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
string | language
string | language_score
float64 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://www.physicsforums.com/threads/linear-algebra-find-the-kernel-and-image.473242/ | 1,620,677,909,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991759.1/warc/CC-MAIN-20210510174005-20210510204005-00169.warc.gz | 994,280,310 | 14,606 | # Linear Algebra: find the Kernel and Image
## Homework Statement
(a)Find the Kernel and Image of each of the following linear transformations.
...
(iv)$$\varphi : V\rightarrow V$$ given $$\varphi(f)=f'+f$$ where V is the subspace of the space of smooth functions $$\Re\rightarrow\Re$$ spanned by sin and cos, and f' denotes the derivative.
...
## The Attempt at a Solution
first I consider an arbitrary vector $$f_1 \in V$$. then $$f_1$$ has the following form:
$$f_1=\lambda_1 sinx + \lambda_2 cosx$$ with$$\lambda_1\lambda_2\in \Re$$
Then
$$\varphi(f_1)=\lambda_1 cosx-\lambda_2 sinx + \lambda_1 sinx+\lambda_2 cosx$$
$$=(\lambda_1+\lambda_2)cosx+(lambda_1-lambda_2)sinx$$ call (*)
By definition,
$$Ker(\varphi)={f\in V: \varphi (f)=0}$$
but from the above, $$\varphi (f)=0$$ iff $$(\lambda_1+\lambda_2)=0, (\lambda_1-\lambda_2)=0$$ since this only occurs when both lambdas are zero, then
$$Ker(\varphi)={0}$$
Originally, by (*) i wanted to say that the image of phi was the whole subpsace V, but this isnt true since the coefficients depend on each other (lambda1-lambda 2 and lambda1+lambda2). Rather the image would be all $$g\in V$$ such that $$g(x)=(\lambda_1+\lambda_2)cosx+(\lambda_1-\lambda_2)sinx, \lambda_1,\lambda_2\in \Re$$
Does this seem right? I am skeptical for some reason.
accordingly a basis for the image would be $${cosx+sinx, cosx-sinx}$$
Can someone check this for me and let me know if if my argument seems right? | 465 | 1,449 | {"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} | 4.0625 | 4 | CC-MAIN-2021-21 | longest | en | 0.765883 |
https://www.numbersaplenty.com/333232 | 1,718,310,091,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861488.82/warc/CC-MAIN-20240613190234-20240613220234-00542.warc.gz | 839,746,772 | 3,240 | Search a number
333232 = 2459353
BaseRepresentation
bin1010001010110110000
3121221002221
41101112300
541130412
611050424
72555344
oct1212660
9557087
10333232
112083a9
12140a14
13b88a3
1489624
1568b07
hex515b0
333232 has 20 divisors (see below), whose sum is σ = 658440. Its totient is φ = 163328.
The previous prime is 333227. The next prime is 333233. The reversal of 333232 is 232333.
It is a happy number.
It is a super-2 number, since 2×3332322 = 222087131648, which contains 22 as substring.
It is a Harshad number since it is a multiple of its sum of digits (16).
It is not an unprimeable number, because it can be changed into a prime (333233) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 768 + ... + 1120.
It is an arithmetic number, because the mean of its divisors is an integer number (32922).
2333232 is an apocalyptic number.
It is an amenable number.
333232 is a deficient number, since it is larger than the sum of its proper divisors (325208).
333232 is a wasteful number, since it uses less digits than its factorization.
333232 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 420 (or 414 counting only the distinct ones).
The product of its digits is 324, while the sum is 16.
The square root of 333232 is about 577.2625052781. The cubic root of 333232 is about 69.3291006621.
Adding to 333232 its reverse (232333), we get a palindrome (565565).
It can be divided in two parts, 3332 and 32, that added together give a square (3364 = 582).
The spelling of 333232 in words is "three hundred thirty-three thousand, two hundred thirty-two". | 496 | 1,702 | {"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-2024-26 | latest | en | 0.893381 |
https://www.concepts-of-physics.com/mechanics/viscosity.php | 1,716,262,492,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058383.61/warc/CC-MAIN-20240521025434-20240521055434-00073.warc.gz | 627,413,809 | 7,614 | # Viscosity, Laminar Flow and Poiseuille's Law
A flow of viscous fluid is laminar if particles of the fluid move in parallel layers (laminas). All particles in a layer move with the same speed. There is a relative motion between particles of adjacent layers. The relative velocity between layers is described by velocity gradient $\mathrm{d}v/\mathrm{d}z$. The velocity gradient is the change in velocity as one moves unit distance in a direction perpendicular to the layers.
Each layer applies a viscous force on its adjacent layer. The viscous force on a layer of area $A$ located at a place having velocity gradient $\mathrm{d}v/\mathrm{d}z$ is given by \begin{align} F=-\eta A \frac{\mathrm{d}v}{\mathrm{d}z} \end{align} where $\eta$ is called viscosity or coefficient of viscosity of the fluid. It is a fluid's property with SI unit is N-s/m2 and CGS unit poise (1 poise = 0.1 N-s/m2). The dimensional formula of viscosity is ML-1T-1. The negative sign in the above equation indicates the resistive nature of the viscous force (it opposes relative motion between the layers).
The viscosity of the liquids is very high as compared to that of gases. In liquids, the viscosity is due to cohesive forces among molecules whereas in gases, it is due to diffusion. The viscosity is very sensitive to the temperature. The viscosity of liquids decreasese with an increase in temperature whereas the viscosity of gases increases with an increase in temperature. The viscosity of liquids is measured by a viscometer.
Consider the laminar flow of a fluid of viscosity $\eta$ through a pipe of radius $r$ and length $l$. If the pressure difference between two ends of the pipe is $P$ then volume flow rate (volume V of the fluid flow in time t) through the pipe is given by Poiseuille's law \begin{align} \frac{V}{t}=\frac{\pi P r^4}{8\eta l} \end{align} Poiseuille's formula can be derived by dimensional analysis. Note that the volume flow rate depends on the 4th power of the tube's radius.
## Solved Problems from IIT JEE
### Problem from IIT JEE 2003
A liquid of density 900 kg/m3 is filled in a cylindrical tank of upper radius 0.9 m and lower radius 0.3 m. A capillary tube of length $l$ is attached at the bottom of the tank as shown in the figure. The capillary has outer radius 0.002 m and inner radius $a$. When pressure $p$ is applied at the top of the tank volume flow rate of the liquid is ${8\times{10}^{-6}}\;\mathrm{m^3/s}$ and if capillary tube is detached, the liquid comes out from the tank with a velocity of ${10}\;\mathrm{m/s}$. Determine the coefficient of viscosity of the liquid. (Given : $\pi a^2={{10}^{-6}}\;\mathrm{m^2}$ and $a^2/l={2\times{10}^{-6}}\;\mathrm{m}$.)
Solution: Let $r_1={0.9}\mathrm{m}$ be the radius at the top, $v_1$ be the velocity at the top, $r_2=0.3\;\mathrm{m}$ be the radius at the bottom, and $v_2={10}\;{m/s}$ be the velocity at the bottom (when capillary tube is detached). The continuity equation, $\pi r_1^2 v_1=\pi r_2^2 v_2$, gives $v_1={{10}/{9}}\;{m/s}$. Let $p_0$ be the atmospheric pressure. Apply Bernoulli's equation between the top and bottom, \begin{align} p_0+p+\rho g H+\tfrac{1}{2}\rho v_2^2=p_0+0+\tfrac{1}{2}\rho v_1^2, \end{align} to get, \begin{align} p+\rho g H=\tfrac{1}{2}\rho(v_2^2-v_1^2)={\tfrac{4}{9}\times{10}^{5}}\mathrm{N/m^2}. \end{align} Note:Bernoulli's theorem is valid for non-viscous fluids. In this problems, the fluid is viscous (we are asked to find its viscosity). We assume that viscous effects are small when fluid flow through a large hole of radius 0.3 m (i.e., when capillary is not connected). We applied Bernoulli's equation with this assumption.
The pressure difference across the capillary is, \begin{align} \Delta p=(p_0+p+\rho gH)-p_0=p+\rho g H. \end{align} By Poiseuille's equation, volume flow rate in a capillary is, \begin{align} \label{glb:eqn:4} Q=\frac{\pi \Delta p a^4}{8\eta l}=\frac{\Delta p (\pi a^2) (a^2/l)}{8\eta}. \end{align} Substitute the values in above equation to get $\eta={\frac{1}{720}}\,\mathrm{N\,s/m^2}$.
### Problem from IIT JEE 2018
Consider a thin square plate floating on a viscous liquid in large tank. The height $h$ of the liquid in the tank is much less than the width of the tank. The floating plate is pulled horizontally with a constant velocity $u_0$. Which of the following statements is (are) true?
1. The resistive force of liquid on the plate is inversely proportional to $h$.
2. The resistive force of liquid on the plate is independent of the area of the plate.
3. The tangential (shear) stress on the floor of the tank increases with $u_0$.
4. The tangential (shear) stress on the plate varies linearly with the viscosity $\eta$ of the liquid.
Solution: In a viscous liquid, there is no relative motion between the solid surface and the liquid layer in contact with it i.e., the liquid layer moves with the speed of the solid surface. Thus, the liquid layer in contact with the plate moves with a speed $u_0$ and the liquid layer in contact with the floor of the tank is stationary. The velocity gradient is given by \begin{align} {\mathrm{d}v}/{\mathrm{d}z}\approx {(u_0-0)}/{h}={u_0}/{h}, \nonumber \end{align} where we have used the fact that $h$ is very small in comparison to the tank width. Consider a liquid layer which is in contact with the plate. The viscous drag force (backward) on this layer, due to the layer just below it, is \begin{align} F_\text{on top layer}=\eta A ({\mathrm{d} v}/{\mathrm{d}z})=\eta A ({u_0}/{h}). \nonumber \end{align}
The top layer moves with a constant speed $u_0$. Thus, the plate should apply an equal but opposite force (forward) on the top layer to balance viscous drag on it. In turn, by Newton's third law, the top layer should apply equal but opposite force (backward) force on the plate. Thus, the resistive force of liquid on the plate is \begin{align} F_\text{on plate}=F_\text{on top layer}=\eta A ({u_0}/{h}). \nonumber \end{align} The tangential (shear) stress on the plate is given by \begin{align} \sigma_\text{on plate}=F_\text{on plate}/A=\eta ({u_0}/{h}). \nonumber \end{align} Apply similar argument on liquid layer in contact with the floor of the tank. The resistive force of liquid on the floor of the tank is $F_\text{on floor}=\eta A ({u_0}/{h})$ and the tangential (shear) stress on the floor of the tank is \begin{align} \sigma_\text{on floor}=F_\text{on floor}/A=\eta ({u_0}/{h}). \nonumber \end{align}
Thus, correct options are (A), (C) and (D).
## Questions on Viscosity and Poiseuille's Law
Question 1: A layer of oil 1.5 mm thick is placed between two microscope slides. A force of 5.5×10−4 N is required to glide one slide over the other at a speed of 1 cm/s when their contact area is 6 cm2. The coefficient of viscosity of the oil is?
A. 0.183 N-s/m2
B. 0.138 N-s/m2
C. 0.128 N-s/m2
D. 0.248 N-s/m2
Question 2: A capillary tube of length l and radius r is joined to another capillary tube of length l/4 and radius r/2. A liquid flows through this series combination. If the pressure difference between the ends of the first tube is P, that between the ends of the second tube is
A. 8 P
B. 4 P
C. P
D. P/2
Question 3: Two capillary tubes of the same length but different radii r1 and r2 are fitted in parallel to the bottom of a vessel. The pressure head is P. What should be the radius of a single tube of the same length that can replace the two tubes so that the rate of flow is the same as before
A. r1 + r2
B. (r1 + r2)/2
C. r1 r2/ (r1 + r2)
D. none of these
Question 4: The viscosities of a liquid and a gas varies with increase in temperature as
A. liquid decrease, gas increase
B. liquid increase, gas decrease
C. liquid increase, gas increase
D. liquid decrease, gas decrease | 2,233 | 7,724 | {"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": 11, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-22 | latest | en | 0.891995 |
https://stackoverflow.com/questions/15347174/python-finding-prime-factors/15347389 | 1,539,789,343,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511175.9/warc/CC-MAIN-20181017132258-20181017153758-00433.warc.gz | 801,084,041 | 37,523 | # Python Finding Prime Factors
Two part question...
1) Trying to determine the largest prime factor of 600851475143, found this program online that seems to work, the problem is im having a hard time figuring out how it works exactly (i understand the basics of what the program is doing)...also if you could shed some light on any method you may know of finding prime (perhaps without testing every number) and how you method works.
The code that i found online for prime factor
``````n = 600851475143
i = 2
while i * i < n:
while n % i == 0:
n = n / i
i = i + 1
print (n)
``````
2) Why is the code so much faster than this code (the code is just to test the speed and has no real purpose other than that)
``````i = 1
while i < 100:
i += 1
``````
• are you saying the latter takes 3 seconds to iterate from 1 to 100? – Hoopdady Mar 11 '13 at 19:46
• 2nd one takes `15.3 us` on my system. – Ashwini Chaudhary Mar 11 '13 at 19:49
• how are you measuring the time? – Hoopdady Mar 11 '13 at 19:50
• For primes generator look here – f p Mar 11 '13 at 20:06
• – poke Mar 11 '13 at 20:15
This question was the first link that popped up when I googled `"python prime factorization"`. As pointed out by @quangpn88, this algorithm is wrong (!) for perfect squares such as `n = 4, 9, 16, ...` However, @quangpn88's fix does not work either, since it will yield incorrect results if the largest prime factor occurs 3 or more times, e.g., `n = 2*2*2 = 8` or `n = 2*3*3*3 = 54`.
I believe a correct, brute-force algorithm in Python is:
``````def largest_prime_factor(n):
i = 2
while i * i <= n:
if n % i:
i += 1
else:
n //= i
return n
``````
Don't use this in performance code, but it's OK for quick tests with moderately large numbers:
``````In [1]: %timeit largest_prime_factor(600851475143)
1000 loops, best of 3: 388 µs per loop
``````
If the complete prime factorization is sought, this is the brute-force algorithm:
``````def prime_factors(n):
i = 2
factors = []
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return factors
``````
• should stop when i*i > n. – Will Ness Apr 2 '14 at 20:15
• @WillNess: Agreed. In the meanwhile, I believe I found a way to achieve both correctness and early termination. Updated my answer. – Stefan Apr 2 '14 at 22:27
• great. you can get rid of `max` call if you'd turn the inner `while` into a simple `if (n%i==0): n //= i; else: i+=1`. – Will Ness Apr 3 '14 at 6:09
• @WillNess: Yes, this works and (somewhat surprisingly) is faster, at the expense of 1 more line of code in standard Python formatting. Updated my answer again. Next big thing to trade lines of code into speed would be to separate out the case `i = 2` and then use `i += 2` from `i = 3` on. But I'll leave it there. I was just concerned that the first link on Google yielded a wrong algorithm... – Stefan Apr 3 '14 at 6:42
• For odd numbers, you could do `i += 2` instead of 1, and start with `i = 3` instead of 2. Don't know how much of a performance difference that would make. – rvighne Nov 11 '14 at 6:27
Ok. So you said you understand the basics, but you're not sure EXACTLY how it works. First of all, this is a great answer to the Project Euler question it stems from. I've done a lot of research into this problem and this is by far the simplest response.
For the purpose of explanation, I'll let `n = 20`. To run the real Project Euler problem, let `n = 600851475143`.
``````n = 20
i = 2
while i * i < n:
while n%i == 0:
n = n / i
i = i + 1
print (n)
``````
This explanation uses two `while` loops. The biggest thing to remember about `while` loops is that they run until they are no longer `true`.
The outer loop states that while `i * i` isn't greater than `n` (because the largest prime factor will never be larger than the square root of `n`), add `1` to `i` after the inner loop runs.
The inner loop states that while `i` divides evenly into `n`, replace `n` with `n` divided by `i`. This loop runs continuously until it is no longer true. For `n=20` and `i=2`, `n` is replaced by `10`, then again by `5`. Because `2` doesn't evenly divide into `5`, the loop stops with `n=5` and the outer loop finishes, producing `i+1=3`.
Finally, because `3` squared is greater than `5`, the outer loop is no longer `true` and prints the result of `n`.
Thanks for posting this. I looked at the code forever before realizing how exactly it worked. Hopefully, this is what you're looking for in a response. If not, let me know and I can explain further.
• 'because the largest prime factor will never be larger than the square root of n' - why? largest prime factor of 10 is 5, and 5 is greater than the square root of 10 – Mathai Aug 14 '13 at 20:13
• What about the case when `n=4`? This seems like it would print 4 as a prime – alxbl Aug 20 '13 at 6:01
• @Mathai I'm guessing Will meant the smallest prime factor, see: math.stackexchange.com/questions/102755/… – tsiki Dec 10 '13 at 14:11
• By this, the largest prime factor of 8 is 1! – Skylar Saveland Feb 24 '14 at 21:36
• @Mathai because we divide the divisors out of the number, we can stop when i*i > n. Then the last `n` is the biggest factor of the original number (if we replace the inner `while` with an `if`: `if n%i==0: n=n/i else: i=i+1`). – Will Ness Apr 2 '14 at 20:34
It looks like people are doing the Project Euler thing where you code the solution yourself. For everyone else who wants to get work done, there's the primefac module which does very large numbers very quickly:
``````#!python
import primefac
import sys
n = int( sys.argv[1] )
factors = list( primefac.primefac(n) )
print '\n'.join(map(str, factors))
``````
For prime number generation I always use `Sieve of Eratosthenes`:
``````def primes(n):
if n<=2:
return []
sieve=[True]*(n+1)
for x in range(3,int(n**0.5)+1,2):
for y in range(3,(n//x)+1,2):
sieve[(x*y)]=False
return [2]+[i for i in range(3,n,2) if sieve[i]]
In [42]: %timeit primes(10**5)
10 loops, best of 3: 60.4 ms per loop
In [43]: %timeit primes(10**6)
1 loops, best of 3: 1.01 s per loop
``````
You can use Miller-Rabin primality test to check whether a number is prime or not. You can find its Python implementations here.
Always use `timeit` module to time your code, the 2nd one takes just `15us`:
``````def func():
n = 600851475143
i = 2
while i * i < n:
while n % i == 0:
n = n / i
i = i + 1
In [19]: %timeit func()
1000 loops, best of 3: 1.35 ms per loop
def func():
i=1
while i<100:i+=1
....:
In [21]: %timeit func()
10000 loops, best of 3: 15.3 us per loop
``````
• `gmpy2` also has a fast Miller-Rabin implementation – John La Rooy Dec 21 '15 at 23:43
Isn't largest prime factor of 27 is 3 ?? The above code might be fastest,but it fails on 27 right ? 27 = 3*3*3 The above code returns 1 As far as I know.....1 is neither prime nor composite
I think, this is the better code
``````def prime_factors(n):
factors=[]
d=2
while(d*d<=n):
while(n>1):
while n%d==0:
factors.append(d)
n=n/d
d+=1
return factors[-1]
``````
• Doesn't work for 1, 2, or 3 – mabraham Jun 13 '17 at 20:10
• @mabraham As I have mentioned above, 1 is neither prime nor composite !! And it doesn't work for 2,3 because d starts from 2 !! so we can add an if condition there !! – m0rpheu5 Jun 14 '17 at 20:34
• I know all these things. You didn't seem to know the code does not work. ;-) – mabraham Jun 14 '17 at 20:52
The code is wrong with 100. It should check case i * i = n:
I think it should be:
``````while i * i <= n:
if i * i = n:
n = i
break
while n%i == 0:
n = n / i
i = i + 1
print (n)
``````
• Unfortunately, this still doesn't work if the largest prime factor occurs 3 or more times (e.g. `n = 8`). See my answer for a fix. – Stefan Apr 2 '14 at 22:35
Another way of doing this:
``````import sys
n = int(sys.argv[1])
result = []
for i in xrange(2,n):
while n % i == 0:
#print i,"|",n
n = n/i
result.append(i)
if n == 1:
break
if n > 1: result.append(n)
print result
``````
sample output :
python test.py 68
[2, 2, 17]
``````def find_prime_facs(n):
list_of_factors=[]
i=2
while n>1:
if n%i==0:
list_of_factors.append(i)
n=n/i
i=i-1
i+=1
return list_of_factors
``````
• Please provide some explanation for your code – thumbtackthief Oct 19 '17 at 14:57
• my code gives factors not multiples – Perviz Mamedov Oct 31 '17 at 8:26
You shouldn't loop till the square root of the number! It may be right some times, but not always!
Largest prime factor of 10 is 5, which is bigger than the sqrt(10) (3.16, aprox).
Largest prime factor of 33 is 11, which is bigger than the sqrt(33) (5.5,74, aprox).
You're confusing this with the propriety which states that, if a number has a prime factor bigger than its sqrt, it has to have at least another one other prime factor smaller than its sqrt. So, with you want to test if a number is prime, you only need to test till its sqrt.
• wrong. you should loop for i=2... and stop when i*i > n. You just need to adjust what you return in which case. This works for your examples either because we divide out each divisor from the number. – Will Ness Apr 2 '14 at 20:16
My code:
``````# METHOD: PRIME FACTORS
def prime_factors(n):
'''PRIME FACTORS: generates a list of prime factors for the number given
RETURNS: number(being factored), list(prime factors), count(how many loops to find factors, for optimization)
'''
num = n #number at the end
count = 0 #optimization (to count iterations)
index = 0 #index (to test)
t = [2, 3, 5, 7] #list (to test)
f = [] #prime factors list
while t[index] ** 2 <= n:
count += 1 #increment (how many loops to find factors)
if len(t) == (index + 1):
t.append(t[-2] + 6) #extend test list (as much as needed) [2, 3, 5, 7, 11, 13...]
if n % t[index]: #if 0 does else (otherwise increments, or try next t[index])
index += 1 #increment index
else:
n = n // t[index] #drop max number we are testing... (this should drastically shorten the loops)
f.append(t[index]) #append factor to list
if n > 1:
return num, f, f'count optimization: {count}'
``````
Which I compared to the code with the most votes, which was very fast
`````` def prime_factors2(n):
i = 2
factors = []
count = 0 #added to test optimization
while i * i <= n:
count += 1 #added to test optimization
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return factors, f'count: {count}' #print with (count added)
``````
TESTING, (note, I added a COUNT in each loop to test the optimization)
``````# >>> prime_factors2(600851475143)
# ([71, 839, 1471, 6857], 'count: 1472')
# >>> prime_factors(600851475143)
# (600851475143, [71, 839, 1471, 6857], 'count optimization: 494')
``````
I figure this code could be modified easily to get the (largest factor) or whatever else is needed. I'm open to any questions, my goal is to improve this much more as well for larger primes and factors.
Another way that skips even numbers after 2 is handled:
``````def prime_factors(n):
factors = []
d = 2
step = 1
while d*d <= n:
while n>1:
while n%d == 0:
factors.append(d)
n = n/d
d += step
step = 2
return factors
``````
``````"""
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
"""
from sympy import primefactors
print primefactors(600851475143)[-1]
``````
Do the following:
``````def primefactors(n):
return filter(lambda x:n%x==0, range(2, int(floor(sqrt(n)))))
``````
• This doesn't work. `primefactors(6)` returns `[]` and `primefactors(32)` returns `[2, 4]`, both of which are wrong. It should be `[2, 2]` and `[2, 2, 2, 2, 2]`. – Elmar Peise Dec 5 '16 at 23:19
``````n=int(input("Enter the number"))
if n==1 : #because the below logic doesn't work on 1
print(n)
for i in range(2 , n+1):
if n%i==0 :
n1=i #get factor
for b in range(2,n+1): #check if it is prime
if ((n1%b)==0) & (n1==b):
print(n1)
elif (n1%b)==0 or n1<b: #if not then pass
break
``````
i am sure this is the worst logic but it's all the knowledge i have in .py this program will get a number from user and prints all of it's factors numbers that are prime like for 12 it will give 2,3
• Can you explain your code a bit, so that others can understand it better? – Robert Mar 18 '17 at 20:33
• just edited if still not clear i can try harder – Muhammad Yusuf Mar 18 '17 at 20:35 | 3,885 | 12,475 | {"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} | 3.75 | 4 | CC-MAIN-2018-43 | latest | en | 0.88552 |
http://cran.usthb.dz/web/packages/adoptr/vignettes/working-with-priors.html | 1,590,354,913,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347385193.5/warc/CC-MAIN-20200524210325-20200525000325-00547.warc.gz | 31,160,451 | 4,713 | # Working with priors
Internally, adoptr is built around the joint distribution of a test statistic and the unknown location parameter of interest given a sample size, i.e. $\mathcal{L}\big[(X_i, \theta)\,|\,n_i\big]$ where $$X_i$$ is the stage-$$i$$ test statistic and $$n_i$$ the corresponding sample size. The distribution class for $$X_i$$ is defined by specifying a DataDistribution object, e.g., a normal distribution
library(adoptr)
datadist <- Normal()
To completely specify the marginal distribution of $$X_i$$, the distribution of $$\theta$$ must also be specified. The classical case where $$\theta$$ is considered fixed, emerges as special case when a single parameter value has probability mass 1.
### Discrete priors
The simplest supported prior class are discrete PointMassPrior priors. To specify a discrete prior, one simply specifies the vector of pivot points with positive mass and the vector of corresponding probability masses. E.g., consider an example where the point $$\delta = 0.1$$ has probability mass $$0.4$$ and the point $$\delta = 0.25$$ has mass $$1 - 0.4 = 0.6$$.
disc_prior <- PointMassPrior(c(0.1, 0.25), c(0.4, 0.6))
For details on the provided methods, see ?DiscretePrior.
### Continuous priors
adoptr also supports arbitrary continuous priors with support on compact intervals. For instance, we could consider a prior based on a truncated normal via:
cont_prior <- ContinuousPrior(
pdf = function(x) dnorm(x, mean = 0.3, sd = 0.2),
support = c(-2, 3)
)
For details on the provided methods, see ?ContinuousPrior.
### Conditioning
In practice, the most important operation will be conditioning. This is important to implement type one and type two error rate constraints. Consider, e.g., the case of power. Typically, a power constraint is imposed on a single point in the alternative, e.g. using the constraint
Power(Normal(), PointMassPrior(.4, 1)) >= 0.8
#> -E[Pr[x2>=c2(x1)]]<Normal<two-armed>;PointMass<0.40>> <= -0.8
If uncertainty about the true response rate should be incorporated in the design, it makes sense to assume a continuous prior on $$\theta$$. In this case, the prior should be conditioned for the power constraint to avoid integrating over the null hypothesis:
Power(Normal(), condition(cont_prior, c(0, 3))) >= 0.8
#> -E[Pr[x2>=c2(x1)]]<Normal<two-armed>;ContinuousPrior<[0,3]>> <= -0.8 | 618 | 2,368 | {"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": 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} | 3.65625 | 4 | CC-MAIN-2020-24 | latest | en | 0.797811 |
http://mathhelpforum.com/algebra/159161-finding-smallest-positive-integer-prime-factors-print.html | 1,519,065,167,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812758.43/warc/CC-MAIN-20180219171550-20180219191550-00182.warc.gz | 207,348,860 | 4,290 | # Finding the smallest positive integer with prime factors.
• Oct 11th 2010, 08:34 AM
weldon29
Finding the smallest positive integer with prime factors.
First I would like to make it clear that I'm not asking this to help with my school work, as I don't go to school and I'm self-studying. So I would appreciate it if people would help me.
So I just started doing some past examination questions and there are a few questions that I'm not sure if I'm using the right method as it took me a long time to finish them.
Example:
The numbers 168 and 324, written as the products of their prime factors, are 168 = 2^3 x 3 x 7, 324 = 2^2 x 3^4.
Find
the smallest positive integer value of n for which 168n is a multiple of 324.
------------------
Expressed as the product of prime factors, 198 = 2 x 3^2 x 11 and 90 = 2 x 3^2 x 5. Use these results to find
the smallest integer, k, such that 198k is a perfect square.
Question ens.
For the second question, I don't even know how to solve it, as my book didn't really explain it fully.
• Oct 11th 2010, 08:49 AM
Pim
So, for the first question, what you want is the smallest number, which prime factorisation contains both the factors of 168 and the factors of 324. (So, for every factor, you take whichever of the two numbers has a higher power for it and merge them)
this gives 2^3 (from 168) * 3^4 (from 324) * 7 (from 168) gives 4536. Do you understand my reasoning and why this always works?
For the second question, try and take the square root of a prime factorisation. What condition is true if and only if the number factorised is a perfect square?
• Oct 11th 2010, 06:50 PM
weldon29
Why do I have to multiply with 7 if it is not the highest power? Or am I suppose to take the two highest power if there are more than two numbers to choose from?
And I still don't understand how to solve the second question. When you say take the square root of a prime factorization, do you mean take the square root of 3^2? And I don't really understand your last sentence either.
• Oct 11th 2010, 11:26 PM
Pim
You can view the factorisation as following:
168 = 2^3 x 3^1 x 7^1, 324 = 2^2 x 3^4. What I meant by taking the highest power is that for every number in one or both of the factorisations (in this case, 2, 3, 7) you raise this number to the highest power in either of the two (168, 324) numbers.
I'll demonstrate why 2^3 * 3^4*7 is the correct answer. The best way to do so is divide by the prime factorisations of the two numbers.
$\frac{2^3*3^4*7}{2^3*3^1*7^1} = 3^3$
$\frac{2^3*3^4*7}{2^2*3^4} = 2^1*7$
Do you see the pattern and do you understand why the algebra is valid?
As for the second problem
What I meant by taking the square root of a factorisation is that you take the root of the whole factorisation. e.g. \sqrt{198} = \sqrt{2 * 3^2 * 11}
I'll give you the basis from which you can deduce "what condition must be true" for a number to be a perfect square.
- Taking the square root of a number is the same as raising it to the power 1/2
- $(a^p*b^q)^c = a^{pc}*b^{qc}$
- A number is integer if and only if it is represented by a prime factorisation with integer powers.
If you find what condition must be true, it should be possible to find the answer to your second question. Don't hesitate to ask if you really don't get it.
• Oct 15th 2010, 09:19 PM
weldon29
Sorry for not replying, but I seriously still don't understand so I was kinda avoiding looking at this page.(the only thing I really understood was taking the highest power thing.)
For one what is the pattern you mentioned? And what is the connection between 3^3 and 2^1 * 7 to 4536?
And I still don't understand the second question. :(
• Oct 16th 2010, 12:08 AM
Pim
The pattern is that 3^3 and 2*7 do not have any common divisors. Perhaps you should try and read this, Fundamental theorem of arithmetic - Wikipedia, the free encyclopedia
For the second question, lets try an example. 6084 is a perfect square. It's prime factorisation is $2^2*3^2*13^2$
Let's try taking the square root of that.
$\sqrt{2^2*3^2*13^2}$
$(2^2*3^2*13^2)^{\frac{1}{2}}$
As per the formula in my post above:
$2^1*3^1*13^1$
Do you see why for a number to be a perfect square, the powers of the primes in the prime factorisation have to be even? | 1,237 | 4,253 | {"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": 0, "codecogs_latex": 7, "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 | 0.968693 |
https://www.easycalculation.com/medical/diagnostic-test-calculator.php | 1,679,377,452,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943625.81/warc/CC-MAIN-20230321033306-20230321063306-00763.warc.gz | 814,174,475 | 7,553 | English
# Diagnostic Test Calculator
The diagnostic test evaluation is a method to evaluate the chances of being affected with diseases, calculated based on the present health conditions. In the below Diagnostic Test Calculator, enter the values for the true positive and false negative values for Disease Present category and False positive and true negative for disease absent category and click calculate to evaluate the medical diagnosis factors such as Sensitivity, Specificity and other values.
## Medical Diagnosis Evaluation | Calculator for Sensitivity
Disease
TestPresentAbsentTotal
PositiveTrue Positive False Positive
NegativeFalse Negative True Negative
Total
%
%
%
%
%
%
%
The diagnostic test evaluation is a method to evaluate the chances of being affected with diseases, calculated based on the present health conditions. In the below Diagnostic Test Calculator, enter the values for the true positive and false negative values for Disease Present category and False positive and true negative for disease absent category and click calculate to evaluate the medical diagnosis factors such as Sensitivity, Specificity and other values.
#### Formula:
Sensitivity = a / (a + b) x 100 Specificity = c / (d + c) x 100 Positive Likelihood Ratio = (Sensitivity / 100) / (1-(Specificity / 100)) Negative Likelihood Ratio = (1- (Sensitivity / 100)) / (Specificity / 100) Disease Prevalence = (a + b) / (a + b + d + c) x 100 Positive Predictive Value = Sensitivity / (Sensitivity + d) x 100 Negative Predictive Value = c / (b + c) x 100
Where, a = True Positive b = False Negative c = True Negative d = False Positive
#### Example
When a patient have disease number with true positive value of 10, false negative value of 5, non disease number with false positive value of 40 and true negative value of 45. His/her medical diagnostic test evaluation will be as follows,
Sensitivity = 10 / (10 + 5) x 100
= 66.67%
Specificity = 45 / (40 + 45) x 100
= 52.95
Positive Likelihood Ratio = (66.67 / 100) / (1-(52.9 / 100))
= 1.42%
Negative Likelihood Ratio = (1- (66.67 / 100)) / (52.9 / 100)
= 0.63
Disease Prevalence = (10 + 5) / (10 + 5 + 40 + 45) x 100
= 15%
Positive Predictive Value = 66.67 / (66.67 + 40) x 100
= 20%
Negative Predictive Value = 45 / (5 + 45) x 100
= 90%
## Related Calculators:
Note :This statistics calculator is presented for your own personal use and is to be used as a guide only. Medical and other decisions should NOT be based on the results of this calculator. Although this calculator has been tested, we cannot guarantee the accuracy of its calculations or results. | 649 | 2,610 | {"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-2023-14 | latest | en | 0.783641 |
https://bumpercarfilms.com/qa/question-can-an-object-have-0-momentum.html | 1,620,284,161,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988741.20/warc/CC-MAIN-20210506053729-20210506083729-00000.warc.gz | 187,948,349 | 9,106 | # Question: Can An Object Have 0 Momentum?
## What is the momentum of an object of mass m?
∴Momentum, P=mv..
## What is the relationship between mass and momentum?
Mass and velocity are both directly proportional to the momentum. If you increase either mass or velocity, the momentum of the object increases proportionally. If you double the mass or velocity you double the momentum.
## Can a single object have kinetic energy but zero momentum?
No , if a object has zero momentum then it doesn’t have kinetic energy because kinetic energy is half of linear momentum. where KE is kinetic energy and M is mass and V is velocity of the object.
## Is angular momentum conserved?
The conserved quantity we are investigating is called angular momentum. The symbol for angular momentum is the letter L. Just as linear momentum is conserved when there is no net external forces, angular momentum is constant or conserved when the net torque is zero.
## Where is angular momentum conserved?
Angular momentum, like energy and linear momentum, is conserved. This universally applicable law is another sign of underlying unity in physical laws. Angular momentum is conserved when net external torque is zero, just as linear momentum is conserved when the net external force is zero.
## What unit is momentum in?
The standard units for momentum are k g ⋅ m / s \mathrm{kg \cdot m/s} kg⋅m/sk, g, dot, m, slash, s, and momentum is always a vector quantity. This simple relationship means that doubling either the mass or velocity of an object will simply double the momentum.
## Is the momentum of an object always positive?
Momentum is a vector. That means that momentum is a quantity that has a magnitude (or size) and a direction. The above problem is a one dimensional problem, so the object is moving along a straight line. In situations like this the momentum is usually stated to be positive to the right, and negative to the left.
## Can you have negative momentum?
Momentum can be negative. Momentum is a vector quantity, meaning it has both magnitude and direction.
## How can you tell if momentum is conserved?
If momentum is conserved during the collision, then the sum of the dropped brick’s and loaded cart’s momentum after the collision should be the same as before the collision. The momentum lost by the loaded cart should equal (or approximately equal) the momentum gained by the dropped brick.
## What is the formula for total momentum?
Solution: The momentum, p, of the object is simply the product of its mass and its velocity: p = mv. Because no direction is specified, we are only interested in determining the magnitude of p, or p.
## Is angular momentum conserved in circular motion?
The uniform circular motion is characterized by constant speed. Hence, speed is conserved. … The particle has constant angular velocity (ω) and constant moment of inertia (I) about the axis of rotation. Hence, angular momentum (Iω) is conserved.
## Can an impulse be negative?
People forget what the sign of impulse means. Impulse is a vector, so a negative impulse means the net force is in the negative direction. Likewise, a positive impulse means the net force is in the positive direction.
## What is difference between momentum and impulse?
The momentum change of an object is the mass•velocity change. The impulse equals the momentum change.
## What does it mean when momentum is zero?
An object is said to have zero momentum when either the mass or the velocity is zero. Since the mass of the object cannot be zero, the velocity of the object must be zero for momentum to be zero. Velocity is zero for an object at rest.
## Is momentum conserved in a closed system?
1) Closed system – A closed system does not interact with its environment so there is no net external impulse. The total momentum of a closed system is conserved. That is, the total momentum of the system remains constant.
## What is the difference between linear and angular momentum?
Angular momentum is inertia of rotation motion. Linear momentum is inertia of translation motion. The big difference is that the type of motion which is related to each momentum is different. It is important to consider the place where the force related to rotation applies, which is appears as ‘r’ in the formula.
## Can an object have momentum if mechanical energy is zero?
Even if its mechanical energy is zero, the body has momentum. The potential energy and total mechanical energy of a body can be negative. … Mechanical energy comprises both potential energy and kinetic energy. Momentum is zero which means velocity is zero.
## Is it possible that the net kinetic energy for two objects be zero while the net momentum is zero?
Since kinetic energy is12mv2and momentum ismv, an object that has kinetic energy mustalso have momentum.It is possible, however, for a system of objects to have nonzero kineticenergy but zero momentum.
## Does Momentum have direction?
Momentum is a derived quantity, calculated by multiplying the mass, m (a scalar quantity), times velocity, v (a vector quantity). This means that the momentum has a direction and that direction is always the same direction as the velocity of an object’s motion.
## Can you have negative mass?
In theoretical physics, negative mass is a type of exotic matter whose mass is of opposite sign to the mass of normal matter, e.g. −1 kg. … Currently, the closest known real representative of such exotic matter is a region of negative pressure density produced by the Casimir effect.
## Is momentum conserved when there is gravity?
Momentum is conserved in case of motion under gravity because a object is pulled by the earth and the earth of pulled by the object. … Gravity is an internal force, so it cannot change the net mechanical energy but it can transform KE into PE and PE into KE. | 1,197 | 5,850 | {"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-2021-21 | latest | en | 0.893385 |
https://gateoverflow.in/321579/mean-value-theorem | 1,597,489,997,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439740838.3/warc/CC-MAIN-20200815094903-20200815124903-00513.warc.gz | 321,253,819 | 16,767 | # Mean Value Theorem
1 vote
175 views
f(x) is a differentiable function that satisfies 5 ≤ f′(x) ≤ 14 for all x. Let a and b be the maximum and minimum values, respectively, that f(11)−f(3) can possibly have, then what is the value of a+b?
in Calculus
0
can possibly have?
What?
Is the question complete?
0
Yes the question is complete. You can check it here:
https://brilliant.org/wiki/mean-value-theorem/
1
Solution:
Since f(x) is differentiable on all intervals, we can choose any two points (Here we take 11 and 3). So from the mean value theorem, we have some c such that
$f'(c) = (f(11)-f(3))/11-3$)
Now given $5 \leqslant f'(c)\leqslant14$ therefore $5 \leqslant (f(11)-f(3))/8 \leqslant14$
=> $40 \leqslant (f(11)-f(3)) \leqslant112$
Hence a = 112 and b = 40 which gives a+b = 152 as the answer.
0
@Nirmal Gaur
1
Done :)
1 vote
Since $f(x)$ is differentiable on all intervals, we can choose any two points $($ Here we take $11$ and $3)$. So from the mean value theorem, we have some $c$ such that:
$f'\left ( c \right )=\dfrac{\bigg(f(11)-f(3)\bigg)}{(11-3)}$
Now given $5\leqslant f'\left ( c \right )\leqslant 14$ therefore:
$\implies 5\leqslant \dfrac{\bigg(f(11)-f(3)\bigg)}{8}\leqslant 14$
$\implies 40\leqslant \bigg(f(11)-f(3)\bigg)\leqslant 112$
Hence $a = 112$ and $b = 40$ which gives $a+b = 152$ as the answer.
edited
## Related questions
1
251 views
Find a point on the graph of y=x3(cube) where the tangent is parallel to the chord joining (1,1) and (3,27).
If f ' (x) =$\frac{8}{x^{}2+3x+4}$ and f(0) =1 then the lower and upper bounds of f(1) estimated by Langrange 's Mean Value Theorem are ___
Question: The value of $\zeta$ of $f(b) - f(a) = (b-a) \cdot f'(\zeta)$ for the function $f(x) = Ax^2 + Bx +C$ in the interval $[a,b]$ is _______ My Attempt : STEP 1: $f(x)$ is polynomial function therefore continuous STEP 2: $f(x)$ is polynomial function therefore differentiable ... How to proceed further? | 692 | 1,960 | {"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} | 4.375 | 4 | CC-MAIN-2020-34 | latest | en | 0.702206 |
https://ccssmathanswers.com/eureka-math-grade-5-module-4-lesson-2/ | 1,726,546,675,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651739.72/warc/CC-MAIN-20240917040428-20240917070428-00898.warc.gz | 145,486,452 | 58,723 | ## Engage NY Eureka Math 5th Grade Module 4 Lesson 2 Answer Key
### Eureka Math Grade 5 Module 4 Lesson 2 Problem Set Answer Key
Question 1.
Draw a picture to show the division. Write a division expression using the unit form. Then, express your answer as a fraction. The first one is partially done for you.
a. 1 ÷ 5 = 5 fifths ÷ 5 = 1 fifth = $$\frac{1}{5}$$
1/5
Explanation:
5 fifths ÷ 5 = 1/5
b. 3 ÷ 4
3/4
Explanation:
12 fourths ÷ 4 = 3 fourths = 3/4
c. 6 ÷ 4
1 1/2
Explanation:
24 fourths ÷ 4 = 6 fourths = 6/4 = 1 2/4 = 1 1/2
Question 2.
Draw to show how 2 children can equally share 3 cookies. Write an equation, and express your answer as a fraction.
Explanation:
3 ÷ 2 = 6 halves ÷ 2 = 3 halves
3/2
1 1/2
Each child gets 1 1/2 cookies.
Question 3.
Carly and Gina read the following problem in their math class:
Seven cereal bars were shared equally by 3 children. How much did each child receive?
Carly and Gina solve the problem differently. Carly gives each child 2 whole cereal bars and then divides the remaining cereal bar among the 3 children. Gina divides all the cereal bars into thirds and shares the thirds equally among the 3 children.
a. Illustrate both girls’ solutions.
    Â
b. Explain why they are both right.
Both girls are right.
Explanation:
Here they used different strategies to solve the division problem and still ended up with the same answer.
7 ÷ 3 = 7/3 = 2 1/3 this is what carly did.
7 ÷ 3 =21 thirds ÷ 3 = 7 = 7/3 = 2 1/3 thirds that’s what gina did.
Question 4.
Fill in the blanks to make true number sentences.
a. 2 ÷ 3 =
2/3
Explanation:
The true number sentence for the given question is 2/3.
b. 15 ÷ 8 =
15/8
Explanation:
The true number sentence for the given question is 15/8.
c. 11 ÷ 4 =
11/4
Explanation:
The true number sentence for the given question is 11/4.
d. $$\frac{3}{2}$$ = _____ ÷ ______
3/2
Explanation:
The true number sentence for the given question is 3/2.
e. $$\frac{9}{13}$$ = _____ ÷ _____
9/13
Explanation:
The true number sentence for the given question is 9/13.
f. 1$$\frac{1}{3}$$ = ______ ÷ ____
1 1/3 that is 4/3
Explanation:
The true number sentence for the given question is 4/3.
### Eureka Math Grade 5 Module 4 Lesson 2 Exit Ticket Answer Key
Question 1.
Draw a picture that shows the division expression. Then, write an equation and solve.
a. 3 ÷ 9
3/9
b. 4 ÷ 3
4/3
Question 2.
Fill in the blanks to make true number sentences.
a. 21 ÷ 8 =
21/8
Explanation:
The true number sentence for the given question is 21/8.
b. $$\frac{7}{4}$$ = ______ ÷ ______
7/4
Explanation:
The true number sentence for the given question is 7/4.
c. 4 ÷ 9 =
4/9
Explanation:
The true number sentence for the given question is 4/9.
d. 1$$\frac{2}{7}$$ = ______ ÷ ______
1 2/7
Explanation:
The true number sentence for the given question is 9/7.
### Eureka Math Grade 5 Module 4 Lesson 2 Homework Answer Key
Question 1.
Draw a picture to show the division. Express your answer as a fraction.
a. 1 ÷ 4
1/4
Explanation:
b. 3 ÷ 5
3/5
Explanation:
3/5
c. 7 ÷ 4
7/4
Explanation:
7 ÷ 4 = 7/4 = 1 3/4
Question 2.
Using a picture, show how six people could share four sandwiches. Then, write an equation and solve.
4/6 = 2/3
Explanation:
Question 3.
Fill in the blanks to make true number sentences.
a. 2 ÷ 7 =
2/7
Explanation:
The true number sentence for the given question is 2/7.
b. 39 ÷ 5 =
39/5
Explanation:
The true number sentence for the given question is 39/5.
c. 13 ÷ 3 =
13/3
Explanation:
The true number sentence for the given question is 13/3.
d. $$\frac{9}{5}$$ = ______ ÷ ______
9/5
Explanation:
The true number sentence for the given question is 9/5.
e. $$\frac{19}{28}$$ = ______ ÷ ______
f. 1$$\frac{3}{5}$$ = ______ ÷ ______ | 1,230 | 3,791 | {"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} | 4.9375 | 5 | CC-MAIN-2024-38 | latest | en | 0.890147 |
https://mcqslearn.com/mba/statistics/quiz/quiz-questions-and-answers.php?page=2 | 1,713,789,319,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818293.64/warc/CC-MAIN-20240422113340-20240422143340-00277.warc.gz | 330,036,158 | 20,161 | MBA Business Statistics Certification Exam Tests
MBA Business Statistics Practice Test 2
# Chow-test Model MCQ (Multiple Choice Questions) PDF - 2
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## Chow-test Model Questions and Answers : Quiz 2
MCQ 6: To test the stability of parameters and estimating the source of change i.e. in slope or in intercept is possible through
1. Binomial coefficient approach
2. Phi-coefficient approach
3. Dummy variable approach
4. Accurate Chow approach
MCQ 7: An element in the distribution of one or more regressors included in the model that generates heteroscedasticity is known as
1. Skewness
2. Residual term
3. Variance
4. Correlation
MCQ 8: In the exponential model, if independent variable 'x' increases by 1%, y will increase by
1. 0.01
2. 1 unit
3. β2 %
4. β2 units
MCQ 9: The time series increases in
1. Decreasing fashion
2. Linear fashion
3. Nonlinear fashion
4. Residual fashion
MCQ 10: For data set x equals to 95, 85, 80, 70, 60 and y equals to 85, 95, 70, 70, 65, what's the variance of intercept(β1) is?
1. 931.3559
2. 932.3559
3. 933.238
4. 934.238
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Human Resource Management (MBA) App (Android & iOS) | 651 | 2,661 | {"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-2024-18 | latest | en | 0.763544 |
https://www.kopykitab.com/blog/vtu-previous-exam-papers-be-cs-4th-semester-analysis-and-design-of-algorithms-july-2006/ | 1,621,164,377,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991269.57/warc/CC-MAIN-20210516105746-20210516135746-00170.warc.gz | 886,608,374 | 21,457 | # VTU Previous Exam Papers BE CS 4th Semester Analysis and Design of Algorithms July 2006
VTU CSE 4th Semester Analysis and Design of Algorithms Question Paper July 2006: You should have depth knowledge in every subject. You can boost your preparation by solving previous year question papers. We will give you information about the important chapters and concepts to be covered in all chapters.
Here we are providing you the complete guide on VTU CSE 4th Semester Analysis and Design of Algorithms Question Paper July 2006.
## VTU CSE 4th Semester Analysis and Design of Algorithms Question Paper July 2006
You must have Analysis and Design of Algorithms Question Paper along with the latest Computer Science 4th sem Syllabus to enhance your semester exam preparation.
Here you can check the VTU CSE 4th Semester Analysis and Design of Algorithms Question Paper July 2006
Note: 1. Answer any FIVE full questions.
1 a. With a neat diagram, briefly explain the design and explain the design and analysis of an algorithm.
b. Explain what property of the adjacency matrix of an undirected graph indicates that the graph is complete has loop / has an isolated vertex.
c. Design a recursive algorithm for computing 2n for a non negative integer n, based on the formula 2n = 2n1+2111. Set up a recurrence relation for the number of additions made by the algorithm and solve it. For n = 5, draw a tree of recursive calls for this algorithm and count the number of calls.
2 a. Explain how we can compare the order of growth of 2 functions using limits. Compare order of growth of log2ii and 4n .
b. Define asymptotic notations 0, 0, Q and prove that ^ n{n -1) e \${n2).
c. Sort the list ‘QUESTION5 in alphabetical order using the Bubble Sort algorithm.
3 a. Write down an algorithm to search for a key in a given array, using linear search. Find its best, worst and average case efficiencies.
b. State whether li e following are true or false: n2 + n + 5 <e j, n2 +1 e 0(10000«), n2 + 5 e \$(n2) n2 +1 e Q(w). (02 Marks)
c. Apply Quick Sort to sort the list “QUESTION5 in alphabetical order. Draw the tree of the recursive calls made.
4 a. Write down a recursive algorithm to compute the number of leaves in a binary tree.
b. Prove the correctness of the above algorithm in Q4(a).
c. Write an algorithm for DFS and explain how it can be used to solve topological sorting problem, with an example.
5 a. Design a presorting — based algorithm to find the mode and determine its efficiency class.
b. Construct an AVL tree by inserting the elements successively, for 3,6,5,7,1,2,8,4, staring from an empty tree.
c. Write down an algorithm to construct a heap by bottom-up method. Trace your algorithm for the list 1,8,6,5,3,7,4.
6 a. Construct a shift table for the pattern BAOBAB, and search for the same in the text BESS-KNEW-ABOUT-BAOBABS, using Horspoo Ps algorithm.
b. Briefly explain the dynamic programming technique using Floyd’s algorithm for the problem of all-pairs, shortest path as an example.
c. What are the requirements to be satisfied to apply greedy technique? Explain Prim ’ s algorithm with a suitable example. : ?
7 a. Construct a Huffman code for the following data: Character A B C D E Probability 0.4 0.1 0.2 0.14 0.16 Encode the text ABACABAD using the above code. Decode the text whose encoding is 100010111001010, using the above Huffman code
b. What is back tracking? Apply back tracing algorithm to solve the instance of the sum-of-subset problem S={1,3,4,5} and d~l 1.
c. With the help of a state-space tree, solve the following instance of the Knapsack problem by the branch-and-bound algorithm.
Item Weight Value
1 10 100 2 7 63 3 8 56 4 /j 12
8 a. What is the need for approximate algorithms? Explain, with a suitable example, the nearest neighbor algorithm.
b. Write down the decision tree for the three-element insertion sort.
c. Define the following: Decision problem, Class of NP problems, NP-Complete problem and Polemically reducible problems.
We have covered VTU CSE 4th Semester Analysis and Design of Algorithms Question Paper July 2006. Feel free to ask us any questions in the comment section below. | 1,053 | 4,188 | {"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-2021-21 | latest | en | 0.873561 |
https://fixya.net/support/t3343485-binomial_function_in_calculator_fx_911es | 1,579,467,527,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250595282.35/warc/CC-MAIN-20200119205448-20200119233448-00512.warc.gz | 451,587,461 | 38,588 | Question about Casio FX-115ES Scientific Calculator
# Binomial function in calculator fx-911ES
How to use binomial function in calculator fx-911ES
Posted by on
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• Casio Master
Hello,
The binomial probability distribution is defined as
P(r;p;n) =(nCr)(p^r)*(1-p)^(n-r)
where n is the number of trials, p the probability of success, and r the expected result.
Let n=20, r=7, p=0.15 ( I do not know wether this exemple has any meaning in the context of your problem, but you have to enter values that mean something to you. I am only showing you the key strokes
To enter 20C7 you press 20 [SHIFT][nCr]7 ;
To enter 0.15 to the power 7 you type 0.15[X to ] 7 the key is between
[x²] and [log]
To enter (1-0.15) to power 20-7, you type 0.85 [X to] 13
With [*] standing for multiplication key , and [X to] the raise to power key, the exemple above can be entered as
( 20 [SHIFT][nCr] 7) [*] ( 0.15 [X to] 7 ) [*] ( 0.85 [X to] 13 ) [=]
Here is a screen capture to show you what it looks like. However on this calculator the combination 20 [SHIFT][nCr] 7 is represented as nCr(20,7).
Hope it helps
Posted on Nov 15, 2009
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• Contributor
HEY I AM USING FX911ES,
i ant to know steps for linear interpolation,
given= X Y
3.0 5800
3.2 6000
no to find value of 3.1... in my old calculator it was.
(fx911 MS) STEPS WERE.....
mode(2times)- REG- linear
3.0,5800 m+ (n1)
3.2,6000 m+ (n2)
3.1 (reg 2times) KITES sign =
now please show steps of fx911MS...
Posted on Apr 16, 2011
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An expert who has achieved level 1.
New Friend:
An expert that has 1 follower.
• Contributor
How to solve this in my fx911ES calc
attached in the picture...i want to get the value of RB
Posted on Jul 31, 2011
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## Related Questions:
### Perform factoring, e.g binomials, trinomials, by grouping
This is a scientific calculator that can perform many operations on numbers but not on letter symbols. It does not have a Computer Algebra System.
Oct 27, 2015 | Casio Office Equipment & Supplies
### How am i supposed to solve a quadratic equation ?
Use the quadratic formula, or factor the quadratic polynomial. Once factored into a product of two first degree binomials, the roots are obtained by setting (in TURN) each binomial factor equal to zero.
