer in A5 of the 3-cycle (345).
2. (15 points) List all possible groups of order 175 (up to isomorphism).
3. (10 points) Let R be a ring, S a subring of R, and I an ideal of R. Suppose
that R = S + I and S ∩ I = 0.
(a) Prove that each element of R may be written r = s + i for unique elements
s ∈ S and i ∈ I.
(b) Prove that R/I is isomorphic to S by giving a ring isomorphism from R/I
to S.
4. (10 points) Let R1 and R2 be integral domains with fields of fractions (i.e.
quotient rings) F1 and F2 , respectively. Suppose that there is an injective ring
homomorphism φ : R1 → R2 . Prove that φ extends to F1 , i.e. there is an
injective ring homomorphism ψ : F1 → F2 such that ψ(r) = φ(r) for all r ∈ R.
5. (10 points) Let R be a ring. Let P1 and P2 be R-modules. Prove that P1 ⊕ P2
is a projective R-module if, and only if, both P1 and P2 are projective.
6. (10 points) Let A be a finitely generated abelian group.
(a) If A is finite, prove that A ⊗Z Q = 0.
(b) If A is infinite, prove that A ⊗Z Q ∼
= Qr , an isomorphism of Z-modules,
for some positive integer r.
7. (15 points)
√
√
(a) Q
Express the minimal polynomial f (x) of 2 + 3 over Q as a product
(x − αi ), for complex numbers αi .
√
√
(b) Q
Express the minimal polynomial g(x) of 3 2 + 3 4 over Q as a product
(x − βi ), for complex numbers βi .
(c) What is the Galois group of f (x)g(x) over Q?
√ √
8. (15 points) Let
W
=
Q(
2, −2).
√ √
(a) Extend { 2, −2} to a basis of W as a vector
√ space over Q.
(b) Let f : W → W be defined by f (α) = (1 + 2)α. Verify that f is a linear
transformation of W as a vector space over Q and find its characteristic
polynomial.
(c) Find a basis of eigenvectors for f of C ⊗Q W .
``` | 797 | 2,628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2020-50 | latest | en | 0.90875 |
https://www.cut-the-knot.org/exchange/sliders2.shtml | 1,721,309,061,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514831.13/warc/CC-MAIN-20240718130417-20240718160417-00837.warc.gz | 643,267,035 | 5,251 | Subject: Re: Sliding puzzles
Date: Wed, 01 Jan 1997 14:42:43 -0500
From: Alex Bogomolny
Dear Vivienne:
I'd like to help you very much. However, the slider puzzles are not all the same. Given that your grandson is five years old, I assume you are talking of the Fifteen puzzle which is the only one I am aware of that has been implemented as a physical gadget. If your grandson plays a computer then the possibilities are too numerous for me to discuss in a short letter. Please be more specific.
As to the Fifteen, it's a simple matter to compose the first two rows just by sliding the counters into their positions. Sometimes, you'll have to shift the counters from their rightful position. For example, assume the first row is 1,2,3,10 while the second row is 12,5,8,4. The task is now to swap 4 and 10. Give room for 1 immediately beneath it. Slide 1,2,3,10 counterclockwise by one square. After which the upper right corner must be empty. Slide 4 into there and 8 to the right into the freed position. Now the empty square will be just below 10 so make good use of this opportunity and remove 10 from the first row. Slide 3,2,1 back. The first row will thus be finished. The second row is tackled in exactly same manner.
The worst thing that may happen with the last two rows is to get
9,10,11,12, and
14,15,13,_
Somehow you must manage to rotate 13 to the first position in the last row. Rotate the two rows clockwise 2 squares:
14,_,9,10
15,13,12,11
Slide 13 up and rotate the two rows back two squares counterclockwise. This must solve the puzzle.
I would guess it matters very little to your grandson that there exists a theory that specifies which starting configurations are solvable and which are not. However, make sure that the starting configuration is obtained backwards from the original 1,2,3... position by sliding counters randomly. For, otherwise, it's a 50/50 chance you'll get an unsolvable position.
Hope it's what you needed.
Sincerely,
Alexander Bogomolny
|Reply| |Up| |Down| |Exchange index| |Contents| |Store| | 518 | 2,048 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-30 | latest | en | 0.937959 |
http://www.shodor.org/interactivate/lessons/IntroStatistics/ | 1,369,143,844,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700074077/warc/CC-MAIN-20130516102754-00080-ip-10-60-113-184.ec2.internal.warc.gz | 697,332,384 | 8,537 | # Introduction to Statistics: Mean, Median, and Mode
Shodor > Interactivate > Lessons > Introduction to Statistics: Mean, Median, and Mode
### Abstract
The goal of this lesson is to introduce the concepts of mean, median and mode and to develop understanding and familiarity with these ideas. The Measures Activity lets students explore mean and median in an efficient way; the Mean, Median and Mode Discussion helps them to formalize their knowledge.
### Objectives
Upon completion of this lesson, students will:
• understand three different measures of "center"
• have been exposed to multiple ways of expressing a set of numbers
• have practiced their arithmetic skills
### Student Prerequisites
• Arithmetic:Students should understand:
• sums, differences, and quotients for all activities.
• Technological::
• Each student or group of students working together will need a computer with a Java-capable browser. Students should be comfortable using the computer and browser. Calculators may be helpful for solving problems that arise in discussions.
### Teacher Preparation
• pencil and paper.
• copies of the following worksheet:
### Key Terms
arithmetic mean See mean average It is better to avoid this sometimes vague term. It usually refers to the (arithmetic) mean, but it can also signify the median, the mode, the geometric mean, and weighted mean, among other things histogram A bar graph such that the area over each class interval is proportional to the relative frequency of data within this interval mean The sum of a list of numbers, divided by the total number of numbers in the list. Also called arithmetic mean median "Middle value" of a list. The smallest number such that at least half the numbers in the list are no greater than it. If the list has an odd number of entries, the median is the middle entry in the list after sorting the list into increasing order. If the list has an even number of entries, the median is equal to the sum of the two middle (after sorting) numbers divided by two. The median can be estimated from a histogram by finding the smallest number such that the area under the histogram to the left of that number is 50% mode For lists, the mode is the most common (frequent) value. A list can have more than one mode. For histograms, a mode is a relative maximum ("bump"). A data set has no mode when all the numbers appear in the data with the same frequency. A data set has multiple modes when two or more values appear with the same frequency. multimodal distribution A distribution with more than one mode. The histogram of a multimodal distribution has more than one "bump" range The range of a set of numbers is the largest value in the set minus the smallest value in the set. Note that the range is a single number, not many numbers total A total is determining the overall sum of numbers or a quantity.
### Lesson Outline
1. Focus and Review
Remind students of what they have learned in previous lessons that will be pertinent to this lesson and/or have them begin to think about the words and ideas of this lesson:
• Does anyone know what "average" means?
2. Objectives
Let the students know what they will be doing and learning today. Say something like this:
• Today, class, we are going to learn about mean, median, and mode.
3. Teacher Input
• Lead a discussion, or the instructor can prepare a "live" discussion, to deepen and formalize the students' intuitive understanding of mean, median, and mode. (10-20 min)
4. Guided Practice
• Introduce and develop the concepts of mean and median with the Measures activity. Students will change parameters and discover patterns related to mean and median. Students can choose their own focus of measure, their own quantity, and their own units. (20 min)
5. Independent Practice
6. Closure
• You may wish to bring the class back together for a discussion of the findings. Once the students have been allowed to share what they found, summarize the results of the lesson.
### Alternate Outline
• Combine this lesson with the Bell Curve Lesson for a look at how means are tied to distributions
### Suggested Follow-Up
This lesson introduced the students to some basic ways of describing sets of data. The next lesson, Histograms and Bar Graphs, introduces histograms, bar graphs, and the concept of class interval. Students will learn to distinguish between bar graphs and histograms and to use each in the appropriate situations. | 913 | 4,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2013-20 | longest | en | 0.906177 |
https://www.wanweibaike.net/wiki-倒易点阵 | 1,632,317,481,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057366.40/warc/CC-MAIN-20210922132653-20210922162653-00709.warc.gz | 1,102,276,026 | 17,555 | # 倒易点阵
## 数学描述
### 一维晶格
${\displaystyle {\boldsymbol {b}}=2\pi {\frac {\boldsymbol {a}}{a^{2}}}}$
### 二维晶格
${\displaystyle {\boldsymbol {b_{1}}}=2\pi {\frac {{\boldsymbol {a_{2}}}\times {\boldsymbol {n}}}{{\boldsymbol {a_{1}}}\cdot ({\boldsymbol {a_{2}}}\times {\boldsymbol {n}})}}}$
${\displaystyle {\boldsymbol {b_{2}}}=2\pi {\frac {{\boldsymbol {n}}\times {\boldsymbol {a_{1}}}}{{\boldsymbol {a_{2}}}\cdot ({\boldsymbol {n}}\times {\boldsymbol {a_{1}}})}}}$
### 三维晶格
${\displaystyle {\boldsymbol {b_{1}}}=2\pi {\frac {{\boldsymbol {a_{2}}}\times {\boldsymbol {a_{3}}}}{{\boldsymbol {a_{1}}}\cdot ({\boldsymbol {a_{2}}}\times {\boldsymbol {a_{3}}})}}}$
${\displaystyle {\boldsymbol {b_{2}}}=2\pi {\frac {{\boldsymbol {a_{3}}}\times {\boldsymbol {a_{1}}}}{{\boldsymbol {a_{2}}}\cdot ({\boldsymbol {a_{3}}}\times {\boldsymbol {a_{1}}})}}}$
${\displaystyle {\boldsymbol {b_{3}}}=2\pi {\frac {{\boldsymbol {a_{1}}}\times {\boldsymbol {a_{2}}}}{{\boldsymbol {a_{3}}}\cdot ({\boldsymbol {a_{1}}}\times {\boldsymbol {a_{2}}})}}}$
### 倒晶格与正晶格的关系
${\displaystyle {\boldsymbol {a_{i}}}\cdot {\boldsymbol {b_{j}}}=2\pi \delta _{ij}={\begin{cases}2\pi ,&i\ =\ j\\0,&i\ \neq \ j\end{cases}}}$
${\displaystyle \mathbf {G} =h{\boldsymbol {b_{1}}}+k{\boldsymbol {b_{2}}}+l{\boldsymbol {b_{3}}}}$
${\displaystyle \mathbf {|G_{hkl}|} ={\frac {2\pi }{d_{hkl}}}}$
${\displaystyle \mathbf {R} =c_{1}{\boldsymbol {a_{1}}}+c_{2}{\boldsymbol {a_{2}}}+c_{3}{\boldsymbol {a_{3}}}}$
${\displaystyle \mathrm {e} ^{\mathrm {i} \mathbf {G\cdot R} }=1}$
${\displaystyle \Omega _{G}={\frac {(2\pi )^{3}}{\Omega }}}$
## 倒晶格的物理意义
${\displaystyle \psi _{\boldsymbol {k}}({\boldsymbol {x}})=\mathrm {e} ^{\mathrm {i} {\boldsymbol {k}}\cdot {\boldsymbol {x}}}u_{\boldsymbol {k}}({\boldsymbol {x}})}$
${\displaystyle {\boldsymbol {G}}=n{\boldsymbol {b}},\ n=0,1,2,\cdots }$
${\displaystyle {\boldsymbol {b}}=2\pi {\frac {\boldsymbol {a}}{a^{2}}}}$
${\displaystyle {\boldsymbol {G}}\cdot {\boldsymbol {a}}=2\pi n}$
{\displaystyle {\begin{alignedat}{2}u_{\boldsymbol {k+G}}({\boldsymbol {x}})&=\mathrm {e} ^{-\mathrm {i} {\boldsymbol {G}}\cdot {\boldsymbol {x}}}u_{\boldsymbol {k}}({\boldsymbol {x}})\\u_{\boldsymbol {k+G}}({\boldsymbol {x+a}})&=\mathrm {e} ^{-\mathrm {i} {\boldsymbol {G}}\cdot {\boldsymbol {x}}}\mathrm {e} ^{-\mathrm {i} {\boldsymbol {G}}\cdot {\boldsymbol {a}}}u_{\boldsymbol {k}}({\boldsymbol {x+a}})\\&=\mathrm {e} ^{-\mathrm {i} {\boldsymbol {G}}\cdot {\boldsymbol {x}}}u_{\boldsymbol {k}}({\boldsymbol {x+a}})\\\end{alignedat}}}
${\displaystyle u_{\boldsymbol {k}}({\boldsymbol {x+a}})=u_{\boldsymbol {k}}({\boldsymbol {x}})}$
${\displaystyle u_{\boldsymbol {k+G}}({\boldsymbol {x+a}})=u_{\boldsymbol {k+G}}({\boldsymbol {x}})}$
{\displaystyle {\begin{aligned}\psi _{\boldsymbol {k}}({\boldsymbol {x}})&=\mathrm {e} ^{\mathrm {i} {\boldsymbol {k}}\cdot {\boldsymbol {x}}}u_{\boldsymbol {k}}({\boldsymbol {x}})\\&=\mathrm {e} ^{\mathrm {i} ({\boldsymbol {k+G}})\cdot {\boldsymbol {x}}}\mathrm {e} ^{-\mathrm {i} {\boldsymbol {G}}\cdot {\boldsymbol {x}}}u_{\boldsymbol {k}}({\boldsymbol {x}})\\&=\mathrm {e} ^{\mathrm {i} ({\boldsymbol {k+G}})\cdot {\boldsymbol {x}}}u_{\boldsymbol {k+G}}({\boldsymbol {x}})\\&=\psi _{\boldsymbol {k+G}}({\boldsymbol {x}})\end{aligned}}}
## 倒晶格与晶体衍射
{\displaystyle {\begin{alignedat}{2}2d\sin \theta =n\lambda \\2\times {\frac {2\pi }{\lambda }}\sin \theta ={\frac {2\pi }{d_{n}}}\\\end{alignedat}}}
${\displaystyle {\begin{array}{lcl}|{\boldsymbol {k}}|={\cfrac {2\pi }{\lambda }}\\\mathbf {|G_{hkl}|} ={\cfrac {2\pi }{d_{hkl}}}\\2|{\boldsymbol {k}}|\sin \theta =|\mathbf {G} |\\\end{array}}}$
${\displaystyle {\boldsymbol {\Delta k}}={\boldsymbol {k_{o}}}-{\boldsymbol {k_{i}}}=\mathbf {G} }$
## 常见布拉菲晶格的倒晶格
### 简单立方晶体
${\displaystyle {\boldsymbol {a_{1}}}=a{\hat {x}}}$
${\displaystyle {\boldsymbol {a_{2}}}=a{\hat {y}}}$
${\displaystyle {\boldsymbol {a_{3}}}=a{\hat {z}}}$
${\displaystyle {\boldsymbol {a_{1}}}\cdot {\boldsymbol {a_{2}}}\times {\boldsymbol {a_{3}}}=a^{3}}$
${\displaystyle {\boldsymbol {b_{1}}}={2\pi \over a}{\hat {x}}}$
${\displaystyle {\boldsymbol {b_{2}}}={2\pi \over a}{\hat {y}}}$
${\displaystyle {\boldsymbol {b_{3}}}={2\pi \over a}{\hat {z}}}$
### 面心立方晶体(FCC)
${\displaystyle {\boldsymbol {a_{1}}}={a \over 2}\left({\hat {y}}+{\hat {z}}\right)}$
${\displaystyle {\boldsymbol {a_{2}}}={a \over 2}\left({\hat {z}}+{\hat {x}}\right)}$
${\displaystyle {\boldsymbol {a_{3}}}={a \over 2}\left({\hat {x}}+{\hat {y}}\right)}$
${\displaystyle {\boldsymbol {a_{1}}}\cdot {\boldsymbol {a_{2}}}\times {\boldsymbol {a_{3}}}={a^{3} \over 4}}$
${\displaystyle {\boldsymbol {b_{1}}}={2\pi \over a}\left(-{\hat {x}}+{\hat {y}}+{\hat {z}}\right)}$
${\displaystyle {\boldsymbol {b_{2}}}={2\pi \over a}\left(+{\hat {x}}-{\hat {y}}+{\hat {z}}\right)}$
${\displaystyle {\boldsymbol {b_{3}}}={2\pi \over a}\left(+{\hat {x}}+{\hat {y}}-{\hat {z}}\right)}$
### 体心立方晶体(BCC)
${\displaystyle {\boldsymbol {a_{1}}}={a \over 2}\left(-{\hat {x}}+{\hat {y}}+{\hat {z}}\right)}$
${\displaystyle {\boldsymbol {a_{2}}}={a \over 2}\left(+{\hat {x}}-{\hat {y}}+{\hat {z}}\right)}$
${\displaystyle {\boldsymbol {a_{3}}}={a \over 2}\left(+{\hat {x}}+{\hat {y}}-{\hat {z}}\right)}$
${\displaystyle {\boldsymbol {a_{1}}}\cdot {\boldsymbol {a_{2}}}\times {\boldsymbol {a_{3}}}={a^{3} \over 2}}$
${\displaystyle {\boldsymbol {b_{1}}}={2\pi \over a}\left({\hat {y}}+{\hat {z}}\right)}$
${\displaystyle {\boldsymbol {b_{2}}}={2\pi \over a}\left({\hat {z}}+{\hat {x}}\right)}$
${\displaystyle {\boldsymbol {b_{3}}}={2\pi \over a}\left({\hat {x}}+{\hat {y}}\right)}$ | 2,288 | 5,603 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 58, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2021-39 | latest | en | 0.367575 |
https://technoteshelp.com/python-numpy-division-with-runtimewarning-invalid-value-encountered-in-double_scalars/ | 1,656,924,139,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104364750.74/warc/CC-MAIN-20220704080332-20220704110332-00110.warc.gz | 600,127,074 | 17,103 | # python – numpy division with RuntimeWarning: invalid value encountered in double_scalars
## python – numpy division with RuntimeWarning: invalid value encountered in double_scalars
You cant solve it. Simply `answer1.sum()==0`, and you cant perform a division by zero.
This happens because `answer1` is the exponential of 2 very large, negative numbers, so that the result is rounded to zero.
`nan` is returned in this case because of the division by zero.
Now to solve your problem you could:
• go for a library for high-precision mathematics, like mpmath. But thats less fun.
• as an alternative to a bigger weapon, do some math manipulation, as detailed below.
• go for a tailored `scipy/numpy` function that does exactly what you want! Check out @Warren Weckesser answer.
Here I explain how to do some math manipulation that helps on this problem. We have that for the numerator:
``````exp(-x)+exp(-y) = exp(log(exp(-x)+exp(-y)))
= exp(log(exp(-x)*[1+exp(-y+x)]))
= exp(log(exp(-x) + log(1+exp(-y+x)))
= exp(-x + log(1+exp(-y+x)))
``````
where above `x=3* 1089` and `y=3* 1093`. Now, the argument of this exponential is
`-x + log(1+exp(-y+x)) = -x + 6.1441934777474324e-06`
For the denominator you could proceed similarly but obtain that `log(1+exp(-z+k))` is already rounded to `0`, so that the argument of the exponential function at the denominator is simply rounded to `-z=-3000`. You then have that your result is
``````exp(-x + log(1+exp(-y+x)))/exp(-z) = exp(-x+z+log(1+exp(-y+x))
= exp(-266.99999385580668)
``````
which is already extremely close to the result that you would get if you were to keep only the 2 leading terms (i.e. the first number `1089` in the numerator and the first number `1000` at the denominator):
``````exp(3*(1089-1000))=exp(-267)
``````
For the sake of it, lets see how close we are from the solution of Wolfram alpha (link):
``````Log[(exp[-3*1089]+exp[-3*1093])/([exp[-3*1000]+exp[-3*4443])] -> -266.999993855806522267194565420933791813296828742310997510523
``````
The difference between this number and the exponent above is `+1.7053025658242404e-13`, so the approximation we made at the denominator was fine.
The final result is
``````exp(-266.99999385580668) = 1.1050349147204485e-116
``````
``````1.105034914720621496.. × 10^-116 # Wolfram alpha.
``````
and again, it is safe to use numpy here too.
You can use `np.logaddexp` (which implements the idea in @gg349s answer):
``````In [33]: d = np.array([[1089, 1093]])
In [34]: e = np.array([[1000, 4443]])
In [36]: log_res
Out[36]: -266.99999385580668
In [37]: res = exp(log_res)
In [38]: res
Out[38]: 1.1050349147204485e-116
``````
Or you can use `scipy.special.logsumexp`:
``````In [52]: from scipy.special import logsumexp
In [53]: res = np.exp(logsumexp(-3*d) - logsumexp(-3*e))
In [54]: res
Out[54]: 1.1050349147204485e-116
`````` | 893 | 2,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2022-27 | longest | en | 0.868521 |
https://convertoctopus.com/61-meters-to-millimeters | 1,638,493,645,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362571.17/warc/CC-MAIN-20211203000401-20211203030401-00438.warc.gz | 256,319,709 | 7,722 | ## Conversion formula
The conversion factor from meters to millimeters is 1000, which means that 1 meter is equal to 1000 millimeters:
1 m = 1000 mm
To convert 61 meters into millimeters we have to multiply 61 by the conversion factor in order to get the length amount from meters to millimeters. We can also form a simple proportion to calculate the result:
1 m → 1000 mm
61 m → L(mm)
Solve the above proportion to obtain the length L in millimeters:
L(mm) = 61 m × 1000 mm
L(mm) = 61000 mm
The final result is:
61 m → 61000 mm
We conclude that 61 meters is equivalent to 61000 millimeters:
61 meters = 61000 millimeters
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 millimeter is equal to 1.6393442622951E-5 × 61 meters.
Another way is saying that 61 meters is equal to 1 ÷ 1.6393442622951E-5 millimeters.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that sixty-one meters is approximately sixty-one thousand millimeters:
61 m ≅ 61000 mm
An alternative is also that one millimeter is approximately zero times sixty-one meters.
