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https://www.jiskha.com/display.cgi?id=1375561731 | 1,511,549,297,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808742.58/warc/CC-MAIN-20171124180349-20171124200349-00380.warc.gz | 827,938,795 | 4,000 | # math
posted by .
if 8 men can do a job in days, what is the percentage increase in number of days required to do the job when 2 men are released?
A.16 2/3%
B.25%
C. 33 1/3%
D.40%
E.48%
• math -
6 men is 3/4 of 8 men
So, they can only do 3/4 as much work in a day.
So, it takes 4/3 as many days, or
(C)
• math -
suppose it took n days
number of man-days = 8n
if done by 6 men
number of days = 8n/6 = 4n/3
change in days = 4n/3 - n = (1/3)n
percentage change = (1/3)n / n = 1/3 or 33 1/3%
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More Similar Questions | 735 | 2,554 | {"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} | 3.90625 | 4 | CC-MAIN-2017-47 | latest | en | 0.947939 |
https://pybullet.org/Bullet/phpBB3/viewtopic.php?f=4&t=4557 | 1,610,752,363,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703497681.4/warc/CC-MAIN-20210115224908-20210116014908-00005.warc.gz | 535,534,764 | 9,963 | ## Solving two contacts at once
Please don't post Bullet support questions here, use the above forums instead.
Dennis
Posts: 12
Joined: Sat Jun 23, 2007 9:13 pm
Location: San Francisco
Contact:
### Solving two contacts at once
Hey folks, I'm tinkering with a new physics engine as a side project. I always thought solving two contacts at once in the same manifold using a direct 2x2 matrix solver was a terrific idea. Erin Catto also seems to have success in this in Box2D as reported on gphysics.com. However, I cannot really get much improvement when trying it. I reorder my contacts in the manifold so that the points solved together are the ones farthest away from each other. I solve pairwise until done, or only one more contact in that manifold (which I solve in isolation), and I fall back to standard SI if the accumulated impulse turns negative, etc.
It works, but I cannot measure or visually confirm that this is any better than good ol' SI. It doesn't give the same solution, of course, just different. I was expecting this to really improve stacking of long objects, but I can't see much difference if any.
Did anyone else try this in 3D and got a significant benefit out of it?
Cheers,
Dennis
Erwin Coumans
Posts: 4232
Joined: Sun Jun 26, 2005 6:43 pm
Location: California, USA
Contact:
### Re: Solving two contacts at once
Hey Dennis,
Good to see you again, and happy new year (I've barely visited this section because of all the Bullet support and development)
I've tried it before with solving 4 points between rigid bodies at once, and it improved stability.
Doom 3 also does this, and the code for solving up to 5x5 matrices is in the Doom 3 SDK, as far as I remember (several years ago though).
If you are using solving only 2 points at a time and there are 4 points, I would use the diagonal axis, A,C and B,D in the picture (top view)
contact.png (5.49 KiB) Viewed 4819 times
Improving the friction model improves stability and performance quite a lot: you could use central (rotational friction) friction and 2 linear friction axis (top view)
friction.png (7.21 KiB) Viewed 4819 times
Cheers,
Erwin
Dennis
Posts: 12
Joined: Sat Jun 23, 2007 9:13 pm
Location: San Francisco
Contact:
### Re: Solving two contacts at once
Hi Erwin, yes that's basically what I'm doing, but my problem is that I see no real benefit. I suspect that warm-starting might cancel out the effect of pair-wise solving, since contacts are likely to be initially separating, so it will fall back to SI very often. I'm not sure exactly what's going on. Is there a reason this is not in bullet if you tried it and saw an improvement?
I'm using central friction already How come Bullet does not use that? I think it could improve both solver performance and stability quite a bit.
Erwin Coumans
Posts: 4232
Joined: Sun Jun 26, 2005 6:43 pm
Location: California, USA
Contact:
### Re: Solving two contacts at once
Have you tried to solve 4 points simulaneously? And have you tried it with a single box resting on the ground without friction and only 1 iteration? Then you surely should see a difference, especially when you solve 4 points at a time.
[...] Is there a reason this is not in bullet [...] How come Bullet does not use that?
There are many things not in Bullet yet, that I tried before, it is a matter of priorities I suppose. I haven't worked on improving solver quality or performance much for Bullet, there are just too many areas to look into. Right now we are experimenting with OpenCL and GPU constraint solver has even fewer features than the original Bullet solver. Same for the PlayStation SPU version. Then, I'm also looking in better demos using Ogre and Irrlicht in this new GameKit project, to combine physics with animation etc.
If you have a patch or pseudo code for any improvement, I might apply it
Thanks,
Erwin
Dennis
Posts: 12
Joined: Sat Jun 23, 2007 9:13 pm
Location: San Francisco
Contact:
### Re: Solving two contacts at once
Yes, in the very simple case of one iteration and one body where no contact velocity is negative it does help. It is in a more realistic scenario, with multiple iterations, warm-starting and stacking that I cannot see any benefit.
I understand there are many priorities for bullet Kepp up the good work!
Dirk Gregorius
Posts: 875
Joined: Sun Jul 03, 2005 4:06 pm
Location: Kirkland, WA
### Re: Solving two contacts at once
Hey Dennis,
I tried this as well and can second your observation. There was no noticeable improvement for me. If you looked at the Box2D implementation you might have noticed that I implemented this in collaboration with Erin. Did you shift variables to account for the clamping against the accumulated impulse? The LCP has basically the following form and can be solved very easy using Direct Enumeration (see Murty):
w = A * dP - b >= 0 and ( P + dP ) >= 0 and w * ( P + dP ) = 0
Here dP is the delta impulse and P is the accumulated impulse. So to make this all work you have to add A * P on both sides of w:
w = A * P + A * dP - b - A * P = A * ( P + dP ) - ( b + A * P ) >= 0
Now substitute: P' = P + dP and b' = b + A * P and you are done. It is easy to oversee this and work in the wrong coordinates. Anyway, I am sure you did all this
I did also expect improved behavior since in theory this should cancel out rotational effects and I expected imroved stacking similar to a sphere stack. Let us know what you find out . It is really a long time ago I tried this and maybe there was a bug in the implementation.
Cheers,
-Dirk
Erin Catto
Posts: 324
Joined: Fri Jul 01, 2005 5:29 am
Location: Irvine
Contact:
### Re: Solving two contacts at once
The block solver gives a big stability increase to Box2D. I have not implemented this in 3D, but I suspect you would need to solve all points (up to 4) in block form to get a similar benefit. This gets to be a bit expensive because 4 points has 16 permutations that must be tested and you need to solve a different matrix equation for each permutation.
There are some implementation details in Box2D that must be followed precisely in terms of warm starting and clamping. Dirk and I had to do a bit of derivation to make sure it was correct.
Dennis
Posts: 12
Joined: Sat Jun 23, 2007 9:13 pm
Location: San Francisco
Contact:
### Re: Solving two contacts at once
Interesting you came to the same conclusion Dirk. Given the circumstances, and especially the computational overhead I think I'll give up on simultaneous solving in 3D, but I'll let you if I make any progress | 1,608 | 6,529 | {"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} | 2.5625 | 3 | CC-MAIN-2021-04 | latest | en | 0.959721 |
https://socratic.org/questions/a-25-0-ml-sample-of-sulfuric-acid-is-completely-neutralized-by-32-8-ml-of-0-116- | 1,579,802,621,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250611127.53/warc/CC-MAIN-20200123160903-20200123185903-00322.warc.gz | 650,585,757 | 6,244 | # A 25.0 mL sample of sulfuric acid is completely neutralized by 32.8 mL of 0.116 mol/L ammonia solution. Ammonium sulfate is formed What is the concentration of the sulfuric acid?
Dec 19, 2016
Approx. $0.08 \cdot m o l \cdot {L}^{-} 1$ with respect to sulfuric acid.
#### Explanation:
We need (i) a stroichiometric equation:
${H}_{2} S {O}_{4} \left(a q\right) + 2 N {H}_{3} \left(a q\right) \rightarrow 2 N {H}_{4}^{+} + S {O}_{4}^{2 -}$
At the 2nd equivalence point, the solution will stoichiometric in $\text{ammonium sulfate}$. An appropriate indicator must be used.
And (ii) a calculation,
$\left[{H}_{2} S {O}_{4}\right] = \frac{1}{2} \times \frac{32.8 \times {10}^{-} 3 \cdot L \times 0.116 \cdot m o l \cdot {L}^{-} 1}{25.0 \times {10}^{-} 3 \cdot L}$
=??*mol*L^-1
Why did we use a $\frac{1}{2}$ factor in the calculation? | 302 | 841 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "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} | 3.640625 | 4 | CC-MAIN-2020-05 | longest | en | 0.718887 |
https://www.quizover.com/course/section/real-world-applications-rational-functions-by-openstax | 1,548,173,759,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583857993.67/warc/CC-MAIN-20190122161345-20190122183345-00473.warc.gz | 897,701,017 | 27,341 | # 5.6 Rational functions (Page 11/16)
Page 11 / 16
$f\left(x\right)=\frac{x}{2x+1}$
Local behavior: $\text{\hspace{0.17em}}x\to -{\frac{1}{2}}^{+},f\left(x\right)\to -\infty ,x\to -{\frac{1}{2}}^{-},f\left(x\right)\to \infty \text{\hspace{0.17em}}$
End behavior: $\text{\hspace{0.17em}}x\to ±\infty ,f\left(x\right)\to \frac{1}{2}$
$f\left(x\right)=\frac{2x}{x-6}$
$f\left(x\right)=\frac{-2x}{x-6}$
Local behavior: $\text{\hspace{0.17em}}x\to {6}^{+},f\left(x\right)\to -\infty ,x\to {6}^{-},f\left(x\right)\to \infty ,\text{\hspace{0.17em}}$ End behavior: $\text{\hspace{0.17em}}x\to ±\infty ,f\left(x\right)\to -2$
$f\left(x\right)=\frac{{x}^{2}-4x+3}{{x}^{2}-4x-5}$
$f\left(x\right)=\frac{2{x}^{2}-32}{6{x}^{2}+13x-5}$
Local behavior: $\text{\hspace{0.17em}}x\to -{\frac{1}{3}}^{+},f\left(x\right)\to \infty ,x\to -{\frac{1}{3}}^{-},\text{\hspace{0.17em}}$ $f\left(x\right)\to -\infty ,x\to {\frac{5}{2}}^{-},f\left(x\right)\to \infty ,x\to {\frac{5}{2}}^{+},f\left(x\right)\to -\infty$
End behavior: $x\to ±\infty ,\phantom{\rule{0.2em}{0ex}}f\left(x\right)\to \frac{1}{3}$
For the following exercises, find the slant asymptote of the functions.
$f\left(x\right)=\frac{24{x}^{2}+6x}{2x+1}$
$f\left(x\right)=\frac{4{x}^{2}-10}{2x-4}$
$y=2x+4$
$f\left(x\right)=\frac{81{x}^{2}-18}{3x-2}$
$f\left(x\right)=\frac{6{x}^{3}-5x}{3{x}^{2}+4}$
$y=2x$
$f\left(x\right)=\frac{{x}^{2}+5x+4}{x-1}$
## Graphical
For the following exercises, use the given transformation to graph the function. Note the vertical and horizontal asymptotes.
The reciprocal function shifted up two units.
The reciprocal function shifted down one unit and left three units.
The reciprocal squared function shifted to the right 2 units.
The reciprocal squared function shifted down 2 units and right 1 unit.
For the following exercises, find the horizontal intercepts, the vertical intercept, the vertical asymptotes, and the horizontal or slant asymptote of the functions. Use that information to sketch a graph.
$p\left(x\right)=\frac{2x-3}{x+4}$
$q\left(x\right)=\frac{x-5}{3x-1}$
$s\left(x\right)=\frac{4}{{\left(x-2\right)}^{2}}$
$r\left(x\right)=\frac{5}{{\left(x+1\right)}^{2}}$
$f\left(x\right)=\frac{3{x}^{2}-14x-5}{3{x}^{2}+8x-16}$
$g\left(x\right)=\frac{2{x}^{2}+7x-15}{3{x}^{2}-14+15}$
$a\left(x\right)=\frac{{x}^{2}+2x-3}{{x}^{2}-1}$
$b\left(x\right)=\frac{{x}^{2}-x-6}{{x}^{2}-4}$
$k\left(x\right)=\frac{2{x}^{2}-3x-20}{x-5}$
$w\left(x\right)=\frac{\left(x-1\right)\left(x+3\right)\left(x-5\right)}{{\left(x+2\right)}^{2}\left(x-4\right)}$
$z\left(x\right)=\frac{{\left(x+2\right)}^{2}\left(x-5\right)}{\left(x-3\right)\left(x+1\right)\left(x+4\right)}$
For the following exercises, write an equation for a rational function with the given characteristics.
Vertical asymptotes at $\text{\hspace{0.17em}}x=5\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=-5,\text{\hspace{0.17em}}$ x -intercepts at $\text{\hspace{0.17em}}\left(2,0\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(-1,0\right),\text{\hspace{0.17em}}$ y -intercept at $\text{\hspace{0.17em}}\left(0,4\right)$
$y=50\frac{{x}^{2}-x-2}{{x}^{2}-25}$
Vertical asymptotes at $\text{\hspace{0.17em}}x=-4\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=-1,\text{\hspace{0.17em}}$ x- intercepts at $\text{\hspace{0.17em}}\left(1,0\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(5,0\right),\text{\hspace{0.17em}}$ y- intercept at $\text{\hspace{0.17em}}\left(0,7\right)$
Vertical asymptotes at $\text{\hspace{0.17em}}x=-4\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=-5,\text{\hspace{0.17em}}$ x -intercepts at $\text{\hspace{0.17em}}\left(4,0\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(-6,0\right),\text{\hspace{0.17em}}$ Horizontal asymptote at $\text{\hspace{0.17em}}y=7$
$y=7\frac{{x}^{2}+2x-24}{{x}^{2}+9x+20}$
Vertical asymptotes at $\text{\hspace{0.17em}}x=-3\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=6,\text{\hspace{0.17em}}$ x -intercepts at $\text{\hspace{0.17em}}\left(-2,0\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(1,0\right),\text{\hspace{0.17em}}$ Horizontal asymptote at $\text{\hspace{0.17em}}y=-2$
Vertical asymptote at $\text{\hspace{0.17em}}x=-1,\text{\hspace{0.17em}}$ Double zero at $\text{\hspace{0.17em}}x=2,\text{\hspace{0.17em}}$ y -intercept at $\text{\hspace{0.17em}}\left(0,2\right)$
$y=\frac{1}{2}\frac{{x}^{2}-4x+4}{x+1}$
Vertical asymptote at $\text{\hspace{0.17em}}x=3,\text{\hspace{0.17em}}$ Double zero at $\text{\hspace{0.17em}}x=1,\text{\hspace{0.17em}}$ y -intercept at $\text{\hspace{0.17em}}\left(0,4\right)$
For the following exercises, use the graphs to write an equation for the function.
$y=4\frac{x-3}{{x}^{2}-x-12}$
$y=-9\frac{x-2}{{x}^{2}-9}$
$y=\frac{1}{3}\frac{{x}^{2}+x-6}{x-1}$
$y=-6\frac{{\left(x-1\right)}^{2}}{\left(x+3\right){\left(x-2\right)}^{2}}$
## Numeric
For the following exercises, make tables to show the behavior of the function near the vertical asymptote and reflecting the horizontal asymptote
$f\left(x\right)=\frac{1}{x-2}$
$x$ 2.01 2.001 2.0001 1.99 1.999 $y$ 100 1,000 10,000 –100 –1,000
$x$ 10 100 1,000 10,000 100,000
$y$ .125 .0102 .001 .0001 .00001
Vertical asymptote $\text{\hspace{0.17em}}x=2,\text{\hspace{0.17em}}$ Horizontal asymptote $\text{\hspace{0.17em}}y=0$
$f\left(x\right)=\frac{x}{x-3}$
$f\left(x\right)=\frac{2x}{x+4}$
$x$ –4.1 –4.01 –4.001 –3.99 –3.999 $y$ 82 802 8,002 –798 –7998
$x$ 10 100 1,000 10,000 100,000 $y$ 1.4286 1.9331 1.992 1.9992 1.999992
Vertical asymptote $\text{\hspace{0.17em}}x=-4,\text{\hspace{0.17em}}$ Horizontal asymptote $\text{\hspace{0.17em}}y=2$
$f\left(x\right)=\frac{2x}{{\left(x-3\right)}^{2}}$
$f\left(x\right)=\frac{{x}^{2}}{{x}^{2}+2x+1}$
$x$ –.9 –.99 –.999 –1.1 –1.01 $y$ 81 9,801 998,001 121 10,201
$x$ 10 100 1,000 10,000 100,000 $y$ 0.82645 0.9803 .998 .9998
Vertical asymptote $\text{\hspace{0.17em}}x=-1,\text{\hspace{0.17em}}$ Horizontal asymptote $\text{\hspace{0.17em}}y=1$
## Technology
For the following exercises, use a calculator to graph $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$ Use the graph to solve $\text{\hspace{0.17em}}f\left(x\right)>0.$
$f\left(x\right)=\frac{2}{x+1}$
$f\left(x\right)=\frac{4}{2x-3}$
$\left(\frac{3}{2},\infty \right)$
$f\left(x\right)=\frac{2}{\left(x-1\right)\left(x+2\right)}$
$f\left(x\right)=\frac{x+2}{\left(x-1\right)\left(x-4\right)}$
$\left(-2,1\right)\cup \left(4,\infty \right)$
$f\left(x\right)=\frac{{\left(x+3\right)}^{2}}{{\left(x-1\right)}^{2}\left(x+1\right)}$
## Extensions
For the following exercises, identify the removable discontinuity.
$f\left(x\right)=\frac{{x}^{2}-4}{x-2}$
$\left(2,4\right)$
$f\left(x\right)=\frac{{x}^{3}+1}{x+1}$
$f\left(x\right)=\frac{{x}^{2}+x-6}{x-2}$
$\left(2,5\right)$
$f\left(x\right)=\frac{2{x}^{2}+5x-3}{x+3}$
$f\left(x\right)=\frac{{x}^{3}+{x}^{2}}{x+1}$
$\left(–1,\text{1}\right)$
## Real-world applications
For the following exercises, express a rational function that describes the situation.
A large mixing tank currently contains 200 gallons of water, into which 10 pounds of sugar have been mixed. A tap will open, pouring 10 gallons of water per minute into the tank at the same time sugar is poured into the tank at a rate of 3 pounds per minute. Find the concentration (pounds per gallon) of sugar in the tank after $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ minutes.
A large mixing tank currently contains 300 gallons of water, into which 8 pounds of sugar have been mixed. A tap will open, pouring 20 gallons of water per minute into the tank at the same time sugar is poured into the tank at a rate of 2 pounds per minute. Find the concentration (pounds per gallon) of sugar in the tank after $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ minutes.
$C\left(t\right)=\frac{8+2t}{300+20t}$
For the following exercises, use the given rational function to answer the question.
The concentration $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ of a drug in a patient’s bloodstream $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ hours after injection in given by $\text{\hspace{0.17em}}C\left(t\right)=\frac{2t}{3+{t}^{2}}.\text{\hspace{0.17em}}$ What happens to the concentration of the drug as $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ increases?
The concentration $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ of a drug in a patient’s bloodstream $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ hours after injection is given by $\text{\hspace{0.17em}}C\left(t\right)=\frac{100t}{2{t}^{2}+75}.\text{\hspace{0.17em}}$ Use a calculator to approximate the time when the concentration is highest.
For the following exercises, construct a rational function that will help solve the problem. Then, use a calculator to answer the question.
An open box with a square base is to have a volume of 108 cubic inches. Find the dimensions of the box that will have minimum surface area. Let $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ = length of the side of the base.
A rectangular box with a square base is to have a volume of 20 cubic feet. The material for the base costs 30 cents/ square foot. The material for the sides costs 10 cents/square foot. The material for the top costs 20 cents/square foot. Determine the dimensions that will yield minimum cost. Let $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ = length of the side of the base.
$A\left(x\right)=50{x}^{2}+\frac{800}{x}.\text{\hspace{0.17em}}$ 2 by 2 by 5 feet.
A right circular cylinder has volume of 100 cubic inches. Find the radius and height that will yield minimum surface area. Let $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ = radius.
A right circular cylinder with no top has a volume of 50 cubic meters. Find the radius that will yield minimum surface area. Let $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ = radius.
$A\left(x\right)=\pi {x}^{2}+\frac{100}{x}.\text{\hspace{0.17em}}$ Radius = 2.52 meters.
A right circular cylinder is to have a volume of 40 cubic inches. It costs 4 cents/square inch to construct the top and bottom and 1 cent/square inch to construct the rest of the cylinder. Find the radius to yield minimum cost. Let $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ = radius.
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al | 4,854 | 13,018 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 120, "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} | 4.3125 | 4 | CC-MAIN-2019-04 | latest | en | 0.411823 |
https://origami.photobrunobernard.com/origami/how-to-divide-a-rectangle-paper-into-quarters/ | 1,675,844,199,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500719.31/warc/CC-MAIN-20230208060523-20230208090523-00198.warc.gz | 456,729,507 | 12,248 | How To Divide A Rectangle Paper Into Quarters. Draw three lines parallel to one of the. Fold twice to obtain quarter markings at the paper bottom.
Here's an easy way to fold any rectangle of paper into a triangle with three sides the same length. Place the left end of the ruler at the left edge of the sheet of paper. She split the larger rectangle into four smaller rectangles.
### She Split The Larger Rectangle Into Four Smaller Rectangles.
Divide into equal parts below are pattern sequences that can be used to split a square or rectangle into an exact number of equally sized parts without doing any measurements at all. How to divide a square into thirds. Lets say you have a 12 pages document and youd like to split it into 3 documents each containing 4.
### Measure 2 18 Inches From The Left.
Fold twice to obtain quarter markings at the paper bottom. Copy it, then paste 7 times. Draw a rectangle 3.33 wide and 3.75 tall.
### However, Dividing Into Thirds, Fifths And Other Odd Fractions Is Less Easy.
Draw three lines parallel to one of the. You need to divide a rectangle into 3 equal parts along it’s length. To split a page to 4 parts, you can insert a table to deal with the job.
### Fold The Other Side On Top Of The Other.
I want to make a powerpoint slide where the slide is divided into quarters and the content is repeated in each quarter. Move the right end of the ruler in a diagonal position across the paper so that the right end is at the right edge of the sheet of paper. Fold a bottom corner up to touch the fold line, making a sharp point on the other corner.
### We Could Divide Our Rectangle Into Two Equal Parts, Or Halves.
Natalie’s perhaps chosen the way that you might decide to split a rectangle into quarters. We shade in one of these four sections to show that it is a quarter. Then, we can halve each half again until we have four equal parts. | 421 | 1,893 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-06 | latest | en | 0.904234 |
https://www.polytechforum.com/control/what-sort-of-regulator-controller-is-this-8275-.htm | 1,685,552,293,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646937.1/warc/CC-MAIN-20230531150014-20230531180014-00337.warc.gz | 1,048,846,608 | 9,285 | # What sort of regulator/controller is this?
• posted
Hello control community,
I have adopted some code (it is in LabView, but this shouldn't matter) that is regulating a voltage. I would like to know what kind of controller this could be. It seems to be sort of PID controller, but I can't rewrite the equations for the algorithm to get a PID equation.
This is the algorithm:
Voltage Control (Chip heating):
K = (V_d - V_d_aim) / V_d_old
V = V_old+K*V_d-K*{1-[ T(A) / T(n) ]} * V_d_old
Interpretation:
The voltage of a chip heating (V) is controlled such that the diffusion voltage of a temperature diode (V_d) reaches the prescribed value (V_d_aim). (Actually the chip temperature should be controlled, but I can calculate V_d_aim from the chip temperature, so, that shouldn't matter).
I don't know, what K, T(A), and T(n) are supposed to mean, but T(A) and T(n) seem to be constants (i.e. they are hardcoded and probably found by trial and error). Obviously, K is *not* constant.
I know two forms of discretized PID algoritm:
form 1:
u[n] = K_c * e[n] + K_c * T_s / T_i * sum(i=0..n; e[i]) + K_c * T_d * ( e[n] - e[n-1] ) / T_s * u[0]
and form 2:
u[n] = u[n-1] + K_c * (e[n] - e[n-1]) + K_c * T_s / T_i * e[n] + K_c * T_d / T_s * ( e[n]-2e[n-1]+e[n-2])
I can get from one form to the other so they are equivalent.
However, I can't get any PID form from my algorithm. e[n] is the error signal, this should be V_d - V_d_aim in my case. T_s, T_i, T_d, and K_c are the sampling rate and the PID constants, respectively.
When I try to come to form 2 with my algorithm, I just get:
V = V_old + e[n] * V_d / V_d_old - e[n] + T(A)/T(n) *e[n]
But probably this is an algorithm for another controller, no PID controller. I would like to use it (since it works in an old project) and to extend it to control the chip voltage according to a prescribed chip power.
Any ideas ?
Regards
Johannes
• posted
This would be a PID if the calculation of K were
K = (V_d - V_d_aim) * B where B is some constant.
That is, K is some multiple of the loop error. But the calculation for K being as it is makes K a kind of non-linear error term. That is, it is the loop error where the gain is higher for lower values of V_d_old. For any steady-state condition we can think of V_d_old as kind of a constant. I don't know why someone would want the loop gain to be a function of the process variable value. But that's what this equation does. Perhaps there are factors in the way that the chip is heated that creates a non-linear relationship between applied voltage and chip temperature, and this non-linear scaling of the error term might be cancelling out that variation.
-Robert Scott Ypsilanti, Michigan (Reply through newsgroups, not by direct e-mail, as automatic reply address is fake.)
• posted
For a constant resistor, power is proportional to voltage squared. We definitely have a non-linear process here. If the quantity we want to control is heat transfer, we have a process gain proportional to voltage. To keep a constant loop gain we would need a controller gain inversely proportional to voltage. I.e. higher gain for lower values.
Walter.
• posted
Hello Robert and Walter,
Walter,
So, the algorithm makes sense to you. Is it a 'well-known' type of controller?
Also, I have to extend the system (well that's almost all software developing is about ;)) to control the chip heating voltage according to a prescribed chip power. It would be nice to use the same algorithm with other parameters. Then it would be:
Voltage Control (Chip heating):
K = (P - P_aim) / P_old
V = V_old+K*P-K*{1-[ T(A) / T(n) ]} * P_old
Any ideas about how to determine/tune T(A) and T(n) ?
Regards
Johannes
PolyTech Forum website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners. | 1,027 | 3,893 | {"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} | 2.984375 | 3 | CC-MAIN-2023-23 | latest | en | 0.935572 |
http://popflock.com/learn?s=Earth_ellipsoid | 1,606,648,744,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141197593.33/warc/CC-MAIN-20201129093434-20201129123434-00174.warc.gz | 77,427,260 | 17,806 | Earth Ellipsoid
Get Earth Ellipsoid essential facts below. View Videos or join the Earth Ellipsoid discussion. Add Earth Ellipsoid to your PopFlock.com topic list for future reference or share this resource on social media.