May 02, 2014 | Casio fx-300ES Calculator
### Texas 30XIIB binomial cdf
The only known equation for the cumulative binomial distribution is the sum of the individual binomial probabilities. Some more sophisticated (and more expensive) calculators have that equation built in, but the 30xii does not.
If n>30 and n*p>5 and n*(1-p)>5 then you can approximate the cumulative binomial with the normal probability function, but again the 30xii does not have that built in.
Apr 14, 2014 | Texas Instruments TI-30 XIIS Calculator
### How do you do permutation problems, combination problems, and binomial probability problems on it? What keys do u press?
Use the key next to which is the marking nPr, for permutations, and the key nCr for combinations. In all probability, the functions are accessed by pressing the SHIFT key first. To use these functions, enter the larger number (n), press the key or key sequence then enter the smaller number (r).
I cannot say about the binomial function (I do not have the hardware with me), and do not really want to second guess you: you do not specify the type of problems you have in mind.
Aug 04, 2011 | Casio FX-300MS Calculator
### When i enter the binomial probability formula into the casio fx-115es it will not operate. it says that the problem lies in the (n-X)!X! part of the formula. I'm writing everything exactly the way it is in...
You may be writing the expression as they appear in the formula, but the problem stems from the fact the factorial function increase rather rapidly and you cannot calculate the factorials of numbers larger that 69 which is the limit of the calculator.
However if you use the built in Combination function nCr, you will avoid the problem. In the binomial function, the n!/(r!(n-r)!) factor can be replaced by nCr or nC(n-r).
Do not use the explicit form with the factorials because you will get an overflow.
Nov 09, 2010 | Casio FX-115ES Scientific Calculator
### How do I compute binomial Probability distribution on a casio fx 115 es
Mar 07, 2010 | Casio FX-115ES Scientific Calculator
### Binomial function
Hello,
The binomial probability distribution is defined as
P(r;p;n) =(nCr)(p^r)*(1-p)^(n-r)
where n is the number of trials, p the probability of success, and r the expected result.
Let n=20, r=7, p=0.15 ( I do not know wether this exemple has any meaning in the context of your problem, but you have to enter values that mean something to you. I am only showing you the key strokes
To enter 20C7 you press 20 [SHIFT][nCr]7 ;
To enter 0.15 to the power 7 you type 0.15[X to ] 7 the key is between
[x²] and [log]
To enter (1-0.15) to power 20-7, you type 0.85 [X to] 13
With [*] standing for multiplication key , and [X to] the raise to power key, the exemple above can be entered as
( 20 [SHIFT][nCr] 7) [*] ( 0.15 [X to] 7 ) [*] ( 0.85 [X to] 13 ) [=]
Here is a screen capture to show you what it looks like. However on this calculator the combination 20 [SHIFT][nCr] 7 is represented as nCr(20,7).
Hope it helps
Nov 12, 2009 | Casio FX-115ES Scientific Calculator
### When i press the MODE button in my CASIO fx - 911ES
Hello,
Just in case : Press [SHIFT][CLR] (9) [3: ALL] to reset the calculator and try [MODE] again.
If that does not work, your calculator may be defective or a fake (not the genuine article). If at all possible, return it to where you bought it and get a refund.
Here is a picture of the FX-115ES, its face looks like the 991ES. Hope I am mistaken.
Oct 18, 2009 | Casio FX-115ES Scientific Calculator
#### Related Topics:
1,822 people viewed this question
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Level 3 Expert | 1,823 | 6,648 | {"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} | 3.796875 | 4 | CC-MAIN-2020-05 | latest | en | 0.880376 |
https://www.crazy-numbers.com/en/2731 | 1,586,244,466,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371675859.64/warc/CC-MAIN-20200407054138-20200407084638-00244.warc.gz | 820,945,015 | 4,025 | Discover a lot of information on the number 2731: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 2731
Is 2731 a prime number? Yes
Is 2731 a perfect number? No
Number of divisors 2
List of dividers 1, 2731
Sum of divisors 2732
## How to write / spell 2731 in letters?
In letters, the number 2731 is written as: Two thousand seven hundred and thirty-one. And in other languages? how does it spell?
2731 in other languages
Write 2731 in english Two thousand seven hundred and thirty-one
Write 2731 in french Deux mille sept cent trente et un
Write 2731 in spanish Dos mil setecientos treinta y uno
Write 2731 in portuguese Dois mil setecentos trinta e um
## Decomposition of the number 2731
The number 2731 is composed of:
1 iteration of the number 2 : The number 2 (two) represents double, association, cooperation, union, complementarity. It is the symbol of duality.... Find out more about the number 2
1 iteration of the number 7 : The number 7 (seven) represents faith, teaching. It symbolizes reflection, the spiritual life.... Find out more about the number 7
1 iteration of the number 3 : The number 3 (three) is the symbol of the trinity. He also represents the union.... Find out more about the number 3
1 iteration of the number 1 : The number 1 (one) represents the uniqueness, the unique, a starting point, a beginning.... Find out more about the number 1
Other ways to write 2731
In letter Two thousand seven hundred and thirty-one
In roman numeral MMDCCXXXI
In binary 101010101011
In octal 5253
In US dollars USD 2,731.00 (\$)
In euros 2 731,00 EUR (€)
Some related numbers
Previous number 2730
Next number 2732
Next prime number 2741
## Mathematical operations
Operations and solutions
2731*2 = 5462 The double of 2731 is 5462
2731*3 = 8193 The triple of 2731 is 8193
2731/2 = 1365.5 The half of 2731 is 1365.500000
2731/3 = 910.33333333333 The third of 2731 is 910.333333
27312 = 7458361 The square of 2731 is 7458361.000000
27313 = 20368783891 The cube of 2731 is 20368783891.000000
√2731 = 52.258970521816 The square root of 2731 is 52.258971
log(2731) = 7.9124231214737 The natural (Neperian) logarithm of 2731 is 7.912423
log10(2731) = 3.4363217001397 The decimal logarithm (base 10) of 2731 is 3.436322
sin(2731) = -0.81688172974646 The sine of 2731 is -0.816882
cos(2731) = -0.57680520074496 The cosine of 2731 is -0.576805
tan(2731) = 1.4162176913305 The tangent of 2731 is 1.416218 | 786 | 2,528 | {"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-2020-16 | latest | en | 0.756014 |
http://www.jiskha.com/display.cgi?id=1337800092 | 1,498,197,136,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320003.94/warc/CC-MAIN-20170623045423-20170623065423-00287.warc.gz | 573,275,482 | 4,176 | # Math/Prob
posted by on .
Claire takes a five-question true-false test. She hasn't studied, so she guesses at random. With her ESP she figures she has a 60% probability of getting any one answer wrong.
A. What is the probability of getting an answer wrong? (Do you think they want me to ignore the ESP nugget or take it into account?)
b. Let P(x) be the probability of getting exactly x of the five answers right. Calculate p(0) through P(5)
c. What is the probability of getting at least three answers right.
• Math/Prob - ,
We will have to assume the ESP estimation is correct.
The resulting distribution is binomial, with 5 questions, and probability of success p=1-0.6=0.4, q=1-p=0.6 (failure).
The probability of getting exactly n question correct (out of 5) is given by the binomial coefficient
C(5,n)p^n q^(5-n)
where C(5,n)=5!/(n!(5-n)!)
Example:
probability of getting 3 answers right
=P(3) = [5!/(3!2!)]0.4^3 0.6^2
=0.2304
To calculate the probability of getting at least three correct would be:
P(3)+P(4)+P(5)=0.317
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### Related Questions
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Post a New Question | 325 | 1,166 | {"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-2017-26 | latest | en | 0.922498 |
http://www.stat.wisc.edu/~larget/math225/notes5.html | 1,512,963,690,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948512121.15/warc/CC-MAIN-20171211033436-20171211053436-00280.warc.gz | 433,546,619 | 2,003 | # Math 225
## Introduction to Biostatistics
### Highlights from Lecture #5
In this lecture I solved several probability problems, some in multiple ways.
### Chapter 3
##### Rabbit problems
1. A box contains six rabbits, one of which has been inoculated and five of which have not. If one rabbit is sampled, what is the chance that it has been inoculated?
Solution: 1/6.
2. A box contains six rabbits, two of which have been inoculated and four of which have not. If two rabbits are sampled at random, what is the probability that both are inoculated?
Solution using events:
```P(both are inoculated)
= P(1st in inoculated and 2nd is inoculated)
= P(1st is inoculated) P(2nd is inoculated | 1st is inoculated)
= (2/6) * (1/5) = 2/30 = 1/15.
```
Solution using combinations:
We must choose two of two inoculated rabbits and zero of four rabbits that have not been inoculated.
P(both are inoculated) = 2C2 * 4C0 / 6C2
= (2!/(2!0!)) * (4!/(0!4!)) / (6!/(2!4!)) = 1/15.
Solution using permutations:
With permutations we consider the order in which the rabbits are selected.
P(both are inoculated) = 2P2 / 6P2
= (2*1) / (6*2) = 1/15.
3. ##### Sensitivity and Specificity (and related definitions)
Consider a two by two table which partitions individuals by a positive or negative screening and presence or absence of a disease.
``` | disease (D) | no disease (~D) | total
-----------------------------------------------------------
positive screen (S) | a | b | a+b
negative screen (~S)| c | d | c+d
-----------------------------------------------------------
total | a+c | b+d | a+b+c+d = N
```
We define the following terms.
prevalence rate = (a+c)/N = P(D).
sensitivity = a/(a+c) = P(S|D).
specificity = d/(b+d) = P(~S|~D).
false positive rate = b/(b+d) = P(S|~D).
false negative rate = c/(a+c) = P(~S|D).
predictive value = a/(a+b) = P(D|S).
Notice that teh false positive rate is 1 - specificity and the false negative rate is 1 - sensitivity.
Bayes' rule is useful for finding the predictive value or yield.
4. Suppose that P(D) = 0.01, P(S|D) = 0.95, and P(~S|~D) = 0.97. What is the predictive value, or P(D|S)?
Solution:
``` S 0.0095
/
0.95/
/
D <
/ \
/ 0.05\
0.01/ \
/ ~S 0.0005
/
\
\ S 0.0297
0.99\ /
\ 0.03/
\ /
~D <
\
0.97\
\
~S 0.9603
```
Thus, P(D|S) = P(D and S) / P(S) = 0.0095 / (0.0095 + 0.0297) = 0.242.
A positive screen (or presence of a symptom) increases the risk of disease from 1% to over 24%. However, it is still true that about three of every four posaitive screens are false, because the disease incidence is so low.
Last modified: January 25, 2001
Bret Larget, [email protected] | 846 | 2,790 | {"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-2017-51 | latest | en | 0.920546 |
https://www.htmlgoodies.com/asp/asp-primer-functional-vbscript/ | 1,726,030,285,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651343.80/warc/CC-MAIN-20240911020451-20240911050451-00304.warc.gz | 753,980,873 | 66,480 | Wednesday, September 11, 2024
# ASP Primer: Functional VBScript
Use these bookmarks to jump around the tutorial:
You’ve gotten a handle on the conditionals. You can do a loop. Now it’s time to learn about some of the built in functions that make life a whole lot easier.
And now for my
disclaimer. Many of the functions that we talk about below will have additional
features that we will not cover in this short tutorial. If you would like more
information on the functions or any of the other topics we cover in this series,
be sure to check out Microsoft’s online documentation at
http://msdn.microsoft.com.
## Do the Math
Even though you probably
won’t be doing much serious math in most of your ASP applications, it’s still a
good idea to review some of the more common math functions. Being that as it
may, here are the more common VBScript mathematical functions:
Mathematics: ABS( ) This stands for ABSolute value. An absolute value, if you remember from math class, is simply taking any number and eliminating the sign so that any number returned here is essentially a positive number. . INT( ) This function will probably get more use than any of the other mathematical functions we list here with maybe the exception of the randomizing functions below. What INT does is simply turn any decimal number into an integer (whole) number. It does this by whacking off everything after the decimal point. That means there is no rounding going on here, it simply throws away the fractional part of any number that you give it. For example, if you give INT( ) the number 7.694, it will return then number 7. LOG( ) This function returns the natural logarithm of the number that you give it. Trigonometry: SIN( ) I’m sure you already guessed that this is sine. Now, I’m not going to go into detail on the uses of sine because you probably already know what sine is used for. If you don’t, you may want to see if you can dig up one of your old math books before you start plugging in this function. COS( ) This function returns the cosine of the angle that you give it. TAN( ) This function returns the tangent of the angle that you give it. ATN( ) This function returns the arctangent of the number that you give it. Miscellaneous: RANDOMIZE This particular math function does not actually return a result. What Randomize does is set a seed for random number generation. In short, it gets the ball rolling so that you can generate random numbers. RND( ) This is the function that actually generates a random number. Unlike our other math functions above, you don’t actually give this function a number to crunch. To see how RANDOMIZE and RND( ) work hand-in-hand, check out our practical example below.
Before we get into the
randomizing functions, let’s take a second to look at how the other mathematical
functions work. Using the mathematical functions above is quite simple. Here’s
an example of how to use TAN( ):
decTangent =
TAN(
intAngle)
It’s as simple as giving
the function a number to crunch and letting it do the rest. The above example
applies to pretty much every function that we mentioned above with the exception
of the randomizing functions.
RANDOMIZE and
RND( )
are a bit different. In order to effectively demonstrate them in use
I’ve come up with this installments practical example, a randomly displayed
<% Option
Explicit %>
<% ‘ Banner Ad
Rotation %>
<% Dim decRandomNumber %>
<% RANDOMIZE %>
<% decRandomNumber = RND( ) %>
<% If decRandomNumber > .5 then %>
<IMG width="468" height="60" src="BannerAd1.jpg">
</a>
<% Else %>
<IMG width="468" height="60" src="BannerAd2.jpg">
</a>
<% End If %>
Now for an explanation
of how the code above works. First, we declare our variable that we are going to
use to store our random number, decRandomNumber.
Next we use the
RANDOMIZE
function. This function sets the randomizing generator in motion
by seeding it. (Random number generators work by using a complex algorithm that
requires a seed number to kick-start the process. That seed number can be taken
any number of ways but usually involves a combination of the current date and
time.) You only need to use the RANDOMIZE function once on each ASP page
since it simply gets the process started. It can be used more than once but it
isn’t necessary.
The next function that
you will see is RND( ). RND( ) is the actual function that goes
out and retrieves your random number. We set our decRandomNumber variable
equal to the RND( ) function. This will set decRandomNumber equal
to a random number between 0 and 1.
Once we have our random
number set, it is a simple process of checking the number to see if it is
greater than .5 or not. That is where the If .. Then statement comes in. If our
number is greater than .5 then we display BannerAd1, otherwise we display
Right now each ad will
be displayed 50% of the time. To give more weight to one ad over another just
change how you make the comparison. For example, if you change the first part of
the If .. Then to say decRandomNumber > .75, BannerAd1 will only be
displayed 25% of the time. You can also add additional banner ads by adding in
an ElseIf or two. (Using the If .. Then, ElseIf and Else statements can be found
here.)
## Messing with Strings
Let’s move on to some
string functions. You will probably find yourself using string functions
infinitely more than the mathematical functions above. The reason is that most
of the data you interact with on the web is in the form of strings, i.e.
people’s names, addresses, web URL’s, e-mail addresses, etc.
Here are some of the
most commonly used string functions. I have tried to give an easy to understand
example of each function. We’ll discuss the practical applications of some of
the functions later. Here they are:
Changing Strings: LCASE( ) This function does exactly what you might think. It takes any string it gets and makes it all lower case. Example: Doing LCASE("Hello World!") will give you "hello world!" back. UCASE( ) This function is, of course, the polar opposite of LCASE( ). UCASE( ) changes any given string to all upper case. Example: Taking the example above, doing UCASE("Hello World!") will give you back "HELLO WORLD!". LTRIM( ) This function gets rid of extra spaces on the left side of any string by just chopping them off. Example: If you give LTRIM( ) the string " There are 5 extra spaces to the left.", it will return a string like this "There are 5 extra spaces to the left." RTRIM ( ) This function does the opposite of LTRIM( ). It removes any extra spaces from the right side of any string. Example: The string "There are 5 spaces to the right of this string. " will be returned as "There are 5 spaces to the right of this string.". TRIM( ) TRIM( ) then is a combination of the two above. It will chop off any extra spaces before and after a given string. Example: The string " There are 5 spaces before and after this string. " will return a result of "There are 5 spaces before and after this string.". REPLACE( ) Learn this one well. REPLACE( ) allows you to replace any given set of characters in a string with another set of characters. Example: This function is a bit more complicated than the rest of the string functions above. It requires more information before it can do its thing. So, with that said, here is a some code to illustrate exactly how the function works: strMySentence = "I like ice cream!" strNewSentence = REPLACE(strMySentence,"like","love") The result of the code above would be that strNewSentence would now equal to "I love ice cream!". REPLACE( ) first wants to know what string it is to perform the replace on, then what word or set of characters it is looking for, and lastly what word or set of characters it is swapping in when it finds a match. If there is more than one match within a string it will replace all of the matches it finds. All About the String: LEN( ) This is one of the few string functions that doesn’t actually return a string. Instead it returns the length of any given string as an integer. In other words, it counts the number of characters, including spaces, and gives you the total. Example: If you do LEN("Test String") you will get the number 11 as the result. LEFT( ) This function allows you to get a selected section of a string. It is done by giving the function the string you want a piece of and the length of that piece. It then counts from the left to a specified length and returns that piece of the string. Example: LEFT("I love ASP!",6) would return the string "I love", which is 6 characters from the left including the space. RIGHT( ) This function does the exact same process as LEFT( ) except from the right. Example: RIGHT("I love ASP!",4) would return the string "ASP!", which is 4 characters from the right. MID( ) As you might expect, MID( ) allows you to take a string out of the middle of another string. For this particular function you will need a starting point and the number of characters that you want. Example: MID("I love ASP!",3,4) will give you "love" as the result. It starts at the third character "l" and snags it plus the next 3 characters for a total of 4. INSTR( ) This function allows you to search for a string within a string. It works much like REPLACE( ) does in that you give it a string to search and a string to search for. The result it returns is the character number of the beginning of the match, if it finds one. If it doesn’t find a match it will return a 0. Example: INSTR("I love ASP!","love") will return the integer 3. If you try INSTR("I love ASP!","like") it will return the integer 0. Here’s an extra little tip about INSTR( ), INSTR("I love ASP!","asp",1) will ignore case-sensitivity. In other words, the number one at the end of the function tells it to find a match ignoring case. The default is to find an exact match. The result of the example above would be 8. The result of this example, INSTR("I love ASP!","asp") would be 0, no match found.
Now that you know some
of the string functions that you have available to you, I thought it might be a
good idea to give you some common uses for them.
UCASE( ) and
LCASE( )
: These are very useful in eliminating case-sensitivity when you are
making comparisons. For example, let’s say someone enters an email address on a
form to look up their account information. You don’t want to tell them you can’t
find their account if they accidentally capitalized a character in their email.
So, to eliminate that possibility you might want to take the email address they
input on the form and transform it to all lower case with the LCASE( )
function before you make a comparison to the accounts in your database.
TRIM( ), LTRIM(
)
and RTRIM( ): These functions are excellent for "cleaning up" a
user’s input. As a rule, I always use TRIM( ) on all input I get from
users. I like to eliminate any extra spaces up front so that they don’t cause me
any problems later. I very rarely use LTRIM( ) and RTRIM( ).
REPLACE( ): This
is one of the handiest functions around. You will find it very necessary for
eliminating problems when interacting with a database using SQL. You will also
find it very useful when trying to translate carriage returns into breaks (<BR>)
in HTML. I will show you specific examples of how REPLACE( ) is a
necessity when we get to some of the database stuff later.
LEN( ): This
function I find most useful when determining whether a user’s input is too long
or too short. If you are storing user input into a database you will often have
size limitations to contend with. This function can check those limits before
you generate an error writing to your database. It is also very useful for
parsing out search strings (we’ll get to that later).
LEFT( ),
RIGHT( )
and MID( ): These functions are great for chopping up user
input into single words or phrases. They are most handy in cooperation with
LEN( )
when separating out words in a user’s search string. I sometimes use
LEFT( ) to cut off user input that is too long.
INSTR( ): I find
this function most useful in searching for illegal words. For example, let’s say
your ASP page is posting user messages to a bulletin board. The bulletin board
is family oriented and you want to ensure that there is no bad language in any
of the postings. You can use INSTR( ) to deny messages where you find
offensive language.
## Nailing Down a Date & Time
As if you haven’t had
your fill of functions yet, here are some common functions used for manipulating
date and time:
DATE( ) This function gets the current date. TIME( ) This function gets the current time. NOW( ) This function gets the current date and time. MONTH( ) This function pulls the month out of any date/time string variable and returns it as a integer. DAY( ) This function pulls the day out of any date/time string variable and returns it as a integer. YEAR( ) This function pulls the year out of any date/time string variable and returns it as a integer. HOUR( ) This function pulls the hours out of any date/time string variable and returns it as a integer. MINUTE( ) This function pulls the minutes out of any date/time string variable and returns it as a integer. SECOND( ) This function pulls the seconds out of any date/time string variable and returns it as a integer. WEEKDAY( ) This function is one of the more handy ones. It takes any date given it and tells you what day of the week it is. It does this by returning an integer number that indicates the day of the week. If a 1 is returned it is a Sunday, a 2 indicates a Monday, etc. DATEDIFF( ) This function allows you get the difference between two days. You can even select how the difference is calculated, i.e. days, hours, minutes, etc. Here’s how it works: dateDay1 = "1/1/2002" DATEDIFF("d",DATE( ),dateDay1) The example above compares today’s date with the first day of 2002 and gives us the number of days between them. The "d" lets the function know to calculate the difference in days. We could have also used "h" for hours, "m" for months, "s" for seconds, etc. to calculate the difference. DATEADD( ) As you would imagine, this function adds two dates together and gives you the result. This function needs some very specific information in order to complete its task. It must know the date it needs to add to, the type of time that is being added (i.e. days, hours, minutes, etc.), and it needs to know how much time to add. Here is an example: DATEADD("h",12,NOW( )) The function above will add 12 hours ("h") to today’s date and time. The result will be a new date and time 12 hours from now.
You probably won’t find
yourself manipulating dates and times as much as strings but you will use these
functions quite a bit. Here are some practical uses for the functions above:
DATE( ), TIME(
)
and NOW( ): You will be surprised how often you need to get the
current date and time. I have used these functions for everything from
displaying the current day to time stamping data that I send to a database. You
will probably find DATE( ) and NOW( ) the most useful.
MONTH( ), DAY(
)
, HOUR( ), MINUTE( ), YEAR( ), SECOND( ): To be
completely honest with you, I will rarely use any of these functions. It is
useful, on occasion, to get the month, day or sometimes year out of a date but I
rarely have used the HOUR( ), MINUTE( ) and SECOND( )
functions.
WEEKDAY( ): This
is a very handy function for displaying a more full and formal date. I actually
use this one quite a bit whenever I need to display a complete date.
DATEDIFF( ) and
DATEADD( ): These functions are both very commonly used. I use
DATEDIFF( )
mostly for determining whether or not to display some data that
has an expiration attached to it. For example, displaying a special message
about a current sale until the sale expires. I use the DATEADD( )
function for determining expiration dates. It is quite handy quickly calculating
expiration dates for cookies and the like.
## What’s Next?
That’s it! This
concludes your crash course in VBScript. Now, there is much more to VBScript
than we have talked about up to this point but you have the basics. We will
cover additional features of VBScript as we come across them later in this
series.
Next it’s on to some
real ASP. You’ll finally begin putting together the pieces and start creating
some full-blown ASP applications.
In the next part of this series
we will:
• Talk to the user with
Response.Write
• Give directions to the
user with Response.Redirect
• Store info about the
### Get the Free Newsletter!
Subscribe to Developer Insider for top news, trends & analysis | 3,888 | 16,603 | {"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} | 3.53125 | 4 | CC-MAIN-2024-38 | latest | en | 0.898817 |
https://brainly.com/question/146870 | 1,484,572,833,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279176.20/warc/CC-MAIN-20170116095119-00356-ip-10-171-10-70.ec2.internal.warc.gz | 791,513,319 | 8,871 | # David has \$538 in his bank account. Kavisha has \$149 less in her bank account. How much money does Kavisha have in her bank account? Does anyone know what operation I should do for this problem
2
by ruthheartbreak
2014-10-09T18:22:12-04:00
The operation you should do for this problem is substraction. If David has \$538 and Kavisha has \$149 less, than Kavisha has:
\$538 - \$149 = \$389 | 113 | 395 | {"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-2017-04 | latest | en | 0.974844 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/appendix-a-exercise-set-page-976/58 | 1,531,788,103,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589536.40/warc/CC-MAIN-20180716232549-20180717012549-00607.warc.gz | 895,153,182 | 11,801 | ## Algebra: A Combined Approach (4th Edition)
$20(x-5)(x-6)$
Step 1: $20x^{2}-220x+600$ Step 2: We factor out the common factor of $20$ from the polynomial: $20(x^{2}-11x+30)$. Step 3: Next, we factor the polynomial by grouping: $20(x^{2}-5x-6x+30)$. Step 4: $20(x(x-5)-6(x-5))$ Step 5: $20(x-5)(x-6)$ | 127 | 302 | {"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} | 4.25 | 4 | CC-MAIN-2018-30 | longest | en | 0.793322 |
https://www.brightstorm.com/math/algebra/exponents/simplifying-expressions-with-exponents-problem-1/ | 1,701,182,298,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099514.72/warc/CC-MAIN-20231128115347-20231128145347-00337.warc.gz | 780,761,583 | 25,195 | Alissa Fong
MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
Thank you for watching the video.
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Simplifying Expressions with Exponents - Problem 1
Alissa Fong
Alissa Fong
MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
Share
To simplify a variable expression with exponents, start by writing out all the exponents. For example, write out x^4 as x x x x. Next, cancel out everything you can from the numerator and denominator. Then multiply the remaining variables and constants. The result is the expression in simplest form.
The most sure fire way to approach a problem with exponents is to write everything out. Like for example this whole fraction’s being squared, so that means times itself. I’m going to write 2x to the 4th over xyz squared times itself.
Next thing I’m going to do is go through and write out every single letter so I can cancel out things that are the same on top and bottom. On top I’ll have 2x to the 4th times 2x to the 4th. On the bottom I’ll have xyzz, and then that same thing again xyzz.
The last step once you write it out is kind of fun. You get to cross out anything that is the same on top and bottom. It’s like reducing a fraction when you have the same factors on top and bottom like that x and that x can go and that x and that x can go, that was it that wasn’t too much fun because I didn’t have any much crossing out to do, but you will.
Okay let’s simplify the top, 2×2=4 dealing with my constants first. Then I have 1, 2, 3, 4, 5, 6 Xs being multiplied together so that’s like x to the 6th. On the bottom I have y times itself and then I have 1, 2, 3, 4 Zs being multiplied. Be really careful that you don’t write that as 4z, z times itself 4 times is written as z to the 4th. This is my final answer, and I know it’s in most simplified form because I don’t have any constants on top and bottom that could be reduced, nor do I have the same letter showing up on top and bottom, and I also don’t have any negative exponents. So this is equal to that just in most simplified form.
Again what I did was write it out twice because it was squared, then I wrote out every single letter so I could cancel things out that were the same on top and bottom of that fraction. | 625 | 2,572 | {"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} | 4.25 | 4 | CC-MAIN-2023-50 | latest | en | 0.92123 |
http://www.onlinemathlearning.com/percentage-increase.html | 1,490,786,075,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218190295.4/warc/CC-MAIN-20170322212950-00296-ip-10-233-31-227.ec2.internal.warc.gz | 635,719,256 | 8,987 | # Percentage Increase Or Decrease
Related Topics:
More Lessons for Arithmetic
Math Worksheets
How to find the percentage increase or decrease
Step 1: Calculate the change in value
Step 2: Divide by the original value (i.e. the value before increase or decrease)
Step 3: Multiply by 100%
Percentage increase =
## Percentage Increase
Example:
What is the percentage increase from 25 to 29.5?
Solution:
Increase = 29.5 – 25 = 4.5
Percentage increase =
Example:
Last month, Zack earned \$500. This month he earned \$520. Calculate the percentage increase in his earnings.
Solution :
Increase = \$520 – \$500 = RM20
Percentage increase =
The percentage increase in his earnings is 4%.
## Percentage Decrease
Example:
What is the percentage decrease from 20 to 15 ?
Solution:
Decrease = 20 – 15 = 4 =
Percentage decrease = = 24%
Example:
Last month, Paul earned \$500. This month he earned \$450. Calculate the percentage decrease in his earnings.
Solution :
Decrease = RM500 – RM450 = RM50
Percentage decrease =
The percentage decrease in his earnings is 10%.
The following video shows more examples of percentage problems
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. | 339 | 1,550 | {"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} | 4.0625 | 4 | CC-MAIN-2017-13 | longest | en | 0.89808 |
http://math.stackexchange.com/questions/127863/pointwise-limit-of-the-sequence-f-n | 1,429,392,946,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246636213.71/warc/CC-MAIN-20150417045716-00015-ip-10-235-10-82.ec2.internal.warc.gz | 171,514,720 | 16,124 | # Pointwise limit of the sequence $\{f_n\}$ [closed]
Determine whether the pointwise limit of the sequence $\{f_n\}$ is uniform on the indicated intervals, where $f_n(x) = x^n$.
a) $[0,1]$
b) $[0,1)$
c) $[0,l]$ where $l \in (0,1)$ is fixed.
-
## closed as off-topic by Jonas Meyer, John, Najib Idrissi, Claude Leibovici, Willie WongApr 2 at 8:25
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jonas Meyer, John, Najib Idrissi, Claude Leibovici, Willie Wong
If this question can be reworded to fit the rules in the help center, please edit the question.
Statz, STOP IT! Stop copying out multiple questions from uncited sources, showing no work of your own! That's not what this site is for! – Gerry Myerson Apr 4 '12 at 4:29
Agreed. SE community appreciates insight as much as it likes giving it. You've shown no genuine interest in the problem or provided data about your attempts to figure it out on your own. If you have - please reformulate the question with more information and provide existing research, where you are stuck and what do you wish to comprehend better. – Domagoj Pandža Apr 4 '12 at 4:33
Some warm up:
First, let's recall the definition of uniform convergence:
A sequence $\{f_m\}$ converges uniformly to to a function $f$ on the set $I$ if for every $\epsilon>0$, there is a positive integer $N$, so that $$\tag{1} |f_n(x)-f(x)|<\epsilon,\quad \text{ for all}\quad n\ge N\ \text{and}\ x\in I.$$
Please note that in the above the value of $N$ does not depend on $x$.
Let's also recall a fact that will be useful here:
Fact: If the sequence of continuous functions $\{f_n\}$ converges uniformly to $f$ on the interval $I$, then $f$ is continuous on $I$.
Now for your sequence let's first find the pointwise limit $f$ on $[0,1]$. We need to find the pointwise limit first, of course, before considering uniform convergence. Towards this end, it would be beneficial to consider the graphs of the $f_n$; below are shown the graphs of several $f_n$:
It is apparent, and can be rigorously proved that, the pointwise limit of $\{f_n\}$ is $$\tag{2} f(x)=\cases{0,&0\le x<1\cr 1,&x=1. }$$
Uniform convergence on $[0,1]$:
In view of the Fact and $(2)$, can $\{f_n\}$ converge uniformly to $f$ on $[0,1]$?
Uniform convergence on $[0,1)$:
It is true that $\{f_n\}$ does not converge uniformly to $f$ on $[0,1)$; but, as the pointwise limit function is continuous here, we cannot use the Fact to show this. However, we can do the following: Look at the graphs above and note that no matter how large $n$ is, we can select a point $x_n\in[0,1)$ such that $f_n(x_n)\ge{1\over2}$ (you could take $x_n=(1/2)^{1/n}$ here).
Do you see why this will show that $\{f_n\}$ does not converge uniformly to $f$ on $[0,1)$? (Note this would also show that $\{f_n\}$ does not converge uniformly on $[0,1]$).
Uniform convergence on $[0,l]$, $0<l<1$:
Now as for the interval $[0,l]$ with $0<l<1$, we cannot find $x_n$ as we did when considering the interval $[0,1)$. Try it...
In fact, notice that if we fix $N$ then for $n>N$ we have $$0\le f_n(x)\le f_N(x)\le f_N(l)$$ for every $x$ with $0\le x\le l$. And since $l<1$, we can make $f_N(l)$ as small as we wish. Do you see why this implies uniform convergence of $\{f_n\}$ on $[0,l]$?
- | 1,059 | 3,652 | {"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} | 3.703125 | 4 | CC-MAIN-2015-18 | latest | en | 0.903144 |
https://mersenneforum.org/showthread.php?s=046ea76dd0683d3455221fb3886dd828&t=1851 | 1,603,246,089,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107874637.23/warc/CC-MAIN-20201021010156-20201021040156-00326.warc.gz | 415,745,781 | 11,053 | mersenneforum.org > Math How do you prove a number is prime?
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2004-01-03, 16:04 #1 Alien Jan 2004 5 Posts How do you prove a number is prime? Hi all, I have a question. How do you prove a number is a prime, but I mean a number with 6 000 000 digits (for example M40)? I read somewhere that if you knew the factors of N +/- 1 you could easily prove that N is prime. How do you do this? Please tell me it's kinda interesting.
2004-01-03, 17:04 #2 GP2 Sep 2003 50258 Posts For numbers as large as 6 million digits like M40, it's only currently possible to prove it's prime if it has some special form (like N = 2P-1). For a random 6-million-digit number, it's impossible with current computer speeds. Note: quite often, when a large number is proven composite (not prime), this is done without actually finding any factor.
2004-01-04, 09:22 #3 Alien Jan 2004 5 Posts OK, that's what I ask. How do you prove that a Mersenne number (in the form N = 2 ^ P - 1) is a prime? I mean you can't just do trial factoring up to sqrt(N) right? I guess that first you try trial factoring up to, let's say, 20 000 000, then if there are no factors you try a Fermat theorem to see if it could be a prime. OK up to here good. But after that what do you do? And is that the way the GIMPS's program work? And again what is that with you knowing the factors of N - 1, then you can say if N is prime.
2004-01-04, 10:34 #4 patrik "Patrik Johansson" Aug 2002 Uppsala, Sweden 23×53 Posts See http://www.mersenne.org/math.htm for how Prime95 works, and http://www.utm.edu/research/primes/n...casLehmer.html for a proof of the Lucas-Lehmer test.
2004-01-04, 16:28 #5 dsouza123 Sep 2002 2×331 Posts Quick info on Prime95 Trial factor, use factors to 2^72 (for the larger mersennes) P-1 factor, can find larger factors. Lucas-Lehmer test, will determine if prime or composite.
2004-01-04, 17:15 #6 Alien Jan 2004 516 Posts So the real primality is proven by a Lucas-Lehmer test (and that only, right? There are no other ways?). But this is funny. I mean S(20996009) is a ridiculously large number. How do they do this? I just can't imagine it. So Prime95 doesn't check if a specific Mersenne number is a probable prime by a-PRPs and a-SPRPs? That's strange - I think it could save some time. But maybe I'm wrong. P.S. Oh now I see S0 = 4 S1 = (4 * 4 - 2) mod 127 = 14 S2 = (14 * 14 - 2) mod 127 = 67 S3 = (67 * 67 - 2) mod 127 = 42 S4 = (42 * 42 - 2) mod 127 = 111 S5 = (111 * 111 - 2) mod 127 = 0 This is how they do it. They don't calculate the whole S(20996009). That makes sense. Last fiddled with by Alien on 2004-01-04 at 17:19
2004-01-04, 18:14 #7 GP2 Sep 2003 258110 Posts M40 exponent = 20996011, by the way Lucas-Lehmer tests are very fast, but only work for numbers of the special form 2P-1. For other numbers of other special forms, there are other kinds of tests. For ordinary random numbers of no special form, it's much more difficult or currently impossible when the number is very large. There's no time gained by doing a preliminary probable-prime test on a Mersenne number instead of just doing the Lucas-Lehmer test.
2004-01-04, 20:55 #8 philmoore "Phil" Sep 2002 Tracktown, U.S.A. 100010111002 Posts To add to GP2's last comment: it would take virtually the same amount of time to do a probable prime test as to do a Lucas-Lehmer test. Since the result of a probable prime test may be ambiguous (i.e., probably prime doesn't necessarily mean the number is actually prime), the Lucas-Lehmer test, giving an unambiguous answer ("composite" or "definitely prime") is preferred. You asked how primality is proved when we know the factors of N-1 rather than N+1. Check out Chris Caldwell's Prime Pages at:http://www.utm.edu/research/primes/prove/
2004-01-05, 13:08 #9
Alien
Jan 2004
5 Posts
Quote:
Originally posted by GP2 M40 exponent = 20996011, by the way
Yes, but the Lucas-Lehmer test requires P - 2 right? So to test if 2 ^ 20996011 - 1 is prime you have to test it with S(P - 2), right? (BTW how do you do the exponent thing, I mean when the exponent is above the number in the way it is written?[sig] or something like that it was)
Quote:
Originally posted by GP2 There's no time gained by doing a preliminary probable-prime test on a Mersenne number instead of just doing the Lucas-Lehmer test.
Of course, I didn't think about it that a sort of a preliminary test is already done by choosing a prime exponent. My mistake :)
One more (hopefully last ) question: How long does it take to perform a Lucas-Lehmer test? For example on the M(40)?
Last fiddled with by Alien on 2004-01-05 at 13:10
2004-01-05, 21:17 #10 dsouza123 Sep 2002 2×331 Posts Lucus-Lehmer does p - 2 iterations, the mod 2^p - 1 would be used except Prime95 uses a special FFT (dwt) which effectively does it for free. In your example 2^7 - 1 = 127 there are 5 iterations (7 - 2 ) of s^2 - 2 mod (2^p - 1).
2004-01-05, 21:30 #11
patrik
"Patrik Johansson"
Aug 2002
Uppsala, Sweden
1101010002 Posts
Quote:
Originally posted by Alien One more (hopefully last ) question: How long does it take to perform a Lucas-Lehmer test? For example on the M(40)?
I ran a double check on M40 just for fun, and it took me ten days and a few hours on my 3.14 GHz P4.
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All times are UTC. The time now is 02:08.
Wed Oct 21 02:08:09 UTC 2020 up 40 days, 23:19, 0 users, load averages: 1.64, 1.64, 1.65 | 1,765 | 5,722 | {"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-2020-45 | latest | en | 0.889641 |
https://metanumbers.com/1856742000000 | 1,643,441,321,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300573.3/warc/CC-MAIN-20220129062503-20220129092503-00070.warc.gz | 439,949,715 | 7,836 | # 1856742000000 (number)
1,856,742,000,000 (one trillion eight hundred fifty-six billion seven hundred forty-two million) is an even thirteen-digits composite number following 1856741999999 and preceding 1856742000001. In scientific notation, it is written as 1.856742 × 1012. The sum of its digits is 33. It has a total of 15 prime factors and 224 positive divisors. There are 495,129,600,000 positive integers (up to 1856742000000) that are relatively prime to 1856742000000.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 13
• Sum of Digits 33
• Digital Root 6
## Name
Short name 1 trillion 856 billion 742 million one trillion eight hundred fifty-six billion seven hundred forty-two million
## Notation
Scientific notation 1.856742 × 1012 1.856742 × 1012
## Prime Factorization of 1856742000000
Prime Factorization 27 × 3 × 56 × 309457
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 15 Total number of prime factors rad(n) 9283710 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,856,742,000,000 is 27 × 3 × 56 × 309457. Since it has a total of 15 prime factors, 1,856,742,000,000 is a composite number.
## Divisors of 1856742000000
1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 25, 30, 32, 40, 48, 50, 60, 64, 75, 80, 96, 100, 120, 125, 128, 150, 160, 192, 200, 240, 250, 300, 320, 384, 400, 480, 500, 600, 640, 800, 960, 1200, 1600, 1920, 2400, 3200, 4800, 9600 Load all the 224 divisors
224 divisors
Even divisors 196 28 14 14
Total Divisors Sum of Divisors Aliquot Sum τ(n) 224 Total number of the positive divisors of n σ(n) 6.1649e+12 Sum of all the positive divisors of n s(n) 4.30816e+12 Sum of the proper positive divisors of n A(n) 2.75219e+10 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.36262e+06 Returns the nth root of the product of n divisors H(n) 67.4642 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,856,742,000,000 can be divided by 224 positive divisors (out of which 196 are even, and 28 are odd). The sum of these divisors (counting 1,856,742,000,000) is 6,164,904,681,960, the average is 275,218,959,01.,607.
## Other Arithmetic Functions (n = 1856742000000)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 495129600000 Total number of positive integers not greater than n that are coprime to n λ(n) 3868200000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 68209350018 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 495,129,600,000 positive integers (less than 1,856,742,000,000) that are coprime with 1,856,742,000,000. And there are approximately 68,209,350,018 prime numbers less than or equal to 1,856,742,000,000.
## Divisibility of 1856742000000
m n mod m 2 3 4 5 6 7 8 9 0 0 0 0 0 6 0 6
The number 1,856,742,000,000 is divisible by 2, 3, 4, 5, 6 and 8.
• Abundant
• Polite
• Practical
• Frugal
## Base conversion (1856742000000)
Base System Value
2 Binary 11011000001001110011100101000010110000000
3 Ternary 20120111120112122102102020
4 Quaternary 123001032130220112000
5 Quinary 220410101423000000
6 Senary 3540550412521440
8 Octal 33011634502600
10 Decimal 1856742000000
12 Duodecimal 25ba23180280
20 Vigesimal 3cabbha000
36 Base36 noz48wao
## Basic calculations (n = 1856742000000)
### Multiplication
n×y
n×2 3713484000000 5570226000000 7426968000000 9283710000000
### Division
n÷y
n÷2 9.28371e+11 6.18914e+11 4.64186e+11 3.71348e+11
### Exponentiation
ny
n2 3447490854564000000000000 6401101064284870488000000000000000000 11885193192302418999630096000000000000000000000000 22067737378261978058211183707232000000000000000000000000000000
### Nth Root
y√n
2√n 1.36262e+06 12290.9 1167.31 284.283
## 1856742000000 as geometric shapes
### Circle
Diameter 3.71348e+12 1.16663e+13 1.08306e+25
### Sphere
Volume 2.68129e+37 4.33224e+25 1.16663e+13
### Square
Length = n
Perimeter 7.42697e+12 3.44749e+24 2.62583e+12
### Cube
Length = n
Surface area 2.06849e+25 6.4011e+36 3.21597e+12
### Equilateral Triangle
Length = n
Perimeter 5.57023e+12 1.49281e+24 1.60799e+12
### Triangular Pyramid
Length = n
Surface area 5.97123e+24 7.54377e+35 1.51602e+12
## Cryptographic Hash Functions
md5 c27019b68ab349da68373b27f6e1c35a aa359ecc763def5befaa79c8aa013e6cf73b64bb 16c8da982a5e254b6b9818cd6dc529c000dd97ea5b56d91a993b9e682798f2bf 32e3d8ce58f9060d2741fd0d817cd22870771029d9aca45235c0719a1d895b592a447efba64bc91ea256bbcc8b70d0ff153dfee24b494ff2c33ec74889781ba0 d1e1e72e3eb7f1c40933093ad46e981d4d664fd9 | 1,905 | 5,055 | {"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-2022-05 | latest | en | 0.7642 |
https://www.analyzemath.com/line/parallel-slope.html | 1,555,862,945,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578531994.14/warc/CC-MAIN-20190421160020-20190421182020-00376.warc.gz | 606,421,575 | 23,656 | # Prove That the Slopes of Two Parallel Lines Are Equal
A detailed tutorial on how to prove that the slopes of two parallel lines are equal.
Problem: Prove that two parallel lines have equal slopes. Solution to Problem: The figure below shows two parallel lines L1 and L2. Points A and B are on the line L1 and points C and D are on the line L2. . Let BM and DN be parallel to the y axis and AM and CN parallel to the x axis. Triangles ABM and CDN have all their corresponding sides parallel are therefore similar. Hence BM / DN = MA / NC The above may be written BM / MA = DN / NC According to the definition of slope, BM / MA is the slope of line L1 and DN / NC is the slope of L2 and are therefore equal. Equations of Line Through Two Points And Parallel and Perpendicular. Match Linear Equations to Graphs | 192 | 812 | {"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} | 3.859375 | 4 | CC-MAIN-2019-18 | longest | en | 0.923808 |
https://testbook.com/question-answer/what-will-come-in-the-place-of-the-question-mark--60e97266fe8a4917810d6fc9 | 1,638,595,128,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362930.53/warc/CC-MAIN-20211204033320-20211204063320-00639.warc.gz | 540,396,941 | 30,875 | # What will come in the place of the question mark ‘?’ in the following question?30% of 50 + 16.66% × 48 – 20 + 13% of 200 = ?
This question was previously asked in
SBI Clerk Prelims: 10 July 2021 Shift 1 - Memory Based Paper
View all SBI Clerk Papers >
1. 29
2. 39
3. 25
4. 33
5. None of these
Option 1 : 29
## Detailed Solution
Concept used:
Follow the BODMAS rule according to the table given below:
Calculation:
⇒ 30% of 50 + 16.66% × 48 – 20 + 13% of 200 = ?
⇒ 15 + 16.66% × 48 – 20 + 26 = ?
⇒ 15 + (1/6) × 48 – 20 + 26 = ? (16.66% = 1/6)
⇒ 15 + 8 – 20 + 26 = ?
⇒ 23 – 20 + 26 = ?
⇒ 29 = ?
∴ The required answer is 29 | 278 | 637 | {"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} | 3.75 | 4 | CC-MAIN-2021-49 | latest | en | 0.512157 |
https://www.hackmath.net/en/math-problem/11921?tag_id=118 | 1,600,765,262,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400204410.37/warc/CC-MAIN-20200922063158-20200922093158-00577.warc.gz | 846,448,086 | 13,536 | # A rhombus
A rhombus has sides of length 10 cm, and the angle between two adjacent sides is 76 degrees. Find the length of the longer diagonal of the rhombus.
Correct result:
d1 = 15.7602 cm
#### Solution:
We would be very happy if you find an error in the example, spelling mistakes, or inaccuracies, and please send it to us. We thank you!
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#### You need to know the following knowledge to solve this word math problem:
We encourage you to watch this tutorial video on this math problem:
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Calculate the size of the angles of the triangle ABC, if it is given by: a = 3 cm; b = 5 cm; c = 7 cm (use the sine and cosine theorem). | 1,025 | 3,660 | {"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} | 4.125 | 4 | CC-MAIN-2020-40 | latest | en | 0.884644 |
https://www.smore.com/efz2y-mrs-rushenberg-s-classroom-blog?ref=my | 1,563,333,587,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525009.36/warc/CC-MAIN-20190717021428-20190717043428-00050.warc.gz | 846,863,786 | 13,502 | # Mrs. Rushenberg's Classroom Blog
## Important Dates
Early Out: February 21
No School: February 23
Science/Art Fair: February 26
Band Solo Night: February 27
Human Growth and Development: March 2-9
Conferences: March 3-7
In reading we have been working on two main ideas and supporting details. We also have started new book clubs and the genre is fantasy. We also just finished researching a space topic with a partner. Everyone did either a poster, Google Slides, picture book etc. After we finished, we shared out with the class.
By: Yrr Einarsdottir and Andrea Chisolm
## Writing
In writing, we are currently working on opinion writing. For an activity, we agreed or disagreed with different opinion topics. It means if you agree with a sentence that the teacher says, you go to the side you feel is your opinion. The right side means yes, the left side is disagree or no. Then someone shares why they chose that. For example, the sentence is that we should get extra recess. 4 people go on the disagree side, and 17 go on the agree side. They share out why they chose to go on agree or disagree. After people listen to that, they decide if they want to change their mind and go to the other side or stay.
By: Poorvi and Mackay
## Poetry Notebook
Right now we are working on Somplace Special in our poetry notebook. It's really good but at the same time it's hard. We have a list of stuff to do. First, we add to table of contents. Then we write 3 words and define them and write in a sentence. Then we write three thick questions. Then we visualize to illustrate.
By: Jack
## Math
In math we are learning about mean, median, mode, range, maximum, and minimum. First we take a group of numbers and put them in order. Then we see what the maximum is which means the highest value. Then we see what the minimum is and that’s the smallest value. We are also doing stem and leaf plots. The stem is the number higher than the ones place. The leaf is the number in the ones place. This helps organize it.
By: Conner H and Isabella
## Science
In science we are learning about space. In the first week, we talked about stars in Star Lab. We were talking about rotation of the Earth and the moon, and the revolution of the Earth around the Sun, and the moon around the Earth. Now we are learning about the moon phases and the average daylight hours of states and countries.
By: Abdullah and Tea
## P.E.
We are working on how to play soccer and how to handle a soccer ball. After that we’re gonna start practicing basketball. We’re going to learn how to shoot and dribble a basketball. Before we were working on soccer, we were working on hockey and how to not stay in one big clump and how to have to pass the ball to our teammates. We also had fun playing each game.
By: Isaac and Rusul
## Computers
We are doing the Olympic research, and we pick our own country. We go to CultureGrams and find our information. We also get pictures of the flag, the pinpoint of the country, how many athletes are participating in the Olympics and info about the country. We also cite our sources. That is what we are doing in Computers.
By: Brayden Williams and Connor Zipprian
## Music
In music we're learning about Opera. We listened to the voices in Opera. We learned the difference between a play and Orchestra.
By: John
## Art
In art we have been doing square 1 art. That's when you draw a picture on a special paper. Then the art teacher sends our drawing to a place where they can put our drawings on multiple objects. Soon we will be starting our clay projects. This is when the art teacher gives us a topic and we make something out of clay that fits into that topic. This years topic is emojis, so we can make any emoji we want!
By: Mei Ling and Mustafa
## Guidance
In Guidance we are learning about how to act in six grade and what to expect in six grade. We learned in six grade you can bring your phone in school. If someone text you a bad picture, you tell an adult or your parents. Another example is if a person says something mean to you have to more mature about it. Overall Guidance is a fun special.
By: Sanja and Enoch
## Band
In band with Mrs. Russell, we are working on multiple tasks at once. One of the tasks being our new big band/concert songs, which we received after our last performance. The songs we are working on are: A Little Bit of Latin, Starsplitter Fanfare, and Shark Attack. These songs will be featured at our next concert/assembly. The other feature we’re working on is our Solo Night songs, where each student gets a time to shine. We learn a song out of 2 choices, and play it as a solo. The student will be accompanied by a pianist, and there will be a judge listening as well.
By: Catie B.
## Chorus
In chorus we are learning three songs: Siyahamba (An African song), Rhythm of Life and Patterns of Sound. They are very fun to learn and it is a great opportunity. We are also doing Honor Choir which is a fancy choir that only ten people can get into.
By: Sarah and Lauren | 1,156 | 5,040 | {"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} | 3.84375 | 4 | CC-MAIN-2019-30 | latest | en | 0.963667 |
https://physics.stackexchange.com/questions/24662/what-is-displacement-position-relative-to-a-reference-point-or-change-of-positi | 1,713,852,002,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818464.67/warc/CC-MAIN-20240423033153-20240423063153-00664.warc.gz | 406,371,056 | 41,230 | # What is displacement? Position relative to a reference point or change of position
What is the "official" or most useful definition of displacement in the context of kinematics? There are two common ones:
1. Displacement is the length and direction of a line from a fixed reference point. (Basically position).
2. Displacement is the change in position.
Textbooks using the first definition frequently define velocity as $v=\Delta s/\Delta t$ (where $s$ is displacement) but then for acceleration give equations such as $s=ut+1/2at^2$ (if they were consistent, they would have to use $\Delta s$). What throws me off is that are the better textbooks.
A similar confusion comes through in Wikipedia: A displacement vector is the straight path between the initial and the final position. But velocity is defined as $v=\Delta d/\Delta t$.
Background: I am writing on a learning tool for students but different textbooks require them to learn contradicting definitions and I would appreciate your help on which is the right definition to learn.
• Displacement is difference (relative vector) between end and start points of a path. Oct 11, 2022 at 14:23
It's possible that the way in which these terms are used varies from person to person, even among professionals in the field. However, in the usage I'm familiar with, displacement is the change in position, period. Definition #2 is correct and #1 is wrong. (The length and direction of a line from a fixed reference point is just called position.)
In this usage the proper form of the constant acceleration kinematic equation would be $$\Delta \vec{x} = \vec{v}t + \frac{1}{2}\vec{a}t^2$$, or $$\vec{x} = \vec{x}_0 + \vec{v}t + \frac{1}{2}\vec{a}t^2$$, where $$\vec{x}$$ is position and $$\vec{v}$$ is initial velocity. It would be valid to write $$\vec{x} = \vec{v}t + \frac{1}{2}\vec{a}t^2$$ if you always choose the origin to be at the initial position, but that seems like an unnecessary restriction.
Alternatively, you could write the equation in terms of displacement. If you use $$\vec{s}$$ for displacement, the equation would be $$\vec{s} = \vec{v}t + \frac{1}{2}\vec{a}t^2$$. That is because $$\vec{s} = \Delta\vec{x}$$ (displacement equals change in position). If these textbooks you're using are using this notation in which $$\vec{s}$$ is displacement, then it seems very strange to write $$\vec{v} = \Delta\vec{s}/\Delta t$$. That is unconventional and probably unclear notation, although it might not necessarily be wrong.
The Wikipedia usage is fine, though, because in that formula the displacement is $$\Delta\vec{d}$$, not just $$\vec{d}$$. In $$\Delta\vec{d} = \vec{d}_f - \vec{d}_i$$, the vectors $$\vec{d}_i$$ and $$\vec{d}_f$$ could be either positions or displacements.
• If we think as a vector difference then we could also take the negative of the difference. Does initial and final imply that we should consider time? Nov 17, 2020 at 17:21
• This answer is absolutely misleading. Displacement is change in position from a fixed reference point. Jul 30, 2022 at 18:21
• @Brethlosze I could just as well say that your comment is absolutely misleading. Do you have a reference or anything to support your assertion? Jul 31, 2022 at 5:35
The official definition is that displacement is the change in position from an original position. Note that "displacement" has other meanings such as the amount of water displaced by an object in a fluid such as the water displaced by a boat.