## Conversion table
### meters to millimeters chart
For quick reference purposes, below is the conversion table you can use to convert from meters to millimeters
meters (m) millimeters (mm)
62 meters 62000 millimeters
63 meters 63000 millimeters
64 meters 64000 millimeters
65 meters 65000 millimeters
66 meters 66000 millimeters
67 meters 67000 millimeters
68 meters 68000 millimeters
69 meters 69000 millimeters
70 meters 70000 millimeters
71 meters 71000 millimeters | 425 | 1,669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2021-49 | latest | en | 0.817347 |
https://artofproblemsolving.com/wiki/index.php/1988_AHSME_Problems/Problem_16 | 1,726,711,020,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651944.55/warc/CC-MAIN-20240918233405-20240919023405-00803.warc.gz | 88,649,864 | 12,903 | # 1988 AHSME Problems/Problem 16
## Problem
$[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair H=origin, B=(1,-(1/sqrt(3))), C=(-1,-(1/sqrt(3))), A=(0,(2/sqrt(3))), E=(2,-(2/sqrt(3))), F=(-2,-(2/sqrt(3))), D=(0,(4/sqrt(3))); draw(A--B--C--A^^D--E--F--D); label("A'", A, N); label("B'", B, SE); label("C'", C, SW); label("A", D, E); label("B", E, E); label("C", F, W); [/asy]$
$ABC$ and $A'B'C'$ are equilateral triangles with parallel sides and the same center, as in the figure. The distance between side $BC$ and side $B'C'$ is $\frac{1}{6}$ the altitude of $\triangle ABC$. The ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$ is
$\textbf{(A)}\ \frac{1}{36}\qquad \textbf{(B)}\ \frac{1}{6}\qquad \textbf{(C)}\ \frac{1}{4}\qquad \textbf{(D)}\ \frac{\sqrt{3}}{4}\qquad \textbf{(E)}\ \frac{9+8\sqrt{3}}{36}$
## Solution
Let $\triangle ABC$ have side length $s$ and $\triangle A'B'C'$ have side length $t$. Thus the altitude of $\triangle ABC$ is $\frac{s\sqrt{3}}{2}$. Now observe that this altitude is made up of three parts: the distance from $BC$ to $B'C'$, plus the altitude of $\triangle A'B'C'$, plus a top part which is equal to the length of the diagonal line from the bottom-left corner of $\triangle ABC$ to the bottom left corner of $\triangle A'B'C'$ (as an isosceles trapezium is formed with parallel sides $AB$ and $A'B'$, and legs $AA'$ and $BB'$). We drop a perpendicular from $B'$ to $BC$, which meets $BC$ at $D$. $\triangle BDB'$ has angles $30^{\circ}$, $60^{\circ}$, and $90^{\circ}$, and the vertical side is that distance from $BC$ to $B'C'$, which is given as $\frac{1}{6} \times \frac{s\sqrt{3}}{2} = \frac{s\sqrt{3}}{12}$, so that by the length relationships in a 30-60-90 triangle,, the length of the diagonal line is $\frac{s\sqrt{3}}{12} \times 2 = \frac{s\sqrt{3}}{6}.$ Thus using the "altitude in three parts" idea, we get $\frac{s\sqrt{3}}{2} = \frac{s\sqrt{3}}{12} + \frac{t\sqrt{3}}{2} + \frac{s\sqrt{3}}{6} \implies \frac{s\sqrt{3}}{4} = \frac{t\sqrt{3}}{2} \implies t = \frac{1}{2}s.$ Thus the sides of $\triangle A'B'C'$ are half as long as $\triangle ABC$, so the area ratio is $(\frac{1}{2}) ^ {2} = \frac{1}{4}$, which is $\boxed{\text{C}}$.
~Johnxyz1(minorEdits) | 813 | 2,234 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 42, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-38 | latest | en | 0.684289 |
https://www.storyofmathematics.com/piecewise-functions | 1,638,417,507,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.69/warc/CC-MAIN-20211202024322-20211202054322-00373.warc.gz | 1,060,550,037 | 26,800 | Piecewise Functions – Definition, Graph & Examples
There are instances where the expression for the functions depends on the given interval of the input values. When this happens, we call these types of functions piecewise-defined functions.
Piecewise functions are defined by different functions throughout the different intervals of the domain.
We actually apply piecewise functions in our lives more than we think so. Tax brackets, estimating our mobile phone plans, and even our salaries (with overtime pay) make use of piecewise functions.
This is why we’ve allotted a special article for this function. In this article, you’ll learn the following:
• Definition of the piecewise function.
• Learning how to evaluate piecewise-defined functions at given intervals.
• Graphing and interpreting piecewise functions.
What is a piecewise function?
To fully understand what piecewise functions are and how we can construct our own piecewise-defined functions, let’s first dive into a deeper understanding of how it works.
Piecewise function definition
A piecewise function is a function that is defined by different formulas or functions for each given interval. It’s also in the name: piece. The function is defined by pieces of functions for each part of the domain.
2x, for x > 0
1, for x = 0
-2x, for x < 0
As can be seen from the example shown above, f(x) is a piecewise function because it is defined uniquely for the three intervals: x > 0, x = 0, and x < 0.
Once we have a given piecewise-defined function, we can interpret it by looking at the given intervals. If we take a look at our example, we can read it as:
• When x > 0, f(x) is equal to 2x.
• When x = 0, f(x) is equal to 1.
• When x < 0, f(x) is equal to -2x.
When given a piecewise function graph, make sure to observe the given intervals where f(x) has varied graphs. But before we try out examples that involve analyzing piecewise function graphs, let’s go ahead and learn how we can evaluate and graph piecewise functions first.
How to solve piecewise functions?
Now that we’ve learned about this unique function, how do we make sure that we return the right value for the function given x? Here are tips to remember when solving for and evaluating piecewise functions:
• Double-check where x lies in the given interval.
• Evaluate the value using the corresponding function.
Let’s say we want to find f(8) using the piecewise function that we’ve shown.
2x, for x > 0
1, for x = 0
-2x, for x < 0
Since 8 is greater than 0, the function we’ll use to evaluate f(8) is f(x) = 2x. Hence, we have f(8) = 2(8) = 16. This also means that f(-6) = -2(-6) = 12 and f(0) = 1.
How to graph piecewise functions?
As we have mentioned before, piecewise functions contain different functions for each of the given intervals. This means that when graphing piecewise functions, expect to graph different functions for each interval as well.
Here are some quick reminders when graphing piecewise functions:
• It helps to identify how each function would look like.
• For inclusive intervals (ie x ≥ 0), including the endpoints.
• For exclusive intervals (ie x < 0), exclude the endpoints by using unfilled dots.
What are the common functions that you may encounter when graphing piecewise functions? Here are some resources, and feel free to check out the links to refresh your knowledge on some of the commonly used graphs:
These are not the only functions that piecewise functions can use, so make sure also to check your textbook’s library of functions whenever you need to. Let’s try graphing the piecewise function given in the first section.
2x, for x > 0
1, for x = 0
-2x, for x < 0
When x > 0 and x < 0, f(x) returns a linear function. Find at least two pairs of points that satisfy each function and use them to construct the two linear graphs.
Since both are exclusive inequalities, we leave the dot at the origin unfilled. Now, we are left with the condition when x = 0. Since the value is constant at f(x) =1, let’s a plot a point at (0,1).
This graph returns the final graph for the given piecewise function. From the graph, we can see that f(x) has a domain of and range of (-∞, ∞) and [0, -∞), respectively.
We’ve covered all the essential properties and techniques we can use with piecewise functions, so it’s time for us to check our knowledge with these examples!
Example 1
Evaluate the given piecewise function at the given values of x as shown below.
√x , for x > 0
5, for x = 0
x/6, for x < 0
a. f(-36)
b. f(0)
c. f(49)
Solution
• When x = -36 (or less than 0), the expression for f(x) is x/6. Let’s evaluate f(-36) using the expression. Hence, we have f(-36) = -36/6 = -6.
• When x = 0, f(x) is a constant. This means that we have f(0) = 5.
• When x = 49 (and consequently, greater 0), the expression for f(x) is x. Let’s evaluate f(49) using the expression. Hence, we have f(49) = 49 = 7.
Example 2
Graph the piecewise function shown below. Using the graph, determine its domain and range.
2x , for x ≠ 0
1, for x = 0
Solution
For all intervals of x other than when it is equal to 0, f(x) = 2x (which is a linear function). To graph the linear function, we can use two points to connect the line. Just make sure that the two points satisfy y = 2x. Make sure to leave the point of origin unfilled.
Since f(x) = 1 when x = 0, we plot a filled point at (0,1). The graph above shows the final graph of the piecewise function.
Since the graph covers all values of x, the domain would be all real numbers or (-∞, ∞). The same reasoning applies to the range of functions. Since it extends in both directions, the range of the function is (-, ) in interval notation.
Example 3
Graph the piecewise function shown below. Using the graph, determine its domain and range.
x2 , for x ≤ 0
5, for 0 < x < 2
x/2 , for x ≥ 2
Solution
Let’s first break down the three intervals and identify how the graph of function would look like:
• When x ≤ 0, f(x) becomes a quadratic function with a parabola that passes through the origin and (-2, 4). Since it only applies for 0 and negative numbers, we will only half of the parabola.
• When 0 < x < 2, f(x) will a represent a constant which is a horizontal line passing through y = 5. Make sure to leave (0,5) and (2,5) unfilled since they are not part of the solution.
• When x ≥ 2, f(x) is a function and will pass through (2, 1) and (6,3).
Using this information, we can now graph f(x).
The image above breaks down the three components of the piecewise function. Let’s go ahead and simplify this graph now so that we can analyze it for its domain and range.
Since all values of x extend in both directions, the domain would be all real numbers or (-∞, ∞). Since the graph only covers the values of y above the x-axis, the range of the function is [0, ) in interval notation.
Example 4
Spoken word poetry is being held at the nearby cafe. They charge $6 per person for a table of 1 to 5 guests. They also offer a fixed fee of$50 for a table with 6 or more people. Write a function that relates the number of people,x, and the cost of attending the event, f(x).
Solution
Let’s go ahead and break down the problem and find the expression of f(x) for each interval:
• For a table of 1 to 5 guests, we can express that as 1 ≤ x ≤ 5 in terms of x. Since it would cost each guest $6, the total for x guests is 6x. • Now, for a table with 6 or more people, we can express the interval as x ≥ 6. For this interval, f(x) will always be equal to 60. We can now summarize this into a piecewise function: 6x, for 1 ≤ x ≤ 5 50, for x ≥ 6 This piecewise function represents the cost of f(x) for x number of guests. Practice Questions 1. Evaluate the given piecewise function at the given values of x as shown below. √(x-1) , for x > 0 5, for x = 0 x+1, for x < 0 a. f(-8) b. f(0) c. f(63) 2. Evaluate the given piecewise function at the given values of x, as shown below. 3x2 , for x ≤ 0 4x – 6, for 0 < x < 2 2x , for x ≥ 2 a. f(-2) b. f(0.75) c. f(7) 3. Graph the piecewise function shown below. Using the graph, determine its domain and range. 2x2 , for x ≠ 0 8, for x = 0 4. Graph the piecewise function shown below. Using the graph, determine its domain and range. √(-4x) , for x ≤ 0 2x, for 0 < x < 9 -x2, for x ≥ 9 5. Suppose that you have a summer job that pays$12 per hour. You are required to work for at least 30 hours per week. The company pays time 1.5 times your hourly rate for overtime.
a. Set up a piecewise function that represents your weekly pay.
b. Graph the piecewise function you have set up.
c. How much would you earn if you worked for 48 hours that week? | 2,304 | 8,643 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.8125 | 5 | CC-MAIN-2021-49 | latest | en | 0.884546 |
https://www.jiskha.com/questions/22470/simplify-by-combining-like-terms-x-sqrt-18-3-sqrt-8x-sqrd-If-means-sqrt-8x-2 | 1,568,675,800,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572964.47/warc/CC-MAIN-20190916220318-20190917002318-00488.warc.gz | 902,166,824 | 5,289 | # algebra
simplify by combining like terms:
x sqrt(18) - 3 sqrt(8x)sqrd
If <sqrt(8x)sqrd> means [sqrt(8x)]^2, then that also equals 8x. So
x sqrt(18) - 3 sqrt(8x)sqrd
= x[sqrt(9*2)-3*8]
= 3x(sqrt 2 -3
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More Similar Questions | 1,049 | 3,087 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2019-39 | latest | en | 0.879281 |
https://www.inchcalculator.com/convert/gram-butter-to-cup-butter/ | 1,718,965,504,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862070.5/warc/CC-MAIN-20240621093735-20240621123735-00414.warc.gz | 733,176,663 | 15,592 | # Convert Grams of Butter to Cups
Enter the butter in grams below to get the value converted to cups.
## Result in Cups:
1 g = 0.004409 c
Do you want to convert cups to grams?
## How to Convert Grams to Cups
To convert a measurement in grams to a measurement in cups, multiply the butter by the following conversion ratio: 0.004409 cups/gram.
Since one gram of butter is equal to 0.004409 cups, you can use this simple formula to convert:
cups = grams × 0.004409
The butter in cups is equal to the butter in grams multiplied by 0.004409.
For example, here's how to convert 500 grams to cups using the formula above.
cups = (500 g × 0.004409) = 2.204617 c
Grams and cups are both units used to measure butter. Keep reading to learn more about each unit of measure.
## What Is a Gram?
Butter has a density of 911 grams per liter, and one stick of butter weighs about 113.4 grams. One gram of butter is equal to 1/1000th of a kilogram.
The gram, or gramme, is an SI unit of weight in the metric system. Grams can be abbreviated as g; for example, 1 gram can be written as 1 g.
Learn more about grams.
## What Is a Cup?
One cup of butter is equal to 2 sticks, or 1/2 pound.
The cup is a US customary unit of volume. Cups can be abbreviated as c, and are also sometimes abbreviated as C. For example, 1 cup can be written as 1 c or 1 C.
Learn more about cups.
## Gram to Cup Conversion Table
Table showing various gram measurements converted to cups.
Grams Cups
1 g 0.004409 c
2 g 0.008818 c
3 g 0.013228 c
4 g 0.017637 c
5 g 0.022046 c
6 g 0.026455 c
7 g 0.030865 c
8 g 0.035274 c
9 g 0.039683 c
10 g 0.044092 c
20 g 0.088185 c
30 g 0.132277 c
40 g 0.176369 c
50 g 0.220462 c
60 g 0.264554 c
70 g 0.308646 c
80 g 0.352739 c
90 g 0.396831 c
100 g 0.440923 c
200 g 0.881847 c
300 g 1.3228 c
400 g 1.7637 c
500 g 2.2046 c
600 g 2.6455 c
700 g 3.0865 c
800 g 3.5274 c
900 g 3.9683 c
1,000 g 4.4092 c | 642 | 1,912 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-26 | latest | en | 0.832739 |
https://quizgecko.com/quiz/geometry-essentials-transversals-parallels-angles-measurement-and-perpendicularity-bjvugd | 1,716,871,707,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059067.62/warc/CC-MAIN-20240528030822-20240528060822-00512.warc.gz | 407,792,271 | 31,415 | # Geometry Essentials: Transversals, Parallels, Angles, Measurement, and Perpendicularity
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## 12 Questions
360 degrees
### What do we call two lines that intersect at a right angle?
perpendicular lines
equal
equal
### How do architects use geometry concepts in their work?
to plan buildings and ensure structural stability
### What is a transversal?
A line that intersects two or more other lines
### What are the points of intersection between a transversal and intersected lines called?
Points of intersection or points of contact
### What are parallel lines?
Two lines that never intersect in a plane
### What is the Correspondence Property of Parallel Lines?
Corresponding angles formed by a transversal are equal
### What is an acute angle?
An angle between 0° and 90°
An angle of 90°
## Geometry Essentials: Transversals, Parallels, Angles, Measurement, and Perpendicularity
Geometry is a foundational subject that explores spatial relationships through rigorous logic and precise measurements. In this article, we'll delve into several fundamental concepts, including transversal lines, parallel lines, types of angles, angle measurement, and perpendicular lines.
### Transversals
A transversal is a line that intersects two or more other lines, called the intersected lines. The points of intersection between the transversal and intersected lines are called points of intersection or points of contact. These points help us determine relationships between angles formed by the transversal.
### Parallel lines
Parallel lines are two lines in a plane that never intersect, no matter how far they are extended. Their corresponding angles formed by a transversal are always equal. This is known as the Correspondence Property of Parallel Lines.
### Types of angles
Angles can be classified into several categories:
1. Acute angle: an angle between (0^\circ) and (90^\circ).
2. Right angle (or right angle): an angle of (90^\circ).
3. Obtuse angle: an angle between (90^\circ) and (180^\circ).
4. Straight angle: an angle of (180^\circ).
5. Reflex angle: an angle greater than (180^\circ) and less than (360^\circ).
6. Right angle (or 90-degree angle): an angle of (90^\circ).
### Angle measurement
Angles are typically measured in degrees, with a full circle being (360^\circ). Angles can also be measured in radians, a unit that is related to the central angle subtended by an arc on a circle.
### Perpendicular lines
Two lines are perpendicular if they intersect at a right angle. Perpendicular lines also have some special properties when a transversal intersects them:
1. Alternate interior angles are equal: When a transversal intersects two perpendicular lines, the alternate interior angles formed are equal.
2. Corresponding angles are equal: The corresponding angles formed by a transversal intersecting two perpendicular lines are equal.
These relationships help us to identify perpendicular lines and understand the connections between them and angles formed by a transversal.
### Applications
Understanding these fundamental concepts of geometry enables us to tackle a wide range of problems in various fields. For instance, architects use these concepts to plan buildings and ensure structural stability. Engineers use them to design bridges and other supporting structures. And artists use them to portray perspective and depth in their works.
By grasping transversals, parallel lines, types of angles, angle measurement, and perpendicular lines, you'll gain a solid foundation in geometry that will serve you well in diverse applications.
Explore fundamental concepts in geometry such as transversals, parallel lines, different types of angles, angle measurement, and perpendicular lines. Learn about the relationships between these concepts and how they are applied in various fields like architecture, engineering, and art.
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Error: | 846 | 4,116 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-22 | latest | en | 0.894461 |
https://mathlake.com/Laws-Of-Exponents | 1,708,647,568,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473871.23/warc/CC-MAIN-20240222225655-20240223015655-00369.warc.gz | 397,172,546 | 6,280 | # Laws Of Exponents
In Mathematics, there are different laws of exponents. All the rules of exponents are used to solve many mathematical problems which involve repeated multiplication processes. The laws of exponents simplify the multiplication and division operations and help to solve the problems easily. In this article, we are going to discuss the six important laws of exponents with many solved examples.
## What are Exponents?
Exponents are used to show repeated multiplication of a number by itself. For example, 7 × 7 × 7 can be represented as 73. Here, the exponent is ‘3’ which stands for the number of times the number 7 is multiplied. 7 is the base here which is the actual number that is getting multiplied. So basically exponents or powers denotes the number of times a number can be multiplied. If the power is 2, that means the base number is multiplied two times with itself. Some of the examples are:
• 34 = 3×3×3×3
• 105 = 10×10×10×10×10
• 163 = 16 × 16 × 16
Suppose, a number ‘a’ is multiplied by itself n-times, then it is represented as an where a is the base and n is the exponent.
Exponents follow certain rules that help in simplifying expressions which are also called its laws. Let us discuss the laws of exponents in detail.
## Rules of Exponents With Examples
As discussed earlier, there are different laws or rules defined for exponents. The important laws of exponents are given below:
• am×an = am+n
• am/an = am-n
• (am)n = amn
• an/bn = (a/b)n
• a0 = 1
• a-m = 1/am
• $$a^{\frac{1}{n}}=\sqrt[n]{a}$$
Also, read: Now let us discuss all the laws one by one with examples here.
### Product With the Same Bases
As per this law, for any non-zero term a,
• am×a= am+n
where m and n are real numbers.
Example 1: What is the simplification of 55 × 51 ?
Solution: 55 × 51 = 55+1 = 56
Example 2: What is the simplification of (−6)-4 × (−6)-7?
Solution: (−6)-4 × (−6)-7 = (-6)-4-7 = (-6)-11
Note: We can state that the law is applicable for negative terms also. Therefore the term m and n can be any integer.
### Quotient with Same Bases
As per this rule,
• am/an = am-n
where a is a non-zero term and m and n are integers.
Example 3: Find the value when 10-5 is divided by 10-3.
Solution: As per the question;
10-5/10-3
= 10-5-(-)3
= 10-5+3
= 10-2
= 1/100
### Power Raised to a Power
According to this law, if ‘a’ is the base, then the power raised to the power of base ‘a’ gives the product of the powers raised to the base ‘a’, such as;
• (am)n = amn
where a is a non-zero term and m and n are integers.
Example 4: Express 83 as a power with base 2.
Solution: We have, 2×2×2 = 8 = 23
Therefore, 83= (23)3 = 29
### Product to a Power
As per this rule, for two or more different bases, if the power is same, then;
• ab= (ab)n
where a is a non-zero term and n is the integer.
Example 5: Simplify and write the exponential form of: 1/8 x 5-3
Solution: We can write, 1/8 = 2-3
Therefore, 2-3 x 5-3 = (2 × 5)-3 = 10-3
### Quotient to a Power
As per this law, the fraction of two different bases with the same power is represented as;
• an/b= (a/b)n
where a and b are non-zero terms and n is an integer.
Example 6: Simplify the expression and find the value:153/53
Solution: We can write the given expression as;
(15/5)3= 33 = 27
### Zero Power
According to this rule, when the power of any integer is zero, then its value is equal to 1, such as;
a= 1
where ‘a’ is any non-zero term.