Earth Ellipsoid
A scale diagram of the oblateness of the 2003 IERS reference ellipsoid. The outer edge of the dark blue line is an ellipse with the same eccentricity as that of Earth, with north at the top. For comparison, the light blue circle within has a diameter equal to the ellipse's minor axis. The red line represents the Karman line 100 km (62 mi) above sea level, while the yellow area denotes the altitude range of the ISS in low Earth orbit.
An Earth ellipsoid is a mathematical figure approximating the Earth's form, used as a reference frame for computations in geodesy, astronomy, and the geosciences. Various different ellipsoids have been used as approximations.
It is an ellipsoid of revolution whose minor axis (shorter diameter), which connects the geographical North Pole and South Pole, is approximately aligned with the Earth's axis of rotation. The ellipsoid is defined by the equatorial axis a and the polar axis b; their difference is about 21 km, or 0.335%.
Many methods exist for determination of the axes of an Earth ellipsoid, ranging from meridian arcs up to modern satellite geodesy or the analysis and interconnection of continental geodetic networks. Amongst the different set of data used in national surveys are several of special importance: the Bessel ellipsoid of 1841, the international Hayford ellipsoid of 1924, and (for GPS positioning) the WGS84 ellipsoid.
## Types
One should distinguish between two types of ellipsoid: mean and reference.
A data set which describes the global average of the Earth's surface curvature is called the mean Earth Ellipsoid. It refers to a theoretical coherence between the geographic latitude and the meridional curvature of the geoid. The latter is close to the mean sea level, and therefore an ideal Earth ellipsoid has the same volume as the geoid.
While the mean Earth ellipsoid is the ideal basis of global geodesy, for regional networks a so-called reference ellipsoid may be the better choice.[1] When geodetic measurements have to be computed on a mathematical reference surface, this surface should have a similar curvature as the regional geoid - otherwise, reduction of the measurements will get small distortions.
This is the reason for the "long life" of former reference ellipsoids like the Hayford or the Bessel ellipsoid, despite the fact that their main axes deviate by several hundred meters from the modern values. Another reason is a judicial one: the coordinates of millions of boundary stones should remain fixed for a long period. If their reference surface changes, the coordinates themselves also change.
However, for international networks, GPS positioning, or astronautics, these regional reasons are less relevant. As knowledge of the Earth's figure is increasingly accurate, the International Geoscientific Union IUGG usually adapts the axes of the Earth ellipsoid to the best available data.
## Historical method of determining the ellipsoid
High precision land surveys can be used to determine the distance between two places at nearly the same longitude by measuring a base line and a chain of triangles. (Suitable stations for the end points are rarely at the same longitude). The distance ? along the meridian from one end point to a point at the same latitude as the second end point is then calculated by trigonometry. The surface distance ? is reduced to ?', the corresponding distance at mean sea level. The intermediate distances to points on the meridian at the same latitudes as other stations of the survey may also be calculated.
The geographic latitudes of both end points, ?s (standpoint) and ?f (forepoint) and possibly at other points are determined by astrogeodesy, observing the zenith distances of sufficient numbers of stars. If latitudes are measured at end points only, the radius of curvature at the midpoint of the meridian arc can be calculated from R = ?'/(|?s-?f|). A second meridian arc will allow the derivation of two parameters required to specify a reference ellipsoid. Longer arcs with intermediate latitude determinations can completely determine the ellipsoid. In practice multiple arc measurements are used to determine the ellipsoid parameters by the method of least squares. The parameters determined are usually the semi-major axis, ${\displaystyle a}$, and either the semi-minor axis, ${\displaystyle b}$, or the inverse flattening ${\displaystyle 1/f}$, (where the flattening is ${\displaystyle f=(a-b)/a}$).
Geodesy no longer uses simple meridian arcs, but complex networks with hundreds of fixed points linked by the methods of satellite geodesy.
## Historical Earth ellipsoids
The reference ellipsoid models listed below have had utility in geodetic work and many are still in use. The older ellipsoids are named for the individual who derived them and the year of development is given. In 1887 the English surveyor Colonel Alexander Ross Clarke CB FRS RE was awarded the Gold Medal of the Royal Society for his work in determining the figure of the Earth. The international ellipsoid was developed by John Fillmore Hayford in 1910 and adopted by the International Union of Geodesy and Geophysics (IUGG) in 1924, which recommended it for international use.
At the 1967 meeting of the IUGG held in Lucerne, Switzerland, the ellipsoid called GRS-67 (Geodetic Reference System 1967) in the listing was recommended for adoption. The new ellipsoid was not recommended to replace the International Ellipsoid (1924), but was advocated for use where a greater degree of accuracy is required. It became a part of the GRS-67 which was approved and adopted at the 1971 meeting of the IUGG held in Moscow. It is used in Australia for the Australian Geodetic Datum and in South America for the South American Datum 1969.
The GRS-80 (Geodetic Reference System 1980) as approved and adopted by the IUGG at its Canberra, Australia meeting of 1979 is based on the equatorial radius (semi-major axis of Earth ellipsoid) ${\displaystyle a}$, total mass ${\displaystyle GM}$, dynamic form factor ${\displaystyle J_{2}}$ and angular velocity of rotation ${\displaystyle \omega }$, making the inverse flattening ${\displaystyle 1/f}$ a derived quantity. The minute difference in ${\displaystyle 1/f}$ seen between GRS-80 and WGS-84 results from an unintentional truncation in the latter's defining constants: while the WGS-84 was designed to adhere closely to the GRS-80, incidentally the WGS-84 derived flattening turned out to be slightly different than the GRS-80 flattening because the normalized second degree zonal harmonic gravitational coefficient, that was derived from the GRS-80 value for ${\displaystyle J_{2}}$, was truncated to eight significant digits in the normalization process.[2]
An ellipsoidal model describes only the ellipsoid's geometry and a normal gravity field formula to go with it. Commonly an ellipsoidal model is part of a more encompassing geodetic datum. For example, the older ED-50 (European Datum 1950) is based on the Hayford or International Ellipsoid. WGS-84 is peculiar in that the same name is used for both the complete geodetic reference system and its component ellipsoidal model. Nevertheless, the two concepts--ellipsoidal model and geodetic reference system--remain distinct.
Note that the same ellipsoid may be known by different names. It is best to mention the defining constants for unambiguous identification.
Reference ellipsoid name Equatorial radius (m) Polar radius (m) Inverse flattening Where used
Maupertuis (1738) 6,397,300 6,363,806.283 191 France
Plessis (1817) 6,376,523.0 6,355,862.9333 308.64 France
Everest (1830) 6,377,299.365 6,356,098.359 300.80172554 India
Everest 1830 Modified (1967) 6,377,304.063 6,356,103.0390 300.8017 West Malaysia & Singapore
Everest 1830 (1967 Definition) 6,377,298.556 6,356,097.550 300.8017 Brunei & East Malaysia
Airy (1830) 6,377,563.396 6,356,256.909 299.3249646 Britain
Bessel (1841) 6,377,397.155 6,356,078.963 299.1528128 Europe, Japan
Clarke (1866) 6,378,206.4 6,356,583.8 294.9786982 North America
Clarke (1878) 6,378,190 6,356,456 293.4659980 North America
Clarke (1880) 6,378,249.145 6,356,514.870 293.465 France, Africa
Helmert (1906) 6,378,200 6,356,818.17 298.3 Egypt
Hayford (1910) 6,378,388 6,356,911.946 297 USA
International (1924) 6,378,388 6,356,911.946 297 Europe
Krassovsky (1940) 6,378,245 6,356,863.019 298.3 USSR, Russia, Romania
WGS66 (1966) 6,378,145 6,356,759.769 298.25 USA/DoD
Australian National (1966) 6,378,160 6,356,774.719 298.25 Australia
New International (1967) 6,378,157.5 6,356,772.2 298.24961539
GRS-67 (1967) 6,378,160 6,356,774.516 298.247167427
South American (1969) 6,378,160 6,356,774.719 298.25 South America
WGS-72 (1972) 6,378,135 6,356,750.52 298.26 USA/DoD
GRS-80 (1979) 6,378,137 6,356,752.3141 298.257222101 Global ITRS[3]
WGS-84 (1984) 6,378,137 6,356,752.3142 298.257223563 Global GPS
IERS (1989) 6,378,136 6,356,751.302 298.257
IERS (2003)[4] 6,378,136.6 6,356,751.9 298.25642 [3]
## References
1. ^ Alexander, J. C. (1985). "The Numerics of Computing Geodetic Ellipsoids". SIAM Review. 27 (2): 241-247. Bibcode:1985SIAMR..27..241A. doi:10.1137/1027056.
2. ^ NIMA Technical Report TR8350.2, "Department of Defense World Geodetic System 1984, Its Definition and Relationships With Local Geodetic Systems", Third Edition, 4 July 1997 [1]
3. ^ a b Note that the current best estimates, given by the IERS Conventions, "should not be mistaken for conventional values, such as those of the Geodetic Reference System GRS80 ... which are, for example, used to express geographic coordinates" (chap. 1); note further that "ITRF solutions are specified by Cartesian equatorial coordinates X, Y and Z. If needed, they can be transformed to geographical coordinates (?, ?, h) referred to an ellipsoid. In this case the GRS80 ellipsoid is recommended." (chap. 4).
4. ^ IERS Conventions (2003) (Chp. 1, page 12) | 2,638 | 10,125 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 11, "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} | 2.890625 | 3 | CC-MAIN-2020-50 | latest | en | 0.867056 |
http://oeis.org/A089730 | 1,563,251,134,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524502.23/warc/CC-MAIN-20190716035206-20190716061206-00172.warc.gz | 117,426,006 | 3,757 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A089730 Greatest prime factor of all differences prime(n)-q, q prime and q
0, 1, 3, 5, 3, 11, 7, 17, 7, 13, 29, 17, 19, 41, 17, 23, 23, 59, 31, 29, 71, 37, 23, 43, 47, 47, 101, 47, 107, 53, 61, 59, 67, 137, 73, 149, 73, 79, 53, 83, 83, 179, 89, 191, 97, 197, 103, 109, 107, 227, 113, 113, 239, 107, 127, 113, 131, 269, 137, 139, 281, 137 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 LINKS EXAMPLE n=10, prime(n)=29: a(10) = Max(Union(Factors(29-prime(i)):i<10)) = Max(Factors(27) u Factors(26) u Factors(24) u Factors(22) u Factors(18) u Factors(16) u Factors(12) u Factors(10) u Factors(6)) = Max({3} u {2,13} u {2,3} u {2,11} u {2,3} u {2} u {2,3} u {2,5} u {2,3}) = Max{2,3,5,11,13} = 13. CROSSREFS Cf. A000040, A030173, A006530. Sequence in context: A177930 A242034 A205708 * A105445 A178095 A177904 Adjacent sequences: A089727 A089728 A089729 * A089731 A089732 A089733 KEYWORD nonn AUTHOR Reinhard Zumkeller, Jan 07 2004 STATUS approved
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Last modified July 15 22:48 EDT 2019. Contains 325061 sequences. (Running on oeis4.) | 572 | 1,433 | {"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} | 3.21875 | 3 | CC-MAIN-2019-30 | latest | en | 0.542199 |
https://brainmass.com/economics/demand-supply/aztec-enterprises-elasticity-of-demand-460617 | 1,477,629,411,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721555.54/warc/CC-MAIN-20161020183841-00273-ip-10-171-6-4.ec2.internal.warc.gz | 812,576,711 | 17,155 | Share
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# Aztec Enterprises - Elasticity of Demand
Text Problem:
Aztec Enterprises depends heavily on advertising to sell its products. Management at Aztec is allowed to spend \$2 million monthly on advertising, but no more than this amount. Each month, Aztec spends exactly \$2 million on advertising. What is Aztec's elasticity of demand for advertising? Can you write the equation for Aztec's demand for advertising?
#### Solution Preview
The Demand Function is very simple: D = 2million/Price
We are going to calculate the price elasticity of demand:
Price elasticity of demand: = (dQ / dP)*(P/Q)
This means we have to calculate dQ/dP (ie, the derivative ... | 153 | 685 | {"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} | 3.03125 | 3 | CC-MAIN-2016-44 | longest | en | 0.912452 |
https://fr.scribd.com/document/423191989/FUNDAMENTALS-OF-AERODYNAMICS | 1,582,446,490,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145747.6/warc/CC-MAIN-20200223062700-20200223092700-00424.warc.gz | 384,090,271 | 99,874 | Vous êtes sur la page 1sur 39
# FSI Introduction to Bridging Course Program
FUNDAMENTALS OF AERODYNAMICS
## A. THE INTERNATIONAL STANDARD ATMOSPHERE
The atmosphere is the mechanical mixture of gases surrounding the earth.
Atmospheric Constituents
## Nitrogen ------------------ 78.03%
Oxygen ------------------- 20.99%
Argon --------------------- 0.94%
Carbon Dioxide ---------- 0.03%
Hydrogen ------------------ 0.01%
Helium --------------------- 0.004%
Neon ----------------------- 0.0012% and a small amount of water vapor and other gases
1. Troposphere
2. Stratosphere
3. Ionosphere
4. Exosphere
Pressure
## PO= 14.7 lb/in2 = 2,116.8 lb/ft2 = 29.92” Hg
= 76 cm Hg = 760 mm Hg = 101,325 Pa = 1 atm
Density
3 3
ρo = 0.002377 slug/ft = 1.225 kg/m
Temperature
## Coefficient of Dynamic Viscosity
µo= 3.7372 x 10-7 slug/ft-sec = 1.7894 x 10-5 kg/m-sec
1
SEA LEVEL UP TO TROPOPAUSE
## 1. TEMPERATURE VARIATION WITH ALTITUDE
T
T = To + ah ; θ =
To
Where:
T
θ= = temperature ratio
To
## T = temperature at any altitude above sea level up to tropopause in oR or K
To = 519 °R or 288K
or -6.51 K/km)
## 2. PRESSURE VARIATION WITH ALTITUDE
5.26
P ah
δ= = 1 +
Po To
Where:
P
δ= = pressure ratio
Po
## 3. DENSITY VARIATION WITH ALTITUDE
4.26
ρ ah
σ= = 1 +
ρo To
Where:
ρ
σ= = density ratio
ρo
## ρ = density at any altitude above sea level up to tropopause in slug/ft3 or kg/m3
3 3
ρ o = 0.002377 slug/ft or 1.225kg/m
## h = any altitude above sea level up to tropopause in ft or m or km
2
ABOVE TROPOPAUSE UP TO STRATOPAUSE
## 2. PRESSURE VARIATION WITH ALTITUDE
English System
P 1.26
δ= =
Po −5
e 4.805x10 h
Where:
2
P = pressure at any altitude above tropopause up to stratopause in lb/ft
2
Po= 2,116.8lb/ft
Metric System
P 1.26
δ= =
Po −4
e1.578x10 h
Where:
Po = 101,325 Pa
## 3. DENSITY VARIATION WITH ALTITUDE
English System
ρ 1.68
σ= = 4.805 x10 −5
ρo e h
Where:
3
ρ= density at any altitude above tropopause up to stratopause in slug /ft
3
ρ0= 0.002377 slug / ft
## h = any altitude above tropopause up to stratopause in feet
Metric System
ρ 1.68
σ= = 1.578 x10 −4
ρ0 e h
3
Where:
3
ρ= density at any altitude above tropopause up to stratopause in kg/m
3
ρ o = 1.225 kg/m
h = any altitude above tropopause up to stratopause in meters
ALTIMETERS
An altimeter is a pressure gauge which indicates an altitude in the standard atmosphere corresponding to
the measured pressure.
## Pressure altitude, hp – is the altitude given by an altimeter set to 29.92 “Hg.
Density altitude, hd– is the altitude corresponding to a given density in the standard atmosphere.
Temperature altitude, hT – is the altitude corresponding to a given temperature in the standard atmosphere.
Problems:
1. Calculate the pressure, density and temperature at 25,000 ft and 60,000 ft altitudes in the standard
atmosphere.
Ans. P25,000ft = 785.50 lb/ft2 , ρ25,000ft = 0.001065 slug/ft3 , T = 429.85 oR
P60,000ft = 149.27 lb/ft2 , ρ60,000ft = 0.000223 slug/ft3 , T= 390.15 oR
2. On a hot day, the measured temperature and pressure are 38°C and 29.0 “Hg, respectively. Calculate
the density and the density ratio.
Ans. ρ = 1.100 kg/m3 , σ = 0.898
3. A standard altimeter reads 4,500 meters when the ambient temperature is 275K. What are the
density altitude and the temperature altitude?
Ans. hd = 5,061.98 m ,hT = 1,996 93 m
4. At a certain altitude, a standard altimeter reads 3,000 meters. If the density altitude is 2,500 meters,
find the true temperature.
Ans. T = 255.18 K
## PRINCIPLE OF MASS CONSERVATION (LAW OF CONTINUITY)
The mass flow of fluid that passing to one section of the tube in one second is equal to the
mass flow of fluid that passing to the other section of the tube in one second.
4
m
o
m = ρAV
Where:
o
m = mass flow of fluid in slug/sec or kg/sec
3 3
ρ = density of fluid in slug/ft or kg/m
2 2
A = cross-sectional area of tube in ft or m
V = velocity of fluid in ft/s or m/s
## a. For incompressible fluid, ρ= constant (M < 0.3 approximately).
A 1V 1 = A 2V 2
or
AV = constant
Differential form:
dV dA
= −
V A
## b. For compressible fluid, ρ ≠ constant (M ≥ 0.3 approximately).
ρ1 A1V1 = ρ 2 A2V2
or
ρAV = constant
Differential form:
dV dA dρ
+ + = 0
V A ρ
5
Problems:
1. A pipe is gradually tapering in size, diminishing by 0.1 sq.ft. per foot run. What is the change in
velocity per foot run where the pipe is 4 sq.ft. in cross section, if the velocity there is 90 ft. per sec?
Is the velocity increasing or decreasing?
Ans. dV/dS = 2.25 fps per ft , increasing
2. A circular pipe, 100 ft. long tapers from 3 ft. in diameter at one end to 2 ft.in diameter at the other.
Fluid is flowing from the bigger toward the smaller end. What is the rate of increase in velocity at
the entrance if the velocity there is 80 ft. per sec?
Ans. dV/dS = 0.444 fps per ft
3. Air having the standard sea level density has a velocity of 100 fps at a section of a wind tunnel. At
another section having an area half as great as the first section the flow velocity is 400 mph. What is
the density at the second section?
Ans. ρ2 = 0.000810 slug per cu ft
## II. INCOMPRESSIBLE BERNOULLI EQUATION – conservation of energy along a streamline.
BERNOULLI’S PRINCIPLE
States that the total energy of a particle in motion is constant at all points on its path in a steady flow.
In a continuous flow of fluid, as velocity increases, pressure decreases; and as velocity decreases,
pressure increases.
## For incompressible fluid, ρ= constant (M < 0.3 approximately).
V2 P
+ = constant
2 ρ
or
V12 P1 V2 2 P2
+ = +
2 ρ 2 ρ
Problems:
1. A horizontal pipe, 1 ft. in diameter, tapers gradually to 8 in. in diameter. If the flow is 500 cu ft. of
water per minute, what is the difference between the pressures at the two sections?
Ans. PA – PB = 443.63lb per sq ft
2. Water flows through a horizontal pipe at a velocity of 50 ft. per sec. Owing to the pipe gradually
expanding to a larger size, the velocity decreases to 35 ft. per sec. What is the difference between the
pressures at two points, one in each size of pipe?
Ans. PB – PA = 1,236. 75 lb per sq ft
3. The diameter of a horizontal tube is 4 in., in which tetrabromoethane (spec. grav. = 0.30) is flowing
at the rate of 0.50 cu ft. per sec. The pressure is 30 lb. per sq. in. (gage). If the tube gradually
decreases to 3 in. in diameter, what is the pressure there?
Ans. PB = 6,416.16 lb per sqft or 44.56 lb per sq in.
6
4. Alcohol (spec. grav. = 0.80) is flowing through horizontal pipe, which is 10 in. in diameter, with a
velocity of 40 ft. per sec. at a smaller section of the pipe, there is 6 lb. per sq. in. less pressure.
Assuming that the flow is smooth, what is the diameter there?
Ans. dB = 0.73 ft or 8.76 in.
5. Air is flowing horizontally at a speed of 100 mph through a duct 4 sq.ft. in cross section. The duct
gradually narrows down to a throat section. If a U-tube shows a difference in pressure between the
throat and main sections of 7 in. of water, what is the cross-sectional area of the throat? (Assume that
the air is non-compressible and has a density of 0.002377 slug per cu ft.)
Ans. AB = 2.57 sq ft
6. A horizontal water pipe is reduced in size from 18in. in diameter at point A to 6 in. in diameter at
point B. The flow in the pipe is 10 cu ft. per sec, and the pressure at A is 20 lb. per sq. in. (gage). If it
is assumed that there is no loss in energy due to friction, what is the pressure at B?
Ans. PB = 2,511.85 lb per sq ft or 17.44 lb per sq in.
7. Air flows through a horizontal pipe at the rate of 3,000 cu ft. per sec. If the pressure is 30 lb per sq.
in. (gage) where the diameter is 3 ft., what is the pressure where the diameter is 2 ft.?
Ans. PB = 5,567.10 lb per sq ft or 38.66 lb per sq in.
8. Water flows through a horizontal pipe at the rate of 800 gal per min. What is the difference in
pressure between a point where the diameter is 2 in. and a point where the diameter is 1 in.?
Ans. PA – PB = 96,856.38 lb per sq ft or 672.61 lb per sq in
9. A horizontal pipeline enlarges from a diameter of 6 in. at point A to a diameter of 12 in. at point B.
The flow of water is 20 cu ft. per sec, and the pressure at A is 10 lb. per sq. in. What is the pressure
at B?
Ans. 3,995.32 lb per sqft or 27.75 lb per sq in.
10. The diameter at section (1) is 0.3m. The diameter at section (2) is 0.15m. What is the flow rate of a
substance (sp. gr. = 0.85) if the pressure between sections (1) and (2) is 12.7 cm Hg?
Ans. Q = 0.115 cu meter per sec.
11. Consider a low-speed subsonic wind tunnel with a 12/1 contraction area ratio for the nozzle. If the
flow in the test section is at a standard sea level conditions with a velocity of 50 m /s, calculate the
height difference in a U-tube mercury manometer with one side connected to the nozzle inlet and the
other to the test section. ρHg = 1.36 x 103 kg/ m3.
Ans. ∆h = 0.114 m
12. The wind tunnel shown in Fig. below has a smallest section (test section) measuring 1.22 m by 1.0m,
and a largest section of 2 m square. At a certain tunnel speed the manometer reading is 0.72m. The
manometer liquid has a specific gravity of 0.85. Calculate the airspeed in the test section. Assume
incompressible flow and standard sea-level conditions.
Ans. VB = 73.50 meters per sec.
7
13. A pitot-static tube is used to measure the airspeed at the test section of a wind tunnel. If the pressure
difference across the pitot-static tube is 0.11 m of water, what is the airspeed at the test section? If
the ratio of the cross-sectional area between the largest section and the test section is 100:1, what is
the airspeed at the largest section? Assume incompressible flow at standard sea level conditions.
Ans. V2 = 41.97 meters per sec.
V1 = 0.4197 meter per sec.
14. Consider water flowing through a smooth pipe whose diameter is decreasing. At one location, the
diameter is 12 cm. If the velocity there is 10 m/s.
(a) Find the mass flow rate.
(b) At a station farther down the pipe, the diameter is 4cm. Find the velocity at this station.
o
Ans. m = 113. 10 kg/sec. , VB = 90 meters per sec.
Venturi Tube
The Venturi tube is a convergent–divergent tube with a short cylindrical throat or constricted
section. This device determines the rate of flow of fluid through the tube by measuring the difference in
pressure between the throat section and the entrance section.
## Practical application is made of Bernoulli Equation.
P1 V12 P2 V2 2
+ = +
ρ 2 ρ 2
1
(
P1 − P2 = ρ V22 − V12
2
)
By the law of continuity, Q being the rate of flow, for non-compressible fluids,
Q = A1 V1 = A2 V2
Q2 Q2
V12 = ; V22 =
A12 A 22
Then,
2 2
ρ Q Q
P1 − P2 = −
2 A 2 A1
8
2
2
ρ Q A2
P1 − P2 = 1−
2 A 2 A1
2
Q
= 2( P1 − P2 )
2
A2 A 2
ρ 1−
A1
2( P1 − P2 )
Q = A2
A 2
ρ 1 − 2
A1
πd 22 2( P1 − P2 )
Q=
4 d 4
ρ 1 − 2
d1
Where:
3 3
Q = rate of flow in ft /s or m /s
2
P1 = pressure at section 1 in lb /ft or Pa
2
P2 = pressure at section 2 in lb/ft or Pa
2 2
A1 = cross-sectional area of section 1 in ft or m
2 2
A2 = cross-sectional area of section 2 in ft or m
d1 = diameter of section 1 in ft or m
d2 = diameter of section 2 in ft or m
3 3
ρ = density in slug/ft or kg/m
Problems:
1. A Venturi tube narrows down from 4 in. in diameter to 2 in. in diameter. What is the rate of flow of
water if the pressure at the throat is 2 lb. per sq. in. less than at the larger section?
Ans. Q = 0.388 cu ft per sec.
2
2. A Venturi tube in 6 in. in diameter at the entrance, where the pressure is 10 lb/in (gage). The
2
throat is 4 in. in diameter; there the pressure is 6 lb/in (gage). What is the flow of water?
Ans. Q = 2.73 cu ft per sec.
3. Consider a Venturi with a throat-to-inlet area ratio of 0.8 mounted in a flow at standard sea level
conditions. If the pressure between the inlet and the throat is 335 Pa, calculate the velocity of the
flow at the inlet.
Ans. V1 = 31.18 m/s
9
III. COMPRESSIBILITY EFFECTS
When the air is regarded as compressible, the internal energy of the air mass must be
considered.
## A. ISENTROPIC EQUATION OF STATE
P
= cons tan t
ργ
γ
P1 P2 P2 ρ
= , = 2
ρ1γ ρ2 γ P1 ρ1
## THERMODYNAMIC PARAMETER RELATIONSHIP
γ 2γ
γ γ Va γ −1
P2 ρ2 T2 γ −1 w
= = = 2 = 2
P1 ρ1 T1 w1 Va
1
Problems:
3
1. Air at standard pressure and temperature, has density of 1.225 kg/m . If the air is compressed
adiabatically to 3 atm, what are the specific weight, density and the temperature?
