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https://www.khanacademy.org/math/calculus-1/cs1-limits-and-continuity/cs1-limits-by-direct-substitution/v/limits-of-piecewise-functions | 1,606,753,660,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141216897.58/warc/CC-MAIN-20201130161537-20201130191537-00515.warc.gz | 718,978,541 | 33,810 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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Main content
# Limits of piecewise functions
AP.CALC:
LIM‑1 (EU)
,
LIM‑1.D (LO)
,
LIM‑1.D.1 (EK)
## Video transcript
- [Instructor] Let's think a little bit about limits of piecewise functions that are defined algebraically like our f of x right over here. Pause this video and see if you can figure out what these various limits would be, some of them are one-sided, and some of them are regular limits, or two-sided limits. Alright, let's start with this first one, the limit as x approaches four, from values larger than equaling four, so that's what that plus tells us. And so when x is greater than four, our f of x is equal to square root of x. So as we are approaching four from the right, we are really thinking about this part of the function. And so this is going to be equal to the square root of four, even though right at four, our f of x is equal to this, we are approaching from values greater than four, we're approaching from the right, so we would use this part of our function definition, and so this is going to be equal to two. Now what about our limit of f of x, as we approach four from the left? Well then we would use this part of our function definition. And so this is going to be equal to four plus two over four minus one, which is equal to 6 over three, which is equal to two. And so if we wanna say what is the limit of f of x as x approaches four, well this is a good scenario here because from both the left and the right as we approach x equals four, we're approaching the same value, and we know, that in order for the two side limit to have a limit, you have to be approaching the same thing from the right and the left. And we are, and so this is going to be equal to two. Now what's the limit as x approaches two of f of x? Well, as x approaches two, we are going to be completely in this scenario right over here. Now interesting things do happen at x equals one here, our denominator goes to zero, but at x equals two, this part of the curve is gonna be continuous so we can just substitute the value, it's going to be two plus two, over two minus one, which is four over one, which is equal to four. Let's do another example. So we have another piecewise function, and so let's pause our video and figure out these things. Alright, now let's do this together. So what's the limit as x approaches negative one from the right? So if we're approaching from the right, when we are greater than or equal to negative one, we are in this part of our piecewise function, and so we would say, this is going to approach, this is gonna be two, to the negative one power, which is equal to one half. What about if we're approaching from the left? Well, if we're approaching from the left, we're in this scenario right over here, we're to the left of x equals negative one, and so this is going to be equal to the sine, 'cause we're in this case, for our piecewise function, of negative one plus one, which is the sine of zero, which is equal to zero. Now what's the two-sided limit as x approaches negative one of g of x? Well we're approaching two different values as we approach from the right, and as we approach from the left. And if our one-sided limits aren't approaching the same value, well then this limit does not exist. Does not exist. And what's the limit of g of x, as x approaches zero from the right? Well, if we're talking about approaching zero from the right, we are going to be in this case right over here, zero is definitely in this interval, and over this interval, this right over here is going to be continuous, and so we can just substitute x equals zero there, so it's gonna be two to the zero, which is, indeed, equal to one, and we're done. | 926 | 3,922 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2020-50 | longest | en | 0.964103 |
https://calculationcalculator.com/biswa-to-feddan | 1,721,323,606,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514848.78/warc/CC-MAIN-20240718161144-20240718191144-00630.warc.gz | 138,027,952 | 22,524 | # Biswa to Feddan Conversion
## 1 Biswa is equal to how many Feddan?
### 0.0298617 Feddan
##### Reference This Converter:
Biswa and Feddan both are the Land measurement unit. Compare values between unit Biswa with other Land measurement units. You can also calculate other Land conversion units that are available on the select box, having on this same page.
Biswa to Feddan conversion allows you to convert value between Biswa to Feddan easily. Just enter the Biswa value into the input box, the system will automatically calculate Feddan value. 1 Biswa in Feddan? In mathematical terms, 1 Biswa = 0.0298617 Feddan.
To conversion value between Biswa to Feddan, just multiply the value by the conversion ratio. One Biswa is equal to 0.0298617 Feddan, so use this simple formula to convert -
The value in Feddan is equal to the value of Biswa multiplied by 0.0298617.
Feddan = Biswa * 0.0298617;
For calculation, here's how to convert 10 Biswa to Feddan using the formula above -
10 Biswa = (10 * 0.0298617) = 0.298617 Feddan
Biswa Feddan Conversion
1 0.0298617 1 Biswa = 0.0298617 Feddan
2 0.0597234 2 Biswa = 0.0597234 Feddan
3 0.0895851 3 Biswa = 0.0895851 Feddan
4 0.1194468 4 Biswa = 0.1194468 Feddan
5 0.1493085 5 Biswa = 0.1493085 Feddan
6 0.1791702 6 Biswa = 0.1791702 Feddan
7 0.2090319 7 Biswa = 0.2090319 Feddan
8 0.2388936 8 Biswa = 0.2388936 Feddan
9 0.2687553 9 Biswa = 0.2687553 Feddan
10 0.298617 10 Biswa = 0.298617 Feddan
11 0.3284787 11 Biswa = 0.3284787 Feddan
12 0.3583404 12 Biswa = 0.3583404 Feddan
13 0.3882021 13 Biswa = 0.3882021 Feddan
14 0.4180638 14 Biswa = 0.4180638 Feddan
15 0.4479255 15 Biswa = 0.4479255 Feddan
16 0.4777872 16 Biswa = 0.4777872 Feddan
17 0.5076489 17 Biswa = 0.5076489 Feddan
18 0.5375106 18 Biswa = 0.5375106 Feddan
19 0.5673723 19 Biswa = 0.5673723 Feddan
20 0.597234 20 Biswa = 0.597234 Feddan
21 0.6270957 21 Biswa = 0.6270957 Feddan
22 0.6569574 22 Biswa = 0.6569574 Feddan
23 0.6868191 23 Biswa = 0.6868191 Feddan
24 0.7166808 24 Biswa = 0.7166808 Feddan
25 0.7465425 25 Biswa = 0.7465425 Feddan
26 0.7764042 26 Biswa = 0.7764042 Feddan
27 0.8062659 27 Biswa = 0.8062659 Feddan
28 0.8361276 28 Biswa = 0.8361276 Feddan
29 0.8659893 29 Biswa = 0.8659893 Feddan
30 0.895851 30 Biswa = 0.895851 Feddan
31 0.9257127 31 Biswa = 0.9257127 Feddan
32 0.9555744 32 Biswa = 0.9555744 Feddan
33 0.9854361 33 Biswa = 0.9854361 Feddan
34 1.0152978 34 Biswa = 1.0152978 Feddan
35 1.0451595 35 Biswa = 1.0451595 Feddan
36 1.0750212 36 Biswa = 1.0750212 Feddan
37 1.1048829 37 Biswa = 1.1048829 Feddan
38 1.1347446 38 Biswa = 1.1347446 Feddan
39 1.1646063 39 Biswa = 1.1646063 Feddan
40 1.194468 40 Biswa = 1.194468 Feddan
41 1.2243297 41 Biswa = 1.2243297 Feddan
42 1.2541914 42 Biswa = 1.2541914 Feddan
43 1.2840531 43 Biswa = 1.2840531 Feddan
44 1.3139148 44 Biswa = 1.3139148 Feddan
45 1.3437765 45 Biswa = 1.3437765 Feddan
46 1.3736382 46 Biswa = 1.3736382 Feddan
47 1.4034999 47 Biswa = 1.4034999 Feddan
48 1.4333616 48 Biswa = 1.4333616 Feddan
49 1.4632233 49 Biswa = 1.4632233 Feddan
50 1.493085 50 Biswa = 1.493085 Feddan
51 1.5229467 51 Biswa = 1.5229467 Feddan
52 1.5528084 52 Biswa = 1.5528084 Feddan
53 1.5826701 53 Biswa = 1.5826701 Feddan
54 1.6125318 54 Biswa = 1.6125318 Feddan
55 1.6423935 55 Biswa = 1.6423935 Feddan
56 1.6722552 56 Biswa = 1.6722552 Feddan
57 1.7021169 57 Biswa = 1.7021169 Feddan
58 1.7319786 58 Biswa = 1.7319786 Feddan
59 1.7618403 59 Biswa = 1.7618403 Feddan
60 1.791702 60 Biswa = 1.791702 Feddan
61 1.8215637 61 Biswa = 1.8215637 Feddan
62 1.8514254 62 Biswa = 1.8514254 Feddan
63 1.8812871 63 Biswa = 1.8812871 Feddan
64 1.9111488 64 Biswa = 1.9111488 Feddan
65 1.9410105 65 Biswa = 1.9410105 Feddan
66 1.9708722 66 Biswa = 1.9708722 Feddan
67 2.0007339 67 Biswa = 2.0007339 Feddan
68 2.0305956 68 Biswa = 2.0305956 Feddan
69 2.0604573 69 Biswa = 2.0604573 Feddan
70 2.090319 70 Biswa = 2.090319 Feddan
71 2.1201807 71 Biswa = 2.1201807 Feddan
72 2.1500424 72 Biswa = 2.1500424 Feddan
73 2.1799041 73 Biswa = 2.1799041 Feddan
74 2.2097658 74 Biswa = 2.2097658 Feddan
75 2.2396275 75 Biswa = 2.2396275 Feddan
76 2.2694892 76 Biswa = 2.2694892 Feddan
77 2.2993509 77 Biswa = 2.2993509 Feddan
78 2.3292126 78 Biswa = 2.3292126 Feddan
79 2.3590743 79 Biswa = 2.3590743 Feddan
80 2.388936 80 Biswa = 2.388936 Feddan
81 2.4187977 81 Biswa = 2.4187977 Feddan
82 2.4486594 82 Biswa = 2.4486594 Feddan
83 2.4785211 83 Biswa = 2.4785211 Feddan
84 2.5083828 84 Biswa = 2.5083828 Feddan
85 2.5382445 85 Biswa = 2.5382445 Feddan
86 2.5681062 86 Biswa = 2.5681062 Feddan
87 2.5979679 87 Biswa = 2.5979679 Feddan
88 2.6278296 88 Biswa = 2.6278296 Feddan
89 2.6576913 89 Biswa = 2.6576913 Feddan
90 2.687553 90 Biswa = 2.687553 Feddan
91 2.7174147 91 Biswa = 2.7174147 Feddan
92 2.7472764 92 Biswa = 2.7472764 Feddan
93 2.7771381 93 Biswa = 2.7771381 Feddan
94 2.8069998 94 Biswa = 2.8069998 Feddan
95 2.8368615 95 Biswa = 2.8368615 Feddan
96 2.8667232 96 Biswa = 2.8667232 Feddan
97 2.8965849 97 Biswa = 2.8965849 Feddan
98 2.9264466 98 Biswa = 2.9264466 Feddan
99 2.9563083 99 Biswa = 2.9563083 Feddan
100 2.98617 100 Biswa = 2.98617 Feddan
200 5.97234 200 Biswa = 5.97234 Feddan
300 8.95851 300 Biswa = 8.95851 Feddan
400 11.94468 400 Biswa = 11.94468 Feddan
500 14.93085 500 Biswa = 14.93085 Feddan
600 17.91702 600 Biswa = 17.91702 Feddan
700 20.90319 700 Biswa = 20.90319 Feddan
800 23.88936 800 Biswa = 23.88936 Feddan
900 26.87553 900 Biswa = 26.87553 Feddan
1000 29.8617 1000 Biswa = 29.8617 Feddan
Biswa is a land measurement unit, It is mainly used in rural area of India. 1 Biswa is equal to 150 Gaj. Biswa is medium area measurement unit.
A Feddan is a land area measurement unit, mainly used in Egypt, Sudan, Syria, and the Sultanate of Oman. In Classical Arabic, the word means "a yoke of oxen".
The value in Feddan is equal to the value of Biswa multiplied by 0.0298617.
Feddan = Biswa * 0.0298617;
1 Biswa is equal to 0.0298617 Feddan.
1 Biswa = 0.0298617 Feddan.
• 1 biswa = feddan
• biswa into feddan
• biswas to feddans
• convert biswa to feddan
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→ | 2,736 | 6,081 | {"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-2024-30 | latest | en | 0.493397 |
https://www.brookhaveninstruments.com/what-is-particle-size-distribution-weighting/ | 1,716,599,973,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058770.49/warc/CC-MAIN-20240525004706-20240525034706-00185.warc.gz | 582,901,491 | 20,620 | # What Is Particle Size Distribution Weighting? How to Get Fooled About What Was Measured & What it Means
Feb 8, 2023
## Introduction
Some particle size instruments determine size for individual particles. They are single particle counters. Some instruments determine surface area as a function of particle size. Some instruments determine mass or volume vs size. And some instruments determine various functions of scattered light intensity as a function of size. All can produce particle size distributions. And, in principle, one can transform from one type to the other in order to compare results. For example, if a measurement with a single particle counter produced a differential number-weighted size distribution and you wanted to compare the results to a measurement with another type of instrument that produced a differential volume-weighted size distribution, what would you do?
## Transform One Weighting To Another
You could convert the differential number-weighted distribution to a differential volume-weighted distribution or the other way around. How do you do that? It’s easy. For each size class in the discrete Number Fraction/(Size Class) distribution*, multiply the number or count by the diameter cubed (spheres are assumed). In the case of discrete distributions, the diameter to be cubed is most likely the midpoint of the size class. The end result is a discrete Volume Fraction/(Size Class) distribution. And to convert a continuous differential number-weighted distribution into a continuous differential volume-weighted distribution, multiply by D3. The end result is a continuous volume-weighted differential distribution. Of course the initial numerical values will not be normalized, but that is easy to remedy. Here is a simple example using continuous distributions:
The bell-shape curves are the differential distributions weighted by number, dN/dD, or by volume, dV/dD. The sigmoidal curves are the cumulative distributions weighted by number, CN(D), or by volume, CV(D).
Perhaps the initial measurement determined dV/dD, from which by integration CV(D) was determined. Then, to determine the unnormalized dN/dD, use the following formula:
At each y-value of the differential volume-weighted distribution, divide by D3. Find the maximum value in the resulting set of unnormalized numbers. Then determine the factor that will make that value 100. Apply the same factor to all the other unnormalized values. Now the differential number distribution is normalized.
## Relationship Amongst Transformed Distributions
Notice that dV/dD is always shifted to the right, towards larger D’s, than dN/dD. Likewise, the cumulative distributions are shifted in a similar fashion. The modal diameter—corresponding to the peak in the differential distribution—and the median diameter— equal to the diameter at 50% of the cumulative distribution–of the volume distribution is always higher than the corresponding ones in the number distribution.
If these were nonporous particles**, then the surface area-weighted differential distribution, dS/dD, is related to the number-weighted by the following simple equation:
And the curves when normalized and plotted would sit between the number- and volume-weighted for both the differential and cumulative representations.
## First Warning: Which Weighting Was It?
If someone says the median diameter is 10.0 nm and it’s a broad distribution, if you don’t ask what the weighting was, you are missing a lot of information. In the graph above, the number median diameter is 10.0 nm and it is a broad distribution; yet, the volume median diameter is 42.3 nm and a large fraction by volume of the particles are well above 10.0 nm.
## Second Warning: Transformations Can Be Dangerous
Look again at the graph. Notice the two arrows, one at the base of each differential distribution. A relatively small volume of particles in the left-hand tail of dV/dD is responsible for a large portion of the dN/dD distribution; likewise, a relatively small number of large particles in the right-hand tail of dN/dD is responsible for a large portion of the dV/dD distribution. This situation is often a prelude to disaster. The reader probably assumed that all the particles were used in calculating the results as shown. Indeed, the author did exactly that. But in real measurements, this is usually not the case: the information in the tails of the differential distribution is often not accurately known.
In counting experiments, one tends to under count a relatively few large particles. But, these are the very ones that dominate the volume distribution. Therefore, the calculated volume distribution is shifted much too low. And in experiments that use scattered or diffracted light, the small particles don’t contribute much to the signal. Thus, they are under represented in the distribution with the most natural weighting (intensity-weighted in this case). Therefore, the calculated number distribution is often shifted much too high. Transformations, though simple algebraically, may be very inaccurate for these reasons.
## Another Example Using a Bimodal Distribution:
The peak centered on 90 nm has a much smaller relative number of particles than the peak centered on 25 nm. In addition, the 90 nm peak is narrower. When the transformation is made to surface area-weighted, it should not be surprising that the amount by surface area has shifted to the larger sizes: 35.8% by surface area centered on 35 nm mode and 64.2% by surface area centered on 90 nm mode.
Since the 90 nm peak was narrow, it is not surprising that it remains the mode. Whereas, the lower peak is somewhat broad and just like the case of the broad unimodal distribution examined earlier, the entire peak shifts to the right.
Finally, the volume-weighted distribution continues the trend:
The larger peak centered on 90 nm contains, by amount, most of the volume; whereas, it contains the least by number.
Notice the cumulative distributions have a plateau. This is characteristic of multimodal distributions. Either from a corresponding tabular presentation or from the graph, you can read off the amount in each peak by finding the plateau value.
## Which Weighting Should I Use To Best Present My Data?
Sometimes the answer is dictated by the field of you are in: when measuring blood, sperm, or micro-contaminates, the absolute number per unit volume is required. Use a single particle counter and stay with the number distribution. If the particles are used by mass, use the volume distribution (here the assumption is all the particles have the same density; if not the volume and mass distributions are no longer equal).
## Summary
As important as it is to know if the “size” is a true, spherical diameter or radius, or an equivalent spherical size determine by the measurement technique, it is equally important to know if the distribution is weighted by number, surface area, volume-mass, or intensity. Without this information, you really don’t know what emphasis to put on the results.
It can’t be emphasized enough that while the algebra for transforming one distribution to another is simple enough, it is the assumption that all the particles have been measured that is usually wanting. Common errors include missing a significant but relatively small fraction of large particles that carry most of the volume and mass of the distribution while counting particles; and missing a significant but relatively large number of small particle that carry most of the numberweighted information, because they don’t contribute much to the intensity of scattered light.
*For a definition of Fraction/(Size Class), see the application note “What Is A Discrete Particle Size Distribution?”.
**Porous particles that have a lot more surface area. Surfacearea weighted size distributions should not be calculated from either number- or volume-weighted ones unless it is certain the porosity is unimportant. This is the case for liquid droplet particles but not the case for many oxide particles. | 1,594 | 8,058 | {"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-2024-22 | latest | en | 0.877739 |
https://www.indiabix.com/engineering-mechanics/force-vectors/002005 | 1,638,446,699,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362219.5/warc/CC-MAIN-20211202114856-20211202144856-00633.warc.gz | 851,492,837 | 6,013 | # Engineering Mechanics - Force Vectors
### Exercise :: Force Vectors - General Questions
• Force Vectors - General Questions
21.
Determine the magnitude of the x and y components of the 2—kN force.
A. Fx = -1.414 kN, Fy = -1.414 kN B. Fx = -1.000 kN, Fy = -1.732 kN C. Fx = -1.732 kN, Fy = -1.000 kN D. Fx = -4.000 kN, Fy = -2.312 kN
Explanation:
No answer description available for this question. Let us discuss.
22.
Cable BC exerts a force of F = 28 N on the top of the flagpole. Determine the projection of this force along the positive z axis of the pole.
A. F = 24 N B. F = -24 N C. F = 12 N D. F = 12 N
Explanation:
No answer description available for this question. Let us discuss.
23.
Determine the magnitudes of the resultant force and its direction measured from the positive x axis.
A. R = 12.49 kN, = 43.9° CW B. R = 13.6 kN, = 21.5° CW C. R = 14.00 kN, = 60.0° CW D. R = 10.80 kN, = 68.2° CW
Explanation:
No answer description available for this question. Let us discuss.
24.
Determine the design angle ( <90°) bretween the two struts so that the 500-lb horizontal force has a component of 600lb directed from A toward C. That is the component of force acting along member AB?
A. FAB = 321 lb, = 40.0° B. FAB = 383 lb, = 36.9° C. FAB = 171.1 lb, = 20.0° D. FAB = 215 lb, = 52.7°
Explanation:
No answer description available for this question. Let us discuss.
25.
Determine the magnitude and direction of F so that this force has components of 40 lb acting from A toward B and 60lb acting from A toward C on the frame.
A. F = 44.7 lb, = 22.9° B. F = 80.3 lb, = 46.2° C. F = 62.9 lb, = 37.1° D. F = 72.1 lb, = 56.3° | 567 | 1,656 | {"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-2021-49 | latest | en | 0.860636 |
http://www.physicsforums.com/showthread.php?t=107328 | 1,369,348,657,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704007597/warc/CC-MAIN-20130516113327-00013-ip-10-60-113-184.ec2.internal.warc.gz | 624,922,409 | 8,064 | ## help: trigonometric equation
$$x_1(cos\alpha-1) + x_2sin\alpha = 0$$
$$x_1sin\alpha + x_2(-cos\alpha-1) = 0$$
How to solve this equation? Can anyone help me?
PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
Mentor It's a system of equations: 2 equations in 2 unknowns. That means you can solve it. Just solve for one unknown in terms of the other using the first equation, and then subsitute that into the second.
Let me try... Solve equation 1: $$x_2 = \frac{-x_1(cos\alpha - 1)}{sin\alpha}$$ Substitute it to the second: $$x_1sin\alpha + \frac{-x_1(cos\alpha - 1)}{sin\alpha}(-cos\alpha-1) = 0$$ $$x_1sin\alpha + \frac{x_1(cos^2\alpha - 1)}{sin\alpha} = 0$$ $$2x_1sin\alpha = 0$$ What is the solutions for $$2x_1sin\alpha = 0$$? Obviously one is $$x_1=0$$, but if $$sin\alpha = 0$$, then...
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## help: trigonometric equation
Quote by physicsRookie Let me try... Solve equation 1: $$x_2 = \frac{-x_1(cos\alpha - 1)}{sin\alpha}$$ Substitute it to the second: $$x_1sin\alpha + \frac{-x_1(cos\alpha - 1)}{sin\alpha}(-cos\alpha-1) = 0$$ $$x_1sin\alpha + \frac{x_1(cos^2\alpha - 1)}{sin\alpha} = 0$$ $$2x_1sin\alpha = 0$$ What is the solutions for $$2x_1sin\alpha = 0$$? Obviously one is $$x_1=0$$, but if $$sin\alpha = 0$$, then...
Well done. The point is, of course, that $$\alpha$$ is a number (not one of the variables) so these can be solved like any pair of equations for x1 and x2.
Notice, by the way, that if $$sin\alpha= 0$$, your first step, dividing by that, would be invalid. You have to look at this case separately.
If $$sin\alpha= 0$$, then $$cos\alpha$$ is either 1 or -1.
What do your equations look like if $$sin\alpha= 0$$ and $$cos\alpha= 1$$?
What do your equations look like if $$sin\alpha= 0$$ and $$cos\alpha= -1$$?
HallsofIvy, thanks.
Quote by HallsofIvy What do your equations look like if $$sin\alpha= 0$$ and $$cos\alpha= 1$$?
$$2x_1=0 and 0=0 => x_1=0, x_2$$ could be any number
Quote by HallsofIvy What do your equations look like if $$sin\alpha= 0$$ and $$cos\alpha= -1$$?
$$0=0 and -2x_2=0 => x_2=0, x_1$$ could be any number
I just try another solution.
Rewrite the equations:
$$(x_1cos\alpha + x_2sin\alpha) - x_1 = 0$$
$$(x_1sin\alpha - x_2cos\alpha) - x_2 = 0$$
Suppose $$x_1 = cos\frac{\alpha}{2}, x_2 = sin\frac{\alpha}{2}$$, then
$$cos\frac{\alpha}{2}cos\alpha + sin\frac{\alpha}{2}sin\alpha - cos\frac{\alpha}{2} = 0$$
$$cos\frac{\alpha}{2}sin\alpha - sin\frac{\alpha}{2}cos\alpha - sin\frac{\alpha}{2} = 0$$
It works!
I am wondering whether there is some general method to solve $$x_1, x_2$$ depending on $$\alpha$$ or not. | 986 | 2,795 | {"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 | 4 | CC-MAIN-2013-20 | longest | en | 0.778782 |
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https://jeopardylabs.com/print/properties-and-simplifying-algebraic-expressions-99 | 1,652,818,591,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662520817.27/warc/CC-MAIN-20220517194243-20220517224243-00347.warc.gz | 387,493,934 | 3,785 | Properties
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9 * 0 = 0 * 9
What is the commutative property of multiplication?
100
3 and 4 in the expression 3x + 4y + 5
What are coefficients?
100
Simplify 3x + 5x + 2
What is 8x + 2?
100
Simplify 3(x+4)
What is 3x + 12 ?
100
Simplify 3(a + b) + 2b
What is 3a + 5b ?
200
3(x + 4) = 3x + 12
What is the distributive property?
200
the like terms in 3x + 4x + 3
What are 3x and 4x?
200
Simplify 3x - 4y + 7x
What is 10x - 4y?
200
Simplify (3y - 5)(-2)
What is -6y + 10 ?
200
Simplify 6m + 7 (7m + 9)
What is 55m + 63 ?
300
( 4 + 5) + 6 = (5 + 4) + 6
What is the commutative property of addition?
300
the constant in 6x - 7y - 5
What is -5?
300
Simplify -2a + 3b - 4
What is -2a + 3b - 4
300
Use the distributive property to simplify and solve 12(5 + 10)
What is 60 + 120 = 180 ?
300
If x=3 solve 3(6+2x)
What is 36
400
(9 x 5) x 3 = 9 x (5 x 3)
What is the associative property of multiplication?
400
the number of terms in the expression 9a - 5b + 7a - 4
What is 4?
400
Simplify m - 3n + 4m + 5n
What is 5m + 2n?
400
Factor 3x + 42 using the GCF
What is 3(x+14)
400
If x=5 solve: 5x - 2x + 3(2+1)
What is 24
500
9 + (3+2) = (9+3) + 2
What is the associative property of addition?
500
the coefficients in the expression -4x - x + 5
What are -4 and -1?
500
Simplify -(2c - 5b) + b - 3c
What is 6b + -5c
500
Factor 25a+10b using the GCF
What is 5(5a+2b)
500
If p=3 and r=2 solve:
(2p + r) - (5p - 6r)
What is 5
Click to zoom | 680 | 1,573 | {"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-2022-21 | latest | en | 0.831014 |
https://mathhelpboards.com/threads/aux-19-normal-distribution.5572/ | 1,657,072,756,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104655865.86/warc/CC-MAIN-20220705235755-20220706025755-00752.warc.gz | 396,679,306 | 16,749 | # [SOLVED]-aux.19 Normal distribution
#### karush
##### Well-known member
Let $$\displaystyle X$$ be normally distributed with $$\displaystyle \mu =100cm$$ and $$\displaystyle \sigma =5 cm$$
$$\displaystyle (a$$) shade region $$\displaystyle P(X>105)$$
View attachment 1010
(b) Given that $$\displaystyle P(X<d)=P(X>105)$$, find the value of $$\displaystyle d$$.
wasn't sure if this meant that $$\displaystyle d$$ is the left of 105 which would be larger in volume than $$\displaystyle X>105$$
Last edited:
#### Prove It
##### Well-known member
MHB Math Helper
What volume? The probabilities are areas...
Anyway, it's asking you to set the two probabilities equal to each other. So if your area was measured from the left instead of from the right, what value on the x axis would you get to? Hint: The areas are equal and symmetrical about the mean.
#### karush
##### Well-known member
What volume? The probabilities are areas...
Anyway, it's asking you to set the two probabilities equal to each other. So if your area was measured from the left instead of from the right, what value on the x axis would you get to? Hint: The areas are equal and symmetrical about the mean.
yes area not volume
so then $$\displaystyle d=95$$ if $$\displaystyle p(X<d)$$ for the same area as $$\displaystyle P(X>105)$$
#### Prove It
##### Well-known member
MHB Math Helper
yes area not volume
so then $$\displaystyle d=95$$ if $$\displaystyle p(X<d)$$ for the same area as $$\displaystyle P(X>105)$$
Correct
#### karush
##### Well-known member
(c) Given that $$\displaystyle P(X>105)=0.16$$ (correct to $$\displaystyle 2$$ significant figures), find $$\displaystyle P(d<X<105)$$
so that is within $$\displaystyle 68\%$$ within
$$\displaystyle 1$$ standard deviation of the mean
or do just $$\displaystyle (2)0.16 = 0.32$$
not sure??
#### Klaas van Aarsen
##### MHB Seeker
Staff member
(c) Given that $$\displaystyle P(X>105)=0.16$$ (correct to $$\displaystyle 2$$ significant figures), find $$\displaystyle P(d<X<105)$$
so that is within $$\displaystyle 68\%$$ within
$$\displaystyle 1$$ standard deviation of the mean
or do just $$\displaystyle (2)0.16 = 0.32$$
not sure??
Yeah. It's 68% within 1 standard deviation.
But that means that P(d<X<105)=P(95<X<105)=0.68.
#### MarkFL
Staff member
(c) Given that $$\displaystyle P(X>105)=0.16$$ (correct to $$\displaystyle 2$$ significant figures), find $$\displaystyle P(d<X<105)$$
so that is within $$\displaystyle 68\%$$ within
$$\displaystyle 1$$ standard deviation of the mean
or do just $$\displaystyle (2)0.16 = 0.32$$
not sure??
I would write (for clarity):
$$\displaystyle P(d<X<105)=P(d<X<100)+P(100<X<105)$$
$$\displaystyle 0.68=P(d<X<100)+\left(0.5-P(X>105) \right)$$
$$\displaystyle 0.68=P(d<X<100)+0.34$$
$$\displaystyle P(d<X<100)=0.34$$
What do you find?
#### karush
##### Well-known member
I would write (for clarity):
$$\displaystyle P(d<X<105)=P(d<X<100)+P(100<X<105)$$
$$\displaystyle 0.68=P(d<X<100)+\left(0.5-P(X>105) \right)$$
$$\displaystyle 0.68=P(d<X<100)+0.34$$
$$\displaystyle P(d<X<100)=0.34$$
What do you find?
i understand what you have here... but don't know how the 0.16 comes into this.
#### MarkFL
$$\displaystyle P(X>105)=0.16$$
$$\displaystyle 0.5-P(X>105)=0.5-0.16=0.34$$ | 1,027 | 3,282 | {"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-2022-27 | latest | en | 0.87712 |
https://www.hackmath.net/en/math-problems/algebra?page_num=37 | 1,547,981,639,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583705737.21/warc/CC-MAIN-20190120102853-20190120124853-00168.warc.gz | 816,756,308 | 7,244 | # Algebra - problems - page 37
1. Cyclist
Cyclist rode two-fifths of the way the first day. The second day drove 5 km less while traveled three-eighths of route. How many kilometers does he have to finish target?
2. The store
The store received the same number of cans of peas and corn. The first day sold 10 cans of peas and 166 cans ofcorn so that left 5 times more peas than corn cans. How many cans of each kind were in the store?
The glass of 1 mm thickness absorb 4% of UV radiation passing through. How many percent of UV rays absorb glass with a thickness 1.1 cm, made from 1 mm thick those glasses?
4. Catheti
The hypotenuse of a right triangle is 41 and the sum of legs is 49. Calculate the length of its legs.
5. Two numbers
The sum of two numbers is 1. Identify this two numbers if you know that the half of first is equal to the third of second number.
6. Cottages
The summer camp is 41 cottages. Rooms are for three and for four in them. How many of the 140 campers lives of three?
7. Nectar
Nectar collected by bees contains 70% water. From the nectar of the some process produces honey which contains 19% water. How many kg of nectar bees need to collect to make 1 kg of honey?
8. Cuboid walls
Calculate the volume of the cuboid if its different walls have area of 195cm², 135cm² and 117cm².
9. Cuboid - complicatef
Three walls of the same cuboid has content 6 cm2, 10 cm2 and 15 cm2. Calculate the volume of the cuboid.
10. Is right triangle
One angle of the triangle is 36° and the remaining two are in the ratio 3:5. Determine whether triangle is a rectangular triangle.
11. Silver medal
To circular silver medal with a diameter of 10 cm is inscribed gold cross, which consists of five equal squares. What is the content area of silver part?
12. Mine
What is temperature in the mine at a depth of 1160 m, where at depth 9 m is 11°C and every 100 m, the temperature increases by 0.7°C?
13. Children
Margaret and Zdena weighs the same and Petra 3 kg more. Together weigh 156 kg. How much does they weigh?
14. Dices throws
What is the probability that the two throws of the dice: a) Six falls even once b) Six will fall at least once
15. Rectangular trapezium
Calculate the perimeter of a rectangular trapezium when its content area is 576 cm2 and sice a (base) is 30 cm, height 24 cm.
16. Two trains meet
From A started at 7:15 express train at speed 85 km/h to B. From B started passenger train at 8:30 in the direction to A and at speed 55 km/h. The distance A and B is 386 1/4 km. At what time and at what distance from B the two trains meet?
17. Angles of a triangle
In the triangle ABC, the angle beta is 15° greater than the angle alpha. The remaining angle is 30° greater than the sum of the angles alpha and beta. Calculate the angles of a triangle.
18. Square vs rectangle
Square and rectangle have the same area contents. The length of the rectangle is 9 greater and width 6 less than side of the square. Calculate the side of a square.
19. 12 Moons
Good Marry came to ask twelve moons for help with collecting strawberries. All twelve moons 1200 strawberries picked in 20 minutes. But lazy months July and August cease tear after 5 minutes. How many minutes in total will take collect strawberries?
20. Holidays
Of the 35 students of class were 7 on holiday in Germany and just as much in Italy. 5 students visited Austria. In none of these countries was 21 students, all three visited by one student . In Italy and Austria were 2 students and in Austria and Germany.
Do you have an interesting mathematical problem that you can't solve it? Enter it, and we can try to solve it.
To this e-mail address, we will reply solution; solved examples are also published here. Please enter e-mail correctly and check whether you don't have a full mailbox. | 962 | 3,777 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2019-04 | latest | en | 0.939042 |
https://www.brightstorm.com/test-prep/ap-calculus-ab/ap-calculus-videos/volume-disc-method/ | 1,695,443,417,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506479.32/warc/CC-MAIN-20230923030601-20230923060601-00717.warc.gz | 766,671,831 | 30,354 | ###### John Postovit
University of North Dakota
M.Ed.,Stanford University
From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.
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# Volume - Disc Method
John Postovit
###### John Postovit
University of North Dakota
M.Ed.,Stanford University
From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.
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[0:00:00]
Today we are going to be talking about volume. Volume is a subject which shows up all over in AP Calculus. The method we are doing today, unlike the one we've done before with cross sections, is going to be one of the methods for finding volumes of objects that are rounded. We are going to be using the disc method. There are a couple of different methods for doing volumes of round objects. This method and another one called the shell method. The shell method is one that is often covered in calculus classes but we are not going to concentrate much on it because it's not required for the AP test. I know the fact that we are not going to be doing the shell method makes you feel all hollow inside like a shell but you'll get over it. That was a joke. Turn on the laugh light.
No matter how complex the problem looks, one thing is the same every time. What you're always doing in some form is taking a function, rotating it around a line and finding the volume of the shape that you get when you rotate it. Let me show you one. For example if you take a nice little triangle right here. I tipped a couple of pencils in this high tech demonstrator. The pencils are the axes. If I rotate it around this axis, you can see that it's describing a shape. If you spin it faster and faster, it might be possible for you to imagine the shape a little bit better. it's a couple of cones put together. Now, another. Let's say for example that you want to take the function 2² of x and rotate it around the x axis? There is our axis that we are going to rotate on and rotate this.
[0:02:00]
Notice the shape, rotate it around its axis. Can you see that shape now? It's kind of like a bowl. When you rotate something, what shape do you get? Let me see if I can do this nicely. You see the circle that it's describing? That's the heart of this, circles. And there is this line which is the radius of the circle, goes around, it basically crosses all the places that make up the area of the circle. Circles. This is the formula and it really comes from circles. There is pi f(x). f(x) is the function. Sometimes people will call this r(x) instead. The reason they call it r(x) is because it's really the radius. This is r².
There is something funny though, pi r² isn't volume. pi r² is area and that's where the weirdness involved in this happens. You got your curve and you rotate it around its axis, rotate it like so, you get a circle. That circle has got an area. Go a little bit further over and rotate another spot around its center. Go a little further over, rotate around its radius. You can see the shape that's starting to emerge? This is in the front here.
[0:04:00]
There is the cone shape. But the weird thing is again, all you really have is stacks of discs. Each of those discs is infinitely thin, so it's got no volume. But if you add up an infinite number of discs across that region, that are infinitely thin, you'll still get volume. That's where the integration comes in.
The integration has to happen from one end of the shape that's being rotated, that's A to the other end of the shape, that's B. So the limits for this always have to be along the direction that you're rotating on. You can rotate on the x axis, you can rotate around other x lines, or you can rotate on the y axis. If you're rotating on the y axis or a y line, this has to be dy and your limits have to be y numbers.
So let's try it. We're going to rotate the shape that we were just looking at around the x axis. The fact that rotating around the x axis makes this a relatively simple problem. Find the volume of the shape obtained when y equals 2² of x is rotated around the x axis and use the limits between 1 and 5.
Here is one of the limits at 1, here is another limit at 5. If I have a disc at 1 and rotate it, it looks something like that. There is the radius. If I have a disc at 5 and rotate it, it's something like that. There is a couple of representative discs.
You notice that I changed the formula this time on the last limit said f(x) there instead of r(x). But remember what I said, radius is what counts. Radius is the distance from wherever you're rotating, to wherever the function is. Now, don't make the mistake though of thinking that it's always going to be found the same way.
[0:06:00]
There is a lot of variations possible in this, that's where the confusion comes in with finding the volume by discs. This one is relatively simple though, the distance from the x axis to the function just is the function. So, r(x), easy enough, it just is 2² of x. If I wanted to be fancy about this I'd say, it's rotating around the axis y equals 0, which is the x axis. So the distance from there is distance you know by subtracting, there is all subtracting whether it's shown. We're subtracting the function from where it's being measured from, or from the axis. So 2² of x minus 0 and it's just really 2² of x.
Later on we will have some problems where we'll be subtracting and you will get an axis that's different, a radius that is different from the function. We'll set this up now. Remember that you've got to set up the integral with the formula. But I feel like making some mistakes here today. So let's see if you can catch them. There's going to be a couple buried in here. Volume we have to do r{x) and r(x) is 2² of x. Dx, have a look through here, there's two mistakes buried in there. The pi is missing, that's one of them. It's easy to forget it. But you really are just doing areas and adding up areas to get volume. So you have to have pi r². That's the other one, there is no square. It's not pi r² yet. So I have to take this quantity and I have to square it.
So, now if I simplify it a little bit, I have pi times integral from 1 to 5. Squaring 2² of x gives you 4x.
[0:08:00]
4x and there's still the dx. Integrating that gives you 2pi x² integrated from 1 to 5. Remember the integral of x is x², but you need the correction factors. So check it by doing the derivatives. Do it once in a while even if you're confident. Derivative of x² is 2x, times that 2 gives you the 4 that you have in there. A couple of substitutions and we are done.
I can have that 2pi factor pulled out front, why not? It will save me some time. And then I'll substitute the 5 in to the x² so that's 5x². I'm using the first fundamental theorem of calculus which says substitute the first thing in and then substitute the second thing in and then substitute the results. So I've got 2 times pi times 5² minus 1². 5² is 25, 1² is 1. 25 minus 1, 24. 24 times 2 pi gives us our final answer a volume of 48 pi.
Onwards and upwards. The last example we did involved rotating around the x axis because the disc method is really oriented toward rotating around the x axis. It does that naturally easily. Rotating around around the other axes is a little harder but we need to try. It's not that much harder. So we are going to try rotating around the y axis. Why? Why not? The y axis. Yeah that was a math joke. I as rotating around a vertical axis. Find the volume of this shape obtained when y equals 1/3x is rotated around the y axis and use limits of 0 to 15.
[0:10:00]
Rotate around the y-axis. We're going to get a cone shape if we do that. Notice that I changed the volume formula bit. Not a lot but a bit, but I had to. Because the direction that you integrate on has to be the one which the discs are stacked up. These are piled up along they y axis, so you have to integrate along the stacks of discs going along the y axis. So that's why it's r(y) dy. And our next problem is that this isn't a function of y, it's a function of x. So I need to change it before I can find the radius and make it a function of y. That's really easy to do though especially for this one.
Y equals 1/3 of x so if that's the case then x has to be 3y, and you should multiply both sides by 3. x equals 3y and now that's a function of y, because you can find x by using a y number. We need to have the radius though. We have the radius in terms of y and the radius is the distance from the axis to what's being rotated. What's being rotated is this. There is the axis, it's going around that vertical axis. Well, if you're measuring from that axis out of the shape, you really are just using the function.
To be technically right though, I'd say in this case that I'm going to take the function and subtract out the axis from it, and then you get a radius of y equals 3y. I've got to tell you something that you have to remember, because it's going to come up near the end of the end of the episode. On the last two that we did, we had the function and we subtracted the axis. So you might be thinking by now, yeah I know what to do, I'm only going to take the function and I'm going to subtract the axis. No, it's not that simple.
[0:12:00]
Sometimes you have to reverse it in kind of strange ways. There's really no predictable way to say what you are going to do because there is a lot of variations possible. The only thing you can say for certain is radius. It's always radius. And you do whatever adding or subtracting is necessary, in whatever order is necessary to find the distance between your axis and the thing that you're rotating. Let's go back to the problem and I'll set it up now.
The volume is the integral of pi and I feel like making mistakes again, bad me. Of course my purpose in making these mistakes, is to see if you can catch them. Because if you can catch then, you're well in your way of understanding what's going on. There's only one mistake in here. Yeah, it's this. 0 and 15. Those are x numbers. We're integrating on y, we need y numbers. When x is 0, 1/3 of 0 is 0 so by coincidence the y number is the same but when x is 15, y is 1/3 of 15 which is just 5. Let's get rid of that. The real limits are 0 to 5. I want to square before I do the integral and while I'm at it to make things a little simpler I know that when I square 3y I'm going to get 9y². I'll put the 9 out front as a factor. That's a nice trick for doing integrals because it makes it a little less confusing if you don't have these constants in there. So we have to integrate that y², dy, and the integral of y² is one power higher, that's y cubed. But of course there is the correction factor of 1/3.
[0:14:00]
So when it's all said and done, we're going to have 9 times the 1/3 which is 3, times pi times y to the 1/3 power. And this one has to be evaluated from 0 up to 5. And I'll leave that for you to try. It's really a pretty quick thing to do. The final result for this is 375 pi.
Time for one more elaboration. Recent AP tests have included problems as complicated as the one we are abut to do so it's best to be prepared. We're going to add two different things in this time. One is that we're going to add in something called washers. Washers are what they sound like. It's a shape that looks like, this, it's a circle with a circle carved in the center, just like the kind of washers that they use in assembling things. So when you're doing this, you're basically doing one shape and subtracting another one. The other elaboration we are going to add is that the axis we rotate on isn't one of the coordinate axis any more.
Let's look at it. Find the volume of the shape obtained when the region defined by all of these is rotated round the line y equals -1. So it's rotated around a y line, that tells me right there that I have to use the version of the formula that integrates on x. Look at this other stuff here. It's not the same formula as it was. It's now capital R² minus little r². In other words, we're doing the volume of the big one and then doing the volume of the little one, and subtracting them out.
Notice there are squares on both of the rs.
[0:16:00]
If you write this, (R-r)² and integrate that, you won't get the same answer. You really could do the two volumes separately and subtract them. But it's often faster to square them, subtract them and then integrate, so let's get rid of this. Drop in all of the axes.
This problem didn't give you the x or y limits, but what it did was give you enough formulas, that you can lock down what the shape is. We have x equals 0, that's this vertical line. We have x equals 3. That's this vertical line. We have f(x) equals x² + 5, it's this parabola. And g(x) equals x + 2, it's that tilted line. So the region that's trapped in there is shaped like that and it's going to be rotated around the axis which is down here. I'm going to draw in some of the shapes now.
If you rotate the outer function, it gives you something like this. If you rotate the inner function, you get something like this. And then you have a nice little washer. You can have lots of these washers, another one like there and there. And if you integrate all those washers up, you're going to get the volume. So, final result is going to be a shape that looks something like that. We've got to get to work now. Remember the last two problems, for the radius, you just use the functions. That's it. It's more complicated this time. Radius is radius. So radius is the distance from the function to the axis.
[0:18:00]
So it's the function minus the axis. Big one first, capital R. R(x), that's the distance from here the axis, to here, the outer function. It's the bigger radius, that's R. Now, if the axes are just the x axis, the distance would be the function, would be x² + 5. But x² + 5, goes something like this. The axis goes something like this. The function result is that length and then you have to add in a bit more because it's farther from the axis from this axis y equals -1 than it is from the function.
So, function minus axis, the function is x² + 5 and I have to subtract the axis so that's -1. It's further away. It's a -1 further way. So our final result for that radius is x² + 6. We'll save that for the set up. Now, we need to find the smaller radius. Small radius is the same thing, it's just a different function minus the same axis. For this one, the smaller function is x + 2. By smaller I mean closer to the axis. And I have to subtract the distance to the axis, so that's minus -1. There's our small one. It is x + 2 minus -1 which is x + 3.
[0:20:00]
So there we got the two parts that we need to go on to the next part. Volume. Both our radii were the large one. The large one was x² + 6, the small one was x + 3 and now we can set it up. Our limits of integration were defined by that left or right vertical lines, so we're ready to go.
Volume is the integral. One of the vertical lines is at 0, the other one was at 3 and I just feel like making mistakes today, see if you can catch it. See if you can catch them all. Look back on your notes for the formula. What did I forget? Something involving areas, it's the squares. And remember, they have to be squared individually like that, and I should really put a big bracket around the whole thing because the whole thing has to be integrated with dx.
Well now, let's go to town on this. We have to square that, square that, subtract the results. So we have pi, integral from 0 to 3. Nice thing though is that, this is just a binomial so when you square x² + 6, you get the first term squared, that's x to the 4th. First term, second terms doubled, that's 12x². You should really memorize the binomial shortcut if you haven't yet. It's super handy. Plus the last thing, squared which is 36 minus.
[0:22:00]
And I really need to have the next parenthesis because if you just square this and leave the minus in front you'll likely to lose some of the minus signs that you need. Squaring x + 3 gives you the first term squared plus the first term last term double that is 6x plus the last term squared, which is 9 and now we're almost up.
So we've got pi times the integral from 0 to 3, put our like terms together. Before I do that I'm going to take that baby step that I think you really should take, even though it seems silly. Of taking that negative in front of the parenthesis and distributing it to everything inside just so you don't mess up a sign. Plus, minus, minus, minus. So x to the 4th, there is no x to the 4th there. 12x² plus -x² is 11x². We have the the plus -6x, we don't want to forget that. And now the 36. So you get 36 plus -9, it is going to be 25. Look at all that stuff. Again, you need to integrate.
Remember this will integrate to be 1/5 of x to the 5th, this will integrate to be 11/3 of x cubed. This will integrate to be -3x², that will integrate to 25x. And after you've substituted all those numbers, all those powers, all those fractions, it does take a while. And if this is on the free response calculator section, you wouldn't bother with this, you'd just type it into the calculator.
[0:24:00]
But if you have to do this manually, this is the result you should get, 1008/5 pi, which is a really big piece of pi, 201.6. We've got enough time to look at one more thing. Take a look.
This one is asking you to find the volume of the shape that's obtained when the region is defined by all that stuff, is rotated around y equals 4. The reason I threw this one in is because, finding the radius on it can be kind of confusing. Let's make a little sketch for this first.
So x, y axis, I need to have a definition of x equals 0, that's one of our boundaries. We have y equals 16, that's another of our boundaries. That's going to be way out here. And the function is the square root of x, and we need one more definition of this boundary. I only have several bounds on that. I don't have quite enough. I don't have a top or bottom bound on there. So I'm going to add that to the problem. I'm going to add that to the problem.
I'm going to add the bound y equals 0. That's just a fancy way of saying the x axis. So here is the shape. One of these bounds is really unnecessary if you think about it. That x equals 0. Because that's the furthest this function can go anyhow. Now let's draw in the axis, the line y equals 4. Here is where the interesting part comes. See when x is 16, you put that into the function, the square of 16 is 4. So height on this spot right here is 4.
[0:26:00]
That means that the axis just touches it and even worse, the axis is above the shape instead of below it. That can make it trickier. You might think, wait a minute. This little problem here, this is just going to be a disc but actually this one is a washer. Believe it or not. Because there's 2 functions that have to be rotated around the axis. There's this one which gives you a nice big circle like so and then there is the smaller one. Imagine if you will the shape that we are getting here is basically a big cylinder and then it's been hollowed out on the inside. It's been hollowed out by this square wood shape.
In a way, imagine that how that square shape is kind of like the bell of a trumpet and then take the bell of the trumpet and then drop it into a cylinder, the volume we're finding is between the bell of the trumpet and the outside of the cylinder. Let's do the set up. The outer unction is the distance between the axis which is at 4. Distance between the axes and this function. This function is just x equals 0 but the radius is still going to be the distance between the axis and the function so in this case, you don;t have to think a lot about that. The distance between y equals 4 and y equals 4. That's the whole outer function. That's okay. Sometimes, they can be that simple.
The inner one. This one can cause you another bit of confusion. On the inner function, it's the distance between the axis and the function. Let me draw another little sketch here so you can see this a little more clearly.
[0:28:00]
Here is our axis at 4, there is our function and little r has to be between the axis and the function. If you use the function, it's not going to give you that, it's going to give you this. That's the function. This is the little radius. But together the function in the little radius make up a total distance of 4, because that's how the axis was. You can get the radius by taking 4 and subtracting the function. Weird, it can happen. It's not always going to be as simple as just taking the function and using the axis. Sometimes you have to do this. 4 minus the function and the function was the square root of x.
Now, I'm going to have this one written out. I'm going to put it into the supplementary materials with all the steps along with the other problems that I've shown you, so you can see everything that's involved in working out the answers.
Today we covered volumes or rotation, which are the volumes that you get when you rotate a shape defined by a certain region around an axis. The first one we did rotated around the x axis, a relatively simple problem. We moved on to one that was a little bit harder, rotating around the y axis. Rotating around the y axis required us to find out the function in terms of a function of y, instead of a function of x.
Next we went to a further elaboration, where we used the washer method. And our axis was not even along the shape. The axis was removed from the shape giving you something like that. Basically the volume minus the volume.
[0:30:00]
The last one we looked at, was one where the axis was attached to the shape and it wasn't on the x axis. So that one again was just one step harder. Well, you are guaranteed to see volumes of rotation on the AP test. Most likely in the free response section. They may ask you one as difficult as where you need to find the intersection between the two functions in order to find the limits of integration. Look in the bonus materials for a practice problems along those lines, including all the steps.
You might also believe it or not, find one of these in the multiple choice. If you do, it might be as simple as a problem that asks you to choose between 5 different possible set ups for finding the volume, in which case, it's pretty quick, you just need to write it out. I suggest writing it out on your own before you even look at the multiple choice answers, because sometimes seeing a multiple choice answer can influence you in the wrong direction. Again, thanks for tuning in and remember keep turning. | 5,629 | 22,831 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2023-40 | latest | en | 0.950157 |
http://www.onlinemathlearning.com/special-triangles-geometry.html | 1,510,978,534,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804610.37/warc/CC-MAIN-20171118040756-20171118060756-00342.warc.gz | 468,364,764 | 9,258 | # Special Triangles in Geometry
Related Topics:
More Lessons for High School Geometry
Math Worksheets
A series of free, online High School Geometry Video Lessons.
Videos, worksheets, and activities to help Geometry students.
In this lesson, we will learn
• 45-45-90 special triangles
• 30-60-90 special triangles
### 45 45 90 Triangles
A 45 45 90 triangle is a special type of isosceles right triangle where the two legs are congruent to one another and the non-right angles are both equal to 45 degrees. Many times, we can use the Pythagorean theorem to find the missing legs or hypotenuse of 45 45 90 triangles. The ratio of the sides to the hypotenuse is always 1:1:square root of two.
How to find the length of a leg or hypotenuse in a 45-45-90 triangle using the Pythagorean Theorem.
Introduction to 45-45-90 Triangles
### 30 60 90 Triangles
A 30 60 90 triangle is a special type of right triangle. What is special about 30 60 90 triangles is that the sides of the 30 60 90 triangle always have the same ratio. Therefore, if we are given one side we are able to easily find the other sides using the ratio of 1:2:square root of three. This special type of right triangle is similar to the 45 45 90 triangle.
How to find the legs and hypotenuse in 30-60-90 triangles when given: the short leg, the long leg, or the hypotenuse.
This video provides examples of how to solve a 30-60-90 triangle given the length of one side.
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. | 453 | 1,839 | {"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-2017-47 | latest | en | 0.842642 |
http://www.reddit.com/r/math/comments/110ejh/the_egg_bizarre_behavior_of_the_roots_of_a_family/?sort=new | 1,427,492,925,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131296951.54/warc/CC-MAIN-20150323172136-00287-ip-10-168-14-71.ec2.internal.warc.gz | 756,776,072 | 14,011 | [–][deleted] (1 child)
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[–] 9 points10 points (2 children)
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I don't know why, but I do have some guesses for where I would start to look.
First, looking at this animation I might guess that the sequence of polynomials f_n(z) converges to an analytic function on the open unit disk. Now, Jentzsch's theorem says that, for a power series with radius of convergence 1, every point on the circle |z| = 1 is a limit point of the zeros of the partial sums of the power series. So, these polynomials might be converging in a nice enough way so that, even though they're not partial sums of whatever power series, the proof of Jentzsch's theorem could apply to this case with small modification. Based on my experience, I would even hazard a guess that the limit function has a pole at z=1 which causes the roots of f_n(z) to "stay away".
A priori, there's also the fact that the zeros of random polynomials are expected to cluster on the unit circle as the degree goes to infinity. See this paper. For this particular problem, if one could get a good enough handle on the coefficients of the polynomials in question, one might be able to directly apply Theorem 4 from that linked paper to get the result.
Edit: After actually calculating some of these polynomials, it looks it's actually the reciprocal polynomials which are converging on the open unit disk. However there are two limit functions, one for n even and one for n odd. That is, if q_n(z) = zdeg(p_n) p_n(1/z), then it looks like
limit as n->infinity q_(2n)(z) = 1 + 3z + 4z2 + 7z3 + 13z4 + 19z5 + 29z6 + 43z7 + ...
and
limit as n->infinity q_(2n+1)(z) = 2 + 2z + 6z2 + 8z3 + 14z4 + 20z5 + 34z6 + 46z7 + ...
In other words, the original polynomials are converging outside the unit disk. Proving this probably isn't impossible.
[–] 1 point2 points (1 child)
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This is interesting, did you see what I wrote about the coefficients in the last edit? Those numbers look very similar to what I've been seeing. I'm trying to figure out what the correcting terms are during the convergence.
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Yes, and I think looking for recurrence relations is definitely a good place to start.
At this point I'll have to use LaTeX, so if the following doesn't render make sure your greasemonkey plugin is working ok.
If
[; \lim_{n \to \infty} q_{2n}(z) = \sum_{k=0}^{\infty} a_k z^k ;]
then it looks like
[; q_{2n}(z) = \sum_{k=0}^{n} a_k z^k + O(z^{n+1}). ;]
Similarly, if
[; \lim_{n \to \infty} q_{2n+1}(z) = \sum_{k=0}^{\infty} b_k z^k ;]
then it looks like
[; q_{2n+1}(z) = \sum_{k=0}^{n} b_k z^k + O(z^{n+1}). ;]
If you could show that the polynomials q_n(z) look like these expressions (i.e. that they stabilize at the front) then that would at least give you that the converge on the open unit disk (since the error would go to zero as n goes to infinity).