Example 7: What is the value of 50 + 22 + 40 + 71 – 31 ?
Solution: 50 + 22 + 40 + 71 – 31 = 1+4+1+7-3= 10
### Negative Exponent Rule
According to this rule, if the exponent is negative, we can change the exponent into positive by writing the same value in the denominator and the numerator holds the value 1.
The negative exponent rule is given as:
a-m = 1/am
Example 8:
Find the value of 2-2
Solution:
Here, the exponent is a negative value (i.e., -2)
Thus, 2-2 can be written as 1/22
2-2 = 1/22
2-2 = 1/4
In other words, we can say that, if “a” is a non-zero number or non-zero rational number, we can say that a-m is the reciprocal of am.
### Fractional Exponent Rule
The fractional exponent rule is used, if the exponent is in the fractional form. The fractional exponent rule is given by:
$$a^{\frac{1}{n}}=\sqrt[n]{a}$$
Here, a is called the base, and 1/n is the exponent, which is in the fractional form. Thus, a1/n is said to be the nth root of a.
Example 9:
Simplify: 41/2
Solution:
Here, the exponent is in fractional form. (i.e., ½)
According to the fractional exponent rule, 41/2 can be written as √4
(i.e.,) 41/2 = √4
41/2 = 2 (As, the square root of 4 is 2)
Hence, the simplified form of 41/2 is 2.
### Practice Problems on Laws of Exponents
Simplify the following expressions using the laws of exponents:
1. (42)3
2. 42×47
3. 3-3
4. 641/2
5. 70×23
## Frequently Asked Questions on Laws of Exponents
### What are exponents?
The exponents, also called powers, define how many times we have to multiply the base number. For example, the number 2 has to be multiplied 3 times and is represented by 23.
### What are the different laws of exponents?
The different Laws of exponents are:
• am×an = am+n
• am/an = am-n
• (am)n = amn
• an/bn = (a/b)n
• a0 = 1
• a-m = 1/am
### What is Power of a power rule?
In the power of a power rule, we have to multiply the exponent values. For example, (23)2 can be written as 26.
### Explain the zero power rule.
According to the zero power rule, if the exponent is zero, the result is 1, whatever the base value is. It means that anything raised to the power of 0 is 1. For example, 50 is 1.
### Simplify the expression 22.25
In expression 22.25, the base values are the same, So we can add the exponents.
Hence, 22.25 = 22+5
22.25 = 27.
Stay tuned with BYJU’S – The Learning App and download the app to get all the Maths concepts and learn in an easy way. | 1,787 | 5,908 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2024-10 | longest | en | 0.9086 |
becyvyfivuxabub.motorcarsintinc.com | 1,560,747,756,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998376.42/warc/CC-MAIN-20190617043021-20190617065021-00155.warc.gz | 21,261,839 | 3,963 | # Write a quadratic function in standard form for each given set of zeros
So, halfway between eight and two. So, we could say t These are actually the exact same function. And once again, we can graph that really fast. Modelling This is a good question because it goes to the heart of a lot of "real" math. So, is equal to zero. We just substitute as before into the vertex form of our quadratic function.
But as in the previous case, we have an infinite number of parabolas passing through 1, 0. This is going to happen if we subtract two from both sides, when x is negative two, and if we subtract four from both sides, or when x is equal to negative four.
In each column, I write a Quadratic Equation in one of the three forms: But is this the correct answer? Table partners are already in homogeneous pairs based on their work thus far with the Quadratics Function Unit. Divide Coefficient Next, divide the coefficient of the x term inside the parentheses by two.
I want to find the zeros. Simplify the equation by combining the terms at the end. We can write a parabola in "vertex form" as follows: And the line of symmetry is going to be the vertical line x is equal to negative three. Students continue matching dominoes end to end until all of the matches are complete.
Factor Coefficient Factor the coefficient a from the first two terms of the standard form equation and place it outside of the parentheses.
And so, this can be written as, and if you have trouble doing this, I encourage you to watch videos on factoring polynomials, What adds up to six and when you take their product is eight?Graphs of Quadratic Functions. Parts of a Parabola. zeros: In a given function, the values of [latex]x[/latex] at which [latex]y=0[/latex], Each coefficient in a quadratic function in standard form has an impact on the shape and placement of the function.
Dec 20, · Write a quadratic function in standard form for the set of zeroes.? Write a quadratic function in standard form for each given set of zeros. Best ANSWER 10 points!?Status: Resolved.
SOLUTION: write a quadratic function in standard form for the given set of zeros: 5 and 4. Algebra -> Quadratic Equations and Parabolas -> SOLUTION: write a quadratic function in standard form for the given set of zeros: 5 and 4.
To find the unique quadratic function for our blue parabola, we need to use 3 points on the curve. We can then form 3 equations in 3 unknowns and solve them to get the required result.
We can then form 3 equations in 3.
Algebra 2 Honors: Quadratic Functions Semester 1, Unit 2: Activity 10 Derive Verify Advantage Disadvantage Counterexample Quadratic equation Standard form of a quadratic equation Imaginary number Complex number Complex conjugate Completing the square For each set of solutions, write a quadratic equation in standard form.
In the standard form, y = ax 2 + bx + c, a parabolic equation resembles a classic quadratic equation. With just two of the parabola's points, its vertex and one other, you can find a parabolic equation's vertex and standard forms and write .
Write a quadratic function in standard form for each given set of zeros
Rated 5/5 based on 36 review | 720 | 3,182 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2019-26 | latest | en | 0.918692 |
https://www.ecbasis.com/blog/2023/12/29/why-fans-fight-the-curse-of-knowledge-in-movies-and-beyond/ | 1,709,530,168,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476413.82/warc/CC-MAIN-20240304033910-20240304063910-00591.warc.gz | 730,470,247 | 20,897 | 01
There is a very interesting phenomenon on the Internet. When a movie adapted from a novel or comic is released, there will often be a quarrel between the “original party” who have read the original work and the “movie party” who have watched the movie directly.
Fans of the original work will be more picky about the plot, the acting is not up to the mark, and the characters, graphics, and styling are not restored and they will complain. But the movie party feels that the quality is good and not so bad in all aspects.
Both sides feel that the other side is unreasonable, and the reason for the disagreement is that the information between the two sides is not equal.
The original party knows the content of the original work and likes the content of the original work. When the movie adaptation destroys the classic image in their hearts, they will be dissatisfied.
People who understand the story through the movie for the first time simply cannot see the problems that only the original party can’t see, so they will appear more tolerant.
02
In fact, the reason for the quarrel between the original film party and the film party was studied by someone as early as 30 years ago.
In 1990, Elizabeth Newton conducted a simple game study at Stanford University.
In this game, she divided participants into two roles: “tappers” and “listeners.” Each tapper would get a list of 25 famous songs, such as “Happy Birthday to You.” and the American national anthem “The Star-Spangled Banner.”
Each tapper needs to select a song from the song list and tap the rhythm of the song to the audience by tapping on the table, and then let the audience guess the song name based on the rhythm of the tapping.
This game sounds simple, but the actual results are unexpected: the tapper knocked out the rhythm of 120 songs, but the listener only guessed 2.5% of them, which is 3 songs.
What’s even more interesting is that before the listeners guessed the songs, the tappers were asked to predict the probability that the listeners would guess correctly. They predicted a probability of 50%, which means they should guess correctly once for at least two songs.
Why is this?
When a tapper taps a song, he “hears” the song in his head because he knows the name of the song.
To the listener, there is no “song title” in their head, so all they hear is: “Ta, tap, tap, tap.”
But it is impossible for the percussionist to imagine that the percussions heard by the listener are not a piece of music.
It’s called: the curse of knowledge.
The curse of knowledge refers to the fact that once we know something ourselves, we find it difficult to imagine what it would be like not to know it .
Our knowledge curses us, and it becomes difficult for us to share our knowledge with others because we cannot easily understand the state of mind of our listeners.
03
In the Marvel series of movies, there is also mention of the curse of knowledge.
In “Avengers 3″, Thanos once said to Iron Man: ” You are not the only one cursed by knowledge. The two cherish each other. ”
Because Iron Man understood the terror of Thanos and the impending threat to the earth, he invented Ultron, thus losing Jarvis and causing widespread death.
Knowing the expansion of the universe’s population, wars, and a series of problems caused by disease, Thanos decided to snap his fingers and cause the death of half of the people in the universe.
It is precisely because of their rich knowledge that they see things that ordinary people cannot see, so they are all cursed by knowledge.
In daily life, there are also many curses of knowledge.
For example: in recent years, tutoring children with homework has been something that many parents talk about with disdain. There is even a popular word in parent circles – auxophobia, which describes parents who tutor their children with homework as if they are going through a tribulation. Once tutored, they lose control of their emotions and their blood pressure rises. .
For example: When a couple quarreled, the girl said angrily, “You don’t understand me at all, I just want to do whatever I want.” The boy felt particularly wronged, but you didn’t say anything. I don’t know what you were thinking. Therefore, boys often think that girls are losing their temper unreasonably and unreasonably.
For example: During class, the teacher talks endlessly on the podium. The knowledge that is very complicated to the students is very basic in the teacher’s view, so there are many jumps during the explanation, which will eventually lead to the students not learning anything in class. .
04
The famous media person Shen Shuaibo once said, “In China, no matter how well-known things are, at least 100 million people don’t know about them, and most of the time, even 1 billion people don’t know about them.”
Let’s rephrase this sentence: What you take for granted within your major is something that one billion people outside your major don’t know about .
The curse of knowledge is also the reason why it is difficult for a lot of academic knowledge to move from universities to the public.
This problem is extremely obvious in academic sharing. When a great scientist talks about scientific discoveries seriously on the podium, the audience may gradually become confused due to too many professional terms and even a series of formulas. Lost interest in listening.
A popular science article with detailed content may also use unfamiliar words and lack of foreshadowing, making it difficult for ordinary readers to digest and abandon the article directly.
Even within academia, boring and difficult lectures and convoluted documents give colleagues a headache.
The reason for this phenomenon is that ordinary people have neither enough background knowledge nor enough patience. When they cannot tolerate the poor expression of a serious science popularization article, they will ask the media to produce more popular content.
But the problem is that the popular science content interpreted by the media is often not close to the essence of science, and may even have nothing to do with science.
This is also the reason why Chinese chips have been “stuck” a while ago. Although there are many media popularizations about how difficult chip manufacturing is, many netizens still shout that it will surpass Japan in three years and surpass the United States in five years. This strong confidence and the current Chinese chip manufacturing The state of the industry is completely mismatched.
05
How can we avoid the curse of knowledge?
It’s also very simple. The main thing is to regard yourself as the audience, put yourself on the same page as the audience, and then conceive the communication method from the perspective of the learner. The details are as follows:
①Clear the object of learning: the recipient of learning is him, not you
Before communicating, you need to first understand who you are communicating with, understand their gender, age, education and other information, analyze the knowledge gaps between you and them, and consider the range of complexity they can accept before communicating.
② Think about the problem from another person’s perspective: Don’t just think about the problem from the perspective of the individual, but from the perspective of the learner and others.
Think of yourself as a learner and pretend to be a learner. When you want to convey knowledge to them, try to explain it from the perspective of the learner, such as background information, development process, essence of principles, follow-up trends, etc. You can make good use of analogies, splitting, etc. to reduce the complexity of explanations. Difficulty.
③Find communication methods and conceive communication methods from the perspective of learners.
Explaining from the perspective of the learner requires down-to-earth and vernacular language. While the explanation is accurate, it is best to use vernacular language to refine and share complex information.
The second is to make good use of concrete things as the basis for abstraction. For example, when we were young, we learned 5-3=?, and the teacher used the question “If you have 5 candies, and you eat 3, how many are left” to guide you. Learning, this method is called contextual arithmetic , and it can very well solve the problem of explaining abstract concepts.
④Finally, adjust the communication method through observation.
While communicating, you must also be good at observing the audience’s reactions, and adjust your explanation methods based on their reaction signals. A frown means it is still too difficult, a distraction means the explanation is boring, etc.
06
As usual, here’s the summary.
The curse of knowledge, the curse of knowledge, means that once we know something ourselves, we find it difficult to imagine what it would be like not to know it.
She is everywhere in life and “impossible to guard against.” However, for ourselves, we must always pay attention to whether we are in the “curse of knowledge”.
Once you find yourself in one, first calm down your emotions, and then remind yourself to deal with it with empathy, so that you can break the curse.
error: Content is protected !! | 1,863 | 9,221 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-10 | latest | en | 0.965338 |
https://artofproblemsolving.com/wiki/index.php?title=Sum_and_difference_of_powers&oldid=77954 | 1,627,151,848,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150307.84/warc/CC-MAIN-20210724160723-20210724190723-00345.warc.gz | 124,884,530 | 10,527 | Sum and difference of powers
The sum and difference of powers are powerful factoring techniques that, respectively, factor a sum or a difference of certain powers.
Sums of Powers
$a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}-a^{2n-1}b+a^{2n-2}b^2-\ldots-ab^{2n-1}+b^{2n})$
Differences of Powers
If $p$ is a positive integer and $x$ and $y$ are real numbers,
$x^{p+1}-y^{p+1}=(x-y)(x^p+x^{p-1}y+\cdots +xy^{p-1}+y^p)$
For example:
$x^2-y^2=(x-y)(x+y)$
$x^3-y^3=(x-y)(x^2+xy+y^2)$
$x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3)$
Note that the number of terms in the long factor is equal to the exponent in the expression being factored.
An amazing thing happens when $x$ and $y$ differ by $1$, say, $x = y+1$. Then $x-y = 1$ and
$x^{p+1}-y^{p+1}=(y+1)^{p+1}-y^{p+1}$
$=(y+1)^p+(y+1)^{p-1}y+\cdots +(y+1)y^{p-1} +y^p$.
For example:
$(y+1)^2-y^2=(y+1)+y$
$(y+1)^3-y^3=(y+1)^2+(y+1)y+y^2$
$(y+1)^4-y^4=(y+1)^3+(y+1)^2y+(y+1)y^2+y^3$
If we also know that $y\geq 0$ then:
$2y\leq (y+1)^2-y^2\leq 2(y+1)$
$3y^2\leq (y+1)^3-y^3\leq 3(y+1)^2$
$4y^3\leq (y+1)^4-y^4\leq 4(y+1)^3$
$(p+1)y^p\leq (y+1)^{p+1}-y^{p+1}\leq (p+1)(y+1)^p$
Sum of Cubes
$1^3=1^2$
$1^3+2^3 =3^2$
$1^3 +2^3+3^3=6^2$ | 605 | 1,177 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 26, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2021-31 | latest | en | 0.60787 |
https://classroom.thenational.academy/lessons/revision-part-2-6gt3ar | 1,660,580,452,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572192.79/warc/CC-MAIN-20220815145459-20220815175459-00657.warc.gz | 189,986,950 | 27,109 | # Revision (Part 2)
In this lesson we will revisit key content such as pressure, calculating speed and distance-time graphs.
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.What is true about all contact forces?
1/8
Q2.What is the unit of force?
2/8
Q3.Which of the forces below is a non-contact force?
3/8
Q4.Which of the forces below acts upwards on an airplane?
4/8
Q5.What does the length of the arrows in a free body diagram?
5/8
Q6.Chose the correct method to calculate the resultant force
6/8
Q7.What are the two variables that weight depends on?
7/8
Q8.Use the equation W = m x g to calculate the weight of a 30 kg dog on a planet with a gravitational field strength of 3 N/kg
8/8
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.What is true about all contact forces?
1/8
Q2.What is the unit of force?
2/8
Q3.Which of the forces below is a non-contact force?
3/8
Q4.Which of the forces below acts upwards on an airplane?
4/8
Q5.What does the length of the arrows in a free body diagram?
5/8
Q6.Chose the correct method to calculate the resultant force
6/8
Q7.What are the two variables that weight depends on?
7/8
Q8.Use the equation W = m x g to calculate the weight of a 30 kg dog on a planet with a gravitational field strength of 3 N/kg
8/8
# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
Quiz:
# Revision 2
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
Q1.If I increase the area what happens to the pressure?
1/8
Q2.If I increase the force what happens to the pressure?
2/8
Q3.Which of the following are units for pressure (chose 2)
Select two (2) boxes
3/8
Q4.What does a flat horizontal line on a distance - time graph mean?
4/8
Q5.What does a straight line on a distance - time graph mean?
5/8
Q6.What does a curve on a distance - time graph mean?
6/8
Q7.If an object is changing speed, what can we say about the forces?
7/8
Q8.If an object is changing direction, what can we say about the forces?
8/8
Quiz:
# Revision 2
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
Q1.If I increase the area what happens to the pressure?
1/8
Q2.If I increase the force what happens to the pressure?
2/8
Q3.Which of the following are units for pressure (chose 2)
Select two (2) boxes
3/8
Q4.What does a flat horizontal line on a distance - time graph mean?
4/8
Q5.What does a straight line on a distance - time graph mean?
5/8
Q6.What does a curve on a distance - time graph mean?
6/8
Q7.If an object is changing speed, what can we say about the forces?
7/8
Q8.If an object is changing direction, what can we say about the forces?
8/8
# Lesson summary: Revision (Part 2)
## Time to move!
Did you know that exercise helps your concentration and ability to learn?
For 5 mins...
Move around:
Walk
On the spot:
Chair yoga | 962 | 3,663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2022-33 | latest | en | 0.830261 |
http://www.governmentadda.com/quant-quiz-for-ibps-po-ibps-clerk-railway-ibps-so-nabard-rbi-other-exams-6/ | 1,550,396,914,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481832.13/warc/CC-MAIN-20190217091542-20190217113542-00127.warc.gz | 359,607,115 | 30,832 | Sunday , February 17 2019
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# Quant Quiz For IBPS PO | IBPS Clerk | Railway | IBPS SO | NABARD | RBI & Other Exams
Q1. Vishal travels by car and covers 25% of his journey with a speed of 10 km/h, 45% of his journey with a speed of 5 km/h and remaining 30% of his journey with a speed of 15 km/h. what will be the average speed of Vishal for the whole journey?
A) 6.50 km/h B) 7.40 km/h C) 7.80 km/h D) 6.75 km/h E) None of these
Q2. The population of a town is 352800. If it increases at the rate of 5% per annum, then what was its population 2 years ago?
#### Get Best Maths Short Tricks Book With 2500+ Questions (50% Off)> Click Here
A) 300000 B) 350000 C) 400000 D) 320000 E) None of these
Q3. If 2/3 part of an article is sold at 30% profit, 1/4 part at 16% profit and remaining part at 12% profit and finally, there is a profit of Rs. 75, then find the cost price of the article?
A) Rs 300 B) Rs 350 C) Rs 400 D) Rs 450 E) Rs 500
Q4 A shopkeeper marked the price 20% more than its cost price. If he allows a discount of 25%, then find his loss per cent?
A) 12% B) 15% C) 10% D) 20% E) None of these
Q5. For a certain sum, the simple interest in 2 yr at 8% per annum is Rs. 50 more than the simple interest in 1.5yr at the rate of 10% per annum for the same sum. Find the sum?
A) 4000 B) 5000 C) 6000 D) 7000 E) None of these
Q6. The ratio between the number of passengers travelling by 1st and 2nd class between the two railway stations is 1 : 50, whereas the ratio of 1st and 2nd class fares between the same stations is 3 : 1. If an a particular day, Rs. 1325 were collected from the passengers travelling between these stations, then what was the amount collected from the 2nd class passengers?
A) Rs. 750 B) Rs. 850 C) Rs. 1000 D) Rs. 1250 E) None of these
Q7. A container contains 40 L of milk. From this container, 4 L of milk was taken out and replaced by water. This process was further repeated two times. How much milk is now there in the container?
A) 29.16 L B) 30 L C) 32.4 L D) 35 L E) None of these
Q8. A, B and C invested capitals in the ratio of 2 : 3 : 5. At the end of the business terms, they received the profit in the ratio of 5 : 3 : 12. Find the ratio of time for which they contributed their capitals?
A) 25:15:23 B) 25:10:24 C) 23:25:10 D) 23:10:25 E) None of these
Q9. Present age of Priyanka is 5 times the age of Ritu. After 10 yr, Priyanka will be 3 times as old as Ritu. What are the present ages of Priyanka and Ritu respectively?
A) 10 yrs and 50 yrs B) 5 years and 25 years C) 25 years and 5 years D) 50 years and 10 year
E) None of these
Q10. If tap A can fill a tank 5 times faster than tap B and takes 36 min less than tap B to fill the tank. If both the taps are opened simultaneously, then find the time taken to fill the tank?
A) 7 min B) 7.5 min C) 8 min D) 10 min E) None of these
Q11. A sum of money invested at compound interest amounts to Rs. 800 in 2 yr and Rs. 880 in 3 yr. Find the rate of interest per annum and the sum?
A) Rs 661.15 B) Rs 761.15 C) Rs 861.15 D) Rs 961.15 E) None of these
Q12. A can do a piece of work in 10 days while B can do it in 15 days. They begin together but 5 days before the completion of the work, B leaves off. Find the total number of days for the work to be completed?
A) 6 Days B) 5 Days C) 10 Days D) 8 Days E) 9 Days
Q13. A, B and C undertake to do a work for Rs. 240. A and B together do ¼ of the work and rest is done by the C alone. How much should C get?
A) 150 B) 180 C) 160 D) 200 E) None of these
Q14. A person covers half of his journey at 30 km/h, one-third at 40 km/h and rest of his journey at 20 km/h. Find his average speed for the whole journey?
A) 25 km/h B) 35 km/h C) 20 km/h D) 30 km/h E) None of these
Q15. Rahul can row a certain distance downstream in 12 h and can return the same distance in 18 h. If the stream flows at the rate of 6 km/h, then find the speed of Rahul in still water?