Ans. w2 = 26.33 N/m3 , ρ2 = 2.685 kg/m3 , T2 = 394.20 K
## 2. Air at standard pressure and temperature is permitted to expand adiabatically to one-half
atmospheric pressure. What are a) the density and b) the temperature?
Ans. (a) ρ2 = 0.747 kg/m3 (b) T2 = 236.26 K
3. Air at standard pressure and temperature is adiabatically compressed to 50 lb. per sq. in. (gage)
pressure. What is the temperature?
Ans. T2 = 792.59 oR
4. Air at standard pressure and temperature is permitted to expand adiabatically until it is one-half
standard density. a) What is the pressure? b) What is the temperature?
Ans. (a) P2 = 38,395 Pa , (b) T2 = 218.26 K
## B. THE SPEED OF SOUND ,Va
γP
Va =
ρ
English System
Va = γgRT
Where:
γ = 1.4
R = 53.342 ft/ °R
10
Va = 49.02 T
Where:
Va = speed of sound in air in ft/s
T = absolute temperature in °R
Metric System
Va = 20.05 T
Where:
Va = speed of sound in air in m/s
T = absolute temperature in K
Problems:
## 1. Find the speed of sound in air at standard sea level conditions.
Ans. Va = 340.26 m/s
2. Find the speed of sound in air at 7,000 ft. altitude in the standard atmosphere.
Ans. Va = 1,089.57 ft/sec
## C. COMPRESSIBLE BERNOULLI EQUATION
V2 γ P
+ = constant
2 γ −1 ρ
or:
V12 γ P1 V2 2 γ P2
+ = +
2 γ − 1 ρ1 2 γ − 1 ρ2
Problems:
3
1. In an undisturbed airstream the pressure is 101,325 Pa, the density is 1.225 kg/m , and the
velocity is 150 m/s. What is the pressure if the velocity is 190 m/s?
Ans. P = 93,236.71 Pa
2. In an undisturbed airstream, where the pressure is 14.7 lb. per sq. in. and the temperature is 59oF,
the velocity is 520 ft. per sec. Where the velocity is 600 ft. per sec., what is the local pressure?
Ans. P = 2,012.23 lb per sq ft
3. In an undisturbed airstream, where the pressure is 14.7 lb. per sq. in. and the temperature is 59oF,
the velocity is 550 ft. per sec. What is the velocity where the pressure is 13.9 lb. per sq. in.?
Ans. V = 633.83 ft/sec
11
IV. AIRSPEED MEASUREMENT
PVρ
Pt Vt ρ t
## Pitot – Static Tube
Where:
PVρ = parameters of airflow initially
Ptvtρt= parameters of the airflow at the stagnation point
Pitot-static tube- an instrument consisting of two tubes, having their points of origin in the open end
unobstructed airstream. One of the tube ends is closed but slotted on the side in such a manner so that static air
pressure is maintained in the tube and the other tube having an end open to the airstream so that it receives the
full impact pressure of the airstream.
Static pressure (P) – the force per unit area exerted by a fluid on a surface at rest relative to the fluid.
Stagnation pressure or total pressure (Pt) – the pressure at stagnation point wherein the velocity of the
medium is equal to zero.
## Applying the incompressible Bernoulli equation:
V 2 P Vt2 Pt
+ = +
2 ρ 2 ρ
1
Pt − P = ρV 2
2
1
Pt = P + ρV 2
2
Pt = P + q
Where:
1
q = ρV 2 = dynamic pressure
2
2(Pt − P )
V=
ρ
12
Application:
2(Pt − P )
Ve =
ρ0
Where:
## Pt= total pressure or stagnation pressure
P = static pressure
ρo= density at sea level
## Definition of true airspeed V:
Ve
V =
σ
Where:
ρ
σ= , density ratio
ρ0
Problems:
1. An airplane is flying at standard sea level conditions at 45 meters per second. What is the difference
between total and static pressure?
Ans. Pt - P = 1,240.31 Pa
2. An airplane is flying at standard sea level conditions at airspeed of 75 meters per second. What is the
total pressure?
Ans. Pt = 104,770.31 Pa
3. An airplane is flying at standard sea level, the difference between total and static pressure is 1,750
Pa. What is the airspeed in meters per second?
Ans. V = 53.45 m/s
## (b) HIGH-SPEED AIRSPEED INDICATORS (COMPRESSIBLE FLOW)
Mach number (M) – is the ratio of the local velocity of the fluid to the velocity of sound at
that same point.
V
M=
Va
Where:
M = Mach number
V = velocity of fluid
Va= ambient speed of sound
13
Three different regimes of aerodynamic flows
## 1. If M <1, the flow is subsonic.
2. M = 1, the flow is sonic.
3. If M > 1, the flow is supersonic.
## 1. If 0.8 < M < 1.2, the flow is transonic.
2. If M > 5, the flow is hypersonic
## Applying the compressible Bernoulli equation:
V2 γ P Vt2 γ Pt
+ = +
2 γ −1 ρ 2 γ − 1 ρt
V2 γ P Pt ρ
= − 1
2 γ − 1 ρ P ρ t
−
1
V2 γ P Pt Pt γ
= − 1
2 γ − 1 ρ P P
γ −1
V2 γ P Pt γ
= − 1
2 γ −1 ρ P
γ −1
2
P
V2 = Va2 t γ − 1
γ −1 P
γ −1
V2 2 Pt γ
= − 1
Va2 γ − 1 P
γ −1
2 tP − P
M2 = + 1 γ − 1
γ −1 P
## For some applications, it is convenient to solve (Pt – P) in terms of M:
γ
γ − 1 2 γ −1
Pt − P = P 1 + M − 1 applicable if M ≥ 2.24
2
γ 1 1 1
Pt − P = PM 2 1 + M 2 + M 4 + M6 + ⋅ ⋅ ⋅ applicable if 0.3≤M<2.24
2 4 40 1,600
14
Application:
## Definition of calibrated airspeed Vc:
P −P Pt − P
Note that if t is replaced by and Va by Vao , the resulting airspeed is called the
P P
o
calibrated airspeed (Vc ) :
2
γ −1
2 V a P − P γ
Vc 2 = 0
t + 1 − 1
γ − 1 P0
2
γ − 1 Vc
Pt − P = Po 1 + − 1
2 Va o
γ −1
γ γ
γ −1
2
2 1 γ − 1 Vc
M2 = 1+ − 1 + 1 − 1
γ − 1 δ 2 Va o
## For air,γ = 1.4.
0.286
1 2 3.5
2 Vc
M = 5 1 + 0.2 − 1 + 1 − 1
δ 661.5
STREAM FUNCTION
In studying the two-dimensional flow of fluid, it is desirable to have a name for the quantity of fluid
passing between a given point and a reference point. The quantity is termed the stream function ψ of the given
point. In two-dimensional flow, the dimensions of stream function are in square feet per second.
In Figure 2.8a, 0 is the reference point, 0AP and 0BP are any two lines connecting 0 and P is the amount
of fluid crossing line 0AP per second must be the same as the amount of fluid crossing 0BP per second: this
amount per second is the stream function of P.
In Figure 2.8b, points P and P’ are two points on the same streamline. The stream function of P is the
amount of fluid crossing line 0P per second. Since fluid cannot cross line PP’, the quantity of fluid crossing line
0P’ per second is the same as the quantity of fluid crossing the line 0P per second. Therefore points on the same
15
streamline have the same stream function, and conversely, points having the same stream function are the on the
same streamline. Similarly if 0’ is on s streamline passing through 0, the fluid crossing 0P per second is the
same quantity as that crossing 0’P per second, since no fluid crosses 00’. Then the stream function of P with
respect to 0 is the same as the stream function of P with respect to 0’.
## RELATION BETWEEN VELOCITY AND STREAM FUNCTION IN TWO-DIMENSIONAL
FLOW
In two-dimensional flow, the velocity V may be resolved into two components, parallel to the X and Y
axes respectively. The velocity component parallel to the X axis is termed u; its direction is positive if from
lower to greater values of x, i.e., to the right. The component parallel to the Y axis is termed v; it is positive if
the direction is toward increasing positive values of y, i.e., upward.
## Stream function is positive if the flow is clockwise about the origin 0.
If P and P’ are points on adjacent streamlines (see Figure 2.9) and if dψ is the difference between their
respective stream functions and dx and dy, the difference between their coordinates, the quantity of fluid
crossing PP’ per second must equal the quantity crossing PA plus the quantity crossing AP’ per second. Then
dψ = udy − vdx
## From differential calculus, if ψ is a function of x and y,
∂ψ ∂ψ
dψ = dx + dy
∂x ∂y
Then
∂ψ
u=
∂y
∂ψ
v=−
∂x
When the value of ψ is known as a function of the x and y coordinates for all points in a two- dimensional fluid
flow, the velocity may be determined.
Example: If fluid is flowing so that ψ = x 2 − y 2 , determine the velocity at the point (3, 2).
Solution:
∂ψ ∂ 2
u= = (x − y2 )
∂y ∂y
u = −2 y = −(2)(2) = −4 ft / sec
∂ψ ∂
v=− = (x2 − y 2 )
∂x ∂x
v = −2 x = −(2)(3) = −6 ft / sec
V = u 2 + v 2 = (−4) 2 + (−6) 2 = 7.21 ft / sec
16
v
tan θ =
u
v −6
θ = tan −1 = tan −1 = 236.31
0
u −4
Problems:
## 1. What is the velocity at point (-2, 4), if ψ = x 2 − y 2 ?
Ans. V = 8.94 ft/sec
θ = 153.430
2. What is the velocity at point (3, 5), if ψ = x 2 + y 2 ?
Ans. V = 11.66 ft/sec
θ = 329.040
3. For the flow ψ = x 2 + y 2 , what is the velocity at point (4, 5)?
Ans. V = 12.81 ft/sec
θ = 321.340
4. For the flow ψ = x 2 + y 2 , what is the velocity at point (5, 5)?
Ans. V = 14.14 ft/sec
θ = 315.00
5. What is the velocity at point (4, 4), if ψ = ln( x 2 + y 2 ) ?
Ans. V = 0.35 ft/sec
θ = 315.00
## SOURCES AND SINKS
In the mathematical study of fluid flow, a useful concept is a source, or a sink. A source is a point at
which fluid is appearing at a definite, uniform rate: if there is no disturbance to the flow the fluid will travel
radically outward from the source. A sink is a negative source or a point at which fluid is disappearing at a
definite, uniform rate. With no disturbance, the fluid travels radially inward equally from all directions.
The strength m of a source is the quantity of fluid appearing per second; in two-dimensional flow it is
measured in square feet per second. The streamlines from a source are radial lines. If one of these streamlines is
designated as a base or reference line, and the streamline through point P makes an angle θ with these
reference line, θ being measured counterclockwise from the reference line, the stream function for the point P
is
ψ= , θ in radians
ψ= , θ in deg rees
360
## Since he flow is radially outward, the magnitude of the velocity V is
m
V=
2πr
and the direction of V makes the angle the angle θ with the reference line.
## Because stream function is a scalar quantity, stream functions may be added.
The stream function of a source, of strength m, located at the origin is ψ = mθ /(2π ), θ being measured
in radians in a counterclockwise direction from the positive X axis. The stream function, straight-line flow
parallel to the X axis and flowing in a positive direction is ψ = +Uy, where U is the constant velocity.
17
The stream function of these two flows when combined is
ψ = +Uy +
m y
ψ = +Uy + tan −1
2π x
Having the equation for ψ in terms of the constant U and m, and the variables x and y, curves are
plotted for various streamlines, as in Figure 2.10a. The streamline representing ψ = 0 is the X axis plus a curve
resembling a parabola, PP’, shown as a heavy solid line in Figure 2.10a.
As no fluid may cross a streamline, any streamline may be replaced by a solid boundary, so that the
streamlines of Figure 2.10a may be considered as the flow of a horizontal wind striking a hill shaped like PP’ in
Figure 2.10b.
The velocity being ascertainable at any point in the flow, if the pressure is known at some one point,
Bermoulli’s law may be used to find the pressure at any other point on a given streamline.
Example: A source of strength of 600 sq ft per sec located at the origin (0, 0) is superimposed on a uniform
flow of 500 ft per sec parallel to the X axis in a negative direction. If the pressure in the flow remote from this
source is standard atmospheric, what is the pressure at the point (0, 1.5)? Note: This point is on the streamline
ψ = 0.
Given:
m = 600 sq ft per sec
U = - 500 ft/sec
P0 = 2,116.8 lb/ft2
Solution:
∂ψ ∂ m y
u= = + Uy + tan −1
∂y ∂y 2π x
m x 600 0
u =U + 2 = − 500 + ( ) = −404.51 ft / sec
2π x + y 2 2π (0) 2 + (1.5) 2
∂ψ ∂ m y
v=− = − − Uy + tan −1
∂x ∂x 2π x
m y 600 1 .5
v=+ 2 =(
2
) 2 2
= 63.66 ft / sec
2π x + y 2π (0) + (1.5)
V = u 2 + v 2 = (−404.51) 2 + (63.66) 2 = 409.49 ft / sec
18
By Bernoulli’s law,
V 2 P U 2 P0
+ = +
2 ρ0 2 ρ0
1
P = P0 + ρ 0 (U 2 − V 2 )
2
1
[ ]
P = 2,116.8 + ( )(0.002377) (−500) 2 − (409.49) 2 = 2,214.63lb / ft 2
2
Problems:
1. With a source and uniform flow as in the example, find the pressure at the point (0.557, 1.0).
Ans. P = 2,117.9 lb/ft2
2. With a source and uniform flow as in the example, find the pressure at the point (-1.15, 2).
Ans. P = 2,109.3 lb/ft2
3. With a source and uniform flow as in the example, find the pressure at the point (0.995, 0).
Ans. P = 2,128.1 lb/ft2
## SOURCE PLUS SINK PLUS UNIFORM FLOW
If a source with a strength m is located on the X axis at x = c and if a sink with a strength –m is located
on the X axis at x = -c (see Figure 2.11), the stream function of point P (x, y) is
mθ1 mθ 2
ψ= −
2π 2π
but
y
tan θ1 =
x−c
and
y
tan θ 2 =
x+c
By trigonometry,
tan θ1 − tan θ 2
tan(θ1 − θ 2 ) =
1 + tan θ1 − tan θ 2
y y
= x−c x+c
y y
1+
x − c x + c
2cy
= 2
x + y2 − c2
19
mθ1 mθ 2
Then Eqn. ψ = − may be written in the form
2π 2π
m 2cy
ψ= tan −1 2 2
2
2π x + y −c
Superimposing a uniform flow of –U ft per sec parallel to the X axis on the outward flow of the source
and the inward flow of the sink given, for the point P (x, y),
m 2cy
ψ = −Uy + tan −1 2 2
2
2π x + y −c
By assigning values to ψ, streamlines may be plotted for the form of flow as in Figure 2.12. The
streamline for ψ = 0 in the X axis and a symmetrical curve, tan (2πUy)/m=2cy/(x2 +y2 –c2), resembling an
ellipse. The flow may be considered as the two-dimensional flow about a cylindrical surface having for its cross
section the quasi-elliptic shape of the ψ = 0 line.
The component velocities, u and v, may be found at any point by differentiating the stream function
∂ψ
u=
∂y
m x−c x+c
u = −U + 2 − 2
2π r1 r2
∂ψ
v=−
∂x
my 1 1
v=+ 2 − 2
2π r1 r2
Where
r12 = ( x − c) 2 + y 2
r22 = ( x + c) 2 + y 2
The components of the velocity being known, the resultant velocity may be found. The pressure being
known at same point in the flow, the pressure at another point on the same streamline may be calculated by
Bernoulli’s law. Owing to the symmetry of the streamlines, the pressure on the surface, ψ = 0, are symmetric
about the X and Y axes. The sum of the vertical pressures is zero, and the sum of the horizontal pressures is
zero.
Example: A source whose strength is 314. 2 sq ft per sec and a sink, whose strength is -314.2 sq ft per sec are
located at (1.0) and (-1.0) respectively. They are superimposed in a uniform flow of -100 ft per second parallel
to the X axis. If the pressure in the flow remote from the source and sink is atmospheric and if the fluid is air,
what is the pressure at the point (1.05, 0.6)? Note: This point is on the streamline ψ = 0.
20
Solution:
## r12 = ( x − c) 2 + y 2 = (1.05 − 1.0) 2 + (0.6) 2 = 0.3625
r22 = ( x + c) 2 + y 2 = (1.05 + 1.0) 2 + (0.6) 2 = 4.5625
x−c x+c 314.2 1.05 − 1.0 1.05 + 1.0
2 − 2 = −100 +
m
u = −U + − 4.5625
2π r1 r2 2π 0.3625
= −115.57 ft pers sec
my 1 1
2 − 2 =
(314.2)(0.6) 1 1
v=+ −
2π r1 r2 2π 0.3625 4.5625
= 76.19 ft per sec
V = u 2 + v 2 = (−115.57) 2 + (76.19) 2 = 138.42 ft per sec
V 2 P U 2 P0
+ = +
2 ρ0 2 ρ0
1
P = P0 + ρ 0 (U 2 − V 2 )
2
1
[ ]
P = 2,116.8 + ( )(0.002377) (−100) 2 − (138.42) 2 = 2,105.91 lb / ft 2
2
Problems:
1. For the flow described in the example, what is the pressure at point (1.38, 0.2)?
Ans. P = 2,124.4 lb/ft2
2. For the flow described in the example, what is the pressure at point (1.27, 0.4)?
Ans. P = 2,115.3 lb/ft2
3. For the flow described in the example, what is the pressure at point (1.18, 0.5)?
Ans. P = 2,110.2 lb/ft2
4. For the flow described in the example, what is the pressure at point (0.56, 0.8)?
Ans. P = 2,099.2 lb/ft2
5. For the flow described in the example, what is the pressure at point (0.0, 0.86)?
Ans. P = 2,098.6 lb/ft2
## FLOW ABOUT A CIRCULAR CYLINDER
It was shown that the stream function of a source and a sink, each located on the X axis at a distance of
2c apart, is
m 2cy
ψ= tan −1 2 2
2
2π x + y −c
and the streamline for ψ = 0 is the X axis and a curve resembling an ellipse.
If the distance 2c between the source and the sink is made smaller while the strengths m and -m are
made larger, the product, 2cm, always being kept constant, the ratio of the length of the major axis to the length
of the minor axis of the elliptic figure will become less. As the distance 2c approaches zero in value, the figure
will become more nearly circular, when it is combined with a uniform flow. Letting µ = 2cm, Eqn.
m 2cy
ψ= tan −1 2 2
may be written
2
2π x + y − c
2cy
tan −1 2 2
2
x + y −c
ψ=
4πc
µ
As c approaches zero, the right-hand side of the above equation becomes indeterminate, but, by applying
a standard method of calculus (Hospital’s rule), differentiating separately the numerator and the denominator
21
with respect to c, and letting c approach zero, a solution is obtained. Performing this on Equation
2cy
tan −1 2 2
2
x + y − c
ψ= gives, for the source-sink doublet,
4πc
µ
µ y
ψ=
2π x + y 2
2
When this flow is combined with a uniform, negative flow parallel to the X axis, the stream function
becomes
µ y
ψ = −Uy +
2π x + y 2
2
Letting
µ
a2 =
2πU
Ua2 y
ψ = −Uy +
x2 + y 2
The streamline for µ = 0 will be the X axis and a circle whose center is at the origin and whose radius is
a.
## Anywhere in the flow, the component velocities are
∂ y
u= − Uy + Ua 2 2
∂y x + y 2
x2 − y2
u = −U + Ua 2 2 2 2
(x + y )
∂ 1
v = − − Uy + Ua 2 y 2
∂x x + y 2
2 xy
v = +Ua 2 2 2 2
(x + y )
## On the circumference of the circle, ψ = 0, since y 2 = a 2 − x 2
dy x
=−
dx y
Also
V2
u = −2U
a2
xy
v = 2U
a2
22
The resultant velocity V makes an angle tan −1 (v / u) with the X axis, but
2Uxy
v 2
= a 2
u − 2Uy
a2
v x
=−
u y
This shows that the resultant velocity V is tangent to the circumference of the circle.
V 2 = u 2 + v2
2 2
− 2Uy 2 2Uxy
= 2
+ 2
a a
4U 2 y 2
=
a2
2Uy
V=
a
## In polar coordinate, this becomes
V = 2U sin θ
The pressure at any point on the surface of a circular cylinder in a uniform flow is, from Bernoulli’s law,
V 2 P U 2 P0
+ = +
2 ρ0 2 ρ0
1
P = P0 + ρ 0 (U 2 − V 2 )
2
1
[
= P0 + ρ 0 U 2 − (2U sin θ ) 2
2
]
1
P = P0 + ρ 0 (U 2 − 4U 2 sin 2 θ )
2
V = 2U sin θ
23
Where:
V = tangential velocity in ft/s or m/s
U = freestream velocity in ft/s or m/s
θ = angle through the point on the surface of the cylinder with the main
direction of the air flow in deg.
P = pressure at any point on the surface of a circular cylinder in psf or Pa
P0 = freestream pressure in psf or Pa
ρ0 = freestream density in slug/ft3 or kg/m3
Problems:
1. A uniform current of air with a speed of 100ft. per sec flows around a circular cylinder. At a distance
from the cylinder the pressure is atmospheric. What is the pressure at a point on the surface of the
cylinder so located that a radial line through the point makes an angle of 150 with the direction of
airflow?
Ans. P = 2,125.50 lb per sq ft
2. For the flow in the above problem, what is the pressure on the cylinder surface at a 90o arc from the
direction of airflow?
Ans. P = 2,081.14 lb per sq ft
## Lift due to Circulation
Vortex - the circulation of particles of fluid having the same energy content and center of rotation.
Γ Γ
V = 2V∞sinθ V= V = 2 V∞ sinθ +
2π r 2π r
V∞
24
V∞
Where:
sin θ = − dL
dF
dL =-sinθdF
dL =-Prsinθdθ
## By Incompressible Bernoulli Equation:
P V2 P∞ V∞ 2
+ = +
ρ∞ 2 ρ∞ 2
ρ
P = P∞ + ∞ V∞
2
( )
2 − V2
ρ 2 Γ
2
P = P∞ + ∞ V∞ − 2V∞ sin θ +
2
2πr
π ρ∞ 2 Γ
2
L
∫0 dL = ∫−π P∞ + 2 V∞ − 2V∞ sin θ + 2πr r sin θdθ
L =ρ∞V∞Γ
Where:
L = lift per unit span in lb/ft or N/m
ρ ∞ =freestream density in slug/ft3 or kg/m3
V∞ = freestream velocity in fps or mps
Γ = strength of circulation in ft2/s or m2/s (= 2пrV)
V = tangential velocity in fps or mps (= 2пrN)
r = radius of cylinder in ft or m
N = rotational speed in rps
25
For total lift
LT = L x l
Where:
LT = total lift in lb or N
l = length of the cylinder in ft or m
Problems:
## 1. A cylinder 30 in. in diameter rotates in an airstream of 70 mph. If develops 40 lb of lift per
foot of length, what is the rotational speed?
Ans. N = 2.66 rev/sec
2. A cylinder 1.22 m in diameter and 3.5 m long is rotating at 100 rpm in an airstream of 18 m/s.
Determine the total lift of the cylinder at standard sea level conditions.
Ans. LT = 1,889.49 N
## VISCOUSEFFECTS, THE BOUNDARY LAYER AND FLOW SEPARATION
SKIN FRICTION - skin friction is air resistance, and it is the tangential component of force on the surface of a
body due to the friction between the two particles.
STREAMLINE AND TURBULENT FLOW - a streamline flow may be defined as smooth non-turbulent flow.
A turbulent flow is defined as a flow characterized by turbulence that is a flow in which the velocity varies
erratically in both magnitude and direction with time.
LAMINAR FLOW – the word laminar is derived from the Latin word lamina meaning a thin plate of metal or
some other material. Laminar flow employs the concept that air is flowing in thin sheets or layers close to the
surface of a wing with no disturbance between the layers of air.
BOUNDARY LAYER– the boundary layer is that layer of air adjacent to the airfoil surface. The cause of the
boundary layer is the friction between the surface of the wing and the air.
LAMINAR BOUNDARY LAYER – in the laminar boundary layers the flow is steady and smooth. As a
result, the layer is very thin, and so the form drag is very small. Also, the velocity gradient at the walls, though
large enough to give significant viscous stress, is yet only moderate, so that the skin friction, though not
negligible, is also very small.
5.2X
δ=
R Nx
Where:
δ = laminar boundary layer thickness
26
X = transition point
R N x = transition Reynolds number
The “rubbing” of the boundary layer on the flat plate gives rise to friction forces Df: friction drag. The skin-
friction drag coefficient for a plate in laminar flow is given by:
Df 2.656
Cf = =
1 ρV 2 S RN
2
Where:
RN = Reynolds number based on the total plate length
TURBULENT BOUNDARY LAYER – in turbulent boundary layers the flow is unsteady and not smooth, but
eddying.
When the flow is transitioned to turbulent flow, the boundary layer thickness will be increased. In fact, this
phenomenon is often used to determine the location of the transition region. The boundary layer thickness can
be determined by:
0.37 X
δ=
(R Nx )1 / 5
The skin-friction drag coefficient for a flat plate can be calculated with Schlichting’s formula:
Df 0.910
Cf = =
1 ρV 2S (log10 R N )2.58
2
Where:
ρVL ρVc
RN = =
µ µ
ρ = air density in slug /ft3 or kg/m3
V = air velocity in ft/s or m/s
L = characteristic length of the body in the flow direction and is equal to the chord length for
an airfoil in feet or meters.
slug kg
µ = coefficient of dynamic viscosity in or
ft − sec m − sec
For air, µ increases with temperature and can be calculated by the following approximate formula for the
standard atmosphere:
3
2.329 x10−8 T 2 slug
µ= , , T in °R
T + 216 ft − sec
or
3
1.458 x10−6 T 2 kg
µ= , , T in K
T + 110.4 m − sec
At low Reynolds numbers, the flow in the boundary layer is laminar (laminar boundary layer). Above
certain “transition Reynolds numbers”, the flow becomes turbulent (turbulent boundary layer).
27
Transition takes place on a flat plate at point x determined by:
ρVx
(R N x )crit. = = 3.5 x10 5 to 10 6
µ crit.
Problems:
1. Two plates, one having 6 ft. span and 3 ft. chord, the other having 9ft. span and 6 ft. chord, are placed
in different airstreams. The freestream velocity for the smaller plate is 100ft/sec. It is found that the
total skin-friction drag for the two plates is the same. Find the airspeed for the larger plate. Assume
laminar flow at standard sea level conditions.
Ans. V2 = 60.57 ft/sec.
2. Consider the stabilizer on a light airplane as a flat plate for the purpose of determining its skin-friction
drag. If the transition RN is 750,000, what is the skin-friction drag of a rectangular stabilizer having a
span of 1.83 m and a chord of 0.91 m at a speed of 44.7 meters per second? Assume standard sea
level conditions.