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Rounding error.
[–] 5 points6 points (0 children)
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No no, I mean, why the solutions are all approximately of magnitude 1? | 1,000 | 3,506 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2015-14 | latest | en | 0.947175 |
https://www.vedantu.com/question-answer/mark-the-correct-alternative-of-the-following-if-class-10-maths-icse-5ee74e274582936f5421c069 | 1,621,382,240,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989874.84/warc/CC-MAIN-20210518222121-20210519012121-00040.warc.gz | 1,109,389,433 | 49,660 | Question
# Mark the correct alternative of the following.If the number 2345 a 60b is exactly divisible by 3 and 5, then the maximum value of $a+b$is?(a) 12(b) 13(c) 14(d) 15
Hint: We can use the divisibility rule for the numbers 3 and 5 to solve this question. In the case of number 5, the value of b can only be either 0 or 5. Similarly, we can apply conditions for number 3 also and get to a conclusion.
Before proceeding we should know the divisibility rule of 3 as well as 5. These are as follows:
Divisibility rule of 3: A number is divisible by 3 if and only if the sum of the digits of a number is divisible by 3.
Divisibility rule of 5: A number is divisible by 5 if and only if the last digit of a number is either 0 or 5.
Therefore, the number 2345 a 60b is divisible by 5 only if the value of $b=0\text{ or 5}$.
We have to take the value of b as 5 as we have been asked the maximum value of $a+b$.
Then the sum of the digits of the given number 2345 a 60b is $25+a$.
Therefore, the numbers greater than 25 and divisible by 3 are 20, 30, 35.
Hence, the value of a maybe 2, 5, 8 .
By comparing, the possible values of a, we get that the maximum value of $a=8$
Therefore, we have $a=8,b=5$.
Hence, the maximum value of $a+b$ is $13$.
Hence, the answer is option (b).
Note: Always remember we have been asked for the maximum value of $a+b$, therefore for the number to be divisible by 5, the last digit must be 5 not 0 as we have been asked the maximum value of $a+b$. Make sure that after getting the maximum values of $a+b$, check at least once the number is divisible by both 3 and 5. | 469 | 1,608 | {"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} | 4.46875 | 4 | CC-MAIN-2021-21 | longest | en | 0.915499 |
https://www.bartleby.com/essay/Probability-of-Blackjack-FK9ULK43RZZA | 1,547,818,021,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583660139.37/warc/CC-MAIN-20190118131222-20190118153222-00037.warc.gz | 728,758,379 | 12,704 | # Probability of Blackjack
2190 Words Mar 23rd, 2013 9 Pages
Mathematical Studies Project
Probability of Blackjack
Content Page
Page
Introduction 3 - 4
Data collection 5 - 6
The four Blackjack strategies 7 - 15
Conclusion 16
Bibliography 17
This project aims to investigate how mathematics works in one of the most popular card games in the world - Blackjack. It examines the probability in the Blackjack and how the use of mathematics can be
Blackjack strategies
Strategy I Hit when the sum of the first two cards is less than the dealer’s upcard
Let x be the player’s sum of card values of the first two cards.
Let D be the dealer’s upcard(the second card).
If x < D, the player should always hit.
When x < D, it is obvious that x must be less than the dealer’s final value (covered card + upcard), given that the sum of the first two cards of the dealer is[pic]- no busting. Therefore, under this situation, the player must hit. Otherwise, his probability of winning is 0.
Strategy II Hit or Stand
If the hard total is 11 or less, the player should always hit. This is because no matter what card the player is getting, the total hand value will not exceed 21. By using the Hit ot Stand strategy, the house edge can be cut to as low as 2%.[1] Fig. 1 is the decision making chart when using this strategy.
Card | |A |2 |3 |4 |5 |6 |7 |8 |9 |10 |J |Q |K | | |Real value |1 |11 |2 |3 |4 |5 |6 |7 |8 |9 |10 |10 |10 |10 | |A |1 |2 |12 |3 |4 |5 |6 |7 |8 |9 |10 |11 |11 |11 |11 | | |11 |12 |22 |13 |14 |15 |16 |17 |18 |19 |20 |21 |21 |21 |21 | |2 |2 |3 |13 |4 |5 |6 |7 |8 |9 |10 |11 |12 |12 |12 |12 | |3 |3 |4 |14 |5 |6 |7 |8 |9 |10 |11 |12 |13 |13 |13 |13 | |4 |4 |5 |15 |6 |7 |8 |9 |10 |11 |12 |13 |14 |14 |14 |14 | |5 |5 |6 |16 |7 |8 |9 |10 |11 |12 |13 |14 | 571 | 1,773 | {"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} | 4.53125 | 5 | CC-MAIN-2019-04 | longest | en | 0.866121 |
http://mathoverflow.net/revisions/96244/list | 1,369,137,474,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368699924051/warc/CC-MAIN-20130516102524-00064-ip-10-60-113-184.ec2.internal.warc.gz | 171,995,972 | 5,421 | Maybe it would have been more convincing if you had given the definition of $q_K$ in a unimodularly equivariant way from the beginning. That's not at all hard to do, and it's also easy to see how to create many more such (and how to create many more in all dimensions).
For example, suppose $K\subset V$ has vertices $v_0,v_1,\ldots,v_n=v_0$ (cyclically ordered counterclockwise, say). Let $\alpha_i\in V^\ast$ be the unique element that satisfies $\alpha_i(v_i)=\alpha_i(v_{i+1})=1$. Then $$q_K = \sum_{i=0}^{n-1} \Omega(v_i,v_{i+1})\ {\alpha_i}^2$$ (where $\Omega$ is the area form.) This is clearly unimodularly equivariant with respect to $K$ and, since, for $K' = tK$ (with $t>0$), one has $v'_i = tv_i$ and $\alpha_i' = (1/t)\alpha_i$, it follows that $q_{tK} = q_K$. Of course, anything like this would have worked. For example, you could have taken $$\tilde q_K = Area_\Omega(K)\left( \sum_{i=0}^{n-1} {\alpha_i}^2\right),$$ and this would also have had the same equivariance property.
In higher dimensions, dimension $n$, I think that the right formula would be to define, for each face $F$ of $K$, the element $\alpha_F\in V^\ast$ to be the linear function that equals $1$ on $F$, let $\Omega(F)$ denote the volume of the cone with vertex $0\in V$ whose base is $F$, and then set $$q_k q_K = \sum_{F\in\mathcal{F}(K)} \Omega(F)\ {\alpha_F}^2.$$ If you want it to be invariant under scaling, you should take $$q_k q_K = \sum_{F\in\mathcal{F}(K)} \Omega(F)^{2/n}\ {\alpha_F}^2.$$ Perhaps, better, though, would be to take $$q_K = Vol_\Omega(K)^{(2-n)/n}\ \left(\sum_{F\in\mathcal{F}(K)} \Omega(F)\ {\alpha_F}^2\right),$$ since this is also invariant under subdivision of the faces of $K$.
1
Maybe it would have been more convincing if you had given the definition of $q_K$ in a unimodularly equivariant way from the beginning. That's not at all hard to do, and it's also easy to see how to create many more such (and how to create many more in all dimensions).
For example, suppose $K\subset V$ has vertices $v_0,v_1,\ldots,v_n=v_0$ (cyclically ordered counterclockwise, say). Let $\alpha_i\in V^\ast$ be the unique element that satisfies $\alpha_i(v_i)=\alpha_i(v_{i+1})=1$. Then $$q_K = \sum_{i=0}^{n-1} \Omega(v_i,v_{i+1})\ {\alpha_i}^2$$ (where $\Omega$ is the area form.) This is clearly unimodularly equivariant with respect to $K$ and, since, for $K' = tK$ (with $t>0$), one has $v'_i = tv_i$ and $\alpha_i' = (1/t)\alpha_i$, it follows that $q_{tK} = q_K$. Of course, anything like this would have worked. For example, you could have taken $$\tilde q_K = Area_\Omega(K)\left( \sum_{i=0}^{n-1} {\alpha_i}^2\right),$$ and this would also have had the same equivariance property.
In higher dimensions, I think that the right formula would be to define, for each face $F$ of $K$, the element $\alpha_F\in V^\ast$ to be the linear function that equals $1$ on $F$, let $\Omega(F)$ denote the volume of the cone with vertex $0\in V$ whose base is $F$, and then set $$q_k = \sum_{F\in\mathcal{F}(K)} \Omega(F)\ {\alpha_F}^2.$$ If you want it to be invariant under scaling, you should take $$q_k = \sum_{F\in\mathcal{F}(K)} \Omega(F)^{2/n}\ {\alpha_F}^2.$$ | 1,029 | 3,169 | {"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} | 3.6875 | 4 | CC-MAIN-2013-20 | latest | en | 0.922699 |
http://web.eecs.umich.edu/~akamil/teaching/su11/lec/lec15.html | 1,642,919,739,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304134.13/warc/CC-MAIN-20220123045449-20220123075449-00439.warc.gz | 74,853,167 | 4,255 | ```Administrative info
PA2 due Monday
HW5 due Tuesday
Review
Recall that a probability space consists of the following:
(1) A random experiment.
(2) The sample space (set of possible outcomes).
(3) The probability of each possible outcome of the experiment.
Further recall that the probabilities must satisfy the following:
(1) ∀ω∈Ω . 0 <= Pr[ω] <= 1
(2) ∑_{ω∈Ω} Pr[ω] = 1
An event is a subset of the sample space, i.e. a set of outcomes
from the sample space. The probability of an event E is the sum of
the probabilities of the outcomes in E:
Pr[E] = ∑_{ω∈E} Pr[ω].
In the case of a uniform distribution, this simplifies to
Pr[E] = |E|/|Ω|.
Probability Identities
Before we move on, let's note some facts about probability that can
make it easier to compute probabilities.
We defined the complement of an event E as E = &Omega\E. Then,
Pr[E] = 1 - Pr[E]. Proof:
∑_{ω∈Ω} Pr[ω]
= ∑_{ω∈E;} Pr[ω] +
∑_{ω∈Ω\E} Pr[ω]
= Pr[E] + Pr[E] = 1.
Let A and B be events in Ω. Then Pr[A ∪ B] = Pr[A] + Pr[B]
- Pr[A ∩ B]. Writing this out in terms of sums, we get
∑_{ω∈Ω} Pr[ω]
= ∑_{ω∈A;} Pr[ω] +
∑_{ω∈B} Pr[ω] -
∑_{ω∈A ∩ B} Pr[ω]
As in inclusion/exclusion for sets, the first two terms double count
the probabilities of those outcomes in A ∩ B, so we have to
subtract the probability of A ∩ B.
EX: What is the probability that a random integer n between 1 and
100 is divisible by 5 or 7?
ANS: Let A be the event that n is divisible by 5, B be the event
that it is divisible by 7. Pr[A] = 1/5, Pr[B] = 14/100 = 7/50,
and Pr[A ∩ B] = 2/100 = 1/50. So Pr[A ∪ B] = 1/5 + 7/50
- 1/50 = 16/50 = 8/25.
Let A_1, ..., A_n be n mutually disjoint events in Ω. Then
Pr[A_1 ∪ ... ∪ A_n] = Pr[A_1] + ... + Pr[A_n]. This follows
from the above, generalized to n events using induction, and then
removing the intersection terms which are all 0.
EX: Suppose I roll a red and a blue die. What is the probability that
the red die is less than 4?
ANS: Let A_i be the event that the red die is i. Then Pr[A_i] = 1/6
for 1 <= i <= 6, and the A_i are mutually disjoint. Thus,
Pr[A_1 ∪ A_2 ∪ A_3] = Pr[A_1] + Pr[A_2] + Pr[A_3] =
1/2.
Conditional Probability
A pharmaceutical company is marketing a new test for HIV that it
claims is 99% effective, meaning that it will report positive for
99% of people who have HIV and negative for 99% of those who don't
have HIV. Suppose a random person takes the test and gets a positive
test result. What is the probability that the person has HIV?
This is an example of conditional probability. Given some
information about a particular event in the sample space, we want to
compute new probabilities for other events.
Let's start off with simpler examples before coming back to the
above.
EX: Suppose I flip a fair coin twice. The result of the first flip
is heads. What is the probability that I got two heads?
ANS: Let's start by drawing the sample space Ω. There are 4
equally likely outcomes HH, HT, TH, and TT. We are now told
that event A = "the first flip is H" has occurred. Which
outcomes are now possible? There are only 2 outcomes in A, HH
and HT, each of which is equally likely. So we have a new
sample space Ω' that consists of just the outcomes HH and
HT, each with probability 1/2. Let event B = "both flips are
heads." What is the probability of B in this new sample space?
Only one of the two outcomes in Ω' is in B, so Pr[B] =
1/2 in the new sample space. We write this as Pr[B|A], "the
probability of B given A," which is the probability of B
occurring in a new sample space consisting of just those
outcomes in A.
Generalizing the above procedure, suppose we are told an event A
occurs. Then what is the new conditional probability of each outcome
ω, i.e. Pr[ω|A]? For ω ∉ A, this is clearly
0. For ω ∈ A, the relative likelyhood of any two outcomes
in A should remain the same, but we need to renormalize so that we
satisfy the requirement that all probabilities add to 1. By
definition, we had ∑_{ω ∈ A} Pr[ω] = Pr[A], so
if we normalize by dividing by Pr[A], i.e. Pr[ω|A] =
Pr[ω]/Pr[A], we get ∑_{ω ∈ A} Pr[ω|A] =
∑_{ω ∈ A} Pr[ω]/Pr[A] = Pr[A]/Pr[A] = 1.
Now suppose we have another event B. What is Pr[B|A]? The outcomes
in B that are not in A contribute nothing, since their new
conditional probabilities are 0. So only the outcomes in both B and
A contribute any probability, and we get Pr[B|A] = ∑_{ω
∈ B ∩ A} Pr[ω|A] = ∑_{ω ∈ B ∩ A}
Pr[ω]/Pr[A] = Pr[B ∩ A]/Pr[A].
To summarize, when conditioning on an event A, we cross out any
possibilities that are incompatible with A and then renormalize by
1/Pr[A] so that the probabilities of the remaining outcomes add to
1. We can compute the probabilities of events directly in this new
sample space or use the identities above to get the same result.
EX: Suppose I toss a red and a blue die, and I tell you that the
resulting sum is 4. What is the probability that the red die is
1?
ANS: Let A be the event that the sum is 7, B be the event that the
red die is 1. The outcomes (1, 3), (2, 2), and (3, 1) are in A,
so Pr[A] = 1/12. What is Pr[B ∩ A]? Only the outcome (1, 3)
is in B ∩ A, so Pr[B ∩ A] = 1/36. Then Pr[B|A] = Pr[B
∩ A]/Pr[A] = 1/3.
We could also have redefined the sample space to come up with
the same result. Given A, we have a new sample space Ω'
consisting of the outcomes (1, 3), (2, 2), and (3, 1), each
with probability 1/3. Then B has probability 1/3 in this new
sample space. So Pr[B|A] = 1/3.
EX: Suppose I toss a red and a blue die, and I tell you that the
resulting sum is 7. What is the probability that the red die is
1?
ANS: Let A be the event that the sum is 7, B be the event that the
red die is 1. Pr[A] = 1/6 as we computed before. What is Pr[B
∩ A]? Only the outcome (1, 6) is in B ∩ A, so Pr[B
∩ A] = 1/36. Then Pr[B|A] = Pr[B ∩ A]/Pr[A] = 1/6.
EX: Suppose I toss 3 balls into 3 bins (with replacement). Let
A = "1st bin empty," B = "2nd bin empty." What is Pr[A|B]?
ANS: Pr[B] = 2^3/3^3 = 8/27, Pr[A ∩ B] = 1/3^3 = 1/27, so
Pr[A|B] = (1/27)/(8/27) = 1/8.
Thus, the fact that the 2nd bin is empty makes it much less
likely that the 1st one is as well.
EX: Suppose I flip a fair coin 51 times. If the first 50 flips are
heads, what is the probability that the 51st is heads?
ANS: Let A be the event that the first 50 flips are heads, B be the
event that the 51st is heads. There are only 2 outcomes in A
out of 2^51, so Pr[A] = 1/2^50. There are 2^50 outcomes in B,
so Pr[B] = 1/2. Only one outcome is in both A and B, so Pr[A
∩ B] = 1/2^51. Then Pr[B|A] = (1/2^51)/(1/2^50) = 1/2.
So the first 50 flips tell us nothing about the 51st; the
probability of heads is still 1/2.
We have seen multiple examples where Pr[B|A] = Pr[B]. We say that A
and B are "independent" of this is the case. Intuitively, two events
A and B are independent of knowing that one happens does not change
the likelihood of the other happening. So the 51st flip of a fair
coin is independent from what came up before.
If A and B are independent, we get Pr[B|A] = Pr[B ∩ A]/Pr[A] =
Pr[B], so Pr[B ∩ A] = Pr[A] Pr[B]. This is a very useful
identity.
EX: Suppose I flip a coin with probability p of heads n times. What
is the probability of a particular outcome with k heads?
ANS: Each flip is independent, with probability p for heads. The k
heads flips have probability p, and the n-k tails flips have
probability (1-p). So we get p^k (1-p)^(n-k) for an outcome
with k heads.
EX: Suppose a casino advertises the following game. You pick a
number from 1 to 6. The casino rolls three dice, and if your
number comes up, you win. What is your probability of winning?
ANS: It's not 1/2! Let A_i be the event that your number comes up
on the ith die. We want to know
Pr[A_1 ∪ A_2 ∪ A_3]
= 1 - Pr[A_1 ∩ A_2 ∩ A_3]
= 1 - Pr[A_1] Pr[A_2] Pr[A_3]
= 1 - (5/6)^3 ≈ 1 - 0.58 = 0.42.
In the third line above, we used the fact that the results of
each dice are mutually independent. We will come back to the
concept of mutual independence later.
So your probability of winning is less than 1/2.
Suppose you are flying to Las Vegas (in order to play the game
above). Your friend, fearing for your safety, gives you the
following advice: "You know, you should always carry a bomb on an
airplane. The chance of there being one bomb on the plane is pretty
small, but the chance of two bombs is miniscule. So by carrying a
bomb on the airplane, your chances of being blown up are
astronomically reduced." What do you think of his advice?
Let A be the event that you carry a bomb on board, B be the event
that someone else carries a bomb on board. How are A and B related?
They are independent, so Pr[B|A] = Pr[B], and the likelihood that
someone else has a bomb doesn't change one bit if you bring one
aboard.
``` | 2,769 | 8,698 | {"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-2022-05 | latest | en | 0.896044 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-3rd-edition/chapter-7-exponential-functions-7-1-derivative-of-f-x-bx-and-the-number-e-exercises-page-327/16 | 1,701,451,255,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100290.24/warc/CC-MAIN-20231201151933-20231201181933-00218.warc.gz | 877,395,807 | 13,655 | ## Calculus (3rd Edition)
$$y=4x+1.$$
We have $$y'=4e^{4x}\Longrightarrow y'(0)=4.$$ Then the slope of the tangent line at $x=0$ is $m= 4$. Hence the equation of the tangent line is given by $$y=mx+c=4x+c.$$ Since $y(0)=1$, then $c=1$, hence the tangent line at $x=0$ is given by $$y=4x+1.$$ | 111 | 292 | {"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} | 3.9375 | 4 | CC-MAIN-2023-50 | latest | en | 0.817772 |
https://testbook.com/question-answer/1-4-27-256-fill-in-the-blanks--6300c34491335bf7336640dc | 1,679,306,655,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943471.24/warc/CC-MAIN-20230320083513-20230320113513-00137.warc.gz | 622,848,688 | 79,033 | # 1, 4, 27, 256, .............. Fill in the blanks.
This question was previously asked in
GPSC AE CE 2017 Official Paper (Part A - General)
View all GPSC Assistant Engineer Papers >
1. 525
2. 3125
3. 725
4. 825
Option 2 : 3125
Free
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1.5 K Users
200 Questions 200 Marks 120 Mins
## Detailed Solution
Given:
1, 4, 27, 256, ..............
Calculation:
11 = 1
22 = 4
33 = 27
44 = 256
55 = 3125
∴ The next term is 3125. | 176 | 473 | {"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-2023-14 | latest | en | 0.65795 |
https://www.storyboardthat.com/storyboards/b2377285/unknown-story | 1,721,677,588,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517915.15/warc/CC-MAIN-20240722190551-20240722220551-00693.warc.gz | 853,189,045 | 77,887 | • My Storyboards
# Unknown Story
Create a Storyboard
#### Storyboard Text
• My name is Emily and I am going to tour the city! Help me figure out the lengths and angles of different objects in the city using trigonometry.
• Here we have a window and we are trying o the angle of depression from the window. The height of the building is 15 ft tall and the object below is 17 ft away. Here we use the inverse of sine. Sine-1 = 15/17. The angle of depression is 62 degrees.
• x
• 17 ft
• x
• 15 ft
• 37
• x
• If we look here we are trying to find the height of the light. The length from the light to the bench is 5 ft and the angle from the bench to the post is 37 degrees. We are going to use cosine of 37= x/5. The height of the light is 4 ft.
• 5 ft
• The height from the second floor window is 8 ft tall and the hypotenuse is 10 ft long. To find the width let’s use Pythagorean theorem. 8^2 plus b^2 =10^2. So this means B = 6. The width of the building is 6 ft.wide.
• Let’s see how wide this building is!!!
• 8 ft
• b
• 10 ft
• 7 ft
• Ooo look at this staircase let’s see the width of these. So the length of this triangle is 7ft and the angle of elevation is 68 degrees. To find the width ground floor length we will have to use tan68 = 5/x. If we solve this the width of the triangle will be 2.8ft.
• x
• 68
• oh look it's a fire station I wonder what the length would be of this building if we leaned a ladder to the top of the building. The ladder is 12 ft leaning towards the top of the building which has an angle of depression of 36 degrees. To find the length we would use sine 36= x/12. After we solve this we know the fire station 's length is 7.1 ft.
• 36
• 12 ft
• 36
• x
• We are all done, I loved exploring this city and getting to find out new things, I hope you liked it too bye!!!
Over 30 Million Storyboards Created | 693 | 2,024 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-30 | latest | en | 0.873133 |
https://cs.stackexchange.com/questions/138455/bipartite-graph-to-solve-the-wolf-river-crossing-problem/138801#138801 | 1,632,269,675,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057274.97/warc/CC-MAIN-20210921221605-20210922011605-00260.warc.gz | 241,351,256 | 39,364 | Bipartite Graph to solve the wolf river crossing problem
I have just started studying about bipartite graphs and there is an example that bipartite graph can be use to solve the wolf, cabbage, and the sheep river crossing problem, as a kid i had fun solving this question, now I was wondering how can we describe the problem in bipartite graph, as to describe it we need to divide the problem into two disjoint and independent sets, I am getting very confused as how it can be done?
It will be helpful, if someone can help me understand it as there is no information about this wolf river crossing problem on the internet.
• The visualisation on wikipedia shows the problem in the form of a cube. That cube is a bipartite graph. Apr 3 at 11:11
• ok, the cube after crossing out the invalid situation is bipartite
– kiv
Apr 3 at 11:26
• @kiv. The cube is bipartite even without deleting the invalid edges. Apr 3 at 13:51
I have never really imagined solving this problem with a graph, but it is an interesting point of view.
My guess is that the graph you have to create is a bipartite graph with 16 vertices, each vertex representing the state of the situation, that is representing where is each of the different actors of the problem: the wolf, the goat, the cabbage and the boat. Each actor have two positions possibles (each side of the river), so $$2^4 = 16$$ different configurations. You might then delete forbidden configurations, representing a case where the wolf is alone with the goat or the goat alone with the cabbage.
Now, you can add an edge between two vertices $$v$$ and $$w$$ if and only if you can go from $$v$$ to $$w$$ by crossing the river with the boat, with one actor from wolf/goat/cabbage on board.
The graph is indeed bipartite, because there are no edges between two vertices where the boat is on the same side.
Solving the problem now results in finding a path between the two vertices where the four actors are on each side of the river.
• If we do not take boat in the consideration,will we have a graph with 8 vertices?
– kiv
Apr 3 at 11:17
• Well yes, that seems to be also a possibility, but I'm not sure the resulting graph will be bipartite. Apr 3 at 11:18
• How would the final graph with 16 vertices would look like, will it be 8 vertices on each side?
– kiv
Apr 3 at 11:21
• You can imagine a binary encoding $(b_0b_1b_2b_3)$ where the indexes $0, 1, 2, 3$ stand for wolf, goat, cabbage, boat, and $b_i = 0$ if $i$ is on the first side of the river, and $b_i=1$ otherwise. Each side of the graph would then correspond to $b_3 = 0$ and $b_3 = 1$. Apr 3 at 11:27
I think the best bet is to create a set of vertices which are permissible states (on one bank or the other), with the edges representing permitted transitions between them. Note that the presence of the farmer is also part of the state, so the two parts of the bipartite graph are nodes with & without the farmer. So this is a variation on the previous answer that limits the created nodes. Using the state on the starting bank and coding the presence or absence of each component in to a binary tuplet (farmer, wolf, sheep, cabbage) we have the permitted states: $$\begin{array}{c} & \text{with farmer} & \text{without farmer} \\ \text{initial state}& (1,1,1,1) & \\ & (1,1,1,0) & \\ & (1,1,0,1) & (0,1,0,1)\\ & & (0,1,0,0) \\ & (1,0,1,1) & \\ & (1,0,1,0) & (0,0,1,0) \\ & & (0,0,0,1) \\ & & (0,0,0,0) &\text{target state}\\ \end{array}$$
(As you may notice, the no-farmer states govern what is permissible; by inference, the state on the other bank eliminates some with-farmer states). Then the edges represent feasible journeys:
, and the problem is finding a path between the initial and target states. | 1,001 | 3,719 | {"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": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2021-39 | latest | en | 0.954365 |
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# In a sequence of terms in which each term is three times the
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In a sequence of terms in which each term is three times the previous term, what is the fourth term?
(1) The first term is 3.
(2) The second to last term is 3^10.
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30 Oct 2006, 07:35
))tricky question
IMHO A it is
1) is obvious
1st term is 1
second=3
third=9
and fourth =27
2)tricky The second to last term is 3^10.
insuff since we don't know how many terms there are in the sequence
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30 Oct 2006, 07:57
This Seq is defined by
S={ n, 3n, 3^2 n. 3^3 n.....
From S1, n=3, so 4th term will be 81.
From S2, second last term is 3^10...but,since total nbr of terms in seq not given, we cannot say which is the 4th term.
Hence A.
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30 Oct 2006, 23:27
only A.
2 is not suff, because the first could be 3, or 9, or 27 etc.
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30 Oct 2006, 23:40
Yurik79 wrote:
Priyah wrote:
This Seq is defined by
S={ n, 3n, 3^2 n. 3^3 n.....
From S1, n=3, so 4th term will be 81.
From S2, second last term is 3^10...but,since total nbr of terms in seq not given, we cannot say which is the 4th term.
Hence A.
Oops you are right))it is 81
Good catch Yurik. Yes, for statement (2) we don't know the total number of terms..
(A)
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31 Oct 2006, 02:30
You're right
must be A
for B, if first term n=1 then i(nb of terms)= 12 since the value of last term is determined by: n*3^(i-1)
if n=3 then i=11
so we cannot determin the 4th term value
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31 Oct 2006, 04:19
this is a GP(geometric progression)
A stands
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31 Oct 2006, 11:04
Code:
this is a GP(geometric progression)
Trivikram,
Can you post the formulae for GP??
Like what we have for AP - the last term, Sum of n terms etc...
What formulae do we have for GP?
Also, HP is just the sum of reciprocal of terms in an AP.
Do we have formulae for that as well?
31 Oct 2006, 11:04
Display posts from previous: Sort by | 1,232 | 3,812 | {"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} | 3.921875 | 4 | CC-MAIN-2017-04 | latest | en | 0.920417 |
https://education.blurtit.com/3767780/if-y-is-inversely-proportional-to-the-square-of-x-and-y-19-when-x-4-find-y-when-x-12 | 1,620,405,471,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988796.88/warc/CC-MAIN-20210507150814-20210507180814-00150.warc.gz | 248,890,088 | 11,164 | # If y is inversely proportional to the square of x and y =1/9 when x =4, find y when x = 12?
When y is inversely proportional to the square of x, it satisfies the equation
y = k/x^2
for some constant k. Using the given information, we can find k to be
k = y*x^2 = (1/9)*4^2 = 16/9
Then when x=12, we have
y = (16/9)/12^2 = 16/(9*16*9) = 1/81
thanked the writer.
When y is inversely proportional to the square of x, it satisfies the equation
y = k/x^2
for some constant k. Using the given information, we can find k to be
k = y*x^2 = (1/9)*4^2 = 16/9
Then when x=12, we have
y = (16/9)/12^2 = 16/(9*16*9) = 1/81
thanked the writer. | 236 | 645 | {"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-2021-21 | latest | en | 0.840723 |
https://www.sarthaks.com/2585524/oven-dried-sizes-shown-table-taken-hydrometer-analysis-1000-suspension-prepared-filled | 1,679,666,807,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945282.33/warc/CC-MAIN-20230324113500-20230324143500-00355.warc.gz | 1,087,780,938 | 16,088 | # 50 gm of oven dried soil of five sizes shown in the table is taken for hydrometer analysis and 1000 cc of suspension is prepared and filled in a jar u
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50 gm of oven dried soil of five sizes shown in the table is taken for hydrometer analysis and 1000 cc of suspension is prepared and filled in a jar up to 30 cm height.
Particle size (mm) Weight (gm) 0·06 6 0·05 20 0·03 15 0·015 5 0·004 4
If the specific gravity of soil solids is 2.7 and viscosity of soil water suspension is 1× 10-6 kN-sec/m2, the time required for all the particles to settle to the bottom of the jar is
1. 2 hr 16 min
2. 5 hr 37 min
3. 11 hr 55 min
4. 23 hr 50 min
## 1 Answer
0 votes
by (35.4k points)
selected
Best answer
Correct Answer - Option 2 : 5 hr 37 min
Concept
Given,
Height of jar = 30 cm
From stokes law
We know the velocity of settling particle is
${V_s} = \frac{{{\gamma _w} \times {d^2} \times \left( {G - 1} \right)}}{{18 \times \mu }}$
Where,
γw is the unit weight of water
d is the diameter of the particle
G is the specific gravity of the particle
μ is the viscosity of the fluid
Note: If the lighter particle from the group of particles settles at the bottom that means every particle is settled at the bottom.
Particle size (mm) Weight (gm) 0·06 6 0·05 20 0·03 15 0·015 5 0·004 4
∴ Here diameter (d) = 0.004 mm is taken from the group
$Time\;required\;for\;settling\;of\;all\;particle = \frac{{Height\;of\;water\;level\;in\;jar}}{{Settling\;Velocity\;}}$
$t = \frac{{30 \times {{10}^{ - 2}}}}{{14.284 \times {{10}^{ - 6}}}}$
= 20215 sec
= 5.61 hr
= 5 hr 37 min
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1 answer | 574 | 1,707 | {"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} | 4.0625 | 4 | CC-MAIN-2023-14 | latest | en | 0.802657 |
https://www.emmamichaels.com/1117/lcm-of-6-and-2.html | 1,679,582,153,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945168.36/warc/CC-MAIN-20230323132026-20230323162026-00183.warc.gz | 832,011,171 | 14,501 | Breaking News
# Lcm Of 6 And 2
Lcm Of 6 And 2. Find the prime factorization of 6. You can find lcm of 6, 2, 5, 7 by simply giving the inputs in the input field and clicking on the calculator button next to it. For 6 and 2 those factors look like this: Prime factors of 6 are 2,3.
The lcm of 6 and 9 is 18.to find the lcm (least common multiple) of 6 and 9, we need to find the multiples of 6 and 9 (multiples of 6 = 6, 12, 18, 24; Find the prime factorization of 2. Use the lcm by gcf formula to calculate (6×10)/2 = 60/2 = 30;
## What is the lcm of 2 and 6?
Lcm of 2, 3 and 6 using division method. The simplest way to find the lcm of any. 2 × 2 × 2 × 5 × 5 = 200 so, the lcm of 40 and 50 is 200 using the division method.
## Both 6And10 Will Multiply Into 30 , But 10.
We know that the smallest multiple which is exactly divisible by 2 and 6 has to be determined. 6 = 2 x 3. Least common multiple (lcm) of 2 and 6 with primes. Least common multiple can be found by multiplying the highest exponent prime factors of 6 and 2.
### Lcm Of 2, 3 And 6 Using Division Method.
7 rows what is the least common multiple of 6 and 2?
### Kesimpulan dari Lcm Of 6 And 2.
The least common multiple is the smallest positive. In this example, 6×10=60 , which is a common multiple cm of the two numbers. Multiples of 6 = 6, 12, 18,. | 426 | 1,328 | {"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} | 4.28125 | 4 | CC-MAIN-2023-14 | latest | en | 0.886978 |
https://www.hiqualitytutorials.com/product/final-exam-math-221/ | 1,601,410,701,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402088830.87/warc/CC-MAIN-20200929190110-20200929220110-00042.warc.gz | 869,769,600 | 22,572 | # Final Exam MATH 221
\$42.00
Final Exam MATH 221
The table below shows the number of male and female students enrolled in nursing…
SKU: MATH221 Final Exam Statistics Categories: ,
## Final Exam MATH 221
Final Exam MATH 221
A+ Formulas in Excel and in Word are included
1. The table below shows the number of male and female students enrolled in nursing at a university for a certain semester. A student is selected at random. Complete parts (a) through (d) (a)Find the probability that the student is male or a nursing major.
P (being male or being nursing major) =
(b) Find the probability that the student is female or not a nursing major.
P(being female or not being a nursing major) =
(c) Find the probability that the student is not female or a nursing major
P(not being female or not being a nursing major) =
(d) Are the events “being male” and “being a nursing major” mutually exclusive? Explain.
1. An employment information service claims the mean annual pay for full-time male workers over age 25 without a high school diploma is \$22,325. The annual pay for a random sample of 10 full-time male workers over age 25 without a high school diploma is listed. At a = 0.10, test the claim that the mean salary is \$22,325. Assume the population is normally distributed.
20,660 – 21,134 – 22,359 – 21,398 – 22,974, – 16,919 – 19,152 – 23,193 – 24,181 – 26,281
(a) Write the claim mathematically and identify
Which of the following correctly states ?
Final Exam MATH 221
(b) Find the critical value(s) and identify the rejection region(s).
What are the critical values?
Which of the following graphs best depicts the rejection region for this problem?
(c) Find the standardized test statistics.
t =
(d) Decide whether to reject or fail to reject the null hypothesis.
reject because the test statistics is in the rejection region.
a. fail to reject because the test statistic is not in the rejection region.
c. reject because the test statistic is not in the rejection region.
d. fail to reject because the test statistic is in the rejection region.
Final Exam MATH 221
(e) Interpret the decision in the context of the original claim.
a. there is sufficient evidence to reject the claim that the mean salary is \$22,325.
b. there is not sufficient evidence to reject the claim that the mean salary is not \$22,325.
c. there is sufficient evidence to reject the claim that the mean salary is not \$22,325.
d. there is not sufficient evidence to reject the claim that the mean salary is \$22,325.
1. The times per week a student uses a lab computer are normally distributed, with a mean of 6.1 hours and a standard deviation of 1.2 hours. A student is randomly selected. Find the following probabilities.
(a) The probability that the student uses a lab computer less than 5hrs a week.
(b) The probability that the student uses a lab computer between 6-8 hrs a week.
(c) The probability that the student uses a lab computer for more than 9 hrs a week.
(a) =
(b) =
(c) =
1. Write the null and alternative hypotheses. Identify which is the claim.
A study claims that the mean survival time for certain cancer patients treated immediately with chemo and radiation is 13 months.
2. Find the indicated probability using the standard normal distribution.
P(z>) =
3. The Gallup Organization contacts 1323 men who are 40-60 years of age and live in the US and asks whether or not they have seen their family doctor.What is the population in the study?
Answer:What is the sample in the study?
Final Exam MATH 221
1. The ages of 10 brides at their first marriage are given below.
4 32.2 33.6 41.2 43.4 37.1 22.7 29.9 30.6 30.8(a) find the range of the data set.
Range =
(b) change 43.4 to 58.6 and find the range of the new date set.
Range =
1. The following appear on a physician’s intake form. Identify the level of measurement of the data.
(a) Martial Status
(b) Pain Level (0-10)
(c) Year of Birth
(d) Height(a) what is the level of measurement for marital status(b) what is the level of measurement for pain level(c) what is the level of measurement for year of birthWhat is the level of measurement for height
1. To determine her air quality, Miranda divides up her day into 3 parts; morning, afternoon, and evening. She then measures her air quality at 3 randomly selected times during each part of the day. What type of sampling is used?
1. Find the equation of the regression line for the given data. Then construct a scatter plot of the data and draw the regression line. Then use the regression equation to predict the value of y for each of the given x-values, if meaningful. The caloric content and the sodium content (in milligrams) for 6 beef hot dogs are shown in the table below.
• X= 150 calories
• X= 100 calories
• X = 120 calories
• X = 60 calories
Final Exam MATH 221
Find the regression equation.
=
Choose the correct graph below.
(a) predict the value of y for x = 150.
(b) predict the value of y for x = 100.
(c) predict the value of y for x = 120.
(d) predict the value of y for x = 60.
11. A restaurant association says the typical household spends a mean of \$4072 per year on food away from home. You are a consumer reporter for a national publication and want to test this claim. You randomly select 12 households and find out how much each spent on food away from home per year. Can you reject the restaurant association’s claim at a = 0.10? Complete parts a through d.
• Write the claim mathematically and identify. Choose the correct the answer below.
Use technology to find the P-value.
P =
Decide whether to reject or fail the null hypothesis.
Interpret the decision in the context of the original claim. Assume the population is normally distributed. Choose the correct answer below.
12. The table below shows the results of a survey in which 147 families were asked if they own a computer and if they will be taking a summer vacation this year.
Final Exam MATH 221
(a) find the probability that a randomly selected family is not taking a summer vacation year.
Probability =
(b) find the probability that a randomly selected family owns a computer
Probability =
(c) find the probability that a randomly selected family is taking a summer vacation this year and owns a computer
Probability =
(d) find the probability a randomly selected family is taking a summer vacation this year and owns a computer.
Probability =
• Are the events of owning a computer and taking a summer vacation this year independent or dependent events?
• 13. Assume the Poisson distribution applies. Use the given mean to find the indicated probability.
Find P(5) when ᶙ = 4
P(5) =
14. In a survey of 7000 women, 4431 say they change their nail polish once a week. Construct a 99% confidence interval for the population proportion of women who change their nail polish once a week.
A 99% confidence interval for the population proportion is…
15 A random sample of 53 200-meter swims has a mean time of 3.32 minutes and the population standard deviation is 0.06 minutes. Construct a 90% confidence interval for the population mean time. Interpret the results.
The 90% confidence interval is
Interpret these results. Choose the correct answer:
Answer: With 90% confidence, it can be said that the population mean time is between the end points of the given confidence interval.
16. Determine whether the variable is qualitative or quantitative: Weight
Quantitative
Qualitative
17. 32% of college students say that they use credit cards because of the reward program. You randomly select 10 college students and ask each to name the reason he or she uses credit cards. Find the probability that the number of college students who say they use credit cards because of the reward program is (a) exactly two, (b), more than two, and (c), between two and five inclusive.
(a) P(2) =
(b) P(X>2) =
(c) P(2<x<5) =
Final Exam MATH 221
18. A light bulb manufacturer guarantees that the mean life of a certain type of light bulb is at least 950 hours. A random sample of 74 light bulbs has a mean life of 943 hours with a standard deviation of 90 hours. Do you have enough evidence to reject the manufacturer’s claim? Use ᶏ = 0.04
• Identify the critical value(s).(c) identify the standardized test statistic.
z =
(d) decide whether to reject or fail to reject the null hypothesis.A. Reject . There is sufficient evidence to reject the claim that the bulb life is at least 950 hours.
B. Fail to reject . There is not sufficient evidence to reject the claim that the mean bulb life is at least 950 hours.
C. Fail to reject . There is sufficient evidence to reject the claim that mean bulb life is at least 950 hours.
D. Reject . There is not sufficient evidence to reject the claim that mean bulb life is at least 950 hours.19. Use technology to find the sample size, mean, medium, minimum data value, and maximum data value of the data. The data represents the amount (in dollars) made by several families during a community yard sale.
25 67.25 156 134.75 98.25 149.25 124.75 109.75 117 104.75 76The sample size is
The mean is
The medium is
The minimum data value is
The maximum data value is20. A researcher wishes to estimate, with 95% confidence, the proportion of adults who have high-speed internet access. Her estimate must be accurate within 5% of the true proportion.
(a) find the minimum sample size needed, using a prior study that found 54% of the respondents said they have high-speed internet access.
(b) no preliminary estimate is available. Find the minimum sample size needed.(a) what is the minimum sample size needed using a prior study that found that 54% of the respondents said they have high-speed internet access?n =(b) what is the minimum sample size needed assuming that no preliminary estimate is available?n =21. You interview a random sample of 50 adults. The results of the survey show that 50% of the adults said they were more likely to buy a product where there are free samples. At ᶏ = 0.05, can you reject the claim that at least 54% of the adults are more likely to buy a product when there are free samples?State the null and alternative hypotheses. Choose the correct answer below.
Determine the critical value(s).
The critical value(s) is/are
find the z-test statistic.
z =
what is the result of the test?
A. reject . The data provide sufficient evidence to reject the claim.
1. fail to reject . The data provide sufficient evidence to reject the claim.
C. Reject . The data do not provide sufficient evidence to reject the claim.
D. fail to reject . The data do not provide sufficient evidence to reject the claim.
22. The budget (in millions of dollars) and worldwide gross (in millions of dollars) for eight movies are shown below. Complete parts (a) through (c)
Final Exam MATH 221
Budget X 209 203 198 198 179 176 175 168 Gross Y 254 341 453 656 721 1049 1839 1267
(a) display the data in a scatter plot. Choose the correct graph below.
(b) calculate the correlation coefficient r.
r =
(c) make a conclusion about the type of correlation.
The correlation is a …linear correlation.
23. A machine cuts plastic into sheets that are 30 feet (360 inches) long. Assume that the population of lengths is normally distributed. Complete parts a and b.
• The company wants to estimate the mean length the machine is cutting the plastic within 0.125 inch. Determine the minimum sample size required to construct a 95% confidence interval for the population mean. Assume the population standard deviation is 0.25 inch.
n =
Repeat part (a) using an error tolerance of 0.0625 inch.
n =
Which error tolerance requires a larger sample size? Explain.
1. The tolerance E = 0.0625 inch requires a larger sample size. As error size decreases, a larger sample must be taken to ensure the desired accuracy.
2. The tolerance E = 0.125 inch requires a larger sample size. As error size decreases, a larger sample must be taken to ensure the desired accuracy.
3. The tolerance E = 0.125 inch requires a larger sample size. As error size increases, a larger sample must be taken to ensure the desired accuracy.
4. The tolerance E = 0.0.625 inch requires a larger sample size. As error size increases, a larger sample must be taken to ensure the desired accuracy.
MATH221 Final Exam Statistics
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A052965 Expansion of (1-x)/(1-3x-4x^2+4x^3). 0
1, 2, 10, 34, 134, 498, 1894, 7138, 26998, 101970, 385350, 1455938, 5501334, 20786354, 78540646, 296762018, 1121303222, 4236795154, 16008550278, 60487618562, 228549876182, 863565901682, 3262946735526, 12328904308578 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 1036 Index entries for linear recurrences with constant coefficients, signature (3,4,-4) FORMULA G.f.: -(-1+x)/(1-3*x-4*x^2+4*x^3) Recurrence: {a(0)=1, a(1)=2, a(2)=10, 4*a(n)-4*a(n+1)-3*a(n+2)+a(n+3)=0} Sum(-1/158*(-17-49*_alpha+40*_alpha^2)*_alpha^(-1-n), _alpha=RootOf(1-3*_Z-4*_Z^2+4*_Z^3)) MAPLE spec := [S, {S=Sequence(Prod(Union(Z, Z, Sequence(Z)), Union(Z, Z)))}, unlabeled ]: seq(combstruct[count ](spec, size=n), n=0..20); MATHEMATICA CoefficientList[Series[(1-x)/(1-3x-4x^2+4x^3), {x, 0, 40}], x] (* or *) LinearRecurrence[{3, 4, -4}, {1, 2, 10}, 40] (* Harvey P. Dale, Dec 01 2016 *) CROSSREFS Sequence in context: A119193 A124634 A192378 * A108924 A221492 A116898 Adjacent sequences: A052962 A052963 A052964 * A052966 A052967 A052968 KEYWORD easy,nonn,changed AUTHOR encyclopedia(AT)pommard.inria.fr, Jan 25 2000 EXTENSIONS More terms from James A. Sellers, Jun 06 2000 STATUS approved
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The OEIS Community | Maintained by The OEIS Foundation Inc. | 671 | 1,894 | {"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-2016-50 | longest | en | 0.547503 |
https://excelexamples.com/post/how-to-use-regression-analysis-in-excel/ | 1,669,487,163,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446708046.99/warc/CC-MAIN-20221126180719-20221126210719-00115.warc.gz | 282,808,902 | 7,918 | This example teaches you how to run a linear regression analysis in Excel and how to interpret the Summary Output.
Below you can find our data. The big question is: is there a relation between Quantity Sold (Output) and Price and Advertising (Input). In other words: can we predict Quantity Sold if we know Price and Advertising?
1. On the Data tab, in the Analysis group, click Data Analysis.
Note: can’t find the Data Analysis button? Click here to load the Analysis ToolPak add-in.
2. Select Regression and click OK.
3. Select the Y Range (A1:A8). This is the predictor variable (also called dependent variable).
4. Select the X Range(B1:C8). These are the explanatory variables (also called independent variables). These columns must be adjacent to each other.
5. Check Labels.
6. Click in the Output Range box and select cell A11.
7. Check Residuals.
8. Click OK.
Excel produces the following Summary Output (rounded to 3 decimal places).
### R Square
R Square equals 0.962, which is a very good fit. 96% of the variation in Quantity Sold is explained by the independent variables Price and Advertising. The closer to 1, the better the regression line (read on) fits the data.
### Significance F and P-values
To check if your results are reliable (statistically significant), look at Significance F (0.001). If this value is less than 0.05, you’re OK. If Significance F is greater than 0.05, it’s probably better to stop using this set of independent variables. Delete a variable with a high P-value (greater than 0.05) and rerun the regression until Significance F drops below 0.05.
Most or all P-values should be below below 0.05. In our example this is the case. (0.000, 0.001 and 0.005).
### Coefficients
The regression line is: y = Quantity Sold = 8536.214 -835.722 * Price + 0.592 * Advertising. In other words, for each unit increase in price, Quantity Sold decreases with 835.722 units. For each unit increase in Advertising, Quantity Sold increases with 0.592 units. This is valuable information.
You can also use these coefficients to do a forecast. For example, if price equals \$4 and Advertising equals \$3000, you might be able to achieve a Quantity Sold of 8536.214 -835.722 * 4 + 0.592 * 3000 = 6970.
### Residuals
The residuals show you how far away the actual data points are fom the predicted data points (using the equation). For example, the first data point equals 8500. Using the equation, the predicted data point equals 8536.214 -835.722 * 2 + 0.592 * 2800 = 8523.009, giving a residual of 8500 – 8523.009 = -23.009.
You can also create a scatter plot of these residuals. | 654 | 2,623 | {"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-2022-49 | longest | en | 0.834407 |
https://forum.piedao.org/answers/16749999-two-cyclists-112-miles-apart-start-riding-toward-each-other-at-the-same-time-one-cycles-3-times-as-fast | 1,709,218,025,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474843.87/warc/CC-MAIN-20240229134901-20240229164901-00488.warc.gz | 256,221,811 | 7,015 | # Two cyclists, 112 miles apart, start riding toward each other at the same time. One cycles 3 times as fast as the other, and they meet after 4 hours of riding.a. Write an equation using the information as it is given above that can be solved to answer this problem. Use the variable r to represent the speed of the slower cyclist.b. What are the speeds of the two cyclists? Put both values in the answerbox, separated with a comma, and select the appropriate units.
Speed of a= 21 miles/hr
r = Speed of b= 7 miles/hr
Speed of a = 3r
Step-by-step explanation:
The cyclist are 112 miles apart
Time traveled by two = 4 hours
Speed of a = 3 * speed of b
If a cylcles 3 times More than b, then a will cover 3*distance of b
But speed = distance/time
Time = 4hours
Total distance=112
a = 3b
3b + b = 112
4b = 112
b = 112/4
b = 28 miles
a = 3b
a = 3*28
a = 84 Miles
They bought traveled 4 hours
Speed of a = 84miles/4 hours
Speed of a= 21 miles/hr
Speed of b = 28miles/4 hours
Speed of b = 7 miles/hr
The slower cyclist is traveling at a speed of 7 mph, while the faster cyclist is traveling at 21 mph.
### Explanation:
We first need to recognize that the sum of the distances travelled by each cyclist is equal to the total 112 miles. We can let r represent the speed of the slower cyclist, and since the faster cyclist is traveling 3 times the speed of the slower, we can call his speed 3r.
As they each traveled for 4 hours, the slower cyclist traveled a distance of 4r miles, and the faster cyclist traveled a distance of 4 × 3r = 12r miles. Their combined distances should equal the total distance between them, 112 miles. This forms the equation 4r + 12r = 112.
To solve for r, we combine the like terms on the left side of the equation to get 16r = 112. Dividing both sides by 16, we find r = 7 miles per hour. Therefore, the slower cyclist is traveling at 7 mph, and the faster cyclist is traveling 3 times that speed, giving us 21 mph.
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## Related Questions
Find the sum of 2x2 – 8x — 6 and 9x2 – 8.
Hence , the sum is .
## WHAT IS A POLYNOMIAL
In mathematics, a polynomial is an expression consisting of indeterminates (also called variables) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponentiation of variables.
### How to Solve?
Given , we need to add and .
So, we can write it as :
Writing like terms together:
Hence , the answer is .
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(x)=x, g(x)=x3−1f(x)=x4+x3−11x2−5x+30, g(x)=2x−2f(x)=x2, g(x)=7x−1
in short ways the answer is 7x−1
Which statement illustrates why cosine is an even function?cos(pie/4) = -cos(5pie/4)
cos(pie/4) = -cos(4pie/4) cos(-pie/4) = cos(pie/4)
cos(-7pie/6) = cos(5pie/4)
illustrates why cosine is an even function.
### What is an even function?
A function such that f(x)=f(−x) where the value remains unchanged if the sign of the independent variable is reversed.
According to the question
Cosine is an even function
If f(x) is an even function, then f(x) = f(-x)
So, illustrates why cosine is an even function.
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C. cos(-pie/4) = cos(pie/4)
Step-by-step explanation: edge
What are the answers to 7 and 8?
Step-by-step explanation: 7 b and d8
A roll of ribbon was 9 meters long. Kevin cut 8 pieces of ribbon each of length 0.8 meter, to tie some presents. He then cut the remaining ribbon into some pieces, each of length 0.4 meter.a) how many pieces of ribbon, each 0.4 meter in length, did Kevin have?
b) What was the length of ribbon left over?
answer for A is 6 pieces and the answer for B is .2 meters..
How many feet are in half a mile?
Step-by-step explanation:
There are 5280 feet in one mile. So, half a mile is equal to 0.5 x 5280 = 2640 feet.
5280 looked it up on safari
Use the drawing tools to form the correct answer on the graph. Graph this function. - 2 + 8 = Reset ® Delet Undo Drawing Tools Click on a tool to begin drawing. Select Point 10 Line 8 3 6- 4 2 2 4 6 -2 8 10 -4 -10 -8 -6 -2 7071 Frmentum. All rights reserved.
The graph of has a slope of -2, and a y-intercept of 8
The function is given as:
The above function is a linear function.
A linear function is represented as:
Where:
m represents the slope, and c represents the y-intercept.
So, by comparison;
This means that the graph of has a slope of -2, and a y-intercept of 8
See attachment for the graph of the function | 1,295 | 4,495 | {"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} | 4.4375 | 4 | CC-MAIN-2024-10 | latest | en | 0.927888 |
http://contexturesblog.com/archives/2010/01/20/something-fishy-using-the-excel-offset-function/?replytocom=4742 | 1,527,252,875,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867092.48/warc/CC-MAIN-20180525121739-20180525141739-00514.warc.gz | 63,763,396 | 9,132 | # Something Fishy: Using the Excel OFFSET Function
That’s my dad in the picture, proudly holding the catch of the day. He tried to teach me how to fish, but without much success. (Worms…ewwww.)
This week someone asked me to explain the Excel OFFSET function, saying “Please teach me to fish.” That’s when it struck me that using OFFSET is similar to fishing.
• When you’re fishing, you can dip into a pond with a bamboo pole and a small hook, or head out to sea, and cast a large net. Or you can fish the way we do in Canada, through a small hole in the ice, but that’s another story. (There’s a video at the end of this article.)
• With the Excel OFFSET function, you can pull data from a single cell nearby, or a large range of cells off in the distance.
The OFFSET function is useful when you want to make the data selection adjustable. For example, if a February date is entered in cell A2, you can sum the February expense column. If a March date is entered, sum the March expenses instead.
To make the OFFSET function work, you’ll tell it 3 things:
1. The starting point
2. Where to go from there
3. How big a range to capture (optional)
The OFFSET syntax is: OFFSET(reference,rows,cols,height,width)
1. The reference is the starting point.
2. The rows and cols tell OFFSET where to go from the starting point. It can go up or down a specific number of rows, and left or right a specific number of columns.
3. The height and width set the size of the range. It can be as small as 1 row and 1 column (a single cell) or much bigger.
For example, this OFFSET formula would return the January total, in cell B6:
=OFFSET(A1,5,1,1,1)
1. The starting reference is cell A1.
2. From there, it goes down 5 rows, and right one column, to cell B6.
3. The selected range size is 1 row tall and 1 column wide.
### Baiting the Hook
Instead of typing all the values in the formula, you can use one or more cell references, to make the OFFSET formula flexible. In this example, all the totals are in row 5, so that number won’t change.
However, the month number is typed in cell G1, so you could use that cell to set the number of columns to offset. Change the formula so G1 is the cols argument.
=OFFSET(A1,5,G1,1,1)
Now, if you change the month number to 3 in cell G1, the March total will be returned.
### Casting the Net
Instead of pulling the result from a single cell, you could use OFFSET with the SUM function, to select a range with multiple cells, and calculate the total.
For example, this formula would calculate the total for the February expenses.
=SUM(OFFSET(A1,1,G1,4,1))
1. The starting reference is cell A1.
2. From there, it goes down 1 row, and right 2 columns, to cell C2.
3. The selected range size is 4 rows tall and 1 column wide – C2:C5.
### Other Fish to Fry
I like the OFFSET function, and use it to create dynamic ranges in some of my workbooks. There are alternatives to using the Excel OFFSET function, such as the Excel INDEX Function. There’s an interesting discussion of the merits of each function on Dick Kusleika’s Daily Dose of Excel Blog: New Year’s Resolution: No More Offset.
### Ice Cold Fish
I’d rather stay inside and work on OFFSET formulas, but ice fishing is popular here in Canada. This video makes the sport look almost appealing.