A) 25 km/h B) 35 km/h C) 20 km/h D) 30 km/h E) None of these | 1,488 | 4,995 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2019-09 | latest | en | 0.794448 |
https://www.physicsforums.com/threads/minimal-polynomials.50783/ | 1,542,249,819,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742338.13/warc/CC-MAIN-20181115013218-20181115035218-00387.warc.gz | 977,635,724 | 14,713 | # Homework Help: Minimal Polynomials
1. Oct 31, 2004
### AKG
Compute the minimal polynomials for each of the following operators. Determine which of the following operators is diagonalizable.
a) $T : P_2(\mathbb{C}) \to P_2(\mathbb{C})$, where:
$$(Tf)(x) = -xf''(x) + (i + 1)f'(x) - 2if(x)$$.
b) Let $V = M_{k \times k}(\mathbb{R})$.
$$T : V \to V[/itex] by $T(A) = -2A^t$$. For (a), I think the notation (Tf)(x) is confusing me a little. Do they mean T(f(x))? If f = ax² + bx + c, am I right in saying that: (Tf)(x) = x²(-2ia) + x(2i)(a - b) + (b + i(b - 2c))? For (b), I start by finding the characteristic polynomial of T. Let B be the matrix representation of T with respect to some ordered basis. Then, the characteristic polynomial of T is: g(t) = det(B - tI) g(t) = det(BI + B(½tI)) g(t) = det(B((1 + ½t)I)) g(t) = det((1 + ½t)BI) g(t) = det((1 + ½t)(-2I)) g(t) = det(-(2 + t)I) g(t) = (-1)ⁿ(2 + t)ⁿ, where n = dim(V) I'm not sure whether n = k or n = k². Now, either way, the minimal polynomial of T is the same as the minimal polynomial of B, which will be some power of (2 + t). Let's try the first power, so the minimal polynomial is: p(t) = 2 + t Then: p(B) = 2I + B = 2I + BI = 2I - 2I = 0, so the first power seems right. Now, for some [itex]A \in V, A \neq A^t$, we have:
$$0 = p(B)(A) = (2I + B)A = 2A + BA = 2A - 2A^t = 2(A - A^t) \neq 0,$$
a contradiction. Where did I go wrong?
2. Oct 31, 2004
### Hurkyl
Staff Emeritus
What is $P_2(\mathbb{C})$? I'm going to assume it means complex quadratic polynomials.
Notice the domain and range of T: it takes a polynomial and spits out another polynomial. The parentheses are correct: you compute Tf first. Tf is a polynomial, so you can evaluate it at x: (Tf)(x).
Your problem for (b) is that you got your identy matrices confused. In:
g(t) = det(B - tI)
I is the identity operator on V, a k^4 dimensional space. However, you treated it as if it was the identity matrix in V, a k^2 dimensional space, when you went to the next line.
(BTW, did you remember to check that T was linear before you assumed it was a matrix?)
I think diagonalizing first is a better approach for this one. I thought it was fairly easy to find k^2 linearly independent eigenvectors of T. Then, once you have the eigenvalues, you can write down the minimum polynomial.
Last edited: Oct 31, 2004
3. Oct 31, 2004
### AKG
I think you mean k²-dimensional.
I checked that T was linear, so there should be a matrix representation of it. You make it seem that the matrix should be a (k² x k²)-matrix. If B is this matrix, then:
T(v) = Bv for all v in V. But this means that we're multiplying a (k² x k²)-matrix, B, by a (k x k)-matrix v, which isn't possible.
Another thing: assuming all else was right, and I found g(t) correctly (with n = k² according to you), then the rest should still hold, and I should still get that contradiction.
4. Oct 31, 2004
### AKG
Basically, T is linear, so it should have a matrix representation, and if v is in V, then T(v) = Bv, so B must be an (k x k)-matrix. However, when we usually have operators over n dimensional spaces, their matrices are (n x n)-matrices, and since we have a k²-dimensional space here, we would expect a (k² x k²)-matrix. Now, is there any reason why the fact that we don't have such a matrix problematic? Do we really get any contradictions, or is it just unusual? As far as I can tell, it isn't really a problem, what do you think?
5. Oct 31, 2004
### Hurkyl
Staff Emeritus
First thing you have to notice for (b):
V, the vector space of all kxk matrices over R, is a k^2 dimensional vector space.
T is a linear operator on V, so it does have a matrix representation, which you can get by selecting a basis for V...
However, just like for any other vector space, you also have to write the elements of V in terms of the basis vectors -- so if you're writing T as a matrix, you have to write elements of V as k^2-tuples, and T would indeed be written as a k^2 x k^2 matrix.
6. Oct 31, 2004
### AKG
Oh, perfect. Thanks! | 1,272 | 4,043 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-47 | latest | en | 0.859715 |
https://www.physicsforums.com/threads/linear-algebra-proof.158685/ | 1,653,442,940,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662577757.82/warc/CC-MAIN-20220524233716-20220525023716-00169.warc.gz | 1,054,817,902 | 16,136 | # Linear Algebra Proof
Gold Member
## Homework Statement
Prove the following theorm:
Given any nxn matrix A, a mxm matrix B, the m+n x m+n matrix C:
[A|X]
[0|B]
where the top left corner of C is A, the top right (X) is any mxn matrix, the bottom right is a nxm 0 matrix and the bottom right is B.
Prove:
|C| = |A|*|B| (|C| is the determinant of C...)
I found a proof but the book did something different and i want to check if my way is correct.
## Homework Equations
1) any square matrix that can't be reduced to the I matrix has a determinant of 0. (because it can be reduced to a matrix with a 0 row or coloum)
2) multiplying a row or coloum by a scalar changes the determinant by a factor of 1 over the scalar.
2) changing rows changes the sign of the determinant
3) adding the multiple of a row doesn't change the D.
## The Attempt at a Solution
Here's my proof:
Lets say that both A and B can be reduced to the I matrix. if A can be reduced than it can be done with by operating only with the coloums using the 3 basic operations. the process of the reduction will change the D of A (|A|) by some factor that we'll call 'a'. so the determinant of A is just a (|A| = a) - this is because the determinant of I is 1.
The same argument can be made with B - this time operating on the rows giving the determinant of b.
Since we reduced A by operating on the coloums and B with the row, we can do all those operations on C without the operation on one of them effecting the rest of the matrix. Applying these operations to C changes the determinant by a*b and since the resaulting matrix is a trianguler matrix with a diagonal of only ones - the determinant of C is a*b. And so:
|C| = a*b = |A| * |B|.
If either A or B can't be reduced then one of them has a 0 determinant so |A|*|B| = 0
and by an argument similar to the above |C| can also be reduced to a matrix with either a 0 row or 0 coloum so |C| = 0.
Q.E.D
Is this a correct proof?
Thanks.
Homework Helper
If either A or B can't be reduced then one of them has a 0 determinant so |A|*|B| = 0
Really? And what if A = I ?
Try to use induction, I found the proof more elegant, but then again, degustibus.
AKG
Homework Helper
## Homework Statement
Prove the following theorm:
Given any nxn matrix A, a mxm matrix B, the m+n x m+n matrix C:
[A|X]
[0|B]
where the top left corner of C is A, the top right (X) is any mxn matrix, the bottom right is a nxm 0 matrix and the bottom right is B.
Prove:
|C| = |A|*|B| (|C| is the determinant of C...)
I found a proof but the book did something different and i want to check if my way is correct.
## Homework Equations
1) any square matrix that can't be reduced to the I matrix has a determinant of 0. (because it can be reduced to a matrix with a 0 row or coloum)
2) multiplying a row or coloum by a scalar changes the determinant by a factor of 1 over the scalar.
2) changing rows changes the sign of the determinant
3) adding the multiple of a row doesn't change the D.
## The Attempt at a Solution
Here's my proof:
Lets say that both A and B can be reduced to the I matrix. if A can be reduced than it can be done with by operating only with the coloums using the 3 basic operations. the process of the reduction will change the D of A (|A|) by some factor that we'll call 'a'. so the determinant of A is just a (|A| = a) - this is because the determinant of I is 1.
The same argument can be made with B - this time operating on the rows giving the determinant of b.
Since we reduced A by operating on the coloums and B with the row, we can do all those operations on C without the operation on one of them effecting the rest of the matrix. Applying these operations to C changes the determinant by a*b and since the resaulting matrix is a trianguler matrix with a diagonal of only ones - the determinant of C is a*b. And so:
|C| = a*b = |A| * |B|.
If either A or B can't be reduced then one of them has a 0 determinant so |A|*|B| = 0
and by an argument similar to the above |C| can also be reduced to a matrix with either a 0 row or 0 coloum so |C| = 0.
Q.E.D
Is this a correct proof?
Thanks.
Yeah, the proof looks correct. You could clean it up a little though. There is an invertible matrix U such that AU is either identity (if A is reducible) or has a zero column (if A is not). There is also an invertible V such that VB is either identity or has a zero row. The matrix
[U|0]
[0|I]
which we'll call U', clearly has determinant |U|, and
[I|0]
[0|V]
which we'll call V', has determinant |V|. If A has a zero column, then V'CU' does too. If B has a zero row, then V'CU' does too. So if either of A or B has determinant 0, V'CU' does too, so |V'CU'| = 0. Since V' and U' are invertible, this means |C| = 0, as desired. If on the other hand both A and B have non-zero determinant, then it's clear that V'CU' is just
[I|X]
[0|I]
which has determinant 1. So
|V'CU'| = 1
|V'||C||U'| = 1
|C| = (|V'||U'|)-1
|C| = (|V||U|)-1
Since AU = I = VB, U and V are just the inverses of A and B, thus have reciprocal determinants to A and B, i.e. |U| = |A|-1, |V| = |B|-1. This gives the desired results.
Your argument was a really good one, I was just thinking it's better to clean it up instead of saying things like
"the process of the reduction will change the D of A (|A|) by some factor that we'll call 'a'."
which are unclear, and in fact wrong in this case (the determinant of A never changes, but row reducing A creates a new matrix with a new determinant). | 1,553 | 5,483 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2022-21 | latest | en | 0.927323 |
https://www.chelseadatamanagement.com/calculating-averages-in-excel/ | 1,708,683,441,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474377.60/warc/CC-MAIN-20240223085439-20240223115439-00782.warc.gz | 719,742,763 | 14,368 | Select Page
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### The Average() Function
The Excel AVERAGE function returns the average (arithmetic mean) of a range of values. It’s much like the SUM function where you enter the range of values you want summarized. For example A10:A50 or B3:B156. The following example averages the values in the range B4:B9.
### The AverageIF() Function
According to Microsoft, if a range or cell reference argument contains text, logical values, or empty cells, those values are ignored; however, cells with the value zero are included.
So if your range has zeros in it, instead of using =Average(), use =AverageIf(). In the following example, column B shows an average that includes blank cells. In Column D, the same values are included but rather than blanks, some cells contain a zero. Notice the difference in the way the average is calculated. Column B adds up 8 numbers and divides the result by 8. Column D adds up 8 numbers but divides the result by 10.
In column F, the AVERAGEIF function is told to only average the cells in the range that are not equal to zero.
The syntax requires the range of cells and the criteria.
### Weighted Average
In the previous examples all of the numbers were given equal significance. The numbers were added together, and then, divided by the number of numbers. With a Weighted Average, one or more numbers is given a greater significance, or weight.
In the following example a student is enrolled in a course where the final grade is determined based on the following categories:
• Tests – 40%
• Final exam – 25%
• Quizzes – 25%
• Homework – 10%
The student has earned the following scores for each category:
• Tests – 73
• Final exam – 85
• Quizzes – 80
• Homework – 100.
We need to calculate the student’s overall grade. Normally, you would multiply each weight by its score and then add those up.
Or you can use the SUMPRODUCT and the SUM functions to find a Weighted Average, which depends on the weight applied to the values. | 451 | 1,987 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-10 | latest | en | 0.885928 |
http://www.jiskha.com/display.cgi?id=1248989599 | 1,498,156,897,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128319688.9/warc/CC-MAIN-20170622181155-20170622201155-00360.warc.gz | 557,403,305 | 4,070 | # compounded interest
posted by on .
A bank offers a rate of 5.3% compounded semi-annually on its four year GICs(Guaranteed Investment Certificates). What monthly and annually compounded rates should it quote in order to have the same effective interest rate at all three nominal rates?
• compounded interest - ,
5.3% annual rate compounded semi-annually is the same as
(1+.053/2)²
=1.05370
=5.370% compounded annually
compounded monthly (6 times for semi-annual),
(1+.053/2) = (1+r)^6
r=((1.0265)(1/6)-1)*6
=5.242409%
• compounded interest - ,
\$10,000 deposited into savings account day Katie was born 18th birthday 6.5% compounded monthly how much in the account on 18th birthday | 191 | 689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-26 | latest | en | 0.961137 |
https://www.cbsd.org/Page/54034 | 1,591,025,292,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347419056.73/warc/CC-MAIN-20200601145025-20200601175025-00556.warc.gz | 655,198,517 | 53,167 | • # Tuesday, March 31, 2020- Day 9
### 6th Grade Prealgebra Distance Learning Activity, Day 9: Math
Overview: This lesson provides a review of proportion
Estimated Time: Approximately 30 minutes
Explanation: You will solve problems using proportions.
Things to know:
1. Proportion: An equation stating two ratios or rates are equal.
2. Property of Proportion
1. Example 1
Example 2
1. Cross products provide one way to solve for an unknown value in the proportion
1. Review the Things to Know above and make sure you understand the examples
2. Watch the following video, take any notes you find helpful in understanding the concept of solving proportions. The video highlights a few ways to solve a proportion. You only need to understand one way that works well for you.
1. Watch the following video, take any notes you find helpful in understanding how to set up a proportion from a word problem. This video only shows how to set up a proportion that could be used to solve a variety of problems.
1. Complete the odd number problems from the proportion practice I provided to you for this lesson. If you do not have a printer, complete your work on notebook paper. I will provide feedback on this assignment and ask you to correct any problems you did not get correct.
2. EXTRA – Only complete if you want more practice.
Complete the even number problems from the practice sheet.
How is this assignment turned in to the teacher: Seesaw
Overview: Read and identify key elements of the setting.
Estimated Time: Approximately 45 minutes
Explanation of Assignment: Using your Tuck Everlasting packet you will complete the activities for day 7 on the reading guide.
Quick things to know
1. Read Tuck Everlasting chapters 11-13
1. Keep track of elements of realism and fantasy (packet page 7)
2. Jot down a description and supporting text evidence (packet page 8)
3. Keep track of each character’s feelings about living forever (packet page 9)
3.Illustrate a Setting page 24 in your packet
1. There are three important settings, Treegap, The Wood and the Pond. Choose one of them to illustrate. You can refer back to you book (pages 5, 24, or 60)
-Use your Tuck Everlasting book and packet that you brought home in your grab bag.
-If you do not have access to your Tuck Everlasting packet, I have included it as an assignment in Teams under reading. You can just use separate sheets of paper to complete the assignments.
Method of Communication/How this assignment is turned into the teacher:
- Please reach out to me with any questions via email: [email protected] You can access your email account through Outlook in 365.
- When you are finished take a picture of one of the activities above and post to the Reading Day 9 Activity in Seesaw
Writing: Day 9
Overview: Begin good copy of picture book pages.
Estimated Time: Approximately 30 minutes
Explanation of Assignment: Using the packet titled, “Directions for Picture Book Assignment” you will follow Day 7 on the schedule.
Things to Know: Your entire “Illustration and Scene Narration Organizer” should be complete. Today, you are beginning the good copy of your picture book. Please be sure that all words are written first in pencil and then traced in pen or marker. Please draw all pictures in pencil first and then color them in using crayons or colored pencils. All white space should be filled/colored in on all good copy pages. Feel free to let your creativity shine!
2. Follow Day 7 (I’m sorry that I made a mistake on the numbered days and will try to be as specific as possible in the directions) on the schedule page of your “Directions for Picture Book Assignment”
3. Day 7 instructs you to complete the good copy of Scene/Pages 1. You can make the good copy on a piece of plain white computer paper that is in your “grab bag”
6. Remember, all words must be written in pencil first and then traced in pen or marker. All pictures must be drawn in pencil first and then colored in with crayon or colored pencil.
7. If you are finished early and are looking for additional practice, go onto your MobyMax account and work on Language
2. Use your “Directions for Picture Book Assignment” packet that you brought home in your grab bag
3. Use a blank sheet of white computer paper from your grab bag.
Method of Communication/How this assignment is turned into the teacher:
2. The final picture book assignment will be handed-in in-person when we return to school. We will share our final picture books in small groups.
Social Studies
Overview: Begin one pager for Chapter 15
Estimated Time: Approximately 20 minutes
Explanation of Assignment: Using the one pager direction worksheet that was sent home with students in their grab back OR view on the Social Studies Team>Files>Class Materials>One Pager for Social Studies, you will begin to sketch and brainstorm ideas for your one pager.
Things to Know: You will have 3 days (totaling 60 minutes) to complete the one pager. A one-pager project will let you share your thoughts and important information about a topic and include artistic elements. Your final project should resemble a magazine cover, with large clear images, block lettering and a balanced layout. The more creative you allow yourself to be, the more you will inspire others to learn about this topic.
1. Read ‘One Pager for Social Studies’ directions that can be found in you grab bag or on Teams.
2. Using a blank sheet of paper that was provided in the grab bag begin to include the following:
· title of chapter and headings
· important vocabulary defined
· a description of the setting (when and where)
· 5 adjective you would use to describe the time period
· at least 3 drawings, symbols, or images, that represent the time period – use color, be creative, be neat, fill the page!
3. You do not need to post anything on Seesaw today. Final one pager should be posted on Thursday to Social Studies April 2 activity. | 1,309 | 5,959 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2020-24 | longest | en | 0.879223 |
http://masteringelectronicsdesign.com/category/operational-amplifier-formulas/page/2/ | 1,516,530,493,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890514.66/warc/CC-MAIN-20180121100252-20180121120252-00717.warc.gz | 212,399,609 | 19,502 | ## Useful Operational Amplifier Formulas and Configurations
### Non-inverting Amplifier
Note: The proof of this transfer function can be found here: How to Derive the Non-Inverting Amplifier Transfer Function.
### Voltage Follower
Note: This configuration can be considered a subset of the Non-inverting Amplifier. When Rf2 is zero and Rf1 is infinity, the Non-inverting Amplifier becomes a voltage follower. When a resistor has an infinity value, in practice it means it is disconnected.
Read moreUseful Operational Amplifier Formulas and Configurations
## How to Derive the Transfer Function of the Inverting Summing Amplifier
The inverting summing amplifier does exactly what its name says: adds the input signals and inverts the result. This amplifier presents a major advantage versus the non-inverting summing amplifier. The input signals are added with their own gain. The disadvantage is the inversion of the sum, which might not be desirable in some cases.
Figure 1
Figure 1 shows the non-inverting summing amplifier with two inputs. Its transfer function is shown in equation (1).
(1)
As you can see, this is a simple function. Each signal is added with its own gain created by the feedback resistor, Rf, and the corresponding resistor for that signal. But, why is that? Why is this transfer function a lot simpler than the non-inverting summing amplifier? How can we derive this function? What is the transfer function of the inverting summing amplifier with 3, 4, or n inputs? This article answers all these questions.
Read moreHow to Derive the Transfer Function of the Inverting Summing Amplifier
## The Transfer Function of the Non-Inverting Summing Amplifier with “N” Input Signals
In a previous article, How to Derive the Summing Amplifier Transfer Function, I deduced the formula for the non-inverting summing amplifier with two signals in its input. But what if we have 3, 4 or an n number of signals? Can we add them all with one amplifier?
Theoretically, yes. Practically, it is a different story. There is a practical limit on how many signals can be summed up with one amplifier. When the number of input signals grows, each signal component in the sum decreases in value. By the end of this article you will understand why.
Figure 1
We already saw that, for a summing amplifier with two input signals (Figure 1), the transfer function is
(1)
If we need to add 3 signals, the circuit schematic looks like the one in Figure 2. What is the transfer function of this summing amplifier with 3 inputs?
Read moreThe Transfer Function of the Non-Inverting Summing Amplifier with “N” Input Signals
## How to Derive the Summing Amplifier Transfer Function
The summing amplifier, or the non-inverting summing amplifier, is an analog processing circuit with the transfer function (the summing amplifier formula as some say) shown in the following equation.
(1)
The first term of the product is the actual summing, while the second term is a gain due to the R3 and R4 resistors. I prefer this type of summing amplifier as shown in Figure 1, because it is more flexible and allows us to achieve any linear function we want.
Figure 1
Some authors prefer the following schematic,
Figure 2
with the transfer function
(2)
One can see that the summing amplifier in Figure 2 is a subset of my preferred schematic in Figure 1. In Figure 2, R4 is zero, while R3 is infinity (open connection). It performs the analog summation between V1 and V2, with a gain of 1. Therefore, the amplifier in Figure 1 gives us more choices when designing a function with this circuit. If the gain is not needed, this should come up from calculations, as in this article Solving the Summing Amplifier.
If you followed this website, by now you probably figured that I am not a promoter of learning formulas by heart. I like to derive the transfer function if I need it. So, how do we prove this formula?
We will use the Superposition Theorem, which says that, the effect of all the sources in a circuit is equal with the sum of the effects of each source taken separately in the same circuit. Therefore, if we take out one source, V2, and replace it with a wire, we then can find the voltage in each node and the current in each branch of this circuit due to the remaining source V1. Then we do the same with V1 and then sum up the currents on each branch and the voltage levels on each node. We are only interested in Vout, so this should be simple.
We will first make V2 = 0V, by connecting R2 to ground, as in Figure 3.