Ans. Df total = 11.65 N
3. An airplane is flying at a density altitude of 4,500 m at an ambient temperature of 234K. If the wing
chord is 1.83 m and the equivalent airspeed is 103 meters per second, what is the overall Reynolds
number of the wing?
Ans. RN = 12,131,685.15
WIND TUNNEL
A device for testing aircraft and its force components in a controlled airstream under laboratory
conditions.
## 1. Open –circuit tunnel
28
2. Closed circuit tunnel
## a. lift d. pitching moment
b. drag e. yawing moment
c. side force f. rolling moment
Importance of RN
Used in comparison of flow pattern of different bodies which are geometrically similar but not in
dimension.
## 4. both bodies must be positioned at the same eagle of attack
Problems:
1. Find RN, for an airplane wing, 4 ft. chord, moving at 130 mph through standard atmosphere.
Ans. RN = 4,850,847.34
2. Find RN for an airplane wing, 4 ft. chord, moving at 150 mph, air is 400C; barometer reading is
29 in. Hg.
Ans. RN = 4,662.149.66
3. Find the velocity at which test should be run in a wind tunnel on a model wing of 0.10 m chord
in order that RN, shall be the same as for a wing with a 1.22 m chord at 44.7 m/s. Air under
standard conditions in both cases.
Ans. VM = 545.34 m/s
4. In a variable – density wind tunnel, under what pressure should test be run on a model with a 3-
in chord, air velocity being 60 mph, in order that the RN shall be the same as for a full-size wing,
of 4ft chord, moving at 100 mph through the air? Air temperature is the same in each case.
Ans. PM = 26.67PFS
Variable Density Wind Tunnel - a wind tunnel in which the air density can be increased by
means of compressed air.
29
Flat Plates
## (a) Flat plates perpendicular to airstream
ρ
F = 1.28 AV 2
2
Where:
F = force on a flat plate normal to airstream in lb or N
ρ = freestream density in slug/ft3 or kg/m3
A = cross-sectional area of the plate in ft2 or m2
V = freestream velocity in fps or mps
Problems:
## 1. What is the total force of a 45-mph wind on a hangar door 40 ft by 25ft?
Ans. F = 6,626.70 lb
2. An automobile windshield is 40 in. wide by 15 in. high and is vertical. What is the force against
the windshield at 60 mph?
Ans. F = 49.09 lb
3. What is the force against the side of a building 70ft. long and 40 ft. high in a 90 – mph wind?
Ans. F = 74,218.99 lb
4. What force is required to push a flat plate 3 ft. by 2 ft. at a speed of 35 ft/s in a direction
perpendicular to its surface?
Ans. F = 11.18 lb
## (b) Curved Deflecting Surfaces
FH = ρAV 2 (1 − cos ε )
FV = ρAV 2 sin ε
2 2
F = FH + FV
## F = ρAV 2 2(1 − cos ε )
Where:
FH = horizontal component of force F in lb or N
FV = vertical component of force F in lb or N
F = resultant force in lb or N
V = freestream velocity in ft/sec or m/s
ε = angle of deflection in deg.
A = cross –sectional area of airstream in ft2 or m2
Problems:
1. A stream of air 60 ft. wide and 8 ft. high is moving horizontally at a speed of 100 mph. What force
is required to deflect it downward 10o without loss in speed.
Ans. F = 4,278.18 lb
30
2. A stream of air 60 ft. wide and 8 ft. high is moving horizontally at a speed of 75mph. What
forceisrequired to deflect it downward 8o without loss in speed
Ans. F = 1,926.06 lb
AIRFOIL THEORY
An airfoil is a streamlined body which when set at a suitable angle of attack, produces more lift than
drag.
## DEFINITION OF AIRFOIL GEOMETRY
MEAN CAMBER LINE – is the line joining the midpoints between the upper and lower surfaces of an airfoil
and measured normal to the chord line.
CHORD LINE – is the line joining the end points of the mean camber line.
## THICKNESS- is the height of profile measured normal to the chord line.
t
THICKNESS RATIO – is the maximum thickness to chord ratio, .
c
CAMBER – is the maximum distance of the mean camber line from the chord line.
LEADING – EDGE RADIUS – is the radius of a circle, tangent to the upper and lower surfaces, with its center
located on a tangent to the mean camber line drawn through the leading edge of this line.
## Definition of Section Forces and Moment
31
FACTORS AFFECTING THE AERODYNAMIC FORCE (F):
1. Velocity of air, V
2. Air density, p
3. Characteristic area or size, S
4. Coefficient of dynamic viscosity, µ
5. Speed of sound (compressibility effect), Va
6. Angle of attack, α
FORMULAS:
LIFT FORCE
1
l = Cl ρV 2c
2
DRAG FORCE
1
d = Cd ρV 2c = Cd q c
2
PITCHING MOVEMENT
1
m = Cm ρV 2 c 2 = C m q c 2
2
Where:
## 1 = lift force in lb/ft or N/m
d = drag force in lb/ft or N/m
m = pitching moment in ft.lb/ft or N.m/m
cl = sectional lift coefficient
cd = sectional drag coefficient
cm = pitching moment coefficient
1
q = ρV 2 = dynamic pressure in lb/ft2 or N/m2
2
ρ = density in slug/ft3 or kg/m3
V = velocity in ft/sec or m/s
c = airfoil chord in ft or m
## IMPORTANT AIRFOIL CHARACTERISTICS
The following relationships are of fundamental importance to airplane design and airplane analysis:
## LIFT CURVE :clversus α
The linear portion of the lift curve can be represented mathematically by the equation:
c l = a (α − α o )
or:
c l = c l α (α − α o )
32
Where:
## a or c l α = lift-curve slope per radian ( the theoretical value is 2π per radian)
αO= angle of attack for zero lift
## An important parameter representing the aerodynamic efficiency of airfoil is the lift-
drag ratio. The maximum cl/cd magnitude can be obtained by drawing a tangent to the cd –
clcurve.
## PITCHING MOMENT CURVE: Cm versus α or Cm versus Cl
The magnitude of cm depends on the location of the moment center. One important moment
center which is often used is the so-called aerodynamic center (a.c.). It is defined as the point about
which the moment coefficient is independent of α.
## AIRFOIL PRESSURE DISTRIBUTION
The pressure distribution is normally expressed in terms of the pressure coefficient, Cp.
P − P∞ P − P∞
Cp = =
q∞ 1
ρ ∞ V∞ 2
2
2
V
Cp = 1 −
V∞
## CRITICAL VELOCITY, Vcr
1
(γ − 1)M 2 + 2 2
Vcr = Va ∞ ∞
γ +1
## FOR AIR, γ = 1.4.
1
M 2 + 5 2
Vcr = Va ∞ ∞
6
33
CRITICAL PRESSURE, Pcr
γ
(γ − 1)M 2
Pcr = P∞ ∞ + 2 γ −1
γ +1
3.5
M 2 + 5
Pcr = P∞ ∞
6
## CRITICAL PRESSURE COEFFICIENT, C Pcr
Pcr − P∞
C p cr =
q∞
γ
2
2 (γ − 1)M ∞ + 2 γ − 1
C p cr = −1
γM ∞ 2 γ +1
## FOR AIR, γ = 1.4.
3 .5
M 2 + 5
∞ −1
6
C p cr =
0 .7 M ∞ 2
Problems:
1. An airplane is flying at an altitude of 3,500m at an airspeed of 300 m/s. Find the critical speed,
critical pressure and critical pressure coefficient.
Ans. Vcr = 322.32 m/s , Pcr = 59,981.66 Pa ,CPcr = -0.146
2. What is the critical value of the pressure coefficient for an airplane flying at 500 knots in air at 25
0
F?
Ans. CPcr = -0.494
3. For an airplane flying at 270 meters per second at 25, 0000 ft. attitude, find the critical value of the
pressure coefficient.
Ans. CPcr = -0.254
4. An airfoil has a lift-curve slope of 6.3 per radian and angle of zero lift of -20. At what angle of
attack will the airfoil develop a lift of 140lb/ft at 100mph under standard sea level conditions?
Assume c = 8ft.
Ans. α = 4.22 deg.
DESIGN OF AIRFOIL
To design an airfoil for any specific use, the following effects of airfoil geometry should be noted:
1. Camber shape will affect mainly α0 and cm. Any increase in camber will make α0 and cm more negative.
34
2. Thickness distribution will change the value of lift curve slope, a.c. location, and center of pressure
location.
4 = camber 0.04c
## 5 –digit airfoils: Example NACA 23015
2 = camber 0.02c
This design lift coefficient is 0.15 times the first digit for this series.
0.30
30 = position of camber at c = 0.15c
2
15 = maximum thickness 0.15c
## 6 –series airfoils: Example NACA 653-421
6 = series designation
## 5 = minimum pressure at 0.5c
3 = the drag coefficient is near its minimum value over a range of lift coefficient of 0.3 above
and below the design lift coefficient.
## 7- series airfoils: Example NACA 747A315
7 = series designation
4 = favorable pressure gradient on the upper surface from L.E. to 0.4c at the design lift
coefficient.
6 = favorable pressure gradient on the lower surface from L.E. to 0.7c at the design lift
coefficient.
A = a serial letter to distinguish different sections having the same numerical designation
but different mean line or thickness distribution.
## 15 = maximum thickness 0.15c
35
AIRFOIL SELECTION
In selecting an airfoil for an airplane lifting surface (wing, tail, or canard) the following considerations
are important:
## 1. Drag (for example: to obtain the highest possible cruise speed)
2. Lift –to-drag ratio a values of cl important to airplane performance
3. Thickness (to obtain the lowest possible structural weight)
4. Thickness distribution (to obtain favorable span loading and/or high fuel volume)
5. Stall characteristics (to obtain gentle stall characteristics)
6. Drag-rise behavior (associated with item 1)
7.
GEOMETRIC FACTORS AFFECTING AIRFOIL MAXIMUM LIFT AT LOW SPEEDS
The main features of airfoil design which affect wing stall and hence, the maximum lift coefficient, are:
1. THICKNESS RATIO
t
For a given thickness ratio, , C lmax very much depends on the leading-edge radius. It seen
c
that the new NASA LS = (Low speed airfoils, a thickness ratio of about 13% will produce the best
volume of maximum lift. For the newer LS airfoils the maximum lift occurs at around 15%
thickness.
## 2. LEADING EDGE RADIUS
Clmax depends not only on the thickness ratio, but also on the ratio of section thickness at 5%
t t
chord to the maximum thickness . The ratio is indicative of leading–edge radius. Therefore, a
c c
relatively large leading-edge radius is beneficial to producing large C lmax at low speeds.
## 3. CAMBER AND LOCATION OF MAXIMUM THICKNESS
The addition of camber is always beneficial to C lmax and the benefit grow with increasing camber.
The increment to maximum lift due to camber is least for sections with relatively large leading edge
t
radius (i.e., the benefit of camber grows with reducing ; and camber is more effective on thin
c
sections than on thick sections). In addition, a forward position of maximum camber produces higher
values of C lmax For example, the NACA 23012 airfoil (with 2% maximum camber at 0.15 chord) has a
C lmax of 1.79 as compared with 1.67 for NACA 4412 (with 4% camber at 0.4 chord but the same
thickness distribution) at a Reynolds number of 9 x 106
4. REYNOLDS NUMBER
For airfoils with moderate thickness ratio, there is a significant because increase in C lmax with
increasing Reynolds number. On the other hand, for thin airfoils the effect of Reynolds number is
relatively significant. In general, these Reynolds number effects are less for cambered than for
symmetrical sections. At low Reynolds number, the effect of camber is more insignificant. The
opposite is true at Reynolds number greater than 6 x 106, where camber losses some of its effects
36
EFFECT OF HIGH LIFT DEVICES ON AIRFOIL MAXIMUM LIFT
## TRAILING –EDGE FLAPS
Plain Flaps
1. Formed by hinging the rear-most par of the wing section about within the contour.
2. The main effect produced by the flap deflection is an increase in the effective camber of the wing.
3. The optimum flap angle is approximately 0.25.
4. The optimum flap angle is approximately 60o.
5. Leakage through gap 1330th of the chord resulted on a loss of 0.35 in C lmax .
6. The maximum achievable increment is C lmax is approximately 0.9.
Split Flaps
1. The unusual split flaps is formed by deflecting the aft portion of the lower surface about a hinge
point on the surfaces of the forward edge of the deflected portion.
2. The optimum flap chord ratio is approximately 0.3 for 12% thick airfoils, increasing to 0.4 or higher
for thicker airfoils.
3. The optimum flap angle is approximately 70o
4. The maximum achievable increment in C lmax is approximately 0.9.
5. The optimum thickness ratio is approximately 18%.
Slotted Flaps
1. Slotted flaps provide one or more slots between the main portion of the wing section and the
deflected flap.
2. The optimum flap chord ratio is approximately 0.9
3. The optimum flap angle is approximately 40o for single slots and 70o for double-slotted flaps.
4. The optimum thickness ratio is approximately 16%
5. The maximum achievable increment in C lmax is approximately 1.5 for single slots and 1.9 for double
slotted flaps.
Fowler Flaps
The Fowler Flaps uses the same principles as the slotted flap, except that the flap also moves
backwards in addition to a downward deflection. Thus, the effective wing area is increased!
## LEADING –EDGE DEVICES
Slats
Leading –edge slats are airfoils mounted ahead of the leading edge of the wing such as to assist in turning the
air around the leading edge at high angle of attack and thus delay leading-edge stalling. They may be either
fixed in position of retractable. The use of slats may increase C lmax by as much as 0.5.
Slots
When the slots is located near the leading edge, the configuration differs only in detail from the leading edge
slat. Additional slots may be introduced at various chordwise stations. The effectiveness of the slot derives
from its BLC (= Boundary Layer control)effect. At low angle of attack, the minimum profile drag may be
greatly increased with such slots.
37
Leading –Edge Flaps
A leading flap may be formed by bending down the forward portion of the wing section to form a droop. Other
types of leading –edge flaps are formed by extending a surface downward and forward from the vicinity of the
leading edge (Kruger flap) Leading Edge flaps reduce the severity of the pressure peak ordinarily associated
with high angle of attack and thereby delay separation.
## BOUNDARY LAYER CONTROL
Higher maximum lift coefficient can also be achieved by boundary layer control (BLC). The idea may involve
injecting high-speed for parallel to the wall (called “blowing”), removing the low energy boundary layer flow
by “suction”, or both. Blowing is done to re-energize the boundary layer flow to delay the separation, while
suction is equivalent to eliminating the low energy shear layer.
38
39 | 18,264 | 51,293 | {"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} | 3.25 | 3 | CC-MAIN-2020-10 | latest | en | 0.485895 |
http://ieeexplore.ieee.org/xpl/articleDetails.jsp?reload=true&arnumber=1514453 | 1,430,958,475,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430459917448.46/warc/CC-MAIN-20150501055837-00005-ip-10-235-10-82.ec2.internal.warc.gz | 94,608,383 | 19,381 | In an earlier paper, the authors introduced the notion of safety control of stochastic discrete event systems (DESs), modeled as controlled Markov chains. Safety was specified as an upper bound on the components of the state probability distribution, and the class of irreducible and aperiodic Markov chains were analyzed relative to this safety criterion. Under the assumption of complete state observations: 1) the authors identified the set of all state-feedback controllers that enforce the safety specification, for all safe initial probability distributions and 2) for any given state-feedback controller, the authors constructed the maximal invariant safe set (MISS). In this paper, the authors extend the work in several ways: 1) safety is specified in terms of both upper and lower bounds; 2) we consider a larger class of Markov chains that includes reducible and periodic chains; 3) we present a more general iterative algorithm for computing the MISS, which is quite flexible in its initialization; 4) we obtain an explicit upper bound for the number of iterations needed for the algorithm to terminate. Note to Practitioners-The paper studies "safety" control of stochastic systems modeled as Markov chains. Safety is defined as a requirement that the probability distribution in each state remain bounded between an upper and a lower bound. For example, a financial investment policy should be such that the probability of ever being bankrupt is bounded below by a positive number. Prior works on control of Markov chains have addressed optimality but not safety. A condition is obtained under which a controlled Markov chain is guaranteed to be safe at all times. For those chains that do not satisfy such a condition, a maximal subset of the safe set of distributions is computed so that if the chain is initialized with a distribution in that maximal subset, it remains safe all the times. A condition is obtained under which such a maximal set is nonempty. The computation of such a maximal set is iterative and we provide a condition under which the computation terminates in a finite number of iterations. Manufacturing system examples are included to illustrate the results.
Published in:
Automation Science and Engineering, IEEE Transactions on (Volume:2 , Issue: 4 ) | 439 | 2,293 | {"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} | 2.5625 | 3 | CC-MAIN-2015-18 | latest | en | 0.932778 |
https://www.exceldemy.com/excel-countif-wildcard/ | 1,726,075,323,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651390.33/warc/CC-MAIN-20240911152031-20240911182031-00823.warc.gz | 736,321,047 | 51,655 | How to Use the COUNTIF function with Wildcards in Excel -7 Methods
This is an overview.
The sample dataset has 8 rows and 2 columns. All cells are in General format.
Method 1 – Use the COUNTIF function with a Wildcard to Count Text Values in Excel
Steps:
• Select C13.
• Enter the formula below.
`=COUNTIF(B5:B11,"*")`
• Press Enter to see the result.
Method 2 – Using the COUNTIF with a Wildcard to count Numeric Values Only in Excel
Steps:
• Enter the formula in C13.
`=COUNTIF(B5:B11,"<>*")`
• Press Enter to see the result.
Method 3 – Inserting the COUNTIF “Starts with” Wildcard in Excel
Steps:
• Enter the formula in C13.
`=COUNTIF(B5:B11,"GHY*")`
• Press Enter to see the result.
Method 4 – Applying the COUNTIF “Ends with” Wildcard in Excel
To count the cells that end with the characters “GH”:
Steps:
• Enter the formula in C13.
`=COUNTIF(B5:B11,"*GH")`
• Press Enter to see the result.
Method 5 – Using the COUNTIF “Contains” Wildcard in Excel
To count the cells that contain the value “256124FK”:
Steps:
• Enter the formula in C13.
`=COUNTIF(B5:B11,"*256124FK*")`
• Press Enter to see the result.
Method 6 – Applying the COUNTIF “?” Wildcard in Excel
“HJI???GH” will search for values that start with HJI and end with GH.
Steps:
• Enter the formula in C13.
`=COUNTIF(B5:B11,"*HJI???GH")`
• Press Enter to see the result.
Method 7 – Inserting the COUNTIF “~ (tilde)” Character Wildcard in Excel
*~?*” will find values containing a question mark.
Steps:
• Enter the formula in C13.
`=COUNTIF(B5:B11,"*~?*")`
• Press Enter to see the result. | 458 | 1,576 | {"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} | 3.328125 | 3 | CC-MAIN-2024-38 | latest | en | 0.76793 |
http://oeis.org/A192232/internal | 1,582,393,346,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145708.59/warc/CC-MAIN-20200222150029-20200222180029-00140.warc.gz | 107,811,701 | 4,964 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A192232 Constant term of the reduction of n-th Fibonacci polynomial by x^2 -> x+1. (See Comments.) 279
%I
%S 1,0,2,1,6,7,22,36,89,168,377,756,1630,3353,7110,14783,31130,65016,
%T 136513,285648,599041,1254456,2629418,5508097,11542854,24183271,
%U 50674318,106173180,222470009,466131960,976694489,2046447180,4287928678,8984443769,18825088134
%N Constant term of the reduction of n-th Fibonacci polynomial by x^2 -> x+1. (See Comments.)
%C Polynomial reduction: an introduction
%C ...
%C We begin with an example. Suppose that p(x) is a polynomial, so that p(x)=(x^2)t(x)+r(x) for some polynomials t(x) and r(x), where r(x) has degree 0 or 1. Replace x^2 by x+1 to get (x+1)t(x)+r(x), which is (x^2)u(x)+v(x) for some u(x) and v(x), where v(x) has degree 0 or 1. Continuing in this manner results in a fixed polynomial w(x) of degree 0 or 1. If p(x)=x^n, then w(x)=x*F(n)+F(n-1), where F=A000045, the sequence of Fibonacci numbers.
%C In order to generalize, write d(g) for the degree of an arbitrary polynomial g(x), and suppose that p, q, s are polynomials satisfying d(s)<d(q). By the division algorithm, there exists a unique pair t and r of polynomials such that p=q*t+r and d(r)<d(q). Replace q by s to get s*t+r, which is q*u+v for some u and v, where d(v)<d(q). Continue applying q->s in this manner until reaching w such that d(w)<d(q). We call w the reduction of p by q->s.
%C The coefficients of (reduction of p by q->s) comprise a vector of length d(q)-1, so that a sequence p(n,x) of polynomials begets a sequence of vectors, such as (F(n), F(n-1)) in the above example. We are interested in the component sequences (e.g., F(n-1) and F(n)) for various choices of p(n,x).
%C Following are examples of reduction by x^2->x+1:
%C n-th Fibonacci p(x) -> A192232+x*A112576
%C n-th cyclotomic p(x) -> A192233+x*A051258
%C n-th 1st-kind Chebyshev p(x) -> A192234+x*A071101
%C n-th 2nd-kind Chebyshev p(x) -> A192235+x*A192236
%C x(x+1)(x+2)...(x+n-1) -> A192238+x*A192239
%C (x+1)^n -> A001519+x*A001906
%C (x^2+x+1)^n -> A154626+x*A087635
%C (x+2)^n -> A020876+x*A030191
%C (x+3)^n -> A192240+x*A099453
%C ...
%C Suppose that b=(b(0), b(1),...) is a sequence, and let p(n,x)=b(0)+b(1)x+b(2)x^2+...+b(n)x^n. We define (reduction of sequence b by q->s) to be the vector given by (reduction of p(n,x) by q->s), with components in the order of powers, from 0 up to d(q)-1. For k=0,1,...,d(q)-1, we then have the "k-sequence of (reduction of sequence b by q->s)". Continuing the example, if b is the sequence given by b(k)=1 if k=n and b(k)=0 otherwise, then the 0-sequence of (reduction of b by x^2->x+1) is (F(n-1)), and the 1-sequence is (F(n)).
%C ...
%C For selected sequences b, here are the 0-sequences and 1-sequences of (reduction of b by x^2->x+1):
%C b=A000045, Fibonacci sequence (1,1,2,3,5,8,...) yields
%C 0-sequence A166536 and 1-sequence A064831.
%C b=(1,A000045)=(1,1,1,2,3,5,8,...) yields
%C 0-sequence A166516 and 1-sequence A001654.
%C b=A000027, natural number sequence (1,2,3,4,...) yields
%C 0-sequence A190062 and 1-sequence A122491.
%C b=A000032, Lucas sequence (1,3,4,7,11,...) yields
%C 0-sequence A192243 and 1-sequence A192068.
%C b=(A000217, triangular sequence (1,3,6,10,...) yields
%C 0-sequence A192244 and 1-sequence A192245.
%C b=(A000290, squares sequence (1,4,9,16,...) yields
%C 0-sequence A192254 and 1-sequence A192255.
%C More examples: A192245-A192257.
%C ...
%C (1) If s(n,x)=(reduction of x^n by q->s) and
%C p(x)=p(0)x^n+p(1)x^(n-1)+...+p(n)x^0, then
%C (reduction of p by q->s)=p(0)s(n,x)+p(1)s(n-1,x)
%C +...+p(n-1)s(1,x)+p(n)s(0,x). See A192744.
%C (2) For any polynomial p(x), let P(x)=(reduction of p(x)
%C by q->s). Then P(r)=p(r) for each zero r of
%C q(x)-s(x). In particular, if q(x)=x^2 and s(x)=x+1,
%C then P(r)=p(r) if r=(1+sqrt(5))/2 (golden ratio) or
%C r=(1-sqrt(5))/2.
%H Vincenzo Librandi, <a href="/A192232/b192232.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-1,-1).
%F Empirical G.f.: -x*(x^2+x-1)/(x^4+x^3-3*x^2-x+1). - _Colin Barker_, Sep 11 2012
%F The above formula is correct. - _Charles R Greathouse IV_, Jan 08 2013
%F a(n) = A265752(A206296(n)). - _Antti Karttunen_, Dec 15 2015
%F a(n) = A112576(n) -A112576(n-1) -A112576(n-2). - _R. J. Mathar_, Dec 16 2015
%e The first four Fibonacci polynomials and their reductions by x^2->x+1 are shown here:
%e F1(x)=1 -> 1 + 0x
%e F2(x)=x -> 0 + 1x
%e F3(x)=x^2+1 -> 2+1x
%e F4(x)=x^3+2x -> 1+4x
%e F5(x)=x^4+3x^2+1 -> (x+1)^2+3(x+1)+1 -> 6+6x.
%e From these, read A192232=(1,0,1,1,6,...) and A112576=(0,1,1,4,6,...).
%t q[x_] := x + 1;
%t reductionRules = {x^y_?EvenQ -> q[x]^(y/2), x^y_?OddQ -> x q[x]^((y - 1)/2)};
%t t = Table[FixedPoint[Expand[#1 /. reductionRules] &, Fibonacci[n, x]], {n, 1, 40}];
%t Table[Coefficient[Part[t, n], x, 0], {n, 1, 40}]
%t (* A192232 *)
%t Table[Coefficient[Part[t, n], x, 1], {n, 1, 40}]
%t (* A112576 *)
%t (* _Peter J. C. Moses_, Jun 25 2011 *)
%t LinearRecurrence[{1, 3, -1, -1}, {1, 0, 2, 1}, 60] (* _Vladimir Joseph Stephan Orlovsky_, Feb 08 2012 *)
%o (PARI) Vec((1-x-x^2)/(1-x-3*x^2+x^3+x^4)+O(x^99)) \\ _Charles R Greathouse IV_, Jan 08 2013
%Y Cf. A168561, A192233 - A192240, A192744.
%Y Cf. A206296, A265398, A265399, A265752, A265753.
%K nonn,easy
%O 1,3
%A _Clark Kimberling_, Jun 26 2011
%E Example corrected by _Clark Kimberling_, Dec 18 2017
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Last modified February 22 12:42 EST 2020. Contains 332136 sequences. (Running on oeis4.) | 2,292 | 6,017 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2020-10 | latest | en | 0.849525 |
https://money.stackexchange.com/questions/146769/time-value-of-money-interest-paid-on-mortgage | 1,719,002,912,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862157.88/warc/CC-MAIN-20240621191840-20240621221840-00742.warc.gz | 348,010,962 | 38,317 | # Time Value of Money — Interest Paid on Mortgage
You you bought a house worth \$328,000. You paid 25% of the purchase price in cash and arranged a 25 year mortgage with a rate of 4.0% compounded semi-annually for the remaining balance. The mortgage has an amortization period of 25 years. How much interest will you pay in the first 7 years (assuming that the first payment is made at the end of the first month)?