_____________________ | 810 | 3,310 | {"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-2018-22 | latest | en | 0.895522 |
https://educationexpert.net/business/1829932.html | 1,638,457,795,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362230.18/warc/CC-MAIN-20211202145130-20211202175130-00192.warc.gz | 294,157,489 | 7,299 | 28 November, 20:36
# Compute the inventory turnover ratio using the following information: Net sales is \$100,000 for the year, costs of goods sold are \$40,000, last year's assets in place were \$900,000, and this year's assets in place are \$1,100,000. Receivables for both years are \$50,000. Inventory changed from \$30,000 last year to \$10,000 this year.
0
1. 28 November, 20:41
0
2
Explanation:
The formula for calculating inventory turn over:
Costs of goods sold / average inventory
In this case:
Opening stock: \$ 30,000.00
Closing stock: \$ 10,000.00
Costs of goods sold : \$ 40,000.00
Average inventory = Opening stock + closing stock/2
= \$ 30,000 + \$ 10,000/2
=\$ 40,000/2
=\$ 20,00.00
Inventory turnover = \$ 40,000.00/\$20,000.00
=2 | 237 | 763 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2021-49 | latest | en | 0.879191 |
http://www.qacollections.com/How-to-Find-the-Area-Volume-Perimeter | 1,495,958,338,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609605.31/warc/CC-MAIN-20170528062748-20170528082748-00016.warc.gz | 780,263,521 | 5,611 | # How to Find the Area, Volume & Perimeter?
Perimeter, area and volume are essential mathematical building blocks for higher mathematical functions including geometry and calculus. Perimeter is the distance around an object, area is the amou... Read More »
http://www.ehow.com/how_8197626_area-volume-perimeter.html
Top Q&A For: How to Find the Area, Volume & Perimeter
## How to find the area and perimeter of the figure?
area is the length multiplied by the width the perimeter is the length +width+length+width
## How to Find the Area of a Square Using Its Perimeter?
A square is a figure with four equal length sides, and the perimeter of a square is the total distance around the outside of the shape. Calculate the perimeter by adding all four sides together. Th... Read More »
http://www.ehow.com/how_7991359_area-square-using-its-perimeter.html
## How to Find the Area & Perimeter of a Square on a Coordinate Plane?
Finding the area and perimeter of figures on a coordinate plane combines the properties of coordinate planes with the properties of geometric figures. After you understand the process of using the ... Read More »
http://www.ehow.com/how_8282345_area-perimeter-square-coordinate-plane.html
## How to Find the Volume & Area of a Rectangle?
A rectangle has four sides and four right angles, which means the both pairs of sides will be parallel. However, the pairs of sides can be of different lengths --- unless the rectangle is a square,... Read More »
http://www.ehow.com/how_8389206_volume-area-rectangle.html
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vranchiv2
Registered: 29.06.2002
From: OZ - Perth
Posted: Friday 29th of Dec 09:56 Friends , I am having a very tough time with my math prep on Pre Algebra. I was of the opinion that this would be solvable and hence didn’t care to have a look at it till now. When I sat down to solve the problems today, I found it to be rather unsolvable . Can any one help me by providing details on the existing tools that can assist me with brushing up my basics on , and topic-kwds.
ameich
Registered: 21.03.2005
From: Prague, Czech Republic
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Gools
Registered: 01.12.2002
From: UK
Posted: Saturday 30th of Dec 17:35 Hey Dude, Algebrator helped me with my homework last week. I got the Algebrator from https://rational-equations.com/solving-2nd-degree-equations.html. Go ahead, try that and let us know your opinion. I have even recommended Algebrator to a couple of my friends at school .
nedslictis
Registered: 13.03.2002
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Posted: Sunday 31st of Dec 10:39 geometry, complex fractions and adding matrices were a nightmare for me until I found Algebrator, which is truly the best algebra program that I have come across. I have used it frequently through many algebra classes – Remedial Algebra, Pre Algebra and Pre Algebra. Just typing in the math problem and clicking on Solve, Algebrator generates step-by-step solution to the problem, and my algebra homework would be ready. I really recommend the program.
comverr
Registered: 08.05.2006
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Posted: Tuesday 02nd of Jan 07:06 Cool! This sounds extremely useful to me. I was searching such software only. Please let me know where I can purchase this software from?
Sdefom Koopmansshab
Registered: 28.10.2001
From: Woudenberg, Netherlands
Posted: Wednesday 03rd of Jan 17:41 The program can be found at https://rational-equations.com/solving-equations.html. | 1,086 | 4,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2022-49 | latest | en | 0.822507 |
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Lesson Transcript
Instructor: Luke Winspur
Luke has taught high school algebra and geometry, college calculus, and has a master's degree in education.
What is a transformation? Well, it's something that transforms one function into another! To see what I mean and how that looks, check out this lesson!
## Transformations
A lot of people are of the opinion that math doesn't make any sense. Some of my students even seem to think that long ago, we mathematicians met in a secret chamber somewhere and came up with all these crazy rules for no good reason other than to drive people crazy. While I can assure you that this is not the case, I do understand why people can feel this way. Teaching things like 5^0 = 1 does make it especially hard for me to defend my position that math isn't there purely for us teachers to laugh at you when you mess up.
But maybe this will help! It's a topic where we picked a word that accurately describes what it means in math. Transformations kinda, like, transform! You don't have to be an English scholar to know that this word implies one thing changing into another thing. Kinda like how a Camaro can transform into a big evil fighting robot, a mathematical function can transform into... a different mathematical function! Not as cool as the robot, I know, but at least it's simple. If it was transforming into a robot, I bet the math would be way harder.
## Types of Transformations
Anyways, let's take a look at a few of the different kinds of transformations that we can have in math. First off we've got rotations. These simply take a function, maybe this one, and rotate it! We can also have reflections. These are transformations that act like a mirror and reflect a function to a new place, maybe like this, or like this! The next kind of transformation is called a dilation. These stretch or shrink the graph, maybe from something like this to this!
## Translations
But the fourth and final transformation, called a translation, is the kind we'll focus on for this lesson. Translations simply slide the function around, maybe taking a function like this and moving it over here or up there! So the question becomes 'how do you accomplish this translation? How would you take this function and change it into this one?' Well, the translation I've shown you here is working in two ways. It's shifting the function up and to the right. So let's look at those shifts one at a time.
First the shift up: it looks like the new graph is three units higher than the original one. Let's see if we can make our original graph turn into this one. If we can do that, we'll be halfway and can then focus on getting it to slide over as well. Okay, so notice that this graph is really exactly the same except every point is three higher than it used to be. That means all we want to do is make every answer three more than it used to be. Instead of f(0), we want f(0) + 3. Instead of f(-2) we want f(-2) + 3. That means that, instead of f(x), we want f(x) + 3.
Cool! Now that we've gotten the function up here, let's try to slide it over to the right. But now that point needs to be five spots over to the right. So what I used to get when I plugged in 0, I now need to get when I plug in 5. So how can we plug in 5 and end up with 0? Well, what about making the function, instead of f(x), f(x - 5)? That way, when I plug in 5, I actually get f(0), which we know gives me the point I wanted. So that means that f(x) + 3 on the outside shifts it up three, but then doing f(x - 5) on the inside is actually going to shift it five to the right. That means this final graph over here must be f(x - 5) + 3.
Let's summarize what we just learned. Translations are accomplished by adding or subtracting values from the function. Adding outside the f(x) shifts the graph up, which implies that subtracting outside the f(x) would shift it down. Subtracting inside the f(x) shifts the graph to the right, which implies that adding inside the f(x) would shift it left. This is the opposite of what you might think. The way I remember it is by asking myself 'if it says f(x + 2), how would I make it zero?' The answer is, 'by putting in a -2,' thus shifting it to the left.
What this means is that if we have a function f(x - h) + k, h tells us how many units to slide the function left or right, and k tells us how many units to slide it up or down. Let's look a few quick examples.
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Did you know… We have over 95 college courses that prepare you to earn credit by exam that is accepted by over 2,000 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 1,285 | 5,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2017-43 | latest | en | 0.94025 |
https://fun2dolabs.com/simple-interest/ | 1,721,147,253,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.37/warc/CC-MAIN-20240716142214-20240716172214-00622.warc.gz | 237,990,462 | 12,506 | Simple Interest
Learning about money is incredibly interesting because it is a crucial subject that is used throughout life. Children are taught about money denominations, unit conversions, profit and loss, and the unitary method under the topic of money. Simple interest is an important concept that is introduced in the primary grades.
This teaching guide aims at explaining the important concept of simple interest with the help of an animated cartoon story, bright posters, colourful worksheets, and even an interesting activity to better comprehend the concept.
When we borrow money from a bank, we also have to pay a fee. Interest is the term for the additional cost we pay. This is also charged on borrowed funds by some financial organisations and moneylenders.
There are a few terms related to simple interest that are very important to understand the topic well. Following are the terms used to determine this.
### Principal :
The sum that was first borrowed (loaned) from the bank or invested is referred to as the principal. The principal is denoted by P.
### Rate :
Rate is the interest rate at which the main sum is loaned to someone for a specific period of time; it may be 5%, 10%, 13%, etc. The rate is denoted by R.
### Time :
Time is the duration for which the principal amount is given to someone. Time is denoted by T.
Simple interest is determined by using the formula :
Where,
S.I. is simple interest, P is the principal amount, T is the time and R is the rate.
The above formula is further used to determine the variables like principal, amount, time, and rate also.
### Amount :
When someone borrows money from a bank, they are required to pay it back, and the sum of money that is paid back is known as the Amount which includes the money borrowed (principal) and the extra sum of money charged on the loan (interest).
Amount = Principal + Simple Interest
Posters
Teaching simple interest with kid-friendly, clear, and easy-to-understand posters from Uncle Math School by Fun2Do Labs
Stories
Ignite kids’ curiosity with engaging stories for role play and skits, making the learning of this concept an exciting and effective experience. Teaching simple interest through stories from Uncle Math School by Fun2Do Labs :
Text of Stories
Activities
Learning simple interest can be made enjoyable by incorporating interactive games and activities.
### Interest Puzzle Time!
This is an entertaining puzzle game to engage children and teach them how to use the basic interest formula. This activity can be carried out in the following steps :
1. Make puzzle cards, mentioning the principal, rate, and time.
2. Children need to arrange puzzle cards by joining the correct value of the interest (S.I.) for given values of principal (P), rate (R), and time (T) by using the following formula :
Worksheets
Help your kids practise simple interest with interesting and engaging fun worksheets and solutions from Uncle Math by Fun2Do Labs.
Worksheet 309 : Word Problems
Solution 309 : Word Problems
Worksheet 311 : Simple Interest
Solution 311 : Simple Interest
Worksheet 312 : Simple Interest Word Problems
Solution 312 : Simple Interest Word Problems | 658 | 3,198 | {"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-2024-30 | latest | en | 0.970767 |
https://www.mytutor.co.uk/answers/242/GCSE/Maths/Can+you+derive+the+Quadratic+Formula%253F | 1,503,302,702,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107744.5/warc/CC-MAIN-20170821080132-20170821100132-00229.warc.gz | 915,813,958 | 66,708 | 751 views
### Can you derive the Quadratic Formula?
For a general quadratic equation of the form ax2 + bx + c = 0 (1)
We will solve for x via the ‘Completing the Square‘ method:
First we divide equation (1) through by a on both sides, yielding:
x2 +(b/a)x + c/a = 0
Slightly rearranging, we then write (for tidiness):
x2 + (b/a)x = -c/a (2)
Now the key behind completing the square is to try and write the x2 and x terms as the square of some quantity, which is close to:
(x + (b/2a))2 (3)
What we have done is half the coefficient of the x term, as when the brackets are expanded out, you have a certain sum repeated twice, I will demonstrate in fuller detail now what I mean.
Expanding out (3), we have
(x +(b/2a))(x+(b/2a)) = x2+ (b/2a)x + (b/2a)x + (b2/4a2= x2 + (b/a)x + (b2/4a2) , which is almost identical to the left hand side of equation (2), except now we have the extra third term which is (b2/4a2)
We can rearrange this easily to show that x2+ (b/a)x =(x +(b/2a))2-(b2/4a2)
Substituting this identity into equation (2), we now have
(x +(b/2a))2 -(b2/4a2) = -c/a (4)
We now rearrange as follows:
(x+(b/2a))2 = (b2/4a2) - (c/a) = (b2-4ac)/(4a2) ,
where in the third equality I have simply summed over a common denominator by multiplying the second term by 4a over the numerator and denominator.
Finally we take the square root of both sides, and rearrange to arrive at the quadratic formula we were looking for, as follows:
x = (-b ± sqrt(b2-4ac))/2a
Note that we have two solutions to the quadratic equation, resulting from the existence of both positive and negative square roots, hence the ± sign. Also ‘sqrt' denotes taking the square root.
Corollary
Note that the solution of x depends on the quantity sqrt(b2-4ac), which is known as the Discriminant. This is important as it results with 3 distinct cases for the solution which are as follows:
1) b2 - 4ac > 0 , and so it gives one positive and one negative square root, leading to 2 unique solutions.
2) b2 -4ac = 0, which then yields exactly one solution, known as a repeated root, -b/2a .
3) b2 -4ac < 0, which cannot be square rooted within our common number system which we call the Real Numbers. If we extend our number system, to allow for square roots of negative numbers, we call these the Complex Numbers, which go far beyond the scope of the course.
If you want to find out more, be sure to stay on till A Level and take Further Maths!
3 years ago
Answered by Kamran, who tutored GCSE Maths with MyTutor
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https://www.easyelimu.com/kenya-secondary-schools-pastpapers/mocks/2022/item/6339-mathematics-paper-2-questions-and-answers-mincks-group-of-schools-mock-examinations-2022 | 1,680,093,597,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948976.45/warc/CC-MAIN-20230329120545-20230329150545-00639.warc.gz | 826,664,850 | 29,745 | # Mathematics Paper 2 Questions and Answers - Mincks Group of Schools Mock Examinations 2022
INSTRUCTION TO CANDIDATES
1. Write your name and Index number in the spaces provided above.
2. Sign and write the date of examination in the spaces provided above.
3. The paper consists of two sections. Section I and Section II.
4. Answer ALL the questions in Section I and any FIVE questions in Section II.
5. Show all the steps in your calculations, giving your answer at each stage in the spaces provided below each question.
6. Marks may be given for correct working even if the answer is wrong.
7. Non-programmable silent electronic calculators and KNEC Mathematical tables may be used except where stated otherwise.
8. Candidates should answer the questions in English.
9. Candidates must check the question paper to ascertain that all pages are printed as indicated and that no question(s) is/are missing.
FOR EXAMINER’S USE
SECTION I
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Total
SECTION II
17 18 19 20 21 22 23 24 Total
### QUESTIONS
SECTION 1 (50 MARKS)
Answer all questions in this section
1. Solve the equation 1/3 (2x+1)+1/4 (5x-1)= 1/2 (3mks)
2.
1. Expand ( 1 – 2x)4 (1mk)
2. Use your expansion to find the value of (0.96)4 correct to 4 s.f (2mks)
3. The expression 1+x/2 is taken as an approximation for . Find the percentage error in doing so if x = 0.44 (3mks)
4. Solve for x in the equation log8 (x+6) - log8(x-3) = 2/3 (4mks)
5. Make p the subject if the formula (3mks)
6. An auto spare dealer sells two types of lubricants, A and B in his shop. While purchasing, type A cost sh40 per 100ml tin and type be B cost sh.60 per 100ml tin. He has Shs 15,000 to spend on both types of lubricants. Type A should not be less than 3 litres while type B should be atleast a litre . He buys x tins of type A and Y tins of type B. Write down three inequalities which represent the above information. (3mks)
7. In the figure below PT is a tangent to the circle at P and AT passes through the centre of the circle.
If PT =9cm and AT=12cm, find the- radius of the circle correct to 1 d.p. (3mks)
8. Find the radius and the centre of the circle whose equation is (3mks)
3x2 + 3y2 – 6x + 3 + 12y = 0
9. Without using a calculator or mathematical tables express in surd form and simplify (3mks)
10. The table below represents the mass and volume of a liquid measured at room, temperature.
Mass (g) 8 10 20 21 30 37 52 55 70 Volume (cm3) 10 13 20 29 35 43 60 63 70
1. By plotting the points and drawing the line of best fit, estimate the density to the liquid. (2mks)
2. Determine the equation of your graph. (2mks)
11. The population of two towns Kathwana and Siakago for three years was as follow.
1st year 2nd year 3rd year Kathwana 40,000 48,000 56,000 Siakago 40,000 48,000 57,600
Calculate the difference in population of the two after six years. (3mks)
12. Solve for ∝ in the equation 6 sin2∝ - cos ∝ – 5 = 0 (3mks)
For Ɵ≤ ∝ ≤ 300º
13. Arabic coffee costing sh90 per kilogram is mixed with rubusta coffee costing sh.75 per kilogram in the ration 2:3. Calculate the selling price of the mixture per kilogram if a profit of 15% if to be realized to the nearest shilling. (3mks)
14. The gradient of a curve at any point is given by 2x-1. Given that the curve passes through point (1,5). Find the equation of the curve. (3mks)
15.
1. Given that AB=6cm. Construct locus of P such that <APB = 90º. (2mks)
2. Hence determine the area enclosed by the locus of P. (2mks)
16. Given the vectors a = 3j + 2k and b = 4i – 7k
Find p if p = 3a – b (2mks)
SECTION II (50 MARKS)
Answer any five questions from this section
1. Three consecutive terms of aGP are (x+2), (x-2), and (x-5). Find:-
1. The value of x. (3mks)
2. The common ratio (1mk)
3. Given that x+2 is the third term of the G.P
Find;
1. The first term giving your answer as a mixed fraction. (3mks)
2. the sum of the first six terms correct to 2 d.p (3mks)
2. A form 1 stream at Butere mixed day secondary school has 15 boys and 25 girls. The probability of a girl reaching form 4 is 2/5 and that of a boy is 3/5. If a girl reaches form 4 then probability that she gets employed is 2/3 while the probability that a girl gets employed without reaching form 4 is 1/3. If a boy reaches form 4 the probability that he gets employed is ¾ while the probability that a boy gets employed without reaching form 4 is 1/4
1. Draw a tree diagram to illustrate the above information. (2mks)
2. Using your tree diagram, determine:-
1. the probability that a girl gets employment. (2mks)
2. the probability any child who has been to school gets employed. (2mks)
3. the probability that a boy fails to get employment. (2mks)
4. The probability that anybody who has studied upto form 4 fails to get employment. (2mks)
3. The diagram below shows a right pyramid VPQRS with V as the vertex and rectangular base PQRS. PQ=3cm, QR=4cm and the height of the pyramids is 6cm. M and N are the mid-points of PQ and QR respectively.
1. Calculate
1. the length PV (3mks)
2. the angle between face VPQ and the base PQRS (2mks)
2.
1. the slant height VM and VN. (2mks)
2. the surface area of the pyramid. (3mks)
4. Two towns A and B lie on the same parallel of latitude 60ºN. If the longitude of A and B are 42ºW and 29ºE respectively.
1. Find the distance between A and B in nautical miles along the parallel of latitude. (2mks)
2. Find the local time at A if at B is 1.00.p.m. (2mks)
3. Find the distance between A and B in kilometre (Take π=22/7 and R=6370km) (2mks)
4. if C is another town due south of A and 1001km away from A, find the co-ordinate of C. (4mks)
5. The table below shows marks scored by students in a mathematics test.
Marks 30-39 40-49 50-59 60-69 70-79 80-89 90-99 No of students 2 3 8 12 15 6 4
1. Draw a cumulative frequency curve to represent the information. (4mks)
2. Use you graph to find:-
1. The median mark (1mk)
2. The quartile deviation (3mks)
3. the pass mark if 60% of the students passed. (2mks)
6. The initial velocity of a particle was 1m/s and it acceleration is given by (2-t) m/s2 every second after the start.
1.
1. Determine the equation representing its velocity. (2mks)
2. Find the velocity of the particle during the third second. (2mks)
2.
1. Find the equation representing its distance t seconds after the start. (2mks)
2. What was the distance covered by the particle during the first three seconds. (2mks)
3. Determine the time when the particle was momentarily at rest. (2mks)
7.
1. using trapezoidal rule estimate the area bounded by the curve y=3x2 – 2 and the lines x = -4, x=4 and x-axis using 8 trapezia (2mks)
2. Find the actual area bounded by the curve y=3x2 -2 and the lines x= -4, x=4 and x-axis. (3mks)
3. Calculate the percentage error when trapezoidal rule is used. (3mks)
8. The vertices of triangle PQR are P(1,1), Q(4,1) and R(5,4). A transformation represented by a matrix maps triangle PQR onto triangle PIQIRI.
A second transformation represented by maps triangle PIQIRI onto triangle PIIQIIRII
1. On the same axis draw the three triangles PQR, PIQIRI and PIIQIIRII and state the co-ordinated of triangle PIQIRI and PIIQIIRII . (6mks)
2. Describe a single transformation which maps triangle PQR onto triangle PIIQIIRII and find its…..matrix. (4mks)
### MARKING SCHEME
1. +1) +3(5x-1) =6
8x+4+15x-3=6
23x + 1=6
23x=5
X=5/23
2.
1. 1-4(2x) +6(2x)2- 4(2x)3+(2x)4
1-8x+24x2-32x3+16x4
2. 1-2x =0.96
0.04=2x
0.02=x
1-8(0.02) +24(0.02)2-32(0.02)3+16(0.02)4
1-0.16+0.00096-0.000256+0.00000256
=0.849858556
=0.8499
3. True value √(1+n)= √1.44 = 1.2
Approx. value =1+n/2=1+ 0.044/2 =1.22
=1.22-1.2
=0.02
= 0.02/1.2 ×100=1.67%
4. 3log8(x+6)-3log8(x-3) =2
log8(x+6)3-log8(x-3)3=log864
log8[((x+6)3)/(x-3)3]=log864
log8[(x+6)/(x-3)]3=log864
3log8((X+6)/(X-3))=log864
log8((x-6)/(x-3))=1/3log864
log 8((x-6)/(x-3))log8641/3
log8 [(x+6)/(x-3) ]= log84
(X+6)/(X-3)=4
x+6=4x-12
18=3x
x=6
5. . x ≥ 30
2x+3y ≤ 75
Y ≥ 10
6. AT × BT = (PT)2
12BT=92
BT= ( 81)/( 12)
BT = 6.5
AB=12-6.75
AB=5.25
7. x2+y2-2x+4y=-1
X2+2x+((-2)/2 )2+y2+4y+(4/2 )2=-1+((-2)/2)2+(4/2)2
(x-1)2+(y+2)2=-1+1+4
(x-1)2+(y+2)2=4
8. (√3)/(1-√3/2) × (1+ (√3)/2)/(1+ (√3)/2)
(√3+3/2)/(1-3/4) = (√3+3/2)/(1/4)
=4(√3+3/2 )
=4√3+6
9.
1. . Graph attached
2. g= (52-30)/(60-35)= 22/25
(y-30)/(x-35)=22/25
Y=22/25x -20/( 25)
10. Kathwana d=8000, a=40000
40000 +(5×8000) =80000
Siakago r=1.2
40000(1.2)5=40000×2.488
=99533
Difference =99533-80000
=9533
11. 6(1-cos2x)-cosx-5=0
6-6 cos2x- cosx-5=0
6a2+a-1=0
6a2+3a-2a-1=0
3a(2a+1)-1(2a+1)=0
3a=1 2a= -1
a=( 1)/3 a=-1/2
a=1/3 and - 1/2
=120° and 240°
70.53° 289°
12.
1.
A R 90 75 2 3 180 225
2. 180+225
5
SH. 81 Per Kg
115/100 × 81
= sh.93.00
13. dy/dx =2x-1
Y= ∫(2x-1)dx
Y=x2-x+c
5=12-1+C
C=5
Y=x2-x+5
14.
1. A=πr2
=22/7× 32
=28.2857
15.
1. a (x-5)/(x-2) = (x-2)/(x+2)
(x-5) (x-2)2= (x-2)2
X2-3x-10=x2+4x+4
-3x+4x=14
X=14
2.
1. 14 - 5 = 9 = 3
14 - 2 12 4
2. t3 = ar2
16=a (3/4 )2
A= 16×16
9
256 =284/9
9
3. a= 259/9
r=3/4
sum = (a(1-rn))/(1-rn)
=(259/9(1-(3/4 )6)/(1-3/4)
259/9(1-(3/4 )6 4
259/9(1-729/4069)4
=93.53
16.
1.
2.
1. (25/40×2/5×2/3) or (25/40×3/5×1/3)
1/6+1/8
=7/24
2. (15/40×3/5 ×3/4)+ (15/40×2/5 ×1/4)
(25/40×2/5×2/3)+( 25/40×3/5×1/3)
27/160+3/80+1/6+1/8 = 239/480
3. ( 15/40×3/5 ×1/4) +( 15/40×2/5 ×3/4)
9/(160 )+9/80
=27/160
4. (15/40×3/5 ×1/4) +( 25/40×2/5×1/3)
9/160+1/12
=67/480
17.
1.
1. OP =1/2(√(32+42)
= 2.5
PV = √(62+ 2.52)
=6.964
2. tanθ=6/2
=3
θ= tan -1(3)
71.56°
2.
1. Slant height vm
VM =√(62+22)
=6.325
VN =√(62+ 1.522)
=6.185
2. (3×4)+(1/2×2×6.325×3)+ 2×1/2×60185×4)
12+18.98+24.74
=55.72
18.
1. 71×60×cos60
2130nm
2. 71×4min =284min
=4hrs 44min
1340hrs = 4hrs 44min
=8:16 am
3. 71/360×2×6370 cos 60×22/7
=3948.39km
4. θ/360×2×22/7×6370=1001
θ=(1001×7×360)/(2×22×6370)
=90°
30°s,42°w
19.
1.
2.
1. median =69.5±0.5
2. Q3 = 77.5± 0.5
Q1 = 58.5 ± 0.5
Quartile deriation = (77.5-58.5)/2
19/2=9.5
3. percentage fail =40%
No of student who
Failed = 40/100×50
=20
Pass mark = 65.5± 0.5
20.
1.
1. a=2-t
v=2t-t2 + c
v = 1, t =0
v= 2t- t2+1
2. v= 2t-t2+1 32
2(3)-32+1 - 2(2)-22+1
6-9+1- 4-4+1
-2-1
-3m/s
2.
1. S=t2-t3/3+t+c ,s=0 t=0
S=t2-t3/3+t
2. S=t2-t3/3+t t=3
S = (32)- 33/3+3
9-9+3
=3m
3. . t^2-2t-1=0
t=2±√((-2)2-(4×1×-1)
2
=2± 2.828
2
=2.414
21.
1. x -4 -3 -2 -1 0 1 2 3 y 46 25 10 1 -2 1 10 25
1/2×1(46+46+(25+10+1+ 2+10+25)
= 1/2(92+148)
=120 of units
2. .∫-44 (3x2 -2)= x2-2x
=56-(-56)
=112 of units
3. (116-112)/112×100
22.
1. P` (-1,1), Q` (-4,1) and R` (-5,4)
P`` (-1, -1), Q`` (-4, -1) and R`` (-5, -4)
2. it’s a reflection in the line y=-x or y+x=0
it’s a rotation through +180°or-180°
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# How does the gcf relate two or more whole numbers?
Updated: 9/26/2023
Wiki User
6y ago
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The greatest common factor, or GCF, is the largest positive integer that will divide evenly with no remainder into all the members of a given set of numbers.
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### What 2 or more numbers for a gcf represent?
The GCF is the largest whole number that is a factor for both of the numbers of interest. GCF is useful when simplifying fractions.
### How does the greatest common factor relate two or not whole numbers?
All numbers have factors. Some numbers have some of the same factors as other numbers. These are known as common factors. The largest of these is known as the greatest common factor, or GCF.
### What is the gcf of 45 and 2.1?
GCF refers to whole numbers, not decimals.
### What is the greatest whole number that is a factor of two or more nonzero whole numbers?
That's the greatest common factor, or GCF.
### The greatest whole number that is a factor of two or more nonzero whole numbers?
That's the greatest common factor, or GCF.
### What is the gcf of 8.4 and 18.6?
The GCF is a relationship between whole numbers, not decimals.
### Is there a GCF of 50 and 1000?
Yes. Any group of positive whole numbers has a GCF. For 50 and 1,000, the GCF is 50 .
### Why can a greatest common factor always be found for a set of whole numbers?
Because every set of whole numbers has a GCF.
### What is the GCF of 14.3584?
The greatest common factor (GCF) refers to a factor that is COMMON to two or more numbers. You have only one number in the question! Furthermore, the concept of GCF makes sense only in the context of integers. Once you allow fractions, every fraction is divisible by every non-zero fraction.
### What does GCF mean in fractions?
GCF doesn't apply to fractions, only to whole numbers. When those numbers are the numerator and denominator of a fraction, the GCF can be used to reduce it to its simplest form.
### What is the GCF of 156?
You need at least two numbers to find a GCF.
### Can two numbers have more than one GCF?
No; two numbers can have only 1 GCF. | 585 | 2,368 | {"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} | 3.96875 | 4 | CC-MAIN-2023-40 | latest | en | 0.936988 |
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Waves are for SURFING… yeah!
Waves are for SURFING… yeah!. Monterey, CA. Cayucos, CA. Get outta there. Ghost tree. Q15.1. If you double the wavelength l of a wave on a string, what happens to the wave speed v and the wave frequency f?. A. v is doubled and f is doubled. B. v is doubled and f is unchanged.
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Waves are for SURFING… yeah!
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1. Waves are for SURFING… yeah! Monterey, CA Cayucos, CA
2. Get outta there Ghost tree
3. Q15.1 If you double the wavelength l of a wave on a string, what happens to the wave speed v and the wave frequency f? A. v is doubled and f is doubled. B. v is doubled and f is unchanged. C. v is unchanged and f is halved. D. v is unchanged and f is doubled. E. v is halved and f is unchanged.
4. A15.1 If you double the wavelength l of a wave on a string, what happens to the wave speed v and the wave frequency f? A. v is doubled and f is doubled. B. v is doubled and f is unchanged. C. v is unchanged and f is halved. D. v is unchanged and f is doubled. E. v is halved and f is unchanged.
5. Q15.2 Which of the following wave functions describe a wave that moves in the –x-direction? A. y(x,t) = A sin (–kx – wt) B. y(x,t) = A sin (kx + wt) C. y(x,t) = A cos (kx + wt) D. both B. and C. E. all of A., B., and C.
6. A15.2 Which of the following wave functions describe a wave that moves in the –x-direction? A. y(x,t) = A sin (–kx – wt) B. y(x,t) = A sin (kx + wt) C. y(x,t) = A cos (kx + wt) D. both B. and C. E. all of A., B., and C.
7. A new swell A swell arrived last night and the buoys indicated an 18 second period with 3m height. Sweet!! The wave velocity where the buoys are located is 28 m/s. What is the amplitude, angular frequency, wavelength and wave number of the wave? Write a wave function describing the wave. What is the velocity of the buoy 5 s after the peak of the wave passes under it? Rincon
8. The speed of a transverse wave • In the first method we will consider a pulse on a string. • Figure 15.11 will show one approach.
9. Q15.3 y A wave on a string is moving to the right. This graph of y(x, t) versus coordinate x for a specific time t shows the shape of part of the string at that time. At this time, what is the velocity of a particle of the string at x = a? 0 x a A. The velocity is upward. B. The velocity is downward. C. The velocity is zero. D. not enough information given to decide
10. A15.3 y A wave on a string is moving to the right. This graph of y(x, t) versus coordinate x for a specific time t shows the shape of part of the string at that time. At this time, what is the velocity of a particle of the string at x = a? 0 x a A. The velocity is upward. B. The velocity is downward. C. The velocity is zero. D. not enough information given to decide
11. Q15.4 y A wave on a string is moving to the right. This graph of y(x, t) versus coordinate x for a specific time t shows the shape of part of the string at that time. At this time, what is the acceleration of a particle of the string at x = a? 0 x a A. The acceleration is upward. B. The acceleration is downward. C. The acceleration is zero. D. not enough information given to decide
12. A15.4 y A wave on a string is moving to the right. This graph of y(x, t) versus coordinate x for a specific time t shows the shape of part of the string at that time. At this time, what is the acceleration of a particle of the string at x = a? 0 x a A. The acceleration is upward. B. The acceleration is downward. C. The acceleration is zero. D. not enough information given to decide
13. Q15.5 y A wave on a string is moving to the right. This graph of y(x, t) versus coordinate x for a specific time t shows the shape of part of the string at that time. At this time, what is the velocity of a particle of the string at x = b? 0 x b A. The velocity is upward. B. The velocity is downward. C. The velocity is zero. D. not enough information given to decide
14. A15.5 y A wave on a string is moving to the right. This graph of y(x, t) versus coordinate x for a specific time t shows the shape of part of the string at that time. At this time, what is the velocity of a particle of the string at x = b? 0 x b A. The velocity is upward. B. The velocity is downward. C. The velocity is zero. D. not enough information given to decide
15. The speed of a transverse wave II • What happens to wave velocity, frequency and wavelength of waves travelling up the rope? • v, f, l • v, f , l • v, f =, l • v, f, l = • Not enough information
16. The speed of a transverse wave II • Nylon rope is tied to a stationary support at the top of a vertical mine shaft 80 m deep. The rope is stretched taut by a 20 kg box of mineral samples at the bottom. The mass of the rope is 2 kg. The geologist at the bottom signals by jerking the rope sideways. • What is the speed of the transverse wave on the rope? • If the rope is given SHM with frequency 2 Hz, how many cycles (wavelengths) are there in the rope’s length?
17. Q15.8 The four strings of a musical instrument are all made of the same material and are under the same tension, but have different thicknesses. Waves travel A. fastest on the thickest string. B. fastest on the thinnest string. C. at the same speed on all strings. D. not enough information given to decide
18. A15.8 The four strings of a musical instrument are all made of the same material and are under the same tension, but have different thicknesses. Waves travel A. fastest on the thickest string. B. fastest on the thinnest string. C. at the same speed on all strings. D. not enough information given to decide
19. The speed of waves on a string and speed of sound • A note on a bass guitar produces sound waves. The frequency of the string’s oscillations is 20 Hz and the wavelength was 10m. What happens to the velocity, frequency and wavelength of the waves when they leave the guitar and propagate through the air? • v, f, l • v, f , l • v, f =, l • v, f, l = • Not enough information
20. Q33.1 When light passes from vacuum (index of refraction n = 1) into water (n = 1.333), A. the wavelength increases and the frequency is unchanged. B. the wavelength decreases and the frequency is unchanged. C. the wavelength is unchanged and the frequency increases. D. the wavelength is unchanged and the frequency decreases. E. both the wavelength and the frequency change.
21. A33.1 When light passes from vacuum (index of refraction n = 1) into water (n = 1.333), A. the wavelength increases and the frequency is unchanged. B. the wavelength decreases and the frequency is unchanged. C. the wavelength is unchanged and the frequency increases. D. the wavelength is unchanged and the frequency decreases. E. both the wavelength and the frequency change.
22. A light wave travels from vacuum, through a transparent material, and back to vacuum. What is the index of refraction of this material? Explain v = c v = c
23. Wave intensity • Go beyond the wave on a string and visualize, say … a sound wave spreading from a speaker. That wave has intensity dropping as 1/r2.
24. Thumping bass • You are 3m from your friend’s car, who is pumping obnoxiously loud, bass-heavy music through his subwoofer-equipped car, causing the entire car to rattle. He yells to you that he’s got a 800W subwoofer in his car and its cranked to full volume. Unimpressed, you measure a peak sound intensity of 4.4 W/m2 on your handy Radioshack soundmeter. How many watts is your friend’s subwoofer putting out? • What is the sound intensity for your friend in the car, at 1 m away from the subwoofer?
25. The logarithmic decibel scale of loudness • Our ears can comfortably hear sound over 12 orders of magnitude in intensity!
26. Thumping bass • Your friend has fixed the “problem” with their subwoofer and is now pumping 800W of obnoxiously loud, bass-heavy music through their subwoofer-equipped car, causing the entire car to rattle. If the bass note is vibrating at 40 Hz, what is the maximum displacement of air at 1 m away from the subwoofer? • How many dB’s is this? • Why does your friend have an 800W subwoofer instead of an 800W amplifier for all sound frequencies?
27. The Doppler Effect II—moving listener, moving source • As the object making the sound moves or as the listener moves (or as they both move), the velocity of sound is shifted enough to change the pitch perceptively.
28. Q16.8 On a day when there is no wind, you are moving toward a stationary source of sound waves. Compared to what you would hear if you were not moving, the sound that you hear has A. a higher frequency and a shorter wavelength. B. the same frequency and a shorter wavelength. C. a higher frequency and the same wavelength. D. the same frequency and the same wavelength.
29. A16.8 On a day when there is no wind, you are moving toward a stationary source of sound waves. Compared to what you would hear if you were not moving, the sound that you hear has A. a higher frequency and a shorter wavelength. B. the same frequency and a shorter wavelength. C. a higher frequency and the same wavelength. D. the same frequency and the same wavelength.
30. Q16.9 On a day when there is no wind, you are at rest and a source of sound waves is moving toward you. Compared to what you would hear if the source were not moving, the sound that you hear has A. a higher frequency and a shorter wavelength. B. the same frequency and a shorter wavelength. C. a higher frequency and the same wavelength. D. the same frequency and the same wavelength.
31. A16.9 On a day when there is no wind, you are at rest and a source of sound waves is moving toward you. Compared to what you would hear if the source were not moving, the sound that you hear has A. a higher frequency and a shorter wavelength. B. the same frequency and a shorter wavelength. C. a higher frequency and the same wavelength. D. the same frequency and the same wavelength.
32. A double Doppler shift • What is the frequency of the 300 Hz siren that’s heard by (a) a person in the warehouse and (b) in the police car as it travels towards the warehouse?
33. Doppler shift from several sources You are standing at x = 0m, listening to seven identical sound sources. At t = 0, all seven are at x = 343 m and moving shown below. The sound from all seven will reach your ear at t = 1s. 50 m/s speeding up 50 m/s, steady speed 50m/s, slowing down At rest 1 2 3 4 50 m/s speeding up 50 m/s, steady speed 50m/s, slowing down 5 6 7 Rank in order, from highest to lowest, the seven frequencies f1 to f7 that you hear at t = 1s.
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https://www.convert-measurement-units.com/convert+Kilonewton+to+Poundal.php | 1,632,680,885,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057913.34/warc/CC-MAIN-20210926175051-20210926205051-00433.warc.gz | 743,104,134 | 12,716 | Convert kN to pdl (Kilonewton to Poundal)
## Kilonewton into Poundal
numbers in scientific notation
https://www.convert-measurement-units.com/convert+Kilonewton+to+Poundal.php
## How many Poundal make 1 Kilonewton?
1 Kilonewton [kN] = 7 233.013 851 209 9 Poundal [pdl] - Measurement calculator that can be used to convert Kilonewton to Poundal, among others.
# Convert Kilonewton to Poundal (kN to pdl):
1. Choose the right category from the selection list, in this case 'Force'.
2. Next enter the value you want to convert. The basic operations of arithmetic: addition (+), subtraction (-), multiplication (*, x), division (/, :, ÷), exponent (^), brackets and π (pi) are all permitted at this point.
3. From the selection list, choose the unit that corresponds to the value you want to convert, in this case 'Kilonewton [kN]'.
4. Finally choose the unit you want the value to be converted to, in this case 'Poundal [pdl]'.
5. Then, when the result appears, there is still the possibility of rounding it to a specific number of decimal places, whenever it makes sense to do so.
With this calculator, it is possible to enter the value to be converted together with the original measurement unit; for example, '252 Kilonewton'. In so doing, either the full name of the unit or its abbreviation can be usedas an example, either 'Kilonewton' or 'kN'. Then, the calculator determines the category of the measurement unit of measure that is to be converted, in this case 'Force'. After that, it converts the entered value into all of the appropriate units known to it. In the resulting list, you will be sure also to find the conversion you originally sought. Alternatively, the value to be converted can be entered as follows: '44 kN to pdl' or '18 kN into pdl' or '53 Kilonewton -> Poundal' or '64 kN = pdl' or '28 Kilonewton to pdl' or '77 kN to Poundal' or '47 Kilonewton into Poundal'. For this alternative, the calculator also figures out immediately into which unit the original value is specifically to be converted. Regardless which of these possibilities one uses, it saves one the cumbersome search for the appropriate listing in long selection lists with myriad categories and countless supported units. All of that is taken over for us by the calculator and it gets the job done in a fraction of a second.
Furthermore, the calculator makes it possible to use mathematical expressions. As a result, not only can numbers be reckoned with one another, such as, for example, '(27 * 20) kN'. But different units of measurement can also be coupled with one another directly in the conversion. That could, for example, look like this: '252 Kilonewton + 756 Poundal' or '55mm x 69cm x 46dm = ? cm^3'. The units of measure combined in this way naturally have to fit together and make sense in the combination in question.
If a check mark has been placed next to 'Numbers in scientific notation', the answer will appear as an exponential. For example, 6.444 521 546 293 1×1027. For this form of presentation, the number will be segmented into an exponent, here 27, and the actual number, here 6.444 521 546 293 1. For devices on which the possibilities for displaying numbers are limited, such as for example, pocket calculators, one also finds the way of writing numbers as 6.444 521 546 293 1E+27. In particular, this makes very large and very small numbers easier to read. If a check mark has not been placed at this spot, then the result is given in the customary way of writing numbers. For the above example, it would then look like this: 6 444 521 546 293 100 000 000 000 000. Independent of the presentation of the results, the maximum precision of this calculator is 14 places. That should be precise enough for most applications. | 923 | 3,749 | {"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-2021-39 | latest | en | 0.836419 |
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Measuring odds requires you to use a very particular formula designed for this exact purpose. Measure odds with help from a longtime mathematics educator in this free video clip.
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Hi, I'm Jimmy Chang, and we're here to talk about how to measure odds. Now, in terms of measuring odds, there's actually a formula associated with that and that the odds of an event happening, you're taking the number of outcomes for the event E and you're pitting it, and there's a colon there, the number of outcomes against the event happening. So, the idea here is if you're flipping a coin and you want to find out the odds of a head, okay; basically, you're thinking about the number of ways you can get a head, if you're flipping the coin once, it'll just be one. And then, the number of ways you cannot get a head which is tail and that would be one; so, the odds of flipping a coin and getting a head is one to one. Now, if you have cards for example and you want to find out the odds of drawing a jack, you want to think about how many ways can you draw a jack and that's going to be four and the odds, and the number of ways you don't draw a jack will be 48, if you're talking about, of a 52-card deck. So, the odds of getting a jack is four to 48. Now, odds can be reduced if you can reduce both numbers and so four to 48 can be reduced as one to 12. So, when it comes to measuring odds, as long as you follow this formula and are aware of different numbers involving different models, you should be fine. So, I'm Jimmy Chang, and that's a brief description on how to measure odds.
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## Presentation on theme: "Walker, Chapter 23 Magnetic Flux and Faraday’s Law of Induction"— Presentation transcript:
Walker, Chapter 23 Magnetic Flux and Faraday’s Law of Induction
Magnetic Induction Demonstrations
Ammeter for overhead projector which measures the current in a coil. Under what circumstances is a current induced in the coil? How do we get the largest current? Disk launcher with Al ring Slit ring Fe ring Bakelite ring coils with bulbs
Electric Currents produce Magnetic Fields
Chapter 22: Electric currents (in a wire, in a plasma, in a fluid solution, inside an atom) produce a disturbance in the surrounding space called the magnetic field. This magnetic field produces forces on any other macroscopic or microscopic currents. Example: MRI: Magnetic field (several Tesla) from superconducting solenoid induces a net alignment of the microscopic currents inside each and every proton at the center of the Hydrogen atoms in your body
Induced emf (Voltage) from changing Magnetic Flux
Electric currents produce magnetic fields. 19th century puzzle: Can magnetic fields produce currents? A static magnet will produce no current in a stationary coil. Faraday: If the magnetic field changes, or if the magnet and coil are in relative motion, there will be an induced emf (and therefore current) in the coil. Key Concept: The magnetic flux through the coil must change. This will induce an emf e in the coil, which produces a current I = e/R in the coil. Such a current is said to be induced by the varying B-field.
Magnetic Flux For a “loop” of wire (not necessarily circular) with area A, in an external magnetic field B, the magnetic flux is: A = area of loop q = angle between B and the normal to the loop SI units of Magnetic Flux: 1 T·m2 = 1 weber = 1 Wb
Current Loop Reminder: Current in a loop generates a magnetic field (and therefore magnetic flux). The magnetic field generated by this current is into the page inside the loop, and out of the page outside the loop. RHR: Point your (right-hand) thumb along the direction of the current. Your fingers point in the direction of the magnetic field (and the magnetic flux). OR Curl your fingers around the loop in the direction of the current. Your (right-hand) thumb points in the direction of the magnetic field this current generates through the loop.
Walker Problem 3, pg. 778 A magnetic field is oriented at an angle of 32º to the normal of a rectangular area 5.5 cm by 7.2 cm. If the magnetic flux through this surface has a magnitude of 4.8 10-5 T·m2, what is the strength of the magnetic field?
Faraday’s Law: The instantaneous emf in a circuit (w/ N loops) equals the rate of change of magnetic flux through the circuit: The minus sign indicates the direction of the induced emf. To calculate the magnitude:
Examples of Induced Current
Any change of current in primary induces a current in secondary.
Induced Current The current in the primary polarizes the material of the core. The magnetic field of the primary solenoid is enhanced by the magnetic field produced by these atomic currents. This magnetic field remains confined in the iron core, and only fans out and loops back at the end of the core. Any change in the current in the primary (opening or closing switch) produces a change in the magnetic flux through the secondary coil. This induces a current in the secondary.
Induction by Relative Motion
When a permanent magnet moves relative to a coil, the magnetic flux through the coil changes, inducing an emf in the coil. In a) the magnitude of the flux is increasing In c) the flux is decreasing in magnitude. In a) and c) the induced current has opposite sign. v
Induction by Rotational Motion
As a coil rotates in a constant magnetic field (uniform or not) the flux through the loop changes, inducing an emf in the coil.
Walker Problem 10, pg. 778 This is a plot of the magnetic flux through a coil as a function of time. At what times shown in this plot does (a) the magnetic flux and (b) the induced emf have the greatest magnitude?
Walker Problem 9, pg. 778 A 0.25 T magnetic field is perpendicular to a circular loop of wire with 50 turns and a radius 15 cm. The magnetic field is reduced to zero in 0.12 s. What is the magnitude of the induced emf?
Lenz’s Law Lenz’s Law: An induced current always flows in a direction that opposes the change that caused it. N S Induced current Induced B field Magnet moving down toward loop In this example the magnetic field in the downward direction through the loop is increasing. So a current is generated in the loop which produces an upward magnetic field inside the loop to oppose the change.
Walker Problem 24, pg. 779 The figure shows a circuit containing a resistor and an uncharged capacitor. Pointing into the plane of the circuit is a uniform magnetic field B. If the magnetic field increases in magnitude with time, which plate of the capacitor (top or bottom) becomes positively charged?
Motional emf x x + x x x x + L x x x x x x v - x x x - x x x
An emf will also be produced if a conductor moves through a magnetic field. The emf comes from the motion of charges, which are free to move in the conductor. In this example, why does the top of the rod become positively charged? x x + x x x x + L x x x x x x v - x x x - x x x
If the moving conductor is part of a circuit, the flux through the circuit will change with time and a current will be induced (Area of loop = Ls): x v L R s
Walker Problems 30-31, pg. 780 The figure shows a zero-resistance rod sliding to the right on two zero-resistance rails separated by the distance L = 0.45 m. The rails are connected by a 12.5 W resistor, and the entire system is in a uniform magnetic field with a magnitude of 0.75 T. (a) If the velocity of the bar is 5.0 m/s to the right, what is the current in the circuit? (b) What is the direction of the current in the circuit? (c) What is the magnetic force on the bar? (d) What force must be applied to keep the bar moving at constant velocity?
Eddy Currents When a conductor is moved in a magnetic field, there is a force on the electrons, which then move in the metal. This movement is called an eddy current. The induced currents cause magnetic fields which tend to oppose the motion of the metal.
The magnetic flux through the loop changes, so an emf is induced.
Generators A generator converts mechanical energy to electrical energy. Consider a current loop which rotates in a constant magnetic field: The magnetic flux through the loop changes, so an emf is induced. If a loop of area A with N turns rotates with angular speed w (period of rotation = 2p/w) in a constant B field, then the instantaneous induced emf is: = NBAw sin(wt) If this loop is part of a circuit, this emf will induce an Alternating Current (AC) in the circuit.
Generator A coil of wire turns in a magnetic field. The flux in the coil is constantly changing, generating an emf in the coil.
Self-Inductance If you try to change the current in a circuit instantaneously, the response will instead be gradual. This is because the circuit produces a self-induced emf to initially oppose any changes as prescribed by Lenz’s Law. This effect is known as self-induction. This does not violate the Newtonian principle of no-self-forces, because in effect individual electrons in the current are exerting forces on the other electrons in the same circuit.
Inductance The self induced emf is given by:
where L is called the inductance of the circuit. The magnetic flux through the loop, produced by current in the loop, is proportional to the current. The inductance L is the constant of proportionality. The unit of inductance is the Henry 1 H = 1 T·m2/A = 1 (T·m2/s) (s/A) = 1 V·s/A Note that inductance, like capacitance, is purely geometrical.
Inductance of a Solenoid
A solenoid has inductance given by L = inductance of the solenoid N = # of turns in solenoid l = length of solenoid A = cross sectional area of solenoid n = # of turns per unit length
Walker Problem 41, pg. 780 The inductance of a solenoid with 450 turns and a length of 24 cm is 7.3 mH. (a) What is the cross-sectional area of the solenoid? (b) What is the induced emf in the solenoid if its current drops from 3.2 A to 0 in 55 ms?
RL Circuits We can construct a circuit from inductors and resistors. The circuit will behave just like an RC circuit, with a time constant given by: t = L/R
Walker Problem 45, pg. 780 (a) How long does it take for the current in an RL circuit with R = 130 W and L = 63 mH to reach half its final value? (b) If the emf in the circuit is 10 V, what is the current in this circuit two characteristic time intervals after closing the switch?
Energy Stored in an Inductor
Just as energy can be stored in a capacitor (recall that U= ½CV2), energy can also be stored in an inductor: U = ½LI2 Whereas energy in a capacitor is stored in the electric field between the plates, energy in an inductor is stored in the magnetic field within the inductor.
Transformers A transformer is a device used to change the voltage in a circuit. AC currents must be used. 120 V in your house 75,000 V in the power lines p = primary s = secondary
Walker Problem 57, pg. 781 A disk drive plugged into a 120-V outlet operates on a voltage of 9.0 V. The transformer that powers the disk drive has 125 turns on its primary coil. (a) Should the number of turns on the secondary coil be greater than or less than 125? (b) Find the number of turns on the secondary coil. | 2,274 | 9,560 | {"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} | 3.609375 | 4 | CC-MAIN-2022-05 | latest | en | 0.865822 |
https://math.stackexchange.com/questions/2253490/how-to-linearize-a-not-linear-program-with-excel | 1,571,824,640,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987833089.90/warc/CC-MAIN-20191023094558-20191023122058-00019.warc.gz | 597,379,552 | 31,536 | # How to linearize a not linear program with excel?
I want to use Excel solver to solve the following integer linear program:
A company wants to select $p$ locations among a set of $m$ possible sites for constructing polluting plants in a contemporary world. The $m$ candidate sites are located on a territory containing different cities. We have:
• $d_{ij}$ the distance between city $i$ and site $j$
• $P_i$ the population (in thousand of inhabitants) of city $i$
I imagined experts that thought that a city was threatened if there was a polluting plant located less than 2 km from it.
The authorities' point of view wants to minimize nuisance. They want to minimize the number of inhabitants threatened by the $p$ selected plants.
• At most 5% of the population of $n$ cities are threatened (constraint imposed by the authorities).
• The demand of a city is delivered by a single plant
• The company minimizes its total transportation cost
# My attempt
$$\begin{cases} \min & \sum_i b_{i}p_i\\ %&s_{i}=b_i\times P_i\mbox{ number of people of city i endangered by site j}\\ &b_i\le b_j\\ &\sum_{j=1}^mb_j=p\\ &2- d_{ij} \le M. b_{i} \\ %\mbox{ constraint over the distance to flag an endangered population}\\ %&\sum_i b_i= \mbox{ number of sites we want}\\ &b_i,b_j \in \{0,1\},\forall i \in \mathbb{N}^* \end{cases}$$
• $b_i$ a boolean variable telling me if city $i$ is threatened.
• $b_j$ a boolean variable telling me if location $j$ is used to settle a plant.
• $M$ is a very large number
We want to minimize the sum of $\sum b_i p_i$, the number of people endangered. I thought about creating a variable $b_i$, standing for "take into account population $i$" with the following test :
$$2- d_{ij} \le M. b_i$$
$$\begin{cases} b_i = 0 \mbox{ if we don't have to take the population of city i into account}\\ b_i = 1 \mbox{ otherwise} \end{cases}$$
But I think it is too much variables. I don't think that $b_j$ is necesary because if a site wasn't used to settle a plant we won't have the distances. Would the following program be better ?
$$\begin{cases} \min & \sum_i b_{i}p_i\\ %&s_{i}=b_i\times P_i\mbox{ number of people of city i endangered by site j}\\ &2- d_{ij} \le M. b_{i} \\ %\mbox{ constraint over the distance to flag an endangered population}\\ %&\sum_i b_i= \mbox{ number of sites we want}\\ &b_i,b_j \in \{0,1\},\forall i \in \mathbb{N}^* \end{cases}$$
• $b$ means two different things. The alphabet has 20+ letters. Use them. – Rodrigo de Azevedo Apr 27 '17 at 9:39
• What do you mean by better? – Michiel uit het Broek Jun 27 '17 at 8:25 | 763 | 2,580 | {"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-2019-43 | latest | en | 0.908385 |
https://www.calculatorbit.com/en/length/10-terameter-to-hectometer | 1,723,036,764,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640694449.36/warc/CC-MAIN-20240807111957-20240807141957-00703.warc.gz | 551,046,814 | 7,750 | 10 Terameter to Hectometer Calculator
Result:
10 Terameter = 100000000000 Hectometer (hm)
Rounded: ( Nearest 4 digits)
10 Terameter is 100000000000 Hectometer (hm)
10 Terameter is 10000000000km
How to Convert Terameter to Hectometer (Explanation)
• 1 terameter = 10000000000 hm (Nearest 4 digits)
• 1 hectometer = 1e-10 Tm (Nearest 4 digits)
There are 10000000000 Hectometer in 1 Terameter. To convert Terameter to Hectometer all you need to do is multiple the Terameter with 10000000000.
In formula distance is denoted with d
The distance d in Hectometer (hm) is equal to 10000000000 times the distance in terameter (Tm):
Equation
d (hm) = d (Tm) × 10000000000
Formula for 10 Terameter (Tm) to Hectometer (hm) conversion:
d (hm) = 10 Tm × 10000000000 => 100000000000 hm
How many Hectometer in a Terameter
One Terameter is equal to 10000000000 Hectometer
1 Tm = 1 Tm × 10000000000 => 10000000000 hm
How many Terameter in a Hectometer
One Hectometer is equal to 1e-10 Terameter
1 hm = 1 hm / 10000000000 => 1e-10 Tm
terameter:
The terameter (symbol: Tm) is a unit of length in the metric system equal to 1000000000000 meters that is 10^12 meters. 1 terameter is equal to 1 billion kilometers or 6.7 astronomical units. 1 terameter is also equal to 1/1000 petameter.
hectometer:
The hectometer (symbol: hm) is unit of length in the International System of Units (SI), equal to 100 meters. Hectometer word is combination of two words 'hecto'+'meter', 'hecto' means 'hundread'. The hectare(ha) is common metric unit for land area that is equal to one square hectomter(hm^2).