Figure 3
The Op Amp is considered an ideal component, so that the input bias currents are negligible. If the current in the non-inverting input is zero, R1 and R2 make a voltage divider for V1. The non-inverting input voltage V1n, can be written as
(3)
and, based on the non-inverting amplifier transfer function, Vout1 is
(4)
By replacing V1n in (4), the output voltage is
(5)
In the second part of my demonstration, based on the Superposition Theorem, R2 is connected back to V2 and V1 = 0, by connecting R1 to ground. Following the same train of thought Vout2 can be written as
(6)
Now we have to add Vout1 to Vout2 to complete the third step of the Superposition Theorem. After factorizing the gain component 1+R4/R3, the summing amplifier transfer function becomes the mathematical relation shown in (7).
(7)
Q.E.D.
>>> <<<
This formula shows that this sum is a weighted sum between V1 and V2. This is better than a direct sum V1 plus V2, because, again, brings flexibility in design. Together with the differential amplifier, this circuit brings another treat in the art of electronics design.
## How to Derive the Differential Amplifier Transfer Function
The transfer function of the differential amplifier, also known as difference amplifier, can be found in articles, websites, formula tables, but where is it coming from? Why is the differential amplifier transfer function as in the following mathematical relation?
(1)
where the resistors are those shown in Figure 1.
Figure 1
First, an important remark: This formula applies only for an ideal operational amplifier. This means that the amplifier has a large gain, so large that it can be considered infinity, and the input offset sufficiently small, so that it can be considered zero. Also, the input bias currents are sufficiently small so that they can be considered zero. I was once asked “but what is sufficiently small?” A voltage or current in electronics is considered sufficiently small, when its numerical value is 1/100 or less versus the dominant voltages or currents in the circuit. For example, if the input voltage levels, in the circuit in Figure 1, are around a few volts, and the operational amplifier input offset is millivolts or sub-millivolts, then we can neglect the input offset and consider it zero.
Having said that, do we need to know this formula by heart? Of course not. All we need to know is how to derive it. This is not difficult at all.
The transfer function can be derived with the help of the Superposition Theorem. This theorem says that the effect of all sources in a linear circuit is the algebraic sum of all of the effects of each source taken separately, in the same circuit. In other words (back at Figure 1), if we remove V1, and replace it with a short circuit to ground and calculate the output voltage, and then we do the same with V2, the output voltage of the differential amplifier is the sum of both output voltages as they were calculated with each source separately.
Let’s first remove V1. R1 cannot be left unconnected, because in the initial circuit there was current flowing through it. Therefore, let’s ground R1 (see Figure 2).
Figure 2
We can see that our amplifier becomes an inverter, which has its non-inverting input connected to ground through R1 and R2. Vout2 is given in equation (2).
(2)
Read MasteringElectronicsDesign.com: How to Derive the Inverting Amplifier Transfer Function for a proof of this function.
Now let’s remove V2 and ground R3 (see Figure 3).
Figure 3
This is a non-inverting amplifier. For an ideal operational amplifier, Vout1 is a function of V, which is the voltage referred to ground at the non-inverting input of the operational amplifier.
(3)
The resistors R1 and R2 are an attenuator for V1, so that V can be determined as in the following relation.
(4)
By replacing V in equation (3), Vout1 becomes:
(5)
Now that we have Vout1 and Vout2, and using the Superposition Theorem, Vout is the algebraic sum of Vout1 and Vout2,
(6)
which is the differential amplifier transfer function. (Q.E.D.)
## Solving the Differential Amplifier – Part 1
### Design a Differential Amplifier Based on the Input and Output Voltage Level Requirements
The differential amplifier, also known as the difference amplifier, is a universal linear processing circuit in the analog domain. Why? Because you can achieve any linear transfer function with it. It can be reduced to a simple inverter, a voltage follower or a gain circuit. It can also be transformed in a summing amplifier.
Figure 1
So, what is the common usage of the differential amplifier in Figure 1? When the resistor ratios are equal
the amplifier transfer function is
and the circuit amplifies the difference between the input signals.
However, there are times when the electronics designer is faced with the following design requirements: Given an input range of, say, -0.5V to 5.5V, the output has to swing between, say -1.25V and +2.365V. It is clear that this requires an amplifier with a certain gain and an offset different than zero. How can we design the differential amplifier to achieve such a function?
Read moreSolving the Differential Amplifier – Part 1
' | 2,255 | 9,811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2018-05 | latest | en | 0.870767 |
https://gobeyondskool.com/questions/1-meter-equals-how-many-centimeters/ | 1,653,332,030,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662560022.71/warc/CC-MAIN-20220523163515-20220523193515-00504.warc.gz | 334,726,505 | 110,107 | # 1 meter equals how many centimeters?
One meter is equal to 100 centimeters.
## Answer: To convert 1 meter (m) into centimeters (cm), we multiply the quantity in meters by 100.
Let’s convert 1m into cm.
Explanation:
The metre is the SI system’s foundation unit for calculating length. It’s abbreviated as m.
We increase the number in metres by 100 to convert metres to centimetres.
1 m = 100 cm is the mathematical representation of the same.
10 m = 10 100 cm = 1000 cm, for example.
### This means Quantity in centimeters = 100 × Quantity in meters
#### Make Math Stronger with Logimath - Personlized Live Math Classes
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## Why is it really important for a child to learn about political causes at an early age?
A child’s basic terminologies of the world that’s outside of... | 210 | 874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2022-21 | longest | en | 0.870116 |
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TITLE : Unit 11: Essential Fractional Understandings SUGGESTED DURATION : 5 days
#### Unit Overview
Introduction
This unit bundles student expectations that address constructing, describing, and naming fractional parts of a whole using models. According to the Texas Education Agency, mathematical process standards including application, a problem-solving model, tools and techniques, communication, representations, relationships, and justifications should be integrated (when applicable) with content knowledge and skills so that students are prepared to use mathematics in everyday life, society, and the workplace.
Prior to this Unit
In Unit 05, students decomposed two-dimensional figures and partitioned objects into two, four, or eight equal parts. Students discovered and explained the relationship between the number of fractional parts used to make a whole and the size of the parts. Using concrete models, students counted fractional parts beyond one whole.
During this Unit
Students revisit and solidify essential understandings of fractions. Students partition objects or sets of objects into equal parts and name the parts, including halves, fourths, and eighths, using words rather than symbols (e.g., one half or 1 out of 2 equal parts rather than ). Students use direct comparison to understand that the equality of fractional parts of an object is determined based on the size in area of the parts, meaning fractional parts of an object are the same size in area and may or may not be the same shape. Students may also experience fractional parts of a set of objects when the set is defined as the whole. Through repeated practice modeling and naming fractions, students recognize the inverse relationship between the number of parts and the size of each part and explain this relationship using appropriate mathematical language. Students determine how many parts it takes to equal one whole and use this understanding to count fractional parts. Students further their understanding of hierarchical inclusion (each prior number in the counting sequence is included in the set as the set increases) as they apply their understanding of counting whole numbers to counting fractional parts beyond one whole using words.
Other considerations: Reference the Mathematics COVID-19 Gap Implementation Tool Grade 2
After this Unit
In Grade 3, students will extend their understanding of fractions to include representing fractions using symbolic notation. Students will compose and decompose fractions and represent equivalent fractions with denominators 2, 3, 4, 6, and 8 using a variety objects and pictorial models, including strip diagrams and number lines. Students will solve problems involving partitioning an object or a set of objects among two or more recipients. Grade 3 students will also decompose two congruent two-dimensional figures into parts with equal areas and recognize that equal shares of identical wholes need not have the same shape.
In Grade 2, constructing, describing, naming, and counting fractions are all included in Grade 2 Texas Response to Curriculum Focal Points (TxRCFP): Applying knowledge of two-dimensional shapes and three-dimensional solids, including exploration of early fraction concepts. This unit is supporting the development of the Texas College and Career Readiness Standards (TxCCRS): I. Numeric Reasoning A1, B1, B2; II. Algebraic Reasoning D1, D2; V. Statistical Reasoning A1, C2; VII. Problem Solving and Reasoning A1, A2, A3, A4, A5, B1, C1, D1, D2; VIII. Communication and Representation A1, A2, A3, B1, B2, C1, C2, C3; IX. Connections A1, A2, B1, B2, B3.
Research
Understanding fraction concepts and relationships has posed a considerable challenge to students. In order to establish a mathematically sound foundation, Van de Walle’s (2006) research suggests, the first goal in the development of fractions should be to help children construct the “idea of fractional parts of the whole – the parts that result when the whole or unit has been partitioned into equal-sized portions or fair shares. Children seem to understand the idea of separating a quantity into two or more parts to be shared fairly among friends. They eventually make connections between the idea of fair shares and fractional parts” (p. 131-132). “In the discussion of sharing, all of the tasks involved sharing something that could be cut into smaller parts. The fractions are based on parts of an area or region. This is a good place to begin and is almost essential when doing sharing tasks” (p.134). In Grade 2, continuing exploration of fractional parts relies on the use of concrete models to count parts and create wholes. The concept of “one,” the original whole unit, is critical to understanding the fractional part. Without this solid understanding, students will not be able to correctly write fractional notations or understand the meaning of the parts. The National Council of Teachers of Mathematics (2011) states, “The concept of unit is fundamental to the interpretation of rational numbers. Why is the unit so important? To describe the size of some quantity with a rational number, the first step is to determine what serves as the unit, the whole. Consequently, the rational number that results must be interpreted with respect to that unit” (p. 19). Students need an organized andconsistent approach to working with fractional parts as numbers.Students need to first be successful with naming fractional parts prior to using the numerical form to name fractions. Bergeson (2000) supports this gradual building to fraction notation. He states, “Students need to work first with the verbal form of fractions (e.g., two-fourths) before they work with the numerical form (e.g., 2/4), as students’ informal language skills can enhance their understanding of fractions” (p. 8). These foundational fractional relationships: partitioning into equal parts, recognizing the unit, and naming fractions are the building blocks for fractional relationships and successful mathematical understanding.
Bergeson, T. (2000). Teaching and Learning Mathematics Using Research to Shift From the ‘Yesterday’ Mind to the ‘Tomorrow’ Mind. Retrieved from: https://www.k12.wa.us/research/pubdocs/pdf/MathBook.pdf
National Council of Teachers of Mathematics. (2011). Developing essential understanding of rational numbers grades 3 – 5. Reston, VA: National Council of Teachers of Mathematics, Inc.
Texas Education Agency & Texas Higher Education Coordinating Board. (2009). Texas college and career readiness standards. Retrieved from http://www.thecb.state.tx.us/index.cfm?objectid=E21AB9B0-2633-11E8-BC500050560100A9
Texas Education Agency. (2013). Texas response to curriculum focal points for kindergarten through grade 8 mathematics. Retrieved from https://www.texasgateway.org/resource/txrcfp-texas-response-curriculum-focal-points-k-8-mathematics-revised-2013
Van de Walle, J., & Lovin, L. (2006). Teaching student-centered mathematics grades 3 – 5. Boston, MA: Pearson Education, Inc.
Numeracy requires the ability to work flexibly with quantities in order to recognize, reason, and solve situations of varying contexts in everyday life, society, and the work place. How is numeracy like literacy? What are some examples of numeracy in everyday life, society, and the work place? How does context influence understanding of a quantity? Why is the ability to work flexibly with quantities essential to developing the foundations of numeracy?
Unit Understandings
and Questions
Overarching Concepts
and Unit Concepts
Performance Assessment(s)
• A thorough understanding of counting involves integrating different skills or characteristics of numbers and is foundational and essential for continued work with numbers (whole numbers and fractions).
• Why is it important to determine the unit or whole before counting fractional parts?
• What strategies can be used to determine the number of fractional parts in the whole?
• How is counting fractions …
• similar
• different
… to counting whole numbers?
• What patterns exist when counting fractions?
• What relationships exist between …
• fractions
• whole numbers and fractions
… in the proper counting sequence?
• The ability to recognize and represent numbers in various forms develops the understanding of equivalence and allows for working flexibly with numbers in order to communicate and reason about the value of the number (halves, fourths, eighths).
• Fractions are an extension of the number system used to represent quantities that exist when a whole is partitioned into equal parts.
• When might fractions be used in real life?
• Why is it important to be able to identify or work with fractional parts of a whole?
• Why is it important to determine the unit or whole when working with fractions?
• How does changing the size of the whole affect the size or amount of a fractional partition?
• What strategies can be used to …
• partition a whole into equal parts?
• determine if partitions of a whole are equal or unequal?
• Why can equal partitions of identical wholes look different?
• How are names of fractional parts determined?
• How can a fraction be named in more than one way?
• How is the fractional name of the partition related to the number of equal partitions in the whole?
• How could representing a fraction using …
• words
• concrete models
• pictorial models
… improve understanding and communicating about the value of a fraction?
• What relationship exists between the number of partitions in a whole and the size of the partition?
• Number and Operations
• Number
• Counting (natural) numbers
• Whole numbers
• Fractions
• Number Recognition and Counting
• Sequence
• Cardinality
• Conservation of set
• Hierarchical inclusion
• Magnitude
• Number Representations
• Word form
• Associated Mathematical Processes
• Application
• Problem Solving Model
• Tools and Techniques
• Communication
• Representations
• Relationships
• Justification
Assessment information provided within the TEKS Resource System are examples that may, or may not, be used by your child’s teacher. In accordance with section 26.006 (2) of the Texas Education Code, "A parent is entitled to review each test administered to the parent’s child after the test is administered." For more information regarding assessments administered to your child, please visit with your child’s teacher.
#### MISCONCEPTIONS / UNDERDEVELOPED CONCEPTS
Misconceptions:
• Some students may think one eighth must be greater than one fourth of the same whole because eight is greater than four rather than understanding that the greater the number of parts of a whole, the smaller the parts.
• Some students may think all fractional parts named “one half” must be equal rather than understanding that the size of the half depends on the size of the whole.
• Some students may think partitioning a shape into any 2 parts means that these parts are halves rather than understanding that parts of a shape must be two equal parts in order to be halves.
• Some students may think the first part of a fraction names the number of parts being considered, and the second part of the fraction names the number of parts remaining rather than the second part of the fraction naming the total number of parts (e.g., 2 parts out of 8 total incorrectly named two sixths because there are 6 parts remaining, etc.).
• Some students may think the term “one quarter” refers to 1 out of 25 parts due to the connection between a quarter and 25¢ rather than 1 out of 4 parts.
• Some students may think fractions can only be represented using commercial manipulatives rather than applying the concept of fractions to other models (e.g., a students may be familiar with a green triangle representing one sixth of a hexagon using pattern blocks but may struggle identifying one sixth of a rectangle).
• Some students may think fractional parts can only be counted up to one whole rather than recognizing the counting sequence of fractional parts beyond one whole.
#### Unit Vocabulary
• Equal parts – fractional parts that are the same size in area and may or may not be the same shape
• Fraction – a number that can be used to name part of an object or part of a set of objects
• Partition – separation or division of an object into parts
Related Vocabulary:
Counting sequence Eighths Fourths Halves Quarters Unequal parts Whole
Unit Assessment Items System Resources Other Resources
Show this message:
Unit Assessment Items that have been published by your district may be accessed through Search All Components in the District Resources tab. Assessment items may also be found using the Assessment Center if your district has granted access to that tool.
System Resources may be accessed through Search All Components in the District Resources Tab.
Texas Higher Education Coordinating Board – Texas College and Career Readiness Standards
Texas Education Agency – Texas Response to Curriculum Focal Points for K-8 Mathematics Revised 2013
Texas Education Agency – Mathematics Curriculum
Texas Education Agency – STAAR Mathematics Resources
Texas Education Agency Texas Gateway – Revised Mathematics TEKS: Vertical Alignment Charts
Texas Education Agency Texas Gateway – Mathematics TEKS: Supporting Information
Texas Education Agency Texas Gateway – Interactive Mathematics Glossary
Texas Education Agency Texas Gateway – Resources Aligned to Grade 2 Mathematics TEKS
TAUGHT DIRECTLY TEKS
TEKS intended to be explicitly taught in this unit.
TEKS/SE Legend:
• Knowledge and Skills Statements (TEKS) identified by TEA are in italicized, bolded, black text.
• Student Expectations (TEKS) identified by TEA are in bolded, black text.
• Portions of the Student Expectations (TEKS) that are not included in this unit but are taught in previous or future units are indicated by a strike-through.
Specificity Legend:
• Supporting information / clarifications (specificity) written by TEKS Resource System are in blue text.
• Unit-specific clarifications are in italicized, blue text.
• Information from Texas Education Agency (TEA), Texas College and Career Readiness Standards (TxCCRS), Texas Response to Curriculum Focal Points (TxRCFP) is labeled.
• A Partial Specificity label indicates that a portion of the specificity not aligned to this unit has been removed.
TEKS# SE# TEKS SPECIFICITY
2.1 Mathematical process standards. The student uses mathematical processes to acquire and demonstrate mathematical understanding. The student is expected to:
2.1A Apply mathematics to problems arising in everyday life, society, and the workplace.
Apply
MATHEMATICS TO PROBLEMS ARISING IN EVERYDAY LIFE, SOCIETY, AND THE WORKPLACE
Including, but not limited to:
• Mathematical problem situations within and between disciplines
• Everyday life
• Society
• Workplace
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxRCFP:
• Developing proficiency in the use of place value within the base-10 numeration system
• Using place value and properties of operations to solve problems involving addition and subtraction of whole numbers within 1,000
• Measuring length
• Applying knowledge of two-dimensional shapes and three-dimensional solids, including exploration of early fraction concepts
• TxCCRS:
• VII.D. Problem Solving and Reasoning – Real-world problem solving
• VII.D.1. Interpret results of the mathematical problem in terms of the original real-world situation.
• IX.A. Connections – Connections among the strands of mathematics
• IX.A.1. Connect and use multiple key concepts of mathematics in situations and problems.
• IX.A.2. Connect mathematics to the study of other disciplines.
• IX.B. Connections – Connections of mathematics to nature, real-world situations, and everyday life
• IX.B.1. Use multiple representations to demonstrate links between mathematical and real-world situations.
• IX.B.2. Understand and use appropriate mathematical models in the natural, physical, and social sciences.
• IX.B.3. Know and understand the use of mathematics in a variety of careers and professions.
2.1B Use a problem-solving model that incorporates analyzing given information, formulating a plan or strategy, determining a solution, justifying the solution, and evaluating the problem-solving process and the reasonableness of the solution.
Use
A PROBLEM-SOLVING MODEL THAT INCORPORATES ANALYZING GIVEN INFORMATION, FORMULATING A PLAN OR STRATEGY, DETERMINING A SOLUTION, JUSTIFYING THE SOLUTION, AND EVALUATING THE PROBLEM-SOLVING PROCESS AND THE REASONABLENESS OF THE SOLUTION
Including, but not limited to:
• Problem-solving model
• Analyze given information
• Formulate a plan or strategy
• Determine a solution
• Justify the solution
• Evaluate the problem-solving process and the reasonableness of the solution
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxRCFP:
• Developing proficiency in the use of place value within the base-10 numeration system
• Using place value and properties of operations to solve problems involving addition and subtraction of whole numbers within 1,000
• Measuring length
• Applying knowledge of two-dimensional shapes and three-dimensional solids, including exploration of early fraction concepts
• TxCCRS:
• I.B. Numeric Reasoning – Number sense and number concepts
• I.B.1. Use estimation to check for errors and reasonableness of solutions.
• V.A. Statistical Reasoning – Design a study
• V.A.1. Formulate a statistical question, plan an investigation, and collect data.
• VII.A. Problem Solving and Reasoning – Mathematical problem solving
• VII.A.1. Analyze given information.
• VII.A.2. Formulate a plan or strategy.
• VII.A.3. Determine a solution.
• VII.A.4. Justify the solution.
• VII.A.5. Evaluate the problem-solving process.
• VII.D. Problem Solving and Reasoning – Real-world problem solving
• VII.D.2. Evaluate the problem-solving process.
2.1C
Select tools, including real objects, manipulatives, paper and pencil, and technology as appropriate, and techniques, including mental math, estimation, and number sense as appropriate, to solve problems.
Select
TOOLS, INCLUDING REAL OBJECTS, MANIPULATIVES, PAPER AND PENCIL, AND TECHNOLOGY AS APPROPRIATE, AND TECHNIQUES, INCLUDING NUMBER SENSE AS APPROPRIATE, TO SOLVE PROBLEMS
Including, but not limited to:
• Appropriate selection of tool(s) and techniques to apply in order to solve problems
• Tools
• Real objects
• Manipulatives
• Paper and pencil
• Technology
• Techniques
• Number sense
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxRCFP:
• Developing proficiency in the use of place value within the base-10 numeration system
• Using place value and properties of operations to solve problems involving addition and subtraction of whole numbers within 1,000
• Measuring length
• Applying knowledge of two-dimensional shapes and three-dimensional solids, including exploration of early fraction concepts
• TxCCRS:
• I.B. Numeric Reasoning – Number sense and number concepts
• I.B.1. Use estimation to check for errors and reasonableness of solutions.
• V.C. Statistical Reasoning – Analyze, interpret, and draw conclusions from data
• V.C.2. Analyze relationships between paired data using spreadsheets, graphing calculators, or statistical software.
2.1D Communicate mathematical ideas, reasoning, and their implications using multiple representations, including symbols, diagrams, graphs, and language as appropriate.
Communicate
MATHEMATICAL IDEAS, REASONING, AND THEIR IMPLICATIONS USING MULTIPLE REPRESENTATIONS, INCLUDING SYMBOLS, DIAGRAMS, GRAPHS, AND LANGUAGE AS APPROPRIATE
Including, but not limited to:
• Mathematical ideas, reasoning, and their implications
• Multiple representations, as appropriate
• Symbols
• Diagrams
• Graphs
• Language
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxRCFP:
• Developing proficiency in the use of place value within the base-10 numeration system
• Using place value and properties of operations to solve problems involving addition and subtraction of whole numbers within 1,000
• Measuring length
• Applying knowledge of two-dimensional shapes and three-dimensional solids, including exploration of early fraction concepts
• TxCCRS:
• II.D. Algebraic Reasoning – Representing relationships
• II.D.1. Interpret multiple representations of equations, inequalities, and relationships.