The distance covered by an object in particular direction is called Displacement | 892 | 3,549 | {"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": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-18 | latest | en | 0.916054 |
https://gomathanswerkeys.com/mcgraw-hill-math-grade-8-lesson-3-1-answer-key/ | 1,723,478,370,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641045630.75/warc/CC-MAIN-20240812155418-20240812185418-00112.warc.gz | 222,682,672 | 42,296 | # McGraw Hill Math Grade 8 Lesson 3.1 Answer Key Changing Improper Fractions to Mixed Numbers
Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 3.1 Changing Improper Fractions to Mixed Numbers to secure good marks & knowledge in the exams.
## McGraw-Hill Math Grade 8 Answer Key Lesson 3.1 Changing Improper Fractions to Mixed Numbers
Exercises Convert to a Mixed Number
Question 1.
$$\frac{64}{3}$$
21$$\frac{1}{3}$$,
Explanation:
Given to convert $$\frac{64}{3}$$ to a mixed number,
As numerator is greater than denominator so we write in
mixed fraction as $$\frac{21 X 3 + 1}{3}$$,
therefore we get 21$$\frac{1}{3}$$.
Question 2.
$$\frac{101}{4}$$
25$$\frac{1}{4}$$,
Explanation:
Given to convert $$\frac{101}{4}$$ to a mixed number,
As numerator is greater than denominator so we write in
mixed fraction as $$\frac{25 X 4 + 1}{4}$$,
therefore we get 25$$\frac{1}{4}$$.
Question 3.
$$\frac{15}{2}$$
7$$\frac{1}{2}$$,
Explanation:
Given to convert $$\frac{15}{2}$$ to a mixed number,
As numerator is greater than denominator so we write in
mixed fraction as $$\frac{7 X 2 + 1}{2}$$,
therefore we get 7$$\frac{1}{2}$$.
Question 4.
$$\frac{52}{3}$$
17$$\frac{1}{3}$$,
Explanation:
Given to convert $$\frac{52}{3}$$ to a mixed number,
As numerator is greater than denominator so we write in
mixed fraction as $$\frac{17 X 3 + 1}{3}$$,
therefore we get 17$$\frac{1}{3}$$.
Question 5.
$$\frac{66}{12}$$
5$$\frac{6}{12}$$,
Explanation:
Given to convert 5$$\frac{6}{12}$$ to a mixed number,
As numerator is greater than denominator so we write in
mixed fraction as $$\frac{5 X 12 + 6}{12}$$,
therefore we get 5$$\frac{6}{12}$$.
Question 6.
$$\frac{137}{11}$$
12$$\frac{5}{11}$$,
Explanation:
Given to convert $$\frac{137}{11}$$ to a mixed number,
As numerator is greater than denominator so we write in
mixed fraction as $$\frac{12 X 11 + 5}{11}$$,
therefore we get 12$$\frac{5}{11}$$.
Question 7.
$$\frac{176}{16}$$
11,
Explanation:
Given to convert $$\frac{176}{16}$$ to a mixed number,
As numerator is greater than denominator so we write in
mixed fraction as $$\frac{11 X 16}{16}$$ = 11.
Question 8.
$$\frac{61}{8}$$
7$$\frac{5}{8}$$,
Explanation:
Given to convert $$\frac{61}{8}$$ to a mixed number,
As numerator is greater than denominator so we write in
mixed fraction as $$\frac{7 X 8 + 5}{8}$$,
therefore we get 7$$\frac{5}{8}$$.
Question 9.
$$\frac{121}{21}$$
5$$\frac{16}{21}$$,
Explanation:
Given to convert $$\frac{121}{21}$$ to a mixed number,
As numerator is greater than denominator so we write in
mixed fraction as $$\frac{5 X 21 + 16}{21}$$,
therefore we get 5$$\frac{16}{21}$$.
Question 10.
$$\frac{53}{2}$$
26$$\frac{1}{2}$$,
Explanation:
Given to convert $$\frac{53}{2}$$ to a mixed number,
As numerator is greater than denominator so we write in
mixed fraction as $$\frac{26 X 2 + 1}{2}$$,
therefore we get 26$$\frac{1}{2}$$.
Question 11.
$$\frac{49}{11}$$
4$$\frac{5}{11}$$,
Explanation:
Given to convert $$\frac{49}{11}$$ to a mixed number,
As numerator is greater than denominator so we write in
mixed fraction as $$\frac{4 X 11 + 5}{11}$$,
therefore we get 4$$\frac{5}{11}$$.
Question 12.
$$\frac{312}{19}$$
16$$\frac{8}{19}$$,
Explanation:
Given to convert $$\frac{312}{19}$$ to a mixed number,
As numerator is greater than denominator so we write in
mixed fraction as $$\frac{16 X 19 + 8}{19}$$,
therefore we get 16$$\frac{8}{19}$$.
Question 13.
$$\frac{98}{8}$$
12$$\frac{2}{8}$$,
Explanation:
Given to convert $$\frac{98}{8}$$ to a mixed number,
As numerator is greater than denominator so we write in
mixed fraction as $$\frac{12 X 8 + 2}{8}$$,
therefore we get 12$$\frac{2}{8}$$.
Question 14.
$$\frac{87}{7}$$
12$$\frac{3}{7}$$,
Explanation:
Given to convert $$\frac{87}{7}$$ to a mixed number,
As numerator is greater than denominator so we write in
mixed fraction as $$\frac{12 X 7 + 3}{7}$$,
therefore we get 12$$\frac{3}{7}$$.
Question 15.
$$\frac{159}{12}$$
13$$\frac{3}{12}$$,
Explanation:
Given to convert $$\frac{159}{12}$$ to a mixed number,
As numerator is greater than denominator so we write in
mixed fraction as $$\frac{13 X 12 + 3}{12}$$,
therefore we get 13$$\frac{3}{12}$$.
Question 16.
$$\frac{360}{16}$$
22$$\frac{8}{16}$$,
Explanation:
Given to convert $$\frac{360}{16}$$ to a mixed number,
As numerator is greater than denominator so we write in
mixed fraction as $$\frac{22 X 16 + 8}{16}$$,
therefore we get 22$$\frac{8}{16}$$.
Question 17.
$$\frac{74}{3}$$
24$$\frac{2}{3}$$,
Explanation:
Given to convert $$\frac{74}{3}$$ to a mixed number,
As numerator is greater than denominator so we write in
mixed fraction as $$\frac{24 X 3 + 2}{3}$$,
therefore we get 24$$\frac{2}{3}$$.
Question 18.
$$\frac{71}{4}$$
17$$\frac{3}{4}$$,
Explanation:
Given to convert $$\frac{71}{4}$$ to a mixed number,
As numerator is greater than denominator so we write in
mixed fraction as $$\frac{17 X 4 + 3}{4}$$,
therefore we get 17$$\frac{3}{4}$$.
Question 19.
Gerrie collects honey from a few beehives. She scoops out the honey with a small jar that holds $$\frac{1}{3}$$ of a cup.
Over the last two weeks Gerrie has filled this jar 158 times. How many cups of honey has she collected?
52$$\frac{2}{3}$$ cups of honey,
Explanation:
Given Gerrie collects honey from a few beehives.
She scoops out the honey with a small jar that holds $$\frac{1}{3}$$ of a cup. Over the last two weeks Gerrie has filled this jar 158 times.
So many cups of honey has she collected are
158 X $$\frac{1}{3}$$ = $$\frac{158}{3}$$
numerator is greater than denominator so we write in
mixed fraction as $$\frac{52 X 3 + 2}{3}$$,
therefore we get 52$$\frac{2}{3}$$.
Question 20.
To finish sewing her tapestry, Petra needs 142 strips of cloth that are each one quarter of a yard. How many yards of cloth is that?
35$$\frac{2}{4}$$ yards of cloth,
So many yards of cloth is that 142 X $$\frac{1}{4}$$ = $$\frac{142}{4}$$ numerator is greater than denominator,
so we write in mixed fraction as $$\frac{35 X 4 + 2}{4}$$,
therefore we get 35$$\frac{2}{4}$$. | 2,025 | 6,023 | {"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} | 4.8125 | 5 | CC-MAIN-2024-33 | latest | en | 0.654138 |
https://www.kaysonseducation.co.in/questions/p-span-sty_468 | 1,653,691,912,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663006341.98/warc/CC-MAIN-20220527205437-20220527235437-00278.warc.gz | 957,846,725 | 12,637 | In a game called “odd man out man out”, m(m > 2) persons toss a coin to determine who will buy refreshments for the entire group. A person who gets an outcome different from that of the rest of the members of the group is called the odd man out. The probability that there is a loser in any game is : Kaysons Education
# In A Game Called “odd Man Out Man Out”, m(m > 2) Persons Toss A Coin To Determine Who Will Buy Refreshments For The Entire Group. A Person Who Gets An Outcome Different From That Of The Rest Of The Members Of The Group Is Called The Odd Man Out. The Probability That There Is A Loser In Any Game Is
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## Question
### Solution
Correct option is
m /2m–1
Let A denote the event that there is an odd man out in a game. The total number of possible cases is 2m. A person is odd man out if he is alone in getting a head or a tail. The number of ways in which there is exactly one tail (head) and the rest are heads (tails) is
.
Thus, the number of favorable ways is m + 2m. Therefore,
#### SIMILAR QUESTIONS
Q1
In a hurdle race, a runner has probability p of jumping over a specific hurdle. Given that in 5 trials, the runner succeeded 3 times, the conditional probability that the runner had succeeded in the first trial is
Q2
Three integers are chosen at random without replacement from the first 20 integers. The probability that their product is even 2/19.
Q3
A box contains tickets numbered 1 to Nn tickets are drawn from the box with replacement. The probability that the largest number on the tickets is is
Q4
A box contain N coins, m of which are fair and rest are biased. The probability of getting a head when a fair coin is tossed is 1/2, when a baised coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. The probability that the coin drawn is fair is
Q5
Given that AB and C are events such that P(A) = P(B) = P(C) = 1/5, P(A B) = P(B ∩ C) = 0 and P(A ∩ C) = 1=10. The probability that at least one of the events AB or C occurs is …….
Q6
Let A and B be two events such that
Q7
A and B toss a coin alternatively till one of them gets a head and wins the game. If A begins the game, the probability B wins the game is
Q8
Suppose X B(np) and P(X = 5). If p > 1/2, then
Q9
A person is known to speak the truth 4 time out of 5. He throws a dia and reports that it is a ace. The probability that it is actually a ace is
Q10
The chance of an event happening is the square of the chance of happening of second event but the odds against the first are the cube of the odds against the second. The chance of the events: | 823 | 3,235 | {"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-2022-21 | latest | en | 0.895786 |
https://www.maths.tcd.ie/undergraduate/modules1213/MA3425.php | 1,540,228,860,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583515352.63/warc/CC-MAIN-20181022155502-20181022181002-00035.warc.gz | 981,093,639 | 6,004 | # Module MA3425: Partial differential equations I
Credit weighting (ECTS)
5 credits
Semester/term taught
Michaelmas term 2012-13
Contact Hours
11 weeks, 3 lectures including tutorials per week
Lecturer
Prof. John Stalker
Learning Outcomes
On successful completion of this module, students will be able to:
• Give and use basic definitions, e.g. order, linear, etc. Apply the concepts of symmetries and in variant solutions, at least for the Wave, Heat and Laplace Equations;
• State correctly and apply to examples the basic facts about the Wave Equation in one space dimension: Energy conservation (differential, local and global forms), existence and uniquess of solutions, finite speed of propagation. Solve the initial value problem for given data using the explicit solution. Students should also able to employ the method of reflection to solve simple boundary value problems;
• State correctly and apply to examples the basic facts about the Heat Equation in one space dimension: Maximum Principle (local and global versions), Existence and uniqueness of bounded solutions, smoothing, decay of solutions. Solve the initial value problem for given data using the explicit solution. Students should also be able to employ the method of reflection to solve simple boundary value problems;
• State correctly and apply to examples the basic facts about the Heat Equation in one space dimension: Maximum Principle (local and global versions), Existence and uniqueness of bounded solutions, smoothing, decay of solutions. Solve the initial value problem for given data using the explicit solution. Students should also be able to employ the method of reflection to solve simple boundary value problems;
• State correctly and apply to examples the basic facts about the Laplace Equation in two space dimensions: Maximum Principle (local and global versions), Existence and uniqueness of solutions to the Dirichlet problem. Solve boundary value problems using the Poisson formulae;
Module Content
• Module Content;
• Classification of partial differential equations
• Wave, Heat and Laplace Equations in low dimensions
Module Prerequisite
MA2223 - Metric Spaces, MA2224 - Lebesgue Integral, MA2326 - Ordinary Differential Equations
Assessment Detail
This module will be examined jointly with MA3426 in a 3-hour examination in Trinity term, except that those taking just one of the two modules will have a 2 hour examination. However there will be separate results for MA3425 and MA3426. Continuous assessment will contribute 10% to the final grade for the module at the annual examination session. | 519 | 2,597 | {"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} | 3.515625 | 4 | CC-MAIN-2018-43 | latest | en | 0.83374 |
http://mathforum.org/library/drmath/sets/mid_equations.html?start_at=1&s_keyid=38074759&f_keyid=38074760&num_to_see=40 | 1,475,063,544,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661367.29/warc/CC-MAIN-20160924173741-00014-ip-10-143-35-109.ec2.internal.warc.gz | 171,455,149 | 7,680 | Ask Dr. Math Middle School Archive
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One chemical contains 15% sodium (Na) and the other chemical contains 30% sodium (Na). Once they mix the two samples the resulting chemical contains 22% sodium (Na). How many milliliters (ml) of each sample must be mixed to obtain 600 ml of the new chemical? 3 Equations, 3 Variables [01/26/1998] How do you solve a system of three equations in three variables? Algebra as a Metaphor for Life [06/11/2002] I've been given 'proofs' that -1 = 1, and that 2 > 4. Can you show me why they aren't true? Algebra/Point-Slope [01/09/1998] Can you explain how to do point-slope equations? Algebra - Solving Equations [11/03/1997] I am having a hard time solving some of my algebra problems... Averaging Speeds of Individual Laps [12/12/2003] A driver drives four laps around a track at 10, 20, 30, and 60 km per hour. What is his average speed? Baseballs, Buckets, and Milk Cartons [10/09/2001] The mass of a baseball is 50g. What is the mass of a bucket when 2 buckets = 6 blocks; 1 bucket + 1 block = 2 milk cartons; and 2 baseballs = 1 milk carton. Changing Equations To Slope-Intercept Form [07/17/1998] Can you explain how to change equations into slope-intercept form? For example, 12(2x-1) - 5(3y+2) = 8? Clearing Fractions in an Equation [01/13/2005] How do I clear the fractions in 1/4(8y + 4) - 17 = -1/2(4y - 8)? I know I need to multiply by 4, but I get confused with all the parentheses involved in doing that. Combining Rates of Work [08/30/1998] It takes me 3 hours to paint a house. It takes you 5 hours to paint a house. How long will it take for both of us to paint a house? Comparing Running Rates [01/20/2005] Jack and Jane ran the 100 yard dash. When Jane won, Jack was 10 yards behind her. They raced again, but this time Jane started 10 yards behind the starting line. If they both ran at the same rate as in the first race, who won the second race? 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http://mathforum.org/ | 2,160 | 7,652 | {"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-2016-40 | longest | en | 0.813007 |
http://mathhelpforum.com/calculus/34386-l-hopitals-rule.html | 1,527,247,197,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867085.95/warc/CC-MAIN-20180525102302-20180525122302-00146.warc.gz | 185,103,262 | 10,686 | 1. ## L'Hopitals Rule
I have these 3 questions, and I think I need to use L'Hopitals rule. But I Have no idea how to go about that. Anything I've read about it confuses me.
a)
lim x->1 (ln x)/(x-1)
b) lim x-> infinity x/ln(1+2e^x)
c) lim x->0 (cot2x)(sin6x)
2. ## Ok
Originally Posted by dangerous_dave
I have these 3 questions, and I think I need to use L'Hopitals rule. But I Have no idea how to go about that. Anything I've read about it confuses me.
a)
lim x->1 (ln x)/(x-1)
b) lim x-> infinity x/ln(1+2e^x)
c) lim x->0 (cot2x)(sin6x)
L'hopital's is the EASIEST way to do this....i will do the hardest one for you $\displaystyle \lim_{x \to {\infty}}\frac{x}{ln\bigg(1+2e^{x}\bigg)}\Rightarr ow{\lim_{x \to {\infty}}\frac{1}{\frac{2e^{x}}{1+2e^{x}}}}\Righta rrow{1}$
3. Could you please explain how L'Hopitals works? I don't understand at all...
4. Originally Posted by dangerous_dave
Could you please explain how L'Hopitals works? I don't understand at all...
If you have a limit which takes the form of $\displaystyle \pm \frac \infty \infty$ or $\displaystyle \frac 00$ then the limit will be equal to the derivative of the numerator over the derivative of the denominator.
For example $\displaystyle \lim_{x\to 0} \frac {sin~x}{x}$ takes the form of 0/0 so we apply L'Hospital's rule:
$\displaystyle \lim_{x\to 0} \frac {sin~x}{x} ~~~~=~~~~ \lim_{x\to 0} \frac {\frac d{dx} sin~x}{\frac {dx}{dx}} ~~~~=~~~~ \lim_{x\to 0} \frac {cos~x}{1}~~~~=~~~~cos~0~~~~=~~~~1$
5. Wow thats a whole lot easier than other sites like to make out. Thanks heaps
6. Does "Dangerous Dave" come from the 1990 computer game?
Originally Posted by dangerous_dave
a)
lim x->1 (ln x)/(x-1)
This is nothing to do with L'Hopital rule. This is the definition of derivative of $\displaystyle \ln x$ at $\displaystyle x=1$.
c) lim x->0 (cot2x)(sin6x)
$\displaystyle \cot 2x \sin 6x = \cos 2x \left( \frac{\sin 6x}{\sin 2x} \right)$.
Note, $\displaystyle \cos 2x \to 1$.
While, $\displaystyle \frac{\sin 6x}{\sin 2x} = \frac{\sin 6x}{6x} \cdot \frac{2x}{\sin 2x}\cdot 3\to 1\cdot 1\cdot 3 = 3$ | 748 | 2,086 | {"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} | 3.96875 | 4 | CC-MAIN-2018-22 | latest | en | 0.773562 |
https://www.brainkart.com/article/The-Law-of-Accidental-Errors-and-Principles-of-Least-Squares-in-Survey_4645/ | 1,723,128,837,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640728444.43/warc/CC-MAIN-20240808124926-20240808154926-00406.warc.gz | 564,111,803 | 6,480 | Home | | Surveying | The Law of Accidental Errors and Principles of Least Squares in Survey
# The Law of Accidental Errors and Principles of Least Squares in Survey
Investigations of observations of various types show that accidental errors follow a definite law, the law of probability.
THE LAW OF ACCIDENTAL ERRORS
Investigations of observations of various types show that accidental errors follow a definite law, the law of probability. This law defines the occurrence of errors and can be expressed in the form of equation which is used to compute the probable value or the probable precision of a quantity. The most important features of accidental errors which usually occur are:
(i) Small errors tend to be more frequent than the large ones; that is they are the most probable.
(ii) Positive and negative errors of the same size happen with equal
frequency ; that is, they are equally probable.
(iii) Large errors occur infrequently and are impossible.
PRINCIPLES OF LEAST SQUARES
It is found from the probability equation that the most probable values of a series of errors arising from observations of equal weight are those for which the sum of the squares is a minimum. The fundamental law of least squares is derived from this. According to the principle of least squares, the most probable value of an observed quantity available from a given set of observations is the one for which the sum of the squares of the residual errors is a minimum. When a quantity is being deduced from a series of observations, the residual errors will be the difference between the adopted value and the several observed values,
Let V1, V2, V3 etc. be the observed values x = most probable value
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
Civil Surveying : Survey Adjustments : The Law of Accidental Errors and Principles of Least Squares in Survey | | 399 | 1,960 | {"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} | 3.671875 | 4 | CC-MAIN-2024-33 | latest | en | 0.890997 |
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Students can Download Maths Chapter 2 Percentage and Simple Interest Ex 2.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.
## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Ex 2.3
Question 1.
14 out of the 70 magazines at the bookstore are comedy magazines. What percentage of the magazines at the bookstore are comedy magazines?
Solution:
Total number of magazines in the bookstore = 100 m
Number of comedy magazines = 14
Percentage of comedy magzines = $$\frac { 14 }{ 70 }$$ × 100% = 20%
20% of the magazines are comedy magazines.
Question 2.
A tank can hold 50 litres of water. At present, it is only 30% full. How many litres of water will fill the tank, so that it is 50% full?
Solution:
Capacity of the tank = 50 litres
Amount of water filled = 30% of 50 litres = $$\frac { 30 }{ 100 }$$ × 50 = 15 litres
Amount of water to be filled = 50 – 15 = 35 litres
Question 3.
Karun bought a pair of shoes at a sale of 25%. If the amount he paid was ₹ 1000, then find the marked price.
Solution:
Let the marked price of the raincoat be ₹ P
Amount he paid at a discount of 25% = ₹ 1000
(Marked Price) – (25% of P) = 1000
P – ($$\frac { 25 }{ 100 }$$ × P) = 1000
P – $$\frac { 1 }{ 4 }$$ × P = 1000
P (1 – $$\frac { 1 }{ 4 }$$) = 1000
$$\frac { 3 }{ 4 }$$ P = 1000
P = 1000 × $$\frac { 4 }{ 3 }$$
= $$\frac { 4000 }{ 3 }$$
P = 1333.33
∴ Marked price of the shoes = ₹ 1333
Question 4.
An agent of an insurance company gets a commission of 5% on the basic premium he collects. What will be the commission earned by him if he collects ₹ 4800?
Solution:
Premium collected = ₹ 4800
Commission earned = 5% of basic premium
Commission earned for ₹ 4800 = 5% of 4800
= $$\frac { 5 }{ 100 }$$ × 4800
= ₹ 240
Commission earned = ₹ 240
Question 5.
A biology class examined some flowers in a local Grass land. Out of the 40 flowers they saw, 30 were perennials. What percentage of the flowers were perennials?
Solution:
Number of flowers examined = 40
Number of perennials = 30
Percentage = $$\frac { 30 }{ 40 }$$ × 100%
= 75%
75% of the flowers were perennials.
Question 6.
Ismail ordered a collection of beads. He received 50 beads in all. Out of that 15 beads were brown. Find the percentage of brown beads?
Solution:
Number of beads received = 50
Number of brown beads = 5
Percentage of brown beads = $$\frac { 15 }{ 50 }$$ × 100 %
= 10 %
10% of the beads was brown
Question 7.
Ramu scored 20 out of 25 marks in English, 30 out of 40 marks in Science and 68 out of 80 marks in mathematics. In which subject his percentage of marks is best?
Solution:
Ramu’s score in English = 20 out of 25
Percentage scored in English = $$\frac { 20 }{ 25 }$$ × 100 % = 80 %
Ramu’s Score in Science = 30 out of 40
Percentage scored in Science = $$\frac { 30 }{ 40 }$$ × 100 % = 75%
Ramu’s score in Mathematics = 68 out of 80
Percentage scored in Maths = $$\frac { 68 }{ 80 }$$ × 100 % = 85 %
85% > 80% > 75%.
∴ In Mathematics his percentage of marks is the best.
Question 8.
Peter requires 50% to pass. If he gets 280 marks and falls short by 20 marks, what would have been the maximum marks of the exam?
Solution:
Peters score = 280 marks
Marks needed for a pass = 20
∴ Total marks required to get a pass = 280 + 20 = 300
i.e. 50% of total marks = 300
$$\frac { 50 }{ 100 }$$ × Total marks = 300
$$\frac { 1 }{ 2 }$$ × Total Marks = 300
Total Marks = 300 × 2 = 600
Total marks of the exam = 600
Question 9.
Kayal scored 225 marks out of 500 in revision test 1 and 265 out of 500 marks in revision test 2. Find the percentage of increase in her score.
Solution:
Marks scored in revision I = 225
Marks scored in revision II = 265
Change in marks = 265 – 225 = 40
Percentage of increase in marks = 8%
Question 10.
Roja earned ₹ 18,000 per month. She utilized her salary in the ratio 2 : 1 : 3 for education, savings and other expenses respectively. Express her usage of income in percentage.
Solution:
Amount of Salary = ₹ 18,000
(i) Total number of parts of salary = 2 + 1 + 3 = 6
Salary is divided into 3 portions as $$\frac { 2 }{ 6 }$$,$$\frac { 1 }{ 6 }$$ and $$\frac { 3 }{ 6 }$$
Portion of salary used for education = $$\frac { 2 }{ 6 }$$
Salary used for education = $$\frac { 2 }{ 6 }$$ × 18,000 = ₹ 6,000
Percentage for Education = $$\frac { 6000 }{ 18000 }$$ × 100 = 33.33%
(ii) Usage of salary for savings = $$\frac { 1 }{ 6 }$$ × 18,000 = ₹ 3,000
Percentage for savings = $$\frac { 3000 }{ 18000 }$$ × 100 = 16.67 %
(iii) Usage of salary for other expenses = $$\frac { 3 }{ 6 }$$ × 18,000 = ₹ 9,000
Percentage for other expenses = $$\frac { 9000 }{ 18000 }$$ × 100 = 50 % | 1,558 | 4,856 | {"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} | 4.75 | 5 | CC-MAIN-2024-30 | latest | en | 0.881272 |
http://mrcompsci.uk/teaching-topics/review-1-lists-and-functions/ | 1,603,533,081,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107882103.34/warc/CC-MAIN-20201024080855-20201024110855-00169.warc.gz | 77,026,772 | 10,433 | Home » Teaching topics » Functional Programming » Review 1: Lists and functions
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Review 1: Lists and functions
When teaching programming theory it is best to apply some of the concepts as practical examples. Often teachers neglect one area or the other but pupils need to see how both elements relate. Teaching content from any of the previous posts should be punctuated with appropriate examples but it is now a good point to try and draw lists and functions together into more concrete examples that apply the skills learnt so that we can solve problems.
The activities below get progressively more difficult and the solutions provided represent only one method that could be employed (and perhaps not always the best way of doing it!) The examples consider only what has been covered so far and this is intentional as we want reinforce and embed these basic concepts before moving on.
1. Over a 5 hour period the number of people riding on a roller coaster is recorded. After each hour the results are recorded as 12, 33, 9, 2 and 53. Create a function that accepts a list of these values and returns the first value (without using the head function).
2. Modify the function so that it checks for an empty list and returns an error using the command error “No one rode the roller coaster”.
3. Create a second function that returns the amount of people who rode on the roller coaster in the final hour.
4. Create another function that returns the average number of people riding the roller coaster each hour using the data in the list. Hint: sum adds all elements of a list and division will need the numerator and denominator converting to fractional numbers (the / function only works on these) using fromIntegral eg: fromIntegral(theNumVariable).
5. The roller coaster has a regular maintenance hour where no people can ride on it for an hour. Create a function that can appropriately update the list when called during this hour.
6. The average for Saturday and Sunday is calculated together. Create a function that takes two lists of results and finds the average. | 421 | 2,135 | {"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} | 3.671875 | 4 | CC-MAIN-2020-45 | latest | en | 0.901736 |
http://mathematica.stackexchange.com/questions/21007/how-to-simplify-a-complicated-sum-in-terms-of-power-sums?answertab=oldest | 1,462,424,016,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860125897.19/warc/CC-MAIN-20160428161525-00020-ip-10-239-7-51.ec2.internal.warc.gz | 187,615,019 | 25,345 | # How to simplify a complicated Sum in terms of power Sums?
For example, I have:
$a=\sum _{r=1}^n x_r \left(\left(\sum _{i=1}^n x_i-x_r\right){}^2-\sum _{i=1}^n x_i^2\right)$
a =
Sum[Subscript[x, r]*
((Sum[Subscript[x, i], {i, 1, n}] - Subscript[x, r])^2 -
Sum[Subscript[x, i]^2, {i, 1, n}]),
{r, 1, n}]
and $\sum _{i=1}^n x_i=s_1, \sum _{i=1}^n x_i^2=s_2, \sum _{i=1}^n x_i^3=s_3$
c2 = Sum[Subscript[x, i]^2, {i, 1, n}] == s2
I would like to represent $a$ by $s_1, s_2, s_3$, e.g. $a=s_1^3+s_3-3s_1s_2$ . How should I do it? I tried Solve or Eliminate, but couldn't find a way.
-
Could you provide the Mathematica code of your definitions and by a bit clearer about what kind of output you expect? – Sjoerd C. de Vries Mar 10 '13 at 23:29
Related question on SO – Michael E2 Mar 11 '13 at 12:57
@colinfang I changed the title. The question didn't seem to me to have to do with subscripts, but with manipulating sums. If you don't like the new title, you can change it back, or change it to something better. – Michael E2 Mar 11 '13 at 13:30
As bound variables, $r$ and $i$ must play no role in the expansion and so they shouldn't even appear in our solution. It should be equally evident that $n$ is just along for the ride as a placeholder for the upper limit; we could call it anything, and therefore we may call it nothing and ignore it except in expressions where it will necessarily appear in the output. Notice, too, that $x$ disappears under the summation: it really just serves as a placeholder for arguments of other expressions being summed.
Consequently, expressions like "$\sum_{i=1}^n f(x_i)$" should be understood as operations on $f$ itself. What properties should they have? Only the obvious ones implied by the ring operations $+$ and $\times$ in the algebra of polynomials in the $x_i$. This gives just four simple rules, after we establish the unique roles of the symbols "$x$", "$s$", and "$n$":
ClearAll[sum, x, s, n];
1. Converting sums of powers of $x$ into the power sum variables $s_i$:
sum[x] := Subscript[s, 1]; sum[x^p_] := Subscript[s, p];
sum[a_ + b_] := sum[a] + sum[b];
3. Linearity (with respect to "scalars" which do not have any $x_i$ in them):
sum[Times[a_, y__]] /; FreeQ[a, x] := a sum[Times[y]];
4. The effect of summing constant values (this is the only place $n$ need appear):
sum[a_] /; FreeQ[a, x] := n a;
That should do it. But to perform the algebra, we need to expand algebraic combinations of everything possible into powers of $x$ (and otherwise leave everything else alone):
expand[a_] /; ! FreeQ[a, sum] := Map[Expand[#, x] &, a, Infinity];
expand[a_] := a;
### Examples
The question:
sum[x ((sum[x] - x)^2 - sum[x^2])] // expand
$s_1^3-3 s_1 s_2+s_3$
Variance:
sum[(x - sum[x]/n)^2] // expand
$-\frac{s_1^2}{n}+s_2$
Skewness:
sum[(x - sum[x]/n)^3] / sum[(x - sum[x]/n)^2] ^(3/2) // expand
$\frac{\frac{2 s_1^3}{n^2}-\frac{3 s_1 s_2}{n}+s_3}{\left(-\frac{s_1^2}{n}+s_2\right){}^{3/2}}$
etc.
We might worry about misapplications, such as to non-polynomial functions. Not to fear:
sum[Exp[(x + 1)^2]] // expand
$\text{sum}\left[e^{1+2 x+x^2}\right]$
The expansion proceeds insofar as it can, but sum does not know how to go any further, and so stops. sum is also ignorant of other non-polynomial objects, but when they can be converted to polynomials, it works:
sum[Series[Exp[x], {x, 0, 4}] // Normal] // expand
$n+s_1+\frac{s_2}{2}+\frac{s_3}{6}+\frac{s_4}{24}$
(This is a series expansion through order four of $\sum_{i=1}^n e^{x_i}$.)
### Edit
Michael E2, in an extension to his answer, reminds us of the value of simplifying multiple sums. His generalization further clarifies the nature of these operations. Because the variable name "$x$" does not explicitly appear in the $s_i$ notation, the variable name is immaterial. What matters are the index names, only insofar as they are used to connect an $x$ with its enclosing summation: as I remarked at the outset, as bound variables they must disappear at the end.
Thus, for example, we could write sum[i^3, i] for $\sum_{i=1}^n x_i^3$: the first argument to sum is a polynomial (or rational function, even) and the second one is a symbol to indicate what is varying over the summation.
A comparable extension is trivial to make: just include the second argument explicitly in the definition of sum. So that we can reproduce the previous solution (where x was the only variable), we can have the second argument default to x if it's missing. Here is the entire (generalized) solution:
ClearAll[sum, s, n];
sum[a_] := sum[a, x];
sum[x_, x_] := Subscript[s, 1]; sum[x_^p_, x_] := Subscript[s, p];
sum[Times[a_, y__], x_] /; FreeQ[a, x] := a sum[Times[y], x];
sum[a_ + b_, x_] := sum[a, x] + sum[b, x];
sum[a_, x_] /; FreeQ[a, x] := n a;
expand[a_] /; ! FreeQ[a, sum] := Map[Expand, a, Infinity]; expand[a_] := a;
For example, compute $\sum_{k=1}^n\sum_{j=1}^n\sum_{i=1}^n (x_i+x_j+x_k - 3\bar{x})^2$, where $\bar{x} = \sum_{i=1}^n x_i / n$ is the average:
sum[sum[sum[(i + j + k - 3 sum[i, i]/n)^3, i], j], k]
$6 s_1^3-9 n s_1 s_2+3 n^2 s_3$
(Notice that it was no problem for i to appear in two summations.) You do have to be a little careful not to go overboard and stretch the use of sum too far:
sum[x^2, x^2] // expand
$s_1$
This surprising result is correct: the symbol x^2 in the second argument is being used to represent the variable summed over; in terms of this pattern, the first argument x^2 is the first power, whence the result is indeed $s_1$.
One can rather easily go further with generalizations, depending on the need. The next big step would be to implement multivariate sums, such as $\sum_{i=1}^n\sum_{j=1}^n x_i y_j^2$, perhaps to be written $s_1(x)s_2(y)$ or--assuming a specific ordering $(x,y)$ of the variables--even as $s_{1,2}$. As we appear to have gone beyond the scope of the present question, I leave these generalizations to interested readers. I hope that the examples given here have shown how stripping the notation down to its essentials can lead to simple and clear solutions.
-
+1 Here I was trying to formulate it as a general problem, but wasn't able to articulate it coherently – Dr. belisarius Mar 12 '13 at 18:20
+1 Nice and clear. Glad if any hints were helpful -- oops, you meant belisarius. That's ok :) – Michael E2 Mar 12 '13 at 18:22
One thing I am stuck on: in principle it is possible to format these sum expressions into exactly the original notation used in the question. For instance, Format[sum[x^p_]] := \!$$\*UnderoverscriptBox[\(\[Sum]$$, $$i = 1$$, $$n$$] \*SuperscriptBox[ SubscriptBox[$$x$$, $$i$$], $$p$$]\) handles simple cases like sum[x^3]. But how to extend this to the general case? (Among other things, we need a way to introduce different bound variable names such as "i" and "r", or perhaps $i_1, i_2,$ etc.) – whuber Mar 12 '13 at 18:28
Not entirely clear what you mean, but here is a solution that works if the structure of your equation fulfills certain constraints, but it will not be looking for mathematical equivalence in general.
Sum[Subscript[x, r] ((Sum[Subscript[x, i], {i, 1, n}] - Subscript[x, r])^2 -
Sum[Subscript[x, i]^2, {i, 1, n}]), {r, 1, n}]
/. {Sum[Subscript[_, i], {i, 1, n}] -> s1, Sum[Subscript[_, i]^2, {i, 1, n}] -> s2}
-
plz see my updates, I would like to have an equation without any x – colinfang Mar 10 '13 at 23:53
@colinfang You provided a substitution rule for the literal sums over i only. You can't use the same substitution for the sum over the r part, as that can't be equal to s1. – Sjoerd C. de Vries Mar 10 '13 at 23:59
hmm, ye, So I am asking how I can use mathmatica to help me solve/eliminate this maths equations, rather than myself doing it on paper – colinfang Mar 11 '13 at 0:09
a = Sum[x[r] ((Sum[x[i], {i, n}] - x[r])^2 - Sum[x[i]^2, {i, n}]), {r, n}];
Module[{t0, t1},
rule = Sum[p1___ x[a_]^i__ p2___, {a_, n}] :> p1 p2 s[i];
a /. x[k_] -> t0 x[k]^t1
/. Sum[a__, l_List] :> (Sum[#, l] & /@ Expand@a)
//. rule /. t0 -> 1 /. t1 -> 1]
(*s[1]^3 - 3 s[1] s[2] + s[3]*)
-
More succinct than mine. And now you have s[1] instead of s1. I guess I'll post mine. But there's one part I'm not proud of...at least. – Michael E2 Mar 11 '13 at 2:56
@MichaelE2 It took me six iterations to get here, and I think it isn't robust enough – Dr. belisarius Mar 11 '13 at 2:59
I guess you have to teach Mathematica some rules of algebra for sums:
a=Sum[x[r] ((Sum[x[i], {i, n}] - x[r])^2 - Sum[x[i]^2, {i, n}]), {r, n}];
rules ={Sum[x[i_], {i_, n}] -> s[1], Sum[x[i_]^(k_), {i_, n}] :> s[k], (* OP's terms *)
(* some rules of algebra: *)
Sum[(a_Plus)*(b_), {i_, n}] :> Sum[Expand[a*b], {i, n}],
Sum[(a_)*(b_), {i_, n}] :>
(Sum[a*b, {i, n}] /. Sum[(c_)?(FreeQ[#1, i] & )*(d_), {i, n}] :> c*Sum[d, {i, n}]),
Sum[(a_) + (b_), {i_, n}] :> Sum[a, {i, n}] + Sum[b, {i, n}]};
a //. rules
(* -> s[1]^3 - 3 s[1] s[2] + s[3] *)
Using Simplify
[Edit: I added some more rules to handle some multiple sums and powers]
One can add some algebraic transformation rules and a special complexity function to Simplify and Mathematica will expand the sums as far as possible. The complexity function (normally just LeafCount) is altered to strong favor more sums and updated to favor evaluating constant sums to n times the summand. The coefficients -10 and -20 need to be large enough but seem to be otherwise arbitrary based on experimentation; however, I suspect that with more summation limits (in double sums etc.), the more one has to compensate for the greater number of leaves in the Sum.
sRules = {Sum[x[i_], {i_, n}] -> s[1], Sum[x[i_]^(k_), {i_, n}] :> s[k]};
(* factorRule factors out constant factors *)
factorRule = HoldPattern[Sum][summand_Times, lims__] :>
Select[summand, ! InternalDependsOnQ[#, First /@ {lims}] &]*
Sum[Select[summand, InternalDependsOnQ[#, First /@ {lims}] &], lims];
(* expandSum expands the summand and distributes the Sum *)
expandRule = HoldPattern[Sum][summand : _Times | _Power, lims__] :>
Sum[Expand[summand], lims];
expandSum[e_] := Distribute[e /. expandRule]; (* break sums apart *)
simplifySum[Sum[summand_, lims__]] := (* for multiple sums *)
Fold[Sum, mySimplify[summand], {lims}];
complexity[e_] := -10 Count[e, _Sum, {0, Infinity}] + (* favors more Sums *)
-20 Count[e, n, {0, Infinity}] + (* favors more sums and evaluated constant sums *)
Count[e, Sum[_Times, __], {0, Infinity}] + (* favors factoring out constants *)
LeafCount[e]; (* favors smaller expression *)
Simplify[a,
TransformationFunctions -> {Automatic, # /. factorRule &, expandSum},
ComplexityFunction -> complexity] /. sRules
Compared to first: plus -- uses Simplify; minus -- undocumented internal function, explicit algebra rules.
More examples
Summand is a function of two indices:
b = Sum[x[r] Sum[(x[i] - x[r])^2, {i, n}], {r, n}];
mySimplify[b] /. sRules
(* -> -2 s[1]^2 + 2 n s[2] *)
Double sum:
c = Sum[(x[i] - x[r])^2, {r, n}, {i, n}];
mySimplify[c] /. sRules
(* -> -2 s[1]^2 + 2 n s[2] *)
As whuber alluded to in a comment to his answer, you can format the s[k] with subscripts as follows:
Format[s[k_]] := Subscript[s, k]
By the way, the rule for factorRule was found on an archive of the newsgroup sci.math.symbolic (19 Nov 2007). I don't know about Internal functions except what I've learned on this site.
-
+1 But I really don't like none of our solutions. There must be an easier way – Dr. belisarius Mar 11 '13 at 3:06
It would be nice if someone found one. None of the usual functions seem to simplify indefinite sums. – Michael E2 Mar 11 '13 at 3:11
@belisarius Of course, you can get the answer like this: Eliminate[{y == a, s[3] == Sum[x[i]^3, {i, n}], s[2] == Sum[x[i]^2, {i, n}], s[1] == Sum[x[i], {i, n}]} /. n -> 3, Table[x[i], {i, 3}]], for any n >= 3. – Michael E2 Mar 11 '13 at 3:23
Yep, but the whole thing is about keeping n symbolic ... – Dr. belisarius Mar 11 '13 at 3:25
@belisarius You give good hints :-). I have proposed a solution based on them. – whuber Mar 12 '13 at 17:40
If you are willing to work with specific integer values of n then you can use SymmetricReduction. Here is an example with n=4.
poly[n_] :=
Sum[Subscript[x,
r]*((Sum[Subscript[x, i], {i, 1, n}] - Subscript[x, r])^2 -
Sum[Subscript[x, i]^2, {i, 1, n}]), {r, 1, n}]
First[
SymmetricReduction[poly[4], Table[Subscript[x, i], {i, 4}],
Table[s[j], {j, 4}]]]
(* Out[146]= -s[1]^3 + 3 s[1] s[2] + 3 s[3] *)
-
However, it resolves into elementary symmetric polynomials, not the ones I expect. – colinfang Mar 11 '13 at 15:18
Daniel's approach can be saved, if what you want is to express your expression entirely in terms of power sums. Since the power sums are always expressible in terms of the elementary symmetric polynomials (by Newton-Girard), you should be able to always process the output of SymmetricReduction[] using GroebnerBasis[]. To wit,
poly[n_Integer] := Sum[Subscript[x, r]*((Sum[Subscript[x, i], {i, 1, n}] -
Subscript[x, r])^2 - Sum[Subscript[x, i]^2, {i, 1, n}]), {r, 1, n}]
With[{n = 3},
First @ GroebnerBasis[
Prepend[Table[
p[k] - First[SymmetricReduction[Sum[Subscript[x, i]^k, {i, n}],
Table[Subscript[x, i], {i, n}], Table[s[j], {j, n}]]],
{k, n}],
First[SymmetricReduction[poly[n],
Table[Subscript[x, i], {i, n}], Table[s[j], {j, n}]]]],
Table[p[j], {j, n}], Table[s[j], {j, n}]]]
p[1]^3 - 3 p[1] p[2] + p[3]
`
- | 4,366 | 13,365 | {"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": 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} | 3.5 | 4 | CC-MAIN-2016-18 | latest | en | 0.877087 |
https://www.teacherspayteachers.com/Product/Heart-Shaped-Triangle-Puzzle-Solving-One-Step-Equations-with-Positive-Numbers-2369992 | 1,547,811,217,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583660070.15/warc/CC-MAIN-20190118110804-20190118132804-00225.warc.gz | 946,464,176 | 20,402 | # Heart Shaped Triangle Puzzle - Solving One-Step Equations with Positive Numbers
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HEART SHAPED TRIANGLE PUZZLE - SOLVING ONE-STEP EQUATIONS WITH POSITIVE NUMBERS
Are you looking for something fun to do for valentine's day? Try this triangle puzzle in the shape of a heart.
Note: The pieces are all triangles, but are different shapes/sizes.
................................................................................................................................
WHAT IS INCLUDED?
• Information about how the puzzle works and how you can use it in your class.
• Puzzle Pieces (18 triangles)
• Optional Corresponding Worksheet Mat
• Optional Corresponding Student Worksheet (32 problems)
................................................................................................................................
HOW DO YOU SOLVE THE PUZZLE?
How to solve the puzzle:
1. Solve each one-step equations to get a solution.
2. Match up each equation with the correct solution.
3. When matches are found, place the puzzle pieces and work on the next part.
4. Make sure that all shared lines represent the same number.
5. Once everything matches up, you are finished! Congrats!
6. Glue the puzzle pieces to the worksheet.
................................................................................................................................
CHECK OUT SOME HEART MYSTERY PICTURES HERE:
Heart Mystery Picture - Plotting Points in One Quadrant
Heart Mystery Picture - Graphing on the Coordinate Plane with Four Quadrants
Heart Mystery Picture - Graphing and Reflecting Points and Lines
Broken Heart Mystery Picture - Graphing on the Coordinate Plane
................................................................................................................................
CHECK OUT EVEN MORE MYSTERY PICTURES:
Check out some other plotting points and reflecting activities:
Pi Symbol Mystery Picture - Graphing on the Coordinate Plane (Two Options)
Snowflake Mystery Picture Two – Graphing and Reflecting on the Coordinate Plane
Snowflake Mystery Picture Two – Graphing on the Coordinate Plane ONLY
Thanksgiving Turkey Mystery Picture – Graphing on the Coordinate Plane
Thanksgiving Pilgrim Hat – Coordinate Plane Mystery Picture (Two Options)
................................................................................................................................
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\$3.00 | 674 | 3,744 | {"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} | 3.859375 | 4 | CC-MAIN-2019-04 | latest | en | 0.638423 |
https://mathchat.me/2011/04/26/math-cafe-gives-students-options-for-math-success/?replytocom=91583 | 1,653,090,690,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662534693.28/warc/CC-MAIN-20220520223029-20220521013029-00031.warc.gz | 449,226,586 | 37,316 | ## Kiss those Math Headaches GOODBYE!
### “Math Cafe” Gives Students Options for Math Success
Sometimes when I tutor I tell students that they are “hanging the in Math Cafe.”
I explain that when I tutor, I like to offer students a “menu of math options.”
Math Cafe, Open 24/7 = 3.42857 ...
Instead of showing students just one way to work a math problem, I try to present a “menu” of approaches, and I tell students that they need to listen so they can decide which approach works best for them.
Students tend to perk up when they hear that they are getting a variety of options. Since they will get a chance to choose the approach that best suits them, they become more interested in what I’m saying. Getting this choice boosts student interest and involvement, I’ve found.
Offering a variety of approaches helps students in other ways, too. For one, students start to learn that each approach has advantages and disadvantages. So students start to see that it might make sense to use Approach #1 in one situation, but better sense to use Approach #2 in a different problem. Encouraging children to evaluate the relative merits of different approaches engages them in higher-order thinking skills. And that in itself is a good thing.
What is more, encouraging flexibility of thought helps students develop other skills that serve them well not only in math, but in other subject areas, too. It could be argued that this helps them in life in general. Finally, presenting a variety of approaches helps more students find success with the math skill. That is because when students have a menu of options, they choose the option that works best for them. As a result, more students will do the math skill successfully.
I try not to worry about why one approach works better for a particular student than other approaches, although I am confident that this would be interesting to study in a controlled way somehow. I just am glad that when I present a variety of approaches, more students succeed in solving the problem.
I’d like to present one example of what it looks like to present a menu of math options. So here’s a simple example of three options for teaching students how to subtract two-digit numbers, with the possibility of re-gropuing.
OPTION 1) First I teach the “standard approach,” in which students re-group, trading one group of 10 for 10 ones. Then I show students how to move those 10 ones to the 1s place, and then subtract. This is the way I learned to subtract with re-grouping (called “borrowing,” back then in the ‘old days’).
For example, in a problem such as 42 – 27, students first demote the 4 of 42 into a 3, and then transfer 10 to the 2 of 42, making the 2 into a 12. Students then subtract, first getting: 12 – 7 = 5. Then, after that they subtract in the 10s place, getting: 3 – 2 = 1. Joining the 1 to the 5, they get their answer: 15.
OPTION 2) Subtract by “adding up.” So in a problem like 42 – 27, I teach students to think of the problem as representing a journey along the number line, with the numbers that end in 10 serving as “towns” along the road where they can stop and get a refreshment. In the problem of 42 – 27, students first travel 3 miles, from 27 to 30, where they get a tall drink. Then they travel 10 miles, from 30 to 40, where they eat lunch. Then they travel 2 more miles, from 40 to 42, at which point they reach their destination. Students just add up the three distances traveled: 3 + 10 + 2, and they get their answer of 15. This approach lends itself much more to “mental math” —doing the problem mentally, in one’s mind, without paper and pencil.
OPTION 3) For my third menu option, I have children subtract from left to right instead of the usual right to left approach. I also have students use negative numbers in a basic way. First students subtract in the 10s place: 40 – 20 = 20. Then they subtract in the 1s place: 2 – 7 = – 5. Putting the two partial answers together, they get 20 and – 5. They learn that this means that they should subtract 5 from 20, to get 15, the answer.
Some people might think that elementary students are too young to work with negative numbers. For the record, I have found that many elementary students are quite ready to work with negative numbers. I don’t even have to explain much about these numbers. I just tell them that when we take more than what we have, the answer is “negative.” I’ve been rather amazed at how many students seem to intuitively grasp this notion.
In any case, those are the three approaches that I usually present for this kind of problem. Of course, there are still other ways to subtract these numbers from each other, and all ways that work are valid.
The thought I’d like to leave you with here is that presenting a menu of options — a range of strategies — can really help students find math class more interesting. Give it a try if you have not yet done so. You might want to start with two options. See how it goes. As you continue with this approach, you will find more and more ways to teach the same math skill to your students, and your repertoire of approaches will grow.
#### Comments on: "“Math Cafe” Gives Students Options for Math Success" (1)
1. Son problemas muy chebres q nos pueden servir para nuetra cotianidad
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20 Questions | Total Attempts: 168
Settings
.
• 1.
What is the simple interest on P10,000 for 1 year at 5% interest per year?
• A.
P500.00
• B.
P5000.00
• C.
P20,000.00
• D.
P200,000.00
• 2.
A desk fan with a power rating of 120 W is used for a period of 8 hours. What is the electric power consumption?
• A.
960 kW-h
• B.
.96 kW-h
• C.
15 kW-h
• D.
.15 kW-h
• 3.
In a kitchen, an oven (1,500 W), a rice cooker (450 W), and a refrigerator (170 W) are used for three hours. What is the total electric power consumption?
• A.
2.12 kW-h
• B.
2 120 kW-h
• C.
6.36 kW-h
• D.
6 360 kW-h
• 4.
A peanut vendor bought 5350 grams of raw peanuts in the market. How much do the peanuts weigh in kilograms?
• A.
5.350
• B.
53.50
• C.
535
• D.
.535
• 5.
Mark weighs 40 kilograms while Ben weighs 89 pounds. Who weighs more?
• A.
Mark
• B.
Ben
• C.
Mark and Ben have equal weights
• D.
Can not be determined
• 6.
The following are the monthly salaries of 10 employees from a small company.Employee 1 P 3,500Employee 2 P 3,500Employee 3 P 3,400Employee 4 P 3,300Employee 5 P 3,300Employee 6 P 3,200Employee 7 P 2,700Employee 8 P 2,600Employee 9 P 2,500Employee 10 P 2,200Compute for the median salary.
• A.
P 2 200.00
• B.
P 2 500.00
• C.
P 3 250.00
• D.
P 3 300.00
• 7.
In a community pool, Rachel is practicing for an upcoming swimming meet. In order to qualify in the eliminations, she needs to clock a time of 1.45 minutes. During her practice, she checks her time and takes note of her progress once in a while. For the first lap, she timed 1.40 minutes; for the second lap, she clocked 1.37 minutes and for the last lap, she clocked 1.49 minutes. Look at the data below: 1st lap = 1.40 minutes 2nd lap = 1.37 minutes 3rd lap = 1.49 minutes What is her average lap time?
• A.
4.26 minutes
• B.
1.42 minutes
• C.
1.40 minutes
• D.
2.13 minutes
• 8.
The following are the scores of 20 students in a math exam.Find the range of their scores.40 39 27 50 4328 16 37 33 4944 34 23 39 2245 19 28 47 38
• A.
33
• B.
701
• C.
35.05
• D.
37
• 9.
Jerbeck got the following final grades.Science 87English 88Reading 90Social studies 93Home economics 95 Suppose she needs an average of 90 in her six subjects in order to be included in the honor roll, what should her grade in math be?
• A.
85
• B.
86
• C.
87
• D.
88
• 10.
Jeff and Marie are studying typing. In six speed tests, Jeff was able to record the following speeds in number of words per minute:56 55 63 62 49 62while Marie had the following speeds:58 63 71 51 59 64Compare their median speeds Who is the faster typist?
• A.
Jeff
• B.
Marie
• C.
Jeff and Marie are both fast
• D.
Neither is fast
• 11.
Dina has two pieces of cloth that are 90 cm and 60 cm long respectively. How many meters of cloth does she have in all?
• A.
5400
• B.
150
• C.
1.5
• D.
30
• 12.
Dindo finished a 1500-meter race. How many kilometers did he run?
• A.
.15
• B.
1.5
• C.
15
• D.
150
• 13.
Ana is interested in buying the vacant lot beside her house. The real estate broker informed her that the lot is 15 meters long and 10 meters wide. What is the perimeter of the lot?
• A.
150
• B.
25
• C.
50
• D.
100
• 14.
A rice field is pentagonal in shape. The lengths of its sides are 205 m, 115 m, 153 m, 187 m, and 165 m respectively. If the owner wants to fence it with three strands of barbed wire, how many meters of wire in all will be needed?
• A.
825
• B.
165
• C.
2475
• D.
1650
• 15.
Mang Pedro’s garden is 28 meters long and 25 meters wide. How many meters of cyclone wire will be needed to fence his garden?
• A.
53
• B.
106
• C.
212
• D.
700
• 16.
A circle has a radius measuring 3 cm. What is the circumference?
• A.
6
• B.
9.42
• C.
18.84
• D.
19.26
• 17.
A farmer would like to fence his circular garden that has a radius of 40 m. How many meters of wire will he need?
• A.
5 024
• B.
80
• C.
125.6
• D.
251.2
• 18.
A circular garden in a park has a diameter of 13 meters. If John jogs around the garden 5 times, what is the total distance that he will cover?
• A.
65
• B.
81.64
• C.
204.1
• D.
408.2
• 19.
How many kilos of rice are there in 7.2 metric tons?
• A.
720
• B.
7200
• C.
72 000
• D.
720 000
• 20.
A candy stall sold 144 ounces of heart-shaped lollipops last Valentine’s Day. How many pounds of heart-shaped lollipops were sold?