Terameter to Hectometer Calculations Table
Now by following above explained formulas we can prepare a Terameter to Hectometer Chart.
Terameter (Tm) Hectometer (hm)
6 60000000000.00001
7 70000000000.00002
8 80000000000.00002
9 90000000000.00002
10 100000000000.00002
11 110000000000.00002
12 120000000000.00002
13 130000000000.00002
14 140000000000.00003
15 150000000000.00003
Nearest 4 digits
Convert from Terameter to other units
Here are some quick links to convert 10 Terameter to other length units.
Convert to Terameter from other units
Here are some quick links to convert other length units to Terameter.
More Hectometer to Terameter Calculations
Converting from one Terameter to Hectometer or Hectometer to Terameter sometimes gets confusing.
Is 10000000000 Hectometer in 1 Terameter?
Yes, 1 Terameter have 10000000000 (Nearest 4 digits) Hectometer.
What is the symbol for Terameter and Hectometer?
Symbol for Terameter is Tm and symbol for Hectometer is hm.
How many Terameter makes 1 Hectometer?
1e-10 Terameter is euqal to 1 Hectometer.
How many Hectometer in 10 Terameter?
Terameter have 100000000000 Hectometer.
How many Hectometer in a Terameter?
Terameter have 10000000000 (Nearest 4 digits) Hectometer. | 915 | 2,833 | {"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-2024-33 | latest | en | 0.672871 |
http://fractioncalculator.pro/fractions-simplifier/Simplify__93/55_ | 1,529,403,594,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267862248.4/warc/CC-MAIN-20180619095641-20180619115641-00229.warc.gz | 116,955,406 | 13,214 | # Simplify 93/55
Reduce / simplify any fraction to its lowest terms by using our Fraction to the Simplest Form Calculator. Find the answer to questions like: Simplify 93/55 or what is 93/55 reduced to the simplest form?
### Fractions Simplifier
Please fill in the two left boxes below:
Inputfraction: Integerpart: Fractionpart: Fraction to Decimal: = = Details: Details...
## How to reduce a fraction
Among different ways simplifying a fraction, we will show the two procedure below:
### Method 1 - Divide by a Small Number When Possible
Start by dividing both the numerator and the denomiator of the fraction by the same number, and repeat this until it is impossible to divide. Begin dividing by small numbers like 2, 3, 5, 7. For example,
### Simplify the fraction 42/98
• First divide both (numerator/denominator) by 2 to get 21/49.
• Dividing by 3 and 5 will not work, so,
• Divide both numerator and denominator by 7 to get 3/7. Note: 21 ÷ 7 = 3 and 49 ÷ 7 = 7
In the fraction 3/7, 3 is only divisible by itself, and 7 is not divisible by other numbers than itself and 1, so the fraction has been simplified as much as possible. No further reduction is possible, so 42/98 is equal to 3/7 when reduced to its lowest terms. This is a PROPER FRACTION once the absolute value of the top number or numerator (3) is smaller than the absolute value of the bottom number or denomintor (7).
### Method 2 - Greatest Common Divisor
To reduce a fraction to lowest terms (also called its simplest form), just divide both the numerator and denominator by the GCD (Greatest Common Divisor).
For example, 3/4 is in lowest form, but 6/8 is not in lowest form (the GCD of 6 and 8 is 2) and 6/8 can be written as 3/4. You can do this because the value of a fraction will remain the same when both the numerator and denominator are divided by the same number.
Note: The Greatest Common Factor (GCF) for 6 and 8, notation gcf(6,8), is 2. Explanation:
• Factors of 6 are 1,2,3,6;
• Factors of 8 are 1,2,4,8.
So, it is ease see that the 'Greatest Common Factor' or 'Divisor' is 2 because it is the greatest number which divides evenly into all of them. | 579 | 2,152 | {"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} | 4.53125 | 5 | CC-MAIN-2018-26 | latest | en | 0.911726 |
https://blog.listcomp.com/math/2017/04/12/digest-of-information-theory-inference-and-learning-algorithms | 1,685,680,945,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648322.84/warc/CC-MAIN-20230602040003-20230602070003-00233.warc.gz | 158,916,913 | 12,984 | Digest of Information Theory, Inference, and Learning Algorithms
Yao Yao on April 12, 2017
1. Introduction to Information Theory
1.1 How can we achieve perfect communication over an imperfect, noisy communication channel?
In some cases, if we transmit data, e.g., a string of bits, over the channel, there is some probability $f$ that the received message will not be identical to the transmitted message. We would prefer to have a communication channel for which this $f$ was 0 – or so close to 0 that for practical purposes it is indistinguishable from 0.
E.g.
• modem $\rightarrow$ phone line $\rightarrow$ modem
• Galileo $\rightarrow$ radio waves $\rightarrow$ Earth
• parent cell $\rightarrow$ $\begin{cases} \text{daughter cell A} \newline \text{daughter cell B} \end{cases}$
• RAM $\rightarrow$ Hard Drive $\rightarrow$ RAM
N.B. 当然也有主动引入 noise 的情况，比如你把一个 720p 转成 480p
• √：$P(y=0 \mid x=0) = 1-f$
• √：$P(y=1 \mid x=1) = 1-f$
• ×：$P(y=1 \mid x=0) = f$
• ×：$P(y=0 \mid x=1) = f$
N.B. This is a Binary Symmetric Channel, A.k.a BSC. In BSC model, a transmitter wishes to send a bit (a zero or a one), and the receiver receives a bit.
A useful disk drive would flip no bits at all in its entire lifetime. If we expect to read and write a gigabyte per day for ten years, we require a bit error probability of the order of 10 −15 , or smaller. There are two approaches to this goal.
• The physical solution： to improve the physical characteristics of the communication channel to reduce its error probability.
• 简单说就是提高 channel 的产品工艺
• 缺点是会提高 channel 的物理成本
• The ‘system’ solution: we accept the given noisy channel as it is and add communication systems to it so that we can detect and correct the errors introduced by the channel.
• We add an encoder before the channel and a decoder after it.
• The encoder encodes the source message $s$ into a transmitted message $t$, adding redundancy to $s$ in some way.
• The channel adds noise to $t$, yielding a received message $r$.
• The decoder uses the known redundancy introduced by the encoder to infer both the original signal $s$ and the added noise.
• $s \xrightarrow[\text{add redundancy}]{\text{encoder}} t \xrightarrow[\text{may add noise}]{\text{channel}} r \xrightarrow[\text{infer}]{\text{decoder}} \hat{s}$
• 新添加了 system 的 computational cost
Information theory is concerned with the theoretical limitations and po- tentials of such systems. ‘What is the best error-correcting performance we could achieve?’
Coding theory is concerned with the creation of practical encoding and decoding systems.
1.2 Error-correcting codes for BSC
Repetition codes
A straightforward idea is to repeat every bit of the message a prearranged number of times – for example, three times:
$s$ $t$
0 000
1 111
We call this repetition code $R_3$.
We can spell out the posterior probability of the two alternatives thus:
\begin{align} P(s=1 \mid r) = \frac{ P(r \mid s = 1) P(s=1)}{P(r)} \newline P(s=0 \mid r) = \frac{ P(r \mid s = 0) P(s=0)}{P(r)} \end{align}
• $P(s)$ is the prior probability of $s$
• $P(r \mid s)$ is the likihood of $s$
• We assume that the prior probabilities are equal: $P(s = 0) = P(s = 1) = 0.5$; then maximizing the posterior probability $P(s \mid r)$ is equivalent to maximizing the likelihood $P(r \mid s)$.
• And we assume that this BSC has noise level $f \le 0.5$,
So the likelihood is
$P(r \mid s) = P(r \mid t) = P(r_1r_2r_3 \mid t_1t_2t_3) = \prod_{i=1}^{3} P(r_i \mid t_i)$
where
$P(r_i \mid t_i) = \begin{cases} 1 - f & \text{ if } r_i = t_i \newline f & \text{ if } r_i \neq t_i \end{cases}$
Thus the likelihood ratio for the two hypotheses is
$\frac{ P(r \mid s = 1) }{ P(r \mid s = 0) } = \prod_{i=1}^{3} \frac{ P(r_i \mid t_i = 1) }{ P(r_i \mid s_i = 0) }$
where each factor $\frac{ P(r_i \mid t_i = 1) }{ P(r_i \mid s_i = 0) }$ equals $\frac{1-f}{f}$ if $r_i = 1$, and $\frac{f}{1-f}$ if $r_i=0$.
We guess $\hat{s} = 1$ if $\frac{ P(r_i \mid t_i = 1) }{ P(r_i \mid s_i = 0) } > 1$ and $\hat{s} = 0$ otherwise.
Let’s define $\gamma \equiv \frac{1-f}{f}$. $\gamma \ge 1$ since $f \le 0.5$. 假设 $r_i = 1$ 的个数为 $k$：
$\frac{ P(r \mid s = 1) }{ P(r \mid s = 0) } = \gamma^k \frac{1}{\gamma^{3-k}} = \gamma^{2k-3}$
Exercise 1.2. Show that the error probability is reduced by the use of $R_3$ by computing the error probability of this code for a BSC with noise level $f$.
Assume the probability of bit error is $p_b$ (注意 $p_b$ 是整个信息传递过程的出错概率，$f$ 仅仅是 channel 的出错概率)
Previous noise level: $p_b = f$
$R_3$ noise level: $p_b = f^3 + 3 f^2 (1-f)$
Exercise 1.3.
The probability of error for $R_N$ is
$p_b = \sum_{n=\lceil \frac{N}{2} \rceil}^{N} { N \choose n } f^n (1-f)^{N-n}$
(待补充)
The repetition code $R_3$ has therefore reduced the probability of error, as desired. Yet we have lost something: our rate of information transfer (姑且理解为 $\frac{len(s)}{len(r)}$) has fallen to $\frac{1}{3}$.
Block codes – the $(7, 4)$ Hamming code
We would like to communicate with tiny probability of error and at a substantial rate. Can we improve on repetition codes? What if we add redundancy to blocks of data instead of encoding one bit at a time? We now study a simple block code.
$\vert s \vert = K, \vert t \vert = N$. $N > k$ because we are adding redundancy.
In a linear block code, the extra $N − K$ bits are linear functions of the original $K$ bits; these extra $N − K$ bits are called parity-check bits. An example of a linear block code is the $(7, 4)$ Hamming code, which transmits $N = 7$ bits for every $K = 4$ source bits.
• parity: (of a number) the fact of being even or odd.
Let’s define
$\operatorname{parity}(x_1,\dots,x_n) = \left ( \sum_{i=1}^{n}x_i \right ) \bmod 2$
N.B. 这个 $\operatorname{parity}()$ 是个 linear function！linear 出来了！
• $t_1 = s_1$
• $t_2 = s_2$
• $t_3 = s_3$
• $t_4 = s_4$
• $t_5 = \operatorname{parity}(s_1, s_2, s_3)$
• $t_6 = \operatorname{parity}(s_2, s_3, s_4)$
• $t_7 = \operatorname{parity}(s_1, s_3, s_4)$
Because the Hamming code is a linear code, it can be written compactly in terms of matrices as follows. The transmitted codeword $t$ is obtained from the source sequence $s$ by a linear operation,
$t = \textbf{G}^\intercal s,$
where $\textbf{G}$ is the generator matrix of the code,
$\textbf{G}^\intercal = \begin{bmatrix} 1 & 0 & 0 & 0 \newline 0 & 1 & 0 & 0 \newline 0 & 0 & 1 & 0 \newline 0 & 0 & 0 & 1 \newline 1 & 1 & 1 & 0 \newline 0 & 1 & 1 & 1 \newline 1 & 0 & 1 & 1 \end{bmatrix}$
and the encoding operation uses modulo-2 arithmetic ($1 + 1 = 0$, $0 + 1 = 1$, etc.).
N.B. $t = \textbf{G}^\intercal s$ 的 dimension 是 $(7,1) = (7,4) \cdot (4,1)$
$\textbf{G}^\intercal(x) = \begin{bmatrix} \text{lambda } x: x[0] \newline \text{lambda } x: x[1] \newline \text{lambda } x: x[2] \newline \text{lambda } x: x[3] \newline \text{lambda } x: x[0] \oplus x[1] \oplus x[2] \newline \text{lambda } x: x[1] \oplus x[2] \oplus x[3] \newline \text{lambda } x: x[0] \oplus x[2] \oplus x[3] \end{bmatrix}$
(待补充)
(待补充)
(待补充)
(待补充)
1.3 What performance can the best codes achieve?
There seems to be a trade-off between the decoded bit-error probability $p_b$ (which we would like to reduce) and the rate $R$ (which we would like to keep large). How can this trade-off be characterized? What points in the $(R, p_b)$ plane are achievable?
Noisy-channel coding theorem: For any channel, there exist codes that make it possible to communicate with arbitrarily small probability of error $p_b$ at non-zero rates.
The maximum rate at which communication is possible with arbitrarily small $p_b$ is called the capacity of the channel. The formula for the capacity of a BSC with noise level $f$ is
$C(f) = 1 - H_2(f) = 1 - \left [ f \log_2 \frac{1}{f} + (1-f) \log_2 \frac{1}{1-f} \right ]$
A channel with noise level $f = 0.1$ has capacity $C \approx 0.53$.
What performance are you trying to achieve? $10^{−15}$? You don’t need sixty disk drives – you can get that performance with just two disk drives (since $1/2$ is less than $0.53$). And if you want $p_b = 10^{−18}$ or $10^{−24}$ or anything, you can get there with two disk drives too!
(待补充)
2. Probability, Entropy, and Inference
This chapter, and its sibling, Chapter 8, devote some time to notation. Just as the White Knight distinguished between
• the song,
• the name of the song, and
• what the name of the song was called (Carroll, 1998),
we will sometimes need to be careful to distinguish between
• a random variable,
• the value of the random variable, and
• the proposition (命题) that asserts that the random variable has a particular value.
2.1 Probabilities and ensembles
An ensemble $X$ is a triple $(x, \mathcal{A}_X, \mathcal{P}_X)$, where the outcome $x$ is the value of a random variable, which takes on one of a set of possible values, $\mathcal{A}_X = \lbrace a_1, a_2, \dots, a_I \rbrace$, having probabilities $\mathcal{P}_X = \lbrace p_1, p_2, \dots, p_I \rbrace$, with $P(x = a_i ) = p_i$, $pi \geq 0$ and $\sum_{a_i \in \mathcal{A}_X} P(x = a_i) = 1$.
• $\mathcal{A}$ means ‘alphabet’.
$P(x = a_i)$ may be written as $P(a_i)$ or $P(x)$.
A joint ensemble $XY$ is an ensemble in which each outcome is an ordered pair $x, y$ (read ‘$x$ and $y$’) with $x \in \mathcal{A}_X = \lbrace a_1, a_2, \dots, a_I \rbrace$ and $y \in \mathcal{A}_Y = \lbrace b_1, b_2, \dots, b_J \rbrace$.
We call $P(x,y)$ the joint probability of $x$ and $y$. Commas are optional when writing ordered pairs, so $xy \iff x, y$.
N.B. In a joint ensemble $XY$ the two variables are not necessarily independent.
We can obtain the marginal probability $P(x)$ from the joint probability $P(x,y)$ by summation:
$P(x = a_i) \equiv \sum_{y \in \mathcal{A}\_Y} P(x = a_i, y)$
Similarly, using briefer notation, the marginal probability of $y$ is:
$P(y) \equiv \sum_{x \in \mathcal{A}\_X} P(x, y)$
N.B. 说到 marginal 的时候一定要记得是先有的 joint 才有的 marginal
Consider the set of bigrams $x,y$ in a corpus:
• The probability $P(y \mid x = \text{‘q’})$ is the probability distribution of the second letter $y$ given that the first letter $x$ is a $\text{‘q’}$.
• The probability $P(x \mid y = \text{‘u’})$ is the probability distribution of the first letter $x$ given that the second letter $y$ is a $\text{‘u’}$.
N.B. 注意这个描述方式！
Product rule or chain rule:
$P(x,y \mid \mathcal{H}) = P(x \mid y, \mathcal{H}) P(y \mid \mathcal{H}) = P(y \mid x, \mathcal{H}) P(x \mid \mathcal{H})$
N.B. $P(x \mid y, \mathcal{H})$ 应该理解为 $P(x \mid \left ( y, \mathcal{H} \right ))$ 而不是 $P(\left ( x \mid y \right ), \mathcal{H})$
2.2 The meaning of probability
Probabilities can be used in two ways.
• Probabilities can describe frequencies of outcomes in random experiments
• but giving noncircular definitions of the terms ‘frequency’ and ‘random’ is a challenge
• Probabilities can also be used, more generally, to describe degrees of belief in propositions that do not involve random variables, for example
• ‘the probability that Mr. S. was the murderer of Mrs. S., given the evidence’
• ‘the probability that Shakespeare’s plays were written by Francis Bacon’
Thus probabilities can be used to describe assumptions, and to describe inferences given those assumptions. This more general use of probability to quantify beliefs is known as the Bayesian viewpoint. It is also known as the subjective (主观的；客观是 objective) interpretation of probability, since the probabilities depend on assumptions. Advocates of a Bayesian approach to data modelling and pattern recognition do not view this subjectivity as a defect, since in their view,
you cannot do inference without making assumptions.
N.B. Readers should be warned that this is not yet a globally held view.
In non-Bayesian methods probabilities are allowed to describe only random variables, while Bayesians also use probabilities to describe inferences.
2.3 Forward probabilities and inverse probabilities
Digress: Generative Models
In probability and statistics, a generative model is a model for randomly generating observable data values, typically given some hidden parameters.
A generative model describes how data is generated, in terms of a probabilistic model.
In the scenario of supervised learning, a generative model estimates the joint probability distribution of data $P(X, Y)$ between the observed data $X$ and corresponding labels $Y$.
Generative model assumes data is created by a particular distribution which is defined by a couple of parameters (Gaussian Distribution) or non-parametric variants…
• Gaussians
• Mixtures of multinomials
• Naive Bayes
• Hidden Markov Models
• Latent Dirichlet Allocation
• Pros:
• We have the knowledge about the data distribution.
• Cons:
• Very expensive to get (a lot of parameters);
• Need lots of data.
Another way of thinking about this is that generative algorithms make some kind of structure assumptions on your model, but discriminative algorithms make fewer assumptions.
Digress: Discriminative Model
Discriminative 向来比 Generative 好理解，还是用 Generative vs. Discriminative Models 的说法吧：
Algorithms aim at learning $P(y x)$ by using probabilistic approaches (e.g., logistic regression), or non-probabilistically by mapping classes from a set of points (e.g., perceptrons and SVMs).
• Pros:
• Easy to model.
• Cons:
• To classify, but not to generate the data.
• Forward probability problems
• Involve a generative model of a process (这里这个 process 的意思就是指生成 examples 的 process)
• The task is to compute the probability distribution.
• 提出假设并求 parameter、计算 expectation、variance 等都算 forward problems
• Inverse probability problems
• Also involve a generative model of a process
• Instead, we compute the conditional probability of one or more of the unobserved variables in the process, given the observed variables.
• 比如 HMM
• This invariably requires the use of Bayes’ theorem.
Terminology of inverse probability
If $\theta$ denotes the unknown parameters, $D$ denotes the data, and $\mathcal{H}$ denotes the overall hypothesis space, the general equation:
$P(\theta \vert D, \mathcal{H}) = \frac{P(D \vert \theta, \mathcal{H}) P(\theta \vert \mathcal{H})}{P(D \vert \mathcal{H})}$
can be written:
$\text{posterior} = \frac{\text{likelihood} \times \text{prior}}{\text{evidence}}$
For a fixed $\theta_1 \in \theta$ and a fixed $D_1 \in D$:
• The conditional probability $P(\theta_1 \vert D_1, \mathcal{H})$ is called the posterior probability of $\theta_1$ given $D_1$
• The conditional probability $P(D_1 \vert \theta_1, \mathcal{H})$ is called the likelihood of parameter $\theta_1$
• also probability of $D_1$ given $\theta_1$
• The marginal probability $P(\theta_1 \vert \mathcal{H})$ is called the prior probability of parameter $\theta_1$
• $P(D \vert \mathcal{H})$ is known as the evidence or marginal likelihood
The likelihood principle
The likelihood principle: given a generative model for data $d$ given parameters $\theta$, $P(d \vert \theta)$, and having observed a particular outcome $d_1$, all inferences and predictions should depend only on the function $P (d_1 \vert \theta)$.
In spite of the simplicity of this principle, many classical statistical methods violate it. | 4,589 | 15,111 | {"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": 2, "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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2023-23 | longest | en | 0.793044 |
http://metamath.tirix.org/mpeuni/cnfldnm | 1,721,639,278,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517833.34/warc/CC-MAIN-20240722064532-20240722094532-00450.warc.gz | 23,167,080 | 2,173 | # Metamath Proof Explorer
## Theorem cnfldnm
Description: The norm of the field of complex numbers. (Contributed by Mario Carneiro, 4-Oct-2015)
Ref Expression
Assertion cnfldnm abs = ( norm ‘ ℂfld )
### Proof
Step Hyp Ref Expression
1 0cn 0 ∈ ℂ
2 eqid ( abs ∘ − ) = ( abs ∘ − )
3 2 cnmetdval ( ( 𝑥 ∈ ℂ ∧ 0 ∈ ℂ ) → ( 𝑥 ( abs ∘ − ) 0 ) = ( abs ‘ ( 𝑥 − 0 ) ) )
4 1 3 mpan2 ( 𝑥 ∈ ℂ → ( 𝑥 ( abs ∘ − ) 0 ) = ( abs ‘ ( 𝑥 − 0 ) ) )
5 subid1 ( 𝑥 ∈ ℂ → ( 𝑥 − 0 ) = 𝑥 )
6 5 fveq2d ( 𝑥 ∈ ℂ → ( abs ‘ ( 𝑥 − 0 ) ) = ( abs ‘ 𝑥 ) )
7 4 6 eqtrd ( 𝑥 ∈ ℂ → ( 𝑥 ( abs ∘ − ) 0 ) = ( abs ‘ 𝑥 ) )
8 7 mpteq2ia ( 𝑥 ∈ ℂ ↦ ( 𝑥 ( abs ∘ − ) 0 ) ) = ( 𝑥 ∈ ℂ ↦ ( abs ‘ 𝑥 ) )
9 eqid ( norm ‘ ℂfld ) = ( norm ‘ ℂfld )
10 cnfldbas ℂ = ( Base ‘ ℂfld )
11 cnfld0 0 = ( 0g ‘ ℂfld )
12 cnfldds ( abs ∘ − ) = ( dist ‘ ℂfld )
13 9 10 11 12 nmfval ( norm ‘ ℂfld ) = ( 𝑥 ∈ ℂ ↦ ( 𝑥 ( abs ∘ − ) 0 ) )
14 absf abs : ℂ ⟶ ℝ
15 14 a1i ( ⊤ → abs : ℂ ⟶ ℝ )
16 15 feqmptd ( ⊤ → abs = ( 𝑥 ∈ ℂ ↦ ( abs ‘ 𝑥 ) ) )
17 16 mptru abs = ( 𝑥 ∈ ℂ ↦ ( abs ‘ 𝑥 ) )
18 8 13 17 3eqtr4ri abs = ( norm ‘ ℂfld ) | 550 | 1,048 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-30 | latest | en | 0.490123 |
https://jp.maplesoft.com/support/help/Maple/view.aspx?path=MathApps/EquationofaPlane3Points | 1,716,702,893,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058868.12/warc/CC-MAIN-20240526043700-20240526073700-00120.warc.gz | 286,428,320 | 23,508 | Equation of a Plane - 3 Points - Maple Help
For the best experience, we recommend viewing online help using Google Chrome or Microsoft Edge.
Equation of a Plane - 3 Points
Main Concept
A plane can be defined by four different methods:
• A line and a point not on the line
• Three non-collinear points (three points not on a line)
• A point and a normal vector
• Two intersecting lines
• Two parallel and non-coincident lines
The Cartesian equation of a plane is , where is the vector normal to the plane.
How to find the equation of a plane using three non-collinear points
Three points (A,B,C) can define two distinct vectors AB and AC. Since the two vectors lie on the plane, their cross product can be used as a normal to the plane.
1 Determine the vectors
2 Find the cross product of the two vectors
3 Substitute one point into the Cartesian equation to solve for d.
Example
Find the equation of the plane that passes through the points .
1 Determine the vectors
2 Determine the normal vector
3 The equation of the plane is
4 Plug in any point to find the value of d
$-d=$ $d=$ $-4$
5 The equation of the plane is
Change the three points on the plane and see how it affects the plane.
Point A Point B ${x}_{B}=$ ${y}_{A}=$ ${y}_{B}=$ ${z}_{A}=$ ${z}_{B}=$
Point C
More MathApps | 334 | 1,333 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 20, "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-2024-22 | latest | en | 0.863139 |
http://www.education.com/science-fair/article/electric-current-movement-charges/ | 1,419,823,668,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447561183.59/warc/CC-MAIN-20141224185921-00089-ip-10-231-17-201.ec2.internal.warc.gz | 200,576,004 | 23,381 | Education.com
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# Electric Current: Movement of Charges
based on 9 ratings
Source:
Author: Janice VanCleave
Electricity is any effect resulting from the presence and/or movement of electrical charges. An electric current is the flow of electric charges. But the reason that an electric current makes a lamp come on instantly when you flip a switch is not because electrons race through the wire to the lamp. Instead, an electrical impulse, passed from electron to electron, moves through the wire to the lamp.
In this project, you will demonstrate the way an electrical impulse moves through a wire. You will determine how voltage affects an electric current. You will also investigate how to test the electrical conductivity or measure of the ability of a material to conduct an electric charge and the effect that the material's resistance has on an electric current passing through it.
### Getting Started
Purpose To demonstrate how an electrical impulse moves through a wire.
### Materials
books
12-inch ruler
1/4 × 16-inch (.31 × 40-cm) dowel
four 11/8-inch (2.8-cm) ceramic disk magnets, each with a hole through the center
### Procedure
1. Stack the books on a table so that you have two equal-size piles that are each at least 11/2 inches (3.75 cm) high.
2. Separate the stacks of books so that they are 12 inches (30 cm) apart. Lay the 12-inch ruler on the table between them.
3. Stick the dowel through the hole in one of the magnets.
4. Stick the dowel through the hole of a second magnet and push the two magnets on the dowel together. If the magnets cling together, remove the second magnet, turn it around, and put it back on the dowel. (1be magnets are to push away from each other.)
5. Repeat step 4 twice, placing the remaining two magnets on the dowel.
6. Support the ends of the dowel on the edges of the books so that the magnets hang above the ruler. Place a book over the ends of the dowel to secure the dowel.
7. Push magnet D (see Figure 16.1) toward the zero end (the left side) of the ruler, which will force the rest of the magnets to move, until magnet A rests against the books and is above the zero end of the ruler. Slowly move your hand away from magnet D. Record the starting position of each of the magnets in a Magnet Movement Data table like Table 16.1. Magnet A's initial position is 0 inches (cm).
8. Push magnet A slowly forward (to the right) 1 inch (2.5 cm).
9. While holding magnet A at its final position of 1 inch (2.5 cm), wait until all the magnets have finished moving, then measure and record the position of each of the magnets.
10. Determine how far each magnet moved by calculating the difference between the starting and the final position of each.
### Results
Magnets B, C, and D appeared to move the instant that magnet A moved. Magnets Band C moved about the same distance as magnet A, which was 1 inch (2.5 cm). Magnet D moved farther than the others.
### Why?
In this experiment the electrons in some solids, particularly metals, are attracted relatively equally to all nearby atoms and are not tightly bound to a single site. These electrons are relatively free to move through the solid, so they are called free electrons. The motion of free electrons results in the transfer of energy from one electron to the next. This transfer of energy is an electrical impulse caused by the repulsive force between negatively charged electrons. When magnet A was pushed forward, magnets Band C moved about the same distance because they each had about the same force pushing from their front and back. Thus they had the same net force acting in a forward direction. The magnets represent free electrons in a metal wire that is part of an electric circuit (the path that electric charges follow). An electric circuit is made of material called an electrical conductor or conductor (material with a large concentration of free electrons). If an electric circuit forms a loop so that the free electrons move in a continuous unbroken path, it is called a closed circuit. If there is a break in the materials forming the circuit so no current can flow, it is called an open circuit. There is no movement of charged particles in an open circuit. This experiment represents only a section of a closed circuit. Magnet D does not have a magnet in front of (i.e., to the right of) it, so it does not simulate the movement of an electron in a closed electric circuit. It moves farther than the rest of the magnets because it does not run into an opposing force, as do the others. (For more on the distance magnet D travels, see ''Try New Approaches" in this chapter.)
Electricity is any effect resulting from the presence and/or movement of electrical charges. Current electricity is the result of moving electric charges. The flow of electric charges through a conductor is called an electric current or current. Energy associated with electricity is called electrical energy.
The electrical energy that causes an electric current to move can be compared to the stored potential energy of two opposing magnets in this experiment, such as magnets A and B. The closer the magnets are, the harder it is to push them together; thus as they move closer together, their potential energy increases. In like manner, the potential energy of two electrons increases when the two charges move closer together. The electrical energy needed to move a charge from one point to another in an electric circuit is called potential difference (difference in electric potential energy between two points).
In current electricity, the motion of the charges is very slow in comparison to the electrical impulse. Free electrons in a metal wire can wander from atom to atom through the metal. Imagine a single row of electrons in a wire. When electron A moves forward, electron B in front of it is pushed forward by the repulsive electrical force of their like charges. Electron C in front of electron B is then pushed forward by the repulsive electrical force, and so on, like the movement of the magnets in this experiment. While the individual electrons move at a speed of about 0.0004 inch (0.001 cm) per second, which is actually a great distance for such a small particle, the electrical impulse that they pass along moves almost as fast as light—186,000 miles (300,000 km) per second. In this experiment, while each magnet moves only a small distance, the magnet moves forward almost instantaneously, representing each transfer of electrical energy through a row of electrons (represented by magnets) by an electrical impulse.
• 1 | 1,393 | 6,594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2014-52 | latest | en | 0.900228 |
https://nus.kattis.com/courses/CS2040/CS2040_S2_AY2324/assignments/k96nfp/problems/zipfsong | 1,721,811,225,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518198.93/warc/CC-MAIN-20240724075911-20240724105911-00108.warc.gz | 362,154,150 | 8,203 | Hide
# Problem AZipf's Song
Photo by suneko on Flickr
Your slightly pointy-bearded boss has assigned you to write software to find the best songs from different music albums. And the software should be finished in an hour. But don’t panic, you don’t have to solve the problem of writing an AI with good taste. At your disposal is the impeccable taste of a vast horde of long-tailed monkeys. Well, at least almost. The monkeys are not very communicative (or rather, you’re not sure which song “Ook!” is supposed to refer to) so you can’t ask them which songs are the best. What you can do however is to look at which songs the monkeys have listened to and use this information to deduce which songs are the best.
At first, you figure that the most listened to songs must be the best songs. However, you quickly realize that this approach is flawed. Even if all songs of the album are equally good, the early songs are more likely to be listened to more often than the later ones, because monkeys will tend to start listening to the first song, listen for a few songs and then, when their fickle ears start craving something else, stop listening. Instead, if all songs are equal, you expect that their play frequencies should follow Zipf’s Law.
Zipf’s Law is an empirical law originally formulated about word frequencies in natural languages, but it has been observed that many natural phenomena, such as population sizes and incomes, approximately follow the same law. It predicts that the relative frequency of the $i$’th most common object (in this case, a song) should be proportional to $1/i$.
To illustrate this in our setting, suppose we have an album where all songs are equally good. Then by Zipf’s Law, you expect that the first song is listened to twice as often as the second song, and more generally that the first song is listened to $i$ times as often as the $i$’th song. When some songs are better than others, those will be listened to more often than predicted by Zipf’s Law, and those are the songs your program should select as the good songs. Specifically, suppose that song $i$ has been played $f_ i$ times but that Zipf’s Law predicts that it would have been played $z_ i$ times. Then you define the quality of song $i$ to be $q_ i = f_ i/z_ i$. Your software should select the songs with the highest values of $q_ i$.
## Input
The first line of input contains two integers $n$ and $m$ ($1 \le n \le 50\, 000$, $1 \le m \le n$), the number of songs on the album, and the number of songs to select. Then follow $n$ lines. The $i$’th of these lines contains an integer $f_ i$ and string $s_ i$, where $0 \le f_ i \le 10^{12}$ is the number of times the $i$’th song was listened to, and $s_ i$ is the name of the song. Each song name is at most $30$ characters long and consists only of the characters ‘a’-‘z’, ‘0’-‘9’, and underscore (‘_’).
## Output
Output a list of the $m$ songs with the highest quality $q_ i$, in decreasing order of quality. If two songs have the same quality, give precedence to the one appearing first on the album (presumably there was a reason for the producers to put that song before the other).
Sample Input 1 Sample Output 1
4 2
30 one
30 two
15 three
25 four
four
two
Sample Input 2 Sample Output 2
15 3
197812 re_hash
78906 5_4
189518 tomorrow_comes_today
39453 new_genious
210492 clint_eastwood
26302 man_research
22544 punk
19727 sound_check
17535 double_bass
18782 rock_the_house
198189 19_2000
13151 latin_simone
12139 starshine
11272 slow_country
10521 m1_a1
19_2000
clint_eastwood
tomorrow_comes_today
Hide
Please log in to submit a solution to this problem | 926 | 3,629 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-30 | latest | en | 0.975593 |
https://dreambiginstitution.com/lcm-and-hcf-questions/ | 1,718,260,828,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861342.74/warc/CC-MAIN-20240613060639-20240613090639-00898.warc.gz | 190,626,021 | 33,591 | # HCF and LCM Formula With Trick for Competitive Exams
0
12675
## HCF and LCM Definition
We know that the factors of a number are exact divisors of that particular number. Let’s proceed to the highest common factor (H.C.F.) and the least common multiple (L.C.M.).
### HCF Definition
The full form of HCF in Maths is Highest Common Factor.
As the rules of mathematics dictate, the greatest common divisor or the gcd of two or more positive integers happens to be the largest positive integer that divides the numbers without leaving a remainder. For example, take 8 and 12. The H.C.F. of 8 and 12 will be 4 because the highest number that can divide both 8 and 12 is 4.
### LCM Definition
The full form of LCM in Maths is Least Common Multiple.
In arithmetic, the least common multiple or LCM of two numbers say a and b, is denoted as LCM (a,b). And the LCM is the smallest or least positive integer that is divisible by both a and b. For example, let us take two positive integers 4 and 6.
Multiples of 4 are: 4,8,12,16,20,24…
Multiples of 6 are: 6,12,18,24….
The common multiples for 4 and 6 are 12,24,36,48…and so on. The least common multiple in that lot would be 12. Let us now try to find out the LCM of 24 and 15
## LCM of Two Numbers
Suppose there are two numbers, 8 and 12, whose LCM we need to find. Let us write the multiples of these two numbers.
8 = 16, 24, 32, 40, 48, 56, …
12 = 24, 36, 48, 60, 72, 84,…
You can see, the least common multiple or the smallest common multiple of two numbers, 8 and 12 is 24.
### Tricks to Find LCM easily
Example 1: Find LCM of 2,4,8,16.
Choose the largest number. In this example, the largest number is 16. Check whether 16 is divisible by all other remaining numbers. 16 is divisible by 2, 4, 8. Hence, the LCM is 16.
Example 2: Find the LCM of 2,3,7,21.
Choose the largest number. The largest number is 21. Check whether 21 is divisible by all other remaining numbers. 21 is divisible by 3 and 7 but not by 2. So multiply 21 and 2. The result is 42. Now, check whether 42 is divisible by 2, 3, 7. Yes, 42 is divisible. Hence, the LCM is 42.
Example 3: Find the LCM of 3,5,15,30
Choose the largest number. The largest number is 30. Check whether 30 is divisible by all other remaining numbers. 30 is divisible by 3, 5, and 15. Hence, the LCM is 30.
Example 4: Find the LCM of 8,24,48,96
Choose the largest number. The largest number is 96. Check whether 96 is divisible by all other remaining numbers. 96 is divisible by 8, 24, and 48. Hence, the LCM is 96.
Example 5: Find the LCM of 12,36,60,108
Choose the largest number. The largest number is 108. Check whether 108 is divisible by all other remaining numbers. 108 is divisible by 12, 36, and 60. Hence, the LCM is 108.
## HCF and LCM Formula
The formula which involves both HCF and LCM is:
Say, A and B are the two numbers, then as per the formula;
A x B = H.C.F.(A, B) x L.C.M.(A, B)
We can also write the above formula in terms of HCF and LCM, such as:
H.C.F. of Two numbers = Product of Two numbers/L.C.M of two numbers
&
L.C.M of two numbers = Product of Two numbers/H.C.F. of Two numbers
NOTE-Â The above relation between H.C.F and L.C.M is not valid for the product of numbers greater than 2. It is only valid for the product of two numbers.
## How to find HCF and LCM?
Here are the methods we can use to find the HCF and LCM of given numbers.
1. Prime factorization method
2. Division method
Let us learn both methods, one by one.
### HCF by Prime Factorization Method
Take an example of finding the highest common factor of 144, 104 and 160.
Now let us write the prime factors of 144, 104 and 160.
144 = 2 × 2 × 2 × 2 × 3 × 3
104 = 2 × 2 × 2 × 13
160 = 2 × 2 × 2 × 2 × 2 × 5
The common factors of 144, 104 and 160 are 2 × 2 × 2 = 8
Therefore, HCF (144, 104, 160) = 8
### HCF by Division Method
Steps to find the HCF of any given numbers;
Frequently Asked Questions on HCF and LCM
### Q1.What is the full form of HCF in Maths? Explain HCF with an example.
The full form of HCF in Maths is the Highest Common Factor. HCF of two or more numbers is the greatest factor that divides the numbers. For example, 2 is the HCF of 4 and 6.
### Q2 What is the full form of LCM in Maths? Explain LCM with an example.
The full form of LCM in Maths is the Least common multiple. LCM is the smallest number which is divisible by two or more given numbers. For example, LCM of 2 & 3 is 6.
Team Dream Big Institution is The Group of Professional Aim to Provide Quality and Affordable Education to Society Our Team is dedicated to helping every aspirant achieve their goal. Our team is experts in handling the querries and have in-depth knowledge of the subject. | 1,416 | 4,728 | {"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} | 4.8125 | 5 | CC-MAIN-2024-26 | latest | en | 0.899641 |
https://www.physicsforums.com/threads/four-4s.98792/ | 1,519,456,623,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815435.68/warc/CC-MAIN-20180224053236-20180224073236-00630.warc.gz | 888,158,362 | 18,963 | # Four 4s
1. Nov 7, 2005
### Joffe
Have you ever heard of that challenge where you have to make each number from four 4s? This list is the best I have been able to come up with, will you help me fill in the gaps?
Any solution that is simpler than one here can take its place, for example if you come up with a solution to 33 that doesnt use a decimal symbol it is simpler.
EDIT: An underline indicates repetition: (i.e: .4 = .44444 / 4/9).
Code (Text):
1 = 4*4/(4*4)
2 = 4/4+4/4
3 = (4+4+4)/4
4 = (4-4)/4+4
5 = 4^(4-4)+4
6 = (4+4)/4+4
7 = 4+4-4/4
8 = 4+4+4-4
9 = 4/4+4+4
10 = (4*4+4!)/4
11 = (4+4!)/4+4
12 = (4-4/4)*4
13 = (4+4!+4!)/4
14 = 4!/4+4+4
15 = 4*4-4/4
16 = 4*4+4-4
17 = 4*4+4/4
18 = (4*4!-4!)/4
19 = 4!-(4+4/4)
20 = (4/4+4)*4
21 = 4!+4/4-4
22 = 4!-(4+4)/4
23 = 4!-4^(4-4)
24 = 4*4+4+4
25 = 4!+(4/4)^4
26 = 4!+4!/4-4
27 = 4!+4-4/4
28 = (4+4)*4-4
29 = 4/4+4!+4
30 = (4*4!+4!)/4
31 = (4+4!)/4+4!
32 = 4^4/(4+4)
33 = (4-.4)/.4+4!
34 = 4!/4+4+4!
35 = (4.4/.4)+4!
36 = (4+4)*4+4
37 = 4/.[U]4[/U]+4+4!
38 = 44-4!/4
39 = (4*4-.4)/.4
40 = (4^4/4)-4!
41 = (4*4+.4)/.4
42 = 4!+4!-4!/4
43 = 44-4/4
44 = 4*4+4+4!
45 = (4!/4)!/(4*4)
46 = (4!-4)/.4 - 4
47 = 4!+4!-4/4
48 = (4*4-4)*4
49 = 4!+4!+4/4
50 = (4*4+4)/.4
51 = 4!/.4-4/.[U]4[/U]
52 = 44+4+4
53 = 44+4/.[U]4[/U]
54 = (4!/4)^4/4!
55 = (4!-.4)/.4-4
56 = 4!+4!+4+4
57 = 4/.[U]4[/U]+4!+4!
58 = (4^4-4!)/4
59 = 4!/.4-4/4
60 = 4*4*4-4
61 = 4!/.4+4/4
62 = (4!+.4+.4)/.4
63 = (4^4-4)/4
64 = 4^(4-4/4)
65 = 4^4+4/4
66 = (4+4!)/.4-4
67 = (4+4!)/.[U]4[/U]+4
68 = 4*4*4+4
69 = (4+4!-.4)/.4
70 = (4^4+4!)/4
71 = (4!+4.4)/.4
72 = (4-4/4)*4!
73 = ([sup].4[/sup]√4+.[u]4[/u])/.[u]4[/u]
74 = (4+4!)/.4+4
75 = (4!/4+4!)/.4
76 = (4!-4)*4-4
77 = (4!-.[u]4[/u])/.[u]4[/u]+4!
78 = (4!x.[u]4[/u]+4!)/.[u]4[/u]
79 = ([sup].4[/sup]√4-.4)/.4
80 = (4*4+4)*4
81 = (4/4-4)^4
82 = 4!/.[U]4[/U]+4!+4
83 = (4!-.4)/.4+4!
84 = (4!-4)*4+4
85 = (4/.4+4!)/.4
86 = (4-.4)x4!-.4
87 = 4!x4-4/.[U]4[/U]
88 = 4^4/4+4!
89
90 = (4!/4)!/(4+4)
91
92 = (4!-4/4)*4
93
94 = (4+4!)/.4 + 4!
95 = 4!*4-4/4
96 = 4!*4+4-4
97 = 4!*4+4/4
98 = (4!+.4)*4+.4
99 = (4!+4!-4)/.[U]4[/U]
100 = 4*4/(.4*.4)
Thanks:
• ceptimus: 71
• LarrrSDonald: 51,53,82,87,79
Last edited: Nov 7, 2005
2. Nov 7, 2005
### ceptimus
71 = (4! + 4.4) / .4
...and if we can use square roots:
37 = (sqrt(4) + 4!)/sqrt(4) + 4!
51 = (4! - sqrt(4)) / .4 - 4
53 = sqrt(4) / .4 + 4! + 4!
57 = (4! - .4) / .4 - sqrt(4)
67 = (sqrt(4) + 4!) / .4 + sqrt(4)
78 = 4 * (4! - 4) - sqrt(4)
79 = (4! - sqrt(4)) / .4 + 4!
82 = 4 * (4! - 4) + sqrt(4)
89 = (sqrt(4) + 4!) / .4 + 4!
91 = 4 * 4! - sqrt(4) / .4
3. Nov 7, 2005
### Joffe
Thanks for 71 ceptimus!
The others I am not as interested in because (as I probably should have stated) I am not considering square roots because they require a 2, I would however consider 4th roots so long as the 4 is counted. No triggonometric functions either, basically whatever techniques I have used so far.
4. Nov 7, 2005
### ArielGenesis
if we can use zero as if in decimal, why couldn't we use 2 as if in square root as well as trigonometry function and, most probabaly also included, logarithm. If you are about using arithmetich signs only, then decimal should not be used.
5. Nov 7, 2005
### ArielGenesis
and also exclemation mark !, as 4! should include other number such 1, 2 and 3
6. Nov 7, 2005
### LarrrSDonald
If you're ok with $$.\bar{4} = \frac{4}{9}$$ then you have:
$$51 = \frac{4!}{4}-\frac{4}{.\bar{4}}$$
$$53 = 44+\frac{4}{.\bar{4}}$$
$$82 = \frac{4!}{.\bar{4}}+4!+4$$
$$87 = 4!*4-\frac{4}{.\bar{4}}$$
Possibly others, I haven't had time to seach around that much.
7. Nov 7, 2005
### RandallB
Should be:
$$51 = \frac{4!}{.4}-\frac{4}{.\bar{4}}$$
8. Nov 7, 2005
### LarrrSDonald
You're right, lost a decimal point. Sorry 'bout that.
[EDIT] If you're willing to consider 4th root, then perhaps you'd consider $$\sqrt[.4]{4} = 32$$ and thus
$$79 = \frac{(\sqrt[.4]{4} - .4)}{.4}$$
Last edited: Nov 7, 2005
9. Nov 7, 2005
### Joffe
LarrrSDonald: I like your ideas! Your repetition idea got me a whole heap of answers including 37! And this latest one, 79 is great lateral thinking, Keep em coming.
ArielGenesis: If I were to include square root signs then it might as well be called the four 4 or 2s challenge, it would make it far too easy. Using decimals is still bending a little (hence the reason to try to try to eliminate some of the solutions that use them) but since it does not make a whole number it keeps the game challenging. And if I were to include trig I could make combinations like: sin-1 ( cos ( 4 ) ) = 86, which is clearly too convenient.
EDIT: only 3 remain!
Last edited: Nov 7, 2005
10. Nov 7, 2005
### Alkatran
4 != 0/4
I'd go for 4 = 4! - 4*4 - 4
11. Nov 7, 2005
### Joffe
(4-4)/4 + 4
= 0/4 + 4
= 0 + 4
= 4
Perhaps you misinterpreted my grouping symbols.
12. Nov 8, 2005
### RandallB
BUT you didn't say what 37 was! Maken us work for it eh?
Did you use:
$$37 = \frac{{4}*{4}+{.\bar{4}}}{.\bar{4}}$$
13. Nov 8, 2005
### ceptimus
He's got 37 = 4 / .4 + 4 + 4! in the table above
(he used an underline to show recurring)
14. Nov 22, 2005
### Joffe
Indeed, here are two more:
89 = (4-.4-4%)/4%
91 = (4-.4+4%)/4%
93 is all that remains!
15. Nov 22, 2005
### ceptimus
I can only do it with the forbidden square root.
$$93 = 4 \times 4! - \sqrt{\frac{4}{.\bar{4}}}$$
16. Nov 27, 2005
### croxbearer
If logarithm is allowed:
93 = 4 x 4! - log(4/.4%)
17. Nov 28, 2005
### Joffe
There is only one problem with that, there is an implied 10. I would definantly accept log in base 4 and maby the natural logarithm of an expression but that (albeit quite novel) just seems to violate in my opinion.
If I don't ever find a better solution I think I will settle for that though croxbearer.
18. Nov 21, 2008
### Markii
Another way to get 93 using an exponent of 0 is 4! x 4 - 4 - 4°
19. Nov 10, 2010
### Fahim555
Yo, I got 89!
20. Aug 5, 2011
### Fang_Lou
5 = (4x4+4)/4 | 2,781 | 5,953 | {"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} | 3.890625 | 4 | CC-MAIN-2018-09 | longest | en | 0.604926 |
http://www.dsplog.com/2012/10/27/gate-2012-ece-q26-electronic-devices/ | 1,675,392,291,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500042.8/warc/CC-MAIN-20230203024018-20230203054018-00624.warc.gz | 69,461,500 | 15,180 | (No Ratings Yet)
# GATE-2012 ECE Q26 (electronic devices)
by on October 27, 2012
Question 26 on Electronic Devices from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.
## Solution
To answer this question, had to dig up the copy of Solid State Electronic Devices, Ben G Streetman, Sanjay Kumar Banarjee (buy from Amazon.com, buy from Flipkart.com) and am referring to the content from Section 3.3 Carrier Concentrations.
Electrons in solids obey Fermi-Dirac statistics. The function $f(E)$ Fermi-Dirac distribution function, gives the probability that an available energy state at $E$ will be occupied by an electron at absolute temperature $T$,
$f(E) = \frac{1}{1+e^{(E-E_F)/kT}}$.
The quantity $E_f$ is the Fermi level and $\begin{array}k&=&\mbox{8.63x10^{-5} eV/K} & = & \mbox{1.38x10^{-23} J/K}\end{array}$ is the Boltzmann’s constant.
The concentration of electrons in the conduction band is,
$n_0 = \int_{E_c}^{\infty}f(E)N(E)dE$, where $N(E)dE$ is the density of states $cm^{-3}$ in the energy range $dE$.
The integral is equivalently stated as,
$n_0 = N_cf(E_c)$, where $N_c$ is the effective density of states at conduction band edge $E_c$.
The Fermi function at $E_c$ is approximately,
$\begin{array}f(E_c)&=&\frac{1}{1+e^{(E_c-E_F)/kT}}&\simeq e^{-(E_c-E_F)/kT}&\end{array}$.
So in this condition the concentration of electrons in the conduction band is,
$n_0 = N_ce^{-(E_c-E_F)/kT}$.
Similarly, the concentration of holes in the valence band is,
$p_0 = N_v$1-f(E_v)$$, where $N_v$ is the effective density of states at valence band edge $E_v$.
The term,
$\begin{array}1-f(E_v)&=&\frac{1}{1-1+e^{(E_v-E_F)/kT}}&\simeq e^{-(E_F-E_v)/kT}&\end{array}$.
So in this condition the concentration of holes in the valence band is,
$p_0 = N_ve^{-(E_F-E_v)/kT}$.
The product of $n_0$ and $p_0$ at equilibrium is a constant for a particular material and temperature even if doping is varied.
$\begin{array}{lll}n_0p_0&=&N_ce^{-(E_c-E_F)/kT}N_ve^{-(E_F-E_v)/kT}\\&=&N_cN_ve^{-E_g/kT}\end{array}$where
$E_g = E_c-E_v$ is the gap between the conduction band and valence band.
Similarly, for an intrinsic material, the product of $n_i$ and $p_i$ is,
$\begin{array}{lll}n_ip_i&=&N_ce^{-(E_c-E_i)/kT}N_ve^{-(E_i-E_v)/kT}\\&=&N_cN_ve^{-E_g/kT}\end{array}$
For an intrinsic material, the electron and hole concentration are equal i.e. $n_i=p_i$.
The constant product of electron and hole concentration can be written as
$\Huge{n_0p_0 = n_i^2}$
With the above understanding, let us try to find the solution to the problem. In our question,
$\begin{array}n_i&=&10^{10}&\mbox{per cm^3\end{array}$ and $\begin{array}n_0&=&10^{19}&\mbox{per cm^3\end{array}$.
The hole concentration is,
$\begin{array}{lll}p_0&=&\frac{n_i^2}{n_0}\\&=&\frac{10^{20}}{10^{10}}\\&=&10\mbox{ per cm^3}\end{array}$.
The volume of the source region is
$\begin{array}{lll}\mbox{volume}&=&10^{-18}\mbox{ m^3}&=&10^{-12}&\mbox{cm^3}\end{array}$.
The number of holes is,
$\begin{array}{lll}\mbox{N_{hole}}&=&10*10^{-12}&=&10^{-11}&\approx&0\end{array}$
Based on the above, the right choice is (D) 0.
## References
[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf
[2] Solid State Electronic Devices, Ben G Streetman, Sanjay Kumar Banarjee (buy from Amazon.combuy from Flipkart.com)
[3] Valence band http://en.wikipedia.org/wiki/Valence_band
[4] Conduction band http://en.wikipedia.org/wiki/Conduction_band
D id you like this article? Make sure that you do not miss a new article by subscribing to RSS feed OR subscribing to e-mail newsletter. Note: Subscribing via e-mail entitles you to download the free e-Book on BER of BPSK/QPSK/16QAM/16PSK in AWGN. | 1,277 | 3,800 | {"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": 43, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2023-06 | latest | en | 0.705637 |
https://text.123doc.org/document/5084338-solution-manual-heat-and-mass-transfer-a-practical-approach-2nd-edition-cengel-ch-2.htm | 1,544,445,604,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823339.35/warc/CC-MAIN-20181210123246-20181210144746-00542.warc.gz | 745,318,326 | 17,793 | # Solution manual heat and mass transfer a practical approach 2nd edition cengel ch 2
Chapter 2 Heat Conduction Equation
Chapter 2
HEAT CONDUCTION EQUATION
Introduction
2-1C Heat transfer is a vector quantity since it has direction as well as magnitude. Therefore, we must
specify both direction and magnitude in order to describe heat transfer completely at a point. Temperature,
on the other hand, is a scalar quantity.
2-2C The term steady implies no change with time at any point within the medium while transient implies
variation with time or time dependence. Therefore, the temperature or heat flux remains unchanged with
time during steady heat transfer through a medium at any location although both quantities may vary from
one location to another. During transient heat transfer, the temperature and heat flux may vary with time
as well as location. Heat transfer is one-dimensional if it occurs primarily in one direction. It is twodimensional if heat tranfer in the third dimension is negligible.
2-3C Heat transfer to a canned drink can be modeled as two-dimensional since temperature differences
(and thus heat transfer) will exist in the radial and axial directions (but there will be symmetry about the
center line and no heat transfer in the azimuthal direction. This would be a transient heat transfer process
since the temperature at any point within the drink will change with time during heating. Also, we would
use the cylindrical coordinate system to solve this problem since a cylinder is best described in cylindrical
coordinates. Also, we would place the origin somewhere on the center line, possibly at the center of the
bottom surface.
2-4C Heat transfer to a potato in an oven can be modeled as one-dimensional since temperature differences
(and thus heat transfer) will exist in the radial direction only because of symmetry about the center point.
This would be a transient heat transfer process since the temperature at any point within the potato will
change with time during cooking. Also, we would use the spherical coordinate system to solve this
problem since the entire outer surface of a spherical body can be described by a constant value of the
radius in spherical coordinates. We would place the origin at the center of the potato.
2-5C Assuming the egg to be round, heat transfer to an egg in boiling water can be modeled as onedimensional since temperature differences (and thus heat transfer) will primarily exist in the radial
direction only because of symmetry about the center point. This would be a transient heat transfer process
since the temperature at any point within the egg will change with time during cooking. Also, we would
use the spherical coordinate system to solve this problem since the entire outer surface of a spherical body
can be described by a constant value of the radius in spherical coordinates. We would place the origin at
the center of the egg.
2-1
Chapter 2 Heat Conduction Equation
2-6C Heat transfer to a hot dog can be modeled as two-dimensional since temperature differences (and thus
heat transfer) will exist in the radial and axial directions (but there will be symmetry about the center line
and no heat transfer in the azimuthal direction. This would be a transient heat transfer process since the
temperature at any point within the hot dog will change with time during cooking. Also, we would use the
cylindrical coordinate system to solve this problem since a cylinder is best described in cylindrical
coordinates. Also, we would place the origin somewhere on the center line, possibly at the center of the hot
dog. Heat transfer in a very long hot dog could be considered to be one-dimensional in preliminary
calculations.
2-7C Heat transfer to a roast beef in an oven would be transient since the temperature at any point within
the roast will change with time during cooking. Also, by approximating the roast as a spherical object, this
heat transfer process can be modeled as one-dimensional since temperature differences (and thus heat
transfer) will primarily exist in the radial direction because of symmetry about the center point.