So far, I have that PV=\$328,000 * 0.75=\$246,000, r=0.00330589 (using effective rate formula: (1+r)^6=(1+0.04/2) ) and n=25 * 12=300. Using the present value of an ordinary annuity:
PV=PMT[(1-(1+r)^-n)/r]
I solved for PMT and got PMT=\$1294.009652 for the monthly payments. The number of payment periods still remaining after 7 years is 18*12=216. The PV of the outstanding balance (FV of 246,000 - FV of 84 PMTs) is \$199,539.6457. However, I don't really know what to do after that. The correct answer is \$62,236.46 but I don't know how they got that. How do I calculate the interest paid in the first 7 years?
• What is the background of this problem? I ask because I am unaware of mortgages that compound semi-annually. Typical is a calculation of interest and payment each month. Is this a homework problem? Commented Dec 7, 2021 at 22:40
• Yes it is a homework problem for a class. I do apologize if this is the wrong place to ask this type of question. Commented Dec 7, 2021 at 22:52
• @JTP-ApologisetoMonica Many fixed-rate mortgages in Canada have semi-annual compounding for the rate, even though payments are typically monthly. Commented Dec 8, 2021 at 12:09
Edit:
Actually, since you already got the Outstandind Principal right after 84 months = \$199,539.6457, you knew that:
Total Principal Paid = 246,000 - 199,539.6457 = \$46,460.3543
Total Interest Paid = 84 x 1,294.009652 - 46,460.3543 = \$62,236.456468
Then use second formula at: https://en.wikipedia.org/wiki/Mortgage_calculator#Total_interest_paid_formula
• P = 246,000
• r = 0.00330589
• c = PMT = 1294.009652
• N = 84
You will get \$62,236.46
You may also play with the relationship:
Total Interest Paid + Total Principal Paid = 84 x 1294.009652
• Very helpful answer (the edit especially)! Thank you very much for your help! Commented Dec 8, 2021 at 21:07
With `s` as the loan amount
``````hse = 328000
s = hse (1 - 0.25) = 246000
``````
and `r` the monthly rate
``````i = 0.04
r = (1 + i/2)^(2/12) - 1 = 0.00330589
n = 25*12 = 300
``````
the payment amount `d` is
``````d = r (1 + 1/((1 + r)^n - 1)) s = 1294.01
``````
The principal balance in month `x` is given by `p(x)` (see link)
``````p(x) = (d + (1 + r)^x (r s - d))/r
``````
and the interest paid in month `x` is given by `int(x)`
``````int(x) = p(x - 1) r
= d + (1 + r)^(x - 1) (r s - d)
``````
The accumulated interest to month `x` is given by `interestsofar(x)`
(formula obtained by induction, summing `int(k)` from `k = 1` to `x`)
``````interestsofar(x) = (d - d (1 + r)^x - r s + r (1 + r)^x s + d r x)/r
interestsofar(7*12) = 62236.46
`````` | 1,003 | 3,028 | {"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-2024-26 | latest | en | 0.94158 |
https://www.physicsforums.com/search/5323772/ | 1,638,678,839,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363135.71/warc/CC-MAIN-20211205035505-20211205065505-00409.warc.gz | 1,018,155,221 | 15,290 | # Search results
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20. ### Which Electromagnet Design Works Best?
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27. ### Electric Car Engine Weight
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29. ### Intermediate Mechanics Practice Test Problem
Hello everyone. The following is from the practice test given to me for my intermediate mechanics final. I'm, at the moment, completely lost on what to do. If you have even just a few ideas of what I should be doing they would be apperciated. If you want to give a detailed solution it would... | 2,069 | 9,167 | {"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} | 2.59375 | 3 | CC-MAIN-2021-49 | latest | en | 0.956081 |
https://neos-guide.org/case-studies/cs/domino-artwork/ | 1,725,962,057,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651241.17/warc/CC-MAIN-20240910093422-20240910123422-00827.warc.gz | 385,031,539 | 19,591 | # Domino Artwork
This Domino Artwork case study describes the optimization model that underlies the NEOS Domino solver, which constructs pictures from target images using complete sets of double-nine dominoes. Complete sets of double-nine dominoes include one domino for each distinct pair of dot values from 0 to 9. The NEOS Domino solver is an implementation of the work of Robert Bosch of Oberlin College.
## Background Information
In his 2006 article “Opt Art”, Robert Bosch of Oberlin College describes some applications of optimization in the area of art. Domino artwork is a type of photomosaic, a picture made up of many smaller pictures. The small pictures can be seen up close but, at a distance, they merge into a recognizable image.
To create a photomosaic, we partition a target image and a blank canvas into a grid. Then, given a set $$F$$ of building-block photographs, we place each photograph from $$F$$ onto one square of the blank canvas grid with the goal of arranging the building blocks to resemble the target image as closely as possible. Optimization is well-suited to assigning the building blocks to the blank canvas grid but choosing the set of building blocks requires an artist’s touch!
Here, we describe domino artwork, which is a special type of photomosaic in which the building blocks are dominoes. The building-block set in the Domino solver is comprised of complete sets of double-nine dominoes; double-nine domino sets contain one domino for each distinct pair of dot values from 0 to 9.
## Mathematical Formulation
In this section, we present Bosch’s integer programming model for constructing domino artwork images. Instead of a set of photographs $$F$$, we have a set $$D=\lbrace d = (d_1, d_2): 0 \leq d_1 \leq d_2 \leq 9\rbrace$$ of double-nine dominoes. Each domino $$d=(d_1, d_2) \in D$$ is black with $$d_1$$ dots on one square and $$d_2$$ dots on the other square. The domino has a brightness value associated with each half; the value corresponds to the number of dots, so it is measured on a scale from 0 to 9, black to white.
Each domino can be positioned horizontally or vertically covering two squares of the canvas. A complete set of double-nine dominoes contains 55 dominoes, and exactly $s$ complete sets must be used. Therefore, we need to partition the target image and the canvas into $$m$$ rows and $$n$$ columns such that $$mn = 110s$$.
The decision variable is a yes-no decision for each possible assignment of a domino $$d$$ to a pair of adjacent squares of the canvas. This is more complicated than a yes-no decision associated with assigning a single photograph to a single square on a canvas. However, we can introduce a binary variable $$x_{dp}$$ for each domino $$d \in D$$ and each pair $$p \in P$$ where $$P$$ is defined as:
$P = \big\{ \lbrace(i,j),(i+1,j) \rbrace : 1\leq i\leq m-1, i \leq j \leq n \big\} \\ \cup \big\{ \lbrace (i,j),(i,j+1)\rbrace : 1\leq i \leq m, 1 \leq j \leq n – 1 \big\}.$
Associated with each assignment of a domino $$d$$ to a pair of squares $$p$$ is a cost $$c_{dp}$$. The cost $$c_{dp}$$ is calculated based on how close the dots of domino $$d$$ match up with the brightness values of the target image. Let $$\beta_{ij}\in[-0.5,9.5]$$ represent the brightness value of square $$(i,j)$$ on the grid representation of target image, and recall that $$d=(d_1,d_2)$$ represents a domino with $$d_1$$ dots on one square and $$d_2\geq d_1$$ dots on the other square. An explicit expression for $$c_{dp}$$ with $$p=\lbrace (i_1,j_1), (i_2,j_2)\rbrace\in P$$ is
$c_{dp} = \min\lbrace \left(d_1-\beta_{i_1j_1}\right)^2 + \left(d_2-\beta_{i_2j_2}\right)^2, \left(d_1-\beta_{i_2j_2}\right)^2 + \left(d_2-\beta_{i_1j_1}\right)^2 \rbrace.$
The cost is calculated by adding the squares of the differences between dot values and actual brightness values for each orientation. $$c_{dp}$$ becomes the minimum of these values.
Then, the domino optimization model can be written as:
$\begin{array}{rrcll} \text{minimize} & \sum_{d\in D}\sum_{p\in P} c_{dp}x_{dp} & & & \\ \text{subject to} & \sum_{p\in P} x_{dp} &= &s & \forall d\in D, \\ & \sum_{d\in D}\sum_{\substack{p\in P: \\ p\ni (i,j)}} x_{dp} & = & 1 & \forall 1\leq i\leq m, 1\leq j\leq n, \\ & x_{dp}\in\{0,1\} & & & \forall d\in D, p\in P. \end{array}$
The objective function measures the total cost of the domino arrangement. The first set of constraints ensures that all of the dominoes are placed on the canvas. The second set of constraints ensures that each square is covered by exactly one domino. The domino optimization model is an assignment problem with side constraints.
Assignment problems (without side constraints) are easy to solve in the sense that they exhibit the integrality property. That is, if we relax the integrality restrictions on the decision variables and solve the problem as a linear program, the optimal solution is guaranteed to be integer-valued. Adding side constraints makes the problem harder to solve in theory, however, many instances still can be solved easily.
## Implementation Details
The main contribution of the NEOS Domino solver is that anyone can create his/her own domino artwork image! Underlying the Domino solver is MATLAB code that processes a target image, creates the GAMS model instance, submits the model to NEOS, reads the solution returned and constructs the image file with only one click of a button. The NEOS Domino solver was developed by Eli Towle as his final project in ISyE/CS 635: Tools and Environments for Optimization at the University of Wisconsin in Madison.
The Domino solver takes as input a JPEG image. There are two optional parameters.
• Quality (low, medium, high): the default quality is medium. Selecting low produces a lower quality output image more quickly. Selecting high produces a high-quality output image but the solver may require a long time to produce the results!.
• Color (black, white): the default domino set is black dominoes with white dots. Selecting white will result in an image with white dominoes with black dots.
The optimization model solved by the NEOS Domino solver includes one change from the original model to improve the speed of constructing the final image. The original model defined the set of valid location assignments $$P$$ as shown above and then defined the cost parameter $$c_{dp}$$ based on either possible orientation of domino $$d$$ assigned to location $$p\in P$$. However, the decision variable $$x_{dp}$$ does not allow for reverse orientations of dominoes. In other words, an optimal assignment of domino $$(0,9)$$ at location $$\lbrace(1,1),(1,2)\rbrace$$ could be flipped so that the 9 comes first, but the solution will not make the distinction between these orientations. Therefore, we introduce parameters in the GAMS optimization model that allow orientations to be inferred based on the value of $$c_{dp}$$. That is, if
\begin{align*} \left(d_1-\beta_{i_1j_1}\right)^2 + \left(d_2-\beta_{i_2j_2}\right)^2 \leq \left(d_1-\beta_{i_2j_2}\right)^2 + \left(d_2-\beta_{i_1j_1}\right)^2, \end{align*} then the proper orientation is $$(d_1,d_2)$$. Otherwise, the orientation is $$(d_2,d_1)$$.
As an alternative, we could expand the set $$P$$ to $$\hat{P}$$ to differentiate between orientations in the solution. In this case, we define
\begin{align*} \hat{P} = & \lbrace (t_1,t_2) : (t_1,t_2)\in P \text{ or } (t_2,t_1)\in P \rbrace, \end{align*} where $$t_1$$ and $$t_2$$ are tuples. The costs could similarly be altered to reflect the cost of each individual orientation rather than being expressed as the minimum of the two. This expansion effectively doubles the number of decision variables in the model, resulting in slower solving times. As an example, we show below a comparison of solution times for an image of Taylor Swift.
Comparison of solution times for two methods of determining domino orientation
Sets Infer from $$c_{dp}$$ Augment $$P$$ to $$\hat{P}$$
18 11.85s 15.94s
72 86.32s 211.70s
162 349.56s 668.46s
With minimal additional effort, calculating the correct orientations by examining $$c_{dp}$$ after the solve is strongly preferred over doubling the number of acceptable orientations.
## Examples
The first image is the target image and the second image is the image created by the NEOS Domino solver (using 896 complete sets of dominoes).
The first image is the target image of Taylor Swift and the second image is the image, created by the NEOS Domino solver, constructed in dominoes! The physical construction of the domino portrait is captured here in a time-lapse video. | 2,257 | 8,589 | {"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": 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} | 3.828125 | 4 | CC-MAIN-2024-38 | latest | en | 0.889401 |
https://www.doubtnut.com/qna/327417004 | 1,719,106,489,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862425.28/warc/CC-MAIN-20240623001858-20240623031858-00183.warc.gz | 664,556,969 | 45,187 | # For H2S,Ka1=10−7,Ka2=10−14,CoS=4×10−21,Ag2S=6.3×10−50 Calculate difference in PH for precipitation of MnS and CoS.
Video Solution
Text Solution
Verified by Experts
## The correct Answer is:5.4
|
Step by step video, text & image solution for For H_(2)S, K_(a_(1))=10^(-7), K_(a_(2))=10^(-14),CoS=4xx10^(-21), Ag_(2)S=6.3xx10^(-50) Calculate difference in PH for precipitation of MnS and CoS. by Chemistry experts to help you in doubts & scoring excellent marks in Class 12 exams.
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## In equalitative analysis, cations of graph II as well as group IV both are precipitated in the form of sulphides. Due to low value of Ksp of group II sulphides, group reagent is H2S in the presence of dil. HC1, and due to high value of Ksp of group IV sulphides, group reagent is H2S in the presence of NH4OH and NH4C1. In a solution containing 0.1M each of Sn2+,Cd2+, and Ni2+ ions, H2Sgas is passed. KspofSnS=8×10−29,KspofCdS=1510−28,KspofNiS−3×10−21,K1ofH2S=1×10−7,K2ofH2S=1×10−14 At what value of pH,NiS will start to precipitate?
A12.76
B7
C1.24
D4
• Question 2 - Select One
## If Ka1>Ka2 of H2SO4 are 10−2 and 10−6 respectively then:
AKa1>Ka2 because it is easy to abstract H+ from H2SO4 and less easy to abstract H+ from HSO4
BKa1 and Ka2 may be measured in acetic acid
CKa1 and Ka2 are measured in H2O
Dthe H+ ion conc. Of 0.01M H2SO4 will be less than 0.02M
• Question 3 - Select One
## If Ka of HCN=4×10−10 , then the pH of 2.4×10−1 molar HCN(aq) is
A2x105
B4.7
C2.0
D5.0
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Algebra Pre-Algebra and Basic Algebra Math Forum
May 18th, 2011, 03:36 PM #1 Newbie Joined: May 2011 Posts: 3 Thanks: 0 How do i find the area of the shaded region? https://skitch.com/am26/fyftf/cam How do I find the area of the shaded region? Thanks
May 18th, 2011, 04:41 PM #2 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,900 Thanks: 1094 Math Focus: Elementary mathematics and beyond Re: How do i find the area of the shaded region? Formula for the area of a circular sector: $A\,=\,\frac{\theta^{o}}{360}\pi r^2$ Make substitutions: $A\,=\,\frac{360^{o}\,-\,147^{o}}{360}\pi (9)^2\,=\,\frac{1917\pi}{40}$
May 18th, 2011, 04:59 PM #3 Newbie Joined: May 2011 Posts: 3 Thanks: 0 Re: How do i find the area of the shaded region? Oh, wow, thanks SO much!
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16728DE CENTRO GRAVIT. SOLID. uel coni portionis axis à centro grauitatis ita diui
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quæ partis, quæ ad baſim, ſit tripla.
Sit pyramis, cuius baſis triangulum a b c; axis d e; & gra
uitatis centrum _K_.
Dico lineam d k ipſius _K_ e triplam eſſe.
trianguli enim b d c centrum grauitatis ſit punctum f; triã
guli a d c centrũ g;
& trianguli a d b ſit h: & iungantur a f,
b g, c h.
Quoniam igitur centrũ grauitatis pyramidis in axe
cõſiſtit:
ſuntq; d e, a f, b g, c h eiuſdẽ pyramidis axes: conue
1117. huíus nient omnes in idẽ punctũ _k_, quod eſt grauitatis centrum.
Itaque animo concipiamus hanc pyramidem diuiſam in
quatuor pyramides, quarum baſes ſint ipſa pyramidis
triangula;
& axis pun-
88[Handwritten note 8] ctum k quæ quidem py-
ramides inter ſe æquales
ſunt, ut demõſtrabitur.
Ducatur enĩ per lineas
d c, d e planum ſecãs, ut
ſit ipſius, &
baſis a b c cõ
munis ſectio recta linea
c e l:
eiuſdẽ uero & triã-
guli a d b ſitlinea d h l.
erit linea a l æqualis ipſi
l b:
nam centrum graui-
tatis trianguli conſiſtit
in linea, quæ ab angulo
ad dimidiam baſim per-
ducitur, ex tertia deci-
ma Archimedis.
quare
221. ſexti. triangulum a c l æquale
eſt triangulo b c l:
& propterea pyramis, cuius baſis trian-
gulum a c l, uertex d, eſt æqualis pyramidi, cuius baſis b c l
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idem uertex. pyramides enim, quæ ab eodẽ
335. duode-
cimi. | 748 | 1,723 | {"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} | 2.875 | 3 | CC-MAIN-2024-10 | latest | en | 0.13658 |
https://www.physicsforums.com/threads/finding-the-electric-field-at-the-origin.886806/ | 1,508,520,546,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824226.31/warc/CC-MAIN-20171020154441-20171020174441-00322.warc.gz | 957,558,298 | 16,439 | # Finding the Electric field at the Origin
1. Sep 26, 2016
### NkaujHelp
1. The problem statement, all variables and given/known data
I'm trying to find the electric field at the origin <0, 0, 0> due to a quarter of a ring with +Q charges on it. The quarter of the ring lies on the +y-axis. Due to symmetry, the x-components of the electric field will cancel out, leaving only the y-component of the electric field pointing in the negative y-direction. So my problem with this is finding the correct limit of integration to use for this problem. At first, I used the limits of integration from 3pi/4 to pi/4. When I used this on the x-component, it didn't give me an exact 0. It gave me like 0.2738 something like that. The answer for the x-components should have been exactly zero. So I don't know if this was the right limits of integration or is there other limits that I can use that I can't think of to make the x-components zero. If my limits of integration that I use doesn't make the x-components zero, then my answer for the electric field for the y-component would be wrong too. I know how to do everything else, but it's the limits that I am stuck on.
2. Relevant equations
E = [(kΔq)/(πr^2)] * r hat
r hat = <-cosσ, -sinσ, 0>
Δq = (2(+Q)Δσ)/π
3. The attempt at a solution
Integrating for the x-component: ∫[2k(+Q)(-cosσdσ)]/[πR^2] = 2k(+Q)/πR^2 ∫-cosσdσ
2. Sep 26, 2016
### TSny
OK, except I think you have a typo in the expression for E. Should there be $\pi$ in the denominator?
This looks correct. (Now there should be the factor of $\pi$ as you have it.)
Why "from 3pi/4 to pi/4" rather than "from pi/4 to 3pi/4"?
Your integral should give exactly zero. Don't forget to put your calculator in radian mode if you are using your calculator to evaluate something like $\sin(\pi/4)$ (which you probably know without needing a calculator). | 520 | 1,861 | {"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} | 3.71875 | 4 | CC-MAIN-2017-43 | longest | en | 0.929046 |
https://hobatuxuqawerir.holidaysanantonio.com/kinematics-classical-mechanics-and-velocity-instantaneous-48718fi.html | 1,621,046,969,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991812.46/warc/CC-MAIN-20210515004936-20210515034936-00253.warc.gz | 323,904,489 | 4,938 | # Kinematics classical mechanics and velocity instantaneous
Therefore, it is convenient to introduce some other, approximate, laws of forces which can in principle be obtained from the fundamental forces. Chu did not utilize a chain model with an extended Kalman filter as Jackson et al. Its solution constitutes the basic problem of dynamics of a mass point.
A horizontal bar rotates with the constant angular velocity. Let us get back to the experiment in which we compared the accelerations of two different bodies subjected to the action of an equally stretched spring. By definition of 3D positional data filtering, the resultant velocities and accelerations will be smoothed curves, which eliminates some of the IMU content that is important for accelerations and decelerations.
This is critical information as it is directly related to joint torques, which has a direct effect on UCL loading rates.
Experience shows that every body "resists" any effort to change its velocity, both in magnitude and direction. In contrast, a quantity that has only magnitude and no direction, such as temperature or time, is called a scalar.
This significant physical fact underlies one of the most fundamental generalizations of Newtonian mechanics, Newton's second law: Table 3 shown below is taken from Lapinski et al. The point was set in motion with the velocity vo.
This approximation is surprisingly accurate. For example, a ball tossed up will be under the influence of a negative downward acceleration due to gravity. The Fundamental Laws of Newtonian Dynamics Investigating various kinds of motion in practice, we discover that in inertial reference frames any acceleration of a body is caused by some other bodies acting on it.
This again is much lower than the typical optical camera based values presented earlier.
The magnitudes of the vectors coo and o' can be easily found from Fig. One particularly important application is the solar system. Inseveral months after Galileo died, Isaac Newton was born. Also called applied kinematics.
Let us see how Eq. The study of motions at relativistic velocities, however, showed that the laws of force should be modified to make the forces dependent on the velocity of a mass point in an intricate way.
Experience shows that in terms of Newtonian mechanics a mass determined that way possesses the following two important properties: Besides, all the other laws of classical mechanics can be derived from Newton's laws.
With a marker at the tip of the acromion as well, they had 4 markers dedicated to measuring scapular kinematics.
Figure 7 shows a segment of a circle of radius L. Find the equation of its path y x. There is clearly a difference between the two systems. While providing a dual system comparison is an excellent methodology for comparing the two systems, it is somewhat limited as no methodology is provided for how they calculated optical camera-based kinematic outputs.
Requires Flash 7; file size is k. Any other reference frame moving rectilinearly and uniformly relative to the heliocentric frame is also inertial. The equations are therefore mathematically identical and have the same solution—i. Angular velocity of precession. Ask Question. Browse other questions tagged newtonian-mechanics rotational-dynamics rotational-kinematics classical-mechanics or ask your own question.
asked. 4 years, 5 months ago. viewed. times. active. 4 years, 5 months ago. kinematic analysis of mechanical system - position, velocity and acceleration analyses; rigid body kinematics; Classical Kinematic Approach. Computational Kinematic Approach.
Formulation of the Driving Constraints. Transcript of Hinges and Catches: The Mechanics of Metalsmithing. kinematics: classical mechanics which analyzes motion of points, and systems of bodies without motion = s/r s is arc length r is radius Angular Velocity: W= / t W is angular velocity T is time Torque: T= r x f T is torque r is position from axis x is the cross product.
The wide range in +/- angular velocity values seems very high given the repeatability of an individual’s pitching mechanics. The majority of the studies in this table used similar methodologies to ASMI which result in mathematical estimations of actual in vivo humeral kinematics.
Topic Disc # Kinematics, Classical Mechanics & Gravitation 01, 02, 03, 04, 05 Oscillations, Waves and Sound. One-Dimensional Kinematics Monday, August 29, Instantaneous Velocity Definition: () This means that we evaluate the average velocity over a shorter and shorter period of time; as that time becomes infinitesimally small, we have the instantaneous velocity.
Kinematics classical mechanics and velocity instantaneous
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Homework
# The United States is becoming more health conscious, and as
Attachment: 2013-10-24_033715_ip_unit_3_question.pdf PART 1 PART 2 The...
Attachment: 2013-10-24_033715_ip_unit_3_question.pdf
PART 1
PART 2
The United States is becoming more health conscious, and as a result, the problem of obesity has gotten more attention. The Body Mass Index (BMI), relates a person’s height and weight, and is often used to determine if someone is overweight. The table below tells the weight status for a given BMI. BMI Weight Status Below 18.5 Underweight 18.5 - 24.9 Normal 24.9 - 29.9 Overweight 29.9 and above Obese The BMI is calculated using the formula: BMI = 703*w / h2 where w is the weight in pounds and h is the height in inches. Solving this formula for h, we see that h = sqrt[703w / BMI] 1. Find the weight of your favorite celebrity. This could be a movie or television personality, athlete, politician or even yourself. 2. Using the weight from part 1, determine the height the celebrity would need to be in order to fall into each of the four weight status categories listed in the table. In other words, select a BMI less than 18.5 (any value, you make it up) and find “h”; then repeat using a new BMI in the range from 18.5 to 24.9, and so on. 3. Using the Internet or other library resource, find the actual height of the celebrity. 4. Determine his or her actual weight status (underweight, normal, overweight or obese) using the original BMI formula at the top of the instructions. 5. How tall would he or she need to be for the normal weight status? 6. Would you consider him or her to actually be in the weight status this formula says based on his or her actual height and weight and considering his or her other physical characteristics? Why or why not? Think about why there may be differences in your calculations and the actual figures. 7. The BMI formula was created by a Belgian Statistician (not a physician), Lambert Adolphe Quetelet, in the early 1830s. Do you think BMI is a fair indication of a person’s weight classification? Why or why not? 8. Summarize your findings in writing using proper style and grammar.
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Ray Atkinson :
Do you have a celebrity in mind?
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Let me know when you read this. I will get to work on the rest of the assignment Thursday morning, and it will not take me long.
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No, anyone is fine. Thank you
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Quite alright. I will get to work on it in the morning. 0031 local time here. I dont want to make mistakes.
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Update: I have the pdf file done. Now to the BMI calculations.
Ray Atkinson :
If you have any trouble reading the file with the formula editor, let me know. I can retype it without the editor, but it will not look nearly as readable.
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Show MoreShow Less | 3,220 | 13,888 | {"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} | 3.4375 | 3 | CC-MAIN-2018-05 | latest | en | 0.927788 |
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THEOREMA XX, PROPOSITIO XXXI. THEOREM XX, PROPOSITION XXXI Si linea recta super horizontalem fuerit utcunque inclinata, planum a dato puncto in horizontali usque ad inclinatam extensum, in quo {252} descensus fit tempore omnium brevissimo, est illud quod bifariam dividit angulum contentum a duabus perpendicularibus a dato puncto extensis, una ad horizontalem lineam, altera ad inclinatam. If a straight line is inclined at any angle to the horizontal and if, from any assigned point in the horizontal, a plane of quickest descent is to be drawn to the inclined line, that plane will be the one which bisects the angle contained {252} between two lines drawn from the given point, one perpendicular to the horizontal line, the other perpendicular to the inclined line. Sit CD linea supra horizontalem ab utcunque inclinata, datoque in horizontali quocunque puncto A, educantur ex eo AC perpendicularis ad AB, AE vero perpendicularis ad CD, et angulum CAE bifariam dividat FA linea: dico, planorum omnium ex quibuslibet punctis lineae CD ad punctum A inclinatorum, extensum {10} per FA esse in quo, tempore omnium brevissimo fiat descensus. Ducatur FG ipsi AE parallela; erunt anguli GFA, FAE coalterni aequales: est autem EAF ipsi FAG aequalis: ergo trianguli latera FG, GA aequalia erunt. Si itaque centro G, intervallo GA, circulus describatur, transibit per F, et horizontalem et inclinatam tanget in punctis A, F; est enim angulus GFC rectus, cum GF ipsi AE sit aequidistans: ex {20} quo constat, lineas omnes usque ad inclinatam ex puncto A productas extra circumferentiam extendi, et, quod consequens est, lationes per ipsas longiori tempore absolvi quam per FA. Quod erat demonstrandum. Let CD be a line inclined at any angle to the horizontal AB; and from any assigned point A in the horizontal draw AC perpendicular to AB, and AE perpendicular to CD; draw FA so as to bisect the angle CAE.Then, I say, that of all the planes which can be drawn through the point A, cutting the line CD at any points whatsoever AF is the one of quickest descent [in quo tempore omnium brevissimo fiat descensus]. Draw FG parallel to AE; the alternate angles GFA and FAE will be equal; also the angle EAF is equal to the angle FAG. Therefore the sides GF and GA of the triangle FGA are equal. Accordingly if we describe a circle about G as center, with GA as radius, this circle will pass through the point F, and will touch the horizontal at the point A and the inclined line at F; for GFC is a right angle, since GF and AE are parallel. It is clear therefore that all lines drawn from A to the inclined line, with the single exception of FA, will extend beyond the circumference of the circle, (Condition 2/06-th-06) (Condition 202C) thus requiring more time to traverse any of them than is needed for FA. Q. E. D.