• A.
12
• B.
16
• C.
2304
• D.
9
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# Solutions for Chapter 4.6: A Summary of Curve Sketching
## Full solutions for Calculus: Early Transcendental Functions | 4th Edition
ISBN: 9780618606245
Solutions for Chapter 4.6: A Summary of Curve Sketching
Solutions for Chapter 4.6
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##### ISBN: 9780618606245
This expansive textbook survival guide covers the following chapters and their solutions. Calculus: Early Transcendental Functions was written by and is associated to the ISBN: 9780618606245. This textbook survival guide was created for the textbook: Calculus: Early Transcendental Functions , edition: 4. Chapter 4.6: A Summary of Curve Sketching includes 99 full step-by-step solutions. Since 99 problems in chapter 4.6: A Summary of Curve Sketching have been answered, more than 39492 students have viewed full step-by-step solutions from this chapter.
Key Calculus Terms and definitions covered in this textbook
• Acute triangle
A triangle in which all angles measure less than 90°
• Base
See Exponential function, Logarithmic function, nth power of a.
• Continuous at x = a
lim x:a x a ƒ(x) = ƒ(a)
• Difference of two vectors
<u1, u2> - <v1, v2> = <u1 - v1, u2 - v2> or <u1, u2, u3> - <v1, v2, v3> = <u1 - v1, u2 - v2, u3 - v3>
• Division
a b = aa 1 b b, b Z 0
• End behavior asymptote of a rational function
A polynomial that the function approaches as.
• Gaussian curve
See Normal curve.
• Linear factorization theorem
A polynomial ƒ(x) of degree n > 0 has the factorization ƒ(x) = a(x1 - z1) 1x - i z 22 Á 1x - z n where the z1 are the zeros of ƒ
• Normal curve
The graph of ƒ(x) = e-x2/2
• Opposite
See Additive inverse of a real number and Additive inverse of a complex number.
• Paraboloid of revolution
A surface generated by rotating a parabola about its line of symmetry.
• Plane in Cartesian space
The graph of Ax + By + Cz + D = 0, where A, B, and C are not all zero.
• Quotient rule of logarithms
logb a R S b = logb R - logb S, R > 0, S > 0
• Regression model
An equation found by regression and which can be used to predict unknown values.
• Repeated zeros
Zeros of multiplicity ? 2 (see Multiplicity).
• Speed
The magnitude of the velocity vector, given by distance/time.
• Standard form of a polynomial function
ƒ(x) = an x n + an-1x n-1 + Á + a1x + a0
• Statute mile
5280 feet.
• Union of two sets A and B
The set of all elements that belong to A or B or both.
• Ymin
The y-value of the bottom of the viewing window.
× | 726 | 2,477 | {"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} | 3.75 | 4 | CC-MAIN-2020-40 | latest | en | 0.848722 |
https://oeis.org/A167868 | 1,585,733,362,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370505550.17/warc/CC-MAIN-20200401065031-20200401095031-00128.warc.gz | 595,271,142 | 4,203 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A167868 a(n) = 3^n*Sum_{ k=0..n } binomial(2*k,k)^3/3^k 5
1, 11, 249, 8747, 369241, 17110731, 840221217, 42944901219, 2260581606657, 121714776747971, 6671749658197129, 371062413164972955, 20887218937200347281, 1187720356043817041843, 68124474120573747125529 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 COMMENTS The expression a(n) = B^n*Sum_{ k=0..n } binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5), A167713 (B=16). The expression a(n) = B^n*Sum_{ k=0..n } binomial(2*k,k)^3/B^k gives A079727 for B=1, A167867 (B=2), A167868 (B=3), A167869 (B=4), A167870 (B=16), A167871 (B=64). LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..200 FORMULA a(n) = 3^n*Sum[ Binomial[2*k,k]^3/3^k, {k,0,n} ]. Recurrence: n^3*a(n) = (67*n^3 - 96*n^2 + 48*n - 8)*a(n-1) - 24*(2*n-1)^3*a(n-2). - Vaclav Kotesovec, Aug 13 2013 a(n) ~ 2^(6*n+6)/(61*(Pi*n)^(3/2)). - Vaclav Kotesovec, Aug 13 2013 MATHEMATICA Table[3^n Sum[Binomial[2k, k]^3/3^k, {k, 0, n}], {n, 0, 20}] (* Vincenzo Librandi, Mar 26 2012 *) CROSSREFS Cf. A079727, A167867, A167868, A167869, A167870, A167872. Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167859. Sequence in context: A219089 A243683 A323255 * A238751 A098672 A056210 Adjacent sequences: A167865 A167866 A167867 * A167869 A167870 A167871 KEYWORD nonn AUTHOR Alexander Adamchuk, Nov 14 2009 EXTENSIONS More terms from Sean A. Irvine, Apr 27 2010 STATUS approved
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Last modified April 1 05:04 EDT 2020. Contains 333155 sequences. (Running on oeis4.) | 791 | 2,002 | {"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-2020-16 | latest | en | 0.468776 |
https://math.stackexchange.com/questions/3399223/how-to-solve-this-nearly-linear-optimization-problem-or-transform-to-quadratic | 1,579,810,244,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250613416.54/warc/CC-MAIN-20200123191130-20200123220130-00548.warc.gz | 548,472,101 | 31,024 | # How to solve this nearly-linear optimization problem, or transform to quadratic form?
A common problem from chemical thermodynamics is to determine the vector $$x_i$$ with dimension $$N$$ that solves
$$\frac{n_i-x_i}{1-\sum x_i} = k_i \frac{x_i}{\sum x_i}$$
for some constant vectors $$n_i$$ & $$k_i$$, subject to the constraints $$\sum n_{i}=1$$, $$1>n_i>x_i>0$$, and $$k_i>0$$, and additionally (to exclude most degenerate cases) $$k_i≠1$$, $$\exists k_i|k_i>1$$, and $$\exists k_i|k_i<1$$. When it exists, $$x_i$$ is unique.
With $$s_i=1$$ and diagonal matrix $$K_{ii}=k_i$$, the problem is
$$\mathbf n = \left(I\, + \, \left(\frac 1 {\mathbf s\cdot \mathbf x} - 1\right)K\right) \cdot \mathbf x.$$
Is there a closed-form expression for $$x_i$$? Can the problem be transformed into standard form for quadratic programming?
Edit:
This corresponds to solving with "volatilities" in chemical language, but I suspect the general case of "saturation pressures" actually has a simpler solution.
This is only a partial answer. Set $$A := ns^T$$. Multiplying your equation with $$s^Tx = \sum x_i$$ gives $$(ns^T)x = Kx + s^Tx(I-K)x,$$ hence $$(I-K)^{-1}(A-K)x = (\sum x_i)x.$$ Thus, we have $$Bx=(\sum x_i)x$$, where $$B := (I-K)^{-1}(A-K)$$. Hence,
Claim: The original system (with constraints) has a solution if and only if $$B$$ has a positive eigenvalue with a corresponding positive eigenvector. In this case, if $$Bu=\lambda u$$ with $$\lambda > 0$$ and $$u>0$$, then $$x = \lambda(\sum u_i)^{-1}u$$ is a solution of the original system.
• Can the related problem (solving with "saturation pressures and finite liquid volumes" in chemical language) be solved similarly, $(n\,s^T - K)x = (s^Tx)x - (v^Tx) K x$ with $v_i>0$ and $0 < v^Tx < 1$? – alexchandel Jan 6 at 22:52 | 565 | 1,782 | {"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": 27, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2020-05 | latest | en | 0.804479 |
https://www.geeksforgeeks.org/peterson-graph/ | 1,696,298,045,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511023.76/warc/CC-MAIN-20231002232712-20231003022712-00777.warc.gz | 845,739,645 | 44,903 | Open In App
A Peterson Graph Problem
The following graph G is called a Petersen graph and its vertices have been numbered from 0 to 9. Some letters have also been assigned to vertices of G, as can be seen from the following picture:
Let’s consider a walk W in graph G, which consists of L vertices W1, W2, …, WL. A string S of L letters ‘A’ – ‘E’ is realized by walking W if the sequence of letters written along W is equal to S. Vertices can be visited multiple times while walking along W.
For example, S = ‘ABBECCD’ is realized by W = (0, 1, 6, 9, 7, 2, 3). Determine whether there is a walk W that realizes a given string S in graph G and if so then find the lexicographically least such walk. The only line of input contains one string S. If there is no walk W which realizes S, then output -1 otherwise, you should output the least lexicographical walk W which realizes S.
Example of a Petersen Graph
Examples:
```Input : s = 'ABB'
Output: 016
Explanation: As we can see in the graph
the path from ABB is 016.
Input : s = 'AABE'
Output :-1
Explanation: As there is no path that
exists, hence output is -1.```
Algorithm for a Peterson Graph Problem:
petersonGraphWalk(S, v):
begin
res := starting vertex
for each character c in S except the first one, do
if there is an edge between v and c in outer graph, then
v := c
else if there is an edge between v and c+5 in inner graph, then
v := c + 5
else
return false
end if
put v into res
done
return true
end
Below is the implementation of the above algorithm:
C++
`// C++ program to find the``// path in Peterson graph``#include ``using` `namespace` `std;` `// path to be checked``char` `S[100005];` `// adjacency matrix.``bool` `adj[10][10];` `// resulted path - way``char` `result[100005];` `// we are applying breadth first search``// here``bool` `findthepath(``char``* S, ``int` `v)``{`` ``result[0] = v + ``'0'``;`` ``for` `(``int` `i = 1; S[i]; i++) {`` ` ` ``// first traverse the outer graph`` ``if` `(adj[v][S[i] - ``'A'``] || adj[S[i] -`` ``'A'``][v]) {`` ``v = S[i] - ``'A'``;`` ``}` ` ``// then traverse the inner graph`` ``else` `if` `(adj[v][S[i] - ``'A'` `+ 5] ||`` ``adj[S[i] - ``'A'` `+ 5][v]) {`` ``v = S[i] - ``'A'` `+ 5;`` ``}` ` ``// if the condition failed to satisfy`` ``// return false`` ``else`` ``return` `false``;` ` ``result[i] = v + ``'0'``;`` ``}` ` ``return` `true``;``}` `// driver code``int` `main()``{`` ``// here we have used adjacency matrix to make`` ``// connections between the connected nodes`` ``adj[0][1] = adj[1][2] = adj[2][3] = adj[3][4] =`` ``adj[4][0] = adj[0][5] = adj[1][6] = adj[2][7] =`` ``adj[3][8] = adj[4][9] = adj[5][7] = adj[7][9] =`` ``adj[9][6] = adj[6][8] = adj[8][5] = ``true``;`` ` ` ``// path to be checked`` ``char` `S[] = ``"ABB"``;`` ` ` ``if` `(findthepath(S, S[0] - ``'A'``) ||`` ``findthepath(S, S[0] - ``'A'` `+ 5)) {`` ``cout << result;`` ``} ``else` `{`` ``cout << ``"-1"``;`` ``}`` ``return` `0;``}`
Java
`// Java program to find the``// path in Peterson graph``class` `GFG``{` ` ``// path to be checked`` ``static` `char` `[]S = ``new` `char``[``100005``];` ` ``// adjacency matrix.`` ``static` `boolean` `[][]adj = ``new` `boolean``[``10``][``10``];`` ` ` ``// resulted path - way`` ``static` `char``[] result = ``new` `char``[``100005``];`` ` ` ``// we are applying breadth first search`` ``// here`` ``static` `boolean` `findthepath(``char``[] S, ``int` `v)`` ``{`` ``result[``0``] = (``char``) (v + ``'0'``);`` ``for` `(``int` `i = ``1``; i<(``int``)S.length; i++)`` ``{`` ` ` ``// first traverse the outer graph`` ``if` `(adj[v][S[i] - ``'A'``] ||`` ``adj[S[i] - ``'A'``][v])`` ``{`` ``v = S[i] - ``'A'``;`` ``}`` ` ` ``// then traverse the inner graph`` ``else` `if` `(adj[v][S[i] - ``'A'` `+ ``5``] ||`` ``adj[S[i] - ``'A'` `+ ``5``][v])`` ``{`` ``v = S[i] - ``'A'` `+ ``5``;`` ``}`` ` ` ``// if the condition failed to satisfy`` ``// return false`` ``else`` ``return` `false``;`` ` ` ``result[i] = (``char``) (v + ``'0'``);`` ``}`` ``return` `true``;`` ``}`` ` ` ``// Driver code`` ``public` `static` `void` `main(String[] args)`` ``{`` ``// here we have used adjacency matrix to make`` ``// connections between the connected nodes`` ``adj[``0``][``1``] = adj[``1``][``2``] = adj[``2``][``3``] = adj[``3``][``4``] =`` ``adj[``4``][``0``] = adj[``0``][``5``] = adj[``1``][``6``] = adj[``2``][``7``] =`` ``adj[``3``][``8``] = adj[``4``][``9``] = adj[``5``][``7``] = adj[``7``][``9``] =`` ``adj[``9``][``6``] = adj[``6``][``8``] = adj[``8``][``5``] = ``true``;`` ` ` ``// path to be checked`` ``char` `S[] = ``"ABB"``.toCharArray();`` ` ` ``if` `(findthepath(S, S[``0``] - ``'A'``) ||`` ``findthepath(S, S[``0``] - ``'A'` `+ ``5``))`` ``{`` ``System.out.print(result);`` ``}`` ``else`` ``{`` ``System.out.print(``"-1"``);`` ``}`` ``}``}` `// This code is contributed by Rajput-Ji`
Python3
`# Python3 program to find the``# path in Peterson graph``# path to be checked` `# adjacency matrix.``adj ``=` `[[``False` `for` `i ``in` `range``(``10``)] ``for` `j ``in` `range``(``10``)]` `# resulted path - way``result ``=` `[``0``]` `# we are applying breadth first search``# here``def` `findthepath(S, v):`` ``result[``0``] ``=` `v`` ``for` `i ``in` `range``(``1``, ``len``(S)):`` ` ` ``# first traverse the outer graph`` ``if` `(adj[v][``ord``(S[i]) ``-` `ord``(``'A'``)] ``or`` ``adj[``ord``(S[i]) ``-` `ord``(``'A'``)][v]):`` ``v ``=` `ord``(S[i]) ``-` `ord``(``'A'``)`` ` ` ``# then traverse the inner graph`` ``else` `if` `(adj[v][``ord``(S[i]) ``-` `ord``(``'A'``) ``+` `5``] ``or`` ``adj[``ord``(S[i]) ``-` `ord``(``'A'``) ``+` `5``][v]):`` ``v ``=` `ord``(S[i]) ``-` `ord``(``'A'``) ``+` `5`` ` ` ``# if the condition failed to satisfy`` ``# return false`` ``else``:`` ``return` `False`` ` ` ``result.append(v)`` ` ` ``return` `True` `# driver code``# here we have used adjacency matrix to make``# connections between the connected nodes``adj[``0``][``1``] ``=` `adj[``1``][``2``] ``=` `adj[``2``][``3``] ``=` `\``adj[``3``][``4``] ``=` `adj[``4``][``0``] ``=` `adj[``0``][``5``] ``=` `\``adj[``1``][``6``] ``=` `adj[``2``][``7``] ``=` `adj[``3``][``8``] ``=` `\``adj[``4``][``9``] ``=` `adj[``5``][``7``] ``=` `adj[``7``][``9``] ``=` `\``adj[``9``][``6``] ``=` `adj[``6``][``8``] ``=` `adj[``8``][``5``] ``=` `True` `# path to be checked``S``=` `"ABB"``S``=``list``(S)``if` `(findthepath(S, ``ord``(S[``0``]) ``-` `ord``(``'A'``)) ``or`` ``findthepath(S, ``ord``(S[``0``]) ``-` `ord``(``'A'``) ``+` `5``)):`` ``print``(``*``result, sep ``=` `"")``else``:`` ``print``(``"-1"``)`` ` `# This code is contributed by SHUBHAMSINGH10`
C#
`// C# program to find the``// path in Peterson graph``using` `System;``public` `class` `GFG``{` ` ``// adjacency matrix.`` ``static` `bool` `[,]adj = ``new` `bool``[10, 10];` ` ``// resulted path - way`` ``static` `char``[] result = ``new` `char``[100005];` ` ``// we are applying breadth first search`` ``// here`` ``static` `bool` `findthepath(String S, ``int` `v)`` ``{`` ``result[0] = (``char``) (v + ``'0'``);`` ``for` `(``int` `i = 1; i < S.Length; i++)`` ``{` ` ``// first traverse the outer graph`` ``if` `(adj[v,S[i] - ``'A'``] ||`` ``adj[S[i] - ``'A'``,v])`` ``{`` ``v = S[i] - ``'A'``;`` ``}` ` ``// then traverse the inner graph`` ``else` `if` `(adj[v,S[i] - ``'A'` `+ 5] ||`` ``adj[S[i] - ``'A'` `+ 5,v])`` ``{`` ``v = S[i] - ``'A'` `+ 5;`` ``}` ` ``// if the condition failed to satisfy`` ``// return false`` ``else`` ``return` `false``;` ` ``result[i] = (``char``) (v + ``'0'``);`` ``}`` ``return` `true``;`` ``}` ` ``// Driver code`` ``public` `static` `void` `Main(String[] args)`` ``{` ` ``// here we have used adjacency matrix to make`` ``// connections between the connected nodes`` ``adj[0,1] = adj[1,2] = adj[2,3] = adj[3,4] =`` ``adj[4,0] = adj[0,5] = adj[1,6] = adj[2,7] =`` ``adj[3,8] = adj[4,9] = adj[5,7] = adj[7,9] =`` ``adj[9,6] = adj[6,8] = adj[8,5] = ``true``;` ` ``// path to be checked`` ``String S = ``"ABB"``;`` ``if` `(findthepath(S, S[0] - ``'A'``) || findthepath(S, S[0] - ``'A'` `+ 5))`` ``{`` ``Console.WriteLine(result);`` ``}`` ``else`` ``{`` ``Console.Write(``"-1"``);`` ``}`` ``}``}` `// This code is contributed by aashish1995`
Javascript
``
Output
`016`
Time complexity: O(N)
The time complexity of the above program is O(N), where N is the length of the given string S. We are applying Breadth First Search here, which runs in linear time.
Space complexity: O(N)
The space complexity of the above program is O(N), where N is the length of the given string S. We are using two auxiliary arrays – result[] and S[] to store the path and the given string, respectively. Both of them require linear space.
This article is contributed by Sunidhi Chaudhary. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks. | 3,770 | 10,006 | {"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} | 3.859375 | 4 | CC-MAIN-2023-40 | latest | en | 0.843396 |
https://socratic.org/questions/58c2302411ef6b7846818017 | 1,571,039,882,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986649841.6/warc/CC-MAIN-20191014074313-20191014101313-00107.warc.gz | 701,935,652 | 6,118 | # Why does the rate of reaction decrease over time?
##### 1 Answer
Mar 10, 2017
Because the rate of reaction is intrinsically dependent on the reactant concentration leftover.
Recall that for the reaction
$a A + b B \to c C + \mathrm{dD}$,
a rate law is written as:
$r \left(t\right) = - \frac{1}{a} \frac{\Delta \left[A\right]}{\Delta t} = - \frac{1}{b} \frac{\Delta \left[B\right]}{\Delta t} = \frac{1}{c} \frac{\Delta \left[C\right]}{\Delta t} = \frac{1}{d} \frac{\Delta \left[D\right]}{\Delta t}$
Thus, a rate of reaction has units of $\text{M"/"s}$ if we are in units of seconds for time.
Since the rate is based on the change in concentration over time, which has to do with how quickly the reactants react, there has to be a relationship between rate and how much concentration of reactants are left.
As the concentration of reactants decreases, the reactants have a harder time trying to find each other to react, so the rate of reaction is slower for lower concentrations.
Thus, as time passes and concentrations of reactants drop, the rate of reaction drops. | 293 | 1,078 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "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-2019-43 | longest | en | 0.912154 |
https://math.stackexchange.com/questions/4109229/a-single-linear-feasibility-formulation-that-exactly-captures-all-the-optimal-so | 1,720,907,781,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514517.94/warc/CC-MAIN-20240713212202-20240714002202-00383.warc.gz | 334,978,492 | 36,843 | # A single linear feasibility formulation that exactly captures all the optimal solutions of both primal and dual.
Given a linear program, we have the primal as follows:
$$\begin{array}{lll} \max: & c^Tx\\ \text{s.t.} & Ax \leq b\\ & x\geq 0\\ \end{array}$$ And we also have the dual as follows: $$\begin{array}{lll} \min: & y^Tb\\ \text{s.t.} & y^TA \geq c\\ & y \geq 0\\ \end{array}$$
Now, the question is (a)to design a single linear feasibility formulation (i.e. Find $$z$$ such that $$Bz \leq k$$) that exactly captures all the optimal solutions of both primal and dual. (b)show that $$(x^*,y^*)$$ is feasible in the formulation if and only if $$x^*$$ is an optimal solution of the primal and $$y^*$$ is an optimal solution of the dual.
Here's what I got so far: The single linear feasibility formulation is the following:
$$\begin{pmatrix}A & 0 \\ 0 & -A^T \end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} \leq \begin{pmatrix}b \\ -c^T\end{pmatrix}$$, where $$A \in \mathbb{R}^{m\times n},x\in\mathbb{R}^{n\times 1}, y\in\mathbb{R}^{m\times 1}$$.
From this formulation, it should capture all the optimal solution of both primal and dual. Now,for part (b), the converse direction is trivial because if $$x^*$$ is an optimal, then it means $$Ax^* \leq b$$ and $$y^*$$ is an optimal solution of the dual means $$-A^Ty^* \leq -c^T$$, so $$z=(x^*,y^*)$$ is feasible in the formulation.
However, I am stuck in the forward direction, which is if $$(x^*,y^*)$$ is feasible in the formulation, then $$x^*$$ is an optimal solution of the primal and $$y^*$$ is an optimal solution of the dual. Can I get some help for that? Do I have to use the strong duality theorem in order to prove that?
You want to impose the duality constraint as well, that is $$c^Tx \ge y^Tb$$.
Also, don't forget the sign constraint, which is $$x \ge 0$$ and $$y \ge 0$$.
Here is the equivalent single linear feasibility formulation which captures optimal solution of primal and dual:
\begin{align} -c^Tx &\le -y^Tb \\ x &\ge 0 \\ y &\ge 0 \\ Ax &\le b \\ -A^Ty &\le -c \end{align}
You just have to write it as a single inequality.
Edit:
$$\begin{bmatrix} -c^T & b^T \\ -I & 0 \\ 0 & -I \\ A & 0 \\ 0 & -A^T\end{bmatrix} \begin{bmatrix}x \\ y \end{bmatrix} \le \begin{bmatrix} 0 \\ 0\\ 0 \\ b \\ -c\end{bmatrix}$$
• You mean imposing the strong duality constraint ($c^Tx = y^Tb$) in the proof of the forward direction or in the original linear feasibility formulation? For the forward direction, are we trying to show that if $(x^*,y^*)$ are feasible, and if we can show $c^Tx^* = y^*Tb$, then we can conclude $(x^*,y^*)$ are optimal? Can you explain more about the forward direction of the proof? Commented Apr 20, 2021 at 7:31
• I mean the formulation. your formulation is not equivalent yet. Commented Apr 20, 2021 at 7:51
• So I was able to get something like this $\begin{pmatrix}A & -1 \\ -1 & -A^T \end{pmatrix}$$\begin{pmatrix}x \\ y\end{pmatrix} \leq \begin{pmatrix}b \\ -c\end{pmatrix}$. But I am stuck at trying to impose those two inequalities $c^Tx \leq y^Tb$ and $-c^Tx \leq -y^Tb$ in the matrix, can you give me some hint how to do that or am I on the right track? Commented Apr 20, 2021 at 9:25
• I have written them in matrix form. Commented Apr 20, 2021 at 9:46
• The weak duality constraint is redundant, so you need to enforce only one side of the equality. Commented Apr 21, 2021 at 2:28 | 1,111 | 3,392 | {"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": 22, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-30 | latest | en | 0.826007 |
http://crystalclearmaths.com/videos-learning-resources/algebra/polynomials/ | 1,566,807,450,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027331228.13/warc/CC-MAIN-20190826064622-20190826090622-00052.warc.gz | 45,625,702 | 18,308 | EASIER THAN YOU THINK...
# Polynomials
The word polynomial means many terms (literally, many names). In modern usage, however, this expression includes monomials (one term, such as 5 or 3x or -7x³), binomials (two terms, such as x² – 3 or 2x³ + 9x), and trinomials (three terms, such as x² + 5x – 3). Each term can have more than one variable, but each variable may only have a power that is a natural number (0, 1, 2, 3, …). No other functions are allowed. So, terms such as -4x³, 7xy², and 9zy³z²/7 are all valid terms (think of them as the ‘nomials’ of which you may have ‘many’).
Because most of the material on this page is designed for high school (and some early university) use, I will be restricting my usage, mostly, to using one variable, so I will be writing of y being a function of x [y = f(x)] or of y being a polynomial in x [y = P(x)]. The examples to the left, where you see the graph of a quadratic and a cubic equation, will be typical. When expressed in this way, such polynomials are also functions. In other words, for any given x value, there is only one y value. These polynomials also have some very nice properties, as we shall see. All these things make them easier to study than other curves, but they still have some nice surprises in store for us! Please note that I am aware that there is a difference in meaning between indeterminates and variables but, for ease of use, I will use the term variable because that is what most high school students will be familiar with.
It is difficult to pin down when this term was first used. Mathematicians had been studying them for many years, but without using this particular term to describe the entire collection. There are claims that the term polynomial was first used by the brilliant Franch mathematician, François Viète (1540-1603), but I have been unable to validate them (please see the first drop-down item below).
[As an aside, the way that we write our algebra has improved immensely since Viète’s day. Robert Recorde had created the = sign in 1557, but it was yet to become popular. Karr, Massey and Gustafson, in their Intermediate Algebra share that, where we would write (x + 1)³ = x³ + 3x² + 3x + 1, Viète would have written cubus aequalis x cubus + x quad. 3 + x in 3 + 1. And you think modern notation is difficult!]
As a more formal definition, a polynomial in one variable (or, more correctly, indeterminate) can be written in the form where are numbers, being elements of a ring, so that the normal four operations (addition, subtraction, multiplication and division) apply to the variable/indeterminate, the coefficients, and the terms. The graphs of polynomials of this kind form a particular family of curves which share some interesting characteristics.
At various times, as you study this material, you may with to refer to a useful site I found which provides a lot of information about the History of Polynomial Equations.
Analysing and Creating Polynomials Using the Babylonian Differencing Technique
How to Graph Polynomials and How to Construct Their Equations From Their Graphs
Notes Concerning the Origin of the Term 'Polynomial'
A number of online sources claim that François Viéta (1540-1603) first used the term polynomial, and quote Cajori 1919, page 139 as a reference:
They also observe that the word is (first) found in English in Samuel Jeake's (1623-1690) Arithmetic (1674): "Those knit together by both Signs are called ... by some Multinomials, or Polynomials, that is, many named." They all provide a reference of OED2 (second edition of the Oxford English Dictionary).
As I mentioned above, althought Viète certainly worked with polynomials, I cannot trace any reference that he called them by that name. I managed to find an online copy of Florian Cajori's (1859-1930) book, A History of the Conceptions of Limits and Fluxions in Great Britain, from Newton to Woodhouse (1919), but could not find any such reference on page 139 or anywhere else in it. So, I remain puzzled.
The reference for the first use of the word polynomial in English (identified by the Oxford English Dictionary) was much easier to trace. Although I could not find a reference to a 1674 publication, Wikipedia states that Jeake's principal work was Arithmetick Surveighed and Reviewed published in four books in 1696. You may view the relevant page 273 (see right) via Google Books (and may download a PDF copy of the entire book by clicking on the [EBOOK - FREE] button). I also located a facsimile of his 1701 book, A Compleat Body of Arithmetick in Four Books (again via Google Books), which is actually the same publication. A facsimile of this 1701 publication is available for purchase if you so desire.
Once again, this is ridiculous. You get no views yet provide some of the most beneficial mathematical videos on the whole of youtube. Please man, keep up the good work. If it means anything, you are really helping me out as I am not the best maths student but have a genuine interest to get better, I only realized it a year ago.
Jack L (on a CCM YouTube video about How to Calculate an Approximate Cube Root for Any Number)
See all Testimonials | 1,255 | 5,184 | {"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-2019-35 | latest | en | 0.95451 |
https://brainly.in/question/135930 | 1,484,757,143,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280308.24/warc/CC-MAIN-20170116095120-00557-ip-10-171-10-70.ec2.internal.warc.gz | 791,307,189 | 9,541 | # Verify that xy[(x+y)(1/x+1/y) -4] = ( x-y)²
1
by samyukthasg
## Answers
2015-06-24T17:25:51+05:30
Taking L.H.S.
=xy[(x+y)(x+y/xy)-4]
=xy[(x+y)^2-4xy]/xy
=(x+y)^2 -4xy
=x^2 +y^2 +2xy-4xy
=x^+y^2 -2xy
=(x-y)^2
=R.H.S.
Hence proved. | 133 | 234 | {"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} | 3.53125 | 4 | CC-MAIN-2017-04 | latest | en | 0.561369 |
https://mathhelpboards.com/threads/canonical-basis-and-standard-basis.7634/ | 1,601,220,818,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400283990.75/warc/CC-MAIN-20200927152349-20200927182349-00041.warc.gz | 532,339,149 | 17,711 | # [SOLVED]Canonical Basis and Standard Basis
#### Sudharaka
##### Well-known member
MHB Math Helper
Hi everyone,
I have a little trouble understanding what Canonical basis means in the following question. I thought that Canonical basis is just another word for the Standard basis. Hope you people could clarify the difference between these two in the given context.
Question:
Find the canonical basis for the orthogonal thransformation $$f:\Re^3\rightarrow \Re^3$$ such that $$A_{f,\,B}=\frac{1}{3}\begin{pmatrix}2&-1&2\\2&2&-1\\-1&2&2\end{pmatrix}$$, $$B$$ being a standard basis of $$\Re^3$$.
#### Bacterius
##### Well-known member
MHB Math Helper
Hi everyone,
I have a little trouble understanding what Canonical basis means in the following question. I thought that Canonical basis is just another word for the Standard basis. Hope you people could clarify the difference between these two in the given context.
Question:
Find the canonical basis for the orthogonal thransformation $$f:\Re^3\rightarrow \Re^3$$ such that $$A_{f,\,B}=\frac{1}{3}\begin{pmatrix}2&-1&2\\2&2&-1\\-1&2&2\end{pmatrix}$$, $$B$$ being a standard basis of $$\Re^3$$.
I'm not sure either, but the only thing I imagine it could mean is the image of the standard basis under the given transformation. I.e. if you have a linear transformation $A : \mathbb{R}^3 \to \mathbb{R}^3$, then the canonical basis for this transformation is:
$$\left \{ A \left [ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right ], A \left [ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right ], A \left [ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right ] \right \}$$
(so it is the standard basis in the coordinate system described by this linear transformation)
#### Sudharaka
##### Well-known member
MHB Math Helper
I'm not sure either, but the only thing I imagine it could mean is the image of the standard basis under the given transformation. I.e. if you have a linear transformation $A : \mathbb{R}^3 \to \mathbb{R}^3$, then the canonical basis for this transformation is:
$$\left \{ A \left [ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right ], A \left [ \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right ], A \left [ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right ] \right \}$$
(so it is the standard basis in the coordinate system described by this linear transformation)
Thanks very much for the answer. I guess so too. But the thing is, for a grad level assignment this is too easy which keeps me thinking whether this is what our prof meant.
#### Bacterius
##### Well-known member
MHB Math Helper
Thanks very much for the answer. I guess so too. But the thing is, for a grad level assignment this is too easy which keeps me thinking whether this is what our prof meant.
Yes, that was my thought as well. It likely means something a bit less trivial, but I haven't seen this used anywhere and I can't find anything like it in my linear algebra book... so this was my best guess. Looking forward to other replies
#### Klaas van Aarsen
##### MHB Seeker
Staff member
The word canonical means obvious. Exactly that.
A canonical basis might be the standard basis, but that is unlikely, since they would have just called it the standard basis then.
If the matrix were diagonalizable with real numbers, I would expect it to be the normalized basis of eigenvectors.
However, your matrix is only diagonalizable with complex numbers.
Can it be that you're supposed to diagonalize it with complex numbers?
It would fit with the other problems you have brought here.
#### smile
##### New member
Hi everyone,
I have a little trouble understanding what Canonical basis means in the following question. I thought that Canonical basis is just another word for the Standard basis. Hope you people could clarify the difference between these two in the given context.
Question:
Find the canonical basis for the orthogonal thransformation $$f:\Re^3\rightarrow \Re^3$$ such that $$A_{f,\,B}=\frac{1}{3}\begin{pmatrix}2&-1&2\\2&2&-1\\-1&2&2\end{pmatrix}$$, $$B$$ being a standard basis of $$\Re^3$$.
I think we need to find the eigenvectors for this matrix, then we can use them to diagonalize this matrix. But the problem is that I am not sure wether we need to normalize those eigenvectors or not. If we do, it is not easy to normalize them.
#### Sudharaka
##### Well-known member
MHB Math Helper
The word canonical means obvious. Exactly that.
A canonical basis might be the standard basis, but that is unlikely, since they would have just called it the standard basis then.
If the matrix were diagonalizable with real numbers, I would expect it to be the normalized basis of eigenvectors.
However, your matrix is only diagonalizable with complex numbers.
Can it be that you're supposed to diagonalize it with complex numbers?
It would fit with the other problems you have brought here.
Yeah, I think for the moment this is the best assumption that we could make. However the problem with this is that in the problem we are dealing with the real number field and diagonalizing the matrix in complex number field would make our canonical basis not so obvious. Isn't?
I think we need to find the eigenvectors for this matrix, then we can use them to diagonalize this matrix. But the problem is that I am not sure wether we need to normalize those eigenvectors or not. If we do, it is not easy to normalize them.
I think we need to normalize the eigenvectors. The reason is otherwise it won't be unique. Am I correct ILSe?
#### smile
##### New member
Yeah, I think for the moment this is the best assumption that we could make. However the problem with this is that in the problem we are dealing with the real number field and diagonalizing the matrix in complex number field would make our canonical basis not so obvious. Isn't?
I think we need to normalize the eigenvectors. The reason is otherwise it won't be unique. Am I correct ILSe?
Yes, you are right, actually the normalization is not hard.
#### Klaas van Aarsen
##### MHB Seeker
Staff member
I think we need to normalize the eigenvectors. The reason is otherwise it won't be unique. Am I correct ILSe?
Since they ask for a canonical basis, I think you can get away with it without normalizing it.
You'd have a canonical basis of eigenvectors.
If they wanted it to be normalized, they should have asked for a normalized basis.
Either way, the problem statement is ambiguous, which is bad in math.
They should know better in my opinion.
#### Sudharaka
##### Well-known member
MHB Math Helper
Since they ask for a canonical basis, I think you can get away with it without normalizing it.
You'd have a canonical basis of eigenvectors.
If they wanted it to be normalized, they should have asked for a normalized basis.
Either way, the problem statement is ambiguous, which is bad in math.
They should know better in my opinion.
Yes, that was my thought as well. It likely means something a bit less trivial, but I haven't seen this used anywhere and I can't find anything like it in my linear algebra book... so this was my best guess. Looking forward to other replies
Thank you ILSe and Bacterius for all the valuable insight that you have provided on this question. It immensely helped me to clarify my doubts. | 1,778 | 7,232 | {"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": 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-2020-40 | longest | en | 0.812487 |
https://www.analystforum.com/t/bayes-theorem/69721 | 1,638,254,733,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358953.29/warc/CC-MAIN-20211130050047-20211130080047-00202.warc.gz | 701,281,040 | 5,178 | # Bayes theorem
Can somebody explain bayes theorem in simple terms?
Of course… Normally, when we talk about probability, we are talking about a future event. For e.g. if today is Monday, we can say that there is a 20% probability that it will rain on Tuesday. This could be conditional on seeing clouds. If the clouds have a 40% probability to show, then (R = Rain, C = Clouds), our equation becomes (R given C) = 20% We can further say that if there are no clouds on Monday, there is only a 5% chance of rain on Tuesday. What Bayes formula does is, it goes into the FUTURE and makes assumptions about the past. So, now, it is already Tuesday and it has rained. Now, we want to know if there were clouds in the sky on Monday. We try to define in a probabilistic manner. P(Rain and Clouds) = P(Rain given Clouds) * P (Clouds) - Multiplication rule But P(Rain and Clouds) is also = P(Clouds given Rain) * P(Rain) Replacing, P(Clouds given rain) * P(Rain) = P(Rain given clouds) * P(Clouds) Since we want to find out P(Clouds given rain), we rearrange to move P(Rain to the right) as P(Clouds given rain) = (P(Rain given clouds) * P(clouds)) / P(rain) Putting in all the values, =(0.2 * 0.4) / (0.4 * 0.2 + 0.6 * 0.05) (total probability rule in the denominator) = 72% This should make sense. If we know that it rained. Seeing that it is VERY likely that it will rain when there are clouds, there should be a high probability that there were clouds when it rained.
lankylint, that’s a good description. thanks
I am digging this one up from the graveyard, but why isnt the denominator simply 0.40 in the last equation like the equation above it demonstrates?
P(Clouds given rain) = (P(Rain given clouds) * P(clouds)) / P(rain)
to
=(0.2 * 0.4) / (0.4 * 0.2 + 0.6 * 0.05) (total probability rule in the denominator)
I wrote an article on Bayes’ formula that may be of some help here: http://financialexamhelp123.com/bayes-formula/
Now I feel like an idiot, thank you. I was overthinking it
My pleasure.
S2000magician Thanks a lot I had alot of trouble remembering Bayes formula, with your explanation it has become crystal clear for me. Keep up the good work!
Have a nice day.
hotmail account is oldschool
My pleasure. | 622 | 2,226 | {"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} | 4.21875 | 4 | CC-MAIN-2021-49 | latest | en | 0.941148 |
https://www.physicsforums.com/threads/eigenvalues-and-eigenfunctions.533840/ | 1,511,578,801,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934809229.69/warc/CC-MAIN-20171125013040-20171125033040-00630.warc.gz | 832,254,250 | 15,486 | # Eigenvalues and eigenfunctions
1. Sep 26, 2011
### c299792458
1. The problem statement, all variables and given/known data
How does one find all the permissible values of $b$ for $-{d\over dx}(-e^{ax}y')-ae^{ax}y=be^{ax}y$ with boundary conditions $y(0)=y(1)=0$?
Thanks.
2. Relevant equations
See above
3. The attempt at a solution
I assume we have a discrete set of $\{b_n\}$ where they can be regarded as eigenvalues? After that how does one find the corresponding $\{y_n\}$? I am sure we substitute the $\{b_n\}$ into the equation, but then I still don't know how this equation is solved. Please help! Perhaps it is easier to find the permissible $b$'s if we write the equation in the form $y''+ay'+(a+b)y=0$?
2. Sep 26, 2011
### HallsofIvy
Staff Emeritus
Well, it would be y''+ ay'- (a+b)y= 0, but yes, that is the simplest way to approach this problem. What is the general solution to that equation? What must b equal in order that there be a solution to this boundary value problem? Remember that the characteristice equation may, depending upon b, have real or complex roots.
3. Sep 26, 2011
### c299792458
@HallsofIvy:
Would the general solution be $y(x)=A\exp[\frac{-a+\sqrt{a^2+4(a+b)}}{2}]+B\exp[\frac{-a-\sqrt{a^2+4(a+b)}}{2}]$
And then the BC's mean that $A+B=0$ and $\exp[\frac{-a+\sqrt{a^2+4(a+b)}}{2}]-\exp[\frac{-a-\sqrt{a^2+4(a+b)}}{2}]=0$, therefore we need $\sqrt{a^2+4(a+b)}=0$ i.e. $b={-1\over 4}(a^2+4a)$? Is this the only permissible $b$?
Thanks.
4. Sep 26, 2011
### HallsofIvy
Staff Emeritus
No. You are assuming, incorrectly, that the solution must be of the form $Ce^{r_1x}+ De^{r_2x}$. That is true only if the characteristic equation has two real roots. In fact, what you give is NOT a "permissible value of b". With that value of b, your characteristic equation reduces to $(r+ a/2)^2= 0$ so that the only solution is $-a/2$. In that case, the general solution to the equation is $y(x)= Ce^{-ax/2}+ Dte^{-ax/2}$. The condition that y(0)= 0 gives C= 0 and then the condition that y(1)= 0 gives $De^{-a/2}= 0$ so that D= 0. That is not a non-trivial solution.
As I said before, look at complex roots to the characteristic equation.
5. Sep 26, 2011
### c299792458
@HallsofIvy:
Thanks. So the general solution is $y(x)=\exp({-ax\over 2})[A\sin({\sqrt{-a-4(a+b)}\over 2}x)+B\cos({\sqrt{-a-4(a+b)}\over 2}x)].$
Then $B=0$ and we need $A\sin({\sqrt{-a-4(a+b)}\over 2}x)=0$ for $A\neq 0$ so ${\sqrt{-a-4(a+b)}\over 2}=n\pi$
Thanks again. | 860 | 2,486 | {"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} | 3.96875 | 4 | CC-MAIN-2017-47 | longest | en | 0.825235 |
https://math.libretexts.org/Courses/Monroe_Community_College/MTH_211_Calculus_II/Chapter_8%3A_Introduction_to_Differential_Equations/8.3%3A_Separable_Differential_Equations | 1,726,457,570,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.29/warc/CC-MAIN-20240916012328-20240916042328-00829.warc.gz | 350,189,810 | 37,047 | # 8.3: Separable Differential Equations
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##### Learning Objectives
• Use separation of variables to solve a differential equation.
• Solve applications using separation of variables.
We now examine a solution technique for finding exact solutions to a class of differential equations known as separable differential equations. These equations are common in a wide variety of disciplines, including physics, chemistry, and engineering. We illustrate a few applications at the end of the section.
## Separation of Variables
##### Definition: Separable Differential Equations
A separable differential equation is any equation that can be written in the form
$y'=f(x)g(y). \label{sep}$
The term ‘separable’ refers to the fact that the right-hand side of Equation \ref{sep} can be separated into a function of $$x$$ times a function of $$y$$. Examples of separable differential equations include
\begin{align} y' &=(x^2−4)(3y+2) \label{eq1} \\[4pt] y' &=6x^2+4x \label{eq2}\\[4pt] y' &=\sec y+\tan y \label{eq3} \\[4pt] y' &=xy+3x−2y−6. \label{eq4} \end{align}
Equation \ref{eq2} is separable with $$f(x)=6x^2+4x$$ and $$g(y)=1$$, Equation \ref{eq3} is separable with $$f(x)=1$$ and $$g(y)=\sec y+\tan y,$$ and the right-hand side of Equation \ref{eq4} can be factored as $$(x+3)(y−2)$$, so it is separable as well. Equation \ref{eq3} is also called an autonomous differential equation because the right-hand side of the equation is a function of $$y$$ alone. If a differential equation is separable, then it is possible to solve the equation using the method of separation of variables.
##### Problem-Solving Strategy: Separation of Variables
1. Check for any values of $$y$$ that make $$g(y)=0.$$ These correspond to constant solutions.
2. Rewrite the differential equation in the form $\dfrac{dy}{g(y)}=f(x)dx. \nonumber$
3. Integrate both sides of the equation.
4. Solve the resulting equation for $$y$$ if possible.
5. If an initial condition exists, substitute the appropriate values for $$x$$ and $$y$$ into the equation and solve for the constant.
Note that Step 4 states “Solve the resulting equation for $$y$$ if possible.” It is not always possible to obtain $$y$$ as an explicit function of $$x$$. Quite often we have to be satisfied with finding y as an implicit function of $$x$$.
##### Example $$\PageIndex{1}$$: Using Separation of Variables
Find a general solution to the differential equation $$y'=(x^2−4)(3y+2)$$ using the method of separation of variables.
###### Solution
Follow the five-step method of separation of variables.
1. In this example, $$f(x)=x^2−4$$ and $$g(y)=3y+2$$. Setting $$g(y)=0$$ gives $$y=−\dfrac{2}{3}$$ as a constant solution.
2. Rewrite the differential equation in the form
$\dfrac{dy}{3y+2}=(x^2−4)\,dx.\nonumber$
3. Integrate both sides of the equation:
$∫\dfrac{dy}{3y+2}=∫(x^2−4)\,dx.\nonumber$
Let $$u=3y+2$$. Then $$du=3\dfrac{dy}{dx}\,dx$$, so the equation becomes
$\dfrac{1}{3}∫\dfrac{1}{u}\,du=\dfrac{1}{3}x^3−4x+C\nonumber$
$\dfrac{1}{3}\ln|u|=\dfrac{1}{3}x^3−4x+C\nonumber$
$\dfrac{1}{3}\ln|3y+2|=\dfrac{1}{3}x^3−4x+C.\nonumber$
4. To solve this equation for $$y$$, first multiply both sides of the equation by $$3$$.
$\ln|3y+2|=x^3−12x+3C\nonumber$
Now we use some logic in dealing with the constant $$C$$. Since $$C$$ represents an arbitrary constant, $$3C$$ also represents an arbitrary constant. If we call the second arbitrary constant $$C_1,$$ where $$C_1 = 3C,$$ the equation becomes
$\ln|3y+2|=x^3−12x+C_1.\nonumber$
Now exponentiate both sides of the equation (i.e., make each side of the equation the exponent for the base $$e$$).
\begin{align*} e^{\ln|3y+2|} &=e^{x^3−12x+C_1} \\ |3y+2| &=e^{C_1}e^{x^3−12x} \end{align*}
Again define a new constant $$C_2= e^{C_1}$$ (note that $$C_2 > 0$$):
$|3y+2|=C_2e^{x^3−12x}.\nonumber$
Because of the absolute value on the left side of the equation, this corresponds to two separate equations:
$3y+2=C_2e^{x^3−12x}\nonumber$
and
$3y+2=−C_2e^{x^3−12x}.\nonumber$
The solution to either equation can be written in the form
$y=\dfrac{−2±C_2e^{x^3−12x}}{3}.\nonumber$
Since $$C_2>0$$, it does not matter whether we use plus or minus, so the constant can actually have either sign. Furthermore, the subscript on the constant $$C$$ is entirely arbitrary, and can be dropped. Therefore the solution can be written as
$y=\dfrac{−2+Ce^{x^3−12x}}{3}, \text{ where }C = \pm C_2\text{ or } C = 0.\nonumber$
Note that in writing a single general solution in this way, we are also allowing $$C$$ to equal $$0$$. This gives us the singular solution, $$y = -\dfrac{2}{3}$$, for the given differential equation. Check that this is indeed a solution of this differential equation!
5. No initial condition is imposed, so we are finished.
##### Exercise $$\PageIndex{1}$$
Use the method of separation of variables to find a general solution to the differential equation
$y'=2xy+3y−4x−6. \nonumber$
Hint
First factor the right-hand side of the equation by grouping, then use the five-step strategy of separation of variables.
$y=2+Ce^{x^2+3x} \nonumber$
##### Example $$\PageIndex{2}$$: Solving an Initial-Value Problem
Using the method of separation of variables, solve the initial-value problem
$y'=(2x+3)(y^2−4),\quad y(0)=−1.\nonumber$
###### Solution
Follow the five-step method of separation of variables.
1. In this example, $$f(x)=2x+3$$ and $$g(y)=y^2−4$$. Setting $$g(y)=0$$ gives $$y=±2$$ as constant solutions.
2. Divide both sides of the equation by $$y^2−4$$ and multiply by $$dx$$. This gives the equation
$\dfrac{dy}{y^2−4}=(2x+3)\,dx.\nonumber$
3. Next integrate both sides:
$∫\dfrac{1}{y^2−4}dy=∫(2x+3)\,dx. \label{Ex2.2}$
To evaluate the left-hand side, use the method of partial fraction decomposition. This leads to the identity
$\dfrac{1}{y^2−4}=\dfrac{1}{4}\left(\dfrac{1}{y−2}−\dfrac{1}{y+2}\right).\nonumber$
Then Equation \ref{Ex2.2} becomes
$\dfrac{1}{4}∫\left(\dfrac{1}{y−2}−\dfrac{1}{y+2}\right)dy=∫(2x+3)\,dx\nonumber$
$\dfrac{1}{4}\left (\ln|y−2|−\ln|y+2| \right)=x^2+3x+C.\nonumber$
Multiplying both sides of this equation by $$4$$ and replacing $$4C$$ with $$C_1$$ gives
$\ln|y−2|−\ln|y+2|=4x^2+12x+C_1\nonumber$
$\ln \left|\dfrac{y−2}{y+2}\right|=4x^2+12x+C_1.\nonumber$
4. It is possible to solve this equation for $$y.$$ First exponentiate both sides of the equation and define $$C_2=e^{C_1}$$:
$\left|\dfrac{y−2}{y+2}\right|=C_2e^{4x^2+12x}.\nonumber$
Next we can remove the absolute value and let a new constant $$C_3$$ be positive, negative, or zero, i.e., $$C_3 =\pm C_2$$ or $$C_3 = 0.$$
Then multiply both sides by $$y+2$$.
$y−2=C_3(y+2)e^{4x^2+12x}\nonumber$
$y−2=C_3ye^{4x^2+12x}+2C_3e^{4x^2+12x}.\nonumber$
Now collect all terms involving $$y$$ on one side of the equation, and solve for $$y$$:
$y−C_3ye^{4x^2+12x}=2+2C_3e^{4x^2+12x}\nonumber$
$y\big(1−C_3e^{4x^2+12x}\big)=2+2C_3e^{4x^2+12x}\nonumber$
$y=\dfrac{2+2C_3e^{4x^2+12x}}{1−C_3e^{4x^2+12x}}.\nonumber$
5. To determine the value of $$C_3$$, substitute $$x=0$$ and $$y=−1$$ into the general solution. Alternatively, we can put the same values into an earlier equation, namely the equation $$\dfrac{y−2}{y+2}=C_3e^{4x^2+12}$$. This is much easier to solve for $$C_3$$:
$\dfrac{y−2}{y+2}=C_3e^{4x^2+12x}\nonumber$
$\dfrac{−1−2}{−1+2}=C_3e^{4(0)^2+12(0)}\nonumber$
$C_3=−3.\nonumber$
Therefore the solution to the initial-value problem is
$y=\dfrac{2−6e^{4x^2+12x}}{1+3e^{4x^2+12x}}.\nonumber$
A graph of this solution appears in Figure $$\PageIndex{1}$$.
##### Exercise $$\PageIndex{2}$$
Find the solution to the initial-value problem
$6y'=(2x+1)(y^2−2y−8) \nonumber$
with $$y(0)=−3$$ using the method of separation of variables.
Hint
Follow the steps for separation of variables to solve the initial-value problem.
$y=\dfrac{4+14e^{x^2+x}}{1−7e^{x^2+x}} \nonumber$
## Applications of Separation of Variables
Many interesting problems can be described by separable equations. We illustrate two types of problems: solution concentrations and Newton’s law of cooling.
#### Solution concentrations
Consider a tank being filled with a salt solution. We would like to determine the amount of salt present in the tank as a function of time. We can apply the process of separation of variables to solve this problem and similar problems involving solution concentrations.
##### Example $$\PageIndex{3}$$: Determining Salt Concentration over Time
A tank containing $$100$$ L of a brine solution initially has $$4$$ kg of salt dissolved in the solution. At time $$t=0$$, another brine solution flows into the tank at a rate of $$2$$ L/min. This brine solution contains a concentration of $$0.5$$ kg/L of salt. At the same time, a stopcock is opened at the bottom of the tank, allowing the combined solution to flow out at a rate of $$2$$ L/min, so that the level of liquid in the tank remains constant (Figure $$\PageIndex{2}$$). Find the amount of salt in the tank as a function of time (measured in minutes), and find the limiting amount of salt in the tank, assuming that the solution in the tank is well mixed at all times.
###### Solution
First we define a function $$u(t)$$ that represents the amount of salt in kilograms in the tank as a function of time. Then $$\dfrac{du}{dt}$$ represents the rate at which the amount of salt in the tank changes as a function of time. Also, $$u(0)$$ represents the amount of salt in the tank at time $$t=0$$, which is $$4$$ kilograms.
The general setup for the differential equation we will solve is of the form
$\dfrac{du}{dt}=\text{INFLOW RATE − OUTFLOW RATE.} \nonumber$
INFLOW RATE represents the rate at which salt enters the tank, and OUTFLOW RATE represents the rate at which salt leaves the tank. Because solution enters the tank at a rate of $$2$$ L/min, and each liter of solution contains $$0.5$$ kilogram of salt, every minute $$2(0.5)=1$$ kilogram of salt enters the tank. Therefore INFLOW RATE = $$1$$.
To calculate the rate at which salt leaves the tank, we need the concentration of salt in the tank at any point in time. Since the actual amount of salt varies over time, so does the concentration of salt. However, the volume of the solution remains fixed at 100 liters. The number of kilograms of salt in the tank at time $$t$$ is equal to $$u(t)$$. Thus, the concentration of salt is $$\dfrac{u(t)}{100}$$ kg/L, and the solution leaves the tank at a rate of $$2$$ L/min. Therefore salt leaves the tank at a rate of $$\dfrac{u(t)}{100}⋅2=\dfrac{u(t)}{50}$$ kg/min, and OUTFLOW RATE is equal to $$\dfrac{u(t)}{50}$$. Therefore the differential equation becomes $$\dfrac{du}{dt}=1−\dfrac{u}{50}$$, and the initial condition is $$u(0)=4.$$ The initial-value problem to be solved is
$\dfrac{du}{dt}=1−\dfrac{u}{50},u(0)=4.\nonumber$
The differential equation is a separable equation, so we can apply the five-step strategy for solution.
Step 1. Setting $$1−\dfrac{u}{50}=0$$ gives $$u=50$$ as a constant solution. Since the initial amount of salt in the tank is $$4$$ kilograms, this solution does not apply.