2-8C Heat loss from a hot water tank in a house to the surrounding medium can be considered to be a
steady heat transfer problem. Also, it can be considered to be two-dimensional since temperature
differences (and thus heat transfer) will exist in the radial and axial directions (but there will be symmetry
about the center line and no heat transfer in the azimuthal direction.)
2-9C Yes, the heat flux vector at a point P on an isothermal surface of a medium has to be perpendicular to
the surface at that point.
2-10C Isotropic materials have the same properties in all directions, and we do not need to be concerned
about the variation of properties with direction for such materials. The properties of anisotropic materials
such as the fibrous or composite materials, however, may change with direction.
2-11C In heat conduction analysis, the conversion of electrical, chemical, or nuclear energy into heat (or
thermal) energy in solids is called heat generation.
2-12C The phrase “thermal energy generation” is equivalent to “heat generation,” and they are used
interchangeably. They imply the conversion of some other form of energy into thermal energy. The phrase
“energy generation,” however, is vague since the form of energy generated is not clear.
2-13 Heat transfer through the walls, door, and the top and bottom sections of an oven is transient in nature
since the thermal conditions in the kitchen and the oven, in general, change with time. However, we would
analyze this problem as a steady heat transfer problem under the worst anticipated conditions such as the
highest temperature setting for the oven, and the anticipated lowest temperature in the kitchen (the so
called “design” conditions). If the heating element of the oven is large enough to keep the oven at the
desired temperature setting under the presumed worst conditions, then it is large enough to do so under all
conditions by cycling on and off.
Heat transfer from the oven is three-dimensional in nature since heat will be entering through all
six sides of the oven. However, heat transfer through any wall or floor takes place in the direction normal
to the surface, and thus it can be analyzed as being one-dimensional. Therefore, this problem can be
simplified greatly by considering the heat transfer as being one- dimensional at each of the four sides as
well as the top and bottom sections, and then by adding the calculated values of heat transfers at each
surface.
2-14E The power consumed by the resistance wire of an iron is given. The heat generation and the heat
flux are to be determined.
Assumptions Heat is generated uniformly in the resistance wire.
Analysis A 1000 W iron will convert electrical energy into
heat in the wire at a rate of 1000 W. Therefore, the rate of heat
generation in a resistance wire is simply equal to the power
rating of a resistance heater. Then the rate of heat generation in
the wire per unit volume is determined by dividing the total
rate of heat generation by the volume of the wire to be
2-2
q = 1000 W
D = 0.08 in
L = 15 in
Chapter 2 Heat Conduction Equation
g& =
G&
V wire
=
G&
(πD / 4) L
2
=
⎛ 3.412 Btu/h ⎞
7
3
⎟ = 7.820 × 10 Btu/h ⋅ ft
1
W
[π(0.08 / 12 ft) / 4](15 / 12 ft) ⎝
1000 W
2
Similarly, heat flux on the outer surface of the wire as a result of this heat generation is determined by
dividing the total rate of heat generation by the surface area of the wire to be
q& =
1000 W
G&
G&
⎛ 3.412 Btu/h ⎞
5
2
=
=
⎟ = 1.303 × 10 Btu/h ⋅ ft
1W
Awire πDL π (0.08 / 12 ft)(15 / 12 ft) ⎝
Discussion Note that heat generation is expressed per unit volume in Btu/h⋅ft3 whereas heat flux is
expressed per unit surface area in Btu/h⋅ft2.
2-3
Chapter 2 Heat Conduction Equation
2-15E
"GIVEN"
E_dot=1000 "[W]"
L=15 "[in]"
"D=0.08 [in], parameter to be varied"
"ANALYSIS"
g_dot=E_dot/V_wire*Convert(W, Btu/h)
V_wire=pi*D^2/4*L*Convert(in^3, ft^3)
q_dot=E_dot/A_wire*Convert(W, Btu/h)
A_wire=pi*D*L*Convert(in^2, ft^2)
q [Btu/h.ft2]
521370
260685
173790
130342
104274
86895
74481
65171
57930
52137
D [in]
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
550000
500000
450000
400000
2
q [Btu/h-ft ]
350000
300000
250000
200000
150000
100000
50000
0
0.02
0.04 0.06
0.08
0.1
0.12
D [in]
2-4
0.14
0.16 0.18
0.2
Chapter 2 Heat Conduction Equation
2-16 The rate of heat generation per unit volume in the uranium rods is given. The total rate of heat
generation in each rod is to be determined.
g = 7×107 W/m3
Assumptions Heat is generated uniformly in the uranium rods.
Analysis The total rate of heat generation in the rod is
determined by multiplying the rate of heat generation per unit
volume by the volume of the rod
D = 5 cm
L=1m
& rod = g& (πD 2 / 4) L = (7 × 107 W / m3 )[π (0.05 m) 2 / 4](1 m) = 1.374 × 105 W = 137.4 kW
G& = gV
2-17 The variation of the absorption of solar energy in a solar pond with depth is given. A relation for the
total rate of heat generation in a water layer at the top of the pond is to be determined.
Assumptions Absorption of solar radiation by water is modeled as heat generation.
Analysis The total rate of heat generation in a water layer of surface area A and thickness L at the top of the
pond is determined by integration to be
G& =
V
g&dV =
L
x =0
g& 0 e
−bx
e −bx
−b
L
=
0
Ag& 0 (1 − e −bL )
b
2-18 The rate of heat generation per unit volume in a stainless steel plate is given. The heat flux on the
surface of the plate is to be determined.
Assumptions Heat is generated uniformly in steel plate.
Analysis We consider a unit surface area of 1 m2. The total rate of heat
generation in this section of the plate is
g
L
& plate = g& ( A × L) = (5 × 10 6 W / m 3 )(1 m 2 )(0.03 m) = 1.5 × 105 W
G& = gV
Noting that this heat will be dissipated from both sides of the plate, the heat
flux on either surface of the plate becomes
q& =
G&
Aplate
=
1.5 × 10 5 W
2 ×1 m 2
= 75,000 W/m 2
Heat Conduction Equation
2-19 The one-dimensional transient heat conduction equation for a plane wall with constant thermal
∂ 2T g& 1 ∂T
+ =
. Here T is the temperature, x is the space variable, g
conductivity and heat generation is
∂x 2 k α ∂t
is the heat generation per unit volume, k is the thermal conductivity, α is the thermal diffusivity, and t is the
time.
2-20 The one-dimensional transient heat conduction equation for a plane wall with constant thermal
1 ∂ ⎛ ∂T ⎞ g& 1 ∂T
conductivity and heat generation is
. Here T is the temperature, r is the space
⎜r
⎟+ =
r ∂r ⎝ ∂r ⎠ k α ∂t
2-5
Chapter 2 Heat Conduction Equation
variable, g is the heat generation per unit volume, k is the thermal conductivity, α is the thermal diffusivity,
and t is the time.
2-6
Chapter 2 Heat Conduction Equation
2-21 We consider a thin element of thickness Δx in a large plane wall (see Fig. 2-13 in the text). The
density of the wall is ρ, the specific heat is C, and the area of the wall normal to the direction of heat
transfer is A. In the absence of any heat generation, an energy balance on this thin element of thickness Δx
during a small time interval Δt can be expressed as
ΔE element
Q& x − Q& x +Δx =
Δt
where
ΔEelement = Et + Δt − Et = mC (Tt + Δt − Tt ) = ρCAΔx (Tt + Δt − Tt )
Substituting,
T
− Tt
Q& x − Q& x + Δx = ρCAΔx t + Δt
Δt
Dividing by AΔx gives
T
− Tt
1 Q& x + Δx − Q& x
= ρ C t + Δt
A
Δx
Δt
Taking the limit as Δx → 0 and Δt → 0 yields
1 ∂ ⎛ ∂T
⎜ kA
A ∂x ⎝ ∂x
∂T
⎟ = ρC
∂t
since, from the definition of the derivative and Fourier’s law of heat conduction,
Q& x + Δx − Q& x ∂Q ∂ ⎛
∂T ⎞
=
= ⎜ − kA
Δx →0
∂x ∂x ⎝
∂x ⎠
Δx
lim
Noting that the area A of a plane wall is constant, the one-dimensional transient heat conduction equation
in a plane wall with constant thermal conductivity k becomes
∂ 2 T 1 ∂T
=
∂x 2 α ∂t
where the property α = k / ρC is the thermal diffusivity of the material.
2-7
Chapter 2 Heat Conduction Equation
2-22 We consider a thin cylindrical shell element of thickness Δr in a long cylinder (see Fig. 2-15 in the
text). The density of the cylinder is ρ, the specific heat is C, and the length is L. The area of the cylinder
normal to the direction of heat transfer at any location is A = 2πrL where r is the value of the radius at that
location. Note that the heat transfer area A depends on r in this case, and thus it varies with location. An
energy balance on this thin cylindrical shell element of thickness Δr during a small time interval Δt can be
expressed as
ΔE element
Q& r − Q& r +Δr + G& element =
Δt
where
ΔEelement = Et + Δt − Et = mC (Tt + Δt − Tt ) = ρCAΔr (Tt + Δt − Tt )
G& element = g&Velement = g&AΔr
Substituting,
T
− Tt
Q& r − Q& r + Δr + g& AΔr = ρCAΔr t + Δt
Δt
where A = 2πrL . Dividing the equation above by AΔr gives
T
− Tt
1 Q& r + Δr − Q& r
+ g& = ρC t + Δt
A
Δr
Δt
Taking the limit as Δr → 0 and Δt → 0 yields
∂T
1 ∂ ⎛ ∂T ⎞
⎜ kA
⎟ + g& = ρC
A ∂r ⎝
∂t
∂r ⎠
since, from the definition of the derivative and Fourier’s law of heat conduction,
Q& r + Δr − Q& r ∂Q ∂ ⎛
∂T ⎞
=
=
⎜ − kA
Δr → 0
∂ r ∂r ⎝
∂r ⎠
Δr
lim
Noting that the heat transfer area in this case is A = 2πrL and the thermal conductivity is constant, the onedimensional transient heat conduction equation in a cylinder becomes
1 ∂ ⎛ ∂T ⎞
1 ∂T
⎜r
⎟ + g& =
r ∂r ⎝ ∂r ⎠
α ∂t
where α = k / ρC is the thermal diffusivity of the material.
2-8
Chapter 2 Heat Conduction Equation
2-23 We consider a thin spherical shell element of thickness Δr in a sphere (see Fig. 2-17 in the text).. The
density of the sphere is ρ, the specific heat is C, and the length is L. The area of the sphere normal to the
direction of heat transfer at any location is A = 4πr 2 where r is the value of the radius at that location.
Note that the heat transfer area A depends on r in this case, and thus it varies with location. When there is
no heat generation, an energy balance on this thin spherical shell element of thickness Δr during a small
time interval Δt can be expressed as
ΔE element
Q& r − Q& r +Δr =
Δt
where
ΔEelement = Et + Δt − Et = mC (Tt + Δt − Tt ) = ρCAΔr (Tt + Δt − Tt )
Substituting,
T
−T
& Δr = ρCAΔr t + Δt t
Q&r − Q&r + Δr + gA
Δt
where A = 4πr 2 . Dividing the equation above by AΔr gives
T
− Tt
1 Q& r + Δr − Q& r
= ρ C t + Δt
A
Δr
Δt
Taking the limit as Δr → 0 and Δt → 0 yields
∂T
1 ∂ ⎛ ∂T ⎞
⎜ kA
⎟ = ρC
A ∂r ⎝ ∂r ⎠
∂t
since, from the definition of the derivative and Fourier’s law of heat conduction,
Q& r + Δr − Q& r ∂Q ∂ ⎛
∂T ⎞
=
=
⎜ − kA
Δr → 0
∂ r ∂r ⎝
∂r ⎠
Δr
lim
Noting that the heat transfer area in this case is A = 4πr 2 and the thermal conductivity k is constant, the
one-dimensional transient heat conduction equation in a sphere becomes
1 ∂ ⎛ 2 ∂T ⎞ 1 ∂T
⎜r
⎟=
r 2 ∂r ⎝ ∂r ⎠ α ∂ t
where α = k / ρC is the thermal diffusivity of the material.
2-24 For a medium in which the heat conduction equation is given in its simplest by
∂ 2T 1 ∂T
=
:
∂x 2 α ∂t
(a) Heat transfer is transient, (b) it is one-dimensional, (c) there is no heat generation, and (d) the thermal
conductivity is constant.
2-25
1 d
r dr
For a medium in which the heat conduction equation is given in its simplest by
⎛ dT ⎞
⎜ rk
⎟ + g& = 0 :
⎝ dr ⎠
(a) Heat transfer is steady, (b) it is one-dimensional, (c) there is heat generation, and (d) the thermal
conductivity is variable.
2-26 For a medium in which the heat conduction equation is given by
2-9
1 ∂ ⎛ 2 ∂T ⎞ 1 ∂T
⎜r
⎟=
r 2 ∂r ⎝ ∂r ⎠ α ∂ t
Chapter 2 Heat Conduction Equation
(a) Heat transfer is transient, (b) it is one-dimensional, (c) there is no heat generation, and (d) the thermal
conductivity is constant.
2-27 For a medium in which the heat conduction equation is given in its simplest by r
d 2T
dr
2
+
dT
= 0:
dr
(a) Heat transfer is steady, (b) it is one-dimensional, (c) there is heat generation, and (d) the thermal
conductivity is constant.
2-28 We consider a small rectangular element of length Δx, width Δy, and height Δz = 1 (similar to the one
in Fig. 2-21). The density of the body is ρ and the specific heat is C. Noting that heat conduction is twodimensional and assuming no heat generation, an energy balance on this element during a small time
interval Δt can be expressed as
Rate of heat ⎞ ⎛ Rate of heat conduction ⎞ ⎛ Rate of change of ⎞
⎟ ⎜
⎟ ⎜
conduction
at the ⎟ − ⎜
at the surfaces at
⎟ = ⎜ the energy content ⎟
⎜ surfaces at x and y ⎟ ⎜ x + Δx and y + Δy
⎟ ⎜ of the element ⎟
⎠ ⎝
⎠ ⎝
Δ
E
element
or
Q& x + Q& y − Q& x + Δx − Q& y + Δy =
Δt
Noting that the volume of the element is Velement = ΔxΔyΔz = ΔxΔy × 1 , the change in the energy content of
the element can be expressed as
ΔE element = E t + Δt − E t = mC (Tt + Δt − Tt ) = ρCΔxΔy (Tt + Δt − Tt )
Substituting,
T
− Tt
Q& x + Q& y − Q& x + Δx − Q& y + Δy = ρCΔxΔy t + Δt
Δt
Dividing by ΔxΔy gives
− Tt
T
1 Q& x + Δx − Q& x
1 Q& y + Δy − Q& y
= ρC t + Δt
Δy
Δx
Δx
Δy
Δt
Taking the thermal conductivity k to be constant and noting that the heat transfer surface areas of the
element for heat conduction in the x and y directions are Ax = Δy × 1 and Ay = Δx × 1, respectively, and
taking the limit as Δx , Δy , and Δt → 0 yields
∂ 2 T ∂ 2 T 1 ∂T
+
=
∂x 2 ∂y 2 α ∂t
since, from the definition of the derivative and Fourier’s law of heat conduction,
∂T ⎞
∂ ⎛ ∂T ⎞
∂ 2T
1 Q& x + Δx − Q& x
1 ∂Q x
1 ∂ ⎛
=
=
⎟ = −k
⎟ = − ⎜k
⎜ − kΔyΔz
Δx →0 ΔyΔz
∂x ⎠
∂x ⎝ ∂x ⎠
Δx
ΔyΔz ∂x
ΔyΔz ∂x ⎝
∂x 2
lim
&
&
1 Q y + Δy − Q y
1 ∂Q y
1 ∂ ⎛
∂T
⎜ − kΔxΔz
=
=
Δy → 0 ΔxΔz
∂y
Δy
ΔxΔz ∂y
ΔxΔz ∂y ⎜⎝
lim
Here the property α = k / ρC is the thermal diffusivity of the material.
2-10
∂ ⎛ ∂T ⎞
∂ 2T
⎟⎟ = − ⎜⎜ k
⎟⎟ = − k
∂y ⎝ ∂y ⎠
∂y 2
Chapter 2 Heat Conduction Equation
2-29 We consider a thin ring shaped volume element of width Δz and thickness Δr in a cylinder. The
density of the cylinder is ρ and the specific heat is C. In general, an energy balance on this ring element
during a small time interval Δt can be expressed as
Δz
ΔE element
(Q& r − Q& r + Δr ) + (Q& z − Q& z + Δz ) =
Δt
But the change in the energy content of the element can be expressed as
ΔE element = E t + Δt − E t = mC (Tt + Δt − Tt ) = ρC (2πrΔr ) Δz(Tt + Δt − Tt )
rr
r+Δr
Substituting,
(Q& r − Q& r + Δr ) + (Q& z − Q& z + Δz ) = ρC (2πrΔr ) Δz
Tt + Δt − Tt
Δt
Dividing the equation above by ( 2πrΔr ) Δz gives
T
− Tt
1 Q& r + Δr − Q& r
1 Q& z + Δz − Q& z
= ρ C t + Δt
2πrΔz
2πrΔr
Δr
Δz
Δt
Noting that the heat transfer surface areas of the element for heat conduction in the r and z directions are
Ar = 2πrΔz and Az = 2πrΔr , respectively, and taking the limit as Δr , Δz and Δt → 0 yields
1 ∂ ⎛ ∂T ⎞ 1 ∂ ⎛ ∂T ⎞ ∂ ⎛ ∂T ⎞
∂T
⎜k
⎟ + ⎜k
⎜ kr
⎟+
⎟ = ρC
r ∂r ⎝ ∂r ⎠ r 2 ∂φ ⎜⎝ ∂φ ⎟⎠ ∂z ⎝ ∂z ⎠
∂t
since, from the definition of the derivative and Fourier’s law of heat conduction,
∂T ⎞
1 Q& r + Δr − Q& r
1 ∂Q
1 ∂ ⎛
1 ∂ ⎛ ∂T ⎞
=
=
⎟=−
⎜ kr
⎜ − k (2πrΔz )
Δr →0 2πrΔz
∂r ⎠
r ∂r ⎝ ∂r ⎠
2πrΔz ∂r 2πrΔz ∂r ⎝
Δr
lim
∂T
1 Q& z + Δz − Q& z
1 ∂Qz
1 ∂ ⎛
=
=
⎜ − k (2πrΔr )
Δz → 0 2πrΔr
∂z
2πrΔr ∂z
2πrΔr ∂z ⎝
Δz
lim
∂ ⎛ ∂T ⎞
⎟ = − ⎜k
z ⎝ ∂z ⎠
For the case of constant thermal conductivity the equation above reduces to
1 ∂ ⎛ ∂T ⎞ ∂ 2 T 1 ∂T
=
⎟+
⎜r
r ∂r ⎝ ∂r ⎠ ∂z 2 α ∂t
where α = k / ρC is the thermal diffusivity of the material. For the case of steady heat conduction with no
heat generation it reduces to
1 ∂ ⎛ ∂T ⎞ ∂ 2 T
=0
⎟+
⎜r
r ∂r ⎝ ∂r ⎠ ∂z 2
2-11
Chapter 2 Heat Conduction Equation
2-30 Consider a thin disk element of thickness Δz and diameter D in a long cylinder (Fig. P2-30). The
density of the cylinder is ρ, the specific heat is C, and the area of the cylinder normal to the direction of
heat transfer is A = πD 2 / 4 , which is constant. An energy balance on this thin element of thickness Δz
during a small time interval Δt can be expressed as
⎛ Rate of heat ⎞ ⎛ Rate of heat
⎟ ⎜
⎜ conduction at ⎟ − ⎜ conduction at the
⎜ the surface at z ⎟ ⎜ surface at z + Δz
⎠ ⎝
⎞ ⎛ Rate of heat ⎞ ⎛ Rate of change of
⎟ ⎜
⎟ ⎜
⎟ + ⎜ generation inside ⎟ = ⎜ the energy content
⎟ ⎜ the element ⎟ ⎜ of the element
⎠ ⎝
⎠ ⎝
or,
ΔE element
Q& z − Q& z +Δz + G& element =
Δt
But the change in the energy content of the element and the rate of heat
generation within the element can be expressed as
ΔE element = E t + Δt − E t = mC (Tt + Δt − Tt ) = ρCAΔz(Tt + Δt − Tt )
and
& element = gA
& Δz
G& element = gV
Substituting,
T
− Tt
& Δz = ρCAΔz t + Δt
Q& z − Q& z + Δz + gA
Δt
Dividing by AΔz gives
T
− Tt
1 Q& z + Δz − Q& z
+ g& = ρC t + Δt
A
Δz
Δt
Taking the limit as Δz → 0 and Δt → 0 yields
1 ∂ ⎛ ∂T ⎞
∂T
⎜ kA
⎟ + g& = ρC
A ∂ z ⎝ ∂z ⎠
∂t
since, from the definition of the derivative and Fourier’s law of heat conduction,
Q& z + Δz − Q& z ∂Q ∂ ⎛
∂T ⎞
=
=
⎜ − kA
Δz → 0
∂ z ∂z ⎝
∂z ⎠
Δz
lim
Noting that the area A and the thermal conductivity k are constant, the one-dimensional transient heat
conduction equation in the axial direction in a long cylinder becomes
∂ 2 T g& 1 ∂T
+ =
∂z 2 k α ∂t
where the property α = k / ρC is the thermal diffusivity of the material.
2-12
Chapter 2 Heat Conduction Equation
2-31 For a medium in which the heat conduction equation is given by
∂2T
∂x
2
+
∂2T
∂y
2
=
1 ∂T
:
α ∂t
(a) Heat transfer is transient, (b) it is two-dimensional, (c) there is no heat generation, and (d) the thermal
conductivity is constant.
2-32 For a medium in which the heat conduction equation is given by
1 ∂ ⎛ ∂T ⎞ ∂ ⎛ ∂T ⎞
⎟ + ⎜k
⎟ + g& = 0 :
⎜ kr
r ∂r ⎝ ∂ r ⎠ ∂ z ⎝ ∂ z ⎠
(a) Heat transfer is steady, (b) it is two-dimensional, (c) there is heat generation, and (d) the thermal
conductivity is variable.
2-33 For a medium in which the heat conduction equation is given by
1 ∂ ⎛ 2 ∂T ⎞
1
∂ 2T 1 ∂T
+
=
r
:
r 2 ∂r ⎝ ∂r ⎠ r 2 sin 2 θ ∂φ 2 α ∂t
(a) Heat transfer is transient, (b) it is two-dimensional, (c) there is no heat generation, and (d) the thermal
conductivity is constant.
Boundary and Initial Conditions; Formulation of Heat Conduction Problems
2-34C The mathematical expressions of the thermal conditions at the boundaries are called the boundary
conditions. To describe a heat transfer problem completely, two boundary conditions must be given for
each direction of the coordinate system along which heat transfer is significant. Therefore, we need to
specify four boundary conditions for two-dimensional problems.
2-35C The mathematical expression for the temperature distribution of the medium initially is called the
initial condition. We need only one initial condition for a heat conduction problem regardless of the
dimension since the conduction equation is first order in time (it involves the first derivative of temperature
with respect to time). Therefore, we need only 1 initial condition for a two-dimensional problem.
2-36C A heat transfer problem that is symmetric about a plane, line, or point is said to have thermal
symmetry about that plane, line, or point. The thermal symmetry boundary condition is a mathematical
expression of this thermal symmetry. It is equivalent to insulation or zero heat flux boundary condition,
and is expressed at a point x0 as ∂T ( x0 , t ) / ∂x = 0 .
2-37C The boundary condition at a perfectly insulated surface (at x = 0, for example) can be expressed as
−k
∂T (0, t )
=0
∂x
or
∂T (0, t )
=0
∂x
which indicates zero heat flux.
2-38C Yes, the temperature profile in a medium must be perpendicular to an insulated surface since the
slope ∂T / ∂x = 0 at that surface.
2-13
Chapter 2 Heat Conduction Equation
2-39C We try to avoid the radiation boundary condition in heat transfer analysis because it is a non-linear
expression that causes mathematical difficulties while solving the problem; often making it impossible to
obtain analytical solutions.
2-40 A spherical container of inner radius r1 , outer radius r2 , and thermal
conductivity k is given. The boundary condition on the inner surface of the
container for steady one-dimensional conduction is to be expressed for the
following cases:
r1
r2
(a) Specified temperature of 50°C: T ( r1 ) = 50° C
(b) Specified heat flux of 30 W/m2 towards the center: k
dT (r1 )
= 30 W / m 2
dr
(c) Convection to a medium at T∞ with a heat transfer coefficient of h: k
dT (r1 )
= h[T (r1 ) − T∞ ]
dr
2-41 Heat is generated in a long wire of radius r0 covered with a plastic insulation layer at a constant rate
of g& 0 . The heat flux boundary condition at the interface (radius r0 ) in terms of the heat generated is to be
expressed. The total heat generated in the wire and the heat flux at the interface are
go
G& = g& V
= g& (πr 2 L)
0 wire
0
0
Q&
G& g& (πr 2 L) g& 0 r0
q& s = s = = 0 0
=
A
A
(2πr0 ) L
2
D
L
Assuming steady one-dimensional conduction in the radial direction, the heat flux boundary condition can
be expressed as
−k
dT (r0 ) g& 0 r0
=
dr
2
2-42 A long pipe of inner radius r1 , outer radius r2 , and
thermal conductivity k is considered. The outer surface of the
pipe is subjected to convection to a medium at T∞ with a heat
transfer coefficient of h. Assuming steady one-dimensional
conduction in the radial direction, the convection boundary
condition on the outer surface of the pipe can be expressed as
dT (r2 )
−k
= h[T (r2 ) − T∞ ]
dr
2-14
h
T∞
r1
r2
Chapter 2 Heat Conduction Equation
2-43 A spherical shell of inner radius r1 , outer radius r2 , and thermal conductivity k is considered. The
outer surface of the shell is subjected to radiation to surrounding surfaces at Tsurr . Assuming no convection
outer surface of the shell can be expressed as
−k
ε
dT (r2 )
4
= εσ T (r2 ) 4 − Tsurr
dr
k
r1
r2
Tsurr
2-44 A spherical container consists of two spherical layers A and B that are at perfect contact. The radius of
the interface is ro. Assuming transient one-dimensional conduction in the radial direction, the boundary
conditions at the interface can be expressed as
TA (r0 , t ) = TB ( r0 , t )
and
−k A
r
∂TA (r0 , t )
∂T (r , t )
= −kB B 0
∂x
∂x
2-45 Heat conduction through the bottom section of a steel pan that is used to boil water on top of an
electric range is considered (Fig. P2-45). Assuming constant thermal conductivity and one-dimensional
heat transfer, the mathematical formulation (the differential equation and the boundary conditions) of this
heat conduction problem is to be obtained for steady operation.
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to
be constant. 3 There is no heat generation in the medium. 4 The top surface at x = L is subjected to
convection and the bottom surface at x = 0 is subjected to uniform heat flux.
Analysis The heat flux at the bottom of the pan is
q& s =
Q& s
G&
0.85 × (1000 W)
=
=
= 27,056 W / m 2
2
As πD 2 / 4
π (0.20 m) / 4
Then the differential equation and the boundary conditions
for this heat conduction problem can be expressed as
d 2T
=0
dx 2
dT (0)
= q& s = 27,056 W / m2
dr
dT ( L)
−k
= h[T ( L) − T∞ ]
dr
−k
2-15
Chapter 2 Heat Conduction Equation
2-46E A 1.5-kW resistance heater wire is used for space heating. Assuming constant thermal conductivity
and one-dimensional heat transfer, the mathematical formulation (the differential equation and the
boundary conditions) of this heat conduction problem is to be obtained for steady operation.
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to
be constant. 3 Heat is generated uniformly in the wire.
Analysis The heat flux at the surface of the wire is
q& s =
Q& s
1200 W
G&
=
=
= 212.2 W / in 2
As 2πr0 L 2π (0.06 in)(15 in)
Noting that there is thermal symmetry about the center line and there is uniform heat flux at the outer
surface, the differential equation and the boundary conditions for this heat conduction problem can be
expressed as
1 d ⎛ dT ⎞ g&
⎜r
⎟+ = 0
r dr ⎝ dr ⎠ k
2 kW
D = 0.12 in
dT (0)
=0
dr
dT (r0 )
−k
= q& s = 212.2 W / in 2
dr
L = 15 in
2-47 Heat conduction through the bottom section of an aluminum pan that is used to cook stew on top of
an electric range is considered (Fig. P2-47). Assuming variable thermal conductivity and one-dimensional
heat transfer, the mathematical formulation (the differential equation and the boundary conditions) of this
heat conduction problem is to be obtained for steady operation.
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to
be variable. 3 There is no heat generation in the medium. 4 The top surface at x = L is subjected to
specified temperature and the bottom surface at x = 0 is subjected to uniform heat flux.
Analysis The heat flux at the bottom of the pan is
q& s =
Q& s
G&
0.90 × (900 W)
=
=
= 31,831 W / m 2
As πD 2 / 4 π (018
. m) 2 / 4
Then the differential equation and the boundary conditions
for this heat conduction problem can be expressed as
d ⎛ dT ⎞
⎜k
⎟=0
dx ⎝ dx ⎠
−k
dT (0)
= q& s = 31,831 W / m2
dr
T ( L) = TL = 108° C
2-16
Chapter 2 Heat Conduction Equation
2-48 Water flows through a pipe whose outer surface is wrapped with a thin electric heater that consumes
300 W per m length of the pipe. The exposed surface of the heater is heavily insulated so that the entire
heat generated in the heater is transferred to the pipe. Heat is transferred from the inner surface of the pipe
to the water by convection. Assuming constant thermal conductivity and one-dimensional heat transfer, the
mathematical formulation (the differential equation and the boundary conditions) of the heat conduction in
the pipe is to be obtained for steady operation.
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to
be constant. 3 There is no heat generation in the medium. 4 The outer surface at r = r2 is subjected to
uniform heat flux and the inner surface at r = r1 is subjected to convection.
Analysis The heat flux at the outer surface of the pipe is
q& s =
Q& s
Q& s
300 W
=
=
= 734.6 W / m 2
As 2πr2 L 2π (0.065 cm)(1 m)
Noting that there is thermal symmetry about the center line and there is uniform heat flux at the outer
surface, the differential equation and the boundary conditions for this heat conduction problem can be
expressed as
d ⎛ dT ⎞
⎜r
⎟=0
dr ⎝ dr ⎠
Q = 300 W
dT (r1 )
= h[T (ri ) − T∞ ]
dr
dT (r2 )
= q& s = 734.6 W/m 2
k
dr
k
h
T∞
r1
r2
2-49 A spherical metal ball that is heated in an oven to a temperature of Ti throughout is dropped into a
large body of water at T∞ where it is cooled by convection. Assuming constant thermal conductivity and
transient one-dimensional heat transfer, the mathematical formulation (the differential equation and the
boundary and initial conditions) of this heat conduction problem is to be obtained.
Assumptions 1 Heat transfer is given to be transient and one-dimensional. 2 Thermal conductivity is given
to be constant. 3 There is no heat generation in the medium. 4 The outer surface at r = r0 is subjected to
convection.
Analysis Noting that there is thermal symmetry about the midpoint and convection at the outer surface,
the differential equation and the boundary conditions for this heat conduction problem can be expressed as
1 ∂ ⎛ 2 ∂T ⎞ 1 ∂T
⎜r
⎟=
r 2 ∂r ⎝ ∂r ⎠ α ∂ t
∂T (0, t )
=0
∂r
∂T (r0 , t )
−k
= h[T (r0 ) − T∞ ]
∂r
k
r2
T∞
h
Ti
T (r ,0) = Ti
2-50 A spherical metal ball that is heated in an oven to a temperature of Ti throughout is allowed to cool
in ambient air at T∞ by convection and radiation. Assuming constant thermal conductivity and transient
one-dimensional heat transfer, the mathematical formulation (the differential equation and the boundary
and initial conditions) of this heat conduction problem is to be obtained.
2-17
Chapter 2 Heat Conduction Equation
Assumptions 1 Heat transfer is given to be transient and one-dimensional. 2 Thermal conductivity is given
to be variable. 3 There is no heat generation in the medium. 4 The outer surface at r = r0 is subjected to
Analysis Noting that there is thermal symmetry about the midpoint and convection and radiation at the
outer surface and expressing all temperatures in Rankine, the differential equation and the boundary
conditions for this heat conduction problem can be expressed as
1 ∂ ⎛ 2 ∂T ⎞
∂T
⎜ kr
⎟ = ρC
∂r ⎠
∂t
r 2 ∂r ⎝
ε
Tsurr
∂T (0, t )
=0
∂r
∂T (r0 , t )
4
]
−k
= h[T (r0 ) − T∞ ] + εσ [T (r0 ) 4 − Tsurr
∂r
k
r2
T∞
h
Ti
T (r ,0) = Ti
2-51 The outer surface of the North wall of a house exchanges heat with both convection and radiation.,
while the interior surface is subjected to convection only. Assuming the heat transfer through the wall to
be steady and one-dimensional, the mathematical formulation (the differential equation and the boundary
and initial conditions) of this heat conduction problem is to be obtained.
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to
be constant. 3 There is no heat generation in the medium. 4 The outer surface at x = L is subjected to
convection and radiation while the inner surface at x = 0 is subjected to convection only.
Analysis Expressing all the temperatures in Kelvin, the differential equation and
the boundary conditions for this heat conduction problem can be expressed as
d 2T
=0
dx 2
−k
dT (0)
= h1[T∞1 − T (0)]
dx
−k
dT ( L)
4
= h1[T ( L) − T∞ 2 ] + ε 2σ T ( L) 4 − Tsky
dx
Tsky
T∞1
h1
T∞2
h2
L
2-18
x
Chapter 2 Heat Conduction Equation
Solution of Steady One-Dimensional Heat Conduction Problems
2-52C Yes, this claim is reasonable since in the absence of any heat generation the rate of heat transfer
through a plain wall in steady operation must be constant. But the value of this constant must be zero since
one side of the wall is perfectly insulated. Therefore, there can be no temperature difference between
different parts of the wall; that is, the temperature in a plane wall must be uniform in steady operation.
2-53C Yes, the temperature in a plane wall with constant thermal conductivity and no heat generation will
vary linearly during steady one-dimensional heat conduction even when the wall loses heat by radiation
from its surfaces. This is because the steady heat conduction equation in a plane wall is d 2 T / dx 2 = 0
whose solution is T ( x ) = C1 x + C2 regardless of the boundary conditions. The solution function represents
a straight line whose slope is C1.
2-54C Yes, in the case of constant thermal conductivity and no heat generation, the temperature in a solid
cylindrical rod whose ends are maintained at constant but different temperatures while the side surface is
perfectly insulated will vary linearly during steady one-dimensional heat conduction. This is because the
steady heat conduction equation in this case is d 2 T / dx 2 = 0 whose solution is T ( x ) = C1 x + C2 which
represents a straight line whose slope is C1.
2-55C Yes, this claim is reasonable since no heat is entering the cylinder and thus there can be no heat
transfer from the cylinder in steady operation. This condition will be satisfied only when there are no
temperature differences within the cylinder and the outer surface temperature of the cylinder is the equal to
the temperature of the surrounding medium.
2-19
Chapter 2 Heat Conduction Equation
2-56 A large plane wall is subjected to specified temperature on the left surface and convection on the right
surface. The mathematical formulation, the variation of temperature, and the rate of heat transfer are to be
determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3
There is no heat generation.
Properties The thermal conductivity is given to be k = 2.3 W/m⋅°C.
Analysis (a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the left
surface, the mathematical formulation of this problem can be expressed as
d 2T
=0
dx 2
and
k
T1=80°C
A=20 m2
T (0) = T1 = 80° C
−k
dT ( L)
= h[T ( L) − T∞ ]
dx
L=0.4 m
T∞ =15°C
h=24 W/m2.°C
(b) Integrating the differential equation twice with respect to x yields
dT
= C1
dx
x
T ( x ) = C1x + C2
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
x = 0:
T (0) = C1 × 0 + C2
C2 = T1
x = L:
− kC1 = h[(C1 L + C2 ) − T∞ ] →
C1 = −
h(C2 − T∞ )
k + hL
C1 = −
h(T1 − T∞ )
k + hL
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be
T ( x) = −
=−
h(T1 − T∞ )
x + T1
k + hL
(24 W / m 2 ⋅° C)(80 − 15)° C
(2.3 W / m⋅° C) + (24 W / m 2 ⋅° C)(0.4 m)
= 80 − 1311
.x
x + 80° C
(c) The rate of heat conduction through the wall is
h(T1 − T∞ )
dT
= −kAC1 = kA
Q& wall = − kA
dx
k + hL
(24 W/m 2 ⋅ °C)(80 − 15)°C
= (2.3 W/m ⋅ °C)(20 m 2 )
(2.3 W/m ⋅ °C) + (24 W/m 2 ⋅ °C)(0.4 m)
= 6030 W
Note that under steady conditions the rate of heat conduction through a plain wall is constant.
2-20
Chapter 2 Heat Conduction Equation
2-57 The top and bottom surfaces of a solid cylindrical rod are maintained at constant temperatures of
20°C and 95°C while the side surface is perfectly insulated. The rate of heat transfer through the rod is to
be determined for the cases of copper, steel, and granite rod.
Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3
There is no heat generation.
Properties The thermal conductivities are given to be k = 380 W/m⋅°C for copper, k = 18 W/m⋅°C for
steel, and k = 1.2 W/m⋅°C for granite.
Analysis Noting that the heat transfer area (the area normal to
the direction of heat transfer) is constant, the rate of heat
transfer along the rod is determined from
T −T
Q& = kA 1 2
L
T1=25°C
Insulated
D = 0.05 m
T2=95°C
where L = 0.15 m and the heat transfer area A is
A = πD 2 / 4 = π ( 0.05 m) 2 / 4 = 1.964 × 10 −3 m 2
L=0.15 m
Then the heat transfer rate for each case is determined as follows:
(a) Copper:
T −T
(95 − 20)° C
Q& = kA 1 2 = (380 W / m⋅° C)(1.964 × 10 −3 m 2 )
= 373.1 W
L
0.15 m
(b) Steel:
T −T
(95 − 20)° C
Q& = kA 1 2 = (18 W / m⋅° C)(1.964 × 10 −3 m 2 )
= 17.7 W
L
0.15 m
(c) Granite:
T −T
(95 − 20)° C
Q& = kA 1 2 = (12
. W / m⋅° C)(1.964 × 10 −3 m 2 )
= 1.2 W
L
0.15 m
Discussion: The steady rate of heat conduction can differ by orders of magnitude, depending on the
thermal conductivity of the material.
2-21
Chapter 2 Heat Conduction Equation
2-58
"GIVEN"
L=0.15 "[m]"
D=0.05 "[m]"
T_1=20 "[C]"
T_2=95 "[C]"
"k=1.2 [W/m-C], parameter to be varied"
"ANALYSIS"
A=pi*D^2/4
Q_dot=k*A*(T_2-T_1)/L
k [W/m.C]
1
22
43
64
85
106
127
148
169
190
211
232
253
274
295
316
337
358
379
400
Q [W]
0.9817
21.6
42.22
62.83
83.45
104.1
124.7
145.3
165.9
186.5
207.1
227.8
248.4
269
289.6
310.2
330.8
351.5
372.1
392.7
400
350
300
Q [W ]
250
200
150
100
50
0
0
50
100
150
200
250
k [W /m -C]
2-22
300
350
400
Chapter 2 Heat Conduction Equation
2-59 The base plate of a household iron is subjected to specified heat flux on the left surface and to
specified temperature on the right surface. The mathematical formulation, the variation of temperature in
the plate, and the inner surface temperature are to be determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional since the surface area of the base plate is
large relative to its thickness, and the thermal conditions on both sides of the plate are uniform. 2 Thermal
conductivity is constant. 3 There is no heat generation in the plate. 4 Heat loss through the upper part of
the iron is negligible.
Properties The thermal conductivity is given to be k = 20 W/m⋅°C.
Analysis (a) Noting that the upper part of the iron is well insulated and thus the entire heat generated in the
resistance wires is transferred to the base plate, the heat flux through the inner surface is determined to be
q& 0 =
Q& 0
800 W
=
= 50,000 W / m 2
Abase 160 × 10 −4 m 2
Taking the direction normal to the surface of the wall to be the x
direction with x = 0 at the left surface, the mathematical formulation of
this problem can be expressed as
d 2T
dx 2
and
−k
k
T2 =85°C
Q=800 W
A=160 cm2
=0
L=0.6 cm
dT (0)
= q& 0 = 50,000 W / m2
dx
T ( L) = T2 = 85° C
x
(b) Integrating the differential equation twice with respect to x yields
dT
= C1
dx
T ( x ) = C1x + C2
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
x = 0:
− kC1 = q& 0 →
C1 = −
x = L:
T ( L) = C1 L + C2 = T2
q& 0
k
C2 = T2 − C1 L →
C2 = T2 +
q& 0 L
k
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be
q& 0
q& L q& ( L − x )
+ T2
x + T2 + 0 = 0
k
k
k
(50,000 W/m 2 )(0.006 − x )m
=
+ 85°C
20 W/m ⋅ °C
= 2500(0.006 − x ) + 85
T ( x) = −
(c) The temperature at x = 0 (the inner surface of the plate) is
T ( 0) = 2500( 0.006 − 0) + 85 = 100° C
Note that the inner surface temperature is higher than the exposed surface temperature, as expected.
2-23
Chapter 2 Heat Conduction Equation
2-60 The base plate of a household iron is subjected to specified heat flux on the left surface and to
specified temperature on the right surface. The mathematical formulation, the variation of temperature in
the plate, and the inner surface temperature are to be determined for steady one-dimensional heat transfer.
Assumptions 1 Heat conduction is steady and one-dimensional since the surface area of the base plate is
large relative to its thickness, and the thermal conditions on both sides of the plate are uniform. 2 Thermal
conductivity is constant. 3 There is no heat generation in the plate. 4 Heat loss through the upper part of
the iron is negligible.
Properties The thermal conductivity is given to be k = 20 W/m⋅°C.
Analysis (a) Noting that the upper part of the iron is well insulated and thus the entire heat generated in the
resistance wires is transferred to the base plate, the heat flux through the inner surface is determined to be
q& 0 =
Q& 0
1200 W
=
= 75,000 W/m 2
Abase 160 × 10 − 4 m 2
Taking the direction normal to the surface of the wall to be the
x direction with x = 0 at the left surface, the mathematical
formulation of this problem can be expressed as
k
d 2T
=0
dx 2
and
−k
T2 =85°C
Q=1200 W
A=160 cm2
L=0.6 cm
dT (0)
= q& 0 = 75,000 W/m 2
dx
x
T ( L) = T2 = 85° C
(b) Integrating the differential equation twice with respect to x yields
dT
= C1
dx
T ( x ) = C1x + C2
where C1 and C2 are arbitrary constants. Applying the boundary conditions give
x = 0:
− kC1 = q& 0 →
C1 = −
x = L:
T ( L) = C1 L + C2 = T2
q& 0
k
C2 = T2 − C1 L →
C2 = T2 +
q& 0 L
k
Substituting C1 and C2 into the general solution, the variation of temperature is determined to be
q& 0
q& L q& ( L − x )
+ T2
x + T2 + 0 = 0
k
k
k
(75,000 W/m 2 )(0.006 − x )m
=
+ 85°C
20 W/m ⋅ °C
= 3750(0.006 − x ) + 85
T ( x) = −
(c) The temperature at x = 0 (the inner surface of the plate) is
T (0) = 3750(0.006 − 0) + 85 = 107.5°C
Note that the inner surface temperature is higher than the exposed surface temperature, as expected.
2-24
Chapter 2 Heat Conduction Equation
2-61
"GIVEN"
Q_dot=800 "[W]"
L=0.006 "[m]"
A_base=160E-4 "[m^2]"
k=20 "[W/m-C]"
T_2=85 "[C]"
"ANALYSIS"
q_dot_0=Q_dot/A_base
T=q_dot_0*(L-x)/k+T_2 "Variation of temperature"
"x is the parameter to be varied"
0
0.0006667
0.001333
0.002
0.002667
0.003333
0.004
0.004667
0.005333
0.006
100
98.33
96.67
95
93.33
91.67
90
88.33
86.67
85
100
98
96
T [C]
94
92
90
88
86
84
0
0.001
0.002
0.003
0.004
0.005
0.006
x [m ]
2-62E A steam pipe is subjected to convection on the inner surface and to specified temperature on the
outer surface. The mathematical formulation, the variation of temperature in the pipe, and the rate of heat
loss are to be determined for steady one-dimensional heat transfer.
2-25
### Tài liệu bạn tìm kiếm đã sẵn sàng tải về
Tải bản đầy đủ ngay
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www.engineersadvice.com | 1,601,155,579,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400245109.69/warc/CC-MAIN-20200926200523-20200926230523-00407.warc.gz | 794,376,822 | 25,424 | Stefan Boltzmann law | Temperature measurement by Radiation
# Stefan Boltzmann law | Stefan and Boltzmann formula of a black body
## What is Stefan Boltzmann’s law?
Stefan and Boltzmann enunciate a formula about the degree of the radiated temperature of a black body.
Formula: The total amount of radiated temperature in a second from a single area of an ideal black body is proportional to the absolute temperature of the fourth power of the object.
If the radiate temperature in one unit of an ideal black body is E and the absolute temperature is T, according to the formula E∝ T4 or E=KT4
Here, K is a constant, whose value is assumed = 5.7×10-8 W/ T4m2
If the object is not an ideal black body then the equation will be, E=eKT4
Here, e= the value of relative emissivity is 0-1.
In reality, the pyrometer is not only accepting the temperature from a radiation source, but the source has also radiated some temperature, in that case, the radiation of pyrometer is E=K(T24 – T14)
Here,
T2= the temperature of the radiation source.
T1=the temperature of the pyrometer.
## How to Temperature measure by Radiation
Temperature is generated with a different wavelength. Any hot object radiates energy (or heat) around it. As the temperature increased, the intensification of radiation increases and the wavelength of average radiation decreased. Again, if the temperature is decreased, the intensification of radiation is increased. Temperature radiation has mainly consisted of some electricity magnetic wave.
From 0.3 to 0.8 microwave length affects the retina of human eyes and consists of light. The thermal wavelength, which is more or less from it, we cannot see it but it produces heat. The intensification of black body radiation is shown in the following diagram.
In the figure, the radiation intensification of the black body is 500° C. In this temperature, some radiation wave becomes very small and cannot create visible light, but gradually the object becomes red. If we increase the temperature more, the wavelength of light is increased and when the object becomes white, warm then we can notice the radiation of Infrared ray. During this time, there is slight radiation of UV rays too.
#### Omar Faruk
I am Omar Faruk, the owner of Engineers Advice. He is a person who always tries to invent something new & share that others. He loves to write about Mechanical, Electrical and Electronics related content. He has completed his diploma from the Department of Mechanical and graduation from the Department of EEE.
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13.What is the slope of the regression line?Interpret the slope in the context of this problem.14.Explain what the quantity S = 3899.57 measures in the context of this problem.15.Below is the same scatterplot, but with the six intersections in California plotted as circlesand the four in other western states plotted as squares.Comment on how the relationship between average number of vehicles per day and hours ofdelay per year differs between the California intersections and the intersections in otherwestern states.32000030000028000026000024000022000020000028000260002400022000200001800016000140001200010000average vehicles per dayHours of delay per yearScatterplot of annual hours of delay vs vehicles per dayPredictorCoefSE CoefConstant-36297367-0.490.634vehicles per day0.078220.026842.910.017S = 3899.57R-Sq = 48.6%R-Sq(adj) = 42.8%TP
© 2011 BFW PublishersThe Practice of Statistics, 4/e- Chapter 3131An ecologist studying breeding habits of the common crossbill in different years finds that thereis a linear relationship between the number of breeding pairs of crossbills and the abundance ofthe spruce cones.Below are statistics on eight years of measurements, wherex= averagenumber of cones per tree andy= number of breeding pairs of crossbills in a certain forest.MeanStandard deviationx= mean number of cones/tree23.016.2y= number of crossbill pairs18.015.1The correlation betweenxandyisr= 0.96816.Find the equation of the least-squares regression line (withyas the response variable).17.What percentage of the variation in numbers of breeding pairs of crossbills can be accountedfor by this regression?18.Based on these data, can we conclude that the abundance of spruce cones is responsible forthe number of breeding pairs of crossbills?Explain.
Test 3CAP StatisticsName:Part 1:Multiple Choice.Circle the letter corresponding to the best answer.1.On May 11, 50 randomly selected subjects had their systolic blood pressure (SBP) recordedtwice—the first time at about 9:00 a.m. and the second time at about 2:00 p.m. If one were toexamine the relationship between the morning and afternoon readings, then one might expectthe correlation to be
2.If data set A of (x, y) data has correlationr= 0.65, and a second data set B has correlationr= –0.65, then
132The Practice of Statistics, 4/e- Chapter 3© 2011 BFW Publishers
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End of preview. Want to read all 23 pages? | 666 | 2,658 | {"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} | 3.5 | 4 | CC-MAIN-2022-21 | latest | en | 0.80264 |
https://makeup-advice.com/recommendations/how-to-use-binary-code.html | 1,627,480,303,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153729.44/warc/CC-MAIN-20210728123318-20210728153318-00262.warc.gz | 384,119,155 | 15,921 | # How To Use Binary Code?
## How do you read binary code?
To read binary, find a number that you want to read, and remember to count the places from right to left.
Then, multiply each digit by 2 to the power of its place number.
For example, if the 3rd place from the right is a 1, you would multiply 1 by 2 to the power of 3 to get 8.
## What does 01001 mean in binary?
Binary number
0000 0+0+0+0
8 01000 0+8+0+0+0
9 01001 0+8+0+0+1
10 01010 0+8+0+2+0
11 01011 0+8+0+2+1
27 more rows
## What is the letter A in binary?
ASCII – Binary Character Table
Letter ASCII Code Binary
A 065 01000001
B 066 01000010
C 067 01000011
D 068 01000100
22 more rows
## How do I write my name in binary code?
Make sure to put a space in between each binary code and use the correct binary code for upper or lower case letters. For example, the binary combination for the name “Paul” would be: 01010000 01100001 01110101 01101100. Repeat to write your last name in binary numbers.
## What does 1101 mean in binary code?
DECIMAL NUMBERS IN BINARY
10 1010
11 1011
12 1100
13 1101
96 more rows
## What does 11111 mean in binary?
DECIMAL NUMBERS IN BINARY
28 11100
29 11101
30 11110
31 11111
96 more rows
## What does 01101001 mean?
01101001 is the ASCII encoding for the character i . The choice of 01101001 was influenced by the art of ambigramming. While any binary number is an ambigram by virtue of the symmetric shapes of the digits 0 and 1, 01101001 also represents other character strings that have ambigram nature.
We recommend reading: How To Use The Force?
## What is B in binary?
The capital letter A is represented by the number 65 in the ASCII code (65 is 01000001 in binary). The first 65 ASCII codes (0 through 64) are used for an assortment of Control characters and special characters, so capital A ended up at 65. Capital B is 66 (01000010) and so on.
## What does 1110 mean in binary?
Binary Decimal Conversion Chart Table
Binary Decimal
1011 11
1100 12
1101 13
1110 14
12 more rows
## What does 0000 mean in binary?
It’s Binary code most likely and represent numbers. In this case it means: 0000 = 0.
## Who created binary code?
Gottfried Leibniz | 653 | 2,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2021-31 | latest | en | 0.826948 |
https://mariadb.com/kb/en/numeric-iterals/ | 1,656,120,366,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103033925.2/warc/CC-MAIN-20220625004242-20220625034242-00741.warc.gz | 430,264,368 | 6,954 | # Numeric Literals
Numeric literals are written as a sequence of digits from `0` to `9`. Initial zeros are ignored. A sign can always precede the digits, but it is optional for positive numbers. In decimal numbers, the integer part and the decimal part are divided with a dot (`.`).
If the integer part is zero, it can be omitted, but the literal must begin with a dot.
The notation with exponent can be used. The exponent is preceded by an `E` or `e` character. The exponent can be preceded by a sign and must be an integer. A number `N` with an exponent part `X`, is calculated as `N * POW(10, X)`.
In some cases, adding zeroes at the end of a decimal number can increment the precision of the expression where the number is used. For example, `PI()` by default returns a number with 6 decimal digits. But the `PI()+0.0000000000` expression (with 10 zeroes) returns a number with 10 decimal digits.
Hexadecimal literals are interpreted as numbers when used in numeric contexts.
## Examples
```10
+10
-10
```
All these literals are equivalent:
```0.1
.1
+0.1
+.1
```
With exponents:
```0.2E3 -- 0.2 * POW(10, 3) = 200
.2e3
.2e+2
1.1e-10 -- 0.00000000011
-1.1e10 -- -11000000000
``` | 336 | 1,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2022-27 | longest | en | 0.807562 |
https://amishraclassroom.blogspot.com/2013/09/solving-equation-in-mathematics.html | 1,500,605,003,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423681.33/warc/CC-MAIN-20170721022216-20170721042216-00446.warc.gz | 608,787,982 | 27,449 | Solving an Equation in Mathematics
In one general case, we have a situation such as
ƒ (x1,...,xn) = c,
where x1,...,xn are the unknowns, and c is a constant. Its solutions are the members of the inverse image
ƒ −1[c] = {(a1,...,an) ∈ T1×···×Tn | ƒ (a1,...,an) = c},
where T1×···×Tn is the domain of the function ƒ. Note that the set of solutions can be empty (there are no solutions), a singleton (there is exactly one solution), finite, or infinite (there are infinitely many solutions).
For example, an equation such as
3x + 2y = 21z
with unknowns x, y and z, can be solved by first modifying the equation in some way while keeping it equivalent, such as subtracting 21z from both sides of the equation to obtain
3x + 2y − 21z = 0
In this particular case there is not just one solution to this equation, but an infinite set of solutions, which can be written
{(xyz) | 3x + 2y − 21z = 0}.
One particular solution is x = 0, y = 0, z = 0. Two other solutions are x = 3, y = 6, z = 1, and x = 8, y = 9, z = 2. In fact, this particular set of solutions describes a plane in three-dimensional space, which passes through the three points with these coordinates.
The methods for solving equations generally depend on the type of equation, both the kind of expressions in the equation and the kind of values that may be assumed by the unknowns. The variety in types of equations is large, and so are the corresponding methods. Only a few specific types are mentioned below; a comprehensive treatment is not possible.
In general, given a class of equations, there may be no systematic method (algorithm) that is guaranteed to work. This may be due to a lack of mathematical knowledge; some problems were only solved after centuries of effort. But this also reflects that, in general, no such method can exist: some problems are known to be unsolvable by an algorithm, such as Hilbert's tenth problem, which was proved unsolvable in 1970.
For several classes of equations, algorithms have been found for solving them, some of which have been implemented and incorporated in computer algebra systems, but often require no more sophisticated technology than pencil and paper. In some other cases, heuristic methods are known that are often successful but that are not guaranteed to lead to success.
Brute force, trial and error, inspired guess
If the solution set of an equation is restricted to a finite set (as is the case for equations in modular arithmetic, for example), or can be limited to a finite number of possibilities (as is the case with some Diophantine equations), the solution set can be found by brute force, that is, by testing each of the possible values. It may be the case, though, that the number of possibilities to be considered, although finite, is so huge that an exhaustive search is not practically feasible; this is, in fact, a requirement for strong encryption methods.