Discorsi Propositions 2/31-th-20 Discorsi Proposition2/31-th-20 | 780 | 2,939 | {"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} | 2.875 | 3 | CC-MAIN-2022-27 | latest | en | 0.44165 |
https://us.metamath.org/mpeuni/isdivrngo.html | 1,716,317,721,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058512.80/warc/CC-MAIN-20240521183800-20240521213800-00435.warc.gz | 511,409,262 | 7,062 | Mathbox for Jeff Madsen < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > isdivrngo Structured version Visualization version GIF version
Theorem isdivrngo 35266
Description: The predicate "is a division ring". (Contributed by FL, 6-Sep-2009.) (New usage is discouraged.)
Assertion
Ref Expression
isdivrngo (𝐻𝐴 → (⟨𝐺, 𝐻⟩ ∈ DivRingOps ↔ (⟨𝐺, 𝐻⟩ ∈ RingOps ∧ (𝐻 ↾ ((ran 𝐺 ∖ {(GId‘𝐺)}) × (ran 𝐺 ∖ {(GId‘𝐺)}))) ∈ GrpOp)))
Proof of Theorem isdivrngo
Dummy variables 𝑔 𝑥 𝑦 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 df-br 5040 . . . . 5 (𝐺DivRingOps𝐻 ↔ ⟨𝐺, 𝐻⟩ ∈ DivRingOps)
2 df-drngo 35265 . . . . . . 7 DivRingOps = {⟨𝑥, 𝑦⟩ ∣ (⟨𝑥, 𝑦⟩ ∈ RingOps ∧ (𝑦 ↾ ((ran 𝑥 ∖ {(GId‘𝑥)}) × (ran 𝑥 ∖ {(GId‘𝑥)}))) ∈ GrpOp)}
32relopabi 5667 . . . . . 6 Rel DivRingOps
43brrelex1i 5581 . . . . 5 (𝐺DivRingOps𝐻𝐺 ∈ V)
51, 4sylbir 238 . . . 4 (⟨𝐺, 𝐻⟩ ∈ DivRingOps → 𝐺 ∈ V)
65anim1i 617 . . 3 ((⟨𝐺, 𝐻⟩ ∈ DivRingOps ∧ 𝐻𝐴) → (𝐺 ∈ V ∧ 𝐻𝐴))
76ancoms 462 . 2 ((𝐻𝐴 ∧ ⟨𝐺, 𝐻⟩ ∈ DivRingOps) → (𝐺 ∈ V ∧ 𝐻𝐴))
8 rngoablo2 35225 . . . . 5 (⟨𝐺, 𝐻⟩ ∈ RingOps → 𝐺 ∈ AbelOp)
9 elex 3489 . . . . 5 (𝐺 ∈ AbelOp → 𝐺 ∈ V)
108, 9syl 17 . . . 4 (⟨𝐺, 𝐻⟩ ∈ RingOps → 𝐺 ∈ V)
1110ad2antrl 727 . . 3 ((𝐻𝐴 ∧ (⟨𝐺, 𝐻⟩ ∈ RingOps ∧ (𝐻 ↾ ((ran 𝐺 ∖ {(GId‘𝐺)}) × (ran 𝐺 ∖ {(GId‘𝐺)}))) ∈ GrpOp)) → 𝐺 ∈ V)
12 simpl 486 . . 3 ((𝐻𝐴 ∧ (⟨𝐺, 𝐻⟩ ∈ RingOps ∧ (𝐻 ↾ ((ran 𝐺 ∖ {(GId‘𝐺)}) × (ran 𝐺 ∖ {(GId‘𝐺)}))) ∈ GrpOp)) → 𝐻𝐴)
1311, 12jca 515 . 2 ((𝐻𝐴 ∧ (⟨𝐺, 𝐻⟩ ∈ RingOps ∧ (𝐻 ↾ ((ran 𝐺 ∖ {(GId‘𝐺)}) × (ran 𝐺 ∖ {(GId‘𝐺)}))) ∈ GrpOp)) → (𝐺 ∈ V ∧ 𝐻𝐴))
14 df-drngo 35265 . . . 4 DivRingOps = {⟨𝑔, ⟩ ∣ (⟨𝑔, ⟩ ∈ RingOps ∧ ( ↾ ((ran 𝑔 ∖ {(GId‘𝑔)}) × (ran 𝑔 ∖ {(GId‘𝑔)}))) ∈ GrpOp)}
1514eleq2i 2903 . . 3 (⟨𝐺, 𝐻⟩ ∈ DivRingOps ↔ ⟨𝐺, 𝐻⟩ ∈ {⟨𝑔, ⟩ ∣ (⟨𝑔, ⟩ ∈ RingOps ∧ ( ↾ ((ran 𝑔 ∖ {(GId‘𝑔)}) × (ran 𝑔 ∖ {(GId‘𝑔)}))) ∈ GrpOp)})
16 opeq1 4776 . . . . . 6 (𝑔 = 𝐺 → ⟨𝑔, ⟩ = ⟨𝐺, ⟩)
1716eleq1d 2896 . . . . 5 (𝑔 = 𝐺 → (⟨𝑔, ⟩ ∈ RingOps ↔ ⟨𝐺, ⟩ ∈ RingOps))
18 rneq 5779 . . . . . . . . 9 (𝑔 = 𝐺 → ran 𝑔 = ran 𝐺)
19 fveq2 6643 . . . . . . . . . 10 (𝑔 = 𝐺 → (GId‘𝑔) = (GId‘𝐺))
2019sneqd 4552 . . . . . . . . 9 (𝑔 = 𝐺 → {(GId‘𝑔)} = {(GId‘𝐺)})
2118, 20difeq12d 4076 . . . . . . . 8 (𝑔 = 𝐺 → (ran 𝑔 ∖ {(GId‘𝑔)}) = (ran 𝐺 ∖ {(GId‘𝐺)}))
2221sqxpeqd 5560 . . . . . . 7 (𝑔 = 𝐺 → ((ran 𝑔 ∖ {(GId‘𝑔)}) × (ran 𝑔 ∖ {(GId‘𝑔)})) = ((ran 𝐺 ∖ {(GId‘𝐺)}) × (ran 𝐺 ∖ {(GId‘𝐺)})))
2322reseq2d 5826 . . . . . 6 (𝑔 = 𝐺 → ( ↾ ((ran 𝑔 ∖ {(GId‘𝑔)}) × (ran 𝑔 ∖ {(GId‘𝑔)}))) = ( ↾ ((ran 𝐺 ∖ {(GId‘𝐺)}) × (ran 𝐺 ∖ {(GId‘𝐺)}))))
2423eleq1d 2896 . . . . 5 (𝑔 = 𝐺 → (( ↾ ((ran 𝑔 ∖ {(GId‘𝑔)}) × (ran 𝑔 ∖ {(GId‘𝑔)}))) ∈ GrpOp ↔ ( ↾ ((ran 𝐺 ∖ {(GId‘𝐺)}) × (ran 𝐺 ∖ {(GId‘𝐺)}))) ∈ GrpOp))
2517, 24anbi12d 633 . . . 4 (𝑔 = 𝐺 → ((⟨𝑔, ⟩ ∈ RingOps ∧ ( ↾ ((ran 𝑔 ∖ {(GId‘𝑔)}) × (ran 𝑔 ∖ {(GId‘𝑔)}))) ∈ GrpOp) ↔ (⟨𝐺, ⟩ ∈ RingOps ∧ ( ↾ ((ran 𝐺 ∖ {(GId‘𝐺)}) × (ran 𝐺 ∖ {(GId‘𝐺)}))) ∈ GrpOp)))
26 opeq2 4777 . . . . . 6 ( = 𝐻 → ⟨𝐺, ⟩ = ⟨𝐺, 𝐻⟩)
2726eleq1d 2896 . . . . 5 ( = 𝐻 → (⟨𝐺, ⟩ ∈ RingOps ↔ ⟨𝐺, 𝐻⟩ ∈ RingOps))
28 reseq1 5820 . . . . . 6 ( = 𝐻 → ( ↾ ((ran 𝐺 ∖ {(GId‘𝐺)}) × (ran 𝐺 ∖ {(GId‘𝐺)}))) = (𝐻 ↾ ((ran 𝐺 ∖ {(GId‘𝐺)}) × (ran 𝐺 ∖ {(GId‘𝐺)}))))
2928eleq1d 2896 . . . . 5 ( = 𝐻 → (( ↾ ((ran 𝐺 ∖ {(GId‘𝐺)}) × (ran 𝐺 ∖ {(GId‘𝐺)}))) ∈ GrpOp ↔ (𝐻 ↾ ((ran 𝐺 ∖ {(GId‘𝐺)}) × (ran 𝐺 ∖ {(GId‘𝐺)}))) ∈ GrpOp))
3027, 29anbi12d 633 . . . 4 ( = 𝐻 → ((⟨𝐺, ⟩ ∈ RingOps ∧ ( ↾ ((ran 𝐺 ∖ {(GId‘𝐺)}) × (ran 𝐺 ∖ {(GId‘𝐺)}))) ∈ GrpOp) ↔ (⟨𝐺, 𝐻⟩ ∈ RingOps ∧ (𝐻 ↾ ((ran 𝐺 ∖ {(GId‘𝐺)}) × (ran 𝐺 ∖ {(GId‘𝐺)}))) ∈ GrpOp)))
3125, 30opelopabg 5398 . . 3 ((𝐺 ∈ V ∧ 𝐻𝐴) → (⟨𝐺, 𝐻⟩ ∈ {⟨𝑔, ⟩ ∣ (⟨𝑔, ⟩ ∈ RingOps ∧ ( ↾ ((ran 𝑔 ∖ {(GId‘𝑔)}) × (ran 𝑔 ∖ {(GId‘𝑔)}))) ∈ GrpOp)} ↔ (⟨𝐺, 𝐻⟩ ∈ RingOps ∧ (𝐻 ↾ ((ran 𝐺 ∖ {(GId‘𝐺)}) × (ran 𝐺 ∖ {(GId‘𝐺)}))) ∈ GrpOp)))
3215, 31syl5bb 286 . 2 ((𝐺 ∈ V ∧ 𝐻𝐴) → (⟨𝐺, 𝐻⟩ ∈ DivRingOps ↔ (⟨𝐺, 𝐻⟩ ∈ RingOps ∧ (𝐻 ↾ ((ran 𝐺 ∖ {(GId‘𝐺)}) × (ran 𝐺 ∖ {(GId‘𝐺)}))) ∈ GrpOp)))
337, 13, 32pm5.21nd 801 1 (𝐻𝐴 → (⟨𝐺, 𝐻⟩ ∈ DivRingOps ↔ (⟨𝐺, 𝐻⟩ ∈ RingOps ∧ (𝐻 ↾ ((ran 𝐺 ∖ {(GId‘𝐺)}) × (ran 𝐺 ∖ {(GId‘𝐺)}))) ∈ GrpOp)))
Colors of variables: wff setvar class Syntax hints: → wi 4 ↔ wb 209 ∧ wa 399 = wceq 1538 ∈ wcel 2115 Vcvv 3471 ∖ cdif 3907 {csn 4540 ⟨cop 4546 class class class wbr 5039 {copab 5101 × cxp 5526 ran crn 5529 ↾ cres 5530 ‘cfv 6328 GrpOpcgr 28250 GIdcgi 28251 AbelOpcablo 28305 RingOpscrngo 35210 DivRingOpscdrng 35264 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1912 ax-6 1971 ax-7 2016 ax-8 2117 ax-9 2125 ax-10 2146 ax-11 2162 ax-12 2178 ax-ext 2793 ax-sep 5176 ax-nul 5183 ax-pow 5239 ax-pr 5303 ax-un 7436 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3an 1086 df-tru 1541 df-ex 1782 df-nf 1786 df-sb 2071 df-mo 2623 df-eu 2654 df-clab 2800 df-cleq 2814 df-clel 2892 df-nfc 2960 df-ral 3131 df-rex 3132 df-rab 3135 df-v 3473 df-sbc 3750 df-dif 3913 df-un 3915 df-in 3917 df-ss 3927 df-nul 4267 df-if 4441 df-sn 4541 df-pr 4543 df-op 4547 df-uni 4812 df-br 5040 df-opab 5102 df-mpt 5120 df-id 5433 df-xp 5534 df-rel 5535 df-cnv 5536 df-co 5537 df-dm 5538 df-rn 5539 df-res 5540 df-iota 6287 df-fun 6330 df-fn 6331 df-f 6332 df-fv 6336 df-ov 7133 df-1st 7664 df-2nd 7665 df-rngo 35211 df-drngo 35265 This theorem is referenced by: zrdivrng 35269 isdrngo1 35272
Copyright terms: Public domain W3C validator | 3,317 | 5,349 | {"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} | 3.09375 | 3 | CC-MAIN-2024-22 | latest | en | 0.264514 |
https://forum.itascainternational.com/t/natural-frequency-of-oscillation-prediction/650 | 1,670,026,883,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.70/warc/CC-MAIN-20221202215443-20221203005443-00195.warc.gz | 294,865,062 | 7,687 | # Natural frequency of oscillation prediction
3DEC → Options → Damping
I am running “; Block under gravity - 3 critically damped cases”.
I want to use “block mech damp rayleigh stiffness” in my model. When I try the different block size I got different result as I predict. I change the upper block to a smaller block and the frequency seems not good(fig1) as shown in the example(fig2). Is there any suggestion to make it like the one in example?
fig1
fig2
======================================
Here is my code:
model new
model config dynamic
model large-strain on
block create brick 0 10 0 1 0 5
block create brick 3 7 0 1 5 10
block cut j-set or 0 0 5
block fix range pos-z 0 5
model grav 0 0 -10
block prop dens 1000
block contact gen-sub
block contact jmodel assign mohr
block contact property stiffness-normal 5e7 stiffness-shear 5e7
block contact material-table default property stiffness-normal 5e7 …
stiffness-shear 5e7
block mechanical damp rayleigh 0 0
hist interval 1
block his vel-z pos 5 0 10
block his disp-z pos 5 0 10
model his dynamic time-total
;no damping
model solve time 1
model title "Vertical displacement v time (undamped)
ret
I think the problem here is that there is a very coarse, asymmetric discretization of the faces. Try adding the command BLOCK FACE TRI RADIAL before generating the subcontacts. This will give a better, symmetric distribution of subcontacts and the velocity histories should look better.
1 Like
Thank you very much. It works.
But I still get some problem in the frequency part. I have two questions.
# Question 1
I make two block the upper one is 1 * 1 * 1 times and the down one is 15 * 15 * 50 (fixed). In the Options-> Damping mention that the frequency can get form a function.
When I use this function I assume that the frequency show be 11.25cycle/second. But I got around 23 cycle/second in the result(fig1). Is there any misunderstand that I made?
# Question 2
I try to use “block mech damp contact” command. What my input is 'block mech damp contact ‘group 1’ ‘2’ ‘test123’ 0.1 0 0.1 0 0.1 0 0.1 0 0.1 0 0.1 0 '. And make a error. Where can I change?
# Here is the code:
## code for the first question
========
model new
model config dynamic
model large-strain on
block create brick 7 8 7 8 50 51
block group “1”
block hide on
block create brick 0 15 0 15 0 50
block group “2”
block hide off
;block face tri radial range pos-z 0 10
block fix range group “2”
model grav 0 0 -10
block prop dens 2000
block contact gen-sub
block contact jmodel assign mohr
block contact property stiffness-normal 1e7 stiffness-shear 1e7
block contact material-table default property stiffness-normal 1e7 …
stiffness-shear 1e7
;block mechanical damp rayleigh 0 0
hist interval 1
block his vel-z pos 7.5 7.5 51
model his dynamic time-total
;no damping
model solve time 1
ret
======
## code for the second question
=====
model new
model config dynamic
model large-strain on
block create brick 6.5 8.5 6.5 8.5 60 62
block group “1”
block hide on
block create brick 0 15 0 15 0 50
block group “2”
block hide off
block fix range group “2”
model grav 0 0 -10
block prop dens 2000
block contact gen-sub
block contact jmodel assign mohr
block contact property stiffness-normal 1e7 stiffness-shear 1e7
block contact material-table default property stiffness-normal 1e7 …
stiffness-shear 1e7
;block mechanical damp rayleigh 0 0
hist interval 1
block his vel-z pos 7.5 7.5 51
model his time
model his dynamic time-total
block mech damp contact ‘group 1’ ‘2’ ‘test123’ 0.1 0 0.1 0 0.1 0 0.1 0 0.1 0 0.1 0
model solve time 2
ret
Thank you very much
I can answer the second one. See here for the documentation for that command: block mechanical command — 3DEC 7.0 documentation (itascacg.com)
It should look something like:
block mech damp contact group1 ‘1’ group2 ‘2’ betaffn 0.1 betaffs 0 …
Note that in this case you will only have one contact so you don’t need to specify the groups. You also don’t need to give the zero values. So this will work:
block mech damp contact betaffn 0.1 betavvn 0.1 …
I’m not sure what’s going on with the first question. If you make the blocks the same size, then you get the expected frequency. I’ll look into it.
I just want to try different size in the beginning for the frequency. I have tried the upper block size with 2 * 2 * 2. And it suit the estimate value. But when size down to certain value the frequency become very strange.
There is a bug in the calculation of areas when you have rigid blocks with mismatching faces. The pseudo-static stresses and displacements are OK, but since the area is off (by a factor of 2) then the dynamic response is off (frequency is too high by a factor of 2).
I’ll work on a fix. In the short term, you can either use deformable blocks, or ensure the faces match by cutting and joining. For example:
block create brick 7 8 7 8 50 51
block group ‘1’
block hide on
block create brick 0 15 0 15 0 50
; cut and join to match top block
block cut joint-set dip 90 dip-dir 90 or 7 7 50
block cut joint-set dip 90 dip-dir 90 or 8 7 50
block cut joint-set dip 90 dip-dir 0 or 7 7 50
block cut joint-set dip 90 dip-dir 0 or 7 8 50
block join
block group ‘2’
block hide off
I have some questions when I do some simple test. The attached file is that I build a sliding model and try to use “block mech damp contact” command. I know what the friction angle is in the material. But I found the contact is different when I compared to the laboratory test. I want to reduce the normal contact by using “block mesh damp contact”. Here are my questions.
1. It takes a long time when I simulate only one block. Is there any other damping you can suggest me to use? Or is there any way to reduce the calculation time?
2. I have asked the sub-contact identify question before. Would the judgement become strange or error when I use rigid body in this case?
;;;;
model new
model config dy lh
model large-strain on
block tolerance edge 0.00002
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
block create brick -1.9927 0 -0.2 2.3746 -0.2 0.2
;block cut joint-set dip 0 dip-direction 0 ori 0 1 0
;del range y 1 9999999999
;block create brick 310cos,0,thickness, 310sin+30cos, channel wide
block cut joint-set dip 60 dip-direction 90 ori 0 -0.15 0
;thickness 0.5 0.5cos
block delete range plane below dip-direction 90 dip 60 ori 0 -0.15 0
block cut joint-set dip 30 dip-direction 270 ori -1.9927 2.3746 0
block delete range plane above dip-direction 270 dip 30 ori -1.9927 2.3746 0
;hight 50cm 50*sec
block cut joint-set dip 60 dip-direction 90 ori 0 0.7778 0
block delete range plane above dip-direction 90 dip 60 ori 0 0.7778 0
mark region 4
block cut joint-set dip 90 dip-direction 0 ori 0 0 0.15
block cut joint-set dip 90 dip-direction 0 ori 0 0 -0.15
block hide range position-z -0.15 0.15
mark region 5
block hide off
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;11:13
block hide on range region 5
block cut joint-set dip 60 dip-direction 90 ori 0.00000 0.001 0.000000
block hide on range plane above dip 60 dip-direction 90 ori 0.00000 0.001 0.000000
mark region 1
block hide on
;;deposited area
block create brick 0 0.005 0 0.5 -1 1.01
block cut joint-set dip 90 dip-direction 0 ori 0 0 0.15
block cut joint-set dip 90 dip-direction 0 ori 0 0 -0.15
block delete range plane position-z -0.151 0.151
mark region 3
block hide on
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;11:14
block create brick 0 1.5 -0.1 0 -1.00 1.00
mark region 2
block hide on
block create brick 0 1.5 0 0.30 -1.005 -1.00
block create brick 0 1.5 0 0.30 1.00 1.005
mark region 3
block hide on
block hide off
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
block hide on range region 4
block join on
block hide on
block hide off range region 4
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
block cut joint-set dip 90 dip-direction 0 ori 0 0 0.144
block cut joint-set dip 90 dip-direction 0 ori 0 0 -0.144
block del range plane position-z -0.151 -0.144
block del range plane position-z 0.144 0.151
block hide on
block hide off region 4
block join on
block cut joint-set dip 60 dip-direction 90 ori 0 0.063391 0
block delete range plane above dip-direction 90 dip 60 ori 0 0.063391 0
block cut joint-set dip 30 dip-direction 270 ori -0.5000067 0.865987 0
block delete range plane below dip-direction 270 dip 30 ori -0.5000067 0.865987 0
block cut joint-set dip 30 dip-direction 270 ori -0.516069 0.893699 0
block delete range plane above dip-direction 270 dip 30 ori -0.516069 0.893699 0
block cut joint-set dip 90 dip-direction 0 ori 0 0 0.016
block delete range plane above dip-direction 0 dip 90 ori 0 0 0.016
block cut joint-set dip 90 dip-direction 0 ori 0 0 -0.016
block delete range plane below dip-direction 0 dip 90 ori 0 0 -0.016
block hide off
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
model gravity 0 -9.81 0
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
block property density 1170 range region 1,2,3,5,7
block property density 2750 range region 4
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;’
block contact generate-subcontacts
block contact prop stiffness-norm=1e9 stiffness-shear=1e9 friction=17.8 range region 1,2,3,5,7
block contact prop stiffness-norm=1e9 stiffness-shear=1e9 friction=24.2 range region 4
block contact material-table default prop stiffness-norm=1e9 stiffness-shear=1e9 fric=17.8
;;;;;;;;;;;;;;;;;;;;;;;
block hide off
block fix range region 1 2 3 5
block mech damp contact betaffn 0.5 betavvn 0.5 betaven 0.5 betavfn 0.5 betaefn 0.5 betaeen 0.5
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
model solve time 3
program return
The left-handed coordinate system is no longer supported. Please reformulate the problem for a RH system and send to technical support. | 2,884 | 9,716 | {"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} | 2.578125 | 3 | CC-MAIN-2022-49 | latest | en | 0.850537 |
https://www.jiskha.com/display.cgi?id=1337415186 | 1,511,576,195,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934809229.69/warc/CC-MAIN-20171125013040-20171125033040-00148.warc.gz | 840,121,047 | 4,338 | # maths
posted by .
1).Last year, a shopkeeper sold an average of 7 cartons of apples from january to april. He sold an average of 9 cartons of apples from june till end of the year. What is the total number of cartons he sold in the 11 months? if each carton contains 66 apples,what is the average number of apples sold per month?
2) Ameter records voltages between 0 volts and 20 volts. If the average value for three readings on the meter was 16 volts, what is the smallest possible reading, in volts?
3) The average of a set of 50 numbers is 38. The numbers of the set, namely 45 and 55 are discarded. What is the average of the remaining set of numbers?
• maths -
1)
4 months @ 7 = 28 cartons
8 months @ 9 = 72 cartons
(assuming the year has 12 months instead of 11)
average apples sold per month is 100*66/12 = 550
2)
since the sum is 48, and no single reading is over 20, then the smallest must be at least 8.
3)
sum of numbers is 50*38 = 1900
new sum, less discarded values = 1800
new average is 1800/48 = 37.5
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Quiz by Taguitay, Raymond
Math
South Carolina Learning Standards
Feel free to use or edit a copy
includes Teacher and Student dashboards
### Measures 3 skills fromGrade 6MathSouth Carolina Learning Standards
6.EEI.5
6.EEI.6
6.EEI.7
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10 questions
• Q1
Which number is in the solution set of this inequality? x > 45
49
44
45
38
30s
6.EEI.5
• Q2
Which set of numbers is NOT in the solution set of the inequality x > 17?
{19, 20, 28, 29, 31}
{29, 32, 35, 40, 55}
{14, 15, 18, 25, 29}
{18, 21, 25, 27, 33}
30s
6.EEI.5
• Q3
Which equation is true for x = 4?
5x − 18 = 7
6x − 10 = 26
3x + 5 = 17
2x + 5 = 25
30s
6.EEI.5
• Q4
What is the value of x when x - 7 = 21?
-14
-28
3
28
30s
6.EEI.5
• Q5
Jaylon makes twice as much an hour as Sam. Sam makes x dollars an hour. Write an expression describing how much Jaylon makes an hour.
x^2
x - 2
2x
2 + x
30s
6.EEI.6
• Q6
Ervin scores half as many points in a game as Larry. If Larry scored x points, write an expression to represent how many points Ervin scored.
2 + x
x - 2
(1/2) x
2x
30s
6.EEI.6
• Q7
One day each summer, Chandler earns money by having a lemonade stand in his neighborhood. He charges \$0.50 per glass of lemonade, and he made a total of \$24. How many glasses of lemonade did he sell?
36
12
48
24
30s
6.EEI.7
• Q8
A number decreased by 23 is 44. Find the number. Which equation can be used to solve this problem?
x - 23 = 44
x - 44 = 23
44 - x = 23
23 - x = 44
30s
6.EEI.7
• Q9
Tom has saved \$128, and his goal is to save \$500. How much more must he save? Which equation BEST represents this situation?
x + 500 = 128
x - 128 = 500
x + 128 = 500
x - 500 = 128
30s
6.EEI.7
• Q10
Which number is in the solution set of this equation? 7x − 9 = 40
6
15
7
11
30s
6.EEI.5
Teachers give this quiz to your class | 768 | 2,203 | {"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-2024-18 | latest | en | 0.924397 |
http://drhuang.com/science/mathematics/software/mathhandbook.htm | 1,679,927,238,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948632.20/warc/CC-MAIN-20230327123514-20230327153514-00489.warc.gz | 14,515,550 | 4,712 | Fractional Calculus Computer Algebra System, Electrochemistry Software, Massage
+ + + =
Please download Java from www.java.com and change Java security level to Medium from Java Control Panel if you cannot see a following Java Applet calculator.