• II.D.2. Convert among multiple representations of equations, inequalities, and relationships.
• VIII.A. Communication and Representation – Language, terms, and symbols of mathematics
• VIII.A.1. Use mathematical symbols, terminology, and notation to represent given and unknown information in a problem.
• VIII.A.2. Use mathematical language to represent and communicate the mathematical concepts in a problem.
• VIII.A.3. Use mathematical language for reasoning, problem solving, making connections, and generalizing.
• VIII.B. Communication and Representation – Interpretation of mathematical work
• VIII.B.1. Model and interpret mathematical ideas and concepts using multiple representations.
• VIII.B.2. Summarize and interpret mathematical information provided orally, visually, or in written form within the given context.
• VIII.C. Communication and Representation – Presentation and representation of mathematical work
• VIII.C.1. Communicate mathematical ideas, reasoning, and their implications using symbols, diagrams, models, graphs, and words.
• VIII.C.2. Create and use representations to organize, record, and communicate mathematical ideas.
• VIII.C.3. Explain, display, or justify mathematical ideas and arguments using precise mathematical language in written or oral communications.
• IX.B. Connections – Connections of mathematics to nature, real-world situations, and everyday life
• IX.B.1. Use multiple representations to demonstrate links between mathematical and real-world situations.
2.1E Create and use representations to organize, record, and communicate mathematical ideas.
Create, Use
REPRESENTATIONS TO ORGANIZE, RECORD, AND COMMUNICATE MATHEMATICAL IDEAS
Including, but not limited to:
• Representations of mathematical ideas
• Organize
• Record
• Communicate
• Evaluation of the effectiveness of representations to ensure clarity of mathematical ideas being communicated
• Appropriate mathematical vocabulary and phrasing when communicating mathematical ideas
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxRCFP:
• Developing proficiency in the use of place value within the base-10 numeration system
• Using place value and properties of operations to solve problems involving addition and subtraction of whole numbers within 1,000
• Measuring length
• Applying knowledge of two-dimensional shapes and three-dimensional solids, including exploration of early fraction concepts
• TxCCRS:
• VIII.B. Communication and Representation – Interpretation of mathematical work
• VIII.B.1. Model and interpret mathematical ideas and concepts using multiple representations.
• VIII.B.2. Summarize and interpret mathematical information provided orally, visually, or in written form within the given context.
• VIII.C. Communication and Representation – Presentation and representation of mathematical work
• VIII.C.1. Communicate mathematical ideas, reasoning, and their implications using symbols, diagrams, models, graphs, and words.
• VIII.C.2. Create and use representations to organize, record, and communicate mathematical ideas.
2.1F Analyze mathematical relationships to connect and communicate mathematical ideas.
Analyze
MATHEMATICAL RELATIONSHIPS TO CONNECT AND COMMUNICATE MATHEMATICAL IDEAS
Including, but not limited to:
• Mathematical relationships
• Connect and communicate mathematical ideas
• Conjectures and generalizations from sets of examples and non-examples, patterns, etc.
• Current knowledge to new learning
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxRCFP:
• Developing proficiency in the use of place value within the base-10 numeration system
• Using place value and properties of operations to solve problems involving addition and subtraction of whole numbers within 1,000
• Measuring length
• Applying knowledge of two-dimensional shapes and three-dimensional solids, including exploration of early fraction concepts
• TxCCRS:
• VII.A. Problem Solving and Reasoning – Mathematical problem solving
• VII.A.1. Analyze given information.
• VIII.A. Communication and Representation – Language, terms, and symbols of mathematics
• VIII.A.1. Use mathematical symbols, terminology, and notation to represent given and unknown information in a problem.
• VIII.A.2. Use mathematical language to represent and communicate the mathematical concepts in a problem.
• VIII.A.3. Use mathematical language for reasoning, problem solving, making connections, and generalizing.
• VIII.B. Communication and Representation – Interpretation of mathematical work
• VIII.B.1. Model and interpret mathematical ideas and concepts using multiple representations.
• VIII.C. Communication and Representation – Presentation and representation of mathematical work
• VIII.C.1. Communicate mathematical ideas, reasoning, and their implications using symbols, diagrams, models, graphs, and words.
• VIII.C.2. Create and use representations to organize, record, and communicate mathematical ideas.
• VIII.C.3. Explain, display, or justify mathematical ideas and arguments using precise mathematical language in written or oral communications.
• IX.A. Connections – Connections among the strands of mathematics
• IX.A.1. Connect and use multiple key concepts of mathematics in situations and problems.
• IX.A.2. Connect mathematics to the study of other disciplines.
2.1G Display, explain, and justify mathematical ideas and arguments using precise mathematical language in written or oral communication.
Display, Explain, Justify
MATHEMATICAL IDEAS AND ARGUMENTS USING PRECISE MATHEMATICAL LANGUAGE IN WRITTEN OR ORAL COMMUNICATION
Including, but not limited to:
• Mathematical ideas and arguments
• Validation of conclusions
• Displays to make work visible to others
• Diagrams, visual aids, written work, etc.
• Explanations and justifications
• Precise mathematical language in written or oral communication
Note(s):
• The mathematical process standards may be applied to all content standards as appropriate.
• TxRCFP:
• Developing proficiency in the use of place value within the base-10 numeration system
• Using place value and properties of operations to solve problems involving addition and subtraction of whole numbers within 1,000
• Measuring length
• Applying knowledge of two-dimensional shapes and three-dimensional solids, including exploration of early fraction concepts
• TxCCRS:
• VII.A. Problem Solving and Reasoning – Mathematical problem solving
• VII.A.4. Justify the solution.
• VII.B. Problem Solving and Reasoning – Proportional reasoning
• VII.B.1. Use proportional reasoning to solve problems that require fractions, ratios, percentages, decimals, and proportions in a variety of contexts using multiple representations.
• VII.C. Problem Solving and Reasoning – Logical reasoning
• VII.C.1. Develop and evaluate convincing arguments.
• VIII.A. Communication and Representation – Language, terms, and symbols of mathematics
• VIII.A.3. Use mathematical language for reasoning, problem solving, making connections, and generalizing.
• VIII.B. Communication and Representation – Interpretation of mathematical work
• VIII.B.1. Model and interpret mathematical ideas and concepts using multiple representations.
• VIII.B.2. Summarize and interpret mathematical information provided orally, visually, or in written form within the given context.
• VIII.C. Communication and Representation – Presentation and representation of mathematical work
• VIII.C.3. Explain, display, or justify mathematical ideas and arguments using precise mathematical language in written or oral communications.
2.3 Number and operations. The student applies mathematical process standards to recognize and represent fractional units and communicates how they are used to name parts of a whole. The student is expected to:
2.3A Partition objects into equal parts and name the parts, including halves, fourths, and eighths, using words.
Partition
OBJECTS INTO EQUAL PARTS, INCLUDING HALVES, FOURTHS, AND EIGHTHS
Name
THE EQUAL PARTS OF PARTITIONED OBJECTS, INCLUDING HALVES, FOURTHS, AND EIGHTHS, USING WORDS
Including, but not limited to:
• Fraction – a number that can be used to name part of an object or part of a set of objects
• Equal parts – fractional parts that are the same size in area and may or may not be the same shape
• Partition – separation or division of an object into parts
• Determination of the whole
• One object or shape defined as the whole
• Multiple connected shapes or objects defined as the whole
• A set of separate objects defined as the whole
• Whole divided into two, four, or eight equal parts
• Appropriate oral and written mathematical language to name equal parts, including halves, fourths, and eighths
• Fractions written in word form do not include hyphens.
• Two equal parts
• One half, two halves or one whole
• Four equal parts
• One fourth, two fourths, three fourths, four fourths or one whole
• One quarter, two quarters, three quarters, four quarters or one whole
• Eight equal parts
• One eighth, two eighths, three eighths, four eighths, five eighths, six eighths, seven eighths, eight eighths or one whole
• Hyphen is used only when the fraction written in word form modifies another word
• Each fractional part of an object is the same size in area and the same shape.
• Orientation of the partitioned parts does not affect the size in area or shape of the parts.
• A whole may be partitioned different ways.
• Equal parts of a whole that are the same size in area may not be the same shape.
• Direct comparison of parts to justify equal size in area
• Equal parts of non-identical wholes may have the same fractional name but may not have equal size in area or the same shape.
• Concrete models of whole objects
• Linear models
• Cuisenaire rods, fraction bars, linking cube trains, folded paper strips, etc.
• Area models
• Fraction circles or squares, pattern blocks, geoboards, etc.
• Concrete models of a set of objects
• Pattern blocks, color tiles, counters, real-world objects, etc.
• Relationship and distinction between ordinal numbers and the number of parts named in a fraction
Note(s):
• Grade 1 partitioned two-dimensional figures into two and four fair shares or equal parts and described the parts using words.
• Grade 2 is not expected to identify the relationship between equivalent fractions (e.g., two-fourths is the same as one-half, etc.).
• Grade 3 will introduce fraction symbols to represent pictorial models of fractions and fractional parts of a set of objects.
• Various mathematical process standards will be applied to this student expectation as appropriate.
• TxRCFP:
• Applying knowledge of two-dimensional shapes and three-dimensional solids, including exploration of early fraction concepts
• TxCCRS:
• I.B. Numeric Reasoning – Number sense and number concepts
• I.B.2. Interpret the relationships between the different representations of numbers.
• VIII.B. Communication and Representation – Interpretation of mathematical work
• VIII.B.1. Model and interpret mathematical ideas and concepts using multiple representations.
2.3B Explain that the more fractional parts used to make a whole, the smaller the part; and the fewer the fractional parts, the larger the part.
Explain
THE MORE FRACTIONAL PARTS USED TO MAKE A WHOLE, THE SMALLER THE PART; AND THE FEWER THE FRACTIONAL PARTS, THE LARGER THE PART
Including, but not limited to:
• Fraction – a number that can be used to name part of an object or part of a set of objects
• Inverse relationship between the size of the fractional part and the number of equal parts in the whole when given the same size whole
• The greater the number of parts, the smaller the size of the parts
• The smaller the number of parts, the greater the size of the parts
• Whole divided into two, four, or eight equal parts
• Concrete models of whole objects
• Linear models
• Cuisenaire rods, fraction bars, linking cube trains, folded paper strips, etc.
• Area models
• Fraction circles or squares, pattern blocks, geoboards, etc.
• Real-world problem situations
Note(s):
• Grade 1 partitioned two-dimensional figures into two and four fair shares or equal parts and described the parts using words.
• Grade 3 will introduce fraction symbols to represent pictorial models of fractions and fractional parts of a set of objects.
• Grade 3 will explain that the unit fraction represents the quantity formed by one part of a whole that has been partitioned into b equal parts where b is a non-zero whole number.
• Various mathematical process standards will be applied to this student expectation as appropriate.
• TxRCFP:
• Applying knowledge of two-dimensional shapes and three-dimensional solids, including exploration of early fraction concepts
• TxCCRS:
• I.A. Numeric Reasoning – Number representations and operations
• I.A.1. Compare relative magnitudes of rational and irrational numbers, and understand that numbers can be represented in different ways.
• VIII.A. Communication and Representation – Language, terms, and symbols of mathematics
• VIII.A.2. Use mathematical language to represent and communicate the mathematical concepts in a problem.
• VIII.C. Communication and Representation – Presentation and representation of mathematical work
• VIII.C.1. Communicate mathematical ideas, reasoning, and their implications using symbols, diagrams, models, graphs, and words.
2.3C Use concrete models to count fractional parts beyond one whole using words and recognize how many parts it takes to equal one whole.
Use
CONCRETE MODELS TO COUNT FRACTIONAL PARTS BEYOND ONE WHOLE USING WORDS
Recognize
HOW MANY PARTS IT TAKES TO EQUAL ONE WHOLE
Including, but not limited to:
• Fraction – a number that can be used to name part of an object or part of a set of objects
• Relationship between counting whole numbers and counting fractional parts of a whole
• Hierarchical inclusion – concept of nested numbers, meaning each prior number in the counting sequence is included in the set as the set increases (e.g., 18 is 17 increased by 1; 18 decreased by 1 is 17; etc.)
• Appropriate oral and written mathematical language
• Determination of the whole
• One object or shape defined as the whole
• Multiple connected shapes or objects defined as the whole
• A set of separate objects defined as the whole
• Wholes divided into two, four, or eight equal parts
• Recognition of the number of parts that equal one whole
• Two halves equal one whole; four fourths equal one whole; eight eighths equal one whole
• Recognition of the number of parts being considered
• Number of parts being considered within one whole
• Number of parts being considered beyond one whole
• Concrete models of the whole
• Linear models
• Cuisenaire rods, fraction bars, linking cube trains, folded paper strips, etc.
• Area models
• Fraction circles or squares, pattern blocks, geoboards, etc.
• Set models
• Pattern blocks, color tiles, counters, real-world objects, etc.
• Count fractional parts up to one whole using concrete models
• Correct sequence of fractional names
• Two equal parts
• One half, two halves or one whole
• Four equal parts
• One fourth, two fourths, three fourths, four fourths or one whole
• One quarter, two quarters, three quarters, four quarters or one whole
• Eight equal parts
• One eighth, two eighths, three eighths, four eighths, five eighths, six eighths, seven eighths, eight eighths or one whole
• Count fractional parts beyond one whole using concrete models
• Correct sequence of fractional names
• Two equal parts
• One half, two halves, three halves, four halves, five halves, etc.
• One half, one whole, one and one half, two wholes, two and one half, etc.
• Four equal parts
• One fourth, two fourths, three fourths, four fourths, five fourths, six fourths, seven fourths, eight fourths, nine fourths, etc.
• One fourth, two fourths, three fourths, one whole, one and one fourth, one and two fourths, one and three fourths, two wholes, two and one fourth, etc.
• One quarter, two quarters, three quarters, four quarters, five quarters, six quarters, seven quarters, eight quarters, nine quarters, etc.
• One quarter, two- quarters, three quarters, one whole, one and one quarter, one and two quarters, one and three quarters, two wholes, two and one quarter, etc.
• Eight equal parts
• One eighth, two eighths, three eighths, four eighths, five eighths, six eighths, seven eighths, eight eighths, nine eighths, ten eighths, eleven eighths, twelve eighths, thirteen eighths, fourteen eighths, fifteen eighths, sixteen eighths, seventeen eighths, etc.
• One eighth, two eighths, three eighths, four eighths, five eighths, six eighths, seven eighths, one whole, one and one eighth, one and two eighths, one and three eighths, one and four eighths, one and five eighths, one and six eighths, one and seven eighths, two wholes, two and one eighth, etc.
Note(s):
• Grade 3 will represent fractions greater than zero and less than or equal to one with denominators of 2, 3, 4, 6, and 8 using concrete objects and pictorial models, including strip diagrams and number lines.
• Grade 3 will introduce fraction symbols to represent pictorial models of fractions and fractional parts of a set of objects.
• Grade 3 will solve problems involving partitioning an object or a set of objects among two or more recipients using pictorial representations of fractions with denominators of 2, 3, 4, 6, and 8.
• Various mathematical process standards will be applied to this student expectation as appropriate.
• TxRCFP:
• Applying knowledge of two-dimensional shapes and three-dimensional solids, including exploration of early fraction concepts
• TxCCRS:
• I.B. Numeric Reasoning – Number sense and number concepts
• I.B.2. Interpret the relationships between the different representations of numbers.
• VIII.B. Communication and Representation – Interpretation of mathematical work
• VIII.B.1. Model and interpret mathematical ideas and concepts using multiple representations. | 8,142 | 38,680 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2020-24 | latest | en | 0.9143 |
https://www.hackmath.net/en/example/1511 | 1,556,087,157,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578633464.75/warc/CC-MAIN-20190424054536-20190424080536-00223.warc.gz | 710,333,879 | 7,078 | # Train from Brno
First train started from Brno at 8 am at 60 km per hour. Second train started from Opava at 7 o'clock at the speed of 40 kilometers per hour. In how many hours and how far from cities trains meet when the distance of cities is 200 kilometers.
Result
t = 9:36 h
b = 96 km
o = 104 km
#### Solution:
b+o = 200
b = 60(t-8)
o = 40(t-7)
b+o = 200
b-60t = -480
o-40t = -280
b = 96
o = 104
t = 485 = 9.6
Calculated by our linear equations calculator.
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
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#### To solve this example are needed these knowledge from mathematics:
Do you have a system of equations and looking for calculator system of linear equations? Do you want to convert length units?
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https://www.jiskha.com/display.cgi?id=1252797729 | 1,503,074,142,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104704.64/warc/CC-MAIN-20170818160227-20170818180227-00094.warc.gz | 914,566,990 | 4,602 | # physics
posted by .
we can say that the force of gravity is equal to mass times acceleration were the acceleration is equal to gravity sense gravity is an acceleration because of newtons second law force = mass times acceleration
hence Fg = mass times acceleration
Fg = mass times gravity
Fg = mg
however newtons second law states that the net force acting on an object is equal to it's mass times it's acceleration so what allows us to say that
Fg = mg
because certainly not for every single situation the
net forc is going to equal to the force of gravity
• physics -
You are totally wrong in postulating the net force=force of gravity.
Nuts.
Force of gravity applies only to the gravitional field, and for a free falling object, Forcegravity=mass*a
But what about other forces? Any unblanced force = mass*acceleartion, whether or not gravity is involved.
Now what if an object is sitting on a shelf..it is not accelerating.
Fgravity=force upward by shelf. What about weight? The downward force is equal to its mass times the gravitional field strength. What is the gravitional field strength? It is commonly called g, and is equal to 9.8N/kg. That is the force per kilogram when under the influence of gravity. What if the object is dropped? what is its acceleration? It is g, or 9.8m/s^2. Hmmm. Note in units, N/kg=m/s^2
One nitpick of your writing, and is the reason I wrote this. Your statement "the acceleration is equal to gravity sense gravity is an acceleration because of newtons..." is inaccurate.
Gravity is not an acceleration, it is a force.
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1. Oct 27, 2009
### phreak
1. The problem statement, all variables and given/known data
Given $$k\in \mathbb{Z} \setminus \{ 0 \}$$, prove that $$\lim_{n\to \infty} \frac{1}{N} \sum_{n=1}^N e^{2\pi i k n \alpha}=0$$, for all $$\alpha \in \mathbb{R} \setminus \mathbb{Q}$$.
2. Relevant equations
3. The attempt at a solution
Well, I had an idea, but I'm not sure how well it works. Even if it did, I'm not sure how to make it rigorous. Basically, my idea was to say that you work your way around the unit circle, and that whenever we come arbitrarily close to hitting the same point again, the sum (between the two points) gets arbitrarily close to zero. Am I on the right track?
Edit: Alternately, could we use some convergence theorem using counting measure? Also, the limit is supposed to be 0, not 1. I changed it in the code, but it isn't working.
2. Oct 27, 2009
### Hurkyl
Staff Emeritus
This sounds like a reasonable line of attack.
Let's idealize for a bit (which may or may not help). For simplicity, let's fix a particular point, say, 1. What would happen if, after some number P of steps you really did return to 1?
(i.e. what if $\alpha \in \mathbb{Q}$, and $k \alpha \notin \mathbb{Z}$)
Can you prove the limit goes to zero in this special case?
You need to reload the image. Reloading the page might not do that, depending on your browser and its settings.
3. Oct 27, 2009
### phreak
Ok, I proved it. Thanks a lot for the help.
(For the curious: You prove it for \alpha k rational. Then you pick r = p/q \in \Q s.t. |\alpha k - p/q | < \epsilon. You do a lot of clever approximation and you get that 1/N * the sum \le (2\epsilon+q)/N, which goes to 0 as N goes to infinity.) | 517 | 1,749 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-47 | longest | en | 0.901123 |
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# NATIONAL UNIVERSITY OF SINGAPORE EXAMINATION ST2131/MA2216 PROBABILITY
## (Semester 2: AY 20102011) April/May 2011 Time allowed: 2 hours
INSTRUCTIONS TO CANDIDATES 1. This examination paper contains SIX (6) questions and comprises THIRTEEN (13) printed pages. 2. Answer ALL questions. Marks for each question are indicated. The total marks for this paper is 70. 3. Candidates may use any calculator. However, they should lay out systematically the various steps in the calculations. 4. This is a CLOSED BOOK examination. Candidates may bring in ONE (1) two-sided, A4 size help sheet. 5. Write down your matriculation number and seat number neatly in the boxes provided below. Answers should be written on this booklet. This booklet will be collected at the end of the examination.
Matriculation Number
Seat Number
Question
Total
Score
PAGE 2
ST2131/MA2216
Question 1 Players A and B play a sequence of independent games in which the two players take turn in rolling a fair die of 6 faces. The 6 faces are labeled as 1, 2, 3, 4, 5 and 6. Player A rolls the die once in game 1. Then, player A wins and the sequence stops if the die lands on 6, or otherwise they proceed to game 2. In game 2, the die is rolled for the second time, but by player B. Then, player B wins and the sequence stops if the die lands on either 6 or 5, or otherwise they proceed to game 3. In game 3, the die is rolled for the third time, but by player A. Then, player A wins and the sequence stops if the die lands on any one of 6, 5 or 4, or otherwise they proceed to game 4. The sequence continues in this manner until the end of game 6. (i) Find the probability that player A wins at game 5. [4 marks]
... 3
PAGE 3
ST2131/MA2216
(ii) Suppose player A wins the sequence. Find the probability that player A wins at game 3. [4 marks]
... 4
PAGE 4
ST2131/MA2216
Question 2 Suppose that a certain disease is present in 10% of the population in a region, and that there is a screening test designed to detect the presence of the disease in this population. Unfortunately, the test does not work perfectly; it is possible that the test shows a negative result for a person who has the disease, or it shows a positive result for a person who has no disease, or sometimes the result is indecisive (that is, neither positive nor negative). The following table shows the proportion of times that the test produces various results after diagnosing every person in this population. For instance, it shows that 2% of the population have the disease and the test showed a negative result. Positive (P) a b Test Result Indecisive (I ) 0.05 0.01 Negative (N ) 0.02 d [3 marks]
Disease present (D) Disease absent (Dc ) (i) Find the value of a.