As with all kinds of problem solving, trial and error may sometimes yield a solution, in particular where the form of the equation, or its similarity to another equation with a known solution, may lead to an "inspired guess" at the solution. If a guess, when tested, fails to be a solution, consideration of the way in which it fails may lead to a modified guess.
Elementary algebra
Equations involving linear or simple rational functions of a single real-valued unknown, say x, such as
$8x+7=4x+35 , \quad \frac{4x + 9}{3x + 4} = 2 \, ,$
can be solved using the methods of elementary algebra.
Systems of linear equations
Smaller systems of linear equations can be solved likewise by methods of elementary algebra. For solving large systems numerically, algorithms are used that are based on linear algebra.
Polynomial equations
Polynomial equations of degree up to four can be solved exactly using algebraic methods, of which the quadratic formula is the simplest example. Polynomial equations with a degree of five or higher require in general numerical methods (see below) or special functions such as Bring radicals, although some specific cases may be solvable algebraically, for example
4x5x3 − 3 = 0
(by using the rational root theorem), and
x6 − 5x3 + 6 = 0,
(by using the substitution x = z1/3, which simplifies this to a quadratic equation in z).
Diophantine equations
In Diophantine equations the solutions are required to be integers. In some case a brute force approach can be used, as mentioned above. In some other cases, in particular if the equation is in one unknown, it is possible to solve the equation for rational-valued unknowns (see Rational root theorem), and then find solutions to the Diophantine equation by restricting the solution set to integer-valued solutions. For example, the polynomial equation
$2x^5-5x^4-x^3-7x^2+2x+3=0\,$
has as rational solutions x = −1/2 and x = 3, and so, viewed as a Diophantine equation, it has the unique solution x = 3.
In general, however, Diophantine equations are among the most difficult equations to solve.
Inverse functions
In the simple case of a function of one variable, say, h(x), we can solve an equation of the form
h(x) = c, c constant
by considering what is known as the inverse function of h.
Given a function h : AB, the inverse function, denoted h−1, defined as h−1 : BA is a function such that
h−1(h(x)) = h(h−1(x)) = x.
Now, if we apply the inverse function to both sides of
h(x) = c, where c is a constant value in B,
we obtain
h−1(h(x)) = h−1(c)
x = h−1(c)
and we have found the solution to the equation. However, depending on the function, the inverse may be difficult to be defined, or may not be a function on all of the set B (only on some subset), and have many values at some point.
If just one solution will do, instead of the full solution set, it is actually sufficient if only the functional identity
h(h−1(x)) = x
holds. For example, the projection π1 : R2R defined by π1(x, y) = x has no post-inverse, but it has a pre-inverse π1−1 defined by π1−1(x) = (x, 0). Indeed, the equation
π1(x, y) = c
is solved by
(x, y) = π1−1(c) = (c, 0).
Examples of inverse functions include the nth root (inverse of xn); the logarithm (inverse of ax); the inverse trigonometric functions; and Lambert's W function (inverse of xex).
Factorization
If the left-hand side expression of an equation P = 0 can be factorized as P = QR, the solution set of the original solution consists of the union of the solution sets of the two equations Q = 0 and R = 0. For example, the equation
$\tan x + \cot x = 2$
can be rewritten, using the identity tan x cot x = 1 as
$\frac{\tan^2 x -2 \tan x+1}{\tan x} = 0,$
which can be factorized into
$\frac{(\tan x - 1)^2}{\tan x}= 0.$
The solutions are thus the solutions of the equation tan x = 1, and are thus the set
$x = \tfrac{\pi}{4} + k\pi, k = \cdots, -2, -1, 0, 1, 2, \ldots.$
Numerical methods
With more complicated equations in real or complex numbers, simple methods to solve equations can fail. Often, root-finding algorithms like the Newton–Raphson method can be used to find a numerical solution to an equation, which, for some applications, can be entirely sufficient to solve some problem.
Taylor series
One well-studied area of mathematics involves examining whether we can create some simple function to approximate a more complex equation near a given point. In fact, polynomials in one or several variables can be used to approximate functions in this way – these are known as Taylor series.
Matrix equations
Equations involving matrices and vectors of real numbers can often be solved by using methods from linear algebra.
Differential equations
There is a vast body of methods for solving various kinds of differential equations, both numerically and analytically. A particular class of problem that can be considered to belong here is integration, and the analytic methods for solving this kind of problems are now called symbolic integration. | 1,963 | 7,888 | {"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": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2017-30 | longest | en | 0.954639 |
https://www.shaalaa.com/textbook-solutions/c/ncert-solutions-physics-textbook-class-12-chapter-2-electrostatic-potential-and-capacitance_120 | 1,571,178,320,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986660323.32/warc/CC-MAIN-20191015205352-20191015232852-00347.warc.gz | 1,062,526,875 | 22,243 | Share
# NCERT solutions for Class 12 Physics chapter 2 - Electrostatic Potential and Capacitance
## Chapter 2: Electrostatic Potential and Capacitance
#### Chapter 2: Electrostatic Potential and Capacitance solutions [Pages 87 - 92]
Q 1 | Page 87
Two charges 5 × 10−8 C and −3 × 10−8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
Q 2 | Page 87
A regular hexagon of side 10 cm has a charge 5 µC at each of its vertices. Calculate the potential at the centre of the hexagon.
Q 3 | Page 87
Two charges 2 μC and −2 µC are placed at points A and B 6 cm apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface?
Q 4 | Page 87
A spherical conductor of radius 12 cm has a charge of 1.6 × 10−7C distributed uniformly on its surface. What is the electric field
(a) Inside the sphere
(b) Just outside the sphere
(c) At a point 18 cm from the centre of the sphere?
Q 5 | Page 87
A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10−12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
Q 6 | Page 87
Three capacitors each of capacitance 9 pF are connected in series.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?
Q 7 | Page 87
Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Q 8 | Page 87
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10−3 m2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
Q 10 | Page 88
A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
Q 11 | Page 88
A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
Q 12 | Page 88
A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of −2 × 10−9C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm).
Q 14 | Page 88
Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:
(a) at the mid-point of the line joining the two charges, and
(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.
Q 15 | Page 88
A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q.
(a) A charge is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?
(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.
Q 16 | Page 88
a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by
(vecE_2-vecE_1).hatn=sigma/in_0
Where hatn is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of hatn is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ hatn/in_0
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
Q 17 | Page 88
A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?
Q 18 | Page 88
In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:
(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.
(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?
(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?
Q 19 | Page 90
If one of the two electrons of a Hmolecule is removed, we get a hydrogen molecular ion H_2^+. In the ground state of an H_2^+ , the two protons are separated by roughly 1.5 Å, and the electron is roughly 1 Å from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.
Q 20 | Page 90
Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.
Q 21 | Page 90
Two charges −q and +q are located at points (0, 0, − a) and (0, 0, a), respectively.
(a) What is the electrostatic potential at the points?
(b) Obtain the dependence of potential on the distance of a point from the origin when r/>> 1.
(c) How much work is done in moving a small test charge from the point (5, 0, 0) to (−7, 0, 0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?
Q 22 | Page 90
Figure 2.34 shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on for r/>> 1, and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).
Q 23 | Page 90
An electrical technician requires a capacitance of 2 µF in a circuit across a potential difference of 1 kV. A large number of 1 µF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.
Q 24 | Page 90
What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realize from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]
Q 25 | Page 90
Obtain the equivalent capacitance of the network in Fig. 2.35. For a 300 V supply, determine the charge and voltage across each capacitor.
Q 26 | Page 90
The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.
(a) How much electrostatic energy is stored by the capacitor?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between and the magnitude of electric field between the plates.
Q 27 | Page 90
A 4 µF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 µF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?
Q 28 | Page 90
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) QE, where is the charge on the capacitor, and is the magnitude of electric field between the plates. Explain the origin of the factor ½.
Q 29 | Page 90
A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig. 2.36). Show
what the capacitance of a spherical capacitor is given by
C=(4piin_0r_1r_2)/(r_1-r_2)
where r1 and r2 are the radii of outer and inner spheres, respectively.
Q 30 | Page 91
A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 µC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.
(a) Determine the capacitance of the capacitor.
(b) What is the potential of the inner sphere?
(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.
Q 31.1 | Page 91
(a) Two large conducting spheres carrying charges Q1 and Q2 are brought close to each other. Is the magnitude of electrostatic force between them exactly given by Q1 Q2/4π∈02, where is the distance between their centres?
Q 31.3 | Page 91
A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?
Q 31.5 | Page 91
We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?
Q 31.6 | Page 91
What meaning would you give to the capacitance of a single conductor?
Q 31.7 | Page 91
Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).
Q 32 | Page 91
A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 µC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).
Q 33 | Page 91
A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm−1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?
Q 34 | Page 91
Describe schematically the equipotential surfaces corresponding to
(a) a constant electric field in the z-direction,
(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction,
(c) a single positive charge at the origin, and
(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.
Q 35 | Page 92
In a Van de Graaff type generator a spherical metal shell is to be a 15 × 106 V electrode. The dielectric strength of the gas surrounding the electrode is 5 × 107 Vm−1. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.)
Q 36 | Page 92
A small sphere of radius r1 and charge q1 is enclosed by a spherical shell of radius r2 and charge q2. Show that if q1 is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q2 on the shell is.
Q 37.1 | Page 92
The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm−1. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)
Q 37.2 | Page 92
A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m2. Will he get an electric shock if he touches the metal sheet next morning?
Q 37.3 | Page 92
The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?
Q 37.4 | Page 92
What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning? (Hint: The earth has an electric field of about 100 Vm−1 at its surface in the downward direction, corresponding to a surface charge density = −10−9 C m−2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.)
## NCERT solutions for Class 12 Physics chapter 2 - Electrostatic Potential and Capacitance
NCERT solutions for Class 12 Physics chapter 2 (Electrostatic Potential and Capacitance) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Physics Textbook for Class 12 solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com are providing such solutions so that students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Class 12 Physics chapter 2 Electrostatic Potential and Capacitance are Combination of Capacitors, Capacitors and Capacitance, Dielectrics and Polarisation, Free Charges and Bound Charges Inside a Conductor, Conductors and Insulators Related to Electric Field, Electrical Potential Energy of a System of Two Point Charges and of Electric Dipole in an Electrostatic Field, Equipotential Surfaces, Potential Due to a System of Charges, Electric Potential Difference, Potential Due to a Point Charge, Electric Potential, Van De Graaff Generator, Effect of Dielectric on Capacity, The Parallel Plate Capacitor, Electrostatics of Conductors, Potential Energy of a Dipole in an External Field, Potential Energy of a System of Two Charges in an External Field, Potential Energy of a Single Charge, Potential Energy of a System of Charges, Potential Due to an Electric Dipole, Relation Between Electric Field and Electrostatic Potential, Energy Stored in a Capacitor, Capacitance of a Parallel Plate Capacitor with and Without Dielectric Medium Between the Plates.
Using NCERT Class 12 solutions Electrostatic Potential and Capacitance exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in NCERT Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 12 prefer NCERT Textbook Solutions to score more in exam.
Get the free view of chapter 2 Electrostatic Potential and Capacitance Class 12 extra questions for Physics and can use Shaalaa.com to keep it handy for your exam preparation
S | 3,805 | 15,419 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-43 | latest | en | 0.877609 |
http://mathhelpforum.com/calculus/149501-max-area-triangle-ellipse-print.html | 1,506,375,032,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818693363.77/warc/CC-MAIN-20170925201601-20170925221601-00694.warc.gz | 213,834,743 | 3,132 | # Max area for triangle in ellipse...
• Jun 27th 2010, 06:51 AM
ice_syncer
Max area for triangle in ellipse...
This is the question :
there is an ellipse of the form
x^2/a^2 + y^2/b^2 = 1
find max area of isosceles triangle inscribed in the ellipse with its vertex at one end of major axis..
Now this is what I did.
From the figure, I took the semi height as y and the length of base as x+a so the area will be A = y*(x+a)
using the ellipse equation I got expression for y in terms of x and the constants a and b. then I did d/dx (A) = 0 to find the value of x, I got x = -a and x = 2a-1
so that means the length is 2a + 1 and y = (b/a ) * (SQRT(a^2-x^2))
and I get area as (b/a)((2a-1)^3/2) but the answer is b/a((3)^3/2)/4 ...
I want to know whats wrong, I dont want to know any new method because I know one method mentioned in the book that is to take it in parametric form ie x=acost and y = bsint
I want to know whats wrong with my method..
Thanks
• Jun 27th 2010, 07:30 AM
skeeter
$A = y(x+a)$
$\displaystyle A = \frac{b}{a}\sqrt{a^2-x^2} \cdot (x+a)$
$\displaystyle \frac{dA}{dx} = \frac{b}{a}\left(\sqrt{a^2-x^2} - \frac{x(x+a)}{\sqrt{a^2-x^2}}\right) = 0$
$a^2-x^2 = x^2+ax$
$0 = 2x^2 + ax - a^2$
$0 = (2x-a)(x+a)$
$x = \frac{a}{2}$ , $x = -a$
throw out the solution $x = -a$ for the obvious result (triangle with $A = 0$ formed)
$\displaystyle A = \frac{3\sqrt{3} \cdot ab}{4}$ | 510 | 1,403 | {"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": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2017-39 | longest | en | 0.872476 |
https://elteoremadecuales.com/ostrowskis-theorem/ | 1,669,877,948,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710801.42/warc/CC-MAIN-20221201053355-20221201083355-00482.warc.gz | 255,700,163 | 11,974 | # Ostrowski's theorem
Now, let {displaystyle a,nin mathbb {N} } with a > 1. Express bn in base a: {displaystyle b^{n}=sum _{i0.} Hence {displaystyle b^{n}geq a^{m-1},quad } so {displaystyle quad mleq n,{frac {log b}{log a}}+1.} Then we see, by the properties of an absolute value: {displaystyle |b|_{*}^{n}=|b^{n}|_{*}leq sum _{i1,} the above argument shows that {displaystyle |a|_{*}>1} regardless of the choice of a > 1 (otherwise {displaystyle |a|_{*}^{log _{a}!b}leq 1} , implying {displaystyle |b|_{*}leq 1} ). As a result, the initial condition above must be satisfied by any b > 1.
Thus for any choice of natural numbers a, b > 1, we get {displaystyle |b|_{*}leq |a|_{*}^{frac {log b}{log a}},} i.e.
{displaystyle {frac {log |b|_{*}}{log b}}leq {frac {log |a|_{*}}{log a}}.} By symmetry, this inequality is an equality.
Since a, b were arbitrary, there is a constant {displaystyle lambda in mathbb {R} _{+}} for which {displaystyle log |n|_{*}=lambda log n} , i.e. {displaystyle |n|_{*}=n^{lambda }=|n|_{infty }^{lambda }} for all naturals n > 1. As per the above remarks, we easily see that {displaystyle |x|_{*}=|x|_{infty }^{lambda }} for all rationals, thus demonstrating equivalence to the real absolute value.
Case (2) As this valuation is non-trivial, there must be a natural number for which {displaystyle |n|_{*}<1.} Factoring into primes: {displaystyle n=prod _{i
Si quieres conocer otros artículos parecidos a Ostrowski's theorem puedes visitar la categoría Theorems in algebraic number theory.
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Utilizamos cookies propias y de terceros para mejorar la experiencia de usuario Más información | 520 | 1,622 | {"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-2022-49 | longest | en | 0.441623 |
https://www.weegy.com/?ConversationId=6C40C68B | 1,542,321,924,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742963.17/warc/CC-MAIN-20181115223739-20181116005739-00528.warc.gz | 1,074,098,082 | 9,907 | Simplify -3p3 + 5p + (-2p2) + (-4) - 12p + 5 - (-8p3) answer choices are: A. 1 - 7p - 2p2 + 5p3 B. 5p3 - 2p2 - 7p + 1 C. 1 + 7p + 2p2 - 5p3 D. -5p3 + 2p2 + 7p + 1
Updated 6/29/2014 1:44:47 AM
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User: Simplify -3p3 + 5p + (-2p2) + (-4) - 12p + 5 - (-8p3) answer choices are: A. 1 - 7p - 2p2 + 5p3 B. 5p3 - 2p2 - 7p + 1 C. 1 + 7p + 2p2 - 5p3 D. -5p3 + 2p2 + 7p + 1
Weegy: 1-4 is equal to -4+1, they both equal -3
[email protected]|Points 440|
User: answer choices are: A. 1 - 7p - 2p2 + 5p3 B. 5p3 - 2p2 - 7p + 1 C. 1 + 7p + 2p2 - 5p3 D. -5p3 + 2p2 + 7p + 1
Weegy: A.) { x | x < 8 }
User: Find the sum of the given polynomials. 5m + 2n, n - 3m, and 7 - 3n
Updated 6/29/2014 1:44:47 AM
This conversation has been flagged as incorrect.
Flagged by andrewpallarca [6/29/2014 1:43:33 AM]
Rating
3
-3p3 + 5p + (-2p2) + (-4) - 12p + 5 - (-8p3)
= 5p^3 - 2p^2 - 7p + 1
3
(5m + 2n) + (n - 3m) + (7 - 3n)
= (3n + 2m) + (7 - 3n)
= 2m + 7
Questions asked by the same visitor
9a^4 - b^2
Question
Updated 8/7/2014 8:44:05 AM
9a^4 - b^2 = (3a^2 - b) (3a^2 + b)
Confirmed by jeifunk [8/7/2014 8:29:52 AM]
Factor -x2y2 + x4 + 9y2 - 9x2
Weegy: (-4 + x^2) (x^2 - y^2). Please rate the answer. User: Factor 360t + 10t3 - 120t2. (More)
Question
Updated 6/11/2014 12:39:37 AM
-x^2y^2 + x^4 + 9y^2 - 9x^2
= x^2(x^2 - y^2) - 9(x^2 - y^2)
= (x^2 - y^2)(x^2 - 9)
= (x + y)(x - y)(x + 3)(x - 3)
Confirmed by jeifunk [6/11/2014 12:55:28 AM]
360t + 10t^3 - 120t^2
= 10t(36 + t^2 - 12t)
= 10t(t - 6)^2
Confirmed by jeifunk [6/11/2014 12:55:52 AM]
(5x2 + 2x - 3)(x - 1) answer choices are: A. 5x - 2x2 + 3x4 B. 5x2 - 5x - 1 C. 7x3 + 6x2 - 1x D.5x3 - 3x2 - 5x + 3
Question
Updated 310 days ago|1/9/2018 4:58:38 AM
(5x^2 + 2x - 3)(x - 1)
= 5x^3 + 2x^2 - 3x - 5x^2 - 2x + 3
= 5x^3 - 3x^2 - 5x + 3
Added 310 days ago|1/9/2018 4:58:38 AM
(5x2 + 2x - 3)(x - 1)
Weegy: B.x = 11/10 User: Choose the product. (5x2 + 2x - 3)(x - 1) (More)
Question
Updated 3/8/2014 4:57:34 AM
(5x2 + 2x - 3)(x - 1)
= 5x^3 - 3x^2 - 5x + 3
Find the product. (n^3)^2 · (n^5)^4
Question
Updated 5/19/2014 5:56:21 AM
(n^3)^2 · (n^5)^4;
n^6 · n^20
= n^26
27,678,467
*
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 1,746 | 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} | 4 | 4 | CC-MAIN-2018-47 | longest | en | 0.672067 |
http://www.enotes.com/homework-help/how-do-solve-2z-12-14-318874 | 1,477,137,320,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718957.31/warc/CC-MAIN-20161020183838-00472-ip-10-171-6-4.ec2.internal.warc.gz | 435,911,586 | 11,225 | # How do I solve 2z=12-14 for z?
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
Th equation 2z = 12 - 14 has to be solved for z.
2z = 12 - 14
both the terms on the right hand side are numbers and 14 can be subtracted from 12 to give -2.
=> 2z = -2
divide both the sides by 2
=> z = -2/2
=> z = -1
The solution of 2z = 12 - 14 is z = -1.
zumba96 | Student, Grade 11 | (Level 3) Valedictorian
Posted on
2z=12-14
Isolate z first
Since it is already isolated combine like terms
2z=-2
Divide by 2
z=-1
jess1999 | Student, Grade 9 | (Level 1) Valedictorian
Posted on
2z = 12 - 14
To solve this equation for z, first subtract 12 with 14
By doing that your equation should look like
2z = -2 now divide 2 on both sides
By dividing 2 on both sides your equation should look like
Wiggin42 | Student, Undergraduate | (Level 2) Valedictorian
Posted on
2z = 12 - 14
Simplify the right side by combining the like terms. Make note of the signs. :
2z = -2
Isolate the z. To do this, we have to undo the multiplication by using division. Since z is multiplied by 2, we will divide both sides by 2:
z = -1
atyourservice | Student, Grade 11 | (Level 3) Valedictorian
Posted on
2z = 12 - 14 for z
first combine like terms 12 - 14 = -2
2z = - 2
since we are solving for z we have to get z by itself and in order to do that we have to divide by 2
`( 2 z ) / 2 = (- 2 )/2`
z=-1
tvxqx3 | Student, Grade 9 | eNotes Newbie
Posted on
2z = 12 - 14
You can start by 12 minus 14 which equals -2.
Then, you have this equation: 2z = -2.
Finally, you divide by 2 to both sides. (2z/2 = -2/2)
You get: z = -1.
That's one way you can solve this problem.
The other way is like this:
2z = 12 - 14
You can divide by 2 first.
z = 12/2 - 14/2
Since there are fractions you can simplfy it.
If you simplfy it, you get:
z = 6 - 7
Which equals: z = -1
So you have two ways to solve that problem. | 655 | 1,938 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2016-44 | latest | en | 0.925458 |
https://math.stackexchange.com/questions/1166493/barycentric-coordinates-in-a-triangle-proof | 1,718,314,896,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861488.82/warc/CC-MAIN-20240613190234-20240613220234-00736.warc.gz | 346,406,506 | 38,816 | # Barycentric coordinates in a triangle - proof
I want to prove that the barycentric coordinates of a point $P$ inside the triangle with vertices in $(1,0,0), (0,1,0), (0,0,1)$ are distances from $P$ to the sides of the triangle.
Let's denote the triangle by $ABC, \ A = (1,0,0), B=(0,1,0), C= (0,0,1)$.
We consider triangles $ABP, \ BCP, \ CAP$.
The barycentric coordinates of $P$ will then be $(h_1, h_2, h_3)$ where $h_1$ is the height of $ABP$, $h_2 \rightarrow BCP$, $h_3 \rightarrow CAP$
I know that $h_1 = \frac{S_{ABP}}{S_{ABC}}$ and similarly for $h_2, \ h_3$
My problem is that I don't know how to prove that if $P= (p_1, p_2, p_3)$
then $(h_1 + h_2 + h_3)P = h_1 (1,0,0) + h_2 (0,1,0) + h_3 (0,0,1)$
Could you tell me what to do about it?
Thank you!
• This question may be helpful (though it's not exactly a duplicate). Commented Mar 1, 2015 at 19:05
I assume that you are familiar with affine subspaces and affine maps. Briefly, in case you are not: An affine map between vector spaces is a linear map plus a constant and affine spaces can always be thought of as vector spaces (move them to pass through the origin). The key observation is that each barycentric coordinate $h_i(P)$ depends affinely on $P$.
Let $T$ denote the plane containing the triangle $ABC$; it is an affine subspace of $\mathbb R^3$. You can extend the barycentric coordinates naturally to all of $T$, and they are affine functions $h_1,h_2,h_3:T\to\mathbb R$. (If you are outside the triangle, at least one of the barycentric coordinates is negative.)
An affine map from $T$ to any affine space is uniquely determined by its values at three points because $T$ is two dimensional. The sum of the barycentric coordinates is constant (as can be deduced from the fact that the sum is affine and the same at all corners of the triangle). Let this constant be $H$.
Now define $f:T\to\mathbb R^3$ by $$f(P)=\frac1H(h_1(P),h_2(P),h_3(P)).$$ It is easy to see that $f$ is affine. Again, we check the values of this affine map at three points: $f(A)=A$, $f(B)=B$ and $f(C)=C$. Therefore $f$ has to be identity on all of $T$. This means that $$(h_1(P)+h_2(P)+h_3(P))P=(h_1(P),h_2(P),h_3(P))$$ for all $P\in T$. In particular, this holds for all $P$ in the triangle.
• @Hagrid, the sum of barycentric coordinates, $H(P)=h_1(P)+h_2(P)+h_3(P)$ is an affine function (as a sum of affine functions). Therefore it is uniquely determined by its values at three points (since it's defined on something with dimension two). Now $H(A)=H(B)=H(C)$, so $H$ is actually a constant. Commented Mar 2, 2015 at 15:57
• @Hagrid, $H(A)$ is the height of your triangle, which I believe is $\sqrt{3/2}$. A sum of affine maps is an affine map and the zero map is a linear map. An affine map is constant if and only if the corresponding linear map is the zero map. Within the triangle, each barycentric coordinate is the distance from a line, which is affine. (When extended to the whole plane, it becomes a signed distance.) Commented Mar 2, 2015 at 16:06
• @Hagrid, for $H$ yes but for $f$ no. Any two affine maps that agree on three non-collinear points are equal. $H$ equals a constant map at such points so it is constant. $f$ equals the identity map (not the constant map!) at such points so it is identity. Commented Mar 2, 2015 at 16:08
• @Hagrid, sorry for the delay. You defined the barycentric coordinates to be the heights of the heights of the small triangles. I took that as a definition and worked with that. That differs from the usual definition by a factor (as we have seen), but it looks like a reasonable definition to me. Commented Mar 7, 2015 at 4:56
• @Hagrid, if your triangle is not equilateral, then the heights of triangles do not agree with barycentric coordinates up to scaling. It might be possible to show (with suitable assumptions) that $h$s satisfying the equation are a multiple of barycentric coordinates. That is an interesting question, but I think it should be asked as a separate question. Commented Mar 7, 2015 at 14:29
Let $A_i$ $\>(1\leq i\leq3)$ be the vertices of your triangle $\triangle$, and let $P=(p_1,p_2,p_3)$ be an arbitrary point of $\triangle$. Then the cartesian coordinates $p_i$ of $P$ satisfy $p_1+p_2+p_3=1$, and at the same time we can write $$P=p_1A_1+p_2A_2+p_3A_3\ ,$$ which says that the $p_i$ can be viewed as well as barycentric corrdinates of $P$ with respect to $\triangle$.
We now draw the normal $n_3$ from $P$ to the side $A_1A_2$ of $\triangle$. This normal will be orthogonal to $\overrightarrow{A_1A_2}=(-1,1,0)$ and to $s:=(1,1,1)$; the latter because $n_3$ lies in the plane of $\triangle$. It follows that $\overrightarrow{A_1A_2}\times s=(1,1,-2)$ has the proper direction. We now have to intersect $$n_3:\quad t\mapsto (p_1,p_2,p_3) +t(1,1,-2)$$ with the plane $x_3=0$ and obtain $t={p_3\over2}$. Therefore the distance from $P$ to $A_1A_2$ is given by $$h_3={p_3\over2}\sqrt{1+1+4}=\sqrt{3\over2}\>p_3\ .$$ The conclusion is that the barycentric coordinates of $P$ are not equal to the three heights in question, but only proportional to these.
• So the barycentric coordinates are $(\frac{h_1}{H}, \frac{h_2}{H}, \frac{h_3}{H})$? Commented Mar 2, 2015 at 16:17
• @Hagrid, yes, the usual barycentric coordinates are. Your definition seems to differ by a constant factor from the usual one, but it is still a reasonable definition. I commented this also under the other answer. Commented Mar 2, 2015 at 16:19
• Great, thanks a lot! Commented Mar 2, 2015 at 16:22 | 1,736 | 5,511 | {"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} | 3.71875 | 4 | CC-MAIN-2024-26 | latest | en | 0.857609 |
https://homework.cpm.org/category/MN/textbook/cc2mn/chapter/6%20Unit%206A/lesson/CC3:%206.1.1/problem/6-5 | 1,591,304,661,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347458095.68/warc/CC-MAIN-20200604192256-20200604222256-00417.warc.gz | 372,782,620 | 14,900 | ### Home > CC2MN > Chapter 6 Unit 6A > Lesson CC3: 6.1.1 > Problem6-5
6-5.
Copy and simplify the following expressions by combining like terms. Using algebra tiles may be helpful.
1. $3+4x+2+2x+2x$
$8x+5$
1. $8x+4-3-x$
See part (a).
1. $7x^2+3x+4+7x^2+3x+4$
See part (a).
$14x^2+6x+8$
1. $5x+4+x+x^2+1$
See part (a). | 150 | 327 | {"found_math": true, "script_math_tex": 6, "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-2020-24 | latest | en | 0.583146 |
https://www.cram.com/subjects/0 | 1,721,550,481,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517648.65/warc/CC-MAIN-20240721060614-20240721090614-00026.warc.gz | 628,138,322 | 10,435 | # 0
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Page 1 2 3 4 5 6 7 8 9 50 | 1,311 | 5,721 | {"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-2024-30 | latest | en | 0.930465 |
https://tradingsim.com/blog/simple-moving-average/ | 1,591,525,376,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348526471.98/warc/CC-MAIN-20200607075929-20200607105929-00573.warc.gz | 557,018,442 | 162,564 | # Simple Moving Average – Top 3 Trading Strategies
## Quick Intro Video for How to Trade with the Simple Moving Average
Before you dive into the content, check out this video on moving average crossover strategies. The video is a great precursor to the advanced topics detailed in this article.
So, what is the simple moving average? Once you begin to peel back the onion, the simple moving average is anything but simple.
There are a few additional resources I would like to point out before you proceed with the article; (1) our Trading Simulator (you will need to practice what you have learned) and (2) additional moving average posts to get a broader understanding of the averages (Displaced Moving Average, Exponential Moving Average, Triple Exponential Moving Average).
I think we all recognize the simple moving average is a popular technical indicator. Forget technical analysis, we all were likely using moving averages in our grade school math class.
While we are all familiar with the indicator, why do you think it’s so difficult to trade with simple moving averages? If you do a quick Google search, you will likely find dozens of day trading strategies, but how do we know which one will work?
Well, in this post, I am going to show you everything you need to know about simple moving averages to identify the system that will work best for your trading style. I will inform you through various channels, including trade examples, charts, and videos.
Also, I will cover a host of topics; to name a few, the simple moving average formula, popular moving averages (5, 10, 200), real-life examples, crossover strategies, and my personal experience with the indicator.
I hope you find this post useful and it helps you in some fashion on your trading journey.
## Simple Moving Average Formula
The simple moving average formula is the average closing price of a security over the last “x” periods. Calculating the simple moving average is not something for technical analysis of securities. This formula is also a key tenet to engineering and mathematical studies. This detailed article from Wikipedia [1] delves into formulas for the simple moving average, cumulative moving average, weighted moving average, and exponential moving average
Let’s look at a simple moving average example with Microsoft. The last five closing prices for Microsoft are:
28.93+28.48+28.44+28.91+28.48 = 143.24
Quite simply to calculate the simple moving average formula, you divide the total of the closing prices by the number of periods.
5-day SMA = 143.24/5 = 28.65
I love the fact the SMA is just math.
Every indicator is based on math, but the SMA is not some proprietary calculation with trademark requirements.
It is simple addition and division, for the entire world to share.
## Popular Simple Moving Averages
Popular Simple Moving Averages
Make no mistakes about it, in theory; there is an infinite number of simple moving averages.
If you think you will come up with some weird 46 SMA to beat the market -- let me stop you now. It is critical to use the most common SMAs as these are the ones many traders will be using daily.
But don’t get me wrong, while I do not advocate you following everyone else, it is essential to know what other traders are looking at for clues. According to Toni Turner, author of the ‘A Beginner’s Guide to Day Trading Online,’ the major popular moving averages used by most traders are the 10, 20, 50, 100 and 200 [2].
5 -- SMA -- For the hyper trader. The shorter the SMA, the more signals you will receive when trading. The best way to use a 5-SMA is as a trade trigger in conjunction with a longer SMA period.
5-Period SMA
10 SMA
20-SMA -- the last stop on the bus for short-term traders.
Beyond the 20-SMA, you are looking at primary trends.
### Learn to Trade Stocks, Futures, and ETFs Risk-Free
20-SMA
50-SMA -- used by traders to gauge mid-term trends.
50-period simple moving average
200-SMA -- welcome to the world of long-term trend followers. Most investors will look for a cross above or below this average to represent if the stock is in a bullish or bearish trend.
200-period simple moving average
## Basic Rules for Trading with the SMA
Basic Rules for Trading with the SMA
Now that I have given you just enough doubt before even attempting to trade with the simple moving average let’s review a few ways to make money with the SMA.
### Going with the Primary Trend
1. Look for stocks that are breaking out or down strongly
2. Apply the following SMAs 5,10,20,40,200 to see which setting is containing price the best
3. Once you have identified the correct SMA, wait for the price to test the SMA successfully and look for price confirmation that the stock is resuming the direction of the primary trend
4. Enter the trade on the next bar
### Fade the Primary Trend Using Two Simple Moving Averages
1. Locate stocks that are breaking out or down strongly
2. Select two simple moving averages to apply to the chart (ex. 5 and 10)
3. Make sure the price has not touched the 5 SMA or 10 SMA excessively in the last 10 bars
4. Wait for the price to close above or below both moving averages in the counter direction of the primary trend on the same bar
5. Enter the trade on the next bar
## Strategy #1 -- Real-Life Example going with the primary trend using the SMA
The simple moving average is probably the most basic form of technical analysis. Even hardcore fundamental guys will have a thing or two to say about the indicator.
A technical analyst must be careful to avoid analysis paralysis because there is an unlimited number of averages and time frames you can choose from.
Below is a play-by-play for using a moving average on an intraday chart. In the below example, we will cover staying on the right side of the trend after placing a long trade.
The below chart was from TIBCO (TIBX) on June 24, 2011.
I know this is a few years ago, but the market is destined to repeat prior setups; it’s all human nature at the end of the day.
Simple Moving Average Example
Notice how the stock had a breakout on the open and closed near the high of the candlestick. A breakout trader would use this as an opportunity to jump on the train and place their stop below the low of the opening candle. At this point, you can use the moving average to gauge the strength of the current trend created during the opening range. In this chart example, we are using the 10-period simple moving average.
#### Simple Moving Average -- When to Sell
Now looking at the chart above, how do you think you would have known to sell at the \$26.40 level using the simple moving average?
What’s the magic formula?
Far too many traders have tried to use the simple moving average to predict the exact sell and buy points on a chart. A trader might be able to pull this off using multiple averages for triggers, but one average alone will not be enough.
Before we go any further, save yourself the time and headache and use the averages to determine the strength of the move.
Now take another look at the chart pattern. Do you see how the stock is starting to rollover as the average is beginning to flatten out?
A breakout trader would want to stay away from this type of activity. Now again, if you were to sell on the cross down through the average, this may work some of the time, but in the long run, you will end up losing money after you factor in commissions.
Why would you lose money? Well in the majority of cases, a break of the simple moving average just leads to choppy trading activity.
If you don’t believe me, try simply buying and selling based on how the price chart crosses up or under a simple moving average. Remember, if trading were that easy, everyone would be making money hand over fist.
Flat Simple Moving Average
Next, let’s take another look at the simple moving average and the primary trend. I like to call this the holy grail setup.
This is the setup you will see in books and seminars. Simply buy on the breakout and sell when the stock crosses down beneath the price action.
The below is an intraday chart of Sina Corporation (SINA) from June 24, 2011. Look at how the price chart stays cleanly above the 20-period simple moving average.
Simple Moving Average -- Perfect Example
Isn’t that a beautiful chart? You buy on the open at \$80 and sell on the close at \$92.
A quick 15% profit in one day and you didn’t have to lift a finger.
The brain is a funny thing. I remember seeing a chart like this when I first started in trading and then I would buy the setup that matched the morning activity.
I would look for the same type of volume and price action, only to later be smacked in the face by reality when my play did not trend as well.
This is the true challenge with trading, what works well on one chart, will not work well on another. Remember, the 20-SMA worked well in this example, but you cannot build a money-making system off one play.
## Strategy #2 -- Real-Life Example going against the primary trend using the Simple Moving Average
Another way to trade using the simple moving average is to go counter to the trend.
Believe it or not, one of the higher probability plays is to go counter to extreme gap moves.
Regardless of the time in history, (60s flat line, late 90s boom, or volatility of the 2000s), it’s a safe assumption that gaps will fill 50% of the time. So, off the bat no matter how new you are to trading, you at least have a 50% shot of being on the right side of the trade using this approach.
But remember this: another validation a trader can use when going counter to the primary trend is a close under or over the simple moving average. In the example below, FSLR had a solid gap of approximately 4%. After the gap, the stock trended up strongly.
FSLR Short
You must be careful with countertrade setups. If you are on the wrong side of the trade, you and others with the same position will be the fuel for the next leg up.
Let’s fast forward a few hours on the chart.
FSLR Short Trend
Whenever you go short, and the stock does little to recover and the volatility dries up, you are in a good spot. Notice how FSLR continued lower throughout the day; unable to put up a fight. Now let’s jump forward one day to July 1, 2011.
Guess what happened?
You got it, the gap filled.
FSLR Gap Filled
## Strategy #3 -- Simple Moving Average Crossover Strategy
Simple Moving Average Crossover Strategy
Moving averages by themselves will give you a great roadmap for trading the markets.
But what about moving average crossovers as a trigger for entering and closing trades? Let me take a clear stance on this one and say I’m not a fan of this strategy.
First, the moving average by itself is a lagging indicator, now you layer in the idea that you have to wait for a lagging indicator to cross another lagging indicator is just too much delay for me.
If you look around the web, one of the most popular simple moving averages to use with a crossover strategy are the 50 and 200 day. When the 50-simple moving average crosses above the 200-simple moving average, it generates a golden cross.
Conversely, when the 50-simple moving average crosses beneath the 200-simple moving average, it creates a death cross.
I only mention this, so you are aware of the setup, which may be applicable for long-term investing. Since Tradingsim focuses on day trading, let me at least run through some basic crossover strategies.
## Moving Average Crossovers and Day Trading
### Two Simple Moving Average Crossover Strategies
The first thing to know is you want to select two moving averages that are somehow related to one another.
For example, 10 is half of 20. Or the 50 and 200 are the most popular moving averages for longer-term investors.
The second thing is coming to understand the trigger for trading with moving average crossovers. A buy or sell signal is triggered once the smaller moving average crosses above or below, the larger moving average.
According to Clif Droke in the book ‘Moving Averages Simplified’, Clif reviewed a number of charts using the 30-day and 60-day moving average crosses. Clif referred to using two moving averages on a chart as double series moving average.
He later went on to say “the same rules that apply for interpreting the 30-day and 60-day moving average combo apply for all types of double series moving averages; and can be used for all time frames…” [3] (Droke, 2001).
#### Buying on a Cross Up
In the below charting example of Apple from 4/9/2013, the 10-period SMA crossed above the 20-period SMA. You will notice that the stock had a nice intraday run from \$424 up to \$428.50.
Isn’t that just a beautiful chart? The 10-period SMA is the red line, and the blue is the 20-period. In this example, you would have bought once the red line closed above the blue which would have given you an entry point slightly above \$424.
#### Selling a Cross Down
Let’s look when a sell action is triggered. In this example, a sell action was triggered when the stock gapped down on 4/15/2013.
Now in both examples, you will notice how the stock conveniently went in the desired direction with very little friction.
Well, this is the furthest thing from reality. If you look at moving average crossovers on any symbol, you will notice more false and sideways signals than high return ones. This is because most of the time stocks on the surface move in a random pattern.
Remember people; it is the job of the big money players to fake you out at every turn to separate you from your money.
With the rise of hedge funds and automated trading systems, for every clean crossover play I find, I can probably show you another dozen or more that don’t play out well. This again is why I do not recommend the crossover strategy as a true means of making money day trading the markets.
## Simple Moving Average Trading Strategy Case Study Using Cryptocurrencies
If you have been looking at cryptocurrencies over the last six months, you are more than aware of the violent price swings. So, it got me thinking.
Are there any indicators that can give a trader an edge, or is bitcoin so volatile that in the end, everyone loses at some point if you try to actively trade the contract?
This is where I got the bright idea to see how the SMA would hold up against bitcoin.
For this study, I am using the golden cross and death cross strategies, which consists of the 50-period and 200-period simple moving averages. For those of you not familiar with these strategies, the goal is to buy when the 50-period crosses above the 200-period and sell when it crosses below.
To make things more interesting, the study will cover the 15-minute time frame so that we can get more signals.
The study will start on January 26th, 2018 and run through March 29th, 2018.
As you can imagine, there are a ton of buy and sell points on the chart. Now, to be clear, I am not a fan for always staying in the market, because you can get crushed during long periods of low volatility.
BTC-Golden Cross
The golden cross/death cross strategies on a 15-minute chart generated several trade signals in a little under two weeks.
The first trade was a short at 10,765, which we later covered for a loss at 11,270. Herein lies the problem with crossover strategies. If the market is choppy, you will bleed out slowly over time.
### Will you Take Every Trade?
I ask this question before we analyze the massive short trade from 10,500 down to 8,465. A challenging part of trading is you must trade every time your edge presents itself. Sounds easy right? Wrong!
That move down is beautiful, and you would have reaped a huge reward, but what is not reflected on this chart are there some whipsaw trades that occurred before the 26th of January.
Do you think you have what it takes to make every trade regardless of how many losers you have just encountered? Oh, how I love the game!
### You Will Always Feel Like You Were Sold a Lemon
The other telling fact is that on the second position you would have exited the trade 2,450 points off the bottom. Herein lies the second challenge of trading with lagging indicators on a volatile issue.
By the time you get the trade signal, you could be showing up to the party late.
The next move up is one that makes every 18-year-old kid believe they have a future in day trading -- simply fire and forget.
BTC-Golden Cross
After this sell signal, bitcoin had several trade signals leading into March 29th, which are illustrated in the below chart.
BTC-Golden Cross
Notice how bitcoin is not too choppy, but the gains/losses are small. If you go through weeks of trading results like this, it becomes difficult to execute your trading approach flawlessly, because you feel beaten down.
Due to the volatility of bitcoin, it’s apparent that your gainers are far larger than the losers.
#### In Summary
Much to my surprise, a simple moving average allows bitcoin to go through its wild price swings, while still allowing you the ability to stay in your winning position. The below infographic visualizes the details of this case study.
## My Personal Journey Day Trading Simple Moving Averages
Now that you have all the basics let me walk you through my experience day trading with simple moving averages.
You could be saying to yourself, “Why do I care about this guy’s experience? Mine will be different?”
In theory, yes, but there are likely parallels between our paths, and I can hopefully help you avoid some of my mistakes.
### #1 -- Newbie
It was spring 2007, and I was just starting in day trading.
In my mind volume and moving averages were all I needed to keep me safe when trading. I read all the books and browsed tons of articles on the web from top “gurus” about technical analysis.
From what I could see, price respected the 10-period moving average “all” the time.
I didn’t know at this point you see what you want to in charts and for every winning example, there are likely dozens that failed.
If the stock closed below the simple moving average and I was long, I should look to get out. But, if the stock could stay above the average, I should just hold my position and let the money flow to me.
Let’s walk through a few chart examples to get a feel for my delusions of grandeur.
### Learn to Day Trade 7x Faster Than Everyone Else
Riding the Simple Moving Average
I’m not even going to worry about giving you the ticker of the above chart because it’s honestly irrelevant.
The point is that I just saw hundreds and I mean hundreds of charts with this pattern.
The pattern I was fixated on was a cross above the 10-period moving average and then a rally to the moon.
I remember feeling such excitement of how easy it was going to be to make money day trading this simple pattern.
Now, shifting gears for a second; anyone that knows me knows that I have a strong analytical mind.
I will review the numbers and then run them all over again to make sure everything nets out.
Hence my second phase on this journey.
### #2 -- Three Lines
At this point in my journey, it’s about the summer of 2007. I am placing some trades and trying different systems, but nothing with great success.
I am using the 10-period simple moving average in conjunction with Bollinger Bands and a few other indicators.
It’s not quite a spaghetti chart just yet, but it’s a little busy.
So, after reviewing my trades, I, of course, came to the realization that one moving average is not enough on the chart.
The need to put more indicators on a chart is always the wrong answer for traders, but we must go through this process to come out of the other side.
I felt that if I combined a short-term, mid-term and long-term simple moving average, I could quickly validate each signal.
I would use the short=term to pull the trigger when it crossed above or below the mid-term line and the long-term to ensure I was on the right side of the trend.
Did that just confuse you a little?
Let’s illustrate this strategy through the chart.
Three Simple Moving Averages -- My Journey
In the above example, the blue line is a 5-period SMA, the red line is a 10-period SMA, and the purple line is a 20-period SMA.
You are welcomed to use any setting that works best for you, but the point is each moving average should be a multiple or two from one another to avoid chaos on the chart.
I used the shortest SMA as my trigger average. When it crossed above or below the mid-term line, I would have a potential trade.
The sign I needed to pull the trigger was if the price was above or below the long-term moving average.
So, going back to the chart the first buy signal came when the blue line crossed above the red and the price was above the purple line. This would have given us a valid buy signal.
Then after a nice profit, once the short line crossed below the red line, it was our time to get out.
Did this mean we should have gone short?
No. Notice that the price was still above the purple line (long-term), so no short position should have been taken.
The purple (long-term) prevents us from always being in a long or short position like in the cryptocurrency case study mentioned earlier.
Looking back many years later, it sounds a bit confusing, but I do have to compliment myself on just having some semblance of a system.
How do you think this all played out?
Don’t worry; I’m going to tell you now.
### #3 -- Buy and Sell Signals
At this point of my journey, I am still in a good place. That’s what I was hoping to represent with the green smiley faces. The green also represents the expectation of the money flow as well.
It’s around late summer at this point, and I was ready to roll out my new system of using three simple moving averages.
It became apparent to me rather quickly that this was much harder than I had originally anticipated.
First off, it was tough trying to figure out which stocks to pick.
Once I landed on trading volatile stocks, they either gave false entry signals or did not trend all day.
This level of rejection from the market cut deeply. I remember staring at the screen thinking, “Why is this not working?”
Charts began to look like the one below, and there was nothing I could do to prevent this from happening.
Multiple Signals
What do you think I did next?
That’s right, my analytical side kicked in, and I needed to review more data.
### #4 Settings
Anyone that has been trading for longer than a few months using indicators at some point has started tinkering with the settings. Well, I took that concept to an entirely different level.
I was using TradeStation at the time trading US equities, and I began to run combinations of every time period you can imagine.
I would then run TradeStation’s report optimizer to see how things would have worked out.
45 simple moving average
Two-period simple moving average
As you can see, these were desperate times. I was running all sorts of combinations until I felt I landed on one that had decent results.
Now, one point to note, I was running these results against one stock at a time.
The goal was to find an Apple or another high-volume security I could trade all day using these signals to turn a profit.
Similar to my attempt to add three moving averages after first settling with the 10-period as my average of choice, I did the same thing of needing to add more validation checks this time as well.
So, instead of just moving forward with the settings I had discovered based on historical data (which is useless the very next day, because the market never repeats itself), I wanted to outsmart the market yet again.
My path to this trading edge was to displace the optimized moving averages.
### Stop Looking for a Quick Fix. Learn to Trade the Right Way
This must be painful to read; it surely is painful for me to relive this experience.
It’s important to note that I was feeling pretty good after all this analysis. I felt that I had addressed my shortcomings and displacing the averages was going to take me to the elite level.
### #5 Displace
For those of you not familiar with displaced moving averages, it’s a means for moving the average before or after the price action.
You can offset the number of periods higher to give the stock a little more wiggle room.
Conversely, you can go negative on the offset to try and jump the trend.
I’m not going to drain this concept in this article, as the focus of this discussion is around simple moving averages.
The point is, I felt that using the averages as a predictive tool would further increase the accuracy of my signals. This way I could jump into a trade before the breakout or exit a winner right before it fell off the cliff.
To illustrate this point, check out this chart example where I would use the same simple moving average duration, but I would displace one of the averages to jump the trend.
Displaced Moving Average Sell Signal
The reality is that I would jump into trades that would never materialize or exit winners too soon before the real pop.
This, of course, left me feeling completely broken and lost. I don’t say that lightly.
I mean the feeling of despair was so real; you feel like quitting, to be honest.
I think this feeling of utter disgust and wanting to never think about trading again is part of the journey to consistent profits.
Going back to my journey, at this point it was late fall, early winter and I was just done with moving averages. This is reflected in my red unhappy face.
### #6 More Indicators
This is the awful curse of technical analysis.
Technical indicators and systems lead to more indicators to try and crack the ever-elusive stock market.
I too fell victim to this horrible symptom of pain from the markets.
This was by far my darkest period of the journey with moving averages.
Not regarding losses, but just in feeling lost with my trading system and overall confidence.
I would try one system one day and then abandon it for the next hot system. This process went on for years as I kept searching for what would work consistently regardless of the market.
This included me trying every indicator from Bollinger Bands, MACD, slow stochastics -- you name it, I tried it.
If you get anything out of this article, do not make the same mistake I did with years of worthless analysis. You will make some traction, but it’s a far better use of your time to zone in on yourself and how you perceive the market.
### #7 -- 20 Period Simple Moving Average
After many years of trading, I have landed on the 20-period simple moving average. At times I will fluctuate between the simple and exponential, but 20 is my number.
This is because I have progressed as a trader from not only a breakout trader but also a pullback trader.
I use the 20-period moving average to gauge market direction, but not as a trigger for buying or selling.
It all comes down to my ability to size up how a stock is trading in and around the average.
At times a stock will crack right through the average, but I don’t panic that a sell-off is looming. I just wait and see how the stock performs at this level.
It’s funny to think that I have essentially reverted to exactly what I was looking at over ten years ago -- one average.
You may ask “Are you upset that it took you this long to come to this conclusion?”
Absolutely not. It wasn’t all death and gloom along the way, and the simple moving average is just one component of my trading toolkit.
In other words, mastering the simple moving average was not going to make or break me as a trader.
However, understanding how to properly use this technical indicator has positioned me to make consistent profits.
There are three disadvantages that come to mind for me when trading with simple moving averages.
The first two have little to do with trading or technicals. Both disadvantages for me deal with the mental aspect of trading, which is where most traders struggle -- the problem is rarely your system.
### Closing Position Remorse
This is something I touched on briefly earlier in this article, essentially with a lagging indicator, you will never get out at the top or bottom. Thinking back to our cryptocurrency example, there were times where we left over 10% or more in paper profits on the table because we did not exit the position.
Now, you could be thinking, well if we make money that is all that matters. Well, if only your brain worked that way.
You could fall into the trap of doing look backs on your trading activity and anguishing at all the loss revenue from exiting too early.
So, how do you fight this demon? How do you let go of the potential that never was meant to be?
Very simple, you let go. You stop obsessing about what you did not receive and start praising and thanking God for what you have!
Yes, I just went there. Trading becomes spiritual as you stretch beyond profits and losses.
### The Emotional Toll of Letting Winners Run
The other very real disadvantage is the intestinal fortitude required to let your winners run. You are going to feel all kinds of emotions that are telling you to just exit the position. Or that you have made enough. Or that the pullback is going to come, and you will end up giving back many of the gains.
You must find some way of just charging through all of that and letting the security do the hard work for you. We have been conditioned our entire lives to always work hard towards something.
The market is a lot like sports. A lot of the hard work is done at practice and not just during game time.
When you are in a winner, you must let them run.
### The Lag
The obvious bone of contention is the amount of lag for moving averages. This becomes even more apparent when you talk about longer moving averages.
In this Forbes article, ‘If You Want to Time the Market, Ignore Moving Averages, Michael Cannivet highlights the issue with using moving averages [4]. First, Cannivet points to a study by Meb Faber, Cambria Investment Management from 1901 to 2012 exiting stocks when the S&P 500 closed below its 200-day moving average, “would have more than doubled your ultimate returns -- and cut your risks by at least a third” [5].
However, Cannivet highlights that if hedge fund managers bought when the S&P 500 SPDR ETF closed above its 200-day moving average and shorted when it closed below its 200-day moving average, this would have net a loss of 20.4% from the period of June 2014 to June 6, 2019.
So, use the longer averages to gauge if a stock is in a bullish or bearish trend, but with the pace of trading in today’s environment, realize the lag can prove detrimental to your bottom line.
## Simple Moving Average versus Exponential Moving Average
It would be wrong of me to not go into this a little more as the comparison of the simple moving average to the exponential moving average is a common question in the trading community.
The formula for the exponential moving average is more complicated as the simple only considers the last number of closing prices across a specified range.
The exponential moving average, however, adjusts as it moves to a greater degree based on the price action. To learn more about the exponential moving average and its calculations, please visit the article -- ‘Why Professional Traders Prefer Using the Exponential Moving Average‘.
Now shifting our focus back to the comparison of the two averages, the bottom line is the exponential moving average will stay closer to the price action, while the simple moving average has a slower/smoothed arc.
To see an actual example of how the closing price, EMA and SMA differ, check out this article from dummies.com.[6]
It is going to come down to your preference. If you like clean charts, stick to the simple moving average. If you feel that you need to try and capture more of your gains, while realizing you may be shaken out of perfectly good trades- the exponential moving average will suit you better.
Below is a charting example that illustrates how each average responds to price.
SMA vs. EMA
Are you able to guess which line is the exponential moving average? If it’s not obvious, the red line is the EMA. You can tell because even though the SMA and EMA are set to 10, the red line hugs the price action a little tighter as it makes its way up.
As you can see from the chart, the difference in the values isn’t anything to call home about.
The price will ultimately respect the line in the same way whether you are using the SMA or EMA. The only time there is a difference is when the price breaks.
What’s slightly confusing is that when the price does break, it will likely penetrate the SMA first. This is because the SMA is slower to react to the price move and if things have been trending higher for a long period of time, the SMA will have a higher value than the EMA.
I know that sounds a bit confusing so let’s look at a chart example.
Price Closing Above SMA First
As you can see, the EMA (red line) hugs the price action as the stock sells off. But then something happens as the price flattens.
The slower SMA is weighing all the closing prices equally. Therefore, it continues to decline at a faster rate.
Conversely, the EMA accounts for the most recent price movement and begins to climb upwards pulling away from the stock’s price as it is in a bottoming formation.
This pulling away by the EMA ultimately results in price breaking the EMA after closing above the SMA.
So, you may be asking yourself, “Well when will the EMA get me out faster?”. The answer to that question is when a stock goes parabolic. The EMA will stop you out first because a sharp reversal in a parabolic stock will not have the lengthy bottoming formation as depicted in the last chart example.
## So How Do You Trade with the Simple Moving Average?
4 Key Takeaways
I’m hoping at this point in the article you can answer this question.
If you haven’t already figured it out, the simple moving average is not an indicator you can use as a standalone trigger.
Now, that doesn’t mean that the indicator can’t be a great tool for monitoring the direction of a trend or helping you determine when the market is getting tired after an impulsive move.
Think of the SMA as a compass. If you want detailed coordinates, you will need other tools, but you at least have an idea of where you are headed.
I know for you the conceptual points are not enough, so let’s get literal:
1. Only use one simple moving average
2. Do not make buy or sell signals based on the price closing above or below the simple moving average
3. You should use the simple moving average, as the indicator is arguably the most popular technical analysis tool
4. Focus on observing how the stock interacts with the simple moving average, as this is often a head fake tool for algorithms and more sophisticated traders
• Test out the strategies detailed in this article using the Tradingsim platform. We have been able to help countless traders improve their results by providing a risk-free environment to practice trading on the most realistic market replay platform in the world.
• Here is a great article titled ‘How to Profit From Moving Averages‘ which details strategies using the 50 and 200-day moving averages.