# int_0^oo 数学手册 dx = sum_0^oo MathHandbook.com
Copy-n-paste math expression from calculator and press the ENTER key:
You typed:
All keywords are lower case, case-sensitive, e.g. sin is differernt from SIN. Its default variable is small letter x. Please type sin(x) into the Input panel, then click on button : plot -- graph. Double click on graph to zoom. taylor -- Taylor expansion at x = 0. limit -- lim _(x->oo) series -- series representation. d/dx -- differentiation d/dx, f dx -- integration int dx, expand -- expansion. factor -- factorization. solve -- to solve equation = 0. last -- put last output into the input panel. = -- to simplify for symbolic answer. ~= -- for numeric answer.
First line of buttions are easy command, but the second line of buttons are hard command, its result can be checked by its above first line button if it returns origial function or not. E.g. plot sin(x), then click the taylor button and the last button, then plot again to compare Taylor expansion at x=0 on graph.
#### Examples: Please input following math expression, then click on the = button.
For lim_(x->0) sin(x)/x, input:
limit(sin(x)/x,x=0)
For differentiation d/dx x^6, input:
d(x^6)
For derivative d/dx | _(x=2) x^6, input:
d(x^6, x=2)
For second order of differntiation d^2/dx f(x), input:
d(f(x),x,2)
For integration int x^6 dx, input:
int(x^6, x)
For definite integration int_1^2 x^6 dx, input:
int(x^6, x, 1,2)
For Taylor series:
taylor(exp(x))
For big integer:
100!
For numeric, input:
n(1/2)
To define function, then use the defined function, input :
f(x_) := x^2;
f(9)
Solve two equations: y=x^2 and y=x+1, input :
plot(y=x^2), plot(y=x+1)
OR input:
plot(x^2), plot(x+1)
click on the cross of two lines to get solutions on graph.
You input f(x), then click on the "dy/dx" button and the Last button many times for high order of differentiation d^2/dx f(x), then click on the "f dx" button and the Last button many times for high order of integration int int.
It is an online graphic calculator and computer algebra system. It can perform exact, numeric, symbolic and graphic computation, e.g. arbitrary-precision calculation, solve equation, plot data and user-defined functions, linear regression, symbolic differentation and integration, pattern-match. It is a programming language, in which you can define conditional, case, piecewise, recursive, multi-value functions and procedures, derivatives, integrals and rules.
For more functions, click on the "Input" menu to input D(x) into the Input panel, then click the "=" button to get the result.
For plotting a function, type Plot(x^2) into the Input panel, click on the "=" button. Or just type x^2 into the input panel, click on the Plot button. click on the "Zoom In" or "Zoom Out" menu under the Graph menu to zoom.
Find coordinates by clicking on graph, get the initial value to the "x=" field by double-clicking, or zoom by drag, on graphics.
If symbolic integration fails, you should add your integration rules into the library file Int.txt in the vmath/library/ directory.
You can develop your library, and save library files into the vmath/library/ directory. The library source codes are text and auto loaded when need. Please read document visualmath.doc or visualmath.doc.htm.
It runs in any computer, PDA, and mobile phone that supports Java™ technology. If you see it online, it can run offline in your computer, PDA, and mobile phone.
It is all of Java applet, Java application and Window application. Run VisualMath.html as Java applet, run VisualMath.bat as Java application, or run VisualMath.exe as Window application.
For security reason, the Copy, Paste, Open and Save, and Print menu work only when it runs as local application. Please read SymbMath manual SymbMath.doc.
You can copy-and-paste from and to your document by right-button clicking on the text field to pop-up menu. For second way of copy-and-paste, first select the text, then key in Ctrl+C (i.e. press the Ctrl key and the C key at the same time), then point to where you want to paste, and finally key in Ctrl+V (i.e. press the Ctrl key and the V key at the same time). | 1,091 | 4,403 | {"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": 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} | 3.546875 | 4 | CC-MAIN-2023-14 | latest | en | 0.704933 |
https://www.studypool.com/services/262997/managerial-economics-510-final-exam | 1,547,625,796,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583657097.39/warc/CC-MAIN-20190116073323-20190116095323-00177.warc.gz | 940,659,527 | 21,595 | # Managerial Economics 510 Final Exam
Jul 14th, 2016
Studypool Tutor
Sullivan University
Price: \$40 USD
Tutor description
Word Count: 844
Showing Page: 1/6
Question No. 1a.Q = 5000 - .514000 + .17500 + .216000 = 1950At this point of tuition Elasticity of demand:ED == -3.6b.We change the demand function to accommodate for the newly given parameter values:Q = 7500 - .414000 + .157500 + .216000 = 6225 students will enrol next year.The new demand elasticity is ED == -0.9To determine whether the advertising campaign is cost effective, you need to determine, whetherthe increased revenues from the advertising campaign exceed its costs.Revenues are given by R = (6225-1950)\$14000 = 59,850,000 > 4,500,000So campaign is a cost effective one.Question No. 2a.b. Firm K has a dominant strategy in Strategy S. The Nash equilibrium is where Firm K doesStrategy S and Firm L does Strategy Q, with allocation (\$7.5 mill, \$14 mill).Question No. 3a.From my perspective john has to sell the house at a price of \$75000 because his marginal costis \$52000 which will lead to a contribution of \$23000 and which is good for John.Chat convQuestion No. 4Question 4 Production should be allocated so the marginal profitability of yard furniture equalsthat of knick knacks. Why? Because if the profitability of one is greater than the other more ofthe former and less of the latter should be produced. y' (derivative)= 2-.002Ty marginalprofiatbility is 100(2-.002Ty) K'(derivative)= 20-.02Tk marginal profitability is 25(20-.02Tk) Sowe have 100(2-.002Ty)= 25(20-.02Tk) 200-.2Ty= 500-.5Tk .5Tk-.2Ty= 300 Tk+Ty= 1300(cons
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2325 Tutors | 648 | 2,351 | {"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} | 3.25 | 3 | CC-MAIN-2019-04 | latest | en | 0.823864 |
http://openstudy.com/updates/509b0dd4e4b085b3a90e4ff8 | 1,448,748,290,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398454160.51/warc/CC-MAIN-20151124205414-00324-ip-10-71-132-137.ec2.internal.warc.gz | 168,480,434 | 25,326 | ## richyw 3 years ago Set up the double integral giving the volume of the solid bounded by $$y=x^2,\quad z=3y,\quad z=2+y$$
1. richyw
this is messing with my head. I am taking R to be $\left[3y,\,2+y\right]\times\left[ 0, 1\right]$
2. richyw
sorry that is z=3y to z=2+y and y=0 to y=1
3. richyw
It's been awhile since I have done this haha
4. mahmit2012
|dw:1352339571390:dw|
5. mahmit2012
|dw:1352339773621:dw|
6. richyw
double integral...
7. richyw
I guess I could evaluate that triple one though to see if I got the same answer as my attempt
8. mahmit2012
|dw:1352340171989:dw|
9. mahmit2012
for calculate the volume you should use terrible integral.
10. richyw
It's not what I should use. It's what the exam tells me to use haha.
11. mahmit2012
|dw:1352340360984:dw| | 272 | 792 | {"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} | 2.8125 | 3 | CC-MAIN-2015-48 | longest | en | 0.694602 |
http://openstudy.com/updates/4f1dc486e4b04992dd244dca | 1,496,144,095,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463615093.77/warc/CC-MAIN-20170530105157-20170530125157-00466.warc.gz | 342,211,436 | 8,440 | • anonymous
x represents the total number of bags a person checks and each person checks more than two bags, the equation below can be used to determine C, the total fee for all the bags. 20 + 25 + 50(x - 2) = C Arianna paid a total of \$145 to check all her bags on a flight. How many bags did she check in all?
Mathematics
• Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
Looking for something else?
Not the answer you are looking for? Search for more explanations. | 325 | 1,242 | {"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} | 2.671875 | 3 | CC-MAIN-2017-22 | longest | en | 0.399847 |
https://www.businessinsider.com/4-market-valuation-2014-2 | 1,560,810,107,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998580.10/warc/CC-MAIN-20190617203228-20190617225228-00429.warc.gz | 687,033,122 | 43,029 | Here is a summary of the four market valuation indicators I update during the first days of the month. The four indicators are:
• The Crestmont Research P/E Ratio ( more)
• The cyclical P/E ratio using the trailing 10-year earnings as the divisor ( more)
• The Q Ratio, which is the total price of the market divided by its replacement cost ( more)
• The relationship of the S&P Composite price to a regression trendline ( more)
To facilitate comparisons, I've adjusted the two P/E ratios and Q Ratio to their arithmetic means and the inflation-adjusted S&P Composite to its exponential regression. Thus the percentages on the vertical axis show the over/undervaluation as a percent above mean value, which I'm using as a surrogate for fair value. Based on the latest S&P 500 monthly data, the market is overvalued somewhere in the range of 51% to 82%, depending on the indicator, an increase from the previous month's 51% to 80%.
I've plotted the S&P regression data as an area chart type rather than a line to make the comparisons a bit easier to read. It also reinforces the difference between the line charts — which are simple ratios — and the regression series, which measures the distance from an exponential regression on a log chart.
The chart below differs from the one above in that the two valuation ratios (P/E and Q) are adjusted to their geometric mean rather than their arithmetic mean (which is what most people think of as the "average"). The geometric mean weights the central tendency of a series of numbers, thus calling attention to outliers. In my view, the first chart does a satisfactory job of illustrating these four approaches to market valuation, but I've included the geometric variant as an interesting alternative view for the two P/Es and Q. In this chart the range of overvaluation would be in the range of 63% to 95%, up from last month's 63% to 93%.
The Average of the Four Valuation Indicators
The next chart gives a simplified summary of valuations by plotting the average of the four arithmetic series (the first chart above). I've also included a log-scale area chart of the real (inflation-adjusted) S&P Composite along with recessions.
At the end of last month, the average of the four equaled the average in July of 2007, three months before the market hit its all-time high prior to the Great Recession.
As I've frequently pointed out, these indicators aren't useful as short-term signals of market direction. Periods of over- and under-valuation can last for many years. But they can play a role in framing longer-term expectations of investment returns. At present market overvaluation continues to suggest a cautious long-term outlook and guarded expectations. However, at the today's low annualized inflation rate and the extremely poor return on fixed income investments (Treasuries, CDs, etc.) the appeal of equities, despite overvaluation risk, is not surprising. For more on that topic, see my periodic update:
Note: For readers unfamiliar with the S&P Composite index, see this article for some background information. | 647 | 3,080 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-26 | latest | en | 0.942249 |
https://www.scribd.com/presentation/152729101/Electricity-Spider | 1,519,589,725,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816912.94/warc/CC-MAIN-20180225190023-20180225210023-00695.warc.gz | 925,218,338 | 26,112 | Electricity Spider | Series And Parallel Circuits | Electrical Network
# Definition
Symbol
Unit
Unit Symbol
Voltage
Current Charge Energy Resistance Power
Push on the electrons
Rate of flow of electrons Total number of charged particles Amount of energy given to a component How hard it is for electrons to flow How fast energy is transferred
V
Volts
V
• Safety Features: Double Insulation: Outer case is _____ so it will always be safe to touch. Insulation: _____ outside cable safe to touch Earthing: outside casing connected to _____ Circuit Breaker: Trips if too big a ____ flows Fuse: _____ if too big a ____ flows • Wiring a Plug: Fuse on the live Wire Green/Yellow: Earth Brown: ______ Blue: _______
Definitions Equations
Mnemonic Exercise Is Very tiring Peter is a Veggie QuIt Very Important Rabbits Equation E=IVt Units J=AVs
Electricity
Mains
• Hazards: Frayed Cables Damaged Plugs Water Don’t play with sockets Series Circuit Current Voltage Parallel Circuit
Series and Parallel Circuits
IV Graphs
Resistor
I V
Circuits
Diode
I Component Diode V Thermistor Function
Bulb
I V
Symbol
Only lets the current through one way Senses _____ (low ______ high _____)
LDR
Senses _____ (low ______ high _____)
Definition Symbol Unit Unit Symbol Voltage Current Charge Energy Resistance Power Push on the electrons Rate of flow of electrons Total number of charged particles Amount of energy given to a component How hard it is for electrons to flow How fast energy is transferred V I Q E R P Volts ampere coulomb joule ohm watt V A C J Ω W • Safety Features: Double Insulation: Outer case is plastic so it will always be safe to touch. Insulation: plastic outside cable safe to touch Earthing: outside casing connected to earth Circuit Breaker: Trips if too big a current flows Fuse: melts if too big a current flows • Wiring a Plug: Fuse on the live Wire Green/Yellow: Earth Brown: live Blue: neutral Definitions Equations Mnemonic Exercise Is Very tiring Peter is a Veggie QuIt Very Important Rabbits Equation E=IVt P = I V Q = I t V = I R Units J=AVs W = A V C = A s V = A Ω Electricity Mains • Hazards: Frayed Cables Damaged Plugs Water Don’t play with sockets Series Circuit Current Voltage same split Parallel Circuit split same Series and Parallel Circuits IV Graphs Resistor I V Circuits Diode I Component Diode V Thermistor Function Less Cabling If one breaks rest still work and/or same voltage Symbol Bulb I V Only lets the current through one way Senses Temp (low resistance high temp) LDR Senses Light (low resistance high light) . | 594 | 2,566 | {"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} | 2.859375 | 3 | CC-MAIN-2018-09 | latest | en | 0.765332 |
https://gmatclub.com/forum/from-530-q40-v23-to-690-q45-v40-35444.html?fl=similar | 1,490,745,027,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218190134.25/warc/CC-MAIN-20170322212950-00659-ip-10-233-31-227.ec2.internal.warc.gz | 833,048,783 | 61,822 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
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# From 530 (q40 v23) to 690 (q45 v40)
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Manager
Joined: 28 Feb 2003
Posts: 147
Location: Kiev
Followers: 1
Kudos [?]: 9 [0], given: 0
From 530 (q40 v23) to 690 (q45 v40) [#permalink]
### Show Tags
19 Sep 2006, 08:26
Hi,
If you look at the registration date of my account in GMAT Club you may notice that I've been present here for quite a long time. But only today, I really have something to tell.
I passed GMAT three times starting from May 2005:
530 - q40 v23
580 - q43 v27
690 - q45 v40
As you may see, there is a slight progress in Q section and a huge one in Verbal.
Since Verbal is a problem section for many non-natives I would like to suggest my explanation of my success in this area:
1. Study materials. I've got GMATPlus in electonic form. It was downloaded somewhere in this forum. I read and solved questions using my PocketPC, which appeared to be very convenient: in public transport and in home lying in the sofa.
I've done all SC and CR questions, partly RC questions paying specific attention to error log. I classified errors into three main groups:
- negligence (simple and stupid errors)
- hard errors (but understandable after additional analysis)
- concept errors (ones, which I don't understand completely)
Then I took several individual lessons with GMAT tutor and addressed all concept errors (actually it helped me a lot in understanding GMAT logic)
2. Reading. I started to read only in English, i.e. if you want to read a new fancy book (e.g. Dan Brown "The Da Vinci Code") get its English version, if you want to read news - read BBC.com or anything else which is in English. That approach helped me a lot in not getting afraid and distracted by long and complex GMAT passages.
3. Vitamines. A month before exam I started to take vitamin C and homeopathic remedy that improves brain activity.
To my mind, all these measures, combined assisted me in improving my score.
What else can I say. I did it )))))))))))
I do really feel that I can take 700 threshold, but, perhaps, it is worth to concentrate on other areas of my application.
_________________
Too much is not enough...
VP
Joined: 14 May 2006
Posts: 1414
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### Show Tags
19 Sep 2006, 08:41
hey BONO!!! you've made it!!! CONGRATULATIONS!!! now you can rock fans like me
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VP
Joined: 14 May 2006
Posts: 1414
Followers: 5
Kudos [?]: 180 [0], given: 0
### Show Tags
19 Sep 2006, 08:48
In all seriousness, I think your achievement with this whole GMAT thing is another proof of the fact that you can go from 500s to whatever you desire!!! And I think not only you learned some concepts, but also gained some real life experience by creating your own strategy and forcing yourself to follow it!!! EXCELLENT IMPROVEMENT!!! And while some people say they aren't happy with 690, I say YOU ROCK MAN!!! 690 by no means is an ok score, it's a great one!!!
Good luck to you in your application process, and I hope I will reach your point very soon!!!
best,
u2lovergirl
PS what were your GMATPrep scores prior going into to your 4th attempt? and how did you study math from one test to the next?
Director
Joined: 07 Jun 2006
Posts: 513
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### Show Tags
19 Sep 2006, 09:19
Your GMAT scores show a nice solid upward trend, adcomms can't miss that!
Good luck with the apps.
Manager
Joined: 09 Jul 2006
Posts: 164
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Kudos [?]: 2 [0], given: 0
### Show Tags
19 Sep 2006, 10:32
This post inspires me very much
Your first attempt score is pretty much the same as mine
and you rock it!! I wish I could improve like you did someday.
Director
Joined: 10 Oct 2005
Posts: 720
Followers: 3
Kudos [?]: 23 [0], given: 0
### Show Tags
19 Sep 2006, 11:14
Pozdravlau))Ochen horoshiy verbal score)
Congrats awesome score!wish you good luck with applications
_________________
IE IMBA 2010
Director
Joined: 28 Dec 2005
Posts: 755
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### Show Tags
19 Sep 2006, 12:14
the first question that came to my mind was....are you related to u2lovergirl???
awesome job...!!!
CEO
Joined: 20 Nov 2005
Posts: 2909
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 25
Kudos [?]: 279 [0], given: 0
### Show Tags
19 Sep 2006, 12:17
Excellent!!!!!
Thats a great improvement. You are the living proof that GMAT can kicked right on the a&%^&.
_________________
SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008
Senior Manager
Joined: 13 Sep 2006
Posts: 281
Location: New York
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### Show Tags
19 Sep 2006, 13:05
CONGRATULATION! That is so inspiring.
CEO
Joined: 15 Aug 2003
Posts: 3460
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Kudos [?]: 871 [0], given: 781
### Show Tags
19 Sep 2006, 13:59
Thanks for sharing. I am sure this will inspire many.
Praetorian
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1263
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19 Sep 2006, 14:50
Way to go! Good luck in B. School!
SVP
Joined: 31 Jul 2006
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Schools: Darden
Followers: 44
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### Show Tags
19 Sep 2006, 16:38
Wow, that is an awesome and inspirational story. Finally finished after first seeing the GMAT a year and a half ago. Now that is hard work and dedication. Congratulations!
Manager
Joined: 28 Feb 2003
Posts: 147
Location: Kiev
Followers: 1
Kudos [?]: 9 [0], given: 0
### Show Tags
19 Sep 2006, 23:43
Thank you guys !!!
I will save this page and keep it along with my diplomas, job offer letters and other proofs of my success.
What I forgot to say is that there may be no necessity to hire tutors for private GMAT lessons.
The GMATClub is the best source of the knowledge, experience, insipration and support. Therefore, all concept errors can be discussed here
P.S.
There is no relation at the moment between me and u2lover. But I think we should have a chat somewhere outside the forum
P.P.S.
Always remember - TOO MUCH IS NOT ENOUGH !!!!!!!!!!!!!!!
_________________
Too much is not enough...
VP
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20 Sep 2006, 03:50
Bono, congratulations. Your endurance is remarkable. Good luck with apps.
_________________
The path is long, but self-surrender makes it short;
the way is difficult, but perfect trust makes it easy.
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20 Sep 2006, 21:38
bono wrote:
There is no relation at the moment between me and u2lover. But I think we should have a chat somewhere outside the forum
we can easily change that I would be the girl on the left
Attachments
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21 Sep 2006, 07:56
This is such a Bonofied thread.
Congratulations!!! Good luck with your apps!
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21 Sep 2006, 08:49
u2lover
Cool picture. I've never seen Bono in such a lovely environment
_________________
Too much is not enough...
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15 Dec 2006, 16:55
Hi Bono,
I just have one question for you. How long did it take you to make the last leap: from 580 to 690? . My GMAT Score is currently 570 and i aim to make a similar jump as yours. I want to know how long should i realistically expect to make this movement ? I study 2 hours daily and 4-5 hours in the weekend.
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16 Dec 2006, 21:28
That's quite an accomplishment bono. You ran the full GMAT marathon and took a top prize in the end. Are you planning for fall 2007 matriculation? If so, see you in b-school application forum! Once again, congratulations!
Current Student
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16 Dec 2006, 21:31
One more thing, regardless of what anybody else says, with solid undergrad grades, upwardly mobile work experience and compelling essays, you have a fighting chance at some top schools. Good luck!
16 Dec 2006, 21:31
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## Wednesday, September 12, 2012
### Red Swastika Sch 2011 PSLE Math Prelim Paper 2 Q10
Katelyn had some green, red and blue ribbons. She had 25% more green ribbons than red ribbons and 20% less blue ribbons than red ones.
a) What is the ratio of red ribbons to green ribbons to blue ribbons?
b) Katelyn exchanges all her blue ribbons for some red and green ribbons. She now has an equal number of red and green ribbons. How many more green than red ribbons did she have at first if she has 244 red ribbons now?
Solution
(a)
Red ----- 100%
Green ----- 25% more than 100% = 125%
Blue ----- 20% less than 100% = 80%
Red : Green : Blue
100 : 125 : 80
20 : 25 : 16
Answer: 20 : 25 : 16
(b)
Before Katelyn exchanges,
Red : Green : Blue
20 : 25 : 16
Blue exchanged for green and red will mean that 16 units of blue will be given up. Since Katelyn finally has an equal of number red and green, 5 units of the 16 units of blue will be exchanged for red first, with the remaining 11 units divided equally between red and green.
Red will receive ----- 5 units + (11 units divided by 2) = 10.5 units
Green will receive ----- 11 units divided by 2 = 5.5 units
Red ribbons now ----- 20 units + 10.5 units = 30.5 units
30.5 units ----- 244
1 unit ---- 244 divided by 30.5 = 8
In the beginning,
Green - Red
25 units - 20 units = 5 units
5 units ----- 5 x 8 = 40 | 449 | 1,507 | {"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-2017-39 | longest | en | 0.92338 |
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A radial pattern contains randomize(bump) (Message 1 to 10 of 15)
From: And Subject: A radial pattern contains randomize(bump) Date: 26 Aug 2023 05:15:00 Message:
```
{
"@context": "https://schema.org",
"@type": "DiscussionForumPosting",
"@id": "#web.64e9c2855b36d8313a0d3e9baa81652d%40news.povray.org",
"dateCreated": "2023-08-26T09:15:00+00:00",
"datePublished": "2023-08-26T09:15:00+00:00",
"author": {
"@type": "Person",
"name": "And"
}
}
I suddenly solve an important problem. About how to generate a circle associated
pattern.
I need such a pattern for my wood-pattern generator.
For wood heart, I don't have good function in the past. But these series opens a
window.
```
Attachments:
Preview of image 'radial associate function.png'
From: And Subject: Re: A radial pattern contains randomize(bump) Date: 26 Aug 2023 05:20:00 Message:
```
{
"@context": "https://schema.org",
"@type": "DiscussionForumPosting",
"@id": "#web.64e9c33224fd44d93a0d3e9baa81652d%40news.povray.org",
"dateCreated": "2023-08-26T09:20:00+00:00",
"datePublished": "2023-08-26T09:20:00+00:00",
"author": {
"@type": "Person",
"name": "And"
}
}
Another one. This one superposition two such a bump.
```
Attachments:
Preview of image 'radial composite function.png'
From: And Subject: Re: A radial pattern contains randomize(bump) Date: 26 Aug 2023 05:30:00 Message:
```
{
"@context": "https://schema.org",
"@type": "DiscussionForumPosting",
"@id": "#web.64e9c4ec24fd44d93a0d3e9baa81652d%40news.povray.org",
"dateCreated": "2023-08-26T09:30:00+00:00",
"datePublished": "2023-08-26T09:30:00+00:00",
"author": {
"@type": "Person",
"name": "And"
}
}
Another one.
In fact I'm not very understand why these function expression generate such
pattern.
Of course I should try to use 2, 3, 4, ... maybe 6~7 bumps superposition.
```
Attachments:
Preview of image 'radial bump function record3.png'
From: And Subject: Re: A radial pattern contains randomize(bump) Date: 26 Aug 2023 05:35:00 Message:
```
{
"@context": "https://schema.org",
"@type": "DiscussionForumPosting",
"@id": "#web.64e9c73524fd44d93a0d3e9baa81652d%40news.povray.org",
"dateCreated": "2023-08-26T09:35:00+00:00",
"datePublished": "2023-08-26T09:35:00+00:00",
"author": {
"@type": "Person",
"name": "And"
}
}
Another situation is ... looks like near degradation.
Using more large number for coefficient of f_external_field.
#declare f_external_field =
function(x, y) {
2.0*sqrt(x*x + y*y) + 0.2
//^^^
}
This one using 2.0, not to mention 3.0, 4.0,...
The color map:
color_map {
[0.00 color rgb <0.19, 0.41, 0.89>]
[0.20 color rgb <0.22, 0.34, 0.62>]
[0.40 color rgb <0.17, 0.44, 0.55>]
[0.60 color rgb <0.22, 0.47,0.40>]
[0.80 color rgb <0.56, 0.33, 0.24>]
[1.00 color rgb <0.7, 0.15,0.08>]
}
```
Attachments:
Preview of image 'raidal composite degradation.png'
From: And Subject: Re: A radial pattern contains randomize(bump) Date: 27 Aug 2023 02:55:00 Message:
```
{
"@context": "https://schema.org",
"@type": "DiscussionForumPosting",
"@id": "#web.64eaf31624fd44d94b51e394aa81652d%40news.povray.org",
"dateCreated": "2023-08-27T06:55:00+00:00",
"datePublished": "2023-08-27T06:55:00+00:00",
"author": {
"@type": "Person",
"name": "And"
}
}
This one is a more special function.