Step 2. Rewrite the equation as
$\dfrac{du}{dt}=\dfrac{50−u}{50}.\nonumber$
Then multiply both sides by $$dt$$ and divide both sides by $$50−u:$$
$\dfrac{du}{50−u}=\dfrac{dt}{50}.\nonumber$
Step 3. Integrate both sides:
\begin{align*} ∫\dfrac{du}{50−u} &=∫\dfrac{dt}{50} \\ −\ln|50−u| &=\dfrac{t}{50}+C. \end{align*}
Step 4. Solve for $$u(t)$$:
$\ln|50−u|=−\dfrac{t}{50}−C\nonumber$
$e^{\ln|50−u|}=e^{−(t/50)−C}\nonumber$
$|50−u|=C_1e^{−t/50}, \text{ where } C_1 = e^{-C}.\nonumber$
Eliminate the absolute value by allowing the constant to be positive, negative, or zero, i.e., $$C_1 = \pm e^{-C}$$ or $$C_1 = 0$$:
$50−u=C_1e^{−t/50}.\nonumber$
Finally, solve for $$u(t)$$:
$u(t)=50−C_1e^{−t/50}.\nonumber$
Step 5. Solve for $$C_1$$:
\begin{align*} u(0) &=50−C_1e^{−0/50} \\ 4 &=50−C_1 \\ C_1 &=46. \end{align*}
The solution to the initial value problem is $$u(t)=50−46e^{−t/50}.$$ To find the limiting amount of salt in the tank, take the limit as $$t$$ approaches infinity:
\begin{align*} \lim_{t→∞}u(t) &=50−46e^{−t/50} \\ &=50−46(0)=50. \end{align*}
Note that this was the constant solution to the differential equation. If the initial amount of salt in the tank is $$50$$ kilograms, then it remains constant. If it starts at less than $$50$$ kilograms, then it approaches $$50$$ kilograms over time.
##### Exercise $$\PageIndex{3}$$
A tank contains $$3$$ kilograms of salt dissolved in $$75$$ liters of water. A salt solution of $$0.4$$ kg salt/L is pumped into the tank at a rate of $$6$$ L/min and is drained at the same rate. Solve for the salt concentration at time $$t$$. Assume the tank is well mixed at all times.
Hint
Follow the steps in Example $$\PageIndex{3}$$ and determine an expression for INFLOW and OUTFLOW. Formulate an initial-value problem, and then solve it.
Initial value problem:
$\dfrac{du}{dt}=2.4−\dfrac{2u}{25},\, u(0)=3 \nonumber$
$u(t)=30−27e^{−t/50} \nonumber$
## Newton’s Law of Cooling
Newton’s law of cooling states that the rate of change of an object’s temperature is proportional to the difference between its own temperature and the ambient temperature (i.e., the temperature of its surroundings). If we let $$T(t)$$ represent the temperature of an object as a function of time, then $$\dfrac{dT}{dt}$$ represents the rate at which that temperature changes. The temperature of the object’s surroundings can be represented by $$T_s$$. Then Newton’s law of cooling can be written in the form
$\dfrac{dT}{dt}=k(T(t)−T_s) \nonumber$
or simply
$\dfrac{dT}{dt}=k(T−T_s). \nonumber$
The temperature of the object at the beginning of any experiment is the initial value for the initial-value problem. We call this temperature $$T_0$$. Therefore the initial-value problem that needs to be solved takes the form
$\dfrac{dT}{dt}=k(T−T_s) \label{newton}$
with $$T(0)=T_0$$, where $$k$$ is a constant that needs to be either given or determined in the context of the problem. We use these equations in Example $$\PageIndex{4}$$.
##### Example $$\PageIndex{4}$$: Waiting for a Pizza to Cool
A pizza is removed from the oven after baking thoroughly, and the temperature of the oven is $$350°F.$$ The temperature of the kitchen is $$75°F$$, and after $$5$$ minutes the temperature of the pizza is $$340°F$$. We would like to wait until the temperature of the pizza reaches $$300°F$$ before cutting and serving it (Figure $$\PageIndex{3}$$). How much longer will we have to wait?
###### Solution
The ambient temperature (surrounding temperature) is $$75°F$$, so $$T_s=75$$. The temperature of the pizza when it comes out of the oven is $$350°F$$, which is the initial temperature (i.e., initial value), so $$T_0=350$$. Therefore Equation \ref{newton} becomes
$\dfrac{dT}{dt}=k(T−75) \nonumber$
with $$T(0)=350.$$
To solve the differential equation, we use the five-step technique for solving separable equations.
1. Setting the right-hand side equal to zero gives $$T=75$$ as a constant solution. Since the pizza starts at $$350°F,$$ this is not the solution we are seeking.
2. Rewrite the differential equation by multiplying both sides by $$dt$$ and dividing both sides by $$T−75$$:
$\dfrac{dT}{T−75}=k\,dt. \nonumber$
3. Integrate both sides:
\begin{align*} ∫\dfrac{dT}{T−75} &=∫k\,dt \\ \ln|T−75| &=kt+C.\end{align*} \nonumber
4. Solve for $$T$$ by first exponentiating both sides:
\begin{align*}e^{\ln|T−75|} &=e^{kt+C} \\ |T−75| &=C_1e^{kt}, & & \text{where } C_1 = e^C. \\ T−75 &=\pm C_1e^{kt} \\ T−75 &=Ce^{kt}, & & \text{where } C = \pm C_1\text{ or } C = 0.\\ T(t) &=75+Ce^{kt}. \end{align*} \nonumber
5. Solve for $$C$$ by using the initial condition $$T(0)=350:$$
\begin{align*}T(t) &=75+Ce^{kt}\\ T(0) &=75+Ce^{k(0)} \\ 350 &=75+C \\ C &=275.\end{align*} \nonumber
Therefore the solution to the initial-value problem is
$T(t)=75+275e^{kt}.\nonumber$
To determine the value of $$k$$, we need to use the fact that after $$5$$ minutes the temperature of the pizza is $$340°F$$. Therefore $$T(5)=340.$$ Substituting this information into the solution to the initial-value problem, we have
$T(t)=75+275e^{kt}\nonumber$
$T(5)=340=75+275e^{5k}\nonumber$
$265=275e^{5k}\nonumber$
$e^{5k}=\dfrac{53}{55}\nonumber$
$\ln e^{5k}=\ln(\dfrac{53}{55})\nonumber$
$5k=\ln(\dfrac{53}{55})\nonumber$
$k=\dfrac{1}{5}\ln(\dfrac{53}{55})≈−0.007408.\nonumber$
So now we have $$T(t)=75+275e^{−0.007048t}.$$ When is the temperature $$300°F$$? Solving for $$t,$$ we find
$T(t)=75+275e^{−0.007048t}\nonumber$
$300=75+275e^{−0.007048t}\nonumber$
$225=275e^{−0.007048t}\nonumber$
$e^{−0.007048t}=\dfrac{9}{11}\nonumber$
$\ln e^{−0.007048t}=\ln\dfrac{9}{11}\nonumber$
$−0.007048t=\ln\dfrac{9}{11}\nonumber$
$t=−\dfrac{1}{0.007048}\ln\dfrac{9}{11}≈28.5.\nonumber$
Therefore we need to wait an additional $$23.5$$ minutes (after the temperature of the pizza reached $$340°F$$). That should be just enough time to finish this calculation.
##### Exercise $$\PageIndex{4}$$
A cake is removed from the oven after baking thoroughly, and the temperature of the oven is $$450°F$$. The temperature of the kitchen is $$70°F$$, and after $$10$$ minutes the temperature of the cake is $$430°F$$.
1. Write the appropriate initial-value problem to describe this situation.
2. Solve the initial-value problem for $$T(t)$$.
3. How long will it take until the temperature of the cake is within $$5°F$$ of room temperature?
Hint
Determine the values of $$T_s$$ and $$T_0$$ then use Equation \ref{newton}.
Initial-value problem $\dfrac{dT}{dt}=k(T−70),\quad T(0)=450\nonumber$
$T(t)=70+380e^{kt}\nonumber$
Approximately $$114$$ minutes.
## Key Concepts
• A separable differential equation is any equation that can be written in the form $$y'=f(x)g(y).$$
• The method of separation of variables is used to find the general solution to a separable differential equation.
## Key Equations
• Separable differential equation
$$y′=f(x)g(y)$$
• Solution concentration
$$\dfrac{du}{dt}=\text{INFLOW RATE − OUTFLOW RATE}$$
• Newton’s law of cooling
$$\dfrac{dT}{dt}=k(T−T_s)$$
## Glossary
autonomous differential equation
an equation in which the right-hand side is a function of $$y$$ alone
separable differential equation
any equation that can be written in the form $$y'=f(x)g(y)$$
separation of variables
a method used to solve a separable differential equation
This page titled 8.3: Separable Differential Equations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by OpenStax. | 7,635 | 22,880 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2024-38 | latest | en | 0.20126 |
https://www.go4expert.com/forums/help-jacobi-iterative-t2086/ | 1,555,813,057,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578530161.4/warc/CC-MAIN-20190421020506-20190421042506-00454.warc.gz | 699,454,229 | 11,270 | # Help w/ Jacobi Iterative
Discussion in 'C' started by bluelolypop, Dec 1, 2006.
1. ### bluelolypopNew Member
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Hi I can't find how to do this in my book ANYWHERE so i dunno how to do this.. it's my hw and it's due tomorrow at 9 pm... if anyone can help that'd be nice... thanks a lot. These are the instructions but I don't even know how to start it..
/* In void prob1() you are asked to solve the system of equations:
a[0]*x[0]+a[1]*x[1]+...+a[4]*x[4] = b for i=0, 1, ..., 4
using the Jacobi iterative method, where a[5][5] and b[5] are defined
as follows:
double a[5][5]={ 5., 0., 1., 0., -1.,
0., 4., 3., -1., 0.,
0., 0., 3., 1., -5.,
1., -1., 0., 3., 0.,
1., 1., 0., 0., -4.}
double b[5]={-1., -4., 1.,-3., 3.0}, x0[5]={0.}, x1[5];
In the method, one starts with an approximate x0[] and obtains an
improved solution x1[]:
x1 =1/a *(b - (sum on j, except j=i)a[j]*x0[j]))
for i=0,...,4.
Perform six** iterations to find an approximate solution
x1[] starting from the initial guess x0[0]=x0[1]=...=x0[4]=0.0.
After each iteration, print the values of x1[] by using %12.3e.*/ | 437 | 1,136 | {"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} | 3.515625 | 4 | CC-MAIN-2019-18 | longest | en | 0.771428 |
https://curriculum.media.pearsoncmg.com/curriculum/math/digits/TXgrade_8/html_books/homework_helper_vol1/htmls/page_124.html | 1,606,315,231,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141182794.28/warc/CC-MAIN-20201125125427-20201125155427-00606.warc.gz | 246,977,650 | 4,991 | # Unit C: Expressions, Equations, and Relationships
## Topic 4: Using the Pythagorean Theorem
### 4-1: The Pythagorean Theorem
#### Key Concept
In a right triangle, the two shortest sides are legs.
The longest side, which is opposite the right angle, is the hypotenuse.
You can represent the lengths of the legs of the triangle with a and b, and the length of the hypotenuse with c.
In any right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. This equation is known as the Pythagorean Theorem.
#### Part 1
Intro
You can use the Pythagorean Theorem to find the length of the hypotenuse of a right triangle when you know the lengths of the legs.
Example Finding Lengths of Hypotenuses Using the Pythagorean Theorem
What is the length of the hypotenuse of each triangle?
• a.
End ofPage 124 | 219 | 870 | {"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-2020-50 | latest | en | 0.874574 |
https://stats.stackexchange.com/questions/110070/log-uniform-distributions | 1,713,837,647,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00225.warc.gz | 478,015,428 | 43,266 | # Log-uniform distributions
I am having some difficulty understanding what log uniform distributions are.
Suppose that $\log X$ is uniformly distributed on the interval $[1,e]$. How do I describe $P(X=x)$? It seems like there is more probability mass on the lower numbers so that X itself is not uniformly distributed, but I am having difficulty formalizing this argument.
• Change-of-variable formula (with some thought regarding the support of X)?... but $P(X=x)$ is non-zero only for discrete r.v.'s. Is $X$ a discrete r.v.? Jul 31, 2014 at 1:07
• I agree with Alecos that you need the density function. I bet you can figure out the cdf $F(x) = P(X \le x)$, then get the density. Jul 31, 2014 at 1:27
Your definition of $X$ suggests that $X$ is a continuous random variable, but your question $\Pr[X = x]$ suggests you wish to treat it as a discrete variable. If you were asking for the probability density function of $X$, rather than the probability mass function, then we could proceed naturally using a transformation, since $\log$ is a monotone function: if $Y = g^{-1}(X) = \log X$, then $X = g(Y) = e^Y$ and $$f_X(x) = f_Y(g^{-1}(x)) \left| \frac{dg^{-1}}{dx} \right| = \ldots$$ This of course means that $X$ is not uniformly distributed.
• Thanks. When I carry this through I find that the density is some constant over $x$inside the interval and zero outside. Is that correct? Jul 31, 2014 at 14:01
• Did you actually apply @heropup's suggestion? What is the density of $\log X$? What is $\left| \frac{dg^{-1}}{dx} \right|$? And no, the density of $X$ is not a constant density -this would mean that it would also be uniform, and it is not. And what is the support of $X$? Aug 1, 2014 at 3:04
• @AlecosPapadopoulos The OP wrote "the density is some constant over $x$," which I interpreted to mean $f_X(x) = c/x$ for some constant $c$. This is correct. Aug 1, 2014 at 3:35
• Hmmm, yes "something over x" means also that. Aug 1, 2014 at 10:12
• I don't disagree with what you've written, however the steps that @heropup has laid out is all but a full solution. At this point, what's left isn't an exercise in statistical thinking, so much as it is an exercise in plugging in the various pieces and performing the prerequisite calculus... I don't think we're robbing the rest of the world of a valid chance to thoughtfully further inquire into this question by completing the derivation (given how far it's already come). I welcome a valid argument to the contrary, however. Jan 15, 2016 at 21:49
I like @heropup's answer, but am slightly bothered by the fact that he didn't finish the derivation for the OP. To enrich his answer, I'd like to add the following picture, and some comments on the above answer:
If you follow @heropup's derivation, you'll find that
$$f_{X}(x) = \frac{I_{[e, e^e]}(x)}{x(e-1)}$$
More generally, if $Y \sim Unif(a,b)$, such that $Y = log(X)$ for some random variable $X$, then
$$f_{X}(x) = \frac{I_{[e^a, e^b]}(x)}{x(b - a)}$$
For the sake of validating your intuition, I've made the figure of $Y \sim Unif(0,1)$, so that we see the originating random variable is actually defined on $[1, e]$ and looks something like $1/x$. As you pointed out, there is, indeed, more mass at the "beginning" of $X$'s domain and from the picture of the CDF, alone, we can see that $X$ cannot also be uniformly distributed (since the CDF is not a straight line).
I hope this fleshes out a bit of the above answer...
edit: you can find the code to make this picture here.
• What does "$\delta$" mean?
– whuber
Jan 15, 2016 at 18:25
• $\delta(x)$ is the delta function, and takes value 1 when $x=0$ and 0, otherwise. I'm abusing it here to take on the value 1 whenever the statement inside of it evaluates to true. In our case, it is 1 (or "on") whenever $x$ is an element of the stated set, and 0 (or "off") when it isn't. It's just a way to explicitly state the support of a function. Jan 15, 2016 at 21:43
• Thank you for the explanation. Whenever you use notation that is idiosyncratic or might not be well known, it's a good idea to explain it in your answer--especially when the question is focusing on how one "describes" a distribution. A more common notation, especially in statistics, is to use "$I$" as a mnemonic for an indicator function, as in $I_{[e^a,e^b]}(x)$, which is not an abuse of notation at all. (Mathematicians often use "$\chi$" but that could cause confusion in some statistical contexts.)
– whuber
Jan 15, 2016 at 22:17
• As a former mathematics/statistics student, I remember being exposed to a myriad of notations for the indicator function (sometimes $I$, sometimes $\mathbf{1}$, sometimes the hacky $\delta$). I didn't realize this was idiosyncratic to my education... Anyway, the answer has been edited to reflect your convention. Jan 16, 2016 at 12:27
• Personally, I'm not fussy about notation. As a moderator, I only check that questions and answers can be readily understood by most readers and are not open to misinterpretation. $\delta$ would be fine provided it's explained. The "$I$" notation is by far the most prevalent on this site.
– whuber
Jan 16, 2016 at 14:55 | 1,428 | 5,139 | {"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-2024-18 | latest | en | 0.941338 |
http://mathhelpforum.com/trigonometry/29261-trig-identity.html | 1,481,019,328,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541896.91/warc/CC-MAIN-20161202170901-00124-ip-10-31-129-80.ec2.internal.warc.gz | 168,764,954 | 9,669 | 1. ## trig identity
i need help with this identity, could someone tell me where i went wrong & help me?
2. Hello, boonies!
Your simplifying was off . . .
$({\color{blue}\tan\theta\cos\theta} + 2\cos\theta)^2 + \left(2\sin\theta - {\color{red}\frac{\sin\theta}{\tan\theta}}\right)^ 2 \;=\;5$
. . ${\color{blue}\tan\theta\cos\theta \:=\:\frac{\sin\theta}{\cos\theta}\!\cdot\!\cos\th eta \:=\:\sin\theta}$
. . ${\color{red}\frac{\sin\theta}{\tan\theta} \:=\:\frac{\sin\theta}{\frac{\sin\theta}{\cos\thet a}} \:=\:\cos\theta}$
The problem becomes: . $({\color{blue}\sin\theta} + 2\cos\theta)^2 + (2\sin\theta - {\color{red}\cos\theta})^2$
. . $= \;\sin^2\!\theta + 4\sin\theta\cos\theta + 4\cos^2\!\theta + 4\sin^2\!\theta - 4\sin\theta\cos\theta + \cos^2\!\theta$
. . $= \;5\sin^2\!\theta + 5\cos^2\!\theta \;=\;5\underbrace{(\sin^2\!\theta + \cos^2\!\theta)}_{\text{This is 1}} \;=\;5(1) \;=\;5$ | 381 | 901 | {"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": 0, "codecogs_latex": 6, "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-2016-50 | longest | en | 0.425022 |
https://infinitylearn.com/surge/blog/iit-jee/physics-and-measurement/ | 1,723,490,946,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641048885.76/warc/CC-MAIN-20240812190307-20240812220307-00245.warc.gz | 250,776,312 | 27,648 | Physics and Measurement
# Physics and Measurement
Physics and measurements are mostly concerned with our daily lives. The study of nature and its laws is the subject of physics, a branch of science. For example, if we understand Newton’s laws of gravitation and motion, we can explain how the moon orbits the earth, how an apple falls from a tree, and how tides in the sea change on a full moon night. Physics is concerned with the fundamental laws that govern all aspects of life. We use measurement in our daily lives, and it helps us understand the fundamentals of physical quantities. Everyone in his life must have gone to the market to buy groceries and vegetables at some point. If you go to the market to buy potatoes, the vegetable vendor will ask you how many kilograms you want, and you will tell him that you want 5 kg. This is the weight of the potatoes, 5 kg. If you go to the store to buy milk, you will do so in liters. When you buy clothes, you do so in meters. You will also learn how to use dimensional analysis to check equations for errors. This weighing machine can be used to measure mass.
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Similarly, length can be measured with scales, measuring tapes, and other tools. We will be able to determine the units, dimensions, and formulas of physical quantities as well as measure the size of an object. After measuring these values, we will learn how to calculate measurement errors and how to check the accuracy of our calculations.
## Physics And Measurements
The quantity is measured in two parts: the first part indicates how many times the standard unit is used, and the second part represents the name of the unit. In this chapter, we will look at the standard units of fundamental physical quantities such as length, time, mass, and temperature, as well as derived quantities that can be expressed in terms of fundamental quantities. We will learn about the dimensional analysis of quantities and then be able to determine whether or not the given formulas for any quantity are correct. If you measure something and get an error in that measurement, you will be able to find the error in your measurements after studying this chapter. Measurement, like any other scientific subject, is an essential component of Physics. Measurement is an essential component of the human race; without it, there would be no trade or statistics. The philosophy of measurement can be seen in small children who have no idea what math is.
Kids try to compare their height, the size of their candy, the size of their dolls, and the number of toys they own. All of this occurs even before they understand math.
Math is hardwired into our brains long before we begin to learn it. Math is a great way to learn about anything, which is why computers are involved in almost everything because they are good at math.
## Dimensional and Unit Analysis
In mathematics, dimensional analysis is the analysis of the relationships between different physical quantities by considering their dimensions. In physics, the dimensions of a physical quantity are the product of its mass, length, and time units. For example, the dimensions of velocity are mass (M) length (L) per time (T).
Dimensional analysis is used to make calculations easier and to understand the relationships between physical quantities. In particular, it can be used to convert between different units of measure. For example, the dimensions of a liter are mass (M) length (L) time (T) squared. So, to convert from liters to cubic meters, we can use the following equation:
1 cubic meter = 1,000 liters
This equation can be rearranged to solve for liters:
1,000 liters = 1 cubic meter
Unit analysis is a special case of dimensional analysis in which the dimensions of the physical quantities under consideration are all the same. In unit analysis, the units of measure are simply converted into a common unit. For example, the units of acceleration are meters per second squared. To convert meters per second squared to kilometers per hour squared, we can use the following equation:
1 kilometer per hour squared = 1,000 meters per second squared
This equation can be rearranged to solve for meters per second squared:
1,000 meters per second squared = 1 kilometer per hour squared
Dimensional analysis is used to determine whether or not a given equation is dimensionally correct. Dimensional analysis is performed on an equation by removing the numerical components and leaving only the unit types (length, mass, or time). It can also be used to figure out what type of unit is being used for an unknown variable. The force of gravity, for example, may appear as follows:
The unit analysis is similar to the dimensional analysis in that it employs units rather than basic dimensions. The same principle applies; the numbers are removed, and the units on both sides of the equation are verified to be equal.
Formula for Density
Density Formula is the formula for density. d = densities m denotes mass and v denotes volume.
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https://www.hackmath.net/en/problem/2069 | 1,560,630,769,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627997335.70/warc/CC-MAIN-20190615202724-20190615224724-00103.warc.gz | 777,234,712 | 6,806 | # Car factory
A workshop had produced 25 cars. While they produced about 15 fewer than they had planned. How many cars they produced in the hall yet?
How many cars are still to manufacture?
Result
x = 10
y = 15
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
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https://www.jiskha.com/display.cgi?id=1487816416 | 1,532,277,564,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676593378.85/warc/CC-MAIN-20180722155052-20180722175052-00469.warc.gz | 906,704,453 | 4,262 | # Algebra I
posted by Cale
Scott wants to make 6 gal of a 50% sugar solution by mixing together a 60% sugar solution and pure water. How much of each solution must he use?
mixture 1 @ 60% = x gallons
pure water = 100% - x (??? that's my question...is that correct? or is it 0?)
mixture @ 50% = 6 gallons
60x + 100(6-x)= 50 * 6
60x + 600-100x = 300
600 - 40x = 300
-40x = -300
x = 7.5
1. Reiny
I would do it this way:
Amount of pure water needed --- x gal
Amount of the 60% stuff ------ 6-x
.6(6-x) + 0x= .5(6)
3.6 - .6x = 3
-.6x = -.6
x = 1
so use 5 gal of the 60% plus 1 gal of pure water
your answer of 7.5 gal should have made no sense to you. You want to end up with only 6 gal, so how could you add 7.5 ????
2. Cale
Thank you. My answer of 7.5 didn't make sense. i was obviously overthinking...thinking pure would = 100%.
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#### Resources tagged with Games similar to Hello Again:
Filter by: Content type:
Stage:
Challenge level:
### There are 94 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Games
### Conway's Chequerboard Army
##### Stage: 3 Challenge Level:
Here is a solitaire type environment for you to experiment with. Which targets can you reach?
### Diagonal Dodge
##### Stage: 2 and 3 Challenge Level:
A game for 2 players. Can be played online. One player has 1 red counter, the other has 4 blue. The red counter needs to reach the other side, and the blue needs to trap the red.
### Khun Phaen Escapes to Freedom
##### Stage: 3 Challenge Level:
Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom.
### Cubic Net
##### Stage: 4 and 5 Challenge Level:
This is an interactive net of a Rubik's cube. Twists of the 3D cube become mixes of the squares on the 2D net. Have a play and see how many scrambles you can undo!
### Ratio Sudoku 3
##### Stage: 3 and 4 Challenge Level:
A Sudoku with clues as ratios or fractions.
### Going First
##### Stage: 4 and 5
This article shows how abstract thinking and a little number theory throw light on the scoring in the game Go.
### Matching Fractions, Decimals and Percentages
##### Stage: 2 and 3 Challenge Level:
An activity based on the game 'Pelmanism'. Set your own level of challenge and beat your own previous best score.
### Online
##### Stage: 2 and 3 Challenge Level:
A game for 2 players that can be played online. Players take it in turns to select a word from the 9 words given. The aim is to select all the occurrences of the same letter.
### Seasonal Twin Sudokus
##### Stage: 3 and 4 Challenge Level:
This pair of linked Sudokus matches letters with numbers and hides a seasonal greeting. Can you find it?
### Pentanim
##### Stage: 2, 3 and 4 Challenge Level:
A game for 2 players with similaritlies to NIM. Place one counter on each spot on the games board. Players take it is turns to remove 1 or 2 adjacent counters. The winner picks up the last counter.
### Winning Lines
##### Stage: 2, 3 and 4
An article for teachers and pupils that encourages you to look at the mathematical properties of similar games.
### Turnablock
##### Stage: 4 Challenge Level:
A simple game for 2 players invented by John Conway. It is played on a 3x3 square board with 9 counters that are black on one side and white on the other.
### Jam
##### Stage: 4 Challenge Level:
A game for 2 players
### Games Related to Nim
##### Stage: 1, 2, 3 and 4
This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning.
### History Mystery
##### Stage: 2, 3 and 4 Challenge Level:
Can you identify the mathematicians?
### One, Three, Five, Seven
##### Stage: 3 and 4 Challenge Level:
A game for 2 players. Set out 16 counters in rows of 1,3,5 and 7. Players take turns to remove any number of counters from a row. The player left with the last counter looses.
### Ratio Sudoku 2
##### Stage: 3 and 4 Challenge Level:
A Sudoku with clues as ratios.
### Wari
##### Stage: 4 Challenge Level:
This is a simple version of an ancient game played all over the world. It is also called Mancala. What tactics will increase your chances of winning?
### Corresponding Sudokus
##### Stage: 3, 4 and 5
This second Sudoku article discusses "Corresponding Sudokus" which are pairs of Sudokus with terms that can be matched using a substitution rule.
### Pole Star Sudoku
##### Stage: 4 and 5 Challenge Level:
A Sudoku based on clues that give the differences between adjacent cells.
### Twin Corresponding Sudokus II
##### Stage: 3 and 4 Challenge Level:
Two sudokus in one. Challenge yourself to make the necessary connections.
### Advent Calendar 2010
##### Stage: 1, 2, 3 and 4 Challenge Level:
Advent Calendar 2010 - a mathematical game for every day during the run-up to Christmas.
### Twin Corresponding Sudoku III
##### Stage: 3 and 4 Challenge Level:
Two sudokus in one. Challenge yourself to make the necessary connections.
### Nim-7
##### Stage: 1, 2 and 3 Challenge Level:
Can you work out how to win this game of Nim? Does it matter if you go first or second?
### Dominoes
##### Stage: 2, 3 and 4 Challenge Level:
Everthing you have always wanted to do with dominoes! Some of these games are good for practising your mental calculation skills, and some are good for your reasoning skills.
### 18 Hole Light Golf
##### Stage: 1, 2, 3 and 4 Challenge Level:
The computer starts with all the lights off, but then clicks 3, 4 or 5 times at random, leaving some lights on. Can you switch them off again?
### Sprouts
##### Stage: 2, 3, 4 and 5 Challenge Level:
A game for 2 people. Take turns joining two dots, until your opponent is unable to move.
### Low Go
##### Stage: 2, 3 and 4 Challenge Level:
A game for 2 players. Take turns to place a counter so that it occupies one of the lowest possible positions in the grid. The first player to complete a line of 4 wins.
### Behind the Rules of Go
##### Stage: 4 and 5
This article explains the use of the idea of connectedness in networks, in two different ways, to bring into focus the basics of the game of Go, namely capture and territory.
### Spiralling Decimals
##### Stage: 2 and 3 Challenge Level:
Take turns to place a decimal number on the spiral. Can you get three consecutive numbers?
### First Connect Three for Two
##### Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### Squayles
##### Stage: 3 Challenge Level:
A game for 2 players. Given an arrangement of matchsticks, players take it is turns to remove a matchstick, along with all of the matchsticks that touch it.
### Have You Got It?
##### Stage: 3 Challenge Level:
Can you explain the strategy for winning this game with any target?
### Sufficient but Not Necessary: Two Eyes and Seki in Go
##### Stage: 4 and 5
The game of go has a simple mechanism. This discussion of the principle of two eyes in go has shown that the game does not depend on equally clear-cut concepts.
### Spiralling Decimals for Two
##### Stage: 2 and 3 Challenge Level:
Spiralling Decimals game for an adult and child. Can you get three decimals next to each other on the spiral before your partner?
### Patience
##### Stage: 3 Challenge Level:
A simple game of patience which often comes out. Can you explain why?
### Snail Trails
##### Stage: 3 Challenge Level:
This is a game for two players. You will need some small-square grid paper, a die and two felt-tip pens or highlighters. Players take turns to roll the die, then move that number of squares in. . . .
### Learning Mathematics Through Games Series: 4. from Strategy Games
##### Stage: 1, 2 and 3
Basic strategy games are particularly suitable as starting points for investigations. Players instinctively try to discover a winning strategy, and usually the best way to do this is to analyse. . . .
### Making Maths: Snake Pits
##### Stage: 1, 2 and 3 Challenge Level:
A game to make and play based on the number line.
### Diamond Collector
##### Stage: 3 Challenge Level:
Collect as many diamonds as you can by drawing three straight lines.
### Diamond Mine
##### Stage: 3 Challenge Level:
Practise your diamond mining skills and your x,y coordination in this homage to Pacman.
### Domino Magic Rectangle
##### Stage: 2, 3 and 4 Challenge Level:
An ordinary set of dominoes can be laid out as a 7 by 4 magic rectangle in which all the spots in all the columns add to 24, while those in the rows add to 42. Try it! Now try the magic square...
### Quadruple Clue Sudoku
##### Stage: 3 and 4 Challenge Level:
Four numbers on an intersection that need to be placed in the surrounding cells. That is all you need to know to solve this sudoku.
### Tangram Pictures
##### Stage: 1, 2 and 3 Challenge Level:
Use the tangram pieces to make our pictures, or to design some of your own!
### Sliding Puzzle
##### Stage: 1, 2, 3 and 4 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### Intersection Sudoku 1
##### Stage: 3 and 4 Challenge Level:
A Sudoku with a twist.
### Fifteen
##### Stage: 2 and 3 Challenge Level:
Can you spot the similarities between this game and other games you know? The aim is to choose 3 numbers that total 15.
### Nim-interactive
##### Stage: 3 and 4 Challenge Level:
Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The winner is the player to take the last counter.
### Learning Mathematics Through Games Series: 1. Why Games?
##### Stage: 1, 2 and 3
This article supplies teachers with information that may be useful in better understanding the nature of games and their role in teaching and learning mathematics.
### Wallpaper Sudoku
##### Stage: 3 and 4 Challenge Level:
A Sudoku that uses transformations as supporting clues. | 2,234 | 9,312 | {"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-2017-47 | latest | en | 0.87995 |
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Is integer H divisible by 5 ? (1) 24H is divisible by 20 (2)
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Is integer H divisible by 5 ? (1) 24H is divisible by 20 (2) [#permalink]
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Is integer H divisible by 5 ?
(1) 24H is divisible by 20
(2) H+1 is divisible by 6
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14 Sep 2008, 04:18
dancinggeometry wrote:
Is integer H divisible by 5 ?
(1) 24H is divisible by 20
(2) H+1 is divisible by 6
2)not necessarily 5+1,11+1,17+1 etc yield diff results
IMO A
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# To the nearest thousand what is the value of a=
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To the nearest thousand what is the value of a= [#permalink]
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To the nearest thousand what is the value of
a= 3^(1/1998)+ 3 ^(-(1/1998) )
a. 0
b. 2
c. 10
d sqrt3
e. 4
you've got 90 seconds to solve this and unlimited time to explain the
CEO
Joined: 15 Aug 2003
Posts: 3454
### Show Tags
06 Dec 2003, 16:07
stoolfi wrote:
3^0+3^0=2
B?
right again.
do you think ,we needed nearest "thousandth" here, instead
of nearest thousand
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### Show Tags
06 Dec 2003, 16:25
I agree, but better still would be to say which answer is closest to the following equation.
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### Show Tags
07 Dec 2003, 05:49
praetorian123 wrote:
stoolfi wrote:
3^0+3^0=2
B?
right again.
do you think ,we needed nearest "thousandth" here, instead
of nearest thousand
The question should be simply:
The <equation> is approximately:
a
b
c
d
e
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07 Dec 2003, 05:49
Display posts from previous: Sort by | 674 | 2,299 | {"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} | 3.515625 | 4 | CC-MAIN-2017-26 | longest | en | 0.89231 |
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1. ## Use integral calculus to solve differential equation problems
Find the equation of the curve that satisfies the differential equation f"(x) = 6x - 12 and which passes through the points (0, -8) and (1, -1).
2. Originally Posted by LilDragonfly
Find the equation of the curve that satisfies the differential equation f"(x) = 6x - 12 and which passes through the points (0, -8) and (1, -1).
$f(x)=x^3-6x^2+ax+b$
Sooooo:
$f(0)=-8 = 0^3-6*0^2+a*0+b=b$
Thus b = -8.
$f(1)=-1=(1)^3-6(1)^2+a(1)-8$
gives:
$-1=1-6+a-8=-13-a$
or
$a = 12$
-Dan
3. I do not know if you understood what topsquak said, but when you have,
$f''(x)=6x-12$
It means if you integrate both sides you have,
$f'(x)=3x^2-12x+a$ with some constant $a$,
the reason to why f''(x) becomes f'(x) is because when you integrate a derivative of a function you get back the function (definition of integral). So when you have f''(x) integrating once would yield f'(x) so you need one more time:
$f(x)=x^3-6x^2+ax+b$
Then use topsqaurk post to complete the problem. | 358 | 1,102 | {"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": 0, "codecogs_latex": 9, "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} | 4.25 | 4 | CC-MAIN-2017-51 | longest | en | 0.821271 |
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# Problem of the Day - PowerPoint PPT Presentation
Problem of the Day. Solve this equation by moving the numbers: 7 6 = 2 4. Problem of the Day. Solve this equation by moving the numbers: 7 6 = 2 4 7 2 = 4 9. CSC 212 – Data Structures. Lecture 34 5 : Implementing Traversals. Traversing Binary Trees.
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Problem of the Day
• Solve this equation by moving the numbers:
76 = 24
Problem of the Day
• Solve this equation by moving the numbers:
76 = 24
72=49
CSC 212 – Data Structures
### Lecture 345:Implementing Traversals
Traversing Binary Trees
• Trees are another Collection of data
• Positions required in the Tree ADT methods
• ADT uses generic type to specify type of element
• Want to iterate through tree’s elements
• This is not optional; Trees are Iterable
• Could use any traversal algorithm; choose logical one
• But how do we implement traversals & Iterator?
Traversal Methods
• Many traversals, differ in order nodes visited
• Do parent then do each kid in pre-order traversal
Traversal Methods
• Many traversals, differ in order nodes visited
• Do parent then do each kid in pre-order traversal
• Post-order traversal does kids before doing parents
Traversal Methods
• Many traversals, differ in order nodes visited
• Do parent then do each kid in pre-order traversal
• Post-order traversal does kids before doing parents
• Do left kid, parent, then right kid in in-order traversal
Traversal Algorithms
• So, how do we do each one of these?
Traversal Algorithms
• So, how do we do each one of these? implement each traversal?
• Recursion usually answer to questions involving trees
• The algorithm uses same approach for all three
• Where Nodes done only difference; rest is identical
Binary Tree Traversal
Algorithmtraverse(tree, v)
// Do node here for pre-order traversal
traverse(tree, tree.left(v));
// Do node here for in-order traversal
traverse(tree, tree.right(v));
// Do node here for post-order traversal
Important Feature of Any Coder
• Nearly all of the great coders are inherently lazy
• Hope for you all -- you have this part completed
• Unfortunately, true laziness requires lot of work
• Debugging code wastes time could be websurfing
• To make small change, more time lost rewriting code
• Small amount of planning save lots of time later
Template Method Pattern
• Common design pattern used to simplify code
• Implement class of algorithms using an abstract class
• Class leaves all actual work in abstract methods
• Subclasses written for each member algorithm
• To make concrete, each overrides abstract method
• Method performs specific actions for that algorithm
• New algorithms are very quick & easy to write
• Most bugs already found & fixed, so less work here
• Leaves only actions & actions usually are small
Binary Tree Traversal Template
public class EulerTour<E, R> {abstract void visit(Position<E> p);Rtour(BinaryTree<E> t, Position<E> p) {TourResult<R> r = new TourResult<R>(); if (t.hasLeft(p)) {r.left = tour(t, t.left(p)); }/*Process p*/ if (t.hasRight(p)) {r.right = tour(t, t.right(p)); } return r.out;}
Oops…
public class EulerTour<E, R> {abstract void visit(Position<E> p);Rtour(BinaryTree<E> t, Position<E> p) {TourResult<R> r = new TourResult<R>(); if (t.hasLeft(p)) {r.left = tour(t, t.left(p)); }/*Process p*/ if (t.hasRight(p)) {r.right = tour(t, t.right(p)); } return r.out;}
Pre-order traversal needsprocessing here
Oops…
public class EulerTour<E, R> {abstract void visit(Position<E> p);Rtour(BinaryTree<E> t, Position<E> p) {TourResult<R> r = new TourResult<R>(); if (t.hasLeft(p)) {r.left = tour(t, t.left(p)); }/*Process p*/ if (t.hasRight(p)) {r.right = tour(t, t.right(p)); } return r.out;}
Post-order traversal needsprocessing here
Better Solution
public class Traversal<E, R> {// Perform actions before kids processedabstract void visitLeft(Position<E> p);// Perform actions after kids processedabstract void visitRight(Position<E> p);// Perform actions between when kids processedabstract void visitBelow(Position<E> p);
Better Solution
public class EulerTour<E, R> {Rtour(BinaryTree<E> t, Position<E> p) {TourResult<R> r = new TourResult<R>();visitLeft(p); if (t.hasLeft(p)) {r.left = tour(t, t.left(p)); }visitBelow(p); if (t.hasRight(p)) {r.right = tour(t, t.right(p)); }visitRight(p); return r.out;}
Better Solution
class InOrder<E> extends EulerTour<E,String>{public String execute(BinaryTree<E> t) { init(t); return eulerTour(t.root());}void visitLeft(TourResult<String> r,Position<E> p){// Do nothing – we are not Woody Allen (pre-order traversal) }void visitRight(TourResult<String> r,Position<E> p){// Do nothing – we are not Michael Jackson (post-order traversal) }void visitBelow(TourResult<String> r,Position<E> v){r.out = r.left + " " + v.element() +}
Better Solution
class InOrder<E> extends EulerTour<E,String>{public String execute(BinaryTree<E> t) { init(t); return eulerTour(t.root());}void visitLeft(TourResult<String> r,Position<E> p){// Do nothing – we are not Woody Allen (pre-order traversal) }void visitRight(TourResult<String> r,Position<E> p){// Do nothing – we are not Michael Jackson (post-order traversal)}void visitBelow(TourResult<String> r,Position<E> v){r.out = r.left + " " + v.element() +...}
Better Solution
class InOrder<E> extends EulerTour<E,String>{public String execute(BinaryTree<E> t) { init(t); return eulerTour(t.root());}void visitLeft(TourResult<String> r,Position<E> p){// Do nothing – we are not Woody Allen (pre-order traversal) }void visitRight(TourResult<String> r,Position<E> p){// Do nothing – we are not … I have no joke (post-order traversal) }void visitBelow(TourResult<String> r,Position<E> p){r.out = r.left + " " + p.element() +...}
Better Solution
class InOrder<E> extends EulerTour<E,String>{// execute from before is here, but not enough space… void visitLeft(TourResult<String> r,Position<E> p) { // Do nothing – its too early }void visitBelow(TourResult<String> r,Position<E> p) { // Do nothing – its too early }
Better Solution
class InOrder<E> extends EulerTour<E,String>{void visitRight(TourResult<String> r,Position<E> v){r.out = ""; if (getTree().hasLeft(v)) {r.out = r.left + " "; }r.out += p.element(); if (getTree().hasRight(v)) {r.out += " " + r.right; }}
Wait A Minute…
• Lecture began talking about Tree's Iterator
• Implementing traversals discussed for rest of lecture
• I may not know much, but traversals != Iterator
• How to implement next() & hasNext()
• Recursion can't be used, since calls occur over time
• How can we reuse these ideas to make an iterator?
Cheating To Win
• next() & hasNext() very hard to implement
• Much easier for Sequences (which are Iterable)
• In fact, we have already done this for Sequence
• Use that Iteratorby adding elements to Sequence
• Define new subclass for this type of traversal
• Valid to build entire Iterator all at one go
• Add & combine Sequences in visit* methods
• executemethod returns a Sequence
• Get into your groups and complete activity
For Next Lecture
• Read parts of GT 7.1 & 7.3 for Monday
• How do we implement a tree?
• Binary trees are similar; how do we implement them?
• I loved CSC111; can’t we use arrays for binary trees?
• Week #12 posted & due next Tuesday (after break)
• Programming Assignment #3 available today | 2,003 | 7,935 | {"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} | 4.21875 | 4 | CC-MAIN-2018-05 | latest | en | 0.802228 |
https://www.mathcelebrity.com/community/threads/aliyah-had-24-to-spend-on-seven-pencils-after-buying-them-she-had-10-how-much-did-each-pencil-co.4816/ | 1,725,835,928,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651035.2/warc/CC-MAIN-20240908213138-20240909003138-00857.warc.gz | 862,605,053 | 11,478 | Aliyah had \$24 to spend on seven pencils. After buying them she had \$10. How much did each pencil co | MathCelebrity Forum
# Aliyah had \$24 to spend on seven pencils. After buying them she had \$10. How much did each pencil co
#### math_celebrity
Staff member
Aliyah had \$24 to spend on seven pencils. After buying them she had \$10. How much did each pencil cost?
Let p be the number of pencils. We're given the following equation:
7p + 10 = 24
To solve this equation for p, we type it in our math engine and we get:
p = 2 | 141 | 532 | {"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-2024-38 | latest | en | 0.988766 |
http://www.teachmathematics.net/page/10475/meeting-functions | 1,487,755,472,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170925.44/warc/CC-MAIN-20170219104610-00340-ip-10-171-10-108.ec2.internal.warc.gz | 629,939,331 | 18,180 | # Meeting Functions
'Get to the heart of what functions are about by matching these inputs, equations and outputs together - a 'simple' matching activity!'
There are 8 algebraic equations giving the relationship between the variables ‘x’ and ‘y’. There are 8 sets of input values (x values or ‘domain’) and 8 sets of corresponding output values (y values or ‘range’). The task is to make 8 sets of 3 including one of each type (1 set of input values, 1 equation and 1 set of output values). Of course the input values could go into almost all of the equations so the challenge is to put them in to an equation that will generate one of the given sets of output values.
It is easy to study functions and forget what they are all about. Equations where 'y' is dependant on 'x' all work on the principal that for given 'x' values there are corresponding 'y' values. The function describes what you have to do with the 'x' values to generate the 'y' values. These pairs of values then correspond to points on a coordinates grid and together make the graph of the equation or function. This activity involves matching a given set of input values with an equation and a corresponding set of output values to practise this fundamental point. Below is a quick screencast to demonstrate the point of the activity.
### Resources
You need the worksheet Meeting Functions. There are teacher notes Meeting functions available as well. Further descriptions and examples of the activity are available below.
Below are a few photographs of 'Meeting functions' in action
#### The Cards
x = {4,9,16} y = x² y = {7} x = {-3,-2,-1,0,1,2,3} y = x3 y = {-4,-3,-1} x = {-3,-2,-1,0,1,2,3} y = x0.5 y = {-12,-2,0,4 } x = {-2,-1,0,1,2} y = 3x + 1 y = {13,28,49} x = {16,9,1} y = -2x + 7 y = {-25,-11,-9,-1,23,39 } x = {-3,-1,0,1,2,4} y = x²+ 5x + 6 y = {0,1,4,9 } x = {-16,-8,4,8,9,16} y = x3 - x² y = {0,2,6,12,20,30 } x = {-3,-2,-1,0,1,2,3} y = 7 y = {-27,-1,0, 8,64 }
#### What to do?
Above, there are 8 algebraic equations giving the relationship between the variables ‘x’ and ‘y’. There are 8 sets of input values (x values or ‘domain’) and 8 sets of corresponding output values (y values or ‘range’). The task is to make 8 sets of 3 including one of each type (1 set of input values, 1 equation and 1 set of output values). Of course the input values could go into almost all of the equations so the challenge is to put them in to an equation that will generate one of the given sets of output values.
#### An Example
If the set of input values
x = {1, 2, 3}
was put into the equation
y = 2x2
they would generate the corresponding set out output values
y = {2, 8, 18}
#### Over to you
Now create 8 sets of 3 from the cards above.
• You could cut them out and rearrange them
• Copy them out in the correct order
• Copy and paste them into the correct order if working digitally
### Description
• Students are given the 'Meeting Functions' worksheet. The task is quite self explanatory but teachers may want to help by showing an example. There is also a screencast above to help.
• Students should compare solutions with each other to help find errors and justify answers.
• Students should share their conclusions with each other.
All materials on this website are for the exclusive use of teachers and students at subscribing schools for the period of their subscription. Any unauthorised copying or posting of materials on other websites is an infringement of our copyright and could result in your account being blocked and legal action being taken against you. | 955 | 3,569 | {"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} | 4.0625 | 4 | CC-MAIN-2017-09 | latest | en | 0.817585 |
https://betterlesson.com/lesson/538940/equations-for-parallel-and-perpendicular-lines?from=consumer_breadcrumb_dropdown_lesson | 1,540,341,574,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583517628.91/warc/CC-MAIN-20181024001232-20181024022732-00340.warc.gz | 614,876,341 | 25,648 | # Equations for Parallel and Perpendicular Lines.
4 teachers like this lesson
Print Lesson
## Objective
SWBAT write the equation of parallel and perpendicular lines based on the given information.
#### Big Idea
For students to write the equation of a line parallel to or perpendicular to a line and passing through a given point using different representations of linear functions.
## Warm up
10 minutes
In this Warm up I intend for the students to reflect on strategies to write the equation of a line parallel to or perpendicular to a given line. The first two problems are relatively open. Problems 3 and 4 require the parallel or perpendicular line to go through a given point. I plan for this warm up to take 10 minutes.
An important part of this lesson is using graphs to allow students a visual approach to each problem. The students are not required to draw the graph in the warm up. Later, I require students to draw the graph for these problems. Some students will be able to write the equation of the new line without the graph. The Warm Up will provide a good indication of who can and cannot succeed without making a graph.
In this lesson I am building off of the prior knowledge of students knowing the slope of parallel and perpendicular lines and continuing to model multi-step problems with parallel and perpendicular lines.
I set up the warm up problems in a PowerPoint that I review after the warm up is completed. The PowerPoint contains several additional problems that students will work on for Independent Practice.
## Power Point
15 minutes
I review the Parallel or Perpendicular problems from the warm up and the problems students are going to work in the independent practice. I model reviewing the last problem in the power point with the students in the video below:
## Independent Practice
20 minutes
In this Practice, I give students four different types of problems. The problems are the same four types of problems that were modeled at the end of the PowerPoint.
For the Independent Practice I also provide a graph and ask students to draw a picture of the problem. By having students draw the picture and label line 1 and line 2 in each problem, it helps students to distinguish between the original line and the new one.
## Exit Slip
10 minutes
I plan for this Rectangle problem to provide students with an opportunity to apply the concepts and skills from this lesson. I want students to make explicit use of the their knowledge that opposite sides of a rectangle are parallel, and, consecutive sides of a rectangle are perpendicular. Students first prove that this is true by verifying that the opposite sides of a rectangle on a graph are parallel because they have the same slope. Then, they show that consecutive sides are perpendicular because they have slopes that are opposite reciprocals of each other.
I expect some students will work concretely and calculate slope by counting the boxes. If so, I will build off concrete strategies to wrap up this lesson. As much as possible I would like to have students explain strategies that involve calculating slopes. I think that students can learn new concepts and strategies by comparing their existing knowledge to that of their peers.
Instructional Note:
In this problem, the answer is that coordinate C should be at the point (6, -8). Line segments AB and CB both have a slope of 1, and line segments AB and DC both have a slope of -1. | 696 | 3,460 | {"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-2018-43 | latest | en | 0.906447 |
http://slideplayer.com/slide/2553552/ | 1,527,067,321,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865468.19/warc/CC-MAIN-20180523082914-20180523102914-00311.warc.gz | 278,148,396 | 18,161 | # es/swf/bartproblem.html es/swf/bartproblem.html.
## Presentation on theme: " es/swf/bartproblem.html es/swf/bartproblem.html."— Presentation transcript:
http://www.math.harvard.edu/~knill/mathmovi es/swf/bartproblem.html http://www.math.harvard.edu/~knill/mathmovi es/swf/bartproblem.html
Whole numbers and their opposites. Natural Numbers - Natural counting numbers. 1, 2, 3, 4 … Whole Numbers - Natural counting numbers and zero. 0, 1, 2, 3 … Integers - … -3, -2, -1, 0, 1, 2, 3 … Integers, fractions, and decimals. Rational Numbers - Ex:
Venn Diagram: Naturals, Wholes, Integers, Rationals Naturals Wholes Integers Rationals
What are RATIONAL Numbers?
A number that can be expressed as a fraction or ratio (rational). The numerator and the denominator of the fraction are both integers. When the fraction is divided out, it becomes a terminating or repeating decimal. ( Even a big, clunky fraction like 7,324,908/56,003,492 is rational, simply because it can be written as a fraction.)
6 can be written as: 6/1 or 6.0 -2 can be written as: -2/1 or -2.0 ½ can be written as: 0.5 -5/4 can be written as: -1.25 2/3 can be written as: ----.66
What are IRRATIONAL Numbers?
An irrational number can be written as a decimal, but not as a fraction. In decimal form, irrational numbers do not repeat in a pattern nor do they terminate.
= 3.141592654….. = 1.414213562….. .6781011132…
http://web.archive.org/web/20070818074103 /http://regentsprep.org/Regents/Math/rationa l/Prat.htm http://web.archive.org/web/20070818074103 /http://regentsprep.org/Regents/Math/rationa l/Prat.htm | 507 | 1,609 | {"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} | 4.625 | 5 | CC-MAIN-2018-22 | latest | en | 0.751823 |
https://testbook.com/question-answer/ifdfracx22x-1timesdfrac2x--60b7544ad36552e4f9a7f124 | 1,716,676,517,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058858.81/warc/CC-MAIN-20240525222734-20240526012734-00431.warc.gz | 475,375,756 | 47,005 | # If $$\dfrac{(x+2)!}{(2x-1)!}\times\dfrac{(2x+1)!}{(x+3)!}=\dfrac{72}{7}$$, where x ∈ N then find the value of x?
This question was previously asked in
DSSSB Patwari Question Paper 16th June 2019 Shift 1
View all DSSSB Patwari Papers >
1. $$\dfrac{-54}{28}$$
2. 4
3. -4
4. $$\dfrac{54}{28}$$
Option 2 : 4
Free
DSSSB Patwari Mental Ability Sectional Test 1
1 K Users
20 Questions 20 Marks 14 Mins
## Detailed Solution
$$\frac{{(x + 2)!}}{{(2x - 1)!}} \times \frac{{(2x + 1)!}}{{(x + 3)!}} = \frac{{72}}{7}$$
$$\frac{{(x + 2)!}}{{(2x - 1)!}} \times$$$$\frac{{(2x + 1)(2x)(2x - 1)!}}{{(x + 3)(x + 2)!}} = \frac{{72}}{7}$$
$$= \frac{{2x(2x + 1)}}{{x + 3}} = \frac{{72}}{7}$$
$$= \frac{{x(2x + 1)}}{{x + 3}} = \frac{{36}}{7}$$
= 14x2 + 7x = 36x + 108
= 14x2 - 29x - 108 = 0
$$x = \frac{{ - ( - 29) \pm \sqrt {{{( - 29)}^2} + 4 \times 14 \times 108} }}{{2 \times 14}}$$
$$= \frac{{29 \pm 83}}{{28}}$$
$$x = \frac{{29 + 83}}{{28}}$$$$x = \frac{{29 - 83}}{{28}}$$
x = 4, -1.92
Given x ϵ N, N → Natural number
x = 4 | 497 | 1,022 | {"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} | 4.25 | 4 | CC-MAIN-2024-22 | latest | en | 0.472103 |
https://socratic.org/questions/how-do-you-divide-4v-3-6v-2-8v-12-by-2v-3-and-is-it-a-factor-of-the-polynomial | 1,623,985,671,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487634616.65/warc/CC-MAIN-20210618013013-20210618043013-00266.warc.gz | 480,928,736 | 6,033 | # How do you divide 4v^3+6v^2-8v-12 by 2v-3 and is it a factor of the polynomial?
Feb 11, 2017
$\text{ not a factor of the polynomial}$
#### Explanation:
Using some $\textcolor{b l u e}{\text{algebraic manipulation}}$ by substituting the divisor into the numerator as a factor.
$\frac{\textcolor{red}{2 {v}^{2}} \left(2 v - 3\right) + \left(\textcolor{b l u e}{6 {v}^{2}} + 6 {v}^{2}\right) - 8 v - 12}{2 v - 3}$
$= \frac{\textcolor{red}{2 {v}^{2}} \left(2 v - 3\right) \textcolor{red}{+ 6 v} \left(2 v - 3\right) + \left(\textcolor{b l u e}{+ 18 v} - 8 v\right) - 12}{2 v - 3}$
$= \frac{\textcolor{red}{2 {v}^{2}} \left(2 v - 3\right) \textcolor{red}{+ 6 v} \left(2 v - 3\right) \textcolor{red}{+ 5} \left(2 v - 3\right) + \left(\textcolor{b l u e}{+ 15} - 12\right)}{2 v - 3}$
$\Rightarrow 4 {v}^{3} + 6 {v}^{2} - 8 v - 12 = 2 {v}^{2} + 6 v + 5 + \frac{3}{2 v - 3}$
Since the remainder of the division is not zero.