• Below is another video on trading with moving averages. It’s a little dated but covers a few crossover strategies.
## External References
1. Turner, Toni. (2007). ‘A Beginners Guide to Day Trading Online, Second Edition‘. Adams Media. p. 246
2. Moving Average. Wikipedia
3. Droke, Clif. (2001). ‘Moving Averages Simplified‘. Marketplace Books. p. 38
4. Cannivet, Michael. If You Want To Time The Market, Ignore Moving Averages [Blog post]. Forbes.com
5. Faber, Meb. A Quantitative Approach to Tactical Asset Allocation [Study].
6. Griffis, M. & Epstein, L. How to Calculate Exponential Moving Average in Trading [Blog post]. Dummies.com
Al Hill is one of the co-founders of Tradingsim. He has over 18 years of day trading experience in both the U.S. and Nikkei markets. On a daily basis Al applies his deep skills in systems integration and design strategy to develop features to help retail traders become profitable. When Al is not working on Tradingsim, he can be found spending time with family and friends.
• #### pmalik
Thanks for posting this .
⭐⭐⭐⭐⭐
great info!!
• #### Alo ekene
this is a good work!!!!!
• #### Luzzie
Great read ! 🙂 Thank you for the post.
• #### Aleem
Very good articles, Do you use a scanner to find percent gainers or losers to choose a stock to trade?
• #### Rahul katariya
This is best website for beginner….You are a great help to me to understand basics, thank you so much. | 8,034 | 36,776 | {"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-2020-24 | longest | en | 0.92007 |
https://www.storyofmathematics.com/subtracting-exponents | 1,638,416,725,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.69/warc/CC-MAIN-20211202024322-20211202054322-00506.warc.gz | 1,024,634,088 | 25,139 | # Subtracting Exponents – Explanation & Examples
Exponents are powers or indices. An exponential expression consists of two parts, namely the base, denoted as b and the exponent, denoted as n. The general form of an exponential expression is b n.
## How to Subtract Exponents?
The operation of subtracting exponents is quite easy if you have a good understanding of exponents. In this article, you will learn the rules and how to apply them when you need to subtract with exponents.
But before we can embark on subtracting with exponents, let us remind ourselves some of the basic terms about exponents.
What is an exponent?
Well, an exponent or power denotes the number of times a number is repeatedly multiplied by itself. For example, when we encounter a number written as, 53, it simply implies that 5 is multiplied by itself three times. In other words, 53 = 5 x 5 x 5 = 125
The same format of writing exponents applies with variables. Variables are represented by letters and symbols. For instance, when x is multiplied repeated by itself 3 times, then we write this as; x3. Variables are usually accompanied by coefficients. A coefficient is therefore an integer that is multiplied by variable.
For instance, in 2x3, the coefficient is the number 2 and x is the variable. When a variable has no number before it, the coefficient is always 1. This is also true when a number has no exponent. The coefficient of 1 is normally negligible, and therefore cannot be written with a variable.
Subtraction of exponents really does not involve any a rule. If a number is raised to a power. You simply compute the result and then perform the normal subtraction. If both the exponents and the bases are the same, you can subtract them like any other like terms in algebra. For example, 3y – 2xy = x y.
### Subtracting exponents with the same base
Let’s explain this concept with the help of a few examples.
Example 1
• 23– 22 = 8 – 4 = 4
• 53 – 52 = 75 – 25 = 50
• Subtract x 3 y 3 from 10 x 3 y 3
In this case the coefficients of exponents are 10 and 1
The variables are like terms and hence can be subtracted
Subtract the coefficients = 10 – 1
= 9
Thus, 10x 3y 3– x 3y 3 = 9 (xy)3
You can notice that, the subtraction of exponents with like terms is done by finding the difference of their coefficients.
• Subtract 8x2 – 4x2
In this case, the variables 4x2 and 8x2 are like terms and their coefficients are 4 and 8 respectively.
= 8x2 – 4x2
= (8-4) x2.
= 4 x2
• Work out (-7x) – (-3x)
Here, -7x and -3x are like terms
= -7x – (-3x)
= -7x + 3x,
= -4x.
• 15x – 4x – 12y – 3y
Subtract like terms
15x – 4x = 11x
12y – 3y = 9y
Thus, the answer is 11x – 9y.
• Subtract (4x + 3y + z) – (2x + 3y – z).
These variables are like terms
(2x + 3y – z) – (4x + 3y + z)
Open the parenthesis;
= 2x + 3y – z – 4x – 3y – z,
Rearrange the like terms, and perform the subtraction
= 2x – 4x + 3y – 3y – z – z
= -2x + 0 – 2z,
= -2x – 2z
### Subtracting exponents with different base
Exponents with different bases are computed separated and the results subtracted. On the other hand, variable with unlike bases can not be subtracted at all. For, example subtraction of a and b can not be performed and the result is just a -b.
To subtract a positive exponents m and negative exponents n, we just connect both the terms by changing the subtraction sign to a positive sign and write the result in the form of m + n.
Therefore, subtraction of a positive and a negative unlike exponents m and -n = m + n.
Example 2
• 42 – 32 = 16 – 9 =7
• Subtract: 11x – 7y -2x – 3x.
= 11x – 2x – 3x – 7y.
= 6x – 7y
• Evaluate 3x2 – 7y2
In this case, the two exponents 3x 2 and 7y2 are unlike terms and so it will remain as it is.
Here 3x and 7y both are unlike terms so it will remain as it is.
Therefore, the answer is 3x2 – 7y2
• Evaluate 15x – 12y – 11x
= 15x5 – 11x5 – 12y5
= 4x5 – 12y5 | 1,175 | 3,897 | {"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} | 4.71875 | 5 | CC-MAIN-2021-49 | longest | en | 0.914035 |
https://socratic.org/questions/how-do-you-find-the-critical-numbers-for-f-x-x-3-x-2-x-to-determine-the-maximum- | 1,701,968,057,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100677.45/warc/CC-MAIN-20231207153748-20231207183748-00732.warc.gz | 579,899,147 | 6,097 | # How do you find the critical numbers for f(x)= x^3 + x^2 + x to determine the maximum and minimum?
Jun 30, 2017
It doesn't have any.
#### Explanation:
The critical values for a function are x-values where $f ' \left(x\right) = 0$.
Differentiating the function gives:
$f ' \left(x\right) = 3 {x}^{2} + 2 x + 1$
You can either use the quadratic formula or your graphing calculator to find the zeros of this function. In the end, you'll get the fact that $f ' \left(x\right)$ is never equal to zero.
Because of this, $f \left(x\right)$ has no critical values, and no minimum or maximum. | 169 | 592 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2023-50 | latest | en | 0.750991 |
http://www.slideserve.com/hunter/displaying-distributions-with-graphs | 1,506,125,907,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689411.82/warc/CC-MAIN-20170922235700-20170923015700-00122.warc.gz | 571,766,393 | 12,824 | 1 / 10
# Displaying Distributions With Graphs - PowerPoint PPT Presentation
Displaying Distributions With Graphs. Section 1.1. A Picture is worth a thousand words. First three rules for describing data: Make a picture Make a picture Make a picture. Categorical Data.
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Presentation Transcript
### Displaying Distributions With Graphs
Section 1.1
• First three rules for describing data:
• Make a picture
• Make a picture
• Make a picture
• The following table displays the sales figures and market share (percent of total sales) achieved by several major soft drink companies in 1999. That year a total of 9930 million cases of soft drink were sold.
• Label axes
• Horizontal Axis is the categories
• Vertical Axis can be
• Counts (Frequency)
• Proportions (Relative Frequency)
• Percents
• Draw vertical bars to represent counts, proportions, or percents
• The most accurate way to create a pie graph is to use a computer. However pie graphs can be created using estimation, or by using a protractor and using the appropriate number of degrees to create the graph.
Bar Graphs show a bar that represents either the count, proportion, or percentage of each category. Bar graphs may or may not show all categories.
Pie Graphs show how a “whole” divides into categories using either the proportion or percentage. To use a pie graph you must include all the categories that make up the whole.
Further Resources proportion, or percentage of each category. Bar graphs may or may not show all categories.
• Practice of Statistics: pg 8-9
• Homework 1.1: #2 | 516 | 2,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-39 | latest | en | 0.851157 |
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=2142.msg7956 | 1,618,781,408,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038860318.63/warc/CC-MAIN-20210418194009-20210418224009-00220.warc.gz | 153,184,007 | 10,574 | NTNUJAVA Virtual Physics LaboratoryEnjoy the fun of physics with simulations! Backup site http://enjoy.phy.ntnu.edu.tw/ntnujava/
April 19, 2021, 04:41:48 am
Discovery consists of seeing what everybody has seen and thinking what nobody has thought. ..."Albert von Szent-Gyorgyi(1893-1986, 1937 Nobel Prize for Medicine, Lived to 93)"
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Author Topic: Cross a river with a boat (Read 5092 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
ahmedelshfie
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« Embed this message on: March 10, 2011, 06:03:44 pm »
The applet design by prof Hwang, modified layout by Ahmed.
Original applet Cross a river with a boat
You wish to cross a river and arrive at a dock that is directly across from you, but the river's current
will tend to carry you downstream. To compensate, you must steer the boat at an angle. Find the angle θ, given the magnitude, |vWL|, of the water's velocity relative to the land, and the maximum speed, |vBW|, of which the boat is capable relative to the water.
◊ The boat's velocity relative to the land equals the vector sum of its velocity with respect to the water and the water's velocity with respect to the land,
vBL = vBW+ vWL .
If the boat is to travel straight across the river, i.e., along the y axis, then we need to have vBL,x=0.
This x component equals the sum of the x components of the other two vectors,
vBL,x = vBW,x + vWL,x , or 0 = -|vBW| sin θ + |vWL| .
Solving for θ, we find sinθ=|vWL|/|vBW|,
so θ =sin-1 |vWL|/|vBW|.
The following simulation let you play with it. Enjoy!
You can adjust the velocity of the river or the boat with slider.
You can also change it's direction (angle θ=c).
It will remember the last 3 traces.
Embed a running copy of this simulation
Embed a running copy link(show simulation in a popuped window)
Full screen applet or Problem viewing java?Add http://www.phy.ntnu.edu.tw/ to exception site list | 538 | 1,992 | {"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-2021-17 | latest | en | 0.862903 |
https://math.stackexchange.com/questions/3156213/residue-theorem-and-a-two-dimensional-integral-not-working | 1,561,616,646,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000894.72/warc/CC-MAIN-20190627055431-20190627081431-00283.warc.gz | 498,446,457 | 34,645 | # Residue theorem and a two-dimensional integral: not working?
Consider the following integral: \begin{align} \iint_{\mathbb{R}^2}d t\,dT\, \frac{e^{-i(t-T)}e^{-t^2}e^{-{T}^2}}{(t-T-i\epsilon)^2}\,. \end{align} The $$i\epsilon$$ prescription simply tells me that if I were to do this as a contour integral, I should deform the contour slightly from the real line to the lower complex plane near the poles.
Consider the following solution: since the integrand has a pole at $$t=T$$ for any given fixed $$T$$, using the residue theorem I get $$$$\int_{-\infty}^\infty\,dT \left(-2\pi i\right)\text{Res}(e^{-i(t-T)}e^{-t^2}e^{-{T}^2};T) = 2\pi i\int_{-\infty}^\infty dT\,e^{-2T^2}(i+2T) = -\pi\sqrt{2\pi}\,.$$$$ The minus sign follows from the fact that the residue is computed anti-clockwise due to the contour orientation. It can be shown that this is in fact wrong, and the right answer is $$$$\frac{\pi ^{3/2} \text{erfc}\left(\frac{1}{\sqrt{2}}\right)}{\sqrt{2}}-\frac{\pi }{\sqrt{e}}\,.$$$$ This is surprising to me at first, because the numerator looks very harmless and the denominator looks like a very standard second-order pole. The reason I know this is wrong is because there are ways to do this without this technique (in fact, I know two ways to do this, one by some coordinate transformation $$y=t-T,x=t+T$$, and another one numerically --- and I have two ways to do numerically) and show that the residue method above does not work. Curiously enough, if the denominator were instead $$(t-T-C)^2$$ where $$C$$ is a constant, it seems to work.
Would appreciate if there is a transparent explanation on what has gone wrong (or stupid mistakes I should have noticed).
P.S. This question started me thinking whether contour integration can be used as a larger embedded techniques in multidimensional integrals, i.e. using it as part of multi-dimensional integral without reducing the multidimensional integral to single-variable one (e.g. often the case if there is spherical symmetry); even in physics, for example, often contour integrals are only used once you reduce a particular integral to a simple, manageable integral that look amenable to standard complex analysis techniques. | 581 | 2,198 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 9, "wp-katex-eq": 0, "align": 1, "equation": 2, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2019-26 | latest | en | 0.899132 |
http://psychology.wikia.com/wiki/Earnshaw's_theorem?oldid=85644 | 1,454,837,811,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701148758.73/warc/CC-MAIN-20160205193908-00084-ip-10-236-182-209.ec2.internal.warc.gz | 170,130,600 | 24,516 | # Earnshaw's theorem
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Earnshaw's theorem states that a collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges. This was first proven by Samuel Earnshaw in 1842. It is usually referenced to magnetic fields, but originally applied to electrostatic fields. It applies to the classical inverse-square law forces (electric and gravitational) and also to the magnetic forces of permanent magnets and paramagnetic materials (but not diamagnetic materials).
## Explanation Edit
Informally, the case of a point charge in an arbitrary static electric field is a simple consequence of Gauss's law. For a particle to be in a stable equilibrium, small perturbations ("pushes") on the particle in any direction should not break the equilibrium; the particle should "fall back" to its previous position. This means that the force field lines around the particle's equilibrium position should all point inwards, towards that position. If all of the surrounding field lines point towards the equilibrium point, then the divergence of the field at that point must be negative (i.e. that point acts as a sink). However, Gauss's Law says that the divergence of any possible electric force field is zero in free space. In mathematical notation, an electrical force F(r) deriving from a potential U(r) will always be divergenceless (satisfy Laplace's equation):
$\nabla \cdot \mathbf{F} = \nabla \cdot (\nabla U) = \nabla^2 U = 0 \,$.
Therefore, there are no local minima or maxima of the field potential in free space, only saddle points. A stable equilibrium of the particle cannot exist.
To be completely rigorous, there is the possibility that the particle might leave the stable point in one direction but return to the stable point from another direction. This possibility can be dealt with using conservation of energy arguments. Also, strictly speaking, the existence of a stable point does not require that all neighboring force vectors point exactly toward the stable point; the force vectors could spiral in towards the stable point, for example. One method for dealing with this invokes the fact that, in addition to the divergence, the curl of any electric force field in free space is also zero (note that zero curl is more or less equivalent to conservation of energy).
This theorem also states that there is no possible static configuration of ferromagnets which can stably levitate an object against gravity, even when the magnetic forces are stronger than the gravitational forces.
Earnshaw's theorem has even been proven for the general case of extended bodies, and this is so even if they are flexible and conducting, provided they are not diamagnetic.[1][2]
There are, however, several exceptions to the rule's assumptions which allow magnetic levitation.
## Impact on physicsEdit
Earnshaw’s theorem, in addition to the fact that configurations of classical charged particles orbiting one another are also unstable due to electromagnetic radiation mean that even dynamic systems of charges are unstable, long term. This, for quite some time lead to the puzzling question of why matter stays together as much evidence was found that matter was held together electromagnetically.
These questions eventually pointed the way to quantum mechanical explanations of the structure of the atom.
## Proofs for magnetic dipolesEdit
### IntroductionEdit
While a more general proof may be possible, three specific cases are considered here. The first case is a magnetic dipole of constant magnitude that has a fast (fixed) orientation. The second and third cases are magnetic dipoles where the orientation changes to remain aligned either parallel or antiparallel to the field lines of the external magnetic field. In paramagnetic and diamagnetic materials the dipoles are aligned parallel and antiparallel to the field lines, respectively.
### BackgroundEdit
The proofs considered here are based on the following principles.
The energy U of a magnetic dipole with a magnetic dipole moment M in an external magnetic field B is given by
$U = -\mathbf{M}\cdot\mathbf{B} = -(M_x B_x + M_y B_y + M_z B_z).$
The dipole will only be stably levitated at points where the energy has a minimum. The energy can only have a minimum at points where the Laplacian of the energy is greater than zero. That is, where
$\nabla^2 U = {\partial^2 U \over \partial x^2} + {\partial^2 U \over \partial y^2} + {\partial^2 U \over \partial z^2} > 0.$
Finally, because both the divergence and the curl of a magnetic field are zero (in the absence of current or a changing electric field), the Laplacians of the individual components of a magnetic field are zero. That is,
$\nabla^2 B_x = 0, \nabla^2 B_y = 0, \nabla^2 B_z = 0.$
This is proved at the very end of this article as it is central to understanding the overall proof.
### Summary of proofsEdit
For a magnetic dipole of fixed orientation (and constant magnitude) the energy will be given by
$U = -\mathbf{M}\cdot\mathbf{B} = -(M_x B_x + M_y B_y + M_z B_z),$
where $M_x$, $M_y$ and $M_z$ are constant. In this case the Laplacian of the energy is always zero,
$\nabla^2 U = 0,$
so the dipole can have neither an energy minimum or an energy maximum. That is, there is no point in free space where the dipole is either stable in all directions or unstable in all directions.
Magnetic dipoles aligned parallel or antiparallel to an external field with the magnitude of the dipole proportional to the external field will correspond to paramagnetic and diamagnetic materials respectively. In these cases the energy will be given by
$U = -\mathbf{M}\cdot\mathbf{B} = -k\mathbf{B}\cdot\mathbf{B} = -k (B_x^2 + B_y^2 + B_z^2),$
where k is a constant greater than zero for paramagnetic materials and less than zero for diamagnetic materials.
In this case, it will be shown that
$\nabla^2 (B_x^2 + B_y^2 + B_z^2) \geq 0,$
which, combined with the constant k, shows that paramagnetic materials can have energy maxima but not energy minima and diamagnetic materials can have energy minima but not energy maxima. That is, paramagnetic materials can be unstable in all directions but not stable in all directions and diamagnetic materials can be stable in all directions but not unstable in all directions. Of course, both materials can have saddle points.
Finally, the magnetic dipole of a ferromagnetic material (a permanent magnet) that is aligned parallel or antiparallel to a magnetic field will be given by
$\mathbf{M} = k{\mathbf{B} \over |\mathbf{B}|},$
so the energy will be given by
$U = -\mathbf{M}\cdot\mathbf{B} = -k{\mathbf{B} \over |\mathbf{B}|}\cdot\mathbf{B} = -k{(B_x^2 + B_y^2 + B_z^2) \over (B_x^2 + B_y^2 + B_z^2)^{1/2}} = -k(B_x^2 + B_y^2 + B_z^2)^{1/2};$
but this is just the square root of the energy for the paramagnetic and diamagnetic case discussed above and, since the square root function is monotonically increasing, any minimum or maximum in the paramagnetic and diamagnetic case will be a minimum or maximum here as well.
It should be noted, however, there are no known configurations of permanent magnets that stably levitate so there may be other reasons not discussed here why it is not possible to maintain permanent magnets in orientations antiparallel to magnetic fields (at least not without rotation—see Levitron).
### Detailed proofsEdit
Earnshaw's theorem was originally formulated for electrostatics (point charges) to show that there is no stable configuration of a collection of point charges. The proofs presented here for individual dipoles should be generalizable to collections of magnetics dipoles because they are formulated in terms of energy which is additive. A rigorous treatment of this topic, however, is currently beyond the scope of this article.
### Fixed-orientation magnetic dipoleEdit
It will be proven that at all points in free space
$\nabla \cdot (\nabla U) = \nabla^2 U = {\partial^2 U \over {\partial x}^2} + {\partial^2 U \over {\partial y}^2} + {\partial^2 U \over {\partial z}^2} = 0.$
The energy U of the magnetic dipole M in the external magnetic field B is given by
$U = -\mathbf{M}\cdot\mathbf{B} = -(M_x B_x + M_y B_y + M_z B_z).$
The Laplacian will be
$\nabla^2 U = -\left( {\partial^2 (M_x B_x + M_y B_y + M_z B_z) \over {\partial x}^2} + {\partial^2 (M_x B_x + M_y B_y + M_z B_z) \over {\partial y}^2} + {\partial^2 (M_x B_x + M_y B_y + M_z B_z) \over {\partial z}^2} \right).$
Expanding and rearranging the terms (and noting that the dipole M is constant) we have
$\nabla^2 U = -\left( M_x\left({\partial^2 B_x \over {\partial x}^2} + {\partial^2 B_x \over {\partial y}^2} + {\partial^2 B_x \over {\partial z}^2}\right) + M_y\left({\partial^2 B_y \over {\partial x}^2} + {\partial^2 B_y \over {\partial y}^2} + {\partial^2 B_y \over {\partial z}^2}\right) + M_z\left({\partial^2 B_z \over {\partial x}^2} + {\partial^2 B_z \over {\partial y}^2} + {\partial^2 B_z \over {\partial z}^2}=\right) \right)$
or
$\nabla^2 U = -(M_x \nabla^2 B_x + M_y \nabla^2 B_y + M_z \nabla^2 B_z),$
but the Laplacians of the individual components of a magnetic field are zero in free space (not counting electromagnetic radiation) so
$\nabla^2 U = -(M_x 0 + M_y 0 + M_z 0) = 0,$
which completes the proof.
### Magnetic dipole aligned with external field linesEdit
The case of a paramagnetic or diamagnetic dipole is considered first. The energy is given by
$U = -k (B_x^2 + B_y^2 + B_z^2).$
Expanding and rearranging terms,
$\nabla^2 (B_x^2 + B_y^2 + B_z^2) = 2[ | \nabla B_x |^2 + | \nabla B_y |^2 + | \nabla B_z |^2 + B_x \nabla^2 B_x + B_y \nabla^2 B_y + B_z \nabla^2 B_z ];$
but since the Laplacian of each individual component of the magnetic field is zero,
$\nabla^2 (B_x^2 + B_y^2 + B_z^2) = 2[ | \nabla B_x |^2 + | \nabla B_y |^2 + | \nabla B_z |^2 ];$
and since the square of a magnitude is always positive,
$\nabla^2 (B_x^2 + B_y^2 + B_z^2) \geq 0.$
As discussed above, this means that the Laplacian of the energy of a paramagnetic material can never be positive (no stable levitation) and the Laplacian of the energy of a diamagnetic material can never be negative (no instability in all directions).
Further, because the energy for a dipole of fixed magnitude aligned with the external field will be the square root of the energy above, the same analysis applies.
### Laplacian of individual components of a magnetic fieldEdit
It is proved here that the Laplacian of each individual component of a magnetic field is zero. This shows the need to invoke the properties of magnetic fields that the divergence of a magnetic field is always zero and the curl of a magnetic field is zero in free space. (That is, in the absence of current or a changing electric field.) See Maxwell's equations for a more detailed discussion of these properties of magnetic fields.
Consider the Laplacian of the x component of the magnetic field
$\nabla^2 B_x = {\partial^2 B_x \over \partial x^2} + {\partial^2 B_x \over \partial y^2} + {\partial^2 B_x \over \partial z^2} = {\partial \over \partial x} {\partial \over \partial x} B_x + {\partial \over \partial y} {\partial \over \partial y} B_x + {\partial \over \partial z} {\partial \over \partial z} B_x.$
Because the curl of B is zero,
${\partial B_x \over \partial y} = {\partial B_y \over \partial x},$
and
${\partial B_x \over \partial z} = {\partial B_z \over \partial x},$
so we have
$\nabla^2 B_x = {\partial \over \partial x} {\partial \over \partial x} B_x + {\partial \over \partial y} {\partial \over \partial x} B_y + {\partial \over \partial z} {\partial \over \partial x} B_z.$
But since $B_x$ is continuous, the order of differentiation doesn't matter giving
$\nabla^2 B_x = {\partial \over \partial x}\left( {\partial B_x \over \partial x} + {\partial B_y \over \partial y} + {\partial B_z \over \partial z} \right) = {\partial \over \partial x}(\nabla \cdot \mathbf{B}).$
The divergence of B is constant (zero, in fact) so
$\nabla^2 B_x = {\partial \over \partial x}(\nabla \cdot \mathbf{B} = 0) = 0.$
The Laplacian of the y component of the magnetic field $B_y$ field and the Laplacian of the z component of the magnetic field $B_z$ can be calculated analogously.
## References Edit
• Samuel Earnshaw, "On the Nature of the Molecular Forces which Regulate the Constitution of the Luminiferous Ether," Trans. Camb. Phil. Soc., V7, pp. 97-112 (1842).
• W. T. Scott, "Who Was Earnshaw?", American Journal of Physics, V27, p. 418 (1959).
1. Is Magnetic Levitation Possible?
2. Earnshaw, S., On the nature of the molecular forces which regulate the constitution of the luminferous ether., Trans. Camb. Phil. Soc., 7, pp 97-112 (1842) | 3,434 | 12,841 | {"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": 32, "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-2016-07 | longest | en | 0.899635 |
https://www.studypool.com/discuss/389658/algebra-1-word-problem-help-1?free | 1,481,395,256,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543434.57/warc/CC-MAIN-20161202170903-00403-ip-10-31-129-80.ec2.internal.warc.gz | 1,008,913,886 | 14,058 | ##### Algebra 1 word problem help?
Mathematics Tutor: None Selected Time limit: 1 Day
Mrs. White wants to crochet beach hats and baby afghans for a church fund-raising bazaar. She needs 7 hours to make a hat and 3 hours to make an afghan and she has 58 hours available. She wants to make no more than 14 items and no more than 12 afghans. The bazaar will sell the hats for \$13 each and the afghans for \$7 each. How many of each should she make to maximize the income for the bazaar? What is the maximum income?
Feb 16th, 2015
58 hr
no more than 14 items
it takes 3hr to make 1 afghan
no more than 12 afghans
12 afghans x 3hr= 36 hr
58 hr- 36hr= 22hr
7hr to make a hat
she only needs to make 2 hats before she reaches the 14 item limit
2 hats x \$13= \$26
12x \$7= \$84
\$84 + \$36= \$120
Feb 16th, 2015
...
Feb 16th, 2015
...
Feb 16th, 2015
Dec 10th, 2016
check_circle | 286 | 887 | {"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-2016-50 | longest | en | 0.895249 |
http://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-6th-edition/chapter-3-section-3-4-the-slope-of-a-line-exercise-set-page-162/68 | 1,480,804,692,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541140.30/warc/CC-MAIN-20161202170901-00051-ip-10-31-129-80.ec2.internal.warc.gz | 489,672,567 | 40,756 | ## Intermediate Algebra (6th Edition)
We are given two equations of lines. We can transform these equations into the form $y=mx+b$, where m is the slope of the line and the point (0,b) is the y-intercept. $-8x+20y=7$ Add 8x to both sides. $20y=8x+7$ Divide both sides by 20. $y=\frac{2}{5}x+\frac{7}{20}$ Next, $2x-5y=0$ Subtract 2x from both sides. $-5y=-2x$ Divide both sides by -5. $y=\frac{2}{5}x=\frac{2}{5}x+0$ These lines have the same value for m ($m=\frac{2}{5}$), so we know that they have the same slope. However, the first line has a b-value of $\frac{7}{20}$ and the second has a b-value of 0, so they must have different y-intercepts. Therefore, these lines are parallel. | 230 | 686 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2016-50 | longest | en | 0.919244 |
http://mathhelpforum.com/differential-geometry/178008-proof-haussdorff-space-when-i-have-open-surjection-grapf-closed-subset-print.html | 1,526,854,054,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863689.50/warc/CC-MAIN-20180520205455-20180520225455-00476.warc.gz | 182,603,385 | 3,326 | # proof Haussdorff space, when i have open surjection and grapf closed subset...
• Apr 18th 2011, 01:53 PM
tom007
proof Haussdorff space, when i have open surjection and graph closed subset...
Hello. I need help with this:
Let $\displaystyle f:X\rightarrow Y$ be open surjection such that graph $\displaystyle \{(x,f(x)):\mbox{x \in X\}}$ is closed subset of $\displaystyle X\times Y$. Prove that Y is Hausdorff.
Thanks.
• Apr 18th 2011, 06:12 PM
hatsoff
Quote:
Originally Posted by tom007
Hello. I need help with this:
Let $\displaystyle f:X\rightarrow Y$ be open surjection such that graph {(x,f(x)):x in X} is closed subset of $\displaystyle X\times Y$. Prove that Y is Hausdorff.
Thanks.
Let $\displaystyle y_1,y_2\in Y$ be distinct, and put $\displaystyle S=\{(x,f(x)):x\in X\}$. Then there are $\displaystyle x_1,x_2\in X$ distinct with $\displaystyle f(x_1)=y_1,f(x_2)=y_2$, and therefore $\displaystyle (x_1,f(x_2))\in S^c=\{(x,y):f(x)\neq y\}$. Since $\displaystyle S^c$ is an open subset of $\displaystyle X\times Y$, then there must be open sets $\displaystyle U\subseteq X,V\subseteq Y$ such that $\displaystyle U\times V\subseteq S^c$, where $\displaystyle x_1\in U$ and $\displaystyle f(x_2)\in V$; this is because $\displaystyle \{U\times V:U\in\tau_X,V\in\tau_Y\}$ is a basis for $\displaystyle X\times Y$. Notice that $\displaystyle f(U)$ and $\displaystyle V$ must be disjoint, because otherwise we would have $\displaystyle (x,f(x))\in U\times V\subseteq S^c$, a contradiction. So $\displaystyle f(U),V$ are disjoint neighborhoods of $\displaystyle f(x_1)=y_1,f(x_2)=y_2$, respectively. It follows that $\displaystyle Y$ is Hausdorff.
EDIT: The latex software is acting up, but you can quote my post to see the code.
• Apr 18th 2011, 09:24 PM
Drexel28
Quote:
Originally Posted by tom007
Hello. I need help with this:
Let $\displaystyle f:X\rightarrow Y$ be open surjection such that graph {(x,f(x)):x in X} is closed subset of $\displaystyle X\times Y$. Prove that Y is Hausdorff.
Thanks.
Note that it suffices to prove that the diagonal \Delta_Y\subseteq Y\times Y is closed, or that Y\times Y-\Delta_Y is open. Prove though that the map f\oplus\text{id}_Y:X\times Y\to Y\times Y:(x,y)\mapsto (f(x),y) is open since f is. Then, since f is a surjection (and thus so is f\oplus \text{id}_Y) we have that \left(f\oplus\text{id}_Y\right)\left(X\times Y-\Gamma_f\right)=Y\times Y-\left(f\oplus\text{id}_Y\right)\left(\Gamma_f\righ t). But, this latter set must be open (since X\times Y-\Gamma_f is open by assumption and f\oplus\text{id}_Y is open) but it's easy to show that \left(f\oplus\text{id}_Y\right)\left(\Gamma_f\righ t)=\Delta_Y. The conclusion then follows.
• Apr 19th 2011, 08:51 AM
tom007
wow.. guys.. thanks a lot! :) | 890 | 2,752 | {"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} | 3.609375 | 4 | CC-MAIN-2018-22 | latest | en | 0.789517 |
http://mathhelpforum.com/advanced-algebra/131788-prove-ab-divides-c-print.html | 1,508,647,699,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825141.95/warc/CC-MAIN-20171022041437-20171022061437-00708.warc.gz | 216,219,087 | 2,942 | # Prove that ab divides c.
• Mar 2nd 2010, 10:17 PM
rainyice
Prove that ab divides c.
Let a|c and b|c, and (a,b)=1, prove that ab divides c.
At first, I don't think this is right. For instance, let a=1 and b=2, then (a,b)=1. However, if c=3, then 2 can't divides 3 which implies b can't divides c......
• Mar 2nd 2010, 11:42 PM
Swlabr
Quote:
Originally Posted by rainyice
Let a|c and b|c, and (a,b)=1, prove that ab divides c.
At first, I don't think this is right. For instance, let a=1 and b=2, then (a,b)=1. However, if c=3, then 2 can't divides 3 which implies b can't divides c......
$a|c \Rightarrow ax=c$, then as $\text{gcd}(a, b) = 1$ and $b|c$ we have that $b|x$. Thus, $abx^{\prime} = c \Rightarrow ab|c$.
• Mar 3rd 2010, 06:53 AM
rainyice
Quote:
Originally Posted by Swlabr
$a|c \Rightarrow ax=c$, then as $\text{gcd}(a, b) = 1$ and $b|c$ we have that $b|x$. Thus, $abx^{\prime} = c \Rightarrow ab|c$.
where does that b|x come from???
• Mar 3rd 2010, 09:42 AM
Swlabr
Quote:
Originally Posted by rainyice
where does that b|x come from???
Because the gcd of a and b is 1, so it must divide x.
(The $b | c$ in my previous post was meant to be $b \nmid c$). | 432 | 1,175 | {"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": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2017-43 | longest | en | 0.92421 |
http://slideplayer.com/slide/4241728/ | 1,524,436,425,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945660.53/warc/CC-MAIN-20180422212935-20180422232935-00582.warc.gz | 292,757,099 | 24,300 | # University of Bridgeport
## Presentation on theme: "University of Bridgeport"— Presentation transcript:
University of Bridgeport
Introduction to ROBOTICS Forward Kinematics University of Bridgeport
Kinematic Forward (direct) Kinematics Inverse Kinematics
Given: The values of the joint variables. Required: The position and the orientation of the end effector. Inverse Kinematics Given : The position and the orientation of the end effector. Required : The values of the joint variables.
Why DH notation Find the homogeneous transformation H relating the tool frame to the fixed base frame
Why DH notation A very simple way of modeling robot links and joints that can be used for any kind of robot configuration. This technique has became the standard way of representing robots and modeling their motions.
DH Techniques Assign a reference frame to each joint (x-axis and z-axis). The D-H representation does not use the y-axis at all. Each homogeneous transformation Ai is represented as a product of four basic transformations
DH Techniques Matrix Ai representing the four movements is found by: four movements
DH Techniques The link and joint parameters :
Link length ai : the offset distance between the Zi-1 and Zi axes along the Xi axis. Link offset di the distance from the origin of frame i−1 to the Xi axis along the Zi-1 axis.
DH Techniques Link twist αi :the angle from the Zi-1 axis to the Zi axis about the Xi axis. The positive sense for α is determined from zi-1 and zi by the right-hand rule. Joint angle θi the angle between the Xi-1 and Xi axes about the Zi-1 axis.
DH Techniques The four parameters:
ai: link length, αi: Link twist , di : Link offset and θi : joint angle. The matrix Ai is a function of only a single variable qi , it turns out that three of the above four quantities are constant for a given link, while the fourth parameter is the joint variable.
DH Techniques With the ith joint, a joint variable is qi associated where All joints are represented by the z-axis. If the joint is revolute, the z-axis is in the direction of rotation as followed by the right hand rule. If the joint is prismatic, the z-axis for the joint is along the direction of the liner movement.
DH Techniques 3. Combine all transformations, from the first joint (base) to the next until we get to the last joint, to get the robot’s total transformation matrix. 4. From , the position and orientation of the tool frame are calculated.
DH Techniques
DH Techniques
DH Techniques
DH Techniques
Example I The two links arm
Base frame O0 All Z ‘s are normal to the page
Example I The two links arm
Where (θ1 + θ2 ) denoted by θ12 and
Example 2 19 19
Example 2 20 20
Example 3 The three links cylindrical
Example 3 The three links cylindrical
Example 3 The three links cylindrical
Example 3 The three links cylindrical
Example 4 Spherical wrist
Example 4 Spherical wrist
Example 4 Spherical wrist
Example 4 Spherical wrist
Example 5 The three links cylindrical with Spherical wrist
Example 5 The three links cylindrical with Spherical wrist
given by example 2, and given by example 3.
Example 5 The three links cylindrical with Spherical wrist
Example 5 The three links cylindrical with Spherical wrist
Example 5 The three links cylindrical with Spherical wrist
Forward kinematics: 1. The position of the end-effector: (dx ,dy ,dz ) 2. The orientation {Roll, Pitch, Yaw }
Roll Pitch Yaw The rotation matrix for the following operations: X Y Z
Example 4 The three links cylindrical with Spherical wrist
How to calculate Compare the matrix R with Of the matrix
Module 1 RRR:RRR Links α a θ d 1 90 * 10 2 3 -90 4 5 6
HW From Spong book: page 112 3.2, 3.3, 3.4, 3.6 , 3.7, 3.8, 3.9, 3.11
No Class on next Tuesday
Module 1 Where are Roll, Pitch, and Yaw
Representing forward kinematics
Transformation Matrix
Remember For n joints: we have n+1 links. Link 0 is the base
Joints are numbered from 1 to n Joint i connects link i − 1 to link i. Frame i {Xi Yi Zi} is attached to joint i+1. So, frame {O0 X0 Y0 Z0}, which is attached to the robot base (inertial frame) “joint 1”.
Similar presentations | 996 | 4,135 | {"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} | 3.65625 | 4 | CC-MAIN-2018-17 | latest | en | 0.854754 |
https://permanentkisses.com/the-definition-of-work-6/ | 1,721,045,808,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514696.4/warc/CC-MAIN-20240715102030-20240715132030-00275.warc.gz | 399,512,393 | 11,478 | # The Definition of Work
Work is a concept that explains how energy is transferred from an object to another via force and displacement. Often, work is represented as a product of force and displacement. The physical properties of work are the same as those of force and displacement. However, unlike energy, which can be stored in a physical system, work cannot be stored. The only way to store it is by measuring the force required to move a heavy object. When a person lifts a heavy object, she is performing work.
Work can be measured in a number of ways. The most common is the Joule, which is the unit for the measurement of mechanical work. However, the unit of work can also be measured in other units, such as foot-pound, kilogram, and liter-atmosphere. Other commonly used units are the kilowatt-hour, calorie, and erg. These non-SI units are sometimes used as measurement units. For example, if a person lifts a weight of 10 kg, she will be performing work equivalent to 20 newtons.
The definition of work is also important when comparing various types of work. There are different types of work, including manual labor and service jobs. For instance, a physics teacher isn’t exerting any force that would displace an object. But she will not cave in to pressure, even if a student pushes her paper across her desk. So she is not doing any work. She is conserving her energy to be able to continue teaching.
Basically, work is a measure of energy transfer. It is measured in units of joule, a SI unit equivalent to one kilogram and one Newton. When a force acts on a body, it transfers energy to it. If the applied force is opposite to the motion of the object, then it is called negative work. In this case, the energy is taken from the object. If the applied force does not result in a positive change in the displacement, then it is considered negative work.
The work of a force depends on its direction of displacement with respect to the force. If an object moves perpendicular to a force, it experiences zero work. This is the case when an object moves parallel to the force. For example, a coolie is at an angle with the force of gravity. A string exerting a ten-pound mass on its head would cause the ball to move in a circular motion.
The work of a force depends on the direction of displacement with respect to the force. In the case of gravity, a mass moving perpendicular to a force is at a 90-degree angle to gravity. In such a situation, the force causes zero work. A weight that moves parallel to a force will experience negative work. The same is true for an object that moves at an angle with a force. The direction of displacement and gravity will determine the work of the force. | 579 | 2,715 | {"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-2024-30 | latest | en | 0.962863 |
https://socratic.org/questions/how-do-you-write-2-22-times-10-6-in-standard-notation | 1,709,268,645,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00276.warc.gz | 542,972,993 | 6,261 | # How do you write 2.22 times 10 ^ -6 in standard notation?
Jun 11, 2016
In standard notation $2.22 \times {10}^{- 6} = 0.00000222$
#### Explanation:
In scientific notation, we write a number so that it has single digit to the left of decimal sign and is multiplied by an integer power of $10$.
In other words, in scientific notation, a number is written as $a \times {10}^{n}$, where $1 \le a < 10$ and $n$ is an integer and $1 \le a < 10$.
To write the number in normal or standard notation one just needs to multiply (or divide if $n$ is negative). This means moving decimal $n$ digits to right if multiplying by ${10}^{n}$ and moving decimal $n$ digits to left if dividing by ${10}^{n}$ (i.e. multiplying by ${10}^{- n}$).
In the given case, as we have the number as $2.22 \times {10}^{- 6}$, we need to move decimal digit to the left by six points. For this, let us write $2.22$ as $0000002.22$ and moving decimal point six points to left means $0.00000222$
Hence in standard notation $2.22 \times {10}^{- 6} = 0.00000222$ | 312 | 1,035 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 17, "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} | 4.65625 | 5 | CC-MAIN-2024-10 | latest | en | 0.845546 |
https://writersanswers.com/questions/at-min-past-pm-time-needed-by-minute-hand-of-clock/ | 1,669,731,676,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710698.62/warc/CC-MAIN-20221129132340-20221129162340-00181.warc.gz | 656,286,683 | 20,248 | # At Min Past Pm Time Needed By Minute Hand Of Clock Show Was Found To Be Min Less Than T2 Min Find
At t min past 2 pm, the time needed by the minute hand of a clock to show 3 pm was found to be 3 min less than t2/4 min. Find t.
We know that, the time between 2 pm to 3 pm = 1 h = 60 min
Given that, at f min past 2 pm, the time needed by the min hand of a clock to show 3 pm was found to be 3 min less than t2/4 min i,e
Hence, the required value off is 14 min. | 147 | 464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2022-49 | latest | en | 0.904701 |
https://www.bartleby.com/solution-answer/chapter-124-problem-50e-calculus-10th-edition/9781285057095/motion-along-an-involute-of-a-circle-the-figure-shows-a-particle-moving-along-a-path-modeled-by/98fc709b-a5e4-11e8-9bb5-0ece094302b6 | 1,576,500,322,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540565544.86/warc/CC-MAIN-20191216121204-20191216145204-00384.warc.gz | 636,069,446 | 83,353 | Chapter 12.4, Problem 50E
### Calculus
10th Edition
Ron Larson + 1 other
ISBN: 9781285057095
Chapter
Section
### Calculus
10th Edition
Ron Larson + 1 other
ISBN: 9781285057095
Textbook Problem
1 views
# Motion Along an Involute of a Circle The figure shows a particle moving along a path modeled by r ( t ) = 〈 cos π t + π t sin π t , sin π t − π t cos π t 〉 .The figure also shows the vectors v(t) and a(t) for t = 1 and t = 2 .(a) Find a T and a N at t = 1 and t = 2 .(b) Determine whether the speed of the particle is increasing or decreasing at each of the indicated values of t. Give reasons for your answers.
(a)
To determine
To Calculate: The value of the tangential and normal components of acceleration as, aT and aN at t=1 and t=2 if the position vector is,
r(t)= cosπt+πtsinπt, sinπtπtcosπt.
Explanation
Given:
The provided position vector is, r(t)= 〈cosπt+πtsinπt, sinπt−πtcosπt〉.
Formula Used:
The tangential velocity and tangential acceleration are:
v(t)=dr(t)dt and a(t)=dv(t)dt.
Calculation:
The velocity is calculated as shown below,
Now, find the magnitude of the velocity vector to find speed of the particle as shown below,
‖v(t)‖= π2t …… (1)
Now, find the acceleration vector of the particle as calculated below,
Now, find the tangential component of acceleration as shown below,
Since, cos2Ï€+sin2Ï€t=1
The value is found as shown below,
aT=Ï€2
At t=1 and t=2 the tangent vector is aT=Ï€2
(b)
To determine
Whether the speed of particle is increasing or decreasing at each of the indicated value of t from t=1 to t=2 if the position vector is,
r(t)= cosπt+πtsinπt, sinπtπtcosπt.
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#### For what values of p does the series converge?
Study Guide for Stewart's Multivariable Calculus, 8th | 701 | 2,385 | {"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-2019-51 | latest | en | 0.788815 |
http://www.jiskha.com/display.cgi?id=1270476592 | 1,498,609,986,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321961.50/warc/CC-MAIN-20170627235941-20170628015941-00029.warc.gz | 571,240,967 | 3,841 | # physics
posted by .
A light string has its ends tied to two walls separated by a distance equal to five-eighths the length of the string. A 53 kg mass is suspended from the center of the string, applying a tension in the string.
What is the tension in the two strings of length L/2 tied to the wall? The acceleration of gravity is 9.8 m/s^2.
Answer in units of N.
• physics -
Draw the triangles.
half the weight is supported by each side.
Let theta be the angle from the wall horizontal to the string. Then on each side, SinTheta=Weight/(2*tension)
But tan cosTheta= half wall distance/halfstring distance
costheta= 5/8
But sin^2theta+cos^2theta=1
or weight^2/4Tension^2+25/64=1
solve for tension. | 191 | 708 | {"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-2017-26 | latest | en | 0.876965 |
https://math.stackexchange.com/questions/4571057/differentiating-t-mapsto-p-tq-1-at-for-an-invertible-matrix-p-at | 1,726,494,012,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.32/warc/CC-MAIN-20240916112213-20240916142213-00433.warc.gz | 355,841,515 | 39,882 | # Differentiating $t \mapsto$ $(P+$ $tQ)^{-1}$ at for an invertible matrix $P$ at $t=0$
Recently, I have been reading about differentiating matrix-valued functions. I read that when finding the directional derivative of an inverse matrix, it is easiest to do this with the identity matrix (i.e., differentiate $$AA^{-1}=I_n$$ and use an analogue of the product rule to rearrange for the derivative of the inverse of $$A$$). I have tried this on some examples from other problems that I have attempted, and want to know if my understanding here is correct.
Evaluate and simplify: $$\frac{d}{dt} \left( (P+tQ)^{-1} \right)$$ at $$t=0$$, where the matrix $$P$$ is invertible.
My understanding here would be to consider:
$$(P+tQ)(P+tQ)^{-1} = I_n$$
If we differentiate both sides of this, then the derivative of the identity matrix will be the zero matrix, and on the left hand side we apply the product rule. By rearrangement for the derivative of $$(P+tQ)^{-1}$$, I have got that this derivative is equal to:
$$\frac{-Q(P)^{-1}}{P}$$
Is this approach correct or can this be further simplified (or have I made a mistake up to this point?). I would be grateful for any feedback.
• I advise you to not use $/$ when the denominator is a matrix. Use the $\cdot^{-1}$ appropriately Commented Nov 7, 2022 at 10:48
Let us denote $$(P+tQ)^{-1}$$ by $$f(t).$$ Then we have
$$(P+tQ)f(t) = I_n.$$
Differentiation gives
$$Qf(t)+(P+tQ)f'(t),$$
thus
$$Q(P+tQ)^{-1}+(P+tQ)f'(t),$$
for $$t=0$$ this gives $$QP^{-1}+Pf'(0).$$ Left multiplication with $$P^{-1}$$ yields
$$P^{-1}QP^{-1}=-f'(0).$$
• Why not use $= 0$? Commented Nov 7, 2022 at 10:46
It is correct until " I have got that this derivative is equal to" but $$\frac{-Q(P)^{-1}}P$$ should be written $$-P^{-1}QP^{-1}$$ ($$\frac{\cdot}P$$ is ambiguous since the matrix product is not commutative).
More generally, as you probably know since you "have recently been reading about differentiating matrix valued functions": if $$t\mapsto A(t)$$ is differentiable at $$t_0$$ and if $$A(t_0)$$ is invertible, then$$\frac{d A^{-1}}{dt}(t_0)=-A^{-1}(t_0)\frac{d A}{dt}(t_0)A^{-1}(t_0).$$ Note that the differentiability of $$A^{-1}$$ needs to be proved before having the right to write $$A\cdot A^{-1}=I_n\Rightarrow A'A^{-1}+A(A^{-1})'=0.$$ Posts adressing this are Derivative of matrix inverse from the definition and differential inverse matrix.
• Just differentiate the product (as you did before) and use that now, you know the derivative of $(I_n+tH)^{-1}.$ Commented Nov 7, 2022 at 11:11
• No, I just mean $(CBA)'=C'BA+CBA'$ with $A=I_n+tH$ and $C=A^{-1}$, and then use that you know (from your initial post) that $C'=-A^{-1}A'A^{-1}.$ Commented Nov 7, 2022 at 11:22
Let $${\bf M} (t) := ( {\bf P} + t {\bf Q} )^{-1}$$. Hence,
\begin{aligned} {\bf M} (t + {\rm d} t) &= \left( ( {\bf P} + t {\bf Q} ) + {\rm d}t \, {\bf Q} \right)^{-1} \\ &= \left( ( {\bf P} + t {\bf Q} ) \left( {\bf I} + {\rm d}t \, {\bf M} (t) \, {\bf Q} \right) \right)^{-1} \\ &= \left( {\bf I} + {\rm d}t \, {\bf M} (t) \, {\bf Q} \right)^{-1} {\bf M} (t) \\ &= \left( {\bf I} - {\rm d}t \, {\bf M} (t) \, {\bf Q} \right) {\bf M} (t) = {\bf M} (t) - {\rm d}t \, {\bf M} (t) \, {\bf Q} \, {\bf M} (t)\end{aligned}
and, thus,
$${\dot{\bf M}} (t) = \color{blue}{- {\bf M} (t) \, {\bf Q} \, {\bf M} (t)}$$
Let $$\varepsilon > 0$$ such that $$\varepsilon^2 = 0$$. Hence,
$$( {\bf I} + \varepsilon {\bf A} ) ( {\bf I} - \varepsilon {\bf A} ) = {\bf I} - \varepsilon^2 {\bf A}^2 = {\bf I}$$
and, thus, $$( {\bf I} + \varepsilon {\bf A} )^{-1} = {\bf I} - \varepsilon {\bf A}$$. | 1,270 | 3,606 | {"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": 33, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-38 | latest | en | 0.891643 |
http://math-mate.com/chapter19_3.shtml | 1,496,057,428,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612283.85/warc/CC-MAIN-20170529111205-20170529131205-00438.warc.gz | 289,810,804 | 6,117 | ## Not Probabilities
What is a ‘not’ probability? It’s the chance of a certain event not happening. Say there’s a 30% chance that it’s going to rain tomorrow. What is the probability that it won’t rain tomorrow. Common sense tells us that it is 70%.
Think about an event like it raining tomorrow. It’s either going to rain tomorrow or it’s not going to rain tomorrow. One or the other has to happen. This means that if we add the probabilities of either one happening together, we should get 1, or 100%:
You can apply this logic to any event you want. The probability of the event happening plus the probability of the event not happening adds together to give 1:
You can use this to easily solve problems like this:
What is the probability of your bus not crashing on the way to school tomorrow if the probability of it crashing is 0.0001? Solution To solve this, we just need to use the ‘not’ equation: Now we need to decide what the event is. We have a probability of the bus crashing, so let’s make that the event: We’re interested in the probability of the bus not crashing on the way to school tomorrow, which is P(not E). So we can rearrange the equation to get it in the form “P(not E) = something…”:
### Certainties and impossibilities
A certainty is something that has a 100 % chance of happening – it will happen for sure:
If I have a bag containing 6 red marbles, and I pick one marble out, it is a certainty that I will pick a red marble.
Impossibilities are the opposite of certainties. They are events that have no chance whatsoever of happening – they will never, ever occur:
If I have a bag containing 6 red marbles, then picking out a blue marble is an impossibility – there is no way it could happen. | 415 | 1,817 | {"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-2017-22 | longest | en | 0.894329 |
http://slidegur.com/doc/306803/document | 1,568,551,224,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514571360.41/warc/CC-MAIN-20190915114318-20190915140318-00121.warc.gz | 166,754,158 | 8,081 | ### Document
```Can rectangles with a different area have the
same perimeter as this rectangle?
In this lesson you will learn that
rectangles can have the same
perimeter but different area by
building rectangles and organizing
information in a chart.
Let’s Review
Perimeter tells us the distance that goes around a
polygon.
Perimeter
Let’s Review
Width
Length
A Common Mistake
Mixing up area and perimeter.
Area
Perimeter
Core Lesson
Perimeter = 10
units
Perimeter = 10
units
Core Lesson
Area = 6 square
units
Area = 4 square
units
Core Lesson
Rectangle #1
Rectangle
#1
#2
#3
Rectangle #2
Rectangle #3
Perimeter
12 units
Length
1 unit
Width
5 units
Area
5 square units
12 units
12 units
4 units
2 units
3 units
3 units
8 square units
9 square units
Core Lesson
Rectangle #3
Rectangle #1
Rectangle #4
Rectangle #2
Core Lesson
Rectangle
#1
Perimeter
Length
Width
16 units
6 units
2 units
Area
12 square units
#2
16 units
7 units
#3
16 units
5 units
3 units
15 square units
#4
16 units
4 units
4 units
16 square units
1 unit
7 square units
In this lesson, you have learned that
rectangles can have the same
perimeter but different area by
building rectangles and organizing
information in a chart.
Guided Practice
Can rectangles with a different area have the same
perimeter as this rectangle?
Guided Practice
Rectangle
Perimeter
Length
Width
Area
Guided Practice
Rectangle #2
Rectangle #1
Rectangle
Perimeter
Length
Width
Area
#1
8 units
2 units
2 units
4 square units
#2
8 units
3 units
1 unit
3 square units
Extension Activities
Sally asked her father to build a fence around her backyard.
Her father bought 24 feet of fencing. Use square tiles to
make the rectangles and then make a chart to find the
biggest area for Sally’s backyard.
Extension Activities
Using square tiles, create as many rectangles as you can with
a perimeter of 36 units. Circle the rectangle with the biggest
area. Create a chart to show the different areas for the given
perimeter.
Extension Activities
This rectangle has a perimeter of 18 units. Fill in the
chart to show the possible areas.
Perimeter
18 units
18 units
18 units
18 units
Length
Width
Area
Extension Activities
Fill in this chart. You can use square tiles to help you
visualize the rectangles.
Perimeter
Length
20 units
7 units
20 units
Width
21 square units
1 unit
20 units
20 units
20 units
Area
20 square units
6 units
8 units
Quick Quiz
Here are some rectangles with a perimeter of 18 units.
Circle the rectangles with an area greater than 10 square
units.
Quick Quiz
Here are some rectangles with a perimeter of 12 units.
Circle the rectangles with an area greater than 7 square
units.
``` | 696 | 2,637 | {"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} | 4.09375 | 4 | CC-MAIN-2019-39 | latest | en | 0.842761 |
http://library.thinkquest.org/5882/longitude.htm | 1,386,395,451,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163053608/warc/CC-MAIN-20131204131733-00009-ip-10-33-133-15.ec2.internal.warc.gz | 135,809,311 | 2,298 | Longitude and Latitude
Longitude and Latitude are very important to many different people. The easiest way to understand Longitude and Latitude is to know how to work a coordinate grid first.
So here is how to work a coordinate grid:
The first thing to know is that the point is always zero. Then you number it on the vertical side going up and on the horizontal side going to the left as shown.
The next thing that you should learn is that you go over, then up to find the destination, that will be very important to know. Below is an example of how to use one.
Now we hope you know how to use a coordinate grid!
Now that you know how to work a coordinate grid you can now go on to the longitude and latitude.
The first thing that you want to learn is that longitude goes from the tip of the north pole to the very bottom of the south pole. Latitude is something different. Latitude is parallel or in other words the lines never touch each other.
The center line of longitude is the Prime Meridian, and the center line of the latitude is the equator.
To work longitude and latitude you must know that the Prime Meridian and the equator are always 0 degrees and the other lines count by 15 degrees. The longitude lines go all the way up to 180 degrees east or west. Latitude lines go up to 90 degrees north or south.
To work longitude and latitude you might get the coordinates 60 degrees north and 15 degrees east. First, to get 60 degrees north you go to the latitude and go up to the 60 because you are going north. And to get the 15 degrees east you go to the longitude and go right to the 15 degree line because you are going right to the east. Where the two lines meet is the spot you are looking for. You should have ended up in Sweden!