(ii) Suppose the events N and D are independent, show that d = 0.18.
[3 marks]
... 5
PAGE 5
ST2131/MA2216
(iii) If a person selected at random from this population has no disease, what is the probability that the test shows a positive result? [3 marks]
(iv) If the values of a, b and d in the table are all known, can you nd the total number of cases in which the test gives a correct result in detecting the disease in this population? In one short sentence, give a reason to support your answer. [3 marks]
... 6
PAGE 6
ST2131/MA2216
Question 3 An urn contains N balls numbered 1 through N . A sample of n balls (where n < N ) are randomly selected. Let X and Y be the minimum and maximum of the chosen numbers respectively. (i) If the balls are selected without replacement, nd the probability mass function of Y . [3 marks]
(ii) If the balls are selected with replacement, nd the probability mass function of Y . [3 marks]
... 7
PAGE 7
ST2131/MA2216
(iii) If the balls are selected without replacement, nd P(Y X = 5) for N = 50, n = 5. Round the answer to 6 decimal accuracy. [4 marks]
(iv) If the balls are selected with replacement, nd P(Y X = 1) for N = 50, n = 5. Round the answer to 6 decimal accuracy. [4 marks]
... 8
PAGE 8 Question 4 Let R be a random variable with the probability density function f (r) = 0.5642er ,
2
ST2131/MA2216
< r < .
(i) Let W be the absolute value of R. Find the probability density function of W . [3 marks]
## (ii) Let X = R2 . Find the probability density function of X .
[3 marks]
... 9
PAGE 9
ST2131/MA2216
(iii) Let Y be a random variable such that ln(Y ) has the same distribution as R. Find the probability density function of Y . [3 marks]
(iv) Rank the 3 probabilities P(0 < W < u), P(0 < X < u) and P(0 < Y < u) from the smallest to the largest, where u is a constant such that 0 < u < 1. [7 marks]
. . . 10
PAGE 10 Question 5
ST2131/MA2216
(a) Bus A arrives at the Computer Center bus stop at a time that is uniformly distributed between 9:00 am and 10:00 am. Bus B arrives at the same bus stop at a time that is uniformly distributed between 9:00 am and the arrival time of bus A. If you have been waiting for bus B since 9:00 am, how long would you expect to wait? [6 marks]
. . . 11
PAGE 11
ST2131/MA2216
(b) Let X and Y be two continuous random variables with nite expectations. The joint probability density function of X and Y , f (x, y), is dened such that f (x, y) = 0, for x > y and f (x, y) 0, for x y. Show that E (X ) E (Y ). [4 marks]
. . . 12
PAGE 12 Question 6
ST2131/MA2216
(a) The popular restaurant chain Hairy Jones has two branches in the cities of Tam-Pines and Bee-Sam. The probability density function of the weekly prot of each branch, in thousands of dollars, is given by f (x) = x/6, if 2 < x < 4, 0, otherwise.
Assuming that the prot of each branch is independent of the other, what is the probability that one branch makes at least \$1000 more than the other in the coming week? [5 marks]
. . . 13
PAGE 13
ST2131/MA2216
(b) Let X be a discrete random variable with probability mass function given by P(X = k) = 2 1 3
k
k = 1, 2, 3, . . .
END OF PAPER | 1,654 | 5,923 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2019-35 | latest | en | 0.903806 |
http://www.brightstorm.com/qna/question/4409/ | 1,368,991,765,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698017611/warc/CC-MAIN-20130516095337-00025-ip-10-60-113-184.ec2.internal.warc.gz | 365,974,143 | 9,797 | Quick Homework Help
# two dice are roll what the odds on favor that the sum of the dice is less than 5 ⚑ Flag
by Castillo001 at March 21, 2010
There are 36 outcomes for the die. List the ways you can get a sum less than 5. 2 = 1+1 3 = 1+2, 2+1 4 = 1+3, 3+1, 2+2 So there is a 6/36 = 1/6 chance.
Kalifornia March 21, 2010
If you want to answer the question for odds, you must have success over failure, not success over total. So the answer should be 6/30 or 1/5
Nancy095 March 22, 2010 | 174 | 491 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2013-20 | latest | en | 0.931595 |
https://theomegafoundation.org/what-is-the-simplest-form-of-14-and-24/ | 1,721,681,643,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517915.15/warc/CC-MAIN-20240722190551-20240722220551-00880.warc.gz | 496,290,512 | 16,581 | # What is the simplest form of 14 and 24?
Answer and Explanation: When simplified the fraction 14/24 becomes 712 . To simplify this fraction we need to find a common factor of the numerator and the denominator. Since they’re both even numbers we can divide them both by two.
## How do I simplify?
The word simplify means to make something easier to do or understand. So, reducing or simplifying fractions means we make the fraction as simple as possible. We do this by dividing the numerator and the denominator by the largest number that can divide into both numbers exactly.
## What is in its simplest form?
A fraction is in its simplest form when the numerator and the denominator have no common factors besides one.
## What is 14 as a fraction in simplest form?
Answer: 14% can be represented as 7/50 in fraction form.
Step 1: We represent the given number in percentage as a fraction by dividing it by 100.
What is the simplest form of 14 and 24? – Related Questions
## How do you simplify fractions?
Find the highest common factor (HCF) of numerator and denominator of the fraction. Divide both the numerator and the denominator by HCF to get the simplified fraction. Write the whole and the simplified fraction together.
## Which fraction is in simplest form?
The fraction is said to be in its simplest form, when the numerator (top) and denominator (bottom) of the fraction does not have any common factor. For example, the simplest form of 6/12 is ½.
## What is the common fraction of 14%?
Solution: 14% as a fraction is 7/50.
## What is √ 14 as a fraction?
The square root of 14 is expressed as √14 in the radical form and as (14)½ or (14)0.5 in the exponent form.
Square Root of 14 in radical form: √14.
1. What Is the Square Root of 14?
3. How to Find the Square Root of 14?
4. Thinking Out of the Box!
## What can equal to 14?
The prime factorisation of 14 is equal 2 x 7. Therefore, 2 and 7 are the prime factors of 14.
## What is 14 divided by 4 as a fraction?
14 divided by 4 is equal to 3 with a remainder of 2 (14 / 4 = 3 R. 2). This can also be written with a decimal (14 / 4 = 3.5).
## What is 5 4 as a number?
Answer: 5/4 as a decimal is 1.25
Let’s look into the two methods to write 5/4 as a decimal.
## Is 100 divisible by 3 yes or no?
100 divided by 3 is 33.333333. This means that there are 33 numbers between 1 and 100 that are divisible by 3. 100 divided by 2 is 50.
## What is 14 3 as a fraction?
Answer: 14/3 as a mixed number can be written as 4 2/3. Let’s look at how to write 14/3 as a mixed number in the explanation below. Explanation: 14/3 is an improper fraction.
## What is to simplify?
transitive verb. : to make simple or simpler: such as. : to reduce to basic essentials. : to diminish in scope or complexity : streamline.
## How do you divide 14 divided by 3?
14 divided by 3 gives quotient 4 and leaves a remainder 2. 14 divided by 3 further can be written in mixed fraction form which is 4(2/3). Thus, 14 divided by 3 as a fraction can be written as 4(2/3).
## What is 3/4 as a fraction?
The fraction 3/4 or three quarters means 3 parts out of 4. The upper number, 3, is called the numerator and the lower number, 4, is the denominator.
## What is 3/6 as a percent?
So 3/6 is equal to 50%.
## How do you write 7/8 as a decimal?
7 divided by 8 or 7/8 is equal to 7 divided by 8, which is equal to 0.875. But I’ll put a leading 0 here just so it makes it clear that this is where the decimal is. 0.875.
## What is 4/7 as a fraction?
4/7 = 47 ≅ 0.5714286
Spelled result in words is four sevenths.
40% | 981 | 3,584 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2024-30 | latest | en | 0.944466 |
https://17calculus.com/spherical-coordinates/ | 1,656,111,533,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033816.0/warc/CC-MAIN-20220624213908-20220625003908-00000.warc.gz | 120,645,108 | 26,745 | ## 17Calculus - Spherical Coordinates
##### 17Calculus
This page covers the basics of spherical coordinates and reference equations that use spherical coordinates for multi-variable calculus.
Spherical coordinates are very different from rectangular and cylindrical coordinates. There are still three coordinates, usually labeled $$(\rho, \theta, \phi)$$, but they are assigned so that spherical-type objects are easy to express. All distances and angles are measured based on a radial line drawn from the origin to the point. Table 1 describes how they are defined and Figure 1 shows this graphically.
Table 1 $$\rho$$ is the radial distance out from the origin to the point $$\theta$$ is the angle measured in the xy-plane from the positive x-axis to the shadow of the radial line. Note: This is the same angle as $$\theta$$ in cylindrical coordinates. $$\phi$$ is the angle measured from the positive z-axis to the radial line
Note - We have specifically chosen $$(\rho, \theta, \phi)$$ to describe spherical coordinates but not all books, instructors and videos use these same variables. Some use $$(r,\theta,\phi)$$ with $$\theta$$ and $$\phi$$ meaning different things depending on the context. (See the Wiktionary page on spherical coordinates for examples.) Check with your instructor and textbook to see which one they require.
$$(x,y,z) \to (\rho, \theta, \phi)$$
$$x=\rho\sin\phi\cos\theta$$
$$y=\rho\sin\phi\sin\theta$$
$$z=\rho\cos\phi$$
$$\rho^2=x^2+y^2+z^2$$
Table 2
The equations to convert from rectangular to spherical coordinates, in Table 2, are somewhat complicated. So take a few minutes to learn and memorize them (why memorize?). Look for similarities and differences between the equations and see if you can make some sense out of them.
Here is a quick video clip discussing these equations and showing a neat animation that will help you get a feel for spherical coordinates.
### MIP4U - Introduction to Spherical Coordinates [2mins-13secs]
video by MIP4U
Converting between Rectangular and Spherical Coordinates
Rectangular → Spherical
In order to convert from rectangular to spherical coordinates, the equations are Table 2 above are directly applied. The technique is not hard but it will help to watch a couple of quick videos showing how to do this with examples. The first video explains how to convert points from rectangular to spherical. The second video explains how to convert equations from rectangular to spherical. Both videos have plenty of examples.
### Krista King Math - Spherical Coordinates [3mins-53secs]
video by Krista King Math
### MIP4U - Converting Between Spherical and Rectangular Equations [8mins-15secs]
video by MIP4U
To be able to use spherical coordinates in triple integrals, you need to understand the unit vectors in spherical coordinates. Here is a video that will help.
### Michel vanBiezen - Spherical Unit Vector Conversions [9mins-45secs]
video by Michel vanBiezen
Before we go on, let's do a few practice problems.
Practice
Find an equation in spherical coordinates for the rectangular equation $$x^2 + y^2 - z^2 = 0$$.
Problem Statement
Find an equation in spherical coordinates for the rectangular equation $$x^2 + y^2 - z^2 = 0$$.
Solution
### 2209 video solution
Log in to rate this practice problem and to see it's current rating.
Spherical → Rectangular
Sometimes it will be easier to evaluate an integral in rectangular coordinates rather than spherical coordinates. In this case, you may be given an equation in spherical coordinates and be asked to convert to rectangular coordinates.
To do this conversion, there is no set of equations with specific techniques. You just have to use trig identities and intuition to get an equation in rectangular coordinates. However, it sometimes helps to convert tangent, cotangent, secant and cosecant to sines and cosines and use $$\cos^2(t) + \sin^2(t) = 1$$. Try these practice problems to get the idea.
Practice
Unless otherwise instructed, convert these equations in spherical coordinates to rectangular coordinates.
$$\rho = 2 \csc(\phi)$$
Problem Statement
Convert $$\rho = 2 \csc(\phi)$$ to rectangular coordinates.
$$x^2 + y^2 = 4$$
Problem Statement
Convert $$\rho = 2 \csc(\phi)$$ to rectangular coordinates.
Solution
### MIP4U - 2205 video solution
video by MIP4U
$$x^2 + y^2 = 4$$
Log in to rate this practice problem and to see it's current rating.
When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.
DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it. | 1,246 | 5,540 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2022-27 | longest | en | 0.867513 |
http://math.stackexchange.com/questions/70646/sequence-finding | 1,469,737,370,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828322.57/warc/CC-MAIN-20160723071028-00271-ip-10-185-27-174.ec2.internal.warc.gz | 156,843,251 | 18,307 | # Sequence finding
Find all such sequences $(x_1, x_2, x_3, ..., x_{63})$ consisting of different positive integers that for $n=1,2,3,...,62$ the number $x_n$ is a divisor of $x_{n+1}+1$ and $x_{63}$ is a divisor of $x_1+1$.
Suppose we let $x_{63}=k$. Then we know $x_1 = k+62$.
Our condition is that the last term is a factor of the first term plus 1. Another way of saying that is that
$\frac{k+62+1}{k}= \frac{k+63}{k}$ is an integer. So we have to find all the integer solutions for the equation
A better way of doing it is noticing that we can rewrite the equation as: $k(n-1)=63$
So we know that there are 6 such sequences as there are 6 factors of 63.
And here is the problem: are the sequences of 62+k, 62+k-1, 62+k-2,... the only ones which can fulfill the requirements? I mean, look: $\{1,3,14,27,14039,...,some\_positive\_integer\}$ also does while not being either descending nor having any particular "step". So are the ones I found really the only ones?
-
3,2,1,63,62,61,60,...,6,5,4 or any other cyclic shift? – Jyrki Lahtonen Oct 7 '11 at 17:38
The answer is "Yes". The only sequences are of the form 63+k,62+k,61+k,...,1+k where $k+1$ divides $k+64$ or some cyclic permutation of such a sequence.
To see this notice that the most one can drop is by 1: Suppose $x_n>x_{n+1}$ (and $x_n$ divides $x_{n+1}+1$). This implies that $x_n\geq x_{n+1}+1$ and $x_n \leq x_{n+1}+1$. Thus $x_n = x_{n+1}+1$.
So once you've increased past $x_j=\ell$, the only way to get back down below $\ell$ (further out in the sequence) is $...,x_{i-2}=\ell+2,x_{i-1}=\ell+1,x_{i}=\ell$ (you must "hit" $\ell$ on the way back down).
Suppose your sequence has two increases: $x_i<x_{i+1},...,x_j<x_{j+1}$. If $x_i < x_j$. then we somehow got "back down" from $x_{j+1}$ to $x_i$. Along the way we must have "hit" $x_j$. This violates our "distinct integer" assumption unless $i=j$ (similarly if $x_j>x_i$). So each sequence can only have one "increase". Everything else is a step down by one. | 673 | 1,989 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2016-30 | latest | en | 0.881938 |
https://convertoctopus.com/115-3-cubic-centimeters-to-cubic-feet | 1,716,732,624,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058956.26/warc/CC-MAIN-20240526135546-20240526165546-00360.warc.gz | 155,945,528 | 7,459 | ## Conversion formula
The conversion factor from cubic centimeters to cubic feet is 3.5314666572208E-5, which means that 1 cubic centimeter is equal to 3.5314666572208E-5 cubic feet:
1 cm3 = 3.5314666572208E-5 ft3
To convert 115.3 cubic centimeters into cubic feet we have to multiply 115.3 by the conversion factor in order to get the volume amount from cubic centimeters to cubic feet. We can also form a simple proportion to calculate the result:
1 cm3 → 3.5314666572208E-5 ft3
115.3 cm3 → V(ft3)
Solve the above proportion to obtain the volume V in cubic feet:
V(ft3) = 115.3 cm3 × 3.5314666572208E-5 ft3
V(ft3) = 0.0040717810557755 ft3
The final result is:
115.3 cm3 → 0.0040717810557755 ft3
We conclude that 115.3 cubic centimeters is equivalent to 0.0040717810557755 cubic feet:
115.3 cubic centimeters = 0.0040717810557755 cubic feet
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 cubic foot is equal to 245.59277286817 × 115.3 cubic centimeters.
Another way is saying that 115.3 cubic centimeters is equal to 1 ÷ 245.59277286817 cubic feet.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one hundred fifteen point three cubic centimeters is approximately zero point zero zero four cubic feet:
115.3 cm3 ≅ 0.004 ft3
An alternative is also that one cubic foot is approximately two hundred forty-five point five nine three times one hundred fifteen point three cubic centimeters.
## Conversion table
### cubic centimeters to cubic feet chart
For quick reference purposes, below is the conversion table you can use to convert from cubic centimeters to cubic feet
cubic centimeters (cm3) cubic feet (ft3)
116.3 cubic centimeters 0.004 cubic feet
117.3 cubic centimeters 0.004 cubic feet
118.3 cubic centimeters 0.004 cubic feet
119.3 cubic centimeters 0.004 cubic feet
120.3 cubic centimeters 0.004 cubic feet
121.3 cubic centimeters 0.004 cubic feet
122.3 cubic centimeters 0.004 cubic feet
123.3 cubic centimeters 0.004 cubic feet
124.3 cubic centimeters 0.004 cubic feet
125.3 cubic centimeters 0.004 cubic feet | 610 | 2,191 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-22 | latest | en | 0.723202 |
http://www.glennwestmore.com.au/prime-numbers-and-bordered-magic-squares/ | 1,511,094,664,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805578.23/warc/CC-MAIN-20171119115102-20171119135102-00728.warc.gz | 410,791,451 | 10,662 | ##### 8 x 8 Prime Bordered Magic Square
The following prime magic square has a magic sum of exactly 19,000!
In addition, it is also a bordered prime magic square.
The inner 4 x 4 square, (shaded red) is a prime magic square with a magic sum of 9,500, exactly half of the magic sum of the 8 x 8 square.
Furthermore, the inner 6 x 6 square is also a prime magic square with a magic sum of 14,250. Exactly three quarters of the magic sum of the 8 x 8 square.
Journal of Recreational Mathematics ( A.W.Johnson Jr Vol 15:2, 1982-83)
##### 13 x 13 Prime Bordered Magic Square
The following 13 x 13 Prime Magic Square contains an amazing number of features.
1. This is a bordered prime magic square. Each of the 11 x 11, 9 x 9, 7 x 7, 5x 5, and 3 x 3 squares are also prime magic squares.
2. The magic sums are 70,681, 59,807, 48,933, 38,059, 27,185 and 16,311 respectively.
3. There is a common difference between these magic sums of 10,874.
4. This common difference is also the same as that between the magic sum of the 3 x 3 and the centre number, 5437.
5. The centre number is exactly half of the common difference between each magic sum. | 333 | 1,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2017-47 | longest | en | 0.914793 |
https://de.scribd.com/document/320048667/Harder-X2-Volumes-Problems | 1,566,689,845,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321786.95/warc/CC-MAIN-20190824214845-20190825000845-00207.warc.gz | 429,605,653 | 75,516 | You are on page 1of 4
# Further X2 Volumes Problems
(a)
## Figure 1 below shows a scale model of the volcano Mt Rekrap.
The base of the model is elliptical in shape with the axes 60 cm by 40 cm reducing
uniformly to a circle of radius 12 cm at the top.
The hollow core of the model has circular cross sections with a circle of radius 6 cm at the
base rising uniformly to a circle also of radius 12 cm at the top.
The model is 24 cm high.
Figure 2 shows the top view of the cross sectional area of the volcano.
Figure 1
24 cm
h cm
40 cm
60 cm
Figure 2
3
h
4
(i)
## Show that at height h, the length of the semi-major axis is given by a = 30
(ii)
Show that the area of the cross sectional slice at height h is given by
A=
9024 448h + 3h 2
16
You may assume that the area of an ellipse with semi-major axis a and semi-minor
axis b is given by ab .
(iii)
## Find the volume of the scale model of Mt Rekrap.
2
4
2
[6336]
7
7
QUESTION 5. (Continued)
QUESTION 5. (Continued)
HSC 1998
(b)
(b)
Marks
Marks
9
a
a
Z
Z
A
A
a
a
Y
Y
W
W
C
C
R
R
S
S
x
x
P
P
Q
Q
x
x
X
X
b
b
B
B
The diagram shows a sandstone solid with rectangular base ABQP of length
The diagram shows a sandstone solid with rectangular base ABQP of length
b metres and width a metres. The end PQRS is a square, and the other end ABC
b metres and width a metres. The end PQRS is a square, and the other end ABC
is an equilateral triangle. Both ends are perpendicular to the base.
is an equilateral triangle. Both ends are perpendicular to the base.
Consider the slice of the solid with face WXYZ and thickness x metres, as
Consider the slice of the solid with face WXYZ and thickness x metres, as
shown in the diagram. The slice is parallel to the ends and AW = BX = x metres.
shown in the diagram. The slice is parallel to the ends and AW = BX = x metres.
(i) Find the height of the equilateral triangle ABC.
(i) Find the height of the equilateral triangle ABC.
(ii) Given that the triangles CRS and CYZ are similar, find YZ in terms of a,
(ii) Given that the triangles CRS and CYZ are similar, find YZ in terms of a,
b and x .
b and x .
(iii) Let the perpendicular height of the trapezium WXYZ be h metres. Show
(iii) Let the perpendicular height of the trapezium WXYZ be h metres. Show
that
that
((
))
a
x
h = a 3 + 2 3 x .
h = 2 3 + 2 3 b .