$\text{Then "2v-3" is not a factor of the polynomial}$ | 430 | 981 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "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} | 4.625 | 5 | CC-MAIN-2021-25 | latest | en | 0.470429 |
https://www.esaral.com/q/simplified-value-of-31526/ | 1,670,541,757,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711368.1/warc/CC-MAIN-20221208215156-20221209005156-00594.warc.gz | 810,620,103 | 24,433 | Simplified value of
Question:
Simplified value oi $(25)^{\frac{1}{3}} \times 5^{\frac{1}{3}}$ is
(a) 25
(b) 3
(c) 1
(d) 5
Solution:
$(25)^{\frac{1}{2}} \times 5^{\frac{1}{3}}=\left(5^{2}\right)^{\frac{1}{3}} \times 5^{\frac{1}{3}}$
$=5^{\frac{2}{3}} \times 5^{\frac{1}{3}}$
$=5^{\frac{2}{3}+\frac{1}{3}}$
$=5^{\frac{2+1}{3}}$
$=5^{\frac{3}{3}}$
$=5$
$\therefore$ Simplified value of $(25)^{\frac{1}{3}} \times 5^{\frac{1}{3}}$ is $5 .$
Hence, the correct option is (d). | 229 | 483 | {"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} | 4.21875 | 4 | CC-MAIN-2022-49 | latest | en | 0.370279 |
https://www.ibpsguide.com/quantitative-aptitude-questions-21/ | 1,631,963,908,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056392.79/warc/CC-MAIN-20210918093220-20210918123220-00204.warc.gz | 867,211,679 | 72,696 | # LIC AAO/SBI PO Prelims Quantitative Aptitude Questions 2019 (Day-21)
Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for LIC AAO/SBI PO 2019 so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.
[WpProQuiz 5880]
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 Direction (1 – 5): Read the following information carefully and answers the following questions. The table below shows number of pens produced and sold by three different companies and the given years:
Year Company M Company N Company O  Production  Sales Production Sales Production Sales 2006 450 60% 20% 240 2007 600 43% 500 15% 2008 20% 40% 180 2009 480 15% 450 40% 2010 680 30% 600
1) If the total number of pens sold by company M and N in the year 2007 is 400 and the number of pens sold by company N in 2008 is 200 more than pens sold by the same company in 2007 then find the number of pens produced by company N in 2008?
a) 200
b) 400
c) 855
d) 305
e) 350
2) If the ratio of the number of pens sold by company M in 2009 to that in 2010 is 8:9 then find the approximate percentage of pens were sold by company M in 2010?
a) 18
b) 12
c) 25
d) 30
e) 35
3) If company O produced ‘x’ number of pens in 2007 and ‘x+ 200’ number of pens in 2009 and the total number of pens sold by company O in these years is 630 then find how many pens were sold in 2009?
a) 200
b) 250
c) 480
d) 300
e) 350
4) Company N produced 900 pens in 2010 and the number of pens sold by the same company in 2007 is 10 % more than number of pens sold by the company in 2010 then find number of pens produced by the same company in 2008 if the number of pens sold by company N in 2007 and 2008 is in the ratio of 9:8?
a) 200
b) 400
c) 100
d) 660
e) 350
5) If the average number of pens sold by company M in 2006, 2007 and 2008 is 200 then find the production of company M in 2008 is what percentage of the production of company N in 2009?
a) 100 %
b) 80 %
c) 75 %
d) 65 %
e) 55 %
6) A box contains 4 red balls, 3 white balls, 6 black balls. If two balls are drawn out random, then find the probability that both balls are either red or black balls
a) 7/26
b) 11/16
c) 3/56
d) 13/16
e) None of these
7) Incomes of Rajiv and Mohan are in the ratio 5:6 and their expenditures are in the ratio 235:278. If Rajiv saves Rs.1000 and Mohan saves Rs.2000, then find the income of Rajiv?
a) Rs. 52600
b) Rs. 42000
c) Rs. 48000
d) Rs. 57400
e) None of these
8) P, Q and R can fill the tank in 15, 18 and 20 min respectively. All three began to fill the tank together but P and Q left 4 and 5 min respectively before filling the tank. In how much time does R takes to fill the tank?
a) 8 Â (30/31) min
b) 7Â (26/33) min
c) 9Â (30/33) min
d) 6Â (26/31) min
e) None of these
9) The sum of the present ages of the four members of a family is 180 years. 5 years ago the ages of the four members A, B, C and D were in the ratio of 3:5:4:8. After how many years would A be as old as the present age of D?
a) 30 years
b) 35 years
c) 37 years
d) 40 years
e) None of these
10) A boat can cover 9.6 km upstream in 24 minutes. If the speed of the current is 1/4th of the boat in still water, then how much distance (in km) can the boat cover downstream in 36 minutes?
a) 36 km
b) 24 km
c) 48 km
d) 32 km
e) None of these
Directions (1-5):
Number of pen sold by company N in 2007
= 400 – (43*600/100)
= 400 – 258
= 142
Number of pen sold company N in 2008= 142+200= 342
Number of pen produced by company N in 2008
= 342*100/40
= 855 pens
Number of pens sold by company M in 2009
= 480*15/100= 72
Number of pens sold by company M in 2010
= 72*9/8= 81
Required percentage= 81*100/680= 12%
Number of pens sold in 2007
= x*15/100 + (x+200)*40/100= 630
= 15x/100 + 40x/100 + 80= 630
= (15x+ 40x)/100= 630- 80
= 55x= 550*100
= x= 1000 pens
So,
Number of pens sold in 2009
= (1000+200)*40/100
= 480 pens
Number of pens sold by company N in 2010
= 900*30/100
= 270 pens
Number of pens sold by company N in 2007
= 270*110/100
= 297 pens
Number of pens sold by company N in 2008
= 297*8/9
= 264 pens
Number of pens produced by company N in 2008
= 264*100/40
= 660 pens
Total pens produced = 200*3= 600
Number of pens sold M in 2008
= 600- 60% of 450 – 43% of 600
= 600 – 270 – 258
= 72
Pens produced in 2008
= 72*100/20
= 360
Required percentage
= 360*100/450
= 80%
P(E) = n(E)/n(S)
n(S) = 13C2
n(E) = 4C2 or 6C2
P(E) = 4C2 or 6C2 / 13C2
P(E) = (6+15)/78
P(E) = 7/26
Let the income of Rajiv and Mohan are Rs.5x and Rs.6x respectively.
According to the question,
(5x – 1000)/(6x – 2000) = 235/278
=> 1390x – 278000 = 1410x – 470000
=> 192000 = 20x
=> x = 9600
Income of Rajiv = 5x = 5 * 9600 = Rs. 48000
(x-4)/ 15 + (x-5)/18 + x/20 =1
(12x-48+10x-50+9x)/180 =1
31x-98=180
31x= 278
X=278/31 = 8 (30/31) min
Let five years ago, age of A, B, C and D were 3x, 5x, 4x and 8x respectively.
According to the question
3x + 5x + 4x + 8x + 5 * 4 = 180
=> 20x = 180 – 20
=> 20x = 160
=> x = 160/20 = 8
Present age of A = 3 * 8 + 5 = 29 years
Present age of D = 8 * 8 + 5 = 69 years
Required number of years = 69 – 29 = 40 years
Speed of upstream= (9.6 *60) / 24 = 24 km/hr
The speed of the current is 1/4th of the boat in still water.
= > Speed of current: Speed of still water = 1:4
Speed of upstream= Speed of still water- speed of Current
24= 3x =>x=8
Speed of current = 8 km/hr
Speed of boat in still water= 32 km/hr
Speed of downstream= Speed of still water + speed of Current
= > 32+8 = 40 km/hr
Here speed= 40 km/hr, Time= 36 minutes
Distance= 40 *(36/60) = 24 km | 2,146 | 5,956 | {"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-2021-39 | latest | en | 0.901612 |
https://www.mymathtables.com/formula-equation/engg-mathematics/four-rules-of-basic-mathematics.html | 1,716,754,482,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058973.42/warc/CC-MAIN-20240526200821-20240526230821-00220.warc.gz | 798,227,563 | 3,715 | # What are the Four Basic Rules of Mathematics
An online basic mathematics rules printable
### Four Rules of Math
The concepts of addition, subtraction, multiplication and division are complex abstract ideas which are interrelated.
positive + positive = (add) positive
Example: 2 + 1 = 3
negative + negative = (add) negative
Example: −3+ (−5) = −8
negative + positive = (subtract)
Example: 2 + (−10) = −8
Take sign of number with largest absolute value
Example: −14 + 16 = 2
Note: −(−7) means take the opposite of (−7) = 7
## 2. Subtraction Rules:
Change all subtraction to addition and take the opposite sign of the following number then follow the addition rules.
Example: −7 −(9)
means −7 +(−9) = -16
Example: −3 −(−10)
−3 + (10) = 7
Example:
−8 + (−9) −(−1)−2 (change all subtraction signs)
−8 + (−9)+ (1) + (−2) (do addition and subtraction from left to right)
−17 +(1)+ (−2)
−16 +(−2) = −18
Adding and subtracting with negative numbers:
When adding and subtracting positive and negative numbers it is useful to remember the follow-ing rules.
If the operation and the sign are the same, they work like adding a (positive) number,so that
works like ++
If the operation and the sign are different, they work like subtracting a (positive) number, sothat
+ works like −+
## 3 and 4. Multiplication / Division Rules:
The rules for multiplication and division are the same.
positive (✗ or ÷) positive = positive
Example: 10 ÷ 2 = 5
negative (✗ or ÷) negative = positive
Example: −4 ✗(−3) = 12
negative (✗ or ÷) positive = negative
Example: 18 ÷ (−2) −9
Multiplying and dividing with negative numbers:
When multiplying pairs of positive and negative numbers it is helpful to remember the followingrules:When the signs of the numbers are the same the answer is a positive number.
(+) ✗ (+) , answer is (+)
(−) ✗ (−) , answer is (+)
(+) ÷ (+) , answer is (+)
(−) ÷ (−) , answer is (+)
When the signs of the numbers are different the answer is a negative number.
(+) ✗ (−) , answer is (−)
(−) ✗ (+) , answer is (−)
(+) ÷ (−) , answer is (−)
(−) ÷ (+) , answer is (−) | 613 | 2,109 | {"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} | 4.75 | 5 | CC-MAIN-2024-22 | latest | en | 0.821953 |
https://forum.allaboutcircuits.com/threads/2-questions-regarding-a-simple-circuit-with-an-led.114430/ | 1,716,444,185,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058611.55/warc/CC-MAIN-20240523050122-20240523080122-00203.warc.gz | 229,765,029 | 29,645 | # 2 questions regarding a simple circuit with an LED
#### Robert Smith_1437948150
Joined Jul 26, 2015
38
Hello All,
If I had a simple circuit with a 12V battery and an LED that drops 2V and needs 20mA, therefore I'd need a resistor to drop 10V and set the current at 20mA, so I'd do R = 10/0.020 = 500 Ohms.
1. Is it possible to 'use up' the voltage before it reaches the LED. For example if you placed the resistor before the LED, and wanted it to drop the whole 12V, then if I used a 600 Ohm resistor, would it drop 12V and then leave nothing for the LED?
2. If I wanted to 40mA to flow through the LED, then I could use R = 10/0.040 = 250 Ohm resistor. Looking at the voltage-current curve for an LED, to increase the current through it I'd also have to increase the voltage (albeit slightly), but I'd still only have 2V after the resistor. How can the LED have a current of 40mA, when it only has 2V available after the resistor, it goes against the voltage-current curve?
If anyone could help clear things up a bit it would be appreciated.
Thanks.
#### mcgyvr
Joined Oct 15, 2009
5,394
LEDs are current driven devices.
The resistor "sets" the current the LED will receive.. and it doesn't matter at all if the resistor is placed before or after the LED..
Its setting the current for that whole circuit..
Don't think of the resistor like its "dropping voltage" its setting the current in the circuit..
An LED is eating the voltage (its forward voltage drop) but a resistor is just setting the current in the whole circuit.
If you put a bunch of people in a circle and have them go around and around they can only move as fast as the slowest walker.
Doesn't matter where that slow walker is but it effects the speed of the entire circle. The resistor is the slow walker..
The "voltage-current" curve is an approximation of the forward voltage drop an LED will cause in a system at that current.
#### Robert Smith_1437948150
Joined Jul 26, 2015
38
Hi mcgyvr, thanks for the response.
I understand that the current is equal throughout the circuit, but obviously the voltage isn't, and I was thinking if you did place a resistor before the LED, could you have so that it dropped all 12V and left nothing for the LED?
Also, If the LED drops 2V at 20mA, and I use a 250 Ohm resistor then current is 40mA, which means that the LED will drop more than 2V, but if the resistor is dropping 10V then the total dropped is more than the source, breaking KVL.
Can't quite get it to add up.
Regards.
#### dl324
Joined Mar 30, 2015
17,001
1. Is it possible to 'use up' the voltage before it reaches the LED. For example if you placed the resistor before the LED, and wanted it to drop the whole 12V, then if I used a 600 Ohm resistor, would it drop 12V and then leave nothing for the LED?
Yes, but not with resistors. For your example, if you put 10 LEDs in series, there wouldn't be sufficient voltage to light any of them. With a 600 ohm resistor, you would simply reduce the current in the LED. Using your stated forward voltage of 2V for the LED, that would give a current of I=V/R=10/600=16.7mA. So increasing the resistance by 20% just decreased the current in the LED by the same amount. The actual amount would be different because the forward voltage of the LED would be reduced at the lower current.
2. If I wanted to 40mA to flow through the LED, then I could use R = 10/0.040 = 250 Ohm resistor. Looking at the voltage-current curve for an LED, to increase the current through it I'd also have to increase the voltage (albeit slightly), but I'd still only have 2V after the resistor. How can the LED have a current of 40mA, when it only has 2V available after the resistor, it goes against the voltage-current curve?
The voltage across the LED can only vary between 0V and the max forward voltage. You look at the IV curve for the diode to determine what it's forward voltage will be at 40mA. Then you subtract that from the power supply voltage to find the voltage across the resistor. Using your 2V, that would leave 10V across the resistor. R=V/I=10/0.04=250 ohms.
#### dl324
Joined Mar 30, 2015
17,001
I understand that the current is equal throughout the circuit, but obviously the voltage isn't, and I was thinking if you did place a resistor before the LED, could you have so that it dropped all 12V and left nothing for the LED?
Kirchhoff's Voltage Law says the sum of voltages around a loop will be equal to 0. In your example, you have 12V from the power supply, minus 2V from the LED; that leaves 10V across the resistor. The resistor will determine the current in the loop.
If the LED drops 2V at 20mA, and I use a 250 Ohm resistor then current is 40mA, which means that the LED will drop more than 2V, but if the resistor is dropping 10V then the total dropped is more than the source, breaking KVL.
KVL will not be violated. The forward voltage of the LED will increase at 40mA, but it will still be close to 2V.
IV characteristics of LEDs:
As current is increased, forward voltage increases. When the LED is operated at 20mA or more, it's past the "knee" of the curve and the slope of the curve shows that further increases in current will result in smaller incremental increases in forward voltage.
I don't have a datasheet handy that gives LED IV characteristics this way; they're all using a log scale.
#### Robert Smith_1437948150
Joined Jul 26, 2015
38
Thanks dl324,
This is going to sound really stupid, but when calculating the current you did I=10/600, and you used 10V because we can observe the circuit and see the LED has voltage drop of 2V, but if the resistor is before the LED, wouldn't the voltage for the resistor be 12? how does it only drop 10 volts so that it leaves 2 for the LED.
This is why I asked if it mattered if the resistor comes before or after the LED, not because of the current which I know is equal throughout the circuit.
Regards.
#### dl324
Joined Mar 30, 2015
17,001
FWIW, an IV graph with log scale for current:
This shows that for super-red LEDs, the forward voltage at 20mA is about 2.3V. At 40mA it would be about 2.7V.
Last edited:
#### dl324
Joined Mar 30, 2015
17,001
This is going to sound really stupid, but when calculating the current you did I=10/600, and you used 10V because we can observe the circuit and see the LED has voltage drop of 2V, but if the resistor is before the LED, wouldn't the voltage for the resistor be 12? how does it only drop 10 volts so that it leaves 2 for the LED.
This is why I asked if it mattered if the resistor comes before or after the LED, not because of the current which I know is equal throughout the circuit.
You answered your own question. The current in a loop is the same for all components and their voltage drops will be determined by the current. The placement of the LED and resistor is more a matter of preference.
In some cases, one order may be safer than another. For instance, if the LED in this circuit was off board (e.g. on a front panel), it would be better to have the resistor before the LED so that shorting the anode to ground would be current limited. If the resistor was after the LED, shorting the anode to ground could cause damage.
#### Robert Smith_1437948150
Joined Jul 26, 2015
38
Thanks dl324 for all that, it's finally clicked.
As a final example using arbitrary values, if I wanted the LED to be brighter and the current to be 60mA, and it turned out that at this current the voltage drop for the LED was 3V, then I'd do 12V - 3V = 9, 9/0.06 = 150. So I'd use an 150 Ohms resister.
Going from a 500 Ohms resister has increased the voltage drop of the LED from 2V to 3V, but decreased the voltage drop of the resister so that it always equals 12V.
Thanks again dl324 and mcgyvr.
#### AnalogKid
Joined Aug 1, 2013
11,131
The D in LED stands for Diode, and like all diodes, an LED is not a resistor. That is, its relationship between voltage and current is not linear. Ideally, a diode has a forward voltage (Vf) that is independent of current. For a Shottkey diode it is around 0.2 v to 0,3 V, for a silicon signal diode it is around o.6 to 0.7 V, and for a simple green LED it is around 2.0 V. For the purpose of this discussion, that voltage does not change no matter what the current is through the diode (until the diode burns up). So the series resistor can be 50, 500, or 5000 ohms and the voltage across the LED still will be 2 V and the voltage across the resistor still will be 10 V. This is why you can select the resistor to set the current through the LED.
Your question about the resistor "using up" the voltage when placed ahead of the LED does not account for this question: In order for there to be 12 V developed across a resistor, there must be current through it. Where does that current go?
ak
#### Robert Smith_1437948150
Joined Jul 26, 2015
38
Thanks AnalogKid,
That's the bit I'm still unclear with tbh. If you remove the LED from the circuit and just have a 600 Ohms resistor, the current will be 2mA and the resistor will drop 12V.
When you add the LED, how does the resistor 'know' to drop the current so that the there is enough voltage remaining for the LED. The resistor was dropping 12V on it's own, but now because there is something further down the line that needs 2V, the current is reduced.
Regards.
#### dl324
Joined Mar 30, 2015
17,001
As a final example using arbitrary values, if I wanted the LED to be brighter and the current to be 60mA, and it turned out that at this current the voltage drop for the LED was 3V, then I'd do 12V - 3V = 9, 9/0.06 = 150. So I'd use an 150 Ohms resister.
Your calculations are correct, but don't plan on operating most LEDs at 60mA. Many specs are for 20mA. When you operate at higher currents, forward voltage will increase, color will change, and the LED may burn out.
Going from a 500 Ohms resister has increased the voltage drop of the LED from 2V to 3V, but decreased the voltage drop of the resister so that it always equals 12V.
This is as you would expect. The voltage drop of the LED varies with current. For a circuit consisting of a 12V supply, a resistor, and an LED; the voltage across the resistor has to be 12V minus the forward voltage of the LED.
#### dl324
Joined Mar 30, 2015
17,001
When you add the LED, how does the resistor 'know' to drop the current so that the there is enough voltage remaining for the LED. The resistor was dropping 12V on it's own, but now because there is something further down the line that needs 2V, the current is reduced.
The voltage across the resistor will be 12V - 2V = 10V. If you add another LED (in series), the voltage across the resistor will now be 8V.
If you keep adding LEDs, eventually you'll get to the point where none of them will conduct so there will be no voltage drop across the resistor.
#### djsfantasi
Joined Apr 11, 2010
9,182
The resistor doesn't know how to drop the current...
Remember:
1) There is only so much voltage to go around;
2) Voltage does not drive an LED.
3) LED's have a fairly constant voltage drop or Vf.
You have an LED with Vf = 2V. With a 12V supply, there remains 10V across the resistor.
If a 600 Ohm resistor has 10 V across it, the current is (10V / 600 Ohm) A of current or 17mA of current.
When you add the LED, how does the resistor 'know' to drop the current so that the there is enough voltage remaining for the LED. The resistor was dropping 12V on it's own, but now because there is something further down the line that needs 2V, the current is reduced
No, no, no... The current isn't reduced because there is something further down the line; the voltage is reduced because there is something further down the line. Using Ohm's law, you can calculate the current through the resistor and hence the circuit. Don't mix up your currents and voltages.
#### Robert Smith_1437948150
Joined Jul 26, 2015
38
Your calculations are correct, but don't plan on operating most LEDs at 60mA. Many specs are for 20mA. When you operate at higher currents, forward voltage will increase, color will change, and the LED may burn out.
True, I was just using random numbers to make it easier.
This is as you would expect. The voltage drop of the LED varies with current. For a circuit consisting of a 12V supply, a resistor, and an LED; the voltage across the resistor has to be 12V minus the forward voltage of the LED.
Thanks.
#### Robert Smith_1437948150
Joined Jul 26, 2015
38
No, no, no... The current isn't reduced because there is something further down the line; the voltage is reduced because there is something further down the line. Using Ohm's law, you can calculate the current through the resistor and hence the circuit. Don't mix up your currents and voltages.
Hi djsfantasi, I didn't mean the current is reduced in the circuit, I mean't that from the previous circuit with just a resistor, the new circuit that now has an LED uses a lower current.
Thanks to everyone, I understand now how it's calculated and where my initial mistakes were in my original questions. My question now is more what's actually going on physically in the circuit.
I just can't picture how it works when the resistor is first. When the LED is first it goes from 12V at the battery, a 2V drop across the LED and then 10V across the resistor, which make sense. When it's the other way, there is 12V at the 600 Ohms resistor, and rather than setting the current at 0.02mA and dropping the whole 12V, it sets the current at 0.016mA so that the voltage drop is just 10V, and there is 2V available at the LED.
What is physically preventing the resistor from setting the current at 0.02mA?
I'm struggling to word it to be honest so god knows how I expect you to be able to answer it...
Regards.
Last edited:
#### AnalogKid
Joined Aug 1, 2013
11,131
A diode, theoretically, is a constant-voltage device just like a battery. In your case, a 2 V device. So whenever it is in any circuit for any reason, and is forward biased, it has 2 V across it no matter what the current is. Get a grip on that, or this won't make sense. This is the reason a diode is consicered an "active" component rather than a "passive" component like resistors, capacitors, and inductors. Passives are never have a constant voltage or constant current as a part of their physics. they react to both voltage and current. A (perfect) diode does not. In this context, it does not react to current; for any current through it, it allows only 2 V to develop across it. This actually makes it a form of a negative-resistance device, but that's for another day.
So the resistor has no choice but to react to the various combinations of voltage and current sources and drops in the circuit. In your case it is a simple, single loop. Battery, resistor, and LED are connected in a circle. In the classic water analogy, the battery is a water pump, pushing water around the circle of plumbing. It should be fairly obvious that in a closed loop, 100% of the water goes through each device, because it has nowhere else to go. The wires are fat pipes, the resistor is a section of narrow pipe, and the LED is a pressure valve. If the pump pressure is high enough, the valve opens and lets water circulate around the loop. In this loop, I hope it is clear that the order of the three components does not matter. 100% of the water goes through 100% of the devices.
The LED circuit is the same. If the battery is less than the 2 V Vf of the diode, there is no electric current. For any battery voltage greater than 2 V, there is 2 V across the LED and whatever remaining voltage is across the resistor because there is nowhere else for a voltage to develop. Kirchhoff's Voltage Law covers this. The sum of all of the voltages around a loop must be zero. So if the battery has a fixed 12 V across it and there are no other batteries in the loop, then the sum of the voltage drops across each other component must add up to 12 V. Since the diode has a fixed 2 V drop, that leaves 10 V for everything else (the resistor). If the diode is not therre, the entire 12 V appears across the resistor. If the 500 ohm resistor is actually five 100 ohm resistors in series, then there is 2 V across each resistor when the LED is in the circuit. AND - it does not matter what order any of the components are in. The LED can be in the middle of the resistor string, and nobody cares.
ak
#### dl324
Joined Mar 30, 2015
17,001
I just can't picture how it works when the resistor is first. When the LED is first it goes from 12V at the battery, a 2V drop across the LED and then 10V across the resistor, which make sense. When it's the other way, there is 12V at the 600 Ohms resistor, and rather than setting the current at 0.02mA and dropping the whole 12V, it sets the current at 0.016mA so that the voltage drop is just 10V, and there is 2V available at the LED.
Solving circuits is often a matter of making assumptions to reduce the number of unknowns and iterating if it gives results that don't make sense.
In your circuit, you only have 3 elements, so it's easy to solve. The first step is to determine what the forward voltage of the LED is. Since you don't know the current, you have to make an assumption. In this discussion, we've assumed the forward voltage was 2V. Without making that assumption, you can't proceed.
It doesn't matter whether the LED comes first or last. The circuit is a single loop so all elements see the same current regardless of their position.
If you have the components, you can wire it to prove it to yourself.
#### Robert Smith_1437948150
Joined Jul 26, 2015
38
A diode, theoretically, is a constant-voltage device just like a battery. In your case, a 2 V device. So whenever it is in any circuit for any reason, and is forward biased, it has 2 V across it no matter what the current is. Get a grip on that, or this won't make sense. This is the reason a diode is consicered an "active" component rather than a "passive" component like resistors, capacitors, and inductors. Passives are never have a constant voltage or constant current as a part of their physics. they react to both voltage and current. A (perfect) diode does not. In this context, it does not react to current; for any current through it, it allows only 2 V to develop across it. This actually makes it a form of a negative-resistance device, but that's for another day.
So the resistor has no choice but to react to the various combinations of voltage and current sources and drops in the circuit. In your case it is a simple, single loop. Battery, resistor, and LED are connected in a circle. In the classic water analogy, the battery is a water pump, pushing water around the circle of plumbing. It should be fairly obvious that in a closed loop, 100% of the water goes through each device, because it has nowhere else to go. The wires are fat pipes, the resistor is a section of narrow pipe, and the LED is a pressure valve. If the pump pressure is high enough, the valve opens and lets water circulate around the loop. In this loop, I hope it is clear that the order of the three components does not matter. 100% of the water goes through 100% of the devices.
The LED circuit is the same. If the battery is less than the 2 V Vf of the diode, there is no electric current. For any battery voltage greater than 2 V, there is 2 V across the LED and whatever remaining voltage is across the resistor because there is nowhere else for a voltage to develop. Kirchhoff's Voltage Law covers this. The sum of all of the voltages around a loop must be zero. So if the battery has a fixed 12 V across it and there are no other batteries in the loop, then the sum of the voltage drops across each other component must add up to 12 V. Since the diode has a fixed 2 V drop, that leaves 10 V for everything else (the resistor). If the diode is not therre, the entire 12 V appears across the resistor. If the 500 ohm resistor is actually five 100 ohm resistors in series, then there is 2 V across each resistor when the LED is in the circuit. AND - it does not matter what order any of the components are in. The LED can be in the middle of the resistor string, and nobody cares.
ak
Thanks a lot AnalogKid, appreciate you taking the time to write that.
Regarding the water analogy, you say "100% of the water goes through each device", with regards to the circuit, is it not that 100% of the voltage is used throughout the circuit, as KVL states, rather than 100% of the voltage goes through each device?
I do understand the rules and laws, for example it doesn't matter whether the resister is first or last in the circuit, if you have two 250 Ohm resisters they will both drop 5V (in my original example) and the sum of all the voltage drops across the circuit must equal the same as the power source.
That's how it works and I'm not trying to dispute it, I just wonder what's physically going on in the circuit. Going back to our circuit with a 12V battery and 600 Ohms resistor, as the first electrons flow through the circuit and hit the resistor, it's as if the resistor 'knows' whether there is an LED further along the circuit, if there isn't then it drops 12V and if there is then drop only 10V.
Your example with five 100 ohm resistors in series and an LED, it's as if each device can see an overview of the circuit to know how much voltage to drop, if you see what I mean? We say the law stating 'The voltage drop across all devices in the circuit must equal the power source', but the devices don't know that, they can't get together and say "Well, we have a 12V battery, an LED that drops 2V and there are five of us, so if we all agree to drop 2V then we won't break Mr Kirchoff's law" haha. What's physically happening in the circuit that makes the law true?
Again, I understand that's the way it works and I should just accept it, I just find it interesting.
Regards.
Last edited:
#### AnalogKid
Joined Aug 1, 2013
11,131
In order...
No, voltage does not go through anything. Voltage pushes current through conductors. Voltage is pressure, which is why it's analogy is a pump. Water is the analog of current. In a DC circuit, electrons physically move through each component in series just like water through pipes.
The resistor "knows" only one thing, the voltage across its terminals. It allows a current to flow through it that is proportional to that voltage, and the constant of proportionality is called "resistance". It doesn't know there is an LED in the circuit. For all it knows, there is a 1000 V battery and 495 LEDs in series with it. It sees 10 V, and acts accordingly. And since the resistor is in series with one or two constant voltage devices depending on the circuit, from the resistor's point of view the circuit with the LED looks exactly like a 10 V battery, nothing more.
It is the current that is "happening". Each individual device, be it resistor or LED, responds to the current through it. If anything can "see an overview of the circuit", it is the battery. Reworking your statement, the battery sees the total impedance and "knows" how much current to push. More correctly, the battery presents an infinite source of electrons at a fixed pressure, and the impedances in the circuit determines the rate at which they flow. Going back to the five 100 ohm resistors, the circuit behaves exactly the same if the five resistors are not all the same value. If the resistors are 50, 75, 100, 125, and 150 ohms, the total resistance is 500 ohms and the loop current is 20 mA. All of the current goes through all of the components all of the time. That sentence allows tremendous flexibility in circuit design.
ak | 5,848 | 23,658 | {"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} | 3.53125 | 4 | CC-MAIN-2024-22 | latest | en | 0.948787 |
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# Is 75 percent equal to three fourths?
Updated: 4/28/2022
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### What percent does three fourths equal?
Three fourths is 75%
### What percent is equal to three-fourths?
three fourths = 75%
### Three fourth is equal to how much percent?
Three fourths is equal to 75 percent.
### Which is larger three fourths or eighty percent?
Three fourths is equal to 75%, therefore 80% is the larger figure.
### Three fourths is equal to what decimal?
Three fourths, or 3/4, is equal to .75 (75%)
### Does three fourths equal fifty percent?
No that equals 75%. One Half is 50%
75%
75%
### What is 75 of 4?
75 percent is the equivalent of three fourths. Because of this logic, 75 percent of 4 is 3 since 3 is three fourths of 4.
.75
### How are the symbols three fourths point 75 and 75 percent related?
They are all equal to one another: 3/4 = 0.75 = 75%
### Is 75 equal to three fourths?
No. 75 is much bigger than 1, three fourths is smaller than 1. But 75% IS equal to 3/4 | 327 | 1,132 | {"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-2024-26 | latest | en | 0.963833 |
https://eng.libretexts.org/Under_Construction/Book%3A_Fluid_Mechanics_(Bar-Meir)__-_old_copy/11%3A_Compressible_Flow_One_Dimensional/11.4_Isentropic_Flow/11.4.7%3A_The_Impulse_Function | 1,547,944,078,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583688396.58/warc/CC-MAIN-20190120001524-20190120023524-00101.warc.gz | 495,318,459 | 16,560 | # 11.4.7: The Impulse Function
Fig. 11.9 Schematic to explain the significances of the Impulse function.
$F_{net}= \dot{m} (U_2 -U_1) + P_2 A_2 - P_1 A_1 \label{gd:iso:eq:monImp} \tag{64}$
The net force is denoted here as $$F_{net}$$. The mass conservation also can be applied to our control volume
$\dot{m} = \rho_1A_1U_1 = \rho_2A_2U_2 \label{gd:iso:eq:massImp} \tag{65}$
Combining equation (64) with equation (??) and by utilizing the identity in equation (50) results in
$F_{net}= kP_2A_2{M_2}^2 - kP_1A_1{M_1}^2 + P_2 A_2 - P_1 A_1 \label{gd:iso:eq:MomMassImp} \tag{66}$
Rearranging equation (66) and dividing it by $$P_0 A^{*}$$ results in
${F_{net} \over P_0 A^{*}} = \overbrace{P_2A_2 \over P_0 A^{*}}^{f(M_2)} \overbrace{\left( 1 + k{M_2}^2 \right)}^{f(M_2)} - \overbrace{P_1A_1 \over P_0 A^{*}}^{f(M_1)} \overbrace{\left( 1 + k{M_1}^2 \right)}^{f(M_1)} \label{gd:iso:eq:beforeDefa} \tag{67}$
Examining equation (67) shows that the right hand side is only a function of Mach number and specific heat ratio, $$k$$. Hence, if the right hand side is only a function of the Mach number and $$k$$ than the left hand side must be function of only the same parameters, $$M$$ and $$k$$. Defining a function that depends only on the Mach number creates the convenience for calculating the net forces acting on any device. Thus, defining the Impulse function as
$F = PA\left( 1 + k{M_2}^2 \right) \label{gd:iso:eq:impulsDef} \tag{68}$
In the Impulse function when $$F$$ ($$M=1$$) is denoted as $$F^{*}$$
$F^{*} = P^{*}A^{*}\left( 1 + k \right) \label{gd:iso:eq:impulsDefStar} \tag{69}$
The ratio of the Impulse function is defined as
${F \over F^{*}} = {P_1A_1 \over P^{*}A^{*}} {\left( 1 + k{M_1}^2 \right) \over \left( 1 + k \right) } = {1 \over \underbrace{P^{*}\over P_{0} }_ {\left(2 \over k+1 \right)^{k \over k-1}}} \overbrace ParseError: EOF expected (click for details) Callstack: at (Under_Construction/Book:_Fluid_Mechanics_(Bar-Meir)__-_old_copy/11:_Compressible_Flow_One_Dimensional/11.4_Isentropic_Flow/11.4.7:_The_Impulse_Function), /content/body/p[3]/span, line 1, column 10 ^{\hbox{see function qref{gd:iso:eq:beforeDefa}}} {1 \over \left( 1 + k \right) } \label{gd:iso:eq:ImpulseRatio} \tag{70}$
This ratio is different only in a coefficient from the ratio defined in equation (67) which makes the ratio a function of $$k$$ and the Mach number. Hence, the net force is
$F_{net} = P_0 A^{*} (1+k) {\left( k+1 \over 2 \right)^{k \over k-1}} \left( {F_2 \over F^{*} } - { F_1 \over F^{*}}\right) \label{gd:iso:eq:NetForce} \tag{71}$
To demonstrate the usefulness of the this function consider a simple situation of the flow through a converging nozzle.
Example 11.11
Fig. 11.10 Schematic of a flow of a compressible substance (gas) through a converging nozzle for example (??)
Consider a flow of gas into a converging nozzle with a mass flow rate of $$1[kg/sec]$$ and the entrance area is $$0.009[m^2]$$ and the exit area is $$0.003[m^2]$$. The stagnation temperature is $$400K$$ and the pressure at point 2 was measured as $$5[Bar]$$. Calculate the net force acting on the nozzle and pressure at point 1.
Solution 11.11
The solution is obtained by getting the data for the Mach number. To obtained the Mach number, the ratio of $$P_1A_1/A^{*}P_0$$ is needed to be calculated. The denominator is needed to be determined to obtain this ratio. Utilizing Fliegner's equation (59), provides the following
$A^{*} P_0 = \dfrac{\dot{m} \sqrt{R\,T} }{ 0.058} = \dfrac{1.0 \times \sqrt{400 \times 287} }{ 0.058} \sim 70061.76 [N]$
and
$\dfrac{A_2\, P_2 }{ A^{\star}\, P_0} = \dfrac{ 500000 \times 0.003 }{ 70061.76 } \sim 2.1$
Isentropic Flow Input: \dfrac{A\, P }{ A^{\star} \, P_0} k = 1.4 $$M$$ $$\dfrac{T}{T_0}$$ $$\dfrac{\rho}{\rho_0}$$ $$\dfrac{A}{A^{\star} }$$ $$\dfrac{P}{P_0}$$ $$\dfrac{A\, P }{ A^{\star} \, P_0}$$ $$\dfrac{F }{ F^{\star}}$$ 0.27353 0.98526 0.96355 2.2121 0.94934 2.1000 0.96666
With the area ratio of $${A \over A^{\star}}= 2.2121$$ the area ratio of at point 1 can be calculated.
$\dfrac{ A_1 }{ A^{\star}} = \dfrac{A_2 }{ A^{\star}} \dfrac{A_1 }{ A_2} = 2.2121 \times \dfrac{0.009 }{ 0.003} = 5.2227$
And utilizing again Potto-GDC provides
Isentropic Flow Input: \dfrac{A\, P }{ A^{\star} \, P_0} k = 1.4 $$M$$ $$\dfrac{T}{T_0}$$ $$\dfrac{\rho}{\rho_0}$$ $$\dfrac{A}{A^{\star} }$$ $$\dfrac{P}{P_0}$$ $$\dfrac{A\, P }{ A^{\star} \, P_0}$$ $$\dfrac{F }{ F^{\star}}$$ 0.11164 0.99751 0.99380 5.2227 0.99132 5.1774 2.1949
The pressure at point $$1$$ is
$P_1 = P_2 {P_0 \over P_2} { P_1 \over P_0} = 5.0 times 0.94934 / 0.99380 \sim 4.776[Bar]$
The net force is obtained by utilizing equation (71)
$F_{net} &= P_2 A_2 {P_0 A^{*} \over P_2 A_2} (1+k) {\left( k+1 \over 2 \right)^{k \over k-1}} \left( {F_2 \over F^{*} } - { F_1 \over F^{*}}\right) \ & = 500000 \times {1 \over 2.1}\times 2.4 \times 1.2^{3.5} \times \left( 2.1949 - 0.96666 \right) \sim 614[kN]$
### Contributors
• Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license. | 2,021 | 5,111 | {"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": 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} | 3.84375 | 4 | CC-MAIN-2019-04 | latest | en | 0.561111 |
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(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
If the following instructions are carried out by a computer, which value of $X$ will be printed because of instruction $5$?
1. START $X$ AT $3$ AND $S$ AT $0$.
2. INCREASE THE VALUE OF $X$ BY $2$.
3. INCREASE THE VALUE OF $S$ BY THE VALUE OF $X$.
4. IF $S$ IS AT LEAST $10000$,
THEN GO TO INSTRUCTION $5$;
OTHERWISE, GO TO INSTRUCTION $2$.
AND PROCEED FROM THERE.
5. PRINT THE VALUE OF $X$.
6. STOP.
$\text{(A) } 19\quad \text{(B) } 21\quad \text{(C) } 23\quad \text{(D) } 199\quad \text{(E) } 201$
## Solution
Looking at the first few values, it becomes clear that the program stops when $$5+7+9+11+13+\ldots+(x-2)+x\ge 10000$$ which is to say $$1+3+5+7+9+11+13+\ldots+(x-2)+x\ge 10004$$ However, the left hand side is now simply the square $\frac{(x+1)^2}4$. Multiplying out, we get $$x+1\ge \sqrt{40016}\approx 200.039996...$$
So the correct answer is $201$, which is $\fbox{E}$
1990 AHSME (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions | 487 | 1,274 | {"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": 22, "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-2024-18 | latest | en | 0.55863 |
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What is the ideal example of a given figure?
The best image to describe the transformation of the figure is identified in the following. It's ideal for grade 5 and grade 6 children. The figure and the image are in each grid.
What is the main purpose of the multi-grade exercises?
Thanks to the wide-ranging and multi-grade exercises incorporated here, students get a great insight into the three transformations - rotation, reflection, and translation of shapes. The students of grade 2 through grade 8 are recommended to take this class. CCSS:.
What is the line y=0?
You don't need a mirror line to reflect a shape. A shape is reflected in the line. The x axis is marked on the axes by recognizing the line y=0. This can be done with tracing paper if you want to make sure the corners of the shape are the same distance from the mirror line.
What is the x-axis b the y-axis c the?
The image of the point is reflected in the x- axis and the y- axis. Y is for y. x The image of the point is under a in: 2. The y- axis is followed by the x- axis. The line x c and the line x are equal.
What is the practice of Transformations Worksheets?
There are transformations in this picture. There are transformations for translations, reflections, rotation and dilations. There are two ways to check students' answers on the transformational geometry. You can line up the student's page and the answer page and hold them up to the light.
What is a change in the position, size, or shape of a figure?
A transformation is a change in a figure. A rigid motion is the act of moving an object without changing its shape or size. , rotation, translations, and glide are all examples of rigid motions. It is rigid.
What was the name of the Transformation Worksheet?
The name of the transformation is U00b7 ABCD was translated to Au 2019. How many units were moved in the direction of the parallelogram ABCD? 3 units to the right and 3 units to the left. A shape is shown. Which is the shape that was transformed by a ip?
What do translations and dilations of two-dimensional shapes on a coordinate plane 8?
The effects of translations over the x- or y- axis, and the effects of rotation limited to 90, are different from reflections, translations, and dilations of two-dimensional shapes on a coordinate plane.
What is the dilation from Figure A to Figure B?
There is a link between transformation and similarity. Determine if the dilation from Figure A to Figure B is a reduction or an enlargement. Is this from figure A to figure B? The y- axis is 14. The degree of rotation is 90° , 180° , 270° , or 360° .
What is the name of transformations that donu2019t change?
Chapter 7 focuses on the transformations that don't change the shape of figures. Isometries are transformations. In Chapter 11 expansion and contraction are studied. There are three major types of isometries. Translating is simply shifting. Students use translations.
What transformation could be used to show that gure A is congruent to gur?
Which transformation could be used to show that the two objects are not the same?
What is a transformation done to a graph/function that causes it to change in some?
Graph Transformations. A transformation is something that happens to a graph or function. Changing a function has effects on a graph. There are two types of translations, which give you 4 key skills.
What is the single transformation that maps T onto H directly?
Take T in the x-axis and label it U and U in the line. Go back to the beginning and label it G. G in the line y and H.
What is a type of opposite isometry?
Take the Trapezoid MATH from (x, y) U00e0 and put it over the y- axis. A glide reflection is a type of opposite isometry. A reflection and a translation are included. Lesson 1.
What do translations and dilations of two-dimensional shapes on a coordinate plane 8?
The effects of translations over the x- or y- axis, and the effects of rotation limited to 90, are different from reflections, translations, and dilations of two-dimensional shapes on a coordinate plane.
What is the dilation from Figure A to Figure B?
There is a link between transformation and similarity. Determine if the dilation from Figure A to Figure B is a reduction or an increase. Is it the reflection from figure A to figure B? There are 14. The degree of rotation is 90° , 180° , 270° , or 360° From figure A to figure Z.
What is the top 10 resources for Transformation cards?
The top 10 resources. The cards are transformation. Word document and publisher file are used for the age range. There are four activities that focus on one of the transformations. There is a set of dominoes, rotation loop cards, sort card on translations, and a nice spot-the-difference activity to finish.
What is the composition of two reflections?
The composition of those transformations is one of the things that distinguishes them. The chapter considers reflections across parallel lines and reflections across nonparallel lines. This chapter looks at glide in the context of tessellations.
What is the geometric transformation?
A set that has a geometric structure by itself or another set is called a geometric transformation. The appearance of a shape is changed if it is transformed. The shape could be similar to its pre image after that. A change of appearance is what the meaning is.
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Consider Two Concentric Circular Coils, One of Radius R1 and the Other of Radius R2 (R1 < R2) Placed Coaxially with Centres Coinciding with Each Other. Obtain the Expression for the Mutual Inductance of the Arrangement. - CBSE (Science) Class 12 - Physics
Question
Consider two concentric circular coils, one of radius r1 and the other of radius r2 (r1 < r2) placed coaxially with centers coinciding with each other. Obtain the expression for the mutual inductance of the arrangement.
Solution
Coefficient of mutual induction − consider two coils and S. Suppose that a current I is flowing through the coil P at any instant i.e.,
Φ ∝ I
Φ = MI… (i)
If ‘e’ is the induced emf produced in the S-coil, then
e=(dphi)/dt=-d/dt(MI)=-M(dl)/dt
Mutual Inductance of two concentric coils, one of radius r1 and the other of radius r2 (r1 < r2) placed coaxially with centers coinciding with each other:
Consider two circular coil S1 and S2 of same length l, such that coil S2 surrounds coil S1 completely
Let
n1 − Number of turns per unit length of S1
n2 − Number of turns per unit length of S2
I1 − Current passed through solenoid S1
Φ21 − Flux linked with S2 due to current flowing through S1
Φ21 ∝ I1
Φ21 = M21I1
Where M21 is the coefficient of mutual induction of the two coils
When current is passed through S1, an emf is induced in S2.
Magnetic field produced inside S1 on passing current through it,
B1 = μ0n1I1
Magnetic flux linked with each turn of Swill be equal to B1 times the area of the cross-section of S1.
Magnetic flux linked with each turn of the S2 = B1A
Therefore, total magnetic flux linked with the S2,
Φ21 = B1A × n2l = μ0n1I1 × A× n2l
Φ21 = μ0n1n2AlI1
∴ M21 = μ0n1n2Al
Similarly, the mutual inductance between the two coils, when current is passed through coil S2 and induced emf is produced in coil S1, is given by
M12 = μ0n1n2Al
∴M12 = M21 = M (say)
Hence, coefficient of mutual induction between the two coil will be
M=mu_0n_1n_2Al
Is there an error in this question or solution?
Video TutorialsVIEW ALL [2]
Solution Consider Two Concentric Circular Coils, One of Radius R1 and the Other of Radius R2 (R1 < R2) Placed Coaxially with Centres Coinciding with Each Other. Obtain the Expression for the Mutual Inductance of the Arrangement. Concept: Inductance - Mutual Inductance.
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# Two water pumps, working simultaneously at their respective
Author Message
Senior Manager
Joined: 22 Sep 2005
Posts: 262
Two water pumps, working simultaneously at their respective [#permalink]
### Show Tags
Updated on: 12 Jan 2014, 04:44
12
00:00
Difficulty:
15% (low)
Question Stats:
84% (01:23) correct 16% (01:25) wrong based on 237 sessions
### HideShow timer Statistics
Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?
A. 5
B. 16/3
C. 11/2
D. 6
E. 20/3
OPEN DISCUSSION OF THIS QUESTION IS HERE: two-water-pumps-working-simultaneously-at-their-respective-155865.html
Originally posted by netcaesar on 14 Nov 2006, 12:56.
Last edited by Bunuel on 12 Jan 2014, 04:44, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
Manager
Joined: 29 Aug 2006
Posts: 156
### Show Tags
14 Nov 2006, 13:06
1
4
E from me...
R(A) and T(A)= rate and time of 1st pump.
R(B) and T(B)= rate and time of 2nd pump.
R(A)= 1.5 R(B)
Hence T(B)=1.5 T(A)
1/T(A) + 1/T(B)= 1/4
Substituting
T(A)= 20/3
Manager
Joined: 12 Oct 2006
Posts: 81
Location: NJ, US
### Show Tags
14 Nov 2006, 13:51
E for me as well
d=r*t
1 = 7/2r*4
1/10 = r
Faster rate = 1/10*3/2 = 3/20
1 = 3/20t
20/3 = t
Senior Manager
Joined: 22 Sep 2005
Posts: 262
### Show Tags
17 Nov 2006, 11:31
You are right.
OA is E
Senior Manager
Joined: 20 Feb 2006
Posts: 367
### Show Tags
17 Nov 2006, 13:02
3
1
Work = r * t
Work A = 1 * 4
Work B = 3/2 * 4
Total work = (1*4) + (3/2 * 4) = 10
10/rate B = 10/(3/2) or 10*2/3 = 20/3
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 4836
Location: Singapore
### Show Tags
19 Nov 2006, 18:49
2
Let the rate be:
Pump A - 1 pool in A hours
Pump B - 1 pool in B hours
Then Pump A - 1/A pool in 1 hour
And Pump B - 1/B pool in 1 hour
Together: (A+B)/AB pool in 1 hour -> 1 pool in AB/(A+B) hours = 4 hours
4A + 4B = AB -- (1)
Assuming pump B is faster and A = 1.5B, then 4A+4B = AB becomes
1.5B^2 = 10B
B(1.5B - 10) = 0
B = 0 or B = 20/3 hours
So 1 pool in 20/3 hours
Ans E
Intern
Joined: 06 Jan 2014
Posts: 38
Re: Two water pumps, working simultaneously at their respective [#permalink]
### Show Tags
11 Jan 2014, 15:09
I felt like I understood "work" problems until I saw this one on my GMATPrep Exam.
In my head the problem looks like this
4hrs = (3/2)X + X
or
1/4 = (3/2)*(1/X) + (1/X)
I simply dont know how to solve this problem....I'm confused!
I've seen how others have solved this problem, but it just isn't clicking.
Is someone willing to help?
Math Expert
Joined: 02 Sep 2009
Posts: 51121
Re: Two water pumps, working simultaneously at their respective [#permalink]
### Show Tags
12 Jan 2014, 04:47
4
7
TroyfontaineMacon wrote:
I felt like I understood "work" problems until I saw this one on my GMATPrep Exam.
In my head the problem looks like this
4hrs = (3/2)X + X
or
1/4 = (3/2)*(1/X) + (1/X)
I simply dont know how to solve this problem....I'm confused!
I've seen how others have solved this problem, but it just isn't clicking.
Is someone willing to help?
Work rate problems are my weakest area. For whatever reason I get mixed up on these problem. Even knowing the answer to this question I can't figure out how it comes out to that answer. Any help would be great, thanks!
Two water pumps, working simultaneously at their respective constant rates, took exactly four hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at it's constant rate?
A. 5
B. 16/3
C. 11/2
D. 6
E. 20/3
Say the rate of the faster pump is x pool/hour, then the rate of the slower pump would be x/1.5=2x/3 pool/hour.
Since, the combined rate is 1/4 pool/hour, then we have that x+2x/3=1/4 --> x=3/20 pool hour.
The time is reciprocal of the rate, therefore it would take 20/3 hours the faster pump to fill the pool working alone.
All DS work/rate problems to practice: search.php?search_id=tag&tag_id=46
All PS work/rate problems to practice: search.php?search_id=tag&tag_id=66
OPEN DISCUSSION OF THIS QUESTION IS HERE: two-water-pumps-working-simultaneously-at-their-respective-155865.html
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Posts: 9128
Re: Two water pumps, working simultaneously at their respective [#permalink]
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14 Aug 2018, 23:32
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Re: Two water pumps, working simultaneously at their respective &nbs [#permalink] 14 Aug 2018, 23:32
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# Application: Work
8.1
What is Work?
Work, in the physics sense, is usually defined as force acting over a distance.
Work is sometimes force times distance,1 but not always. Work is more subtle than
that. Every time you exert a force, it is not the case that any work is done (even
though it may feel like that to you!
Why? Well, the work/energy equation says that work done (by the net force on
an object) equals the objects change in kinetic energy. More simply:
Work = Change in Kinetic Energy.
## Work will be force times distance in all
of the applications we consider.
1
(8.1)
This means that if an objects kinetic energy doesnt change, then no work has
been done on the objectwhether or not a force has been exerted. In particular, a
force will do work only if the force has a component in the direction that the object
moves.
Heres the distinction: if you push on a big box and it moves in the direction
you push, then work is accomplished and it equals force times distance. If the
box is very big and when you push it nothing happens (it does not move), then
no work is done. If a force is applied to the box in a certain direction and the box
moves, but not in the direction you push (maybe your friend is pushing on another
side) but in some other direction, then work is done, but the amount depends
on the angle the box moves relative to the direction the box moves. So again the
answer is not simply force times distance.
However, to keep things simple, in the context of this section we will assume the
direction the force is applied and the direction of the motion of the object are one
and the same. In this case, assuming that the force is constant,
W = Work = force Distance = F x.
The unit used to measure work vary depending on the system you are inand
they are probably less familiar to you than the units for velocity and acceleration.
Table 8.1 lists the units for three common systems.
System
Force
Distance
Work
British
cgs
SI (international)
pounds
dynes
Newtons
feet
centimeters
meters
foot-pounds (ft-lbs)
ergs
joules
EXAMPLE 8.1. Calculate the work done in lifting a 147 lb object 22 feet (e.g., me walk-
## ing from the first to the third floor in Lansing)? Well,
Work = force Distance = 147 22 = 3, 234 ft-lbs.
## Table 8.1: Units of work.
math 131
application: work
## That was easy.
Now the key thing to notice is that under our assumptions, work is a product.
One of the assumptions is that a constant force is applied. But most forces are not
constant. Consider the following situation.
EXAMPLE 8.2. Suppose we want to hoist a leaking bucket vertically. Because of the
leak, the force applied to lift the bucket decreases, so that
x2
F ( x ) = 60 1
,
0 x 50
5000
where x is measured in feet. Find the work done in lifting the bucket.
So how do we calculate the work done if the force varies? Well, remember that
Riemann sums involve products. So we will use the subdivide and conquer strategy once more. Unlike in the earlier cases there is no natural figure to draw whose
area, volume, or arc length we are trying to calculate. This time we must apply the
theory.
General Situation. Assume that F ( x ) is a variable but continuous force that is a
function of the position x and that it is applied over an interval [ a, b]. Find the
work done. As usual, let P = { x0 , x1 , . . . , xn } be a regular partition of [ a, b] into n
equal width subintervals of length x. Now if the intervals are short enough, since
the force is continuous, the force will be nearly constant on each interval, though
its value will vary from interval to interval. So let Wi denote the work done on the
ith subinterval. Since the length of the ith subinterval is x, then
Wi F ( xi )x.
Consequently, adding up the pieces of work on each subinterval,
Total Work =
Wi F(xi )x.
(8.2)
Notice that we now have a Riemann sum involving the force function! To improve the approximation we do the standard thing: We let the number of subdivisions get large and take the limit. We find
n
Z b
Wi = nlim
F(xi )x =
n
## Total Work = lim
F ( x ) dx.