And that is how to work longitude and latitude! | 393 | 1,806 | {"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-2013-48 | latest | en | 0.905511 |
https://statskey.com/approximate-confidence-interval-estimates-of-u-4314 | 1,638,219,363,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358842.4/warc/CC-MAIN-20211129194957-20211129224957-00068.warc.gz | 613,068,868 | 14,051 | # Approximate Confidence Interval Estimates of U Stats Assignment Homework Help
Approximate Confidence Interval Estimates of U
In the usual situation encountered in practice, we do not know the value of and we are not sure the population is normally distributed. However, we do know (Chapter 9) that for samples of more than 30 (n > 30), the distribution of sample means is nearly normal even though the population is not normal. Second, experimentation has shown t distributions for n > 30 provide good approximations of confidence intervals for skewed (non normal)populations which are uni modal (have one peak). Third, as we saw in Table 10.2, t distributions are nearly normal when n is more than 30. Statisticians have used the foregoing three facts to formulate a rule of thumb, which is as follows: When az is not known, but n > 30, estimate az by sz, but use the.
of sulfur oxides is between 7.2 and 11.8 tons per day.
In the example, if we had known U’z, and it was 3.24 tons per day, we
would have used ZO.025 = 1.96 rather than t = 2.26 in construction of the interval.
The result would have been the more precise (narrower) interval
In general, interval estimates of J.L are more precise when we know U’z than.
when we do not know U’z and have to estimate it from a sample.
EXERCISE See the last example. Suppose that for another random sample,
this time with n = 16 days, the sample mean emission is 10.3 tons per day and
the sample standard deviation is 3 tons per day. Construct a 90 percent confidence
interval estimate of u:
ANSWER 9.0 ~ p: ~ 11.6 tons per day.
Posted on August 27, 2014 in STATISTICAL INFERENCE ESTIMATION | 417 | 1,646 | {"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-2021-49 | latest | en | 0.919179 |
https://tasks.illustrativemathematics.org/content-standards/5/OA/B/3/tasks/1895 | 1,675,664,418,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500304.90/warc/CC-MAIN-20230206051215-20230206081215-00272.warc.gz | 581,182,751 | 10,561 | Sidewalk Patterns
Alignments to Content Standards: 5.OA.B.3
Task
Cora and Cecilia each use chalk to make their own number patterns on the sidewalk. They make each of their patterns 10 boxes long and line their patterns up so they are next to each other.
Cora puts 0 in her first box and decides that she will add 3 every time to get the next number.
Cecilia puts 0 in her first box and decides that she will add 9 every time to get the next number.
1. Complete each girl's sidewalk pattern.
2. How many times greater is Cecilia’s number in the 5th box be than Cora’s number in the 5th box? What about the numbers in the 8th box? The 10th box?
3. What pattern do you notice in your answers for part b? Why do you think that pattern exists?
4. If Cora and Cecilia kept their sidewalk patterns going, what number will be in Cora's box when Cecilia's corresponding box shows 153?
IM Commentary
This purpose of this task is to help students articulate mathematical descriptions of number patterns. Though Cora and Cecilia's patterns are separate, we notice an interesting relationship emerge between corresponding terms in their patterns. For any given term, we will notice that Cecilia's number will always be three times Cora's number. Students may also notice that if we kept both patterns going indefinitely, Cecilia's numbers would always show up in Cora's path, though only some of Cora's numbers will show up in Cecilia's path.
The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable during exploration of a task depend on how instruction unfolds in the classroom. While it is possible that tasks may be connected to several practices, only one practice connection will be discussed in depth. Possible secondary practice connections may be discussed but not in the same degree of detail.
This task helps illustrate Mathematical Practice 8, Look for and express regularity in repeated reasoning. During this activity, fifth graders are actively comparing Celia’s and Cora’s sidewalk patterns and noticing any connections between the two. They are asked to directly compare numbers generated from a repeated pattern in certain boxes in order to draw conclusions and generalize a pattern between the sets of numbers. This explanation of “Why?” draws upon MP.3, Construct viable arguments and critique the reasoning of others. Through this in-depth exploration of these two sidewalk patterns, students are immersed in pattern-recognition and pattern-generalizing which not only develops a better understanding of how numbers behave in our base ten system, but pushes students to generalize by reasoning repeatedly. Students will notice that Celia’s numbers will always appear in Cora’s numbers but the opposite isn’t true. Then they will observe that Cecilia’s terms are always three times Cora’s corresponding term. A full explanation of this thinking is detailed in the solution set. While looking closely at the repeated pattern and making generalizations students are also engaged in examining the structure of the number sets and making use of that structure in their comparisons (MP.7).
Solutions
Solution: 1
a.
b.
We can see that at the 5th, 8th and 10th boxes, Cecilia's number is always 3 times Cora's number.
Cecilia's number will always be three times Cora's number, no matter which pair of corresponding boxes the girls stand on. Cora's box shows 0 and the multiples of 3. Cecilia's box shows 0 and the multiples of 9.
c.
The image above shows us the 3 as a factor in each of Cora's terms in her path. It also shows the 9 as a factor in each of Cecilia's terms in her path, though 9 is being represented as 3 x 3. This allows us to see that Cora's factor of 3 is contained in each of Cecilia's terms, as well as one extra factor of 3. This is why Cecilia's term will always be three times Cora's corresponding term in her path.
d.
If Cecilia's box shows 153, then Cora's corresponding box must have 51. We know that Cecilia's box will always show three times as much as Cora's corresponding box, so Cora's box must show 51.
Solution: | 973 | 4,386 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2023-06 | longest | en | 0.930731 |
https://thebrilliantcosmos.wordpress.com/2016/06/29/the-celestial-sphere/ | 1,532,296,966,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676594018.55/warc/CC-MAIN-20180722213610-20180722233610-00151.warc.gz | 774,030,012 | 21,511 | # The Celestial Sphere
Humans perceive in Euclidean space -> straight lines and planes. But, when distances are not visible (i.e. very large) than the apparent shape that the mind draws is a sphere -> thus, we use a spherical coordinate system for mapping the sky with the additional advantage that we can project Earth reference points (i.e. North Pole, South Pole, equator) onto the sky. Note: the sky is not really a sphere!
From the Earth’s surface we envision a hemisphere and mark the compass points on the horizon. The circle that passes through the south point, north point and the point directly over head (zenith) is called the meridian.
This system allows one to indicate any position in the sky by two reference points, the time from the meridian and the angle from the horizon. Of course, since the Earth rotates, your coordinates will change after a few minutes.
The horizontal coordinate system (commonly referred to as the alt-az system) is the simplest coordinate system as it is based on the observer’s horizon. The celestial hemisphere viewed by an observer on the Earth is shown in the figure below. The great circle through the zenith Z and the north celestial pole P cuts the horizon NESYW at the north point (N) and the south point (S). The great circle WZE at right angles to the great circle NPZS cuts the horizon at the west point and the east point (E). The arcs ZN, ZW, ZY, etc, are known as verticals.
The two numbers which specify the position of a star, X, in this system are the azimuth, A, and the altitude, a. The altitude of X is the angle measured along the vertical circle through X from the horizon at Y to X. It is measured in degrees. An often-used alternative to altitude is the zenith distance, z, of X, indicated by ZX. Clearly, z = 90 – a. Azimuth may be defined in a number of ways. For the purposes of this course, azimuth will be defined as the angle between the vertical through the north point and the vertical through the star at X, measured eastwards from the north point along the horizon from 0 to 360°. This definition applies to observers in both the northern and the southern hemispheres.
It is often useful to know how high a star is above the horizon and in what direction it can be found – this is the main advantage of the alt-az system. The main disadvantage of the alt-az system is that it is a local coordinate system – i.e. two observers at different points on the Earth’s surface will measure different altitudes and azimuths for the same star at the same time. In addition, an observer will find that the star’s alt-az coordinates changes with time as the celestial sphere appears to rotate.
Celestial Sphere:
To determine the positions of stars and planets on the sky in an absolute sense, we project the Earth’s spherical surface onto the sky, called the celestial sphere.
The celestial sphere has a north and south celestial pole as well as a celestial equator which are projected reference points to the same positions on the Earth surface. Right Ascension and Declination serve as an absolute coordinate system fixed on the sky, rather than a relative system like the zenith/horizon system. Right Ascension is the equivalent of longitude, only measured in hours, minutes and seconds (since the Earth rotates in the same units). Declination is the equivalent of latitude measured in degrees from the celestial equator (0 to 90). Any point of the celestial (i.e. the position of a star or planet) can be referenced with a unique Right Ascension and Declination.
The celestial sphere has a north and south celestial pole as well as a celestial equator which are projected from reference points from the Earth surface. Since the Earth turns on its axis once every 24 hours, the stars trace arcs through the sky parallel to the celestial equator. The appearance of this motion will vary depending on where you are located on the Earth’s surface.
Note that the daily rotation of the Earth causes each star and planet to make a daily circular path around the north celestial pole referred to as the diurnal motion. | 864 | 4,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-30 | latest | en | 0.91675 |
http://math.stackexchange.com/questions/107035/how-to-calculate-integral-of-int-0-sqrt34-sqrt-fracx4-x3-2-mathr/107040 | 1,469,478,415,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824345.69/warc/CC-MAIN-20160723071024-00154-ip-10-185-27-174.ec2.internal.warc.gz | 150,816,723 | 19,186 | # How to calculate integral of $\int_0^\sqrt[3]4\!\sqrt\frac{x}{4-x^{3/2}}\,\mathrm{d}x$
Given the following integral:
$$\int_0^\sqrt[3]4\!\sqrt\frac{x}{4-x^{3/2}}\,\mathrm{d}x$$
How to solve it? I thought it may be possible to substitute it, but I didn't find anything to substitute. I tried to solve it with Maple, but the CAS didn't get it therefore I don't know how to carry on with this. Can you give me some hints?
-
First let $x=u^2$, hopefully after that you will see a final substitution that finishes it off. – Ragib Zaman Feb 8 '12 at 12:07
It seems like setting $x=u^{2/3}$ would be more effective. Then $\sqrt{x}dx = u^{1/3} \frac{2}{3} u^{-1/3} du = \frac{2}{3} du$, so you are now trying to solve: $$\frac{2}{3}\int_0^2{\frac{du}{\sqrt{4-u}}}$$ – Thomas Andrews Feb 8 '12 at 14:02
$$\int\sqrt{\frac{x}{4-x^{3/2}}}\,\mathrm dx = -\frac{4\sqrt{x}}{3\sqrt{-\frac{x}{x^{3/2}-4}}}+\mathrm{constant}$$
where you can find the integration steps here by clicking on the button 'Show steps' next to the result.
In your particular case $x\geq 0$ over the whole integration domain such that we may simplify to
$$-\frac{4}{3}\sqrt{4-x^{3/2}}$$
evaluating at $x=0$ and $x=4^{1/3}$ gives the result
$$\int_0^{4^{1/3}}\sqrt{\frac{x}{4-x^{3/2}}}\,\mathrm dx = \frac{4}{3} (2-\sqrt{2})$$
-
What happened to the minus sign in the indefinite integral? Shouldn't that simplify to $-\frac 4 3 \sqrt{4-x^{3/2}}$? – Thomas Andrews Feb 8 '12 at 13:57
Yes, corrected it. – Till Hoffmann Feb 8 '12 at 14:43
I think it is worth mentioning the case of the integration of the differential binomials.
The expression of the form
$$x^m(a+bx^n)^pdx$$ where $m,n,p,a,b$ are constant is called a differential binomial.
THEOREM. (Piskunov)
The integral
$$\int x^m(a+bx^n)^pdx$$
can be reduced if $m,n,p$ are rational numbers, to the integral of a rational function, and can thus be expressed in terms of elementary functions if:
$1.$ $p$ is an integer.
$2.$ $\dfrac{m+1}{n}$ is an integer.
$3.$ $\dfrac{m+1}{n}+p$ is an integer.
PROOF
We transform the integral writing $x^n = z$ so $dx = \frac 1 n z^{\frac 1 n -1}$. Then:
$$\int {{x^m}} {(a + b{x^n})^p}dx = \int {{z^{{{m + 1} \over n} - 1}}} {(a + bz)^p}dz = \int {{z^q}} {(a + bz)^p}dz$$
$1.$ Let $p$ be an integer. Being $q$ a rational number, let it be $\dfrac r s$. This integral then takes the form $$\int {R\left( {{z^{q/s}},z} \right)dz}$$
which can be reduced by substituting $z=t^s$.
$2.$ If $\dfrac{m+1}{n}$ is an integer. then $q=\dfrac{m+1}{n}-1$ is an integer. $p$ is rational $=\dfrac \lambda \mu$. The integral is reduced to $$\int {R\left( {{z^q},{{\left( {a + bz} \right)}^{{\lambda \over \mu }}}} \right)dz}$$ which can be reduced substituting $a+bz=t^\mu$
$3.$ If $\dfrac{m+1}{n}+p$ is an integer then $\dfrac{m+1}{n}+p-1=q+p$ is an integer. We tranform the integral into
$$\int {{z^{q + p}}{{\left( {{{a + bz} \over z}} \right)}^p}dz}$$
where $q+p$ is an integer and $p=\dfrac \lambda \mu$ is rational. The integral is then
$$\int {R\left[ {z,{{\left( {{{a + bz} \over z}} \right)}^{{\lambda \over \mu }}}} \right]dz}$$
which can be reduced using
$${{a + bz} \over z} = {t^\mu }$$
Note. P.L. Chebyshev, a russian mathematician, proved the integrals just analysed can't be expressed in terms of elementary functions if it isn't the case $1$ , $2$ or $3$.
- | 1,207 | 3,340 | {"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} | 4.3125 | 4 | CC-MAIN-2016-30 | latest | en | 0.861546 |
https://www.esaral.com/q/evaluate-the-following-integrals-72938 | 1,722,714,658,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640377613.6/warc/CC-MAIN-20240803183820-20240803213820-00526.warc.gz | 622,251,393 | 11,370 | # Evaluate the following integrals:
Question:
Evaluate the following integrals:
$\int 5^{x+\tan ^{-1} x}\left(\frac{x^{2}+2}{x^{2}+1}\right) d x$
Solution:
Assume $x+\tan ^{-1} x=t$]
$d\left(x+\tan ^{-1} x\right)=d t$
$\Rightarrow 1+\frac{1}{x^{2}+1}=d t$
$\Rightarrow \frac{2+\mathrm{x}^{2}}{\mathrm{x}^{2}+1}=\mathrm{dt}$
Substituting $\mathrm{t}$ and $\mathrm{dt}$
$\Rightarrow \int 5^{t} d t$
$\Rightarrow \frac{5^{t}}{\log 5}+c$
But $\mathrm{t}=\mathrm{x}+\tan ^{-1} \mathrm{x}$
$\Rightarrow \frac{5^{x+\tan ^{-1} x}}{\log 5}+c$ | 222 | 546 | {"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-2024-33 | latest | en | 0.183966 |
https://justaaa.com/statistics-and-probability/528369-an-opinion-poll-based-on-a-sample-of-50-subjects | 1,709,124,554,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474715.58/warc/CC-MAIN-20240228112121-20240228142121-00044.warc.gz | 335,771,720 | 9,894 | Question
An opinion poll based on a sample of 50 subjects estimated p, the proportion of the...
An opinion poll based on a sample of 50 subjects estimated p, the proportion of the population in favor of the proposition, as 0.72.
(i) Estimate the true proportion, θ, with a 95% confidence interval. State any assumptions you may have to make in answering this question.
(ii) If the true population proportion is suspected to be θ =0.8, and the estimate from an opinion poll is to be determined to within ±0.05 with 95% confidence, how
many people, n, should be sampled?
(iii) If the proportion is to be estimated to within the same margin of ±0.05, but with 90% confidence, what is the value of n required? Comment on the effect that
reducing the confidence level has on the sample size, n required to achieve the desired precision.
Given: Sample size = n = 50, sample proportion = p = 0.72
(i) The random variable has a binomial distribution with probability of success as P. As the sample size is sufficiently large, the normal approximation to binomial distribution can be used.
Thus, the required 95% confidence interval is { p - 1.96*sqrt(p*(1-p)/n) , p +1.96*sqrt(p*(1-p)/n) }
i.e. {0.72 - 1.96*sqrt(0.72*0.28/50), 0.72 + 1.96*sqrt(0.72*0.28/50)}
i.e.(0.72-0.1245 , 0.72+0.1245), i.e. (0.5955,0.8445)
(ii) According to the given condition,
1.96*sqrt(p*(1-p)/n) <= 0.05
i.e. 1.96*sqrt(0.8*0.2/n) <= 0.05
i.e. n >= 0.16*(1.96/0.05)^2
i.e. n >= 245.8624
Thus the sample should be of size at least 246.
(iii) 1.64*sqrt(p*(1-p)/n) <= 0.05
i.e. n >= 0.16*(1.64/0.05)^2
i.e. n >= 172.1344
Thus the sample should be of size at least 173.
Thus, for lower confidence, a smaller sample size suffices.
Hope this helps!
Earn Coins
Coins can be redeemed for fabulous gifts. | 584 | 1,782 | {"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-2024-10 | latest | en | 0.853764 |
https://hamnus.com/category/courses/computer-engineering/physics-for-engineers/ | 1,695,879,575,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510358.68/warc/CC-MAIN-20230928031105-20230928061105-00799.warc.gz | 325,398,091 | 23,554 | Introduction to Waves | Physics for Engineers Lesson
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Kinematics is a branch of classical mechanics that deals with the study of motion of objects without considering the forces that cause or result from | 412 | 2,091 | {"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-2023-40 | longest | en | 0.904408 |
https://www.studypug.com/ca/grade5/arithmetic-properties-associative-property | 1,675,192,194,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499890.39/warc/CC-MAIN-20230131190543-20230131220543-00708.warc.gz | 1,015,183,066 | 45,456 | # Arithmetic properties: Associative property
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0/6
##### Intros
###### Lessons
1. Introduction to the associative property of addition and multiplication:
2. Showing that $(a + b) + c = a + (b + c)$
3. Why is it called the "associative" property?
4. Addition shortcuts using the associative property
5. Showing that (a × b) × c = a × (b × c)
6. Multiplication shortcuts using the associative property
7. The general formulas for the associative property
0/12
##### Examples
###### Lessons
1. Associative Property Equations
Use the associative property (for addition and multiplication) to fill in the blanks.
1. (7 + 2) + 8 = 7 + (2 + __)
2. $\large \frac{1}{4} \, \times \, (\frac{1}{6} \, \times \, \frac{1}{5}) = (\frac{1}{4} \, \times \,$ ______ $\large ) \, \times \, \frac{1}{5}$
3. ($x$ + 3 + 8) + __ = (3 + 9 + x) + 8
2. Changing the grouping to add lists of numbers
1. 0.1 + 0.6 + 0.9 + 0.4 =
2. $\large \frac{2}{10} + \frac{3}{10} + \frac{1}{10} + \frac{4}{10} + \frac{9}{10} + \frac{7}{10} + \frac{6}{10} + \frac{8}{10} =$
3. $\large \frac{2}{3} + \frac{5}{9} + \frac{1}{3} + \frac{4}{9} + \frac{6}{7} =$
3. Changing the grouping to multiply lists of numbers
Decide how to group the factors as a shortcut for multiplication. Double check your answer by multiplying without groups.
1. 20 × 15 × 5
2. 0.8 × 0.9 × 0.5
3. 6 × 20 × 5 × 2
4. Associative property of addition word problem
Ryan added these numbers together and his answer is correct. Show another way of adding numbers (with grouping) using the associative property!
1. Associative property of multiplication word problem
Explain which choice is NOT an equal statement to: (6 × 8) × 5
1. 6 × 40
2. 48 × 5
3. 6 × (8 × 5)
4. 14 × 5
1. Associative property and volume
The formula for the volume of a rectangular prism is given by:
Volume = length × width × height
Use the associative property of multiplication to show 2 ways to solve for this prism.
0%
##### Practice
###### Topic Notes
In this lesson, we will learn:
• What is the commutative property of addition?
• What is the commutative property of multiplication?
• How to write the general formulas/equations for the commutative properties
• Changing the order of a list of addends/factors does not change the answer
• How to solve word problems for the commutative property
Notes:
• The associative property means that changing the grouping of numbers in an equation does NOT change the answer when you are performing ONLY addition or ONLY multiplication
• The numbers can be any real number (whole numbers, fractions, decimals, integers, etc.)
• To “associate” can mean to interact with a group of people/friends or to group together.
• No matter how you want to group (using brackets) the numbers in an addition or multiplication equation, it will not change the answer in the end.
• Ex. (1 + 2) + 3 = 1 + (2 + 3) will equal 6 either way
• Because (1 + 2) + 3 = (3) + 3 = 6
• As well, 1 + (2 + 3) = 1 + (5) = 6
• The associative property for addition can make shortcuts for adding whole numbers and decimals by making sums of 10 (i.e. 1 + 9, 2 + 8, 3 + 7, 4 + 6, and 5 + 5)
• Ex. 8 + 6 + 2 + 4 + 5 + $x$
• Group as: (8 + 2) + (6 + 4) + 5 + x = (10) + (10) + 5 + $x$ = 25$x$
• Ex. 0.9 + 0.7 + 0.3 + 0.1
• Group as: (0.9 + 0.1) + (0.7 + 0.3) = (1.0) + (1.0) = 2.0
• Shortcuts for adding fractions is also possible with the associative property by making wholes (i.e. same numerator and denominator; $\large \frac{4}{4}, \frac{2}{2},\frac{10}{10}$)
• Ex. $\large \frac{3}{4} + \frac{2}{4} + \frac{1}{4}$
• Group as: $\large (\frac{3}{4} + \frac{1}{4}) + \frac{2} {4} = \frac{4} {4} + \frac{2} {4} = 1 + \frac{2} {4} = 1 \frac{2}{4}$
• Ex. $\large \frac{2}{9} + \frac{2}{5} + \frac{7}{9} + \frac{3}{5} + \frac{1}{4}$
• Group as: $\large (\frac{2}{9} + \frac{7}{9}) + (\frac{2} {5} + \frac{3} {5}) + \frac{1} {4} = (\frac{9} {9}) + (\frac{5}{5}) + \frac{1}{4} = 1 + 1 + \frac{1}{4} = 2 \frac{1}{4}$
• For multiplication: the grouping of factors does not change the answer
• Ex. (2 × 3) × 4 = 2 × (3 × 4) will equal 24 either way
• Because (2 × 3) × 4 = (6) × 4 = 24
• As well, 2 × (3 × 4) = 2 × (12) = 24
• The associative property for multiplication can make shortcuts for multiplying any real numbers by making multiples of 10 (i.e. 10, 20, 30, 40…)
• Ex. 2 × 8 × 5 × $e$
• Group as: (2 × 5) × 8 × $e$ = (10) × 8 × $e$ = 80 × $e$ = 80$e$
• Ex. 0.9 × 0.5 × 0.6
• Group as: (0.5 × 0.6) × 0.9 = (0.30) × 0.9 = 0.270
• Ex. $\large \frac{5}{2}$ × $\frac{9}{13}$ × $\frac{4}{50}$
• Group as: $\large \frac{5 \, x \, 9 \, x \, 4}{2 \, x \, 13 \, x \, 50}$ = $\large \frac{(5 \, x \, 4) \, x \, 9}{(2 \, x \, 50) \, x \, 13}$ = $\large \frac{(20) \, x \, 9 }{(100) \, x \, 13 } = \frac{180}{1300}$
• The general formulas (where $a$, $b$ and $c$ are variables that represent real numbers) for the associative property are:
Arithmetic Property Of Addition Of Multiplication Associative Property $(a + b) + c = a + (b + c)$ $(a × b) × c = a × (b × c)$ | 1,879 | 5,230 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 30, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2023-06 | latest | en | 0.770472 |
www.brianteaches.com | 1,576,221,471,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540551267.14/warc/CC-MAIN-20191213071155-20191213095155-00446.warc.gz | 168,548,112 | 31,158 | ;nbsp
Hi parents and students! This is an introduction to what you’ll be needing to know by the end of the first Calculus semester. My courses cover all of the common core requirements (the syllabus for semester 1 is here, and semester 2 is here!) and prepare my students to think critically. Not only do I want your student to know the material in my class, I want them to succeed in all future classes and opportunities. Please try out this sample test to see what your child will be learning this semester!
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1. Determine the value of this limit.
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5. Find the instantaneous rate of change of f(x) = 2x² + 5x at x = 3 using the limit definition of the instantaneous derivative.
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6a. Using the formal definition of the derivative, find the slope of the tangent line to the curve given by f(x) = 1/(x – 6).
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6b. Find the slope of the normal line to the curve given by f(x) = 1/(x – 6).
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6c. Find the slope of the normal line to the curve given by f(x) = 1/(x – 6) at x = 0.
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7. Which of the following is (are) correct representation(s) of the derivative of f(x) evaluated at a?
All correct answers must be selected. No partial credit.
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10a. Find the slope of the tangent line to the curve given by f(x) = (x³ + 7)(√2x)³
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11c. Using the functions f(x) and g(x) from the adjacent graph, find p ’(2) where p(x) = 3f(x)g(x).
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14a. Find dy/dx when y = -4cos(11x).
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## An object is moving rectilinearly in time with velocity given by v(t) = 6t – t² with units m/s. The position at t = 0 is 5 meters.
15a. Find the position function of time.
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16. Differentiate implicity to find y’ where 4x³y = 2x + sin(y)
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## Assume f(0) = 2, f'(0) = -1, f”(0) = 1, and g(x) = sec(3x)f(x). What is the equation of the line tangent to g(x) at x = 0.
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## For each x value below, state if it is a critical value or not.
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21g. x = 7
This is a required question
22. Which of the points in the graph of f(x) shown above have both a negative first derivative and also a negative second derivative value?
This is a required question
23. A rectangular beam is cut out of a circular log of radius R. What is the maximum area of the log?
This is a required question
24. What is the derivative of f(x) = ³√(4^(3x) + log₄x) ?
This is a required question
25. Find f ’(x) where f(x) = arcsin(2x³/5).
This is a required question
26. Find the value of the indefinite integral above.
This is a required question
27. What is f(x) when f ’(x) = 1/cos²(x) and f(2π/3 radians) = 2?
This is a required question
## Find f(x) when f ’’(x) = x + sin(5x) , f ’(π/6 radians) = 0, and f(0 radians) = 1.
28a. Find f ‘ (x) by taking the indefinite integral of f’’(x)
This is a required question
28b. Find C₁
This is a required question
28c. Find f(x) by taking the indefinite integral of f'(x)
This is a required question
28d. Find C₂ and indentify the final formula for f(x).
This is a required question
This is a required question | 1,712 | 5,950 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-51 | latest | en | 0.918411 |
https://www.cmtoinchesconvert.com/convert/power/dBm_to_mW.html | 1,708,743,637,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474482.98/warc/CC-MAIN-20240224012912-20240224042912-00594.warc.gz | 725,917,055 | 7,290 | # dBm to mW Conversion
Decibel-milliwatts (dBm) to milliwatts (mW), power conversion calculator and how to convert.
### dBm to mW conversion calculator
Decibel-milliwatts (dBm) to milliwatts (mW), power conversion calculator.
Enter the power in dBm and press the Convert button:
Enter power in dBm: dBm Milliwatts result: mW
mW to dBm conversion calculator ►
## How to convert dBm to mW
So The power conversion of dBm to mW is given by the formula.
P(mW) = 1mW ⋅ 10(P(dBm)/ 10)
So
1dBm = 1.258925mW
#### Example 1
Convert 12dBm to milliwatts:
P(mW) = 1mW ⋅ 10(13dBm/ 10) = 15.84mW
#### Example 2
Convert 14dBm to milliwatts:
P(mW) = 1mW ⋅ 10(13dBm/ 10) = 25.11mW
#### Example 3
Convert 16dBm to milliwatts:
P(mW) = 1mW ⋅ 10(13dBm/ 10) = 39.81mW
#### Example 4
Convert 18dBm to milliwatts:
P(mW) = 1mW ⋅ 10(13dBm/ 10) = 63.09mW
## dBm to mW conversion table
Power (dBm) Power (mW)
-40 dBm 0.0001 mW
-30 dBm 0.001 mW
-20 dBm 0.01 mW
-10 dBm 0.1 mW
0 dBm 1 mW
1 dBm 1.2589 mW
2 dBm 1.5849 mW
3 dBm 1.9953 mW
4 dBm 2.5119 mW
5 dBm 3.1628 mW
6 dBm 3.9811 mW
7 dBm 5.0119 mW
8 dBm 6.3096 mW
9 dBm 7.9433 mW
10 dBm 10 mW
20 dBm 100 mW
30 dBm 1000 mW
40 dBm 10000 mW
50 dBm 100000 mW
mW to dBm conversion ►
## Features of dBm to mW Calculator
dBm to mW Calculator offered by cmtoinchesconvert.com is a free online utility that allows users to converter dBm to mW without any manual efforts. Some of the key features of this dBm to mW Calculator are listed below:
### 100% Free
You don't need to go through any registration process to use this dBm to mW Calculator. You can use this utility for free and do unlimited dBm to mW conversions without any limitations.
#### Easily accessible
You don't need to install any software on your device to access dBm to mW Calculator. You can access and use this online service with any web browser with a stable internet connection.
#### User-friendly interface
dBm to mW Calculator is easy to use interface. use that enable users to convert dBm to mW online in seconds. You don't need to acquire any special skills or follow complicated procedures, to use this dBm to mW Calculator.
#### Fast conversion
This dBm to mW Calculator offers users the fastest conversion. Once the user enters the dBm to mW values in the input field and clicks the Convert button, the utility will start the conversion process and return the results immediately.
##### Accurate Results
The results generated by this dBm to mW Calculator are 100% accurate. The advanced algorithms used by this utility provided users with error-free results. If you ensure the authenticity of the results provided by this utility, you can use any method to verify them.
###### Compatibility
dBm to mW Calculator is compatible with all types of devices. Whether you are using a smartphone, tablet, desktop, laptop or Mac, you can easily use this dBm to mW Calculator. | 958 | 2,902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-10 | latest | en | 0.660431 |
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# CBSE Class 10 Mathematics Previous Year Question Paper 2020 Delhi Set - 3
Does the sound of Class 10 Maths intrigue you? Or, does it scare you away? All of these happen because it's not easy to handle the Board pressure for CBSE Class 10. But, we will surely help you to tackle the syllabus for Class 10 Maths by providing you with the Class 10 Revision NotesClass 10 Textbook SolutionsClass 10 Tests for all of the chapters available in CBSE Class 10 Maths. We know that scoring more marks in CBSE Class 10 Maths has never been this easy before. But, by referring to all of the study materials we provide at TopperLearning, you can easily score more marks in the Class 10 board examination.
The study materials are created by our subject experts that we offer for CBSE Class 10, very well know the syllabus and essential facts of CBSE Class 10. These study materials will help you understand all the CBSE Class 10 Maths concepts as we focus on providing solutions that simplify a subject's complex fundamentals. At TopperLearning, we believe in delivering quality solutions at a low cost, and we strictly follow the latest CBSE Class 10 syllabus. We make sure that these study materials are revised from time to time. Our TopperLearning packages involve all the study resources for CBSE Class 10, such as Solved question papersvideo lessons and revision notes to help you score high marks. We also provide you with the updated NCERT textbook Solutions and RD Sharma textbook solutions, which provide students with step-by-step explanations.
Our study materials have also introduced the Case Study Based Questions for CBSE Class 10 of all the chapters available in Class 10. These questions are designed based on the latest syllabus of CBSE Class 10.
So why wait when you can quickly get the CBSE class 10 plans.
Question numbers 1 to 10 are multiple choice questions of 1 mark each.
Select the correct option.
Q 1. The point P on x-axis equidistant from the points A(–1, 0) and B(5, 0) is
A. (2, 0)
B. (0, 2)
C. (3, 0)
D. (2, 2)
Q 2. The co-ordinates of the point which is reflection of point (–3, 5) in x-axis are
A. (3, 5)
B. (3, -5)
C. (-3, -5)
D. (-3, 5)
Q 3. If the point P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3:1, then the value of y is
A. 4
B. 3
C. 2
D. 1
Q 4. The sum of exponents of prime factors in the prime-factorisation of 196 is
A. 3
B. 4
C. 5
D. 2
Q 5. Euclid’s division Lemma states that for two positive integers a and b, there exists unique integer q and r satisfying a = bq + r, and
A. 0 < r < b
B. 0 < r < b
C. 0 ≤ r < b
D. 0 ≤ r ≤ b
Q 6. The zeroes of the polynomial x2 – 3x – m(m+3) are
A. m, m+3
B. –m, m+3
C. m, –(m+3)
D. –m, –(m+3)
Q 7. The value of k for which the system of linear equations x + 2y = 3, 5x + ky + 7 = 0 is inconsistent is
A.
B.
C. 5
D. 10
Q 8. The roots of the quadratic equation x2 – 0.04 = 0 are
A. ±0.2
B. ±0.02
C. 0.4
D. 2
Q 9. The common difference of the A.P.
A. 1
B.
C. -1
D. 2
Q 10. The nth term of the A.P. a, 3a, 5a, …. Is
A. na
B. (2n – 1)a
C. (2n + 1)a
D. 2na
In Q. Nos. 11 to 15, fill in the blanks. Each question carries 1 mark:
Q 11. In fig. 1, the angles of depressions from the observing positions O1 and O2 respectively of the object A are ______, ______.
Q 12. In ΔABC, AB=6 cm, AC = 12 cm and BC = 6 cm, then ∠B=______.
OR
Two triangles are similar if their corresponding sides are ______.
Q 13. In given Fig. 2, the length PB=______cm.
Q 14. In fig. 3, MN || BC and AM : MB =1 : 2, then
Q 15. The value of sin 32° cos 58° +cos 32° sin 58° is ______.
OR
The value of is ______.
Q Nos. 16 to 20 are short answer type questions of 1 mark each.
Q 16. A die is thrown once. What is the probability of getting a prime number?
Q 17. If a number x is chosen at random from the numbers –3, –2, –1, 0, 1, 2, 3, then find the probability of x< 4.
OR
What is the probability that a randomly taken leap year has 52 Sundays?
Q 18. If sin A + sin2 A = 1, then find the value of the expression (cos2 A + cos4 A).
Q 19. Find the area of the sector of a circle of radius 6 cm whose central angle is 30°. (Take π =3.14).
Q 20. Find the class marks of the classes 20 – 50 and 35 – 60.
Q.Nos.21 to 26 carry 2 marks each.
Q 21. A teacher asked 10 of his students to write a polynomial in one variable on a paper and then to handover the paper.
The following were the answers given by the students:
2x + 3, 3x2 + 7x + 2, 4x3 + 3x2 + 2, x3 + 7, 7x + , 5x3 – 7x + 2, 2x2 + 3 – , 5x – , ax+ bx2 + cx + d, x + .
Answer the following questions:
1. How many of the above ten, are not polynomials?
2. How many of the above ten, are quadratic polynomials?
Q 22 .A child has a die whose six faces show the letters as shown below:
The die is thrown once. What is the probability of getting (i) A, (ii) D?
Q 23. In fig. 4, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that .
OR
In fig.5, if AD ⊥ BC, then prove that AB2 + CD2 = BD2 + AC2.
Q 24. Prove that
OR
Show that tan4θ + tan2θ = sec4θ - sec2θ
Q 25. Find the mode of the following frequency distribution:
Q 26. From a solid right circular cylinder of height 14 cm and base radius 6 cm, a right circular cone of same height and same base radius is removed. Find the volume of the remaining solid.
Q. Nos. 27 to 34 carry 3 marks each.
Q 27. If a circle touches the side BC of a triangle ABC at P and extended sides AB and AC at Q and R, respectively, prove that .
Q 28. The area of a circular play ground is 22176 cm2. Find the cost of fencing this ground at the rate of Rs. 50 per metre.
Q 29. If the mid – point of the line segment joining the points A (3, 4) and B (k, 6) is P(x, y) and x + y – 10 = 0, find the value of k.
OR
Find the area of triangle ABC with A (1, –4) and the mid-points of sides through A being (2, –1) and (0, –1).
Q 30. If Fig.6, if ΔABC ~ ΔDEF and their sides of lengths (in cm) are marked along them, then find the lengths of sides of each triangle.
Q 31. If 2x + y = 23 and 4x – y = 19, find the value of (5y – 2x) and .
OR
Solve for
Q 32. Which term of A.P. 20, 19, 18, 17,.... is the first negative term.
OR
Find the middle term of the A.P. 7, 13, 19, …., 247.
Q 33. Water in a canal, 6m wide and 1.5m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8cm standing water is required?
Q 34. Show that:
Q. Nos. 35 to 40 carry 4 marks each.
Q 35. The mean of the following frequency distribution is 18. The frequency f in the class interval 19 – 21 is missing. Determine f.
OR
The following table gives production yield per hectare of wheat of 100 farms of a village:
Change the distribution to a ‘more than’ type distribution and draw its ogive.
Q 36. From a point on the ground, the angles of elevation of the bottom and the top of a tower fixed at the top of a 20m high building are 45° and 60° respectively. Find the height of the tower.
Q 38. Prove that is an irrational number.
Q 39. Draw a circle of radius 3.5cm. From a point P, 6cm from its centre, draw two tangents to the circle.
OR
Construct a ΔABC with AB=6cm, BC=5cm and ∠B=60°. Now construct another triangle whose sides are times the corresponding sides of ΔABC.
Q 40. A solid is in the shape of hemisphere surmounted by a cone. If the radius of hemisphere and base radius of cone is 7cm and height of cone is 3.5cm, find the volume of the solid. (Take π = ). | 2,431 | 7,572 | {"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 | 4 | CC-MAIN-2023-23 | latest | en | 0.832786 |
https://www.ilmkidunya.com/online-test/ics-part-1-physics-chapter-8 | 1,709,078,574,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474688.78/warc/CC-MAIN-20240227220707-20240228010707-00391.warc.gz | 816,084,682 | 67,190 | ×
8th Chapter
### 11th Class Physics Chapter 8 Test
Here you can prepare 11th Class Physics Chapter 8 Waves Test. Click the button for 100% free full practice test.
## First Year Physics Chapter 8 Online MCQ Test for 1st Year Physics Chapter 8 (Waves)
This online test contains MCQs about following topics:
- Progressive Waves - Periodic Waves - Speed of Sound in Air - Principle of Superposition - Interference - Beats - Reflection of waves - Stationary waves - Stationary waves in a Stretched String - Stationary Waves in Air Columns - Doppler Effect
ICS Part 1 Physics Ch 8 Test
### First Year Physics Chapter 8 Online MCQ Test for 1st Year Physics Chapter 8 (Waves)
1 The pitch of sound deepens upon
• A. Intensity of sound
• B. Loudness of sound
• C. Wavelength of sound
• D. Frequency of sound
2 When the amplitude of a wave become double, its energy becomes
• A. One half
• B. Two times
• C. Three times
• D. Four times
3 The speed of sound in air does not depend upon
• A. Temperature
• B. Pressure
• C. Density
• D. Medium
4 the velocity of sound at 0 <sup>o</sup>C is 332 ms-1, the velocity of sound at 10 <sup>o</sup>C will be
• A. 337.1 ms-1
• B. 338.1 ,ms-1
• C. 342.1 ms-1
• D. 328.1 ms-1
5 In transverse waves the particles of medium vibrate
• A. Along the direction of wave
• B. Opposite to direction of wave
• C. Perpendicular to direction of wave
• D. Slowly
6 10 waves pass through a point in 2 seconds with speed 10 ms-1 the frequency of wave will be
• A. 1 Hz
• B. 2 Hz
• C. 5 Hz
• D. 10 Hz
7 In stationary waves the points which always remain at rest are.
• A. Nodes
• B. Antinodes
• C. Crest
• D. Trough
8 When the amplitude of a wave is increase to doubled is energy.
• A. Remain the same
• B. Increases 4 times
• C. Increases by two times
• D. Decreases by half
9 Wavelength of a wave for closed pipe having length 'l' in the fundamental mode is.
• A. 2 l
• B. 1/2
• C. 4 l
• D. l
10 If a stretched string 4 m long and it has 4 loops of stationary waves, then the wave length is.
• A. 1m
• B. 2 m
• C. 3 m
• D. 4 m
### Top Scorers of 11th Class Part 1 Physics Chapter 8 Test Online
M
#### Maroosha Nisar
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03 Mins 17 Sec
M
#### Maroosha Nisar
Lahore14 - Feb - 2024
13/17
05 Mins 25 Sec
U
Lahore29 - Nov - 2023
12/17
03 Mins 05 Sec
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#### Shayan Haris
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#### Maroosha Nisar
Lahore14 - Feb - 2024
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05 Mins 06 Sec
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#### Ali Saqib
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10/17
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#### Zeemal Malik
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### Chapterwise Lists
Chapter Name Questions PDF File
Physics English Medium Fsc Part 1 Chapter 1 Online MCQs Test 10 Questions
11th Class Part 1 Physics Chapter 2 Test Online 17 Questions
Physics Fsc Part 1 Chapter 2 Online Test 17 Questions
Physics English Medium Fsc Part 1 Chapter 2 Online MCQs Test 10 Questions
11th Class Part 1 Physics Chapter 3 Test Online 17 Questions
Physics Fsc Part 1 Chapter 3 Online Test 17 Questions
Physics English Medium Fsc Part 1 Chapter 3 Online MCQs Test 10 Questions
11th Class Part 1 Physics Chapter 4 Test Online 17 Questions
Physics Fsc Part 1 Chapter 4 Online Test 17 Questions
Physics English Medium Fsc Part 1 Chapter 4 Online MCQs Test 10 Questions
11th Class Part 1 Physics Chapter 5 Test Online 17 Questions
Physics Fsc Part 1 Chapter 5 Online Test 17 Questions
Physics English Medium Fsc Part 1 Chapter 5 Online MCQs Test 10 Questions
11th Class Part 1 Physics Chapter 6 Test Online 17 Questions
Physics Fsc Part 1 Chapter 6 Online Test 17 Questions
Physics English Medium Fsc Part 1 Chapter 6 Online MCQs Test 10 Questions
11th Class Part 1 Physics Chapter 7 Test Online 17 Questions
Physics Fsc Part 1 Chapter 7 Online Test 17 Questions
Physics English Medium Fsc Part 1 Chapter 7 Online MCQs Test 10 Questions
11th Class Part 1 Physics Chapter 8 Test Online 17 Questions
Physics Fsc Part 1 Chapter 8 Online Test 17 Questions
Physics English Medium Fsc Part 1 Chapter 8 Online MCQs Test 10 Questions
11th Class Part 1 Physics Chapter 9 Test Online 17 Questions
Physics Fsc Part 1 Chapter 9 Online Test 17 Questions
Physics English Medium Fsc Part 1 Chapter 9 Online MCQs Test 10 Questions
11th Class Part 1 Physics Chapter 10 Test Online 17 Questions
Physics Fsc Part 1 Chapter 10 Online Test 17 Questions
Physics English Medium Fsc Part 1 Chapter 10 Online MCQs Test 10 Questions
11th Class Part 1 Physics Chapter 11 Test Online 17 Questions
Physics Fsc Part 1 Chapter 11 Online Test 17 Questions
Physics English Medium Fsc Part 1 Chapter 11 Online MCQs Test 10 Questions
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X | 1,490 | 4,828 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-10 | longest | en | 0.768651 |
https://mathlake.com/Important-4-Marks-Questions-for-CBSE-8-Maths | 1,702,159,170,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100972.58/warc/CC-MAIN-20231209202131-20231209232131-00706.warc.gz | 432,036,878 | 8,062 | # Important 4 Marks Questions for CBSE 8 Maths
Important 4 marks questions for CBSE Class 8 Maths are provided here with solutions. The questions are prepared as per latest exam pattern and syllabus (2021-2022). Practicing these questions will students to score well in exams. Class 8th CBSE is a very important stage in the development of the fundamentals of a student. As the basics of Maths and Science help students to understand complex topics which will come in 9th standard.
Most of the subjects in class 8 are vast, where some of them require not only memorization but also requires skill. One such is the subject of Mathematics, which seems to be a nightmare for students and requires a lot of practice to boost confidence. So we at BYJU’S provide students of Class 8 with important 4 marks questions, which can be beneficial for them in their examination.
## Class 8 Maths Important 4 Marks Questions and Solutions
Students preparing for CBSE Class 8 Maths Examination are advised to practice the given questions and verify the solutions as well:
Q.1) Sohan wants to prepare a chart on education of children for a social work campaign. The chart is in shape of a parallelogram HEAR with HE = 5cm, EA = 6cm and $$\angle R = 85^{\circ}$$ .
1. Construct the parallelogram HEAR.
2. Do you think education of children is important for development of a society. Justify your statement.
Solution:
(a)
Rough Figure:
Step (i): Draw a line segment HE of 5 cm and an angle of 85º at point E. As vertex A is 6 cm away from vertex E, cut a line segment EA of 6 cm from this ray.
Step (ii): Vertex R is 6 cm and 5 cm away from vertex H and A respectively. By taking a radius as 6 cm and 5 cm, draw arcs from point H and A respectively. These will be intersecting each other at point R.
Step (iii): Join R to H and A.
(b) Yes, Early childhood education provides children to learn with diversity. This helps in creation of a better society.
Q.2) Divide the sum of $$\frac{7}{8}$$ and $$\frac{15}{24}$$ by their difference. Express your answer in standard form.
Solution:
Sum of 7/8 and 15/24:
(7/8) + (15/24)
= (21 + 15)/24
= 36/24
= 3/2
Difference of 7/8 and 15/24:
(7/8) – (15/24)
= (21 – 15)/24
= 6/24
= 1/4
Now, dividing the sum by difference,
(3/2)/ (1/4)
= (3/2) × 4
= 6
Q.3) The denominator of a rational number is greater than its numerator by 6. If the numerator is decreased by 1 and denominator is increased by 1, the number obtained is $$\frac{1}{3}$$ . Find the rational number.
Solution:
Let x be the numerator of a rational number.
Then, denominator = x + 6
As per the given condition,
(x – 1)/(x + 6 + 1) = 1/3
(x – 1)/(x + 7) = 1/3
3(x – 1) = (x + 7)
3x – 3 = x + 7
3x – x = 7 + 3
2x = 10
x = 5
Therefore, the rational number = 5/(5 + 6) = 5/11
Q. 4) PJUM is a parallelogram. Its diagonal meet at point O. Find x and y.
Solution:
We know that the diagonals of a parallelogram bisect each other.
y + 9 = 18
y = 18 – 9
y = 9
Also, x + y = 20
Substituting y = 9,
x + 9 = 20
x = 20 – 9
x = 11
Therefore, x = 11 and y = 9.
Q. 5) Draw a histogram for the daily earnings of 40 drug stores in following table:
Daily Earnings in Rupees 450-500 500-550 550-600 600-650 650-700 Number of Stores 16 10 8 4 2
Solution:
The class limits are represented along the x-axis and the frequencies along the y-axis on a suitable scale. The rectangles can be drawn to obtain the histogram for the given frequency distribution, taking class intervals as bases and the corresponding frequencies as heights. The histogram is given below:
Q. 6) Sanya took a loan of Rs. 80000 from a bank. If the rate of interest is 10% per annum, find the difference in amount she would pay after $$1\frac{1}{2}$$ years, if the interest is compounded:
• a) Annually
• b) Half Yearly
Solution:
Given,
Principal amount = P = Rs. 80000
Rate of interest = r = 10%
(i) Compounded annually
Amount after 1 year = 80000 + 10% of 80000 = 80000 + 8000 = 88000
Interest for the next 6 months = 88000 x (10/2)% = 4400
Amount after one and half years = Rs. 88000 + Rs. 4400 = Rs. 92400
(ii) Compounded half-yearly
In this case, the rate of interest will become half and time will be 3 half years.
Thus, amount = 80000[1 + (5/100)]3
= 80000[1 + (1/20)]3
= 80000 × (21/20) × (21/20) × (21/20)
= 92610
Now, the difference = Rs. 92610 – Rs. 92400 = Rs. 210
Q. 7) Sarita borrowed Rs. 18000 from Vimla at 3% per annum simple interest for 2 years. If she had borrowed this sum at 3% per annum compound interest, what extra amount would she have to pay?
Solution:
Given,
P = Rs. 18000
Rate of interest = R = 3% p.a.
Time = 2 years
Simple interest = PTR/100
= (18000 × 2 × 3)/100
= 1080
Amount to be returned on simple interest = Rs. 18000 + Rs. 1080 = Rs. 19080
Amount to be returned on compound interest = P[1 + R/100]nt
= 18000[1 + (3/100)]2
= 18000 × (103/100) × (103/100)
= 19096.2
Difference = Rs.19096.2 – Rs. 19080 = Rs. 16.2
Sarita has to pay Rs. 16.2 extra if she had borrowed the amount on compound interest.
Q.8) Simplify and write the answer with positive exponents:
$$\frac{3^{-5}\times 10^{-5}\times 125\times t^{4}}{5^{-7}\times t^{-2}}$$
Solution:
$$\frac{3^{-5} \times 10^{-5}\times 125 \times t^4}{5^{-7}\times t^{-2}}\\ = \frac{3^{-5} \times 5^{-5}\times 2^{-5}\times 5^3 \times t^4}{5^{-7}\times t^{-2}}\\ = 3^{-5} \times 5^{-5}\times 2^{-5}\times 5^3 \times t^4 \times 5^7 \times t^2\\ = 3^{-5} \times 5^{5}\times 2^{-5} \times t^{6}\\ = 6^{-5} \times 5^{5}\times t^{6}\\ = \frac{5^{5}\times t^{6}}{6^5}$$
Q. 9) 500 people were stranded on an island they had enough food to last for 30 days. How long the food would last if there were 200 more people?
Solution:
Given,
Food last for 500 people = 30 days
Now, 200 more people joined.
500 people = 30 days
700 people = n
Here, the data follows an inverse proportion.
Thus, 500 × 30 = 700 × n
n = (500/700) × 30
n = 150/7 days
Q. 10) 6 persons could build a wall in 10 days. 3 persons left before the work started. How long it would take to build the wall now?
Solution:
Given,
6 people could build a wall in 10 days.
3 people left before the work started.
Now, the number of persons = 6 – 3 = 3
Persons 6 3 Days 10 ?
This shows an inverse proportion.
Number of days required to build the wall = (6/3) × 10
= 20 days
Q. 11) Draw a graph for the following table of values, with suitable scale on the axis:
Deposit(in Rs) 1000 2000 3000 4000 5000 Simple Interest(in Rs) 80 160 240 320 400
Solution:
We can draw a linear graph for the given data.
Take “Simple interest” on X-axis and “Deposit” on Y-axis.
Q.12) If each edge of a cube is doubled
• a) How many times will its surface area increase?
• b) How many times will its volume increase?
Solution:
Let “a” be the edge of a cube.
Surface area of cube with edge “a” = 6a2
Volume of cube = a3
New measure of the edge = 2a
(i) New surface area = 6(2a)2
= 6 × 4 × a2
= 4 × (6a2)
= 4 × original surface area
Thus, if the edge of a cube is doubled, then its surface area will be increased by 4 times.
(ii) New volume = (2a)3
= 8 × a3
= 8 × original volume
Thus, if the edge of a cube is doubled, then its volume will be increased by 8 times.
Q.13) A rectangular piece of paper 12 cm x 5 cm is folded without overlapping to make a cylinder of height 4 cm. Find the volume of the cylinder.
Solution:
Given that,
Length of the rectangular paper = 12 cm
Breadth of the rectangular paper = 5 cm
Circumference of circular part of cylinder = 2πr
Also, given that the paper is rolled along its length.
Circumference of circular part of cylinder = Length of the rectangular paper
2πr = 12 cm
2 × (22/7) × r = 12
r = (12 × 7)/44
r = 84/44
r = 21/11 cm
Height of cylinder = Breadth of the rectangular paper = h = 4 cm
Volume of cylinder = πr2h
= (22/7) × (21/11) × (21/11) × 4
= 45.82 cm3
Q. 14) Subtract the sum of 3a + 2b – c and 5b – 6a + 2ab from the sum of 9b -11ab and -4a + 21ab
Solution:
3a + 2b – c + 5b – 6a + 2ab
= (3a – 6a) + (2b + 5b) + 2ab
= -3a + 7b + 2ab….(i)
9b -11ab + (-4a + 21ab)
= 9b + (-11ab + 21ab) – 4a
= 9b + 10ab – 4a….(ii)
Subtracting (i) from (ii),
9b + 10ab – 4a – (-3a + 7b + 2ab)
= 9b + 10ab – 4a + 3a – 7b – 2ab
= 8ab – a + 2b
Q.15) In a cylindrical dome there are 16 pillars. The radius of each pillar is 35 cm and height is 7 m. Find the total cost of painting the curved surface area of all pillars at the rate of Rs. 8.50 per $$m^{2}$$ .
Solution:
Given,
Radius of cylindrical pillar = r = 35 cm = 0.35 m
Height of the pillar = h = 7 m
Curved surface area of cylinder = 2πrh
= 2 × (22/7) × 0.35 × 7
= 15.4 m2
The curved surface area of 16 pillars = 16 × 15.4 m2 = 246.4 m2
Also, given that the cost of painting one m2 = Rs. 8.50
Total cost of painting the curved surface area of 16 pillars = 246.4 × Rs. 8.50 = Rs. 2094.4
Q.16) Water is pouring in a cuboidal reservoir at the rate of 60 liters per minute. If the volume of reservoir is 108 $$m^{3}$$ , find the number of hours it will take to fill the reservoir.
Solution:
Given, volume of reservoir = 108 m3
Rate of pouring water into cuboidal reservoir = 60 liters/minute
= 60/1000 m3 per minute {since 1 liter = (1/1000 )m3}
= (60 × 60)/1000 m3 per hour
Therefore, (60 × 60)/1000 m3 water filled in the reservoir in 1 hour.
Time taken for 1 m3 water to be filled in reservoir = 1000/(60 × 60) hours
Time taken for 108 m3 water to be filled in reservoir = (108 × 1000)/(60 × 60) hours = 30 hours
Q.17) Praful reads 15 pages of a storybook in a day. Raju reads 10 pages of the same book in a day. How many days will Raju take to finish the fiction book if Praful finishes in 13 days?
Solution:
Given,
Number of pages read by Praful in a day = 15
Number of pages read by Raju in a day = 10
Number of days required for Praful to finish the book = 13
That means, total number of pages in the book = 15 × 13 = 195
Hence, the number of days required for Raju to finish the book = 195/10 = 19.5 days
Q.18) Find the least number that must be added to 9911 so as to get a perfect square. Also find the square root of perfect square obtained.
Solution:
Let us find the square root of 9911.
Thus, the square of 100 will be the nearest perfect square number to 9911.
1002 = 10000
Subtracting 9911 from the square of 100,
10000 – 9911 = 89
Therefore, the least number to be added to 9911 so as to get a perfect square is 89.
Q.19) Plot the distance time graph for given values:
Distance (in km) 125 150 175 200 Time (min.) 25 30 35 40
Solution:
Considering the distance on the vertical axis and time on the horizontal axis.
Q.20) Factorise the expressions and divide them;
$$lm(l^{4} – m^{4})\div l(l^{2}+m^{2})$$
Solution:
lm(l4 – m4) ÷ l(l2 + m2)
= lm[(l2)2 – (m2)2]/ [l(l2 + m2)]
Using the identity a2 – b2 = (a + b)(a – b),
= [lm(l2 + m2)(l2 – m2)]/ [l(l2 + m2)]
= m(l2 – m2) | 3,594 | 10,926 | {"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} | 4.875 | 5 | CC-MAIN-2023-50 | latest | en | 0.94241 |
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