2
b
(iv) Hence show that the cross-sectional area of WXYZ is given by
(iv) Hence show that the cross-sectional area of WXYZ is given by
[[((
]]
a22
a
2 3 x + b 3 (b + x ) .
4 b22 2 3 x + b 3 (b + x ) .
4b
(v) Find the volume of the solid.
(v) Find the volume of the solid.
))
9
9
SGS 6A q23
## (b) The region bounded by arc AB and the chord AB is rotated
about the chord AB. Show that the solid formed has volume
103
r 3 cubic units.
6 2
EXTENS ION
## 23. [A First Principles Approach] The region bounded by y = x2
and y = x is rotated about the line y = x to generate a solid.
In this question you will find the volume of this solid by first
CHAPTER 6: Volumes
approximating a small portion with a disk.
y ' x2
y'x
"h
r
&
Let P (x, x2) be a typical point on the parabola. From x to x + x
P#x\$ x %
(b) 4
we will approximate the area between the curve and the line16(a)
with
y
x3
"x
1
a rectangle. One corner of this rectangle is at P and a side is on
the line y Exercise
= x. Let6A
the(Page
dimensions
27) of this rectangle be r h as
"
shown in the
diagram.
!)#
)#
1(a)
(ii) 3
(b)(i) 9
y
14
5
x
(a) Show that h (ii)
= 2 x and find r as a function of x.
!" "
#
(b) When the rectangle is rotated about the line y = x it generates a cylindrical prism
with volume V ."Find V
& .
Chapter Six
y\$x %1
## (d) Hence show that
!" the volume
" (i) x of the solid is
2(a)
30 2
(b) 8
15
6B The Method
ofy \$Cylindrical
Shells
x
#
#
## Inevitably when" tackling yvolume
\$'#'! x problems there will be
SGS
2012
THSC q16
b result
certain
questions
which
in awkward integrals. As
an example, consider the
solid
generated
when the region
x
"
#
2
between y = 4x x and y = x is rotated about the
y-axis, as shown on the right. The volume
can be found
32
(b)(i) 5
3(a)
(ii) 8
by the methods ofySection 6A since each
cross-section is
(ii)
an annulus. However the situation is complicated by the
(
fact that the formula for the outer radius changes. Below
! is router = y whilst above y = 3 it is
router = 2 + 4 y. Hence
(i) the integral must be split
into two parts. (Try this as an exercise.)
## Exercise 6B (Page 32)
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2
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(b) 128
5
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Chapter 6: Volumes
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Question 8 (continued)
HSC 2002 q8
38
CHAPTER 6: Volumes
(b)
## 12. In the diagram on the rightKis a cylindrical wedge ABCD. The N
height of the cylinder is equal to the diameter of its base. Let the
L
M
radius of the base be r units.
(a) Show that the typical triangular cross-section shaded has area
(r 2 x2 ) square units.
(b) Hence show that the wedge has volume 43 r 3 cubic units.
A
(x, y)
B
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!r
## 13. The diagram on the right shows a cylindrical drinking glass of
B glass
interior radius 4 cm and perpendicular height 12 cm. The
a
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bottom of the glass is exposed. Use the methods
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## 15. A rubbish skip
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site KLMN.
has a rectangular base 6 metres by 3 metres, and its
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to top. The open top is a rectangle 7 metres by 4 metres.
(ii) Find the volume of the tetrahedron ABCD.
!
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"
(a) Show that a rectangular cross-section h metres above the base has area h2 + 6 h2 + 3
square metres.
SGS
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4
2
End of paper
16.
C
Q
P
U
T
A
CD
O
S
AB
R
The first diagram shows two identical intersecting cylinders of radius r. The second
diagram is the view from above. Their axes AB and CD intersect at 90 at the point O.
The third diagram shows the solid which is common to both cylinders, bounded at its
widest point by the horizontal square P QRS.
(a) The typical square cross-section T U V W shown is parallel to the square P QRS and
y units above it. Find an expression for the area of this typical cross-section of the
42
CHAPTER 6: Volumes
solid in terms of y.
SGS NOTES
14
(b) Hence find the volume that is common to the two intersecting cylinders.
10(a)
(b) 27
2
y
#
## SGS Mathematics Year 12
Chapter 6: Volumes
volumes
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2
2
3
16(a) 4(r y ) (b) 16
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15(b) | 2,855 | 8,369 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2019-35 | latest | en | 0.883403 |
https://mathematica.stackexchange.com/questions/159127/equation-of-a-line-of-intersection-of-two-planes | 1,686,344,989,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656833.99/warc/CC-MAIN-20230609201549-20230609231549-00521.warc.gz | 452,796,144 | 33,165 | # Equation of a Line of Intersection of Two planes [duplicate]
I have two planes $−11x+20y−23z+25=0$ and $−24x+5y+6z+14=0$. How do I go about finding the equation of the line of intersection? And once I get it, how do i plot both planes and the line?
• Welcome. Is your question related to Mathematica software? What have you tried? You can search the documentation of Solve and Plot or ParametricPlot. Nov 2, 2017 at 17:42
• I have to find an equation of the line of intersection of the two planes and then plot both the planes and line of intersection in mathematica. I tried using "Solve" but the answer was incorrect (I found the answer manually). @anderstood Nov 2, 2017 at 17:47
• Then I suggest you include your attempts in your question (click on "edit") to get adequate help. Nov 2, 2017 at 17:53
First get the solution for the line of intersection:
sol = Solve[{-11 x + 20 y - 23 z + 25 ==
0 && -24 x + 5 y + 6 z + 14 == 0}]
linevec = {x, y, z} /. sol[[1]]
then get the z for both planes,
plane1 = z /.
Solve[-11 x + 20 y - 23 z + 25 == 0, z][[1]]
plane2 = z /. Solve[-24 x + 5 y + 6 z + 14 == 0, z][[1]]
Finally, plot and show them together
g0 = Plot3D[{plane1, plane2}, {x, -3, 3}, {y, -3, 3},
PlotStyle -> Opacity[.7]];
g1 = ParametricPlot3D[linevec, {x, -3, 3},
PlotStyle -> {Red, Thick}];
Show[{g0, g1}]
• This is awesome! I ended up doing it a different way, but this will come in handy in the future! Thank you! @egwenesadai Nov 16, 2017 at 18:36 | 488 | 1,476 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2023-23 | latest | en | 0.883392 |
http://ccckmit.wikidot.com/mt:smtbasics | 1,524,394,295,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945584.75/warc/CC-MAIN-20180422100104-20180422120104-00094.warc.gz | 54,480,388 | 17,633 | # 簡介
原文 (英文) 譯文 (中文) … The man saw a lady in red walked in the park. A little dog followed behind and barking around. … …那個男人看見一個紅衣女郎在公園裡走著，還有一隻小狗跟在後面汪汪的叫。…
對照版 …The (那個) man (男人) saw(看見) a(一個) lady(女郎) in red(紅衣) walked(走著) in the park(在公園裡). A(一隻) little(小) dog(狗) followed(跟在) behind(後面) and barking around (汪汪的叫)….
# 方法
(1)
\begin{align} bts = mt(s) = argmax_t \{ score(t|s) \} \end{align}
(2)
\begin{align} bts = mt(s) = argmax_t \{ P(t|s) \} \end{align}
(3)
\begin{align} P(t|s) = P(t_1|s_1) P(t_2|s_2)... P(t_n|s_n) = \prod_{i=1}^{n} P(t_i|s_i) \end{align}
(4)
\begin{align} bts = mt(s) = argmax_t { \prod_{i=1}^{n} P(t_i|s_i) } \end{align}
(5)
$$P(t|s)=P(s|t) P(t)$$
(6)
\begin{align} P(t) = P(t_1) P(t_2|t_1) ... P(t_{n-1}|t_n) = P(t_1) \prod_{i=2}^n P(t_{i-1}|t_i) \end{align}
(7)
\begin{align} P(t|s) = P(s|t) P(t) = \prod_{i=1}^{n} P(s_i|t_i)*P(t_1) \prod_{i=2}^{n} P(t_{i-1}|t_i) \end{align}
# Facebook
page revision: 20, last edited: 07 Sep 2010 08:23
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# P3 Integration Dude's Hard One! Who Can Crack It?! watch
Announcements
1. question:
integral of (lnx)^2dx
with limits of e and 1
2. (Original post by mabs)
question:
integral of (lnx)^2dx
with limits of e and 1
is the answer 2 by any chance?!
3. e-2
4. (e - 2) ?
5. what method is everyone using - i don't know how to start
6. (Original post by mabs)
question:
integral of (lnx)^2dx
with limits of e and 1
2-ln1
7. (Original post by Silly Sally)
what method is everyone using - i don't know how to start
lol, intergrate by parts..... 2.lnx
8. (Original post by Silly Sally)
what method is everyone using - i don't know how to start
by parts
u=(lnx)^2
dv/dx = 1
...
9. u g2 use parts
10. (Original post by Bhaal85)
2-ln1
ln1 is 0, therefore answer is 2....which is wot i got, but its not right, lol
11. (Original post by Silly Sally)
what method is everyone using - i don't know how to start
lnx^2 can be done by parts, just do integration by parts with 1*lnx^2
12. IT STILL DOESN'T WORK!
U GET 2/X lnx AND THAT WON'T REDUCE
13. (Original post by mabs)
question:
integral of (lnx)^2dx
with limits of e and 1
If you fell up to it you can integrate by parts
u = lnx
u' = 1/x
v = xlnx + x
v' = lnx
integral of question = [ x(lnx)^2 - xlnx ] - [ integral of (lnx -1) ]
=> x(lnx)^2) - xlnx - xlnx +x +x
then simplify and put the limits in and you get e-2. Maybe someone with html can type it out so it looks similar. Im sure theres an easier method but this one works!
14. check P3 ex 4d q.19
15. I got the right answer so here's the method:
Integrate by parts with v = lnx, dv/dx = 1/x, du/dx = lnx, u = x(lnx - 1)
So int (lnx)^2 = xlnx(lnx - 1) - int 1/x*(x(lnx - 1)) = x[(lnx)^2 - lnx] - x(lnx -1) + x = x[ln(x)^2 - 2 lnx + 2], evaluate between 1 and e to give (e - 2)
16. INT lnx^2 dx = INT 1*lnx^2
u = lnx^2 therefore u' = 2/x
v' = 1 therefore v = x
xlnx^2 - INT x*2/x which gave me:
[xlnx^2 - 2x] with limits e and 1
[2e-2e] - [ln1-2]
= 2-ln1
= 2
17. George Explain!!!!!
18. BHAAL, (lnx)^2 means lnx all to the power 2 and u don't get wot u get when u use the chain rule.
19. (Original post by mabs)
BHAAL, (lnx)^2 means lnx all to the power 2 and u don't get wot u get when u use the chain rule.
It does work if you use the chain rule - thats the only bit i can do!!!
20. (Original post by Bezza)
I got the right answer so here's the method:
Integrate by parts with v = lnx, dv/dx = 1/x, du/dx = lnx, u = x(lnx - 1)
So int (lnx)^2 = xlnx(lnx - 1) - int 1/x*(x(lnx - 1)) = x[(lnx)^2 - lnx] - x(lnx -1) + x = x[ln(x)^2 - 2 lnx + 2], evaluate between 1 and e to give (e - 2)
i didnt know u could intergrate 'ln' functions?! how does du/dx + lnx become u = x(lnx - 1)???
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by JeremyWsmith
Tags: interpret, negative, yintercept
P: 10 I'm taking Physics 11 and we're doing a graphing exercise. The graph is Cricket Chirps per minute(as dependent variable) on the Y-Axis versus Degrees Celsius(Independent Variable) Along the x-axis. I've drawn my line of best fit. The exercise asks that I find the y-intercept of the line of best fit. I've found it to be -65 chirps/minute. The next question asks how would one interpret the y-intercept. This doesn't make any sense to me.. I'm positive I've done the graphing and calculations correctly as the equation works find for finding points on the graph. I've asked my physics teacher some questions and he says they're not correct. I've asked: Is it possible the cricket stops chirping and begins processing chirps. He said no, which leaves me at a dead end. I could just put 'undefined' not sure if that'd be correct though. Also, how would I show this on a graph?
Emeritus Sci Advisor PF Gold P: 5,196 Welcome to PF, The bottom line is that for a discrete, countable thing like cricket chirps, it just doesn't make sense to have a negative value. Your result is unphysical, and it indicates either that you have done something wrong, or that the graph should be a piecewise function (meaning that it is linear with some non-zero slope down to the temperature at which the crickets stop chirping altogether, and then linear with zero slope (i.e. flat) below that temperature). So, assuming you have done the fitting right, then maybe the best interpretation of your data is, "below some positive temperature, the crickets stop chirping altogether." EDIT: Another way to express what I'm saying is that the cricket chirp rate is only well-modelled by a linear function over a limited temperature range. Below some temperature, there is a cutoff.
P: 10 So say the cricket chrips 0 times a minute at 10 degrees, could I say: The y-intercept is undefined because the temperature must be above or equal to 10 degrees?
P: 15,319 How to interpret a negative y-intercept? Can you describe your data a little? I'd like to see how your data led to a negative y-intercept.
Emeritus
PF Gold
P: 5,196
Quote by JeremyWsmith So say the cricket chrips 0 times a minute at 10 degrees, could I say: The y-intercept is undefined because the temperature must be above or equal to 10 degrees?
No, there is nothing preventing the temperature from being below 10 degrees. Did you read the edit I made to my post? And the y-intercept is certainly not undefined. After all, if the cricket chirp rate is zero at 10 degrees, then presumably it is zero at every temperature below that as well. What I'm saying is that the true y-intercept cannot be obtained by simply extrapolating the best fit line of the data down to T = 0, because the cricket chirp rate is not linear with a single slope over this entire temperature range. In other words, the model that most accurately describes the cricket chirp rate is not simply a single linear function, in spite of the fact that that is what the limited data set you have would indicate.
Again, all of this ASSUMES you've done the earlier parts of the homework exercise properly, which is not a given. EDIT: Yes, DaveC is right. Taking a look at what you've done so far is the best way of making that determination.
P: 10
Quote by DaveC426913 Can you describe your data a little? I'd like to see how your data led to a negative y-intercept.
Where Temperature is the Independent Variable
Chirps Per Minute : Degrees Celsius
7 : 11
20 : 13
36 : 15
47 :17
61 : 19
69 : 20
Using this data I created a line if best fit. The mathematical equation of the line is:
y = 6.7x - 65
P: 15,319 What cepheid said.
P: 10 Alright, so it would it be fair to say: There is not a sufficient amount of data to accurately determine the true y-intercept, as the rate of chirps per degree Celsius does not grow or decline at a consistent enough rate. So than my equation for finding chirps per minute is only accurate between a given range of temperature. Would this mean I would have to add a limit to the temperature value in the equation? i.e (Mathematically) Y: Chirps/min X: Temperature y = 6.7x - 65 20>x>=10 Would that be correct?
P: 15,319
Quote by JeremyWsmith Alright, so it would it be fair to say: There is not a sufficient amount of data to accurately determine the true y-intercept, as the rate of chirps per degree Celsius does not grow or decline at a consistent enough rate.
Well no, even if it your data were perfectly consistent and your best-fit line were razor-sharp, you still could not report a realistic y-intercept.
P: 10 Ohh, okay. So than one should interpret a negative 65 chirps per minute as 0 chirps per a minute. Or presumably 0 chirps per a minute, though I can't be sure because I haven't been given enough data?
Emeritus | 1,168 | 4,876 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2014-35 | latest | en | 0.944519 |
https://www.gradesaver.com/textbooks/science/physics/college-physics-4th-edition/chapter-5-problems-page-186/24 | 1,723,386,414,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641002566.67/warc/CC-MAIN-20240811141716-20240811171716-00609.warc.gz | 614,434,142 | 13,057 | ## College Physics (4th Edition)
The road should be banked at an angle of $5.08^{\circ}$
We can assume that the frictional force is zero. Let $F_N$ be the normal force exerted by the road on the car. We can make an equation with the vertical components of the forces acting on the car: $\sum F_y = 0$ $F_N~cos~\theta -mg = 0$ $F_N = \frac{mg}{cos~\theta}$ The horizontal component of the normal force provides the centripetal force to keep the car moving around the curve: $F_N~sin~\theta = m~a_r$ $(\frac{mg}{cos~\theta})~sin~\theta = m~\frac{v^2}{r}$ $g~tan~\theta = \frac{v^2}{r}$ $tan~\theta = \frac{v^2}{g~r}$ $\theta = tan^{-1}(\frac{v^2}{g~r})$ $\theta = tan^{-1}\left[\frac{(26.8~m/s)^2}{(9.80~m/s^2)(825~m)}~\right]$ $\theta = 5.08^{\circ}$ The road should be banked at an angle of $5.08^{\circ}$ | 294 | 806 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2024-33 | latest | en | 0.773476 |
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#11, pg.32: Determine if $\vec{b}$ is a linear combination of $\vec{a}_1,\vec{a}_2,$ and $\vec{a}_3$ where $$\vec{a}_1 = \left[ \begin{array}{ll} 1 \\ -2 \\ 0 \end{array} \right], \vec{a}_2=\left[ \begin{array}{ll} 0 \\ 1 \\ 2 \end{array} \right], \vec{a}_3 = \left[ \begin{array}{ll} 5 \\ -6 \\ 8 \end{array} \right], \vec{b}=\left[ \begin{array}{ll} 2 \\ -1 \\ 6 \end{array} \right].$$ Solution: To write $\vec{b}$ as a linear combination of $\vec{a}_1, \vec{a}_2,$ and $\vec{a}_3$ means we must find the weights $x_1,x_2,x_3$ (if they exist) that satisfy $$x_1 \vec{a}_1 + x_2 \vec{a}_2 + x_3\vec{a}_3 = \vec{b}.$$ When the vector algebra is simplified (carry out the vector sums) we get the equation $$\left[ \begin{array}{ll} x_1 + 5x_3 \\ -2x_1 +x_2 - 6x_3 \\ 2x_2 + 8x_3 \end{array} \right] = \left[ \begin{array}{ll} 2 \\ -1 \\ 6 \end{array} \right],$$ which is a system of linear equations. We will solve this system using the augmented matrix $$\begin{array}{ll} \left[ \begin{array}{llll} 1 & 0 & 5 & 2\\ -2 & 1 & -6 & -1 \\ 0 & 2 & 8 & 6 \end{array} \right] &\stackrel{r_2^* = r_2+2r_1}{\sim} \left[ \begin{array}{llll} 1 & 0 & 5 & 2 \\ 0 & 1 & 4 & 3 \\ 0 & 2 & 8 & 6 \end{array} \right] \\ &\stackrel{r_3^* = r_3-2r_2}{\sim} \left[ \begin{array}{ll} 1 & 0 & 5 & 2 \\ 0 & 1 & 4 & 3 \\ 0 & 0 & 0 & 0 \end{array} \right] \\ \end{array}$$ If we interpret this augmented marix as a system, we get $$\left\{ \begin{array}{ll} x_1 + 5x_3 = 2 \\ x_2 + 4x_3 = 3 \\ 0=0 \end{array} \right.$$ or equivalently $$\left\{ \begin{array}{ll} x_1 = 2-5x_3 \\ x_2 = 3-4x_3 \\ 0 = 0 \end{array} \right.$$ We are free to choose any value of $x_3$ so we will choose $x_3=0$ and doing so yields the solution $$\left\{ \begin{array}{ll} x_1 = 2 \\ x_2 = 3 \\ x_3 = 0. \end{array} \right.$$ This implies that we should be able to write the vector $\vec{b}$ as $$2 \vec{a}_1 + 3\vec{a}_2 + 0\vec{a}_3 = \left[ \begin{array}{ll} 2 \\ -4 \\ 0 \end{array} \right] + \left[ \begin{array}{ll} 0 \\ 3 \\ 6 \end{array} \right] = \left[ \begin{array}{ll} 2 \\ -1 \\ 6 \end{array} \right]= \vec{b},$$ as was to be shown.
NOTE: Every different choice of $x_3$ yields a different linear combination that yields $\vec{b}$. Here we call $x_3$ a free variable.
Additional Problem (A): Solve the matrix equation $A \vec{x}=\vec{b}$ where $A = \left[ \begin{array}{ll} 1 & 0 & 1 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{array} \right]$ and $\vec{b}=\left[ \begin{array}{ll} b_1 \\ b_2 \\ b_3 \end{array} \right]$.
NOTE: Many people wrote the matrix in this problem as $\left[ \begin{array}{ll} 1 & 0 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array} \right]$. This may because I made a typo in class...I will not penalize correct answers that start with this matrix instead of the actual problem.
Solution: We know the solution of this equation to be equivalent to the solution of the system whose augmented matrix is $$\begin{array}{ll} \left[ \begin{array}{ll} 1 & 0 & 1 & b_1 \\ 0 & -1 & 1 & b_2 \\ 0 & 0 & -1 & b_3 \end{array} \right] &\stackrel{r_1^* = r_1+r_3}{\stackrel{r_2^*=r_2+r_3}{\sim}} \left[ \begin{array}{ll} 1 & 0 & 0 & b_1+b_3 \\ 0 & -1 & 0 & b_2+b_3 \\ 0 & 0 & -1 & b_3 \end{array} \right] \\ & \stackrel{r_2^*=-r_2}{\stackrel{r_3^*=-r_3}{\sim}} \left[ \begin{array}{llll} 1 & 0 & 0 & b_1+b_3 \\ 0 & 1 & 0 & -b_2-b_3 \\ 0 & 0 & 1 & -b_3 \end{array} \right] \end{array}$$ hence we see that the solution is given by $$\vec{x} = \left[ \begin{array}{ll} b_1+b_3 \\ -b_2-b_3 \\ -b_3. \end{array} \right]$$ | 1,503 | 3,498 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2023-23 | latest | en | 0.522304 |
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