We are certain that this limit of the Riemann sums exists and is, in fact, the definite
integral because we assumed that the force F ( x ) is continuous function of the
position.
THEOREM 8.1 (Work Formula). If F ( x ) is a continuous force that is a function of the position
x that is applied over an interval [ a, b], then the work done over the interval is
Work =
Z b
a
F ( x ) dx.
Stop and Step Back. There are a couple of things I want you to notice. First, Theorem 8.1 amounts to saying that work is the area under the force curve. Thats
probably not how you would first think of it, but thats what the theorem says!
Second, when I was writing these notes, I simply cut-and-pasted the earlier material on arc length and changed the a few words here and there. But it is the
same subdivide and conquer strategy that you have seen several times now. You
should be able to create such arguments for yourself now.
math 131
application: work
EXAMPLE 8.3 (Return to the leaky bucket). Return to the leaking bucket. The force ap-
## plied was continuous and it was applied on the
x2
F ( x ) = 60 1
,
5000
and it was applied on the interval [0, 50] where x is measured in feet. So by Theorem 8.1 the work done in lifting the bucket is
Z b
Z 50
x2
Work =
dx
F ( x ) dx =
1
5000
a
0
50
x3
= 60 x
15, 000 0
125, 000
= 60 50
0
15, 000
= 2500 ft-lbs.
Pretty straightforward!
H
8.2
## Work Done Emptying a Tank
The following tank problems involve pumping liquids from one height to another
and determining the amount of work required to do it. Heres the general question. Given a tank containing a liquid between heights a and b, how much work is
required to pump the liquid to a height H. (Note: H may or may not be the height
of the top of the tank.) See Figure 8.1.
To solve the problem we do the usual thing. We subdivide and conquer. Use
a regular partition of the interval [ a, b] on the vertical y-axis into n equal width
subintervals, each of width y. The subintervals are used to create layers (slices)
of liquid, each of width y. See Figure 8.1. What properties of the layer are important in determining the work done in lifting the layer out? Take a moment before
We need
the volume of the layer;
the density of the layer;
the distance the layer is moved.
The weight of the layer is the force here (e.g., weight is measured in pounds) and
is the density of the liquid times the volume. So the work to move the ith layer to
height H is
Wi = Work to move the ith layer to height H
## = Weight Distance the layer is moved
= (Density Volume) Distance the layer is moved
(8.3)
OK, its time to use a bit of analysis and notation we saw in the section on volume.
Lets denote the density by D (in units such as lbs/ft3 ).
As usual, let yi denote the ith partition point of the interval [ a, b].
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## Figure 8.1: A tank containing a liquid
between levels a and b. The liquid is to
be moved to height H. Also shown is
a representative layer of the liquid at
height yi . Can we determine how much
work is required to lift this layer to
height H? What will this work depend
on?
math 131
application: work
The ith layer of liquid is approximately is approximately a cylinder (see Figure 8.1) so its volume is
Vi = Area of the base height = Cross-sectional area height = A(yi )y,
where A(yi ) denotes the the cross-sectional area of the tank at height yi and the
height of the layer is y.
Finally, since the ith layer is at height yi and has to be moved to height H, then
the ith layer is moved a distance of H yi .
Putting this all together, we can rewrite the work to move the ith layer to height H
in (8.4) as
Wi = (Density Volume) Distance the layer is moved
= D Vi ( H yi )
D [ A(yi )y]( H yi ).
(8.4)
But now you know how the rest of the story goes. To estimate the total work we
add up the work to move each layer for each subinterval,
Total Work =
Wi D[ A(yi )y]( H yi ).
(8.5)
## We now have a Riemann sum.2
To improve the approximation we do the standard thing: We let the number of
subdivisions get large and take the limit. We find
Total Work = lim
Z b
Wi = nlim
D[ A(yi )y]( H yi ) = D
## Take a second to compare (8.2) and
(8.5). Then take another to anticipate
the rest of this argument.
A(y)[ H y] dy.
We are certain that this limit of the Riemann sums exists and is, in fact, the definite
integral as long as we assume that the cross-sectional area A(y) is continuous
function of y.
THEOREM 8.2 (Work Formula for Emptying a Tank). Assume the cross-sectional A(y) is a
continuous function of the position y and that the density of the contents is a constant D.
If the contents of the tank to be moved lie in the interval [ a, b], then the work done to move
this material to a height H is
Work = D
Z b
a
A(y)[ H y] dy.
Caution: The tank may not be full, the contents may be moved to a height H above the tank,
or the entire tank may not be emptied. If the tank is being filled from a source at height H
(either at the bottom of or below the tank), then the contents must be moved to each layer
height y between a and b so the distance moved is y H rather than H y.
EXAMPLE 8.4. Heres a simple first example. An above-ground backyard swimming
pool has the shape of a circular cylinder with a radius of 10 ft and a depth of 8 ft.
Assume the depth of the water in the pool is 5 ft. Find the work done in emptying the
pool by pumping the water over the top edge of the pool. Note: The density of water
is 62.5 lbs/ft3 . See Figure 8.2.
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8
5
SOLUTION. We apply Theorem 8.2. Be attentive to the different heights. The liquid
lies between 0 and 5 feet but has to be moved to a height of H = 8 feet. The crosssections are circles of radius r = 10 feet. So A(y) = r2 = 100. (In most problems
## Figure 8.2: The pool is filled to a depth
of 5 feet and the water is to be pumped
out over the top edge at height 8 feet.
math 131
application: work
## the cross-sections will vary.) So
Work = D
Z b
a
A(y)[ H y] dy = 62.5
Z 5
0
100 [8 y] dy
5
y2
= 6250 8y
2 0
25
= 6250 40
0
2
## = 17, 1875 ft-lbs.
EXAMPLE 8.5. (A revision) Suppose the swimming pool in Example 8.4 above was in-
ground so that the top of the pool was at ground level. How would the work integral
change? Would the work done emptying the pool change?
SOLUTION. Three things change in the set-up. The liquid now lies between 8 and
3 feet and has to be moved to a height of H = 0 feet. The cross-sections are the
same.
Work = D
Z b
a
A(y)[ H y] dy = 62.5
Z 3
8
100 [0 y] dy
3
y2
= 6250
2 8
9
= 6250 + 32
2
## = 17, 1875 ft-lbs.
The work should and does remain the same.
EXAMPLE 8.6. (A more complicated problem) An underground hemispherical tank
with radius 10 ft is filled with oil of density 50 lbs/ft3 . Find the work done pumping
the oil to the surface if the top of the tank is 6 feet below ground.
SOLUTION. It will be easiest to set up the equation of the hemisphere if we think
of the top of the tank at height 0 and then pump the oil to a height of 6 feet. See
Figure 8.3.
The cross-sections are circles. We will be able to determine the cross-sectional area
once we determine the radius of the cross-section. The semi-circle is part of the circle
of radius 10 centered at the origin which has equation x2 + y2 = 10. The radius of a
cross-section is the x-coordinate of the point ( x, y) that lies on the semi-circle in fourth
quadrant. (See Figure 8.4.) Thus,
q
r = x = (10)2 y2 .
6
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10
## Figure 8.3: The tank with its position
re-imagined and a representative layer.
## A(y) = r2 = [102 y2 ] = (100 y2 ).
10
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10
## We apply Theorem 8.2. Remember the liquid is pumped to height H = 6 in our
re-casting of the problem.
Work = D
Z b
a
A(y)[ H y] dy = 50
Z 0
10
Z 0
((10)2 y2 )[6 y] dy
600 100y 6y2 + y3 dy
10
!
0
y4
= 50 600y 50y2 2y3 +
4
10
= 50
## = 50 [0 (6000 5000 + 2000 + 2500)]
= 325, 000 ft-lbs.
10 y
## Figure 8.4: The radiuspof the crosssection at height y is (10)2 y2 .
math 131
application: work
YOU TRY IT 8.1. Set up the new integral for each modification of the example above and
determine the work required.
(a) How would the integral and work change if the tank were only 5 feet below ground?
3 ft-lbs.)
(b) How would the integral and work change if the top of the tank were at ground level?
(Answer: 125, 000 ft-lbs.)
EXAMPLE 8.7. (A tank formed by rotation) A tank is formed by rotating the region
between y = x2 , the y-axis and the line y = 4 in the first quadrant around the y axis.
The tank is filled with oil with density 50 lbs/ft3 .
(a) Find the work done pumping the oil to the top of the tank.
(b) Find the work done pumping the oil to the top of the tank if there is only 1 foot
of oil in the tank.
(c) Suppose the tank is empty and is filled from a hole in the bottom to a depth of
3 feet. Find the work done.
SOLUTION. Figure 8.5 shows a sketch of the tank with a representative cross-section.
The circular cross-section formed by rotating the point ( x, y) about the y-axis has
radius r = x. So the cross-sectional area is
as a function of y.
(a) We apply Theorem 8.2.
Z b
a
A(y)[ H y] dy = 50
= 50
Z 4
0
Z 4
0
(y)[4 y] dy
4y y2 dy
4
y3
= 50 2y2
3 0
64
= 50 32
0
3
1600
ft-lbs.
=
3
(b) If there is only 1 foot of oil in the tank, the only change is that the upper limit of
integration is now 1 instead of 4. (Remember the limits of integration represent
the upper an lower levels of the content of the tank.) So
Work = D
Z b
a
A(y)[ H y] dy = 50
Z 1
0
(y)[4 y] dy
1
y3
= 50 2y2
3
0
1
= 50 2
3
250
ft-lbs.
=
3
(c) The key observation when a tank is filled from the bottom is that each layer
must be moved or lifted from ground level 0 to its final height H = y. Since the
tank is only filled to a depth of 3 feet, the limits of integration are 0 (bottom) to
3 ft. So this time
Work = D
Z b
a
A(y)[ H y] dy = 50
r = x ( x, y)
A(y) = r2 = x2 = y,
Work = D
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Z 3
0
(y)[y 0] dy
3
y3
3 0
1
= 50 2
3
= 50
= 450 ft-lbs.
## Figure 8.5: The circular cross-section
formed by rotating the point ( x, y)
about the y-axis has radius r = x.
math 131
application: work
YOU TRY IT 8.2. How would the integral and work change if the tank were full and the oil
was pumped to a height 3 feet above the top of the tank? (Answer: 5200
3 ft-lbs.)
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YOU TRY IT 8.3. Suppose instead that the tank is formed by rotating the region in the first
quadrant enclosed by y = ( x + 2)2 , the y-axis, and the x-axis about the y-axis (see Figure 8.6).
R4
(a) Find the work done pumping the oil to the top of the tank. (Answer: 50 0 ( y
2)2 (4 y) dy = 1280
3 ft-lbs.)
(b) Find the work done if the oil was pumped to a height 3 feet above the top of the tank?
3 ft-lbs.)
(c) Find the work done pumping the oil to the top of the tank if there is only 1 foot of oil
in the tank. (Answer: 330 ft-lbs.)
( x, y)
## Figure 8.6: The circular cross-section
formed by rotating the point ( x, y)
about the y-axis has radius r = x =
y 2.
EXAMPLE 8.8. (A more complicated problem) A large gasoline storage tank lies 3 feet
## underground. It is in the shape of a half-cylinder (flat side up) 4 ft in radius and 8 ft
long. If the tank is full of gas with density 60 lbs/ft3 , find the work done pumping the
gas to a level 2 ft above ground.
SOLUTION. It will be easiest to set up the equation of the hemisphere if we think of
the top of the tank at height 0. Since the original tank was 3 feet underground and the
contents were to be pumped to 2 feet above ground, in our model we need to pump
the gas to a total of 5 feet above ground. See Figure 8.7.
2
2
The cross-sections
p are rectangles of width 2x and length 8. Since x + y = 4 on the
semi-circle, x = 16 y2 . So the cross-sectional area is
q
A(y) = 2x (8) = 16 16 y2 .
We apply Theorem 8.2. Remember the liquid is pumped to height H = 5 in our
re-casting of the problem.
5
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.....
x2 + y2 = 42
Work = D
Z b
a
## Figure 8.7: Left: Think of the tank top
at ground level and the liquid being
pumped to height H = 5 ft. Right: The
width of
p the cross-section at height y is
2x = 2 42 y2 .
3
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x2 + y2 = 42
Z 0
q
(16 16 y2 [5 y] dy
4
Z 0 q
q
= 800
5 16 y2 y 16 y2 dy
A(y)[ H y] dy = 50
= 4000
Z 0 q
4
16 y2 dy 800
Z 0
4
16 y2 dy.
(8.6)
R0 p
The first integral 4 16 y2 dy represents the area of a quarter circle of radius
2
r = 4 and so its value3 (area) is 4 4 = 4.
The second integral in (8.6) can be done using the substitution u = 16 y2 with
1
2 du = y dy. Be careful to change the limits!
Z 0
4
16 y2 dy =
16
Z 16
1 1/2
1
64
u
du = u3/2 =
.
2
3
3
0
0
## Remember that we have p
not yet developed an antiderivative of 16 y2 , as
we noted in the section on arc length,
so we have to use the geometric interpretation of the definite integral
here.
3
math 131
application: work
## Consequently, (8.6) becomes
Z 0 q
Z 0
16 y2 dy 800
y
4
4
64
= 4000(4 ) 800
3
Work = 4000
16 y2 dy
33172.14913 ft-lbs.
YOU TRY IT 8.4. How would the integral and work change if the tank in the previous problem had the flat side facing down? Hint: Put the center of the semi-circle at the origin. The
tank now lies above the x-axis. What height is the gas pumped to this time? Now work out
ft-lbs.)
3
YOU TRY IT 8.5. A tank in the form of a truncated cone is formed by rotating the segment
between (2, 0) and (4, 4) around the y-axis. It is filled with sludge (density 80 lbs/ft3 ). If
the sludge is pumped 3 feet upwards into a tank truck, how much work was required?
ft-lbs.)
3
YOU TRY IT 8.6 (Rotation about a non-axis line). This time a tank in the form of a truncated
cone is formed by rotating the segment between (0, 4) and (2, 0) around the line x = 4. It
is filled with sludge (density 80 lbs/ft3 ). If the sludge is pumped 3 feet upwards into a tank
ft-lbs.)
truck, how much work was required? (Answer: 40960
3
YOU TRY IT 8.7 (Non-rotation). A triangular trough for cattle is 8 ft long. The ends are trian-
gles with with a base of 3 ft and height of 2 ft (but the vertex points down). See Figure 8.8.
Find the work done by the cattle in emptying just the top foot of water (density 62.5 lbs/ft3 )
over the edge. (Answer: 500 ft-lbs.)
YOU TRY IT 8.8. A cup shaped tank is obtained by rotating the curve y = x3 about the y-axis
where 0 x 2.
lbs/ft3 ?).
## (a) Assume the tank is full of water (density 62.5
How much work is done in
emptying the tank by removing the water over the top edge of the tank? (Answer:
3600 ft-lbs?)
(b) How much work would be done in raising the water 2 feet above the tanks top?
(c) Suppose the depth of the liquid in the tank is 1 foot. Find the work required to pump
the liquid to the top edge of the tank.
YOU TRY IT 8.9. (a) A cone-shaped reservoir has a 10 foot radius across the top and a 15 foot
depth. If the reservoir has 9 feet of oil (density 54 lbs/ft3 ) in it, how much work is
required to empty it by bringing the water to the top of the reservoir? (First figure out
the equation of the line that determines the cone.)
(b) Same question with the reservoir being completely full. (Answer: 101250 ft-lbs?)
YOU TRY IT 8.10 (Expanding Gas). Boyles law says that the pressure (force) exerted by a
gas is inversely proportional to the volume. A quantity of gas with an initial volume of 2
cubic feet and a pressure of 1000 pounds per square foot expands to a volume of 5 cubic
feet. Find the work done. Hint: p = k/V, so first solve for k. Next, the work done by an
expanding gas is
W=
Z V1
V0
p dv.
## Answer: 2000(ln 5 ln 2) ft. lbs.)
YOU TRY IT 8.11 (Extra Credit). A heavy rope is 60 feet long and has a density of 1.5 lbs/ft. It
is hanging over the edge of a building 100 ft tall. Find the work done in pulling the rope to
the top of the building. Hint: Model your answer like a tank problem. Think of the rope in
sections. What force must be applied to move each section of the rope? What distance must
each section be used? OK, set up the integral and do it!
8.3
Problems
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## Figure 8.8: Find the work done in
emptying the top foot of the trough
over the top edge.
math 131
1. Assume 25 ft-lb of work is required to stretch a spring 3 feet beyond its natural length.
Find the work done in stretching the spring from 2 to 4 ft beyond its natural length.
Hint: First find the spring constant. (Answer: k = 50/9 lbs/ft and 100/3 ft-lbs.)
2. Let R be the region in the first quadrant bounded by the x-axis, the y-axis, and the curve
y = 4 x2 . Rotate the R around the y-axis to form a silo tank. If the tank is filled with
wheat (100 lbs per cu. ft) how much work is done in raising the wheat to the top of the
silo. (Ans: 6400
3 ft lbs.)
3. Let R be the region in the first quadrant bounded by the y-axis and the curves y = x2
and y = 6 x. Rotate the R around the y-axis to form a tank. If the tank is filled with
honey (90 lbs per cu. ft) how much work is done in raising the honey to the top of the
tank. Note: you will have to divide the work into two pieces! Be careful of the limits.
(Ans: 2760 ft lbs.)
4. A beer ball with radius 1 foot is located with its center at the origin. It has a tap 0.5 ft
long sticking up out the top. If beer has a density of 60 pounds per cubic foot, find the
work done in emptying the beer ball. Hint: Locate the center of the ball at the origin.
(Ans: 120 ft-lbs)
5. Ants excavate a chamber underground that is described as follows: Let S be the region in
the fourth quadrant enclosed by y = x, y = 1, and the y-axis; revolve S around the
y-axis.
(a) Set up and simplify the integral for the volume of the chamber using the shell
method. (For future reference: Ans: /5)
(b) [From an exam] Suppose that the chamber contained soil which weighed 50 lbs per
cubic foot. How much work did the ants do in raising the soil to ground level? (Ans:
50/6 ft. lbs.)
If you have time, set up each of the following integrals which we will do with Maple.
6. A small farm elevated water tank is in the shape obtained from rotating the region in the
first quadrant enclosed by the curves y = 10 21 x2 , y = 8, and the y-axis about the y-axis.
(a) Find the work lost" if the water (62.5 lbs/ft3 ) leaks onto the ground from a hole in
the bottom of the tank. (Answer: 6500/3 ft-lbs.)
(b) Find the work lost" if the water leaks onto the ground from a hole in the side of the
tank at height 9 feet. (Answer: 1750/3 ft-lbs.)
7. The line segment between the points (1, 3) and (3, 1) is rotated around the y-axis to
form a truncated conical gas storage tank. If the tank has only 1 foot of gasoline (density
60 lbs/ft3 ) in it, set up the integral for the work done to pump all of the gas to ground
level.
application: work | 7,797 | 29,185 | {"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} | 4.125 | 4 | CC-MAIN-2019-35 | latest | en | 0.936364 |
http://firefightermath.org/index.php?option=com_content&view=article&id=19&Itemid=34 | 1,532,150,819,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592387.80/warc/CC-MAIN-20180721051500-20180721071500-00129.warc.gz | 135,012,378 | 5,993 | 1.14 Subtracting Fractions
### Subtracting Fractions
Similar to adding fractions, fractions must have a common denominator (the bottom numbers must be the same) before being subtracted.
There are three steps to subtracting fractions:
1. Ensure that the bottom numbers (denominators) are the same. If they are not, change them so that they are the same (they have a common denominator).
2. Once the denominators are the same, subtract the top numbers (numerators) and place the result over the common denominator.
3. Simplify the fraction (if possible).
### Example 1
Now, let's work through the example below. The questions below will guide you through the process.
Subtract and simplify
First, both numbers must be put in fraction form. In the simplifying fractions section we learned that .
Thus, we are subtracting
Step 1. Are the denominators the same?
The denominators are not the same.
Is one of the denominators multiple of the other?
The denominators are not multiple of each other.
Select the common denominator?
6 is the common denominator and it is obtained by multipling the two denominators: 3x2=6.
Next we need to expand both fractions to have a denominator of 6. Expand both fractions by multiplying them and click done when you are ready. Enter your results in the appropriate boxes.
5 x = 2 x
2 x = 3 x
The expanded fractions are and
Now subtract the expanded fractions
=
Can be simplified?
Because the top number is larger than the bottom number, the fraction can be simplified to a whole number with a fraction remainder.
can be simplified to
### Practice
Subtract and simplify | 355 | 1,626 | {"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} | 4.78125 | 5 | CC-MAIN-2018-30 | latest | en | 0.909645 |
http://www.veritasprep.com/blog/2013/08/quarter-wit-quarter-wisdom-using-the-number-line/ | 1,529,513,952,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863830.1/warc/CC-MAIN-20180620163310-20180620183310-00269.warc.gz | 530,831,729 | 12,353 | # Using the Number Line
By now, you know that we like to discuss visual approaches to problems. A visual tool that we have used before for solving inequality and modulus questions is the number line. The number line is also useful in helping us solve many number properties questions.
A few things to keep in mind when dealing with number line:
1. x < y (in other words, x is less than y) implies x is to the left of y on the number line. x and y could be in any region i.e. negative or positive but x must be to the left of y in any case.
2. ‘x – y > 0’ (in other words, x – y is positive) implies x is to the right of y on the number line. Again, x and y could be in any region of the number line but x will be to the right of y i.e. x will be greater than y in any case.
The importance of these points is not apparent without a couple of questions.
Question 1: If a, b, and c are positive integers, is b between a and c?
Statement 1: b is 3 greater than a, and b is 5 less than c.
Statement 2: c is 5 greater than b, and c is 8 greater than a.
Solution: You might be tempted to use algebra with equations such as b = a + 3, b = c – 5 etc. But the question ‘is b between a and c’ should remind you of the number line. If we can figure out the relative position of ‘a’, ‘b’ and ‘c’ on the number line, we can say whether ‘b’ is between ‘a’ and ‘c’. Many of these ‘is this number less than that number’ questions can be easily done using the number line.
The question ‘Is b between a and c?’ essentially means ‘does b lay between a and c on the number line?’
Statement 1: b is 3 greater than a, and b is 5 less than c.
This means ‘b’ is 3 steps to the right of ‘a’ but 5 steps to the left of ‘c’ on the number line. It must lay between ‘a’ and ‘c’.
This statement alone is sufficient to answer the question.
Statement 2: c is 5 greater than b, and c is 8 greater than a.
‘c’ is 5 steps to the right of ‘b’ which means ‘b’ is 5 steps to the left of ‘c’. ‘c’ is 8 steps to the right of ‘a’ which means ‘a’ is 8 steps to the left of ‘c’. ‘a’ is further to the left of ‘c’ than ‘b’. So ‘b’ must be between ‘a’ and ‘c’.
This statement alone is sufficient to answer the question too.
Hence the answer is (D).
Working with equations would have been far too cumbersome. Don’t take my word for it; try it on your own.
Let’s look at another question based on the same concepts.
Question 2: The points A, B, C and D are on a number line, not necessarily in this order. If the distance between A and B is 18 and the distance between C and D is 8, what is the distance between B and D?
Statement 1: The distance between C and A is the same as the distance between C and B.
Statement 2: A is to the left of D on the number line.
Solution: This question specifically mentions number line.
We are given that distance between A and B is 18. We don’t know how to place A and B on the number line yet:
We don’t know in which region they lay. We can make a similar diagram for C and D. Note that we don’t know how to place these points. All we know is the relative distance between them. We also don’t know which one lays to the left and which one lays to the right.
Statement 1: The distance between C and A is the same as the distance between C and B.
Since distance between C and A is the same as distance between C and B, C must lay in the center of A and B. There are still many different ways of placing B and D so the distance between B and D is not known yet.
This statement alone is not sufficient.
Statement 2: A is to the left of D on the number line.
If the only constraint is that A is to the left of D, there are many different ways of placing A relative to D.
The distance between B and D will be different in different cases. This statement alone is not sufficient.
Let’s consider both the statements together. C is in the middle of A and B and A is to the left of D. There are still two different cases possible.
The distance between B and D will be different in the two cases. Hence, we still cannot say what the distance between the two points is. | 1,054 | 4,080 | {"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} | 4.90625 | 5 | CC-MAIN-2018-26 | latest | en | 0.951851 |
http://www.algebra.com/algebra/homework/Rectangles/Rectangles.faq.question.393831.html | 1,369,241,167,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702019913/warc/CC-MAIN-20130516110019-00023-ip-10-60-113-184.ec2.internal.warc.gz | 302,769,277 | 4,413 | # SOLUTION: what is the proportion with a length of 18 and 9 then a width of 8 and 4
Algebra -> Algebra -> Rectangles -> SOLUTION: what is the proportion with a length of 18 and 9 then a width of 8 and 4 Log On
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Geometry: Rectangles Solvers Lessons Answers archive Quiz In Depth
Click here to see ALL problems on Rectangles Question 393831: what is the proportion with a length of 18 and 9 then a width of 8 and 4Answer by SavannahStar(3) (Show Source): You can put this solution on YOUR website!With these given numbers, here is the proportion that you can set up: 18 8 _=_ 9 4 This is an equal proportion because 72=72. (18 times 4 is 72, 8 times 9 is 72.) | 230 | 854 | {"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-2013-20 | latest | en | 0.859979 |
https://icsecbsemath.com/2021/01/11/class-11-the-straight-line-exercise-23-5/ | 1,611,458,939,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703544403.51/warc/CC-MAIN-20210124013637-20210124043637-00255.warc.gz | 372,834,410 | 33,559 | Note: Equation of a like is also defined as $y - y_1 = m ( x-x_1)$
Question 1: Find the equation of the straight through the following pair of points:
i) $(0, 0)$ and $(2,-2)$ ii) $(a, b)$ and $(a + c \sin \alpha, b + c \cos \alpha)$
iii) $(0, -a)$ and $( b, 0)$ iv) $(a, b)$ and $(a + b , a -b)$
v) $( a t_1, a/t_1)$ and $( a t_2, a/t_2)$ vi) $(a \cos \alpha, a\sin \alpha)$ and $(a \cos \beta, a\sin \beta)$
i) Given points $A(0, 0)$ and $B(2,-2)$
Slope of $AB = m =$ $\frac{-2-0}{2-0}$ $=$ $\frac{-2}{2}$ $= 1$
Substituting in $y - y_1 = m ( x-x_1)$ we get the equation of $AB$ as:
$y - 0 = -1(x-0) \hspace{0.5cm} \Rightarrow y + x = 0$
ii) Given points $A(a, b)$ and $B(a + c \sin \alpha, b + c \cos \alpha)$
Slope of $AB = m =$ $\frac{b+c \cos \alpha - b}{a+c \sin \alpha - a}$ $=$ $\frac{c \cos \alpha}{c \sin \alpha}$ $= \cot \alpha$
Substituting in $y - y_1 = m ( x-x_1)$ we get the equation of $AB$ as:
$y - b = \cot \alpha (x-a) \hspace{0.5cm} \Rightarrow y - x \cot \alpha + a - b = 0$
iii) Given points $A(0, -a)$ and $B( b, 0)$
Slope of $AB = m =$ $\frac{0-(-a)}{b-0}$ $=$ $\frac{a}{b}$
Substituting in $y - y_1 = m ( x-x_1)$ we get the equation of $AB$ as:
$y - (-a) =$ $\frac{a}{b}$ $(x-0)$
$\Rightarrow by + ab = ax$
$\Rightarrow ax-by-ab=0$
iv) Given points $A(a, b)$ and $B(a + b , a -b)$
Slope of $AB = m =$ $\frac{a-b-b}{a+b-a}$ $=$ $\frac{a-2b}{b}$
Substituting in $y - y_1 = m ( x-x_1)$ we get the equation of $AB$ as:
$y - b =$ $\frac{a-2b}{b}$ $(x-a)$
$\Rightarrow yb-b^2=(a-2b)x-a(a-2b)$
$\Rightarrow yb-b^2 = (a-2b)x - a^2+2ab$
$\Rightarrow (a-2b)x - yb + b^2 + 2ab - a^2=0$
v) Given points $A( a t_1, a/t_1)$ and $B( a t_2, a/t_2)$
Slope of $AB = m =$ $\frac{\frac{a}{t_2} - \frac{a}{t_1}}{at_2 - at_1}$ $=$ $\frac{t_1-t_2}{t_1t_2(t_2-t_1)}$ $= -$ $\frac{1}{t_1t_2}$
Substituting in $y - y_1 = m ( x-x_1)$ we get the equation of $AB$ as:
$y -$ $\frac{a}{t_1}$ $=$ $\frac{-1}{t_1t_2}$ $(x-at_1)$
$\Rightarrow t_1t_2y- at_2 = -x + at_1$
$\Rightarrow x + t_1t_2y-a(t_1+t_2)=0$
vi) Given points $A(a \cos \alpha, a\sin \alpha)$ and $B(a \cos \beta, a\sin \beta)$
Slope of $AB = m =$ $\frac{a \sin \beta - a \sin \alpha}{a \cos \beta - a \cos \alpha}$ $=$ $\frac{\sin \beta - \sin \alpha}{ \cos \beta - \cos \alpha}$
Substituting in $y - y_1 = m ( x-x_1)$ we get the equation of $AB$ as:
$y - a \sin \alpha =$ $\frac{\sin \beta - \sin \alpha}{ \cos \beta - \cos \alpha}$ $(x-a \cos \alpha)$
$\Rightarrow y ( \cos \beta - \cos \alpha) - a \sin \alpha ( \cos \beta - \cos \alpha) = x ( \sin \beta - \sin \alpha) - a \cos \alpha ( \sin \beta - \sin \alpha)$
$\Rightarrow y ( \cos \beta - \cos \alpha) - x ( \sin \beta - \sin \alpha) = a \sin \alpha ( \cos \beta - \cos \alpha) - a \cos \alpha ( \sin \beta - \sin \alpha)$
$\Rightarrow y . 2 \sin \Big(\frac{\beta + \alpha}{2} \Big) \sin \Big(\frac{ \alpha - \beta}{2} \Big) - x . 2 \sin \Big(\frac{\beta - \alpha}{2} \Big) \cos \Big(\frac{ \beta + \alpha }{2} \Big) = a ( \sin \alpha \cos \beta - \cos \alpha \sin \beta)$
$\Rightarrow y . 2 \sin \Big(\frac{\beta + \alpha}{2} \Big) \sin \Big(\frac{ \alpha - \beta}{2} \Big) - x . 2 \sin \Big(\frac{\beta - \alpha}{2} \Big) \cos \Big(\frac{ \beta + \alpha }{2} \Big) = 2a \sin \Big(\frac{\alpha-\beta}{2} \Big) \cos \Big(\frac{ \alpha - \beta}{2} \Big)$
$\Rightarrow y \sin \Big(\frac{\beta + \alpha}{2} \Big)+ x \cos \Big(\frac{ \beta + \alpha }{2} \Big) = a \cos \Big(\frac{ \alpha - \beta}{2} \Big)$
$\\$
Question 2: Find the equation to the sides of the triangles the coordinates of whose angular points are respectively: (i) $(1, 4),(2, -3)$ and $(-1, -2)$ ii) $(0, 1), (2, 0)$ and $( -1, -2)$
i) Given points $A(1, 4), B(2, -3)$ and $C(-1, -2)$
Slope of $AB = m_1 =$ $\frac{-3-4}{2-1}$ $=$ $\frac{-7}{1}$ $= -7$
Therefore equation of $AB$:
$y - 4 = - 7 ( x-1) \hspace{0.5cm} \Rightarrow 7x + y - 11 = 0$
Slope of $BC = m_2 =$ $\frac{-2-(-3)}{-1-2}$ $=$ $\frac{1}{-3}$ $=$ $\frac{-1}{3}$
Therefore equation of $BC$:
$y - (-3) =$ $\frac{-1}{3}$ $( x-2) \hspace{0.5cm} \Rightarrow 3y + 9 = - x + 2 \hspace{0.5cm} \Rightarrow x + 3y + 7 = 0$
Slope of $CA = m_3 =$ $\frac{4-(-2)}{1-(-1)}$ $=$ $\frac{6}{2}$ $= 3$
Therefore equation of $CA$:
$y - 4 = 3 ( x-1) \hspace{0.5cm} \Rightarrow 3x-y+1 = 0$
ii) Given points $A(0,1), B(2,0)$ and $C(-1, -2)$
Slope of $AB = m_1 =$ $\frac{0-1}{2-0}$ $=$ $\frac{-1}{2}$
Therefore equation of $AB$:
$y - 1 = \frac{-1}{2} ( x-0) \hspace{0.5cm} \Rightarrow x+2y-2 = 0$
Slope of $BC = m_2 =$ $\frac{-2-0}{-1-2}$ $=$ $\frac{-2}{-3}$ $=$ $\frac{2}{3}$
Therefore equation of $BC$:
$y - 0 =$ $\frac{2}{3}$ $( x-2) \hspace{0.5cm} \Rightarrow 2x-3y-4=0$
Slope of $CA = m_3 =$ $\frac{1-(-2)}{0-(-1)}$ $=$ $\frac{3}{1}$ $= 3$
Therefore equation of $CA$:
$y - 1 = 3 ( x-0) \hspace{0.5cm} \Rightarrow 3x-y+1 = 0$
$\\$
Question 3: Find the equations of the medians of a triangle, the the coordinates of the vertices are $(-1, 6), (- 3, - 9)$ and $(5, -8)$.
Given points $A(-1,6), B(-3,-9)$ and $C(5,-8)$. Please refer to the figure shown.
Let $D, E \ \& \ F$ be the mid point of $AB, BC$ and $CA$ respectively. Therefore,
$D = \Big($ $\frac{-3-1}{2}$ $,$ $\frac{-9+6}{2}$ $\Big ) = \Big( -2,$ $\frac{-3}{2}$ $\Big )$
$E = \Big($ $\frac{5-3}{2}$ $,$ $\frac{-8-9}{2}$ $\Big ) = \Big( 1,$ $\frac{-17}{2}$ $\Big )$
$F = \Big($ $\frac{-1+5}{2}$ $,$ $\frac{6-8}{2}$ $\Big ) = ( 2, -1)$
Slope of $AE = m_1 =$ $\frac{\frac{-17}{2}-6}{1-(-1)}$ $=$ $\frac{-29}{4}$
Therefore equation of $AE$:
$y - 6 =$ $\frac{-29}{4}$ $( x-(-1)) \hspace{0.5cm} \Rightarrow 4y-24=-29x-29 \hspace{0.5cm} \Rightarrow 29x+4y+5 = 0$
Slope of $BF = m_2 =$ $\frac{-1-(-9)}{2-(-3)}$ $=$ $\frac{8}{5}$
Therefore equation of $BF$:
$y - (-9) =$ $\frac{8}{5}$ $( x-(-3)) \hspace{0.5cm} \Rightarrow 5y+45 = 8x+ 24 \hspace{0.5cm} \Rightarrow 8x-5y-21=0$
Slope of $CD = m_3 =$ $\frac{\frac{-3}{2} - (-8)}{-2-5}$ $=$ $\frac{-13}{14}$
Therefore equation of $CD$:
$y - (-8) =$ $\frac{-13}{14}$ $( x-5) \hspace{0.5cm} \Rightarrow 14y + 112 = - 13x +65 \hspace{0.5cm} \Rightarrow 13x + 14y + 47 = 0$
$\\$
Question 4: Find the equations to the diagonals of the rectangle the equations of whose sides are $x = a, x = a', y = b$ and $y = b'$.
Please refer to the figure shown. Therefore the four points are $A(a, b), B ( a', b), C( a', b'), D( a , b')$
Slope of $AC = m_3 =$ $\frac{b'-b}{a'-a}$
Therefore equation of $AC$:
$y - b=$ $\frac{b'-b}{a'-a}$ $( x-a)$
$\Rightarrow (a'-a)y-a'b+ab=(b'-b)x-ab'+ab$
$\Rightarrow (b'-b)x-(a'-a)y - ab'+a'b = 0$
$\Rightarrow (a'-a)y - ( b'-b)x = ba'-ab'$
Slope of $BD = m_3 =$ $\frac{b'-b}{a-a'}$
Therefore equation of $BD$:
$y - b=$ $\frac{b'-b}{a-a'}$ $( x-a')$
$\Rightarrow (a-a')y - ab + ba' = ( b'-b) x - a'b' + a'b$
$\Rightarrow (b-b')x - ( a - a') y + ab - a'b'=0$
$\Rightarrow ( a - a') y + (b-b')x = a'b' - ab$
$\\$
Question 5: Find the equation of the side $BC$ of the $\triangle ABC$ whose vertices are $A(- 1, - 2), B (0, 1)$ and $C (2,0)$ respectively. Also, find the equation of the median through $A ( -1, -2)$.
Given points $A(-1,-2), B(0,1)$ and $C(2,0)$
Slope of $BC = m_1 =$ $\frac{0-1}{2-0}$ $=$ $\frac{-1}{2}$
Therefore equation of $BC$:
$y - 1 =$ $\frac{-1}{2}$ $( x-0) \hspace{0.5cm} \Rightarrow 2y-2=-x \hspace{0.5cm} \Rightarrow x+ 2y - 2 = 0$
Mid point $D$ of $BC = \Big($ $\frac{0+2}{2}$ $,$ $\frac{1+0}{2}$ $\Big)= \Big( 1,$ $\frac{1}{2}$ $\Big)$
Slope of $AD = m_2 =$ $\frac{\frac{1}{2} -(-2)}{1-(-1)}$ $=$ $\frac{5}{4}$
Therefore equation of $AD$:
$y - (-2) =$ $\frac{5}{4}$ $( x-(-1))$
$\Rightarrow y+2 =$ $\frac{5}{4}$ $( x+1) \Rightarrow 4y+ 8 = 5x + 5 \Rightarrow 5x - 4y - 3$
$\\$
Question 6: Using the concept of the equation of a line, prove that the three points $(-2, -2) (8, 2)$ and $( 3, 0)$ are collinear.
Given points $A(-2,-2), B(8,2)$ and $C(3,0)$
Slope of $AC = m_1 =$ $\frac{0-(-2)}{3-(-2)}$ $=$ $\frac{2}{5}$
Therefore equation of $AC$
$y - (-2) =$ $\frac{2}{5}$ $( x-(-2))$
$\Rightarrow y+2 = \frac{2}{5} (x+2) \hspace{0.5cm} \Rightarrow 5y+10 = 2x + 4 \hspace{0.5cm} \Rightarrow 2x - 5y - 6 = 0$
Now check if $B( 8, 2)$ satisfy the equation
$\therefore 2 ( 8) - 5(2) - 6 = 0$
$\Rightarrow 16-10-6 = 0$
Therefore $A, B, C$ are collinear.
$\\$
Question 7: Prove that the line $y -x+2=0$, divides the join of points $(3,-1)$ and $(8,9)$ in the ratio $2:3$.
Let $y -x+2=0$, divides the join of points $(3,-1)$ and $(8,9)$ at a point P in the ratio of $k:1$
$\therefore P = \Big($ $\frac{3+8k}{k+1}$ $,$ $\frac{-1+9k}{k+1}$ $\Big)$
Since $P$ lies on $y - x+2 = 0$, it should satisfy the equation.
$\therefore$ $\frac{-1+9k}{k+1}$ $-$ $\frac{3+8k}{k+1}$ $+ 2 = 0$
$\Rightarrow -1 + 9k - 3 - 8 k + 2 k + 2 = 0$
$\Rightarrow 3k - 2 = 0$
$\Rightarrow k =$ $\frac{2}{3}$
Hence the line $y -x+2=0$, divides the join of points $(3,-1)$ and $(8,9)$ in the ratio $2:3$.
$\\$
Question 8: Find the equation to the straight line which bisects the distance between the points $(a, b), (a', b')$ and also bisects the distance between the points $( -a, b)$ and $(a', -b')$.
Given points $A(a, b), B(a', b')$ and $C( -a, b)$ and $D(a', -b')$
Mid point $P$ of $AB = \Big($ $\frac{a+a'}{2}$ $,$ $\frac{b+b'}{2}$ $\Big)$
Mid point $Q$ of $CD = \Big($ $\frac{-a+a'}{2}$ $,$ $\frac{b-b'}{2}$ $\Big)$
Slope of $PQ = m_1 =$ $\frac{\frac{b-b'}{2} - \frac{b+b'}{2} }{\frac{-a+a'}{2} - \frac{a+a'}{2}}$ $=$ $\frac{b-b'-b-b'}{-a+a'-a-a'}$ $=$ $\frac{-2b'}{-2a}$ $=$ $\frac{b'}{a}$
Therefore equation of $PQ$:
$y -$ $\frac{b+b'}{2}$ $=$ $\frac{b'}{a}$ $\Big( x-$ $\frac{a+a'}{2}$ $\Big)$
$\Rightarrow 2y - b - b' =$ $\frac{b'}{a}$ $( 2x - a - a')$
$\Rightarrow 2ay - ab - ab' = 2b'x-ab'-a'b'$
$\Rightarrow 2b'x-2ay+ab-a'b'=0$
$\Rightarrow 2ay - 2b'x = ab - a'b'$
$\\$
Question 9: In what ratio is the line joining the points $(2, 3)$ and $(4, -5)$ divided by the line passing through the points $(6,8)$ and $(-3, -2)$.
Given points: $(6,8)$ and $(-3, -2)$
Slope of $AB =$ $\frac{-2-8}{-3-6}$ $=$ $\frac{-10}{-9}$ $=$ $\frac{10}{9}$
Therefore equation of $AB$:
$y - 8 =$ $\frac{10}{9}$ $( x - 6) \hspace{0.5cm} \Rightarrow 9y - 72 = 10 x - 60 \hspace{0.5cm} \Rightarrow 10x - 9 y + 12 = 0$
Let $10x - 9 y + 12 = 0$ divides the line joining $(2, 3)$ and $(4, -5)$ in the ratio of $k:1$
$\therefore$ point of intersection $= \Big($ $\frac{4k+2}{k+1}$ $,$ $\frac{-5k+3}{k+1}$ $\Big)$
This point is on line $10x - 9 y + 12 = 0$
$\therefore 10 \Big($ $\frac{4k+2}{k+1}$ $\Big ) - 9 \Big($ $\frac{-5k+3}{k+1}$ $\Big) + 12 = 0$
$\Rightarrow 40k+20+45k-27+12k+12=0$
$\Rightarrow 97k + 5 = 0$
$\Rightarrow k = -$ $\frac{5}{97}$
Hence the required ratio is $5:97$ externally.
$\\$
Question 10: The vertices of a quadrilateral are $A(-2,6),8(1,2),C(10, 4)$ and $D(7, 8)$. Find the equations of its diagonals.
Given points: $A(-2,6), B(1,2) , C(10,4), D(7,8)$
Slope of $AC =$ $\frac{4-6}{10-(-2)}$ $=$ $\frac{-2}{12}$ $=$ $\frac{-1}{6}$
Therefore equation of $AC$:
$y - 6 =$ $\frac{-1}{6}$ $( x - (-2)) \hspace{0.5cm} \Rightarrow 6y-36=-x-2 \hspace{0.5cm} \Rightarrow x+6y-34=0$
Slope of $BD =$ $\frac{8-2}{7-1}$ $=$ $\frac{6}{6}$ $= 1$
Therefore equation of $BD$:
$y - 2 =$ $1 ( x - 1) \hspace{0.5cm} \Rightarrow x-y+1=0$
$\\$
Question 11: The length $L$ (in centimeters) of a copper rod is a linear function of its Celsius temperature $C$. In an experiment, if $L = 124.942$ when $C = 20$, and $L = 125.134$ when $C = 110$, express $L$ in terms of $C$
Given two points $( 124.942, 20)$ and $(125.134, 110)$
Slope of line $=$ $\frac{110-20}{125.134-124.942}$ $=$ $\frac{90}{0.192}$
Therefore the equation of line:
$y - 20 =$ $\frac{90}{0.192}$ $( x - 124.942)$
$\Rightarrow 0.192 y - 3.84 = 90x - 11244.78$
$\Rightarrow 90x = 0.192y + 11240.78$
$\Rightarrow x =$ $\frac{0.192}{90}$ $y +$ $\frac{11240.78}{90}$
$\Rightarrow x =$ $\frac{4}{1875}$ $y + 124.898$
$\\$
Question 12: The owner of a milk store finds that he can sell $980$ liters milk each week at ($14$ per liter and $1220$ liters of milk each week at Rs. $16$ per liter. Assuming a linear relationship between selling price and demand, how many liters can he sell weekly at Rs. $17$ per liter.
Given two points $( 14, 980)$ and $(16, 1220)$
Slope of line $=$ $\frac{1220-980}{16-14}$ $=$ $\frac{240}{2}$ $= 120$
Therefore the equation of line:
$y - 980 = 120(x-14)$ $\hspace{0.5cm} \Rightarrow y = 120x - 700$
When $x = 17, y = 120 \times 17 - 700 = 1340$
Hence the owner of the milk store would be able to sell $1340$ liters of milk at Rs. $17$ / liter.
$\\$
Question 13: Find the equation of the bisector of $\angle A$ of the triangle whose vertices are $A (4,3), B(0,0)$ and $C( 2, 3)$.
Let $AD$ be the bisector of $\angle BAC$.
We know, $\frac{AB}{AC}$ $=$ $\frac{BD}{DC}$
Now $AB = \sqrt{(4-0)^2 + ( 3-0)^2} = \sqrt{25} = 5$
$AC = \sqrt{(4-2)^2 + ( 3-3)^2} = \sqrt{4} = 2$
$\therefore$ $\frac{BD}{DC}$ $=$ $\frac{5}{2}$
Therefore coordinates of $D = \Big($ $\frac{2 \times 5 + 0 \times 2}{5+2}$ $,$ $\frac{3 \times 5 + 0 \times 2}{5+2}$ $\Big) = \Big($ $\frac{10}{7}$ $,$ $\frac{15}{7}$ $\Big)$
Slope of $AD =$ $\frac{\frac{15}{7}-3}{\frac{10}{7}-4}$ $=$ $\frac{15-21}{10-28}$ $=$ $\frac{-6}{-18}$ $=$ $\frac{1}{3}$
Therefore equation of $AD$:
$y - 3 =$ $\frac{1}{3}$ $( x - 4) \hspace{0.5cm} \Rightarrow 3y-9=x-4 \hspace{0.5cm} \Rightarrow x-3y+5=0$
$\\$
Question 14: Find the equations to the straight lines which go through the origin and trisect the portion of the straight line $3x+ y = 12$ which is intercepted between the axes of coordinates
Given line $3x+y =12$
x-intercept $= A ( 4, 0)$
y-intercept $= B ( 0, 12$
$C$ divides $BC$ in the ratio of $2:1$
Coordinates of $C = \Big($ $\frac{2 \times 4 + 1 \times 0}{2+1}$ $,$ $\frac{2 \times 0 + 1 \times 12}{2+1}$ $\Big) = \Big($ $\frac{8}{3}$ $, 4 \Big)$
$D$ divides $BC$ in the ratio of $1:2$
Coordinates of $C = \Big($ $\frac{2 \times 4 + 1 \times 0}{1+2}$ $,$ $\frac{2 \times 0 + 1 \times 12}{1+2}$ $\Big) = \Big($ $\frac{8}{3}$ $, 4 \Big)$
Slope of $OC =$ $\frac{4-0}{\frac{8}{3}-0}$ $=$ $\frac{3}{2}$
Therefore equation of $OC$:
$y - 0 =$ $\frac{3}{2}$ $( x - 0) \hspace{0.5cm} \Rightarrow 2y=3x$
Slope of $OD =$ $\frac{8-0}{\frac{4}{3}-0}$ $= 6$
Similarly, equation of $OD$:
$y - 0 = 6 ( x - 0) \hspace{0.5cm} \Rightarrow y=6x$
$\\$
Question 15: Find the equations of the diagonals of the square formed by the lines $x=0, y=0, x=1$ and $y = 1$.
Slope of $AC =$ $\frac{1-0}{1-0}$ $= 1$
Therefore equation of $AC$:
$y - 0 = 1 ( x - 0) \hspace{0.5cm} \Rightarrow y=x$
Slope of $BD =$ $\frac{1-0}{0-1}$ $= -1$
Similarly, equation of $OD$:
$y - 0 = -1 ( x - 1) \hspace{0.5cm} \Rightarrow x+y = 1$ | 6,393 | 14,747 | {"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": 537, "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} | 4.625 | 5 | CC-MAIN-2021-04 | longest | en | 0.287374 |
https://www.physicsforums.com/threads/simple-harmonic-oscillator-problem.663568/ | 1,532,321,016,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676594886.67/warc/CC-MAIN-20180723032237-20180723052237-00164.warc.gz | 965,064,959 | 14,756 | # Homework Help: Simple Harmonic Oscillator Problem
1. Jan 10, 2013
### Hibbs
1. The problem statement, all variables and given/known data
The position of a mass that is oscillating on a Slinky (which acts as a simple harmonic oscillator) is given by 18.5 cm cos[ 18.0 s-1t]. What is the speed of the mass when t = 0.360 s?
2. Relevant equations
x(t)=Acos(ωt+θ)
v(t)=-Aωsin(ωt+θ)
3. The attempt at a solution
I used the formula, v(t)=-Aωsin(ωt+θ) because you basically have everything you need such as:
A=18.5cm
ω=18.0s^-1
t=0.360s
What I get is:
v(0.360)=-(18.5cm)(18.0)sin(18.0*0.360)
to get -65.1cm/s
Please tell me where I went wrong!
2. Jan 11, 2013
### Staff: Mentor
Speed is never negative.
3. Jan 11, 2013
### rude man
The highlighted formula makes little sense to me. First, the argument of the cos must be dimensionless but here it looks like it's time. 18.0 s must be the phase term but what's with the s? Is 18.0 in radians, deg or ??? The "1" in front of t must be √(k/m), k = spring const. & m = mass, aka ω. The phase is due to the fact that this mass had initial velocity and displacement. Anyway, no way do I see that ω = 18.0.
Weird! I guess you could go
x = 18.5cos(18 - t) cm
x' = -18.5sin(18 - t) cm/s since I guess ω = 1 rad/s;
so x'(t=0.36) = -18.5sin(18 - 0.36) = -18.5sin(17.64) cm/s.
BTW v can be negative. Speed can't.
4. Jan 12, 2013
### ehild
v=-65.1 cm/s is the velocity. The speed is magnitude of velocity.
ehild
5. Jan 12, 2013
### Staff: Mentor
I believe the formula is just misformatted and is intended to be $\cos(18s^{-1}\times t)$ - so there is no problem with the units.
6. Jan 13, 2013
### Hibbs
Thanks a lot! I got it!
Last edited: Jan 13, 2013
7. Jan 13, 2013
### rude man
OK. What can be done to get the OPs to accurately state the problem, I wonder in my oft leisure moments ... | 635 | 1,850 | {"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} | 3.859375 | 4 | CC-MAIN-2018-30 | latest | en | 0.891136 |
http://math.stackexchange.com/questions/44113/whats-the-value-of-sum-limits-k-1-infty-frack2k/44131 | 1,438,728,278,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042992201.62/warc/CC-MAIN-20150728002312-00328-ip-10-236-191-2.ec2.internal.warc.gz | 151,579,026 | 21,008 | # What's the value of $\sum\limits_{k=1}^{\infty}\frac{k^2}{k!}$?
For some series, it is easy to say whether it is convergent or not by the "convergence test", e.g., ratio test. However, it is nontrivial to calculate the value of the sum when the series converges. The question is motivated from the simple exercise to determining whether the series $\sum\limits_{k=1}^{\infty}\frac{k^2}{k!}$ is convergent. One may immediately get that it is convergent by the ratio test. So here is my question:
What's the value of $$\sum_{k=1}^{\infty}\frac{k^2}{k!}?$$
-
The sum is equal to $2e$. First of all, the term $k^2/k!$ may be partly canceled as $k/(k-1)!$. Second, this can be written as $(k-1+1)/(k-1)!$. The term $k-1+1$ is divided to two terms. In the first term, the $k-1$ may be canceled again, giving us $e$. The second term leads to $e$ immediately. So the total sum is $2\exp(1)$.
In a similar way, one may easily calculate the sum even if $k^2$ is replaced by $k^n$, any positive integer power of $k$. The result is always a multiple of $e$.
-
The value of $T_n := \displaystyle\sum_{k=1}^{\infty} \frac{k^n}{k!}$ is $B_n \cdot e$, where $B_n$ is the $n^{th}$ Bell number.
To see this, note that
\begin{align} T_{n+1} = \sum_{k=1}^{\infty} \frac{k^{n+1}}{k!} &= \sum_{k=0}^{\infty} \frac{(k+1)^n}{k!} \\ &= \sum_{k=0}^{\infty} \frac{1}{k!} \sum_{j=0}^n {n \choose j} k^j \\ &= \sum_{j=0}^n {n \choose j} \sum_{k=1}^{\infty} \frac{k^j}{k!} \\ &= \sum_{j=0}^{n} {n \choose j} T_j \end{align}
This is precisely the recursion formula that the Bell numbers follow, except that every term here is being multiplied by $e$.
Edit: I was unaware at the time I posted my answer, but the argument I gave goes by the name of Dobiński's formula.
-
+1,I like this answer. – Eric Naslund Jun 8 '11 at 16:28
+1, nice for providing the background. – Jack Jun 8 '11 at 16:41
@Eric/Jack Thanks! – JavaMan Jun 8 '11 at 16:49
Wolfram Alpha says it is $2e$. Another derivation is to start with $e^x=\sum \limits_{n=0}^\infty \frac{x^n}{n!}$, apply $\frac{d}{dx}x\frac{d}{dx}$ to both sides, and evaluate at $x=1$.
-
$\frac{d}{dx}x\frac{d}{dx}$, typo? – Jack Jun 8 '11 at 17:46
It means: differentiate, then multiply by $x$, then differentiate. – GEdgar Jun 8 '11 at 17:54
I would be curious to know why the downvote. – Ross Millikan Jul 7 '11 at 12:33
Hint 1: $k^2=k(k-1)+k$. Hint 2: simplify the fractions $k(k-1)/k!$ and $k/k!$. Hint 3: write the series expansion around $x=0$ of the function $x\mapsto\exp(x)$.
- | 898 | 2,520 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2015-32 | longest | en | 0.825371 |
https://www.geeksforgeeks.org/ideal-gas-equation-and-absolute-temperature/ | 1,638,419,439,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.69/warc/CC-MAIN-20211202024322-20211202054322-00113.warc.gz | 827,069,766 | 25,185 | # Ideal Gas Equation and Absolute Temperature
• Last Updated : 16 Jun, 2021
A thermometer with a liquid-filled bulb at one end, the most commonly used liquid are Mercury, Toluene, Alcohol, Pentane, Creosote show different readings for temperatures other than the fixed reading because of their different expansion properties. A thermometer that uses a gas, on the other hand, shows the same reading for temperatures. It does not matter which type of gas is used. Experiments show that all gases expand in the same way at low densities.
Pressure (P), volume (V), and temperature (T) where T = t + 273.15 and t is the temperature in °C are the variables that explain the behavior of a given quantity (mass) of gas. The ideal gas law, also known as the universal gas equation, is a state-of-equation for a hypothetical ideal gas. Despite its flaws, the ideal gas law provides a good approximation of the behavior of many gases under a variety of situations. Benoit Paul Émile Clapeyron proposed the ideal gas law in 1834 as a mixture of the empirical Charles’ law, Boyle’s law, Avogadro’s law, and Gay-law. Lussac’s.
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### Ideal Gas Equation
Ideal gas laws are the combination of the observational work of Boyle in the seventeenth century and Charles in the eighteenth century.
Boyle’s law: The gas pressure is inversely proportional to the gas volume for a given amount of gas kept at a fixed temperature i.e. at constant temperature the relation between the pressure and volume of a quantity of gas can be written as,
P ∝ 1 / V
or
PV = Constant
where P is the pressure and V is the volume.
Charles’ law: The gas volume is directly proportional to the gas temperature for a given fixed amount of gas kept at a constant pressure i.e. at constant temperature the relation between the volume and temperature of a quantity of gas can be written as,
V ∝ T
or
V / T = Constant
where T is the Temperature.
These two laws apply to low-density gases and can be grouped into a single relationship. It’s worth noting that,
PV = Constant
and
V/T = Constant
For a particular quantity of gas, then
PV/T should thus be a constant as well.
It can be stated in a more general form that applies to any quantity of any low-density gas, not simply a specific quantity of that gas. This relationship describes the ideal gas law and knows as the ideal gas equation
It can be expressed as,
PV / T = nR
or
PV = nRT
where, n is the number of moles in the sample of gas and R is the universal gas constant.
Note: The universal gas constant (R) has a value of 8.314 kJ/mole in the SI system.
It can also be stated in a more general form that applies to any quantity of any low-density gas, not simply a specific quantity of that gas.
### Derivation of the Ideal Gas Equation
Let P is the pressure exerted by the gas, V is the volume f the gas, T is the Temperature.
According to Boyle’s Law,
P ∝ 1/V
or
V ∝ 1/P ……(1)
According to Charles’ Law,
V ∝ T ……..(2)
When P and T are both constant, the volume of a gas is proportional to the number of moles of gas.i.e.
V ∝ n …….(3)
Compare equation (1), (2) and (3) as,
V ∝ nT/P
or
PV = nRT
where R is the Universal gas constant and it is value of 8.314 J/mol-K
### Absolute Temperature
Thermodynamic temperature is another name for absolute temperature. The thermodynamic energy of a system is lowest at this temperature. Absolute temperature equals zero Kelvin or -273 °C, commonly known as absolute zero. The velocity of the gas particles stops at absolute zero temperature. This signifies that the particles of the gas really aren’t moving. At absolute zero, the volume of the gas is zero. As a result, the volume of a gas is measured by its absolute zero.
The temperature has a direct relationship with pressure and volume i.e.
PV ∝ T
This relationship enables a gas that will be utilized to determine the temperature in a Gas thermometer with a constant volume.
Therefore, at constant volume, the relationship can be written as,
P ∝ T, and Temperature is read in terms of pressure with a constant-volume gas thermometer.
A straight line emerges from a plot of pressure against temperature.
At constant volume, a plot of Pressure versus temperature of a low-density gas
Observations on real gases differ from the values anticipated by the ideal gas law at low temperatures. However, the relationship is linear over a wide temperature range, and it appears that if the gas remained a gas, the pressure would drop to zero with decreasing temperature. Extrapolating the straight line to the axis yields the absolute minimum temperature for an ideal gas. Absolute zero is defined as a temperature of – 273.15 degrees Celsius. The Kelvin temperature scale, often known as absolute scale temperature, is founded on absolute zero.
The same absolute zero temperature in a plot of pressure versus temperature and extrapolation of lines for low-density gases.
On the Kelvin temperature scale, – 273.15 °C is taken as the zero points, that is 0 K. In both the Kelvin and Celsius temperature systems, the unit size is the same. So, the relation between them can be expressed as
T = t + 273.15
where t is the temperature in °C
### Sample Problems
Problem 1: What is the volume occupied by 2.34 grams of carbon dioxide gas at STP?
Solution:
Given,
Weight (m) of the carbon dioxide is 2.34 grams.
At STP, Temperature is 273.0 K.
Pressure is 1.00 atm.
The universal gas constant (R) has a value of 0.08206 L atm mol¯11.
The expression for the number of mole is,
n = m/M
where, n is the number of moles, m is the weight and M is the molar mass of the substance.
Molar mass of the carbon dioxide is 44.0 g mol¯1.
So, the value of n can be calculated as,
n = 2.34 g / 44.0 g mol¯1
= 0.0532 mol
According to the ideal gas equation,
PV = nRT
Rearranging the equation,
V = nRT / P
Substituting all the values,
V = [0.0532 mol) (0.08206 L atm mol¯1 K¯1) (273.0 K)] / 1.00 atm
= 1.19 L
Problem 2: A sample of argon gas at STP occupies 56.2 liters. Determine the number of moles of argon and the mass of argon in the sample.
Solution:
Given,
Volume (V) of the argon gas is 56.2 liters.
At STP, Temperature is 273.0 K.
Pressure is 1.00 atm.
Molar mass of the argon gas is 39.948 g/mol.
According to the ideal gas equation,
PV = nRT
Rearranging the equation,
n = PV / RT
Subtitituting all the values in the above equation,
n = [(1.00 atm) (56.2 L) ] / [ (0.08206 L atm mol¯1 K¯1) (273.0 K)]
= 2.50866 mol
The expression for the number of mole is
n = m/M
Rearranging the equation,
m = nM
Subtitituting all the values in the above equation,
m = (2.50866 mol)×(39.948 g/mol)
= 100 g
Problem 3:At what temperature will 0.654 moles of neon gas occupy 12.30 liters at 1.95 atmospheres?
Solution:
Given,
The Volume (V) of the neon gas is 12.30 liters.
The Pressure is 1.95 atm.
The Number of moles is 0.654 moles.
According to the ideal gas equation,
PV = nRT
Rearranging the equation,
T = PV / nR
Subtitituting all the values in the above equation,
T = [(1.95 atm) ×(12.30 L)] / [(0.654 mol)×(0.08206 L atm mol¯1 1)]
= 447 K
Problem 4: Derive the Ideal Gas Equation?
Solution:
Let P is the pressure exerted by the gas, V is the volume f the gas, T is the Temperature.
According to Boyle’s Law,
P ∝ 1/V
or
V ∝ 1/P ……(1)
According to Charles’ Law,
V ∝ T ……..(2)
When P and T are both constant, the volume of a gas is proportional to the number of moles of gas.i.e.
V ∝ n …….(3)
Compare equation (1), (2) and (3) as,
V ∝ nT/P
or
PV = nRT
where R is the Universal gas constant and it is value of 8.314 J/mol-K
Problem 5: 5.600 g of solid CO2 is put in an empty sealed 4.00 L container at a temperature of 300 K. When all the solid CO2 becomes gas, what will be the pressure in the container?
Solution:
Given,
The Weight (m) of the carbon dioxide is 5.600 g.
The Volume (V) of the carbon dioxide is 4.00 L.
The Temperature is 300 K.
Molar mass of the carbon dioxide is 44.0 g mol¯1
The expression for the number of mole is
n = m/M
Subtitituting all the values in the above equation,
n = (5.600 g) / (44.009 g/mol)
= 0.1272467 mol
According to the ideal gas equation,
PV = nRT
Rearranging the equation,
P = nRT/V
Subtitituting all the values in the above equation,
P = (0.1272467 mol)× (0.08206 L atm mol¯1 1)× (300 K)/ (4.00 L)
= 0.7831 atm
My Personal Notes arrow_drop_up | 2,338 | 9,941 | {"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-2021-49 | latest | en | 0.914946 |
https://www.wikiwand.com/en/Variable-order_Markov_model | 1,571,892,699,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987841291.79/warc/CC-MAIN-20191024040131-20191024063631-00031.warc.gz | 1,101,504,882 | 26,204 | Variable-order Markov model - Wikiwand
# Variable-order Markov model
In the mathematical theory of stochastic processes, variable-order Markov (VOM) models are an important class of models that extend the well known Markov chain models. In contrast to the Markov chain models, where each random variable in a sequence with a Markov property depends on a fixed number of random variables, in VOM models this number of conditioning random variables may vary based on the specific observed realization.
This realization sequence is often called the context; therefore the VOM models are also called context trees.[1] The flexibility in the number of conditioning random variables turns out to be of real advantage for many applications, such as statistical analysis, classification and prediction.[2][3][4]
## Example
Consider for example a sequence of random variables, each of which takes a value from the ternary alphabet {abc}. Specifically, consider the string aaabcaaabcaaabcaaabc...aaabc constructed from infinite concatenations of the sub-string aaabc.
The VOM model of maximal order 2 can approximate the above string using only the following five conditional probability components: {Pr(a | aa) = 0.5, Pr(b | aa) = 0.5, Pr(c | b) = 1.0, Pr(a | c)= 1.0, Pr(a | ca) = 1.0}.
In this example, Pr(c|ab) = Pr(c|b) = 1.0; therefore, the shorter context b is sufficient to determine the next character. Similarly, the VOM model of maximal order 3 can generate the string exactly using only five conditional probability components, which are all equal to 1.0.
To construct the Markov chain of order 1 for the next character in that string, one must estimate the following 9 conditional probability components: {Pr(a | a), Pr(a | b), Pr(a | c), Pr(b | a), Pr(b | b), Pr(b | c), Pr(c | a), Pr(c | b), Pr(c | c)}. To construct the Markov chain of order 2 for the next character, one must estimate 27 conditional probability components: {Pr(a | aa), Pr(a | ab), ..., Pr(c | cc)}. And to construct the Markov chain of order three for the next character one must estimate the following 81 conditional probability components: {Pr(a | aaa), Pr(a | aab), ..., Pr(c | ccc)}.
In practical settings there is seldom sufficient data to accurately estimate the exponentially increasing number of conditional probability components as the order of the Markov chain increases.
The variable-order Markov model assumes that in realistic settings, there are certain realizations of states (represented by contexts) in which some past states are independent from the future states; accordingly, "a great reduction in the number of model parameters can be achieved."[1]
## Definition
Let A be a state space (finite alphabet) of size ${\displaystyle |A|}$.
Consider a sequence with the Markov property ${\displaystyle x_{1}^{n}=x_{1}x_{2}\dots x_{n))$ of n realizations of random variables, where ${\displaystyle x_{i}\in A}$ is the state (symbol) at position i ${\displaystyle \scriptstyle (1\leq i\leq n)}$, and the concatenation of states ${\displaystyle x_{i))$ and ${\displaystyle x_{i+1))$ is denoted by ${\displaystyle x_{i}x_{i+1))$.
Given a training set of observed states, ${\displaystyle x_{1}^{n))$, the construction algorithm of the VOM models[2][3][4] learns a model P that provides a probability assignment for each state in the sequence given its past (previously observed symbols) or future states.
Specifically, the learner generates a conditional probability distribution ${\displaystyle P(x_{i}\mid s)}$ for a symbol ${\displaystyle x_{i}\in A}$ given a context ${\displaystyle s\in A^{*))$, where the * sign represents a sequence of states of any length, including the empty context.
VOM models attempt to estimate conditional distributions of the form ${\displaystyle P(x_{i}\mid s)}$ where the context length ${\displaystyle |s|\leq D}$ varies depending on the available statistics. In contrast, conventional Markov models attempt to estimate these conditional distributions by assuming a fixed contexts' length ${\displaystyle |s|=D}$ and, hence, can be considered as special cases of the VOM models.
Effectively, for a given training sequence, the VOM models are found to obtain better model parameterization than the fixed-order Markov models that leads to a better variance-bias tradeoff of the learned models.[2][3][4]
## Application areas
Various efficient algorithms have been devised for estimating the parameters of the VOM model.[3]
VOM models have been successfully applied to areas such as machine learning, information theory and bioinformatics, including specific applications such as coding and data compression,[1] document compression,[3] classification and identification of DNA and protein sequences,[5] [1][2] statistical process control,[4] spam filtering,[6] haplotyping[7] and others.
## References
1. ^ a b c Rissanen, J. (Sep 1983). "A Universal Data Compression System". IEEE Transactions on Information Theory. 29 (5): 656–664. doi:10.1109/TIT.1983.1056741.
2. ^ a b c d Shmilovici, A.; Ben-Gal, I. (2007). "Using a VOM Model for Reconstructing Potential Coding Regions in EST Sequences". Computational Statistics. 22 (1): 49–69. doi:10.1007/s00180-007-0021-8.
3. Begleiter, R.; El-Yaniv, R.; Yona, G. (2004). "On Prediction Using Variable Order Markov models" (PDF). Journal of Artificial Intelligence Research. 22: 385–421. doi:10.1613/jair.1491.
4. ^ a b c d Ben-Gal, I.; Morag, G.; Shmilovici, A. (2003). "CSPC: A Monitoring Procedure for State Dependent Processes" (PDF). Technometrics. 45 (4): 293–311. doi:10.1198/004017003000000122.
5. ^ Grau J.; Ben-Gal I.; Posch S.; Grosse I. (2006). "VOMBAT: Prediction of Transcription Factor Binding Sites using Variable Order Bayesian Trees" (PDF). Nucleic Acids Research, vol. 34, issue W529–W533. Cite journal requires |journal= (help)
6. ^ Bratko, A.; Cormack, G. V.; Filipic, B.; Lynam, T.; Zupan, B. (2006). "Spam Filtering Using Statistical Data Compression Models" (PDF). Journal of Machine Learning Research. 7: 2673–2698.
7. ^ Browning, Sharon R. "Multilocus association mapping using variable-length Markov chains." The American Journal of Human Genetics 78.6 (2006): 903–913.
[1]
1. ^ Smith, A. R.; Denenberg, J. N.; Slack, T. B.; Tan, C. C.; Wohlford, R. E (August 1985). "Application of a Sequential Pattern Learning System to Connected Speech Recognition" (PDF). Proceedings of the IEEE 1985 International Conference on Acoustics, Speech, and Signal Processing: 1201–1204. | 1,656 | 6,483 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 14, "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} | 3.640625 | 4 | CC-MAIN-2019-43 | longest | en | 0.884998 |
http://www.wyzant.com/resources/answers/16605/find_the_largest_value_of_x | 1,405,356,750,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776441023.78/warc/CC-MAIN-20140707234041-00023-ip-10-180-212-248.ec2.internal.warc.gz | 600,589,499 | 9,864 | Search 75,485 tutors
0 0
## find the largest value of x
find the largest value of x that satisfies:
log2(x2)-log2(x+5)=2
loga b − loga c = loga (b / c)
loga x = b -----> ab = x
~~~~~~~~~~~
log2 x2 − log2 (x + 5) = 2
log2 [x2 / (x + 5)] = 2
x2
-------- = 22
x + 5
x2 = 4(x + 5)
x2 - 4x - 20 = 0
D = (-4)2 +80 = 96
x12 = (4 ± √96) / 2
x1 = 2 + 2√6 > 0
x2 = 2 − 2√6 < 0 (logarithm is not exist for negative numbers and 0)
x = 2(1 + √6) is the answer. | 231 | 469 | {"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} | 3.515625 | 4 | CC-MAIN-2014-23 | latest | en | 0.631223 |
http://www.engineeringexpert.net/Engineering-Expert-Witness-Blog/tag/traction-motor | 1,524,384,716,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945552.45/warc/CC-MAIN-20180422080558-20180422100558-00428.warc.gz | 412,239,486 | 16,129 | ## Posts Tagged ‘traction motor’
### Overcoming Inertia
Monday, February 3rd, 2014
Inertia. It’s the force that keeps us in bed after the alarm has rung. It seems to have a life of its own, and today we’ll see how it comes into play in keeping other stationary objects at rest. Last time we identified a specific point of contact between spur gear teeth in a gear train and introduced the opposing forces, F1 and F 2, generated there. Today we’ll see what these forces represent, identifying one of them as inertia. So where do these forces come from? They’re forces generated by different means that converge at the same point of contact, the point at which gear teeth mesh. They follow a very specific geometric path to meet there, an imaginary straight line referred to as the line of action. F1 is always generated by a source of mechanical energy. In our locomotive example introduced earlier in this blog series that source is an electric traction motor, upon which a driving gear is mounted. When the motor is energized, a driving force F1 is generated, which causes gear teeth on the driving gear to push against gear teeth of the driven gear. Force F2 is not as straightforward to understand, because it’s not generated by a motor. Instead, it’s the resisting force that the weight of a stationary object poses against its being moved from an at-rest position, known as inertia. The heavier the object, the more inertia it presents with. Trains, of course, are extremely heavy, and to get them to move a great deal of inertia must be overcome. Inertia is also a factor in attempting to stop objects already in motion. To get a stationary locomotive to move, mechanical energy must be transmitted from the driving gear that’s attached to its traction motor, then on to the driven gear attached to its axle. At their point of contact, the driving force of the motor, F1, is met by the resisting force of inertia, F2. In order for the train to move, F1 must be greater than F2. If F1 is less than or equal to F2, then the train won’t leave the station. Next week we’ll animate our static image and watch the interplay between gear teeth, taking note of the line of action during their movement. _______________________________________
### Spur Gear Tooth Geometry and the Involute Curve
Sunday, January 19th, 2014
Last time we learned how spur gears mesh together to form a gear train and we examined a train consisting of just two gears, one being the driving gear, the other the driven gear. Today we’ll take a look at the geometry behind the smooth functioning of modern spur gear teeth when we identify their peculiar shape to be that of an involute curve. The curved profile of spur gear teeth conforms to a type of mathematical curve found in geometry known as an involute. The involute profile of a spur gear tooth is shown in red below. The mathematical notion of the involute was first presented in 1673 by Dutch mathematician Christiaan Huygens, in his book, Horologium Oscillatorium. Huygens’ book presents his studies on clock pendulums and the applied mathematics he used in an effort to predict their often erratic motion on ships at sea. His book ultimately dealt with far more than this, resulting in a treatise on the mathematical properties of the involutes of curves. To see how an involute curve is formed, we’ll conduct a simple experiment. One end of string is attached with a tack to a circular object, like the yellow rod shown in the following illustration. The other end of string has a red ball attached to it. Forming An Involute Curve If we grab the ball and pull the string taught while wrapping the string around the rod, the ball’s path will form an involute curve due to the incremental shortening of the string that occurs as it wraps around the rod. Next time we’ll see how the involute profile of gear teeth contributes to efficient mechanical energy transmission in gear trains. _______________________________________
### Gear Trains
Monday, January 13th, 2014
Last time we covered the basic terminology of spur gears. Today we’ll see how they interact with one another to form a gear train, such as the one depicted below. Meshing Spur Gears Form A Gear Train A gear train is formed when the teeth of two or more gears mesh and work together for the purpose of powering a mechanical device. A gear train can consist of as little as two gears, but trains can be so large as to contain dozens of gears, depending on the complexity of the device they are powering. But no matter how many gears are employed, there are certain key features that are shared by every gear train assembly. First, one gear within the train must be attached to a shaft rotated by a source of mechanical energy, such as an engine or electric motor. This gear is called the driving gear. The second requirement of a gear train is that at least one gear other than the driving gear is mounted to the shaft of a rotating machine part. This gear is called the driven gear. Locomotive Gear Train Consisting Of Two Gears The illustration above shows an exploded view of a locomotive gear train assembly consisting of two gears. The driving gear is mounted to the shaft of an electric traction motor. The driven gear is mounted to the locomotive’s axle. When a motor is attached to the axle, the two gears mesh together. The traction motor converts electrical energy into mechanical energy, which is supplied to the driving gear via the spinning motor’s shaft. The teeth of the driving gear then transmit the motor’s mechanical energy to the teeth of the driven gear, which then turn the locomotive’s wheels. It’s just one of countless operations that can be performed with gear train assemblies. Next time we’ll examine the geometry behind modern spur gear tooth design. _______________________________________
### Regenerative Brakes
Sunday, June 13th, 2010
Have you ever been stranded on a subway car? The lights flickered wildly, then the odd humming sounds all came to an abrupt stop, and you sat there exchanging uncomfortable glances with your fellow passengers? “Hey, get this thing going! I’ve got an appointment to make!” someone shouts at the driver, sequestered in his cab. The trouble is, he, like the rest of the passengers, is equally helpless in this situation. You see, streetcars, subway cars, and light rail cars are all forms of electric railway cars, and when their source of electricity goes out, so do they. They’re similar to the diesel-electric locomotive we looked at last week because they use electric traction motors for propulsion, meaning to move forward. But their difference lies in the fact that electric rail cars don’t carry their own source of power and are entirely reliant on an external source, an electrical substation. See Figure 1 below. Figure 1 – Electric Railway Car Propulsion System This substation performs the task of taking the power provided by an electric utility power plant and converting it into a form of power that the electric rail car can use to operate its traction motors. The two are connected via a trolley wire and the two rails that the car runs on. The railway car has a spring-loaded arm called a pantograph on its roof that touches the trolley wire, allowing electrical current to flow into a speed control system housed under the car. This speed control system performs the task of varying the flow of electrical current to the traction motors, enabling the car to move, before it eventually exits the motor through its wheels, then back to the substation where it originated, thus completing an electrical circuit. Many newer electric railway cars couple a regenerative braking system with a mechanical one. Their operation is similar in nature to a dynamic braking system where the traction motors are turned into generators. The difference is that with regenerative braking systems the current from the traction motors is sent to the trolley wire through the pantograph as shown in Figure 2 below. Figure 2 – An Electric Railway Car Using Regenerative Brakes This diagram shows the railcar generating electricity, but it may not be so obvious how its motion is made to slow down, after all, we see no resistor grids like we did in last week’s illustration of a dynamic braking system. So how does it stop? The trick here is that there are other cars running on the rail line at the same time which are using electrical current to move forward. So what does this have to do with stopping it you ask? Let’s take a look at Figure 3 for clarification. Figure 3 – How Regenerative Brakes Help Save Power In this illustration we see that as Car A goes downhill and the operator applies the brakes, the regenerative brake will be caused to start pumping electrical current into the trolley wire. Now, if Car B is on the same rail line going up the other side of the hill, it will need power to climb that hill, and it will need to draw that power from the trolley wire by way of its pantograph. But instead of drawing all its electrical current from the substation, Car B will first draw off the current produced by nearby Car A, and only then will it draw the remainder of its power requirements from the substation. During this braking process the kinetic energy in Car A is converted into electrical energy by its traction motors. Then Car B uses its own traction motors to convert the electrical energy drawn from Car A into mechanical energy, enabling it to climb the hill. Car B has effectively robbed Car A of its energy, so Car A slows down. As we discovered last week during our discussion of dynamic brakes, regenerative brakes become ineffective below a certain minimum speed. This is the reason that electric rail cars need mechanical brakes to complete the job of stopping. We see that the regenerative braking process is actually quite green. It allows for electrical energy that would normally be wasted as heat energy escaping into the atmosphere to be converted into useful energy, taking a significant chunk out of the demand for new energy off of substations. It also helps the electric railway to save money when it comes time to paying the electric bills. You may have noticed a copyright symbol and my name included in this week’s illustrations. That’s because it has recently come to my attention that many of my readers are unaware that I am the CAD (Computer Aided Design) artist who creates the illustrations featured in most of my blogs. CAD is a useful tool in explaining difficult technical subject matter, which is precisely why I use it so often. Next week we’ll explore in more depth the benefits of using this tool. _____________________________________________
### Diesel Locomotive Brakes
Sunday, June 6th, 2010
In the past few weeks we’ve taken a look at both mechanical and dynamic brakes. Now it’s time to bring the two together for unparalleled stopping performance. Have you ever wandered along a railroad track, hopping from tie to tie, daring a train to come roaring along and wondering if you could jump to safety in time? Many have, and many have lost the bet. That’s because a train, once set into motion, is one of the hardest things on Earth to bring to a stop. In this discussion, let’s focus on the locomotive. A large, six-axle variety is shown in Figure 1. Figure 1 – A Six-Axle Diesel-Electric Locomotive These massive iron horses are known in the industry as diesel-electric locomotives, and here’s why. As Figure 2 shows, diesel-electric locomotives are powered by huge diesel engines. Their engine spins an electrical generator which effectively converts mechanical energy into electrical energy. That electrical energy is then sent from the generator through wires to electric traction motors which are in turn connected to the locomotive’s wheels by a series of gears. In the case of a six-axle locomotive, there are six traction motors all working together to make the locomotive move. So how do you get this beast to stop? Figure 2 – The Propulsion System In A Six-Axle Diesel-Electric Locomotive You probably noticed in Figure 2 that there are resistor grids and cooling fans. As long as you’re powering a locomotive’s traction motors to move a train, these grids and fans won’t come into play. It’s when you want to stop the train that they become important. That’s when the locomotive’s controls will act to disconnect the traction motor wires running from the electrical generator and reconnect them to the resistor grids as shown in Figure 3 below. Figure 3 – The Dynamic Braking System In A Six-Axle Diesel Electric Locomotive The traction motors now become generators in a dynamic braking system. These motors take on the properties of a generator, converting the moving train’s mechanical, or kinetic, energy into electrical. The electrical energy is then moved by wires to the resistor grids where it is converted to heat energy. This heat energy is removed by powerful cooling fans and released into the atmosphere. In the process the train is robbed of its kinetic energy, causing it to slow down. Now you may be thinking that dynamic brakes do all the work, and this is pretty much true, up to a point. Although dynamic brakes may be extremely effective in slowing a fast-moving train, they become increasingly ineffective as the train’s speed decreases. That’s because as speed decreases, the traction motors spin more slowly, and they convert less kinetic energy into electrical energy. In fact, below speeds of about 10 miles per hour dynamic brakes are essentially useless. It is at this point that the mechanical braking system comes into play to bring the train to a complete stop. Let’s see how this switch from dynamic to mechanical dominance takes place. A basic mechanical braking system for locomotive wheels is shown in Figure 4. This system, also known as a pneumatic braking system, is powered by compressed air that is produced by the locomotive’s air pump. A similar system is used in the train’s railcars, employing hoses to move the compressed air from the locomotive to each car. Figure 4 – Locomotive Pneumatic Braking System In the locomotive pneumatic braking system, pressurized air enters an air cylinder. Once inside, the air bears against a spring-loaded piston, as shown in Figure 4(a). The piston moves, causing brake rods to pivot and clamp the brake shoes to the locomotive’s wheel with great force, slowing the locomotive. When you want to get the locomotive moving again, you vent the air out of the cylinder as shown in Figure 4(b). This takes the pressure off the piston, releasing the force from the brake shoes. The spring in the cylinder is now free to move the shoes away from the wheel so they can turn freely. We have now returned to the situation present in Figure 2, and the locomotive starts moving again. Next week we’ll talk about regenerative braking, a variation on the dynamic braking concept used in railway vehicles like electric locomotives and subway trains. _____________________________________________ | 3,250 | 15,328 | {"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} | 3.5 | 4 | CC-MAIN-2018-17 | longest | en | 0.948002 |
http://quickbooks.intuit.ca/r/finance-accounting/using-coverage-ratios-understand-business/ | 1,508,304,808,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822747.22/warc/CC-MAIN-20171018051631-20171018071631-00783.warc.gz | 287,274,138 | 12,474 | Coverage ratios are accounting ratios that can help a small business analyze its ability to maintain operations, stay alive, and continue to grow. Learn more about three of the most important coverage ratios for a new small business.
### What are Coverage Ratios?
Coverage ratios help determine if a company has a healthy amount of debt or if it is overextended. Coverage ratios analyze a company’s ability to service its debt and other financial obligations. They help show how well a company can afford to make its interest payments. Relying too heavily on borrowing can destroy a business, and regularly analyzing coverage ratios may prevent this from happening.
### Times Interest Earned Ratio
The times interest earned ratio measures the amount of times income can cover the company’s interest payments on its debt. The more profit a company generates, the more easily it can cover its debt. This ratio paints this picture.
Calculate times interest earned by dividing earnings before interest and taxes by the interest expense.
For example, if a firm has an EBIT of \$500,000 and an interest expense of \$100,000, the times interest earned equals 5. This means the the company earns five times what it has to so it can meet its debt obligations. A higher times interest earned ratio is better. A value of lower than 1.5 typically indicates that the company is having difficulty paying back its creditors.
### Fixed Charge Coverage Ratio
The fixed charge coverage ratio measures a company’s ability to pay all of its financial charges and expenses using income before interest or taxes. This ratio is an extension to the times interest earned ratio. Fixed cost line items such as lease payments, insurance payments, and preferred dividend payments can be added into the calculation.
The formula for this ratio is:Fixed charge coverage ratio = (EBIT + fixed charges before taxes) / (fixed charges before taxes + interest expense)
Expanding on the example above, assume the company also has \$75,000 of fixed charges before taxes. The value of the ratio is:(\$500,000 + \$75,000) / (\$75,000 + \$100,000) = 3.29
This shows that the company is generating 3.29 times more earnings than required to cover its financial obligation. Lenders commonly use the fixed charge coverage ratio to analyze a company’s financial health. A higher ratio indicates that the company is stronger.
### Debt Service Coverage Ratio
The debt service coverage ratio compares a company’s operating income to its total debt service costs to measure its ability to service its current debts. The debt service coverage ratio takes into consideration all debt-related expenses, such as interest, principal, pension obligations, and sinking fund obligations. Because of this, it is a very important ratio for creditors as they look at a company’s ability to meet all of its debt obligations.
Calculate the debt service coverage ratio by dividing the operating income by the total debt service costs.
A higher ratio is always desirable, while a ratio lower than 1 means that the company is not generating enough profits to service debt and must resort to using savings.
## Track Financial Trends with Ratio Analysis
The success of your small business hinges on you staying on top… | 639 | 3,267 | {"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-2017-43 | latest | en | 0.959972 |
https://www.physicsforums.com/threads/current-in-a-conducting-loop.720404/ | 1,521,411,387,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257646176.6/warc/CC-MAIN-20180318204522-20180318224522-00161.warc.gz | 878,316,795 | 16,270 | # Current in a conducting loop
1. Nov 2, 2013
### RenD94
1. The problem statement, all variables and given/known data
A conducting loop with area 0.13m2 and resistance 6.0Ω lies in the x-y plane. A spatially uniform magnetic field points in the z direction. The field varies with time according to Bz=at2−b, where a = 2.8T/s2 and b = 8.0T .
(a) Find the (magnitude of the) loop current when t = 1.3s.
2. Relevant equations
V = I * R
EMF = Δ(BA)/Δt * n
3. The attempt at a solution
I calculated the EMF using the second formula listed, with n = 1 as there is only a single loop.
EMF = Δ(BA)/Δt
Binitial = 2.8(02) - 8 = -8
Bfinal = 2.8(1.32) - 8 = -3.268
=> ΔB = 4.732
=> Δ(BA) = 4.732 * 0.13 = 0.6152
=> EMF = 0.6152 / 1.3 = 0.4732 V
Then I used the first formula to find a value for I...
0.4372 = I * 6
0.0789 A = I
This is incorrect however, and the correct answer given is 0.16 - roughly twice my solution. Am I leaving out a multiplier somewhere? Any help is appreciated, this seemingly simple problem has frustrated me for too long.
2. Nov 2, 2013
### Staff: Mentor
The magnetic field is not changing linearly, so a linear approximation of the change is not going to be too accurate.
Since you have the expression for B(t), I'd suggest using the derivative to find the change in flux since (Faraday's Law):
$$EMF = \frac{d \Phi}{dt}~~~~~~\text{and}~~~~~~\Phi(t) = Area\;B(t)$$
3. Nov 2, 2013
### RenD94
Got it, thanks gneill! | 485 | 1,450 | {"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} | 3.703125 | 4 | CC-MAIN-2018-13 | latest | en | 0.907415 |
https://www.slideserve.com/duman/section-1-4 | 1,519,439,086,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815034.13/warc/CC-MAIN-20180224013638-20180224033638-00147.warc.gz | 939,208,349 | 14,595 | Section 1.4
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Section 1.4 - PowerPoint PPT Presentation
Section 1.4. Fair division. Fair division. Fair division is the concept of dividing something among 2 or more people in such a way that each person finds his/her share to be fair. Division doesn’t mean to be equal; they need to be fair in the eye of the beholder. The divider-chooser method.
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PowerPoint Slideshow about 'Section 1.4' - duman
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Section 1.4
Fair division
Fair division
• Fair division is the concept of dividing something among 2 or more people in such a way that each person finds his/her share to be fair.
• Division doesn’t mean to be equal; they need to be fair in the eye of the beholder.
The divider-chooser method
• It is a practical and simple approach to having two people share an object than can be cut into pieces.
Example
• Let's say that there are three people trying to divide a cake. The lone divider cuts the cake in equal parts.
• The other two people (the "choosers") get to determine which share is fair for them.
• If they choose different parts, the division is simple: each one gets the share he/she choses, while the divider ends up with the remaining part.
• Since for the divider every share is fair, everyone is satisfied.
• It is a method by which two parties can divide a group of items in a fair manner, provided one of the items can be divided.
Example
• Suppose Bob and Carol are getting a divorce and must divide up some of their assets. We assume that they distribute 100 points among the five items as follows:
The method of sealed bids
• Each person writes down the highest amount they would be willing to pay for each item secretly.
• The highest bidder for each item will win it, and the other people will be compensated with money, paid to them by the person who won the item. Bidders are not compensated directly. Instead, if there are n bidders, the winner will place
(n-1)/n * (the winning bid) into a kitty.
• For example, if there are 4 bidders, the winner places ¾ of his bid into the kitty.
• The non-winners will take then 1/nth of their individual bids from the kitty. “ (1/n)*non-winners bid amount”
• After each non-winner has taken the appropriate amount from the kitty, the remaining funds in the kitty are then divided equally among each bidder, including the one who won the item.
Example
• Clark Kent and Lois Lane are no longer going to live together, but they have decided who will get to keep their dog. Obviously, they wont be able to divide the dog, and they do not want to sell it, either. So they will use the secret sealed bid method.
• Clark bids \$400 for the dog
• Lois bid \$600 for the dog
• Since this division involved 2 people, each of them is entitled to ½ of the value of their bid.
• Lois is the higher bidder, will get the dog, and she will have to put ½ of her bid, 300, into the kitty.
• Clark takes ½ of his bid from the kitty, \$200, from the kitty.
• This leaves 100 in the kitty, which is split evenly between the two people. So each will take \$50 more from the kitty.
• That means, to be fair, Lois gets the dog and pays \$250 to Clark.
Example
• For students to try.
Example
• There is unseen film about Star Wars from Episode IV. No one has seen this before except for the directors and the actors. Three friends found it. Now they decided to use the secret ballot bid method.
• James bids \$6000
• Kirk bids \$ 4200
• Tiberius bids \$5100
• Find who gets what?
• James pays-\$3700
• Kirk gets \$1700
• Tiberius gets \$2000 | 992 | 4,075 | {"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-2018-09 | latest | en | 0.932214 |
https://www.jiskha.com/display.cgi?id=1491021033 | 1,527,279,909,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867217.1/warc/CC-MAIN-20180525200131-20180525220131-00147.warc.gz | 754,775,276 | 5,046 | # Math
posted by Tyreice
You plan on borrowing \$14000 at 16% APR for 5 years. Use Web Loan Calculator to find the monthly payment. Find the total interest paid on this loan for the first 20 months.
Set up a spread sheet to answer. Here is an example of what it should look like. All cells in lower portion are formulas. You can use my examples numbers to test your spread sheet before you answer the questions. This is what is known as an amortization of the loan.
What is the monthly payment found on a bank web site?
\$
per month. (rounded to nearest cent)
After 20 months of payments, how much total interest has been paid for this loan? Use a spreadsheet to answer, round to nearest cent.
\$
of interest in the first 20 payments.
1. Reiny
I used
http://www.calculatorweb.com/calculators/loancalc.shtml
and got a payment of \$3404.53 , but I learned nothing about the actuarial math involved.
Present value = 140000
n = 5(12) = 60
i = .16/12 = .013333...
(ridiculously high rate for current rates, time to get a new text)
Payment = p
p(1 - 1.01333..^(-60) )/.01333... = 140000
p(41.12170722..) = 140000
p = 3404.53
value of debt after 20 periods if no payment had been made:
140000(1.01333..)^20 = \$182,462.92
amount of 20 payment at that time
= 3404.53(1.01333..^20 - 1)/.01333.. = \$77,446.22
outstanding balance after 20 periods
= 182462.92 - 77446.22 = \$105,016.70
actual amount paid off = 140,000 - 105,016.70
= \$34,983.30
actual amount paid = 20(3404.53) = \$68,090.60
interest = \$68,090.60 - \$34,983.30 = \$33,107.30
The last paragraph of calculations would be totally invalid and unacceptable to an actuarial mathematician, since you cannot add up monetary values that are not in the same "time spot" on a time-line.
In your spread sheet, there might be a splitup of the monthly payments into "interest" and "repayment".
The sum of the column for "interest" for 20 lines should be 33107.30
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http://prob140.org/textbook/content/Chapter_19/05_Exercises.html | 1,719,261,708,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865482.23/warc/CC-MAIN-20240624182503-20240624212503-00200.warc.gz | 24,073,288 | 10,961 | # 19.5. Exercises#
1. Let $$X_i$$, $$i = 1,2$$ be independent random variables such that $$X_i$$ has the exponential $$(\lambda_i)$$ distribution. Suppose $$\lambda_1 \ne \lambda_2$$.
(a) Use the convolution formula to find the density of the sum $$S = X_1 + X_2$$.
(b) Show by algebra that the expression you got for the density in Part (a) is non-negative for all positive $$\lambda_1 \ne \lambda_2$$.
(c) For $$i=1, 2$$, let $$f_i$$ denote the exponential density of $$X_i$$. Show that the density you got in part (a) is equal to $$c_1f_1 + c_2f_2$$ for two constants $$c_1$$ and $$c_2$$ such that $$c_1+c_2 = 1$$. Are $$c_1$$ and $$c_2$$ both positive?
2. A coin lands heads with probability $$p$$. Let $$X$$ be the number of tosses till the first head appears and let $$Y$$ be the number of tails before the first head.
(a) Find the moment generating function of $$X$$.
(b) Use the answer to (a) to find $$E(X)$$. Note that by now you have found $$E(X)$$ in several ways: by the tail sum formula, by conditioning on the first toss, by the pgf, and now by the mgf.
(c) Use the answer to (a) to find the moment generating function of $$Y$$.
3. Recognizing distributions saves a lot of work. See if you can do the following without explicit integration or differentation.
(a) $$W$$ is a normal $$(0, \sigma^2)$$ random variable. For $$n = 0, 1, 2, 3$$, and $$4$$, find $$E(W^n)$$.
(b) $$X$$ and $$Y$$ are i.i.d. with moment generating function $$M(t) = e^{t + t^2}$$, $$-\infty < t < \infty$$. What is the distribution of $$(X-Y)^2$$?
4. Markov and Chebyshev Bounds on the Poisson-Binomial Upper Tail
For $$j \ge 1$$ let $$I_j$$ be independent indicators such that $$P(I_j = 1) = p_j$$. Let $$X = I_1 + I_2 + \ldots + I_n$$. Then $$X$$ is the number of successes in $$n$$ independent trials that are not necessarily identically distributed.
We say that $$X$$ has the Poisson-binomial distribution with parameters $$p_1, p_2, \ldots, p_n$$. As you saw in lab, the binomial is the special case when all the $$p_j$$’s are equal.
These distributions arise in statistical learning theory, the theory of randomized algorithms, and other areas.
Let $$E(X) = \mu$$. For $$c > 0$$, you are going to find an upper bound on $$P(X \ge (1+c)\mu)$$. That’s the chance that $$X$$ exceeds its mean by some percent.
In the special case of the binomial, $$\mu = np$$ and so $$P(X \ge (1+c)\mu)$$ can be rewritten as $$P(\frac{X}{n} - p \ge cp)$$. That’s the chance that the sample proportion exceeds $$p$$ by some percent.
(a) Find $$\mu = E(X)$$ and $$\sigma^2 = Var(X)$$ in terms of $$p_1, p_2, \ldots, p_n$$.
(b) Find Markov’s bound on $$P(X \ge (1+c)\mu)$$.
(c) Find Chebyshev’s bound on $$P(X \ge (1+c)\mu)$$ in terms of $$\mu$$ and $$\sigma$$.
(d) If all the $$p_j$$’s are equal to $$p$$, what is the value of the bound in (c)?
5. Chernoff Bound on Poisson-Binomial Upper Tail
This exercise continues the previous one and uses the same notation.
(a) Show that the mgf of $$I_j$$ is given by $$M_{I_j}(t) = 1 + p_j(e^t - 1)$$ for all $$t$$.
(b) Use (a) to derive an expression for $$M_X(t)$$, the mgf of $$X$$ evaluated at $$t$$.
(c) An useful exponential bound is that $$e^x \ge 1 + x$$ for all $$x$$. You don’t have to show it but please look at the graphs. Use the fact to show that $$M_X(t) \le \exp\big{(}\mu(e^t -1)\big{)}$$ for all $$t$$. Notice that the right hand side is the mgf of a Poisson random variable that has the same mean as $$X$$.
(d) Use Chernoff’s method and the bound in (c) to show that
$P\big{(}X \ge (1+c)\mu\big{)} ~ \le ~ \Big{(} \frac{\exp(c)}{ (1+c)^{1+c}} \Big{)}^\mu$
Remember that $$\mu = np$$ when all the $$p_j$$’s are equal. If $$g(c) = \exp(c)/(1+c)^{1+c}$$ is small then the bound above will decrease exponentially as $$n$$ gets large. That is the focus of the next exercise.
6. Simplified Chernoff Bounds on Poisson-Binomial Upper Tail
The bound in the previous exercise is a bit complicated. Often, simpler versions are used because they are good enough even though they are weaker.
(a) It is not hard to show that $$\log(1+c) \ge \frac{2c}{2+c}$$ for $$c > 0$$. You don’t have to show it but please look at the graphs. Use the fact to show that $$c - (1+c)\log(1+c) \le -\frac{c^2}{2+c}$$ for $$c > 0$$.
(b) Show that if $$X$$ has a Poisson-binomial distribution with mean $$\mu$$ then
$P\big{(} X \ge (1+c)\mu\big{)} ~ \le ~ \exp\big{(}-\frac{c^2}{2+c}\mu\big{)}, ~~~~ c > 0$
(c) A simpler but weaker version of the bound in (b) is also often used. Show that
$P\big{(} X \ge (1+c)\mu\big{)} ~ \le ~ \exp\big{(}-\frac{c^2}{3}\mu\big{)}, ~~~~ c \in (0, 1)$
7. Let $$N$$ be a non-negative integer valued random variable, and let $$X, X_1, X_2, \ldots$$ be i.i.d. and independent of $$N$$. As before, define the random sum $$S$$ by
\begin{split} \begin{align*} S ~~&=~~0~~ \mbox{if}~N=0 \\ &=~~ X_1 + X_2 + \cdots + X_n ~~ \mbox{if}~N = n > 0 \end{align*} \end{split}
(a) Let $$M$$ be our usual notation for moment generating functions. By conditioning on $$N$$, show that
$M_S(t) ~~=~~ M_N\big{(}\log M_X(t) \big{)}$
You can assume that all the quantities above are well defined.
(b) Let $$N$$ have the geometric $$(p)$$ distribution on $$\{1, 2, 3, \ldots \}$$. Find the mgf of $$N$$. This doesn’t use Part (a).
(c) Let $$X_1, X_2, \ldots$$ be i.i.d. exponential $$(\lambda)$$ variables and let $$N$$ be geometric as in Part (b). Use the results of Parts (a) and (b) to identify the distribution of $$S$$.
8. Let $$U_1, U_2, \ldots$$ be i.i.d. uniform on $$(0, 1)$$. For $$n \ge 1$$, let $$S_n = \sum_{i=1}^n U_i$$.
Let $$f_{S_n}$$ be the density of $$S_n$$. The formula for $$f_{S_n}$$ is piecewise polynomial on the possible values $$(0, n)$$. In this problem we will just focus on the density on the interval $$(0, 1)$$ and discover a nice consequence.
(a) For $$0 < x < 1$$, find $$f_{S_2}(x)$$.
(b) Use Part (a) and the convolution formula to find $$f_{S_3}(x)$$ for $$0 < x < 1$$.
(c) Guess a formula for $$f_{S_n}(x)$$ for $$0 < x < 1$$ and prove it by induction.
(d) Use Part (c) to find $$P(S_n < 1)$$.
(e) Let $$N = \min\{n: S_n > 1\}$$. Use Part (d) to find $$E(N)$$. | 2,082 | 6,157 | {"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} | 4.03125 | 4 | CC-MAIN-2024-26 | latest | en | 0.826582 |
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# m03 Q 22
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m03 Q 22 [#permalink]
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16 Mar 2010, 13:52
Is the product of three integers P, Q, and R even?
1. (p-1) (r+1) is odd
2. Square of (q-r) is odd
a. Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
b. Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
c. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
d. EACH statement ALONE is sufficient
e. Statements (1) and (2) TOGETHER are NOT sufficient
[Reveal] Spoiler: OA
D
Let me know your reasoning too.
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Re: m03 Q 22 [#permalink]
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16 Mar 2010, 21:42
PQR is even if any of P, Q, or R is even and none of them are 0.
A: (p-1)*(r+1) = odd either if p and r are both even (since only even+-odd = odd and only odd*odd = odd) or if p and/or r is 0. NOT SUFFICIENT
B: (q-r)^2 = odd only if either q or r is even and the other is odd or if one of them is 0 and the other is odd. NOT SUFFICIENT
Together: r is restricted to 0 or even. Can't determine PQR.
Not sure why answer is C
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Re: m03 Q 22 [#permalink]
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17 Mar 2010, 08:25
I chose e too with 0 value as deciding factor for my choice but it turns out the 0 is considered an even number. I just googled it and found it out. Hope the understanding is correct.
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Re: m03 Q 22 [#permalink]
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17 Mar 2010, 09:18
Ironduke wrote:
Is the product of three integers P, Q, and R even?
1. (p-1) (r+1) is odd
2. Square of (q-r) is odd
[Reveal] Spoiler:
OA c
Let me know your reasoning too.
First of all zero is an even number.
Second answer can not be C, it should be D.
The product of three integers P, Q, and R is even if either of them is even.
(1) $$(p-1)(r+1)=odd$$. Product of two integers is odd only if both are odd. Hence $$p-1=odd$$ and $$r+1=odd$$, which means $$p$$ and $$r$$ are even. Sufficient.
(2) $$(q-r)^2=odd$$ --> $$q-r=odd$$. Difference of two integers is odd only if one of them is odd and another is even. Hence $$p$$ or $$r$$ is even. Sufficient.
Answer: D.
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Re: m03 Q 22 [#permalink]
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17 Mar 2010, 10:43
Excellent. Good reasoning.
Bunuel wrote:
Ironduke wrote:
Is the product of three integers P, Q, and R even?
1. (p-1) (r+1) is odd
2. Square of (q-r) is odd
[Reveal] Spoiler:
OA c
Let me know your reasoning too.
First of all zero is an even number.
Second answer can not be C, it should be D.
The product of three integers P, Q, and R is even if either of them is even.
(1) $$(p-1)(r+1)=odd$$. Product of two integers is odd only if both are odd. Hence $$p-1=odd$$ and $$r+1=odd$$, which means $$p$$ and $$r$$ are even. Sufficient.
(2) $$(q-r)^2=odd$$ --> $$q-r=odd$$. Difference of two integers is odd only if one of them is odd and another is even. Hence $$p$$ or $$r$$ is even. Sufficient.
Answer: D.
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Re: m03 Q 22 [#permalink]
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17 Mar 2010, 13:21
Ironduke wrote:
I chose e too with 0 value as deciding factor for my choice but it turns out the 0 is considered an even number. I just googled it and found it out. Hope the understanding is correct.
Then it's D. Dang math rules
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Re: m03 Q 22 [#permalink]
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18 Mar 2010, 03:32
The OA is D, not C as was indicated in the first post. I've edited the first post to include a correct OA.
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Re: m03 Q 22 [#permalink]
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19 Mar 2010, 07:36
Ironduke wrote:
Is the product of three integers P, Q, and R even?
1. (p-1) (r+1) is odd
2. Square of (q-r) is odd
a. Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
b. Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
c. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
d. EACH statement ALONE is sufficient
e. Statements (1) and (2) TOGETHER are NOT sufficient
[Reveal] Spoiler: OA
D
Let me know your reasoning too.
product P Q R can be even only if one of them is even.
1. (p-1)(r+1)
both p-1 and r+1 are odd
means p and r are even => pqr is even suff.
2. (q-r)^2 is odd => q-r is odd.
there can be following cases for q-r to be odd
q is odd and r is even or q is even and r is odd
in any case one of either q or r are even so pqr will be even
hence d
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Re: m03 Q 22 [#permalink]
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18 Sep 2012, 08:16
The key here is any multiplication by an even number results in an even.
So therefore we're on the hunt for even numbers:
1) (p-1)(r+1) is odd. Only odd x odd gives = odd, so both p-1 and r+1 are odd and hence both are even. An even number in qpr will given an even results
Sufficient
2)Similar to above, odd x odd = odd. So q-r must be odd, and therefore one of q or r must be even. An even number in qpr makes it eve
Sufficient
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Re: m03 Q 22 [#permalink] 18 Sep 2012, 08:16
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# m03 Q 22
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# Relative Error Formula Wikipedia
## Contents
Retrieved February 15, 2007. ^ Braiker, Brian. "The Race is On: With voters widely viewing Kerry as the debate’s winner, Bush’s lead in the NEWSWEEK poll has evaporated". Later sections will present the standard error of other statistics, such as the standard error of a proportion, the standard error of the difference of two means, the standard error of The standard deviation of the age was 9.27 years. Standard errors provide simple measures of uncertainty in a value and are often used because: If the standard error of several individual quantities is known then the standard error of some https://en.wikipedia.org/wiki/Relative_change_and_difference
## Percent Difference Formula
For more complex survey designs, different formulas for calculating the standard error of difference must be used. In other words, the maximum margin of error is the radius of a 95% confidence interval for a reported percentage of 50%. T-distributions are slightly different from Gaussian, and vary depending on the size of the sample. As an example of the above, a random sample of size 400 will give a margin of error, at a 95% confidence level, of 0.98/20 or 0.049—just under 5%.
and Bradburn N.M. (1982) Asking Questions. When it halves again, it is a -69cNp change (a decrease.) Examples Comparisons Car M costs $50,000 and car L costs$40,000. If σ is not known, the standard error is estimated using the formula s x ¯ = s n {\displaystyle {\text{s}}_{\bar {x}}\ ={\frac {s}{\sqrt {n}}}} where s is the sample Percent Difference Vs Percent Error Standard error of the mean Further information: Variance §Sum of uncorrelated variables (Bienaymé formula) The standard error of the mean (SEM) is the standard deviation of the sample-mean's estimate of a
In this example the cost of car L was considered the reference value, but we could have made the choice the other way and considered the cost of car M as The unbiased standard error plots as the ρ=0 diagonal line with log-log slope -½. A disadvantage of this measure is that it is undefined whenever a single actual value is zero. Thus, if an experimental value is less than the theoretical value, the percent error will be negative.
Retrieved 17 July 2014. Relative Error Definition As a result, there are many options for how to define relative difference and which one is used depends on what the comparison is being used for. Next, consider all possible samples of 16 runners from the population of 9,732 runners. Retrieved 30 December 2013. ^ "NEWSWEEK POLL: First Presidential Debate" (Press release).
## Relative Change Formula
A special case of percent change (relative change expressed as a percentage) called percent error occurs in measuring situations where the reference value is the accepted or actual value (perhaps theoretically https://en.wikipedia.org/wiki/Mean_percentage_error The mean age for the 16 runners in this particular sample is 37.25. Percent Difference Formula These assumptions may be approximately met when the population from which samples are taken is normally distributed, or when the sample size is sufficiently large to rely on the Central Limit Relative Difference Consider a sample of n=16 runners selected at random from the 9,732.
References Sudman, Seymour and Bradburn, Norman (1982). http://wapgw.org/relative-error/relative-percent-error-formula.php This negative result provides additional information about the experimental result. Hyndman: Errors on Percentage Errors Retrieved from "https://en.wikipedia.org/w/index.php?title=Symmetric_mean_absolute_percentage_error&oldid=739280296" Categories: Statistical deviation and dispersionHidden categories: Articles lacking in-text citations from August 2011All articles lacking in-text citations Navigation menu Personal tools Not logged Retrieved 2010-05-05. Absolute Change Formula
ed. The margin of error of an estimate is the half-width of the confidence interval ... ^ Stokes, Lynne; Tom Belin (2004). "What is a Margin of Error?" (PDF). The margin of error of 2% is a quantitative measure of the uncertainty – the possible difference between the true proportion who will vote for candidate A and the estimate of http://wapgw.org/relative-error/relative-error-definition-wikipedia.php Second, an X cNp change in a quantity following a -X cNp change returns that quantity to its original value.
The percent error equation, when rewritten by removing the absolute values, becomes: % Error = Experimental − Theoretical | Theoretical | × 100. {\displaystyle \%{\text{ Error}}={\frac {{\text{Experimental}}-{\text{Theoretical}}}{|{\text{Theoretical}}|}}\times 100.} It is important Relative Error Calculator Perspect Clin Res. 3 (3): 113–116. The proportion or the mean is calculated using the sample. | 1,065 | 4,859 | {"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} | 4.21875 | 4 | CC-MAIN-2018-34 | latest | en | 0.880048 |
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