Just like before, the function "f_freq_control_test" is what the output image
shows. And the color_map the same.
pigment {
function {f_freq_control_test(x,y,z)}
color_map {
[0.00 color rgb <0.19, 0.41, 0.89>]
[0.20 color rgb <0.22, 0.34, 0.62>]
[0.40 color rgb <0.17, 0.44, 0.55>]
[0.60 color rgb <0.22, 0.47,0.40>]
[0.80 color rgb <0.56, 0.33, 0.24>]
[1.00 color rgb <0.7, 0.15,0.08>]
}
}
```
Attachments:
Preview of image 'bump function power 1-x2.png'
From: And Subject: Re: A radial pattern contains randomize(bump) Date: 27 Aug 2023 03:30:00 Message:
```
{
"@context": "https://schema.org",
"@type": "DiscussionForumPosting",
"@id": "#web.64eafae324fd44d94b51e394aa81652d%40news.povray.org",
"dateCreated": "2023-08-27T07:30:00+00:00",
"datePublished": "2023-08-27T07:30:00+00:00",
"author": {
"@type": "Person",
"name": "And"
}
}
Because my mathematics is not so good, so I need spend so many years to think
these composite(I just try something I prepare for other concept).
The result yesterday is surprising me.
```
Attachments:
Preview of image 'ellipse bump function.png'
From: Thomas de Groot Subject: Re: A radial pattern contains randomize(bump) Date: 27 Aug 2023 07:31:00 Message: <64eb33f4@news.povray.org>
```
{
"@context": "https://schema.org",
"@type": "DiscussionForumPosting",
"@id": "#64eb33f4%40news.povray.org",
"dateCreated": "2023-08-27T11:31:00+00:00",
"datePublished": "2023-08-27T11:31:00+00:00",
"author": {
"@type": "Person",
"name": "Thomas de Groot"
}
}
Op 26-8-2023 om 11:14 schreef And:
> I suddenly solve an important problem. About how to generate a circle associated
> pattern.
>
> I need such a pattern for my wood-pattern generator.
> For wood heart, I don't have good function in the past. But these series opens a
> window.
I think you are doing alright with this series of experiments.
Integrating these functions (or combination of functions) in your
wood-pattern generator will be interesting. Have you already done this?
I would be interested to see how that looks.
--
Thomas
```
From: Bald Eagle Subject: Re: A radial pattern contains randomize(bump) Date: 27 Aug 2023 12:30:00 Message:
```
{
"@context": "https://schema.org",
"@type": "DiscussionForumPosting",
"@id": "#web.64eb791424fd44d91f9dae3025979125%40news.povray.org",
"dateCreated": "2023-08-27T16:30:00+00:00",
"datePublished": "2023-08-27T16:30:00+00:00",
"author": {
"@type": "Person",
"name": "Bald Eagle"
}
}
"And" <49341109@ntnu.edu.tw> wrote:
> I suddenly solve an important problem. About how to generate a circle associated
> pattern.
>
> I need such a pattern for my wood-pattern generator.
> For wood heart, I don't have good function in the past. But these series opens a
> window.
Hi And,
These are interesting pattern experiments, and I'm looking forward to seeing
what you come up with.
I find it interesting to see the pattern that you have here, as I didn't
understand how the effect was being generated from a circular base pattern.
I coded your pattern into a scene, and for some reason, did not get the same
result. Perhaps you could post the full code of a working scene.
Also, if you could describe the exact type of pattern that you are looking to
make, it would be easier to help you develop something through a logical
elaboration of a base pattern function.
I started with an atan2 (x, y) function, as that will give you a radial pattern
#version 3.8;
global_settings {assumed_gamma 1.0}
default {finish {diffuse 1}}
#include "functions.inc"
camera {
location <0, 0, -10>
right x*image_width/image_height
up y
look_at <0, 0, 0>
}
light_source {<0, 1, -50> rgb 1.0}
sky_sphere {pigment {rgb 1}}
#declare C = function {0.5 - sqrt (x*x+y*y) + 0.1}
#declare W1 = function {f_bumps (x/C(x, y, z), y/C(x, y, z), z)}
#declare Freq = 20;
#declare Ang = function {atan2 (x, y)/(tau/Freq)}
#declare W2 = function {f_bumps (Ang (x, y, z), Ang (x, y, z), z)}
#declare Wood1 =
color_map {
[0.00 srgb <249, 210, 173>/255]
//[0.25 srgb <249, 210, 173>/255]
[0.25 srgb <252, 156, 173>/255]
//[0.50 srgb <252, 156, 173>/255]
[0.50 srgb <206, 126, 72>/255]
//[0.75 srgb <206, 126, 72>/255]
[0.75 srgb <113, 51, 24>/255]
//[1.00 srgb <113, 51, 24>/255]
}
plane {z, 0 pigment {function {W1(x, y, z)} color_map {Wood1}} }
//plane {z, 0 pigment {function {W2(x, y, z)} color_map {Wood1}} }
Attached is what I get when I try to use the functions posted here.
```
Attachments:
From: Alain Martel Subject: Re: A radial pattern contains randomize(bump) Date: 27 Aug 2023 13:30:41 Message: <64eb8841\$1@news.povray.org>
```
{
"@context": "https://schema.org",
"@type": "DiscussionForumPosting",
"@id": "#64eb8841%241%40news.povray.org",
"dateCreated": "2023-08-27T17:30:41+00:00",
"datePublished": "2023-08-27T17:30:41+00:00",
"author": {
"@type": "Person",
"name": "Alain Martel"
}
}
Le 2023-08-27 à 03:27, And a écrit :
> Because my mathematics is not so good, so I need spend so many years to think
> these composite(I just try something I prepare for other concept).
>
> The result yesterday is surprising me.
That's not surprising for me. That's a formula for an ellipse that you
got there for f_external_field.
```
From: And Subject: Re: A radial pattern contains randomize(bump) Date: 28 Aug 2023 10:10:00 Message:
```
{
"@context": "https://schema.org",
"@type": "DiscussionForumPosting",
"@id": "#web.64eca9e624fd44d989131d8aaa81652d%40news.povray.org",
"dateCreated": "2023-08-28T14:10:00+00:00",
"datePublished": "2023-08-28T14:10:00+00:00",
"author": {
"@type": "Person",
"name": "And"
}
}
Thomas de Groot <tho### [at] degrootorg> wrote:
> I think you are doing alright with this series of experiments.
> Integrating these functions (or combination of functions) in your
> wood-pattern generator will be interesting. Have you already done this?
> I would be interested to see how that looks.
>
> --
> Thomas
Thank you... I have not integrated these function with my wood-pattern generator
yet.
In fact, I'm not sure the shape of such patterns in three dimensions. I need
researching and think more.
``` | 3,149 | 9,449 | {"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} | 2.546875 | 3 | CC-MAIN-2023-40 | latest | en | 0.609397 |
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Question
# Using the Pythagorean Theorem and quadratic equation to find side lengths of a right triangle. />The length of the longer leg of a right triangle is 20 inches more than twice the length of the shorter leg. The length of the hypotenuse is 22 inches more than twice the length of the shorter leg. Find the side lengths of the triangle.Length of the shorter leg:Length of the longer leg:Length of the hypotenuse:
Length of the shorter leg: L = 1 4 Length... View the full answer
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# What is Bijective function with example?
A function f: X→Y is said to be bijective if f is both one-one and onto. Example: For A = {1,−1,2,3} and B = {1,4,9}, f: A→B defined as f(x) = x2 is surjective. Example: Example: For A = {−1,2,3} and B = {1,4,9}, f: A→B defined as f(x) = x2 is bijective. A function is a bijection if it is both injective and surjective.
Besides, How do you show a bijection? According to the definition of the bijection, the given function should be both injective and surjective. In order to prove that, we must prove that f(a)=c and f(b)=c then a=b. Since this is a real number, and it is in the domain, the function is surjective.
What is surjective function example? Surjective function is a function in which every element In the domain if B has atleast one element in the domain of A such that f(A)=B. Let A={1,−1,2,3} and B={1,4,9}. Then, f:A→B:f(x)=x2 is surjective, since each element of B has at least one pre-image in A.
Likewise, What is meant by injective function?
In mathematics, an injective function (also known as injection, or one-to-one function) is a function f that maps distinct elements to distinct elements; that is, f(x1) = f(x2) implies x1 = x2. (Equivalently, x1 ≠ x2 implies f(x1) ≠ f(x2) in the equivalent contrapositive statement.)
In respect to this, How do you prove surjective? To prove a function, f : A → B is surjective, or onto, we must show f(A) = B. In other words, we must show the two sets, f(A) and B, are equal.
## How do you prove injective?
So how do we prove whether or not a function is injective? To prove a function is injective we must either: Assume f(x) = f(y) and then show that x = y. Assume x doesn’t equal y and show that f(x) doesn’t equal f(x).
## Is a rational function a bijection?
The range is , but it is not bijective. A rational function is the ratio of two polynomials. You observed if there’s anything “funky” in the denominator, then certainly R won’t get mapped to (0,1). Now just think about what “funky” really means..
## What is the difference between injective and surjective?
The function is bijective (one-to-one and onto, one-to-one correspondence, or invertible) if each element of the codomain is mapped to by exactly one element of the domain. That is, the function is both injective and surjective.
Bijection, injection and surjection.
surjective non-surjective
non- injective surjective-only general
## Why is x3 injective?
As we all know, this cannot be a surjective function, since the range consists of all real values, but f(x) can only produce cubic values. Also from observing a graph, this function produces unique values; hence it is injective.
## What is meant by Bijective function?
In mathematics, a bijection, also known as a bijective function, one-to-one correspondence, or invertible function, is a function between the elements of two sets, where each element of one set is paired with exactly one element of the other set, and each element of the other set is paired with exactly one element of …
## What are the two types of functions?
Ans. 2 The different types of functions are as follows: many to one function, one to one function, onto function, one and onto function, constant function, the identity function, quadratic function, polynomial function, modulus function, rational function, signum function, greatest integer function and so on.
## What is a Bijective function Class 12?
Bijective. Function : one-one and onto (or bijective) A function f : X → Y is said to be one-one and onto (or bijective), if f is both one-one and onto. Numerical: Let A be the set of all 50 students of Class X in a school. Let f : A →N be function defined by f (x) = roll number of the student x.
## What is bijection in sets?
In mathematics, a bijection, also known as a bijective function, one-to-one correspondence, or invertible function, is a function between the elements of two sets, where each element of one set is paired with exactly one element of the other set, and each element of the other set is paired with exactly one element of …
## How do you check if function is surjective?
Another way to think about it. If f:X→Y is a function then for every y∈Y we have the set f−1({y}):={x∈X∣f(x)=y}. f is injective iff f−1({y}) has at most one element for every y∈Y. f is surjective iff f−1({y}) has at least one element for every y∈Y.
## Which functions are surjective?
In mathematics, a surjective function (also known as surjection, or onto function) is a function f that maps an element x to every element y; that is, for every y, there is an x such that f(x) = y. In other words, every element of the function’s codomain is the image of at least one element of its domain.
## What is injective function example?
Examples of Injective Function
If function f: R→ R, then f(x) = 2x is injective. If function f: R→ R, then f(x) = 2x+1 is injective. If function f: R→ R, then f(x) = x2 is not an injective function, because here if x = -1, then f(-1) = 1 = f(1).
## Are all inverse function bijective?
Functions that have inverse functions are said to be invertible. A function is invertible if and only if it is a bijection. for every y in Y there is a unique x in X with y = f(x).
## Are all functions injective?
If the domain of a function is the empty set, then the function is the empty function, which is injective. If the domain of a function has one element (that is, it is a singleton set), then the function is always injective.
## Are rational functions surjective?
There cannot exist a rational function f:R→R injective, not surjective.
## Are all functions surjective?
Any function induces a surjection by restricting its codomain to the image of its domain. Every surjective function has a right inverse assuming the axiom of choice, and every function with a right inverse is necessarily a surjection. The composition of surjective functions is always surjective.
## How do you check if function is surjective?
Graph. Whenever we are given a graph, the easiest way to determine whether a function is a surjections is to compare the range with the codomain. If the range equals the codomain, then the function is surjective, otherwise it is not, as the example below emphasizes.
## Is COSX Bijective function?
So no, it’s not bijective.
## Is this function surjective?
In mathematics, a surjective function (also known as surjection, or onto function) is a function f that maps an element x to every element y; that is, for every y, there is an x such that f(x) = y. In other words, every element of the function’s codomain is the image of at least one element of its domain. | 1,691 | 6,662 | {"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-2022-27 | latest | en | 0.929624 |
https://www.eastcape.org/decimal-placement-chart/ | 1,591,337,793,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348493151.92/warc/CC-MAIN-20200605045722-20200605075722-00591.warc.gz | 695,165,333 | 8,876 | eastcape.org
# 5 Lovely Collection Of Decimal Placement Chart
Saturday, May 23rd, 2020 - Chart
## 5 Lovely Collection Of Decimal Placement Chart
Decimal Place Value Lesson Plans Task Cards and Quiz from decimal placement chart , image source: pinterest.com
decimal place value chart math salamanders we have a range of different decimal place value charts for you to print each chart es in several different forms so that you can choose the one that most suits your needs the first chart has headings and one space below to record a number there are multiple charts on each page the second chart has headings and enough space for 10 decimals place value enchantedlearning decimals place value enchantedlearning is a user supported site as a bonus site members have access to a banner ad free version of the site with print friendly pages decimals and that is a decimal number we can continue with smaller and smaller values from tenths to hundredths and so on like in this example have a play with decimal numbers yourself and small so our decimal system lets us write numbers as large or as small as we want using the decimal point place value chart examples solutions videos place value chart for decimals for decimals the place value chart will include the decimal point and the positions after the decimal point as illustrated in the following table remember that the place value after the decimal point is tenths decimal place value to thousandths show step by step solutions printable place value charts whole numbers and decimals both files include the types and range of charts described below types of place value charts there are four types of charts for you to choose from just print the type of chart that fits your preference chart with abbreviation and place value colored chart with abbreviation and place value black and white chart with place the decimal place value chart is essential for teaching the decimal place value chart a great tool that helps students as they re learning to read and write decimal numbers we need to make sure kids understand the difference between a whole number and a decimal number decimals place value chart example 1 in this presentation we shall see how to represent decimals using the place value chart
## Pictures gallery of 5 Lovely Collection Of Decimal Placement Chart
5 Lovely Collection Of Decimal Placement Chart | | 4.5 | 440 | 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} | 2.890625 | 3 | CC-MAIN-2020-24 | latest | en | 0.886767 |
https://study.com/academy/topic/graphing-with-functions.html | 1,571,526,247,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986700435.69/warc/CC-MAIN-20191019214624-20191020002124-00082.warc.gz | 597,373,088 | 25,701 | # Ch 11: Graphing with Functions
Watch online math video lessons to learn methods for factoring and graphing polynomial functions. These lessons utilize engaging animations and graphics that bring these concepts to life.
## Graphing with Functions - Chapter Summary and Learning Objectives
Take advantage of the video lessons in this chapter to learn graphing basics for polynomial functions. Our instructors introduce you to some real-world applications for the slopes, tangents and asymptotes depicted in the graphs of quadratic functions, trigonometric functions and more. You can also get an introduction to the basic properties of exponents and the methods used to simplify or factor polynomial expressions. By the time you reach the end of this chapter, you should be able to do the following:
• Simplify exponents in polynomial functions
• Find the slope of a line and identify tangents
• Describe implicit functions and asymptotes
Video Objectives
Graphing Basic Functions Review components of the coordinate plane, including the x- and y-axes, origin and quadrants.
Polynomials Functions: Properties and Factoring Learn what a polynomial is and explore methods for factoring quadratic functions.
Polynomials Functions: Exponentials and Simplifying Explain the addition, multiplication, division and powers properties of exponents. Use them to simplify polynomial functions.
Slopes and Tangents on a Graph Explore methods for calculating slopes and tangents on a graph.
Horizontal and Vertical Asymptotes Learn to identify horizontal and vertical asymptotes.
Implicit Functions Explain the connection between inverse and implicit functions.
6 Lessons in Chapter 11: Graphing with Functions
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### Other Chapters
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Support | 518 | 2,657 | {"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-2019-43 | longest | en | 0.856205 |
https://www.studypug.com/linear-algebra/the-inverse-of-3-x-3-matrices-with-matrix-row-operations | 1,718,608,137,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861698.15/warc/CC-MAIN-20240617060013-20240617090013-00684.warc.gz | 889,652,693 | 59,302 | # The inverse of 3 x 3 matrices with matrix row operations
0/1
##### Intros
###### Lessons
1. The Inverse of 3 x 3 Matrix using matrix row operations
0/5
##### Examples
###### Lessons
1. Finding the Inverse
You are given that . Find the inverse of this matrix.
1. You are given that . Find the inverse of this matrix.
1. You are given that . Find the inverse of this matrix.
1. You are given that . Find the inverse of this matrix.
1. You are given that . Find the inverse of this matrix. | 129 | 489 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 60, "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} | 2.921875 | 3 | CC-MAIN-2024-26 | latest | en | 0.765745 |
https://www.naukri.com/code360/problems/ninja-s-apartment_1466926 | 1,726,819,027,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652138.47/warc/CC-MAIN-20240920054402-20240920084402-00177.warc.gz | 838,352,580 | 44,439 | Code360 powered by Coding Ninjas X Naukri.com. Code360 powered by Coding Ninjas X Naukri.com
Ninja’s Apartment
Easy
0/40
Average time to solve is 15m
Contributed by
9 upvotes
Asked in companies
Problem statement
Ninja is planning to build a new apartment but he wants an apartment in the shape of a rectangle whose length is greater than the breadth of the rectangle. Also, the difference between the length and breadth should be minimum for the area of the apartment.
So your task is to print the length and breadth of the rectangle. You are provided with the ‘AREA’ of the region in which the apartment is to be built.
Example:
``````Suppose given ‘AREA’ is ‘6’. So, possible combinations are [ [ 1, 6 ], [ 2, 3 ], [ 3, 2 ], [ 6, 1 ] ]. We can’t choose [ 1, 6 ], [ 2, 3 ] as in this case breadth is greater than length.
So, we choose [ 3, 2 ] over [ 6, 1 ] as the difference between | 3 - 2 | < | 6 - 1 | so we return | 3, 2 | as the answer.
``````
Note:
``````You are not required to print anything explicitly. It has already been taken care of. Just implement the function.
``````
Detailed explanation ( Input/output format, Notes, Images )
Input Format:
``````The first line of input contains a ‘T’ number of test cases.
The first line of each test case contains an integer ‘AREA’ denoting the area of the rectangle.
``````
Output Format:
``````For each test case, print the length and breadth of the given area.
``````
Note:
``````You do not need to print anything, it has already been taken care of. Just implement the given function.
``````
Constraints:
``````1 <= 'T' <= 10^2
2 <= ‘AREA’ <= 10 ^ 6
Where ‘T’ represents the number of test cases and ‘AREA’ represents the given area of the apartment.
Time Limit: 1sec
``````
Sample Input 1:
``````2
12
8
``````
Sample Output 1:
``````4 3
4 2
``````
Explanation For Sample Input 1:
``````In the first test case,
Given ‘AREA’ is ‘12’ so possible combinations are [ [ 1, 12 ], [ 2, 6 ], [ 3, 4 ], [ 4, 3 ], [ 6, 2 ], [ 12, 1 ] ]. We can’t choose [ 1, 12 ], [ 2, 6 ], [ 3, 4 ] as in this case breadth is greater than length.
We choose [ 4, 3 ] over [ 6, 2 ] and [ 12, 1 ] as the difference between | 4 - 3 | < | 6 - 2 | and | 4 - 3 | < | 12 - 1 | so we print | 4, 3 | as the answer.
In the second test case,
Given ‘AREA’ is ‘8’ so possible combinations are [ [ 1, 8 ], [ 2, 4 ], [ 4, 2 ], [8, 1 ] ]. We cant choose [ 1, 8 ], [ 2, 4 ] as in this case breadth is greater than length.
We choose [ 4, 2 ] over [ 8, 1 ] as the difference between | 4 - 2 | < | 8 - 1 | so we print | 3, 2 | as the answer.
``````
Sample Input 1:
``````2
17
16
``````
Sample Output 1:
``````17 1
4 4
``````
Explanation For Sample Input 2:
``````In the first test case,
Given ‘AREA’ is ‘17’ so possible combinations are [ [ 1, 17 ], [ 17, 1 ] ]. We can’t choose [ 1, 17 ] as in this case breadth is greater than length.
We choose [ 17, 1] and so we print | 17, 1 | as the answer.
In the second test case,
Given ‘AREA’ is ‘16’ so possible combinations are [ [ 1, 16 ], [ 2, 8 ], [ 4, 4], [ 8, 2 ], [16, 1 ] ]. We cant choose [ 1, 16 ], [ 2, 8 ] as in this case breadth is greater than length.
We choose [ 4, 4 ] over [ 8, 2 ] and [16,1] as the difference between | 4 - 4 | < | 8 - 2 | so we print | 4, 4 | as the answer.
``````
Console | 1,102 | 3,297 | {"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} | 3.1875 | 3 | CC-MAIN-2024-38 | latest | en | 0.923199 |
https://www.kiddom.co/content/887-direct-variation-models | 1,519,510,151,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815951.96/warc/CC-MAIN-20180224211727-20180224231727-00644.warc.gz | 882,197,692 | 8,446 | ### Direct Variation Models
Examines how to solve word problems using direct variation models.
Grades 6 through 8 Material from CK-12
#### Why Use This Content?
See when students have begun their work. Or not.
#### Standards
##### 7.EE.B.3
Solve multi?step real? life and mathematical problems posed with positive and negative rational numbers in any form (whole numbers, fractions, and decimals), using tools strategically. Apply properties of operations to calculate with numbers in any form; convert between forms as appropriate; and assess the reasonableness of answers using mental computation and estimation strategies.
##### 7.EE.B.4
Use variables to represent quantities in a real? world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities.
##### 7.RP.A.2.B
Recognize and represent proportional relationships between quantities.
##### 7.RP.A.2.C
Recognize and represent proportional relationships between quantities.
##### 8.EE.B.5
Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. For example, compare a distance?time graph to a distance?time equation to determine which of two moving objects has greater speed.
##### 8.F.B.4
Construct a function to model a linear relationship between two quantities. Determine the rate of change and initial value of the function from a description of a relationship or from two (x, y) values, including reading these from a table or from a graph. Interpret the rate of change and initial value of a linear function in terms of the situation it models, and in terms of its graph or a table of values. | 353 | 1,752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-09 | latest | en | 0.892745 |
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# Volume of an Object by Using Calculus
12/16/2015 2:43 PM
On this picture we see a octagonal dome. I am trying to calculate the volume of this object by integral calculus but I can't find a way. How would you calculate this? https://dl.dropboxusercontent.com/u/17974596/Sk%C3%A6rmbillede%202015-12-17%20kl.%2002.14.48.png
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#1
### Re: Volume of an Object by Using Calculus
12/16/2015 3:01 PM
It is the sum of eight slices. Each slice has a triangle as basis limited on both sides by planes and the 3rd surface is to be defined if it is a cylinder or generated by any other 2nd degree curve translated along the triangle side.
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#2
### Re: Volume of an Object by Using Calculus
12/16/2015 4:11 PM
Calculate it as a hemisphere then deduct ~4%...
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#3
### Re: Volume of an Object by Using Calculus
12/16/2015 4:34 PM
I think you can do this by simply calculating for only one slice of the eight sections. Then multiply that result by eight for the answer.
The base of one slice will form a triangle subtended at 45° and the opposite angles will be 67.5° and 67.5° respectively. Calculating the area is now simple.
Now think of each subsequent horizontal triangular slice stacked in the vertical direction. Volume is simply the sum of the area of each stacked triangle.
Since the top is a hemisphere you should be able to simply calculate the lengths of each side of each subsequent stacked triangle based on the profile of that hemisphere, but the curve will not be a pure curve, rather it will be half of an eclipse due to the rotation of the two vertical edges by 22.5° (as viewed from a god's eye view).
Again, you can calculate the eclipse created and therefore the subsequent sides of each triangle with a little geometry.
You should be able to figure out the geometry. Just don't forget to multiply the result by eight!
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#9
### Re: Volume of an Object by Using Calculus
12/17/2015 4:41 AM
Why must be the top a hemisphere ?
As I mentioned the sides can be also a cylinder surface. Why not ?
And the cylinder must not have its axis at the center !
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#4
### Re: Volume of an Object by Using Calculus
12/16/2015 5:46 PM
It doesn't say it's symmetrical, but let's assume it is. I.e., the 'radius' is the same as the height of the dome.
Let r be the radius of the dome. This will also the be length, r1, of the hypotenuse of the bottom-most octagonal piece.
The area of this triangle is A1 = r12 *cosθsinθ, where θ is 22.5 degrees.
The volume of the first slice will be this area times the thickness t of the slice. (You should make t as thin as possible). V1 = A1*t
The area of the next triangle is A2 = r22 *cosθsinθ.
Here, r2 is given by r2 = √(r2 - t2), and then V2 = A2*t
Continue calculating the slices n number of times, where n is r/t.
Then Sum the triangles to get the volume of one chord of the dome.
V(total of 1 chord) = ∑ (V1 + V2 +... Vn)
V(dome) = 8*V(total of 1 chord)
The sketch is a bit cluttered where I show r1, r2, and r3 along the curve of the dome. I also drew-in, but didn't label r4. Nevertheless, this should be enough to illustrate my approach.
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#5
### Re: Volume of an Object by Using Calculus
12/16/2015 6:04 PM
The angle subtended is 45°, not 22.5° for the triangle.
The base is an octagon, so it will be 45°.
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#6
### Re: Volume of an Object by Using Calculus
12/16/2015 6:57 PM
I should have labeled the angle I'm calling theta.
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#7
### Re: Volume of an Object by Using Calculus
12/16/2015 9:39 PM
Yes, that helps!
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#8
### Re: Volume of an Object by Using Calculus
12/16/2015 9:55 PM
Details are missing on the shape of the red curves, and how high the dome is compared to its width.
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#10
### Re: Volume of an Object by Using Calculus
12/17/2015 5:27 AM
Fill it up completely with water, and measure the volume needed to fill it using a flowmeter on the fill pipe.
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#11
### Re: Volume of an Object by Using Calculus
12/17/2015 5:42 AM
The dome looks like Filippo Brunelleschi's work at the cathedral of Santa Maria del Fiore in Florence (Firenze).
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#12
### Re: Volume of an Object by Using Calculus
12/17/2015 6:01 AM
Using Calculus: Get a formula for circumference C as a function of height h. Then integrate C with respect to h, i.e. ∫C(h)dh.
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#13
### Re: Volume of an Object by Using Calculus
12/17/2015 6:53 AM
You need a maths expression for the shape of the curve of the dome to start with.
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#14
### Re: Volume of an Object by Using Calculus
12/17/2015 7:21 AM
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#15
### Re: Volume of an Object by Using Calculus
12/17/2015 12:27 PM
The figure shows a dome with an octagonal basis. Most probably because of the building constrains the surfaces are generated along a at least second degree guide. The equation for this guide line can be: l= a-P(z) with P(z) = a0*z^2+a1*z+a2
The coefficients can be determined with the boundary conditions:
At z= 0 l=a ; dP/dz= c → a2=0 and c= a1
at z= H i=b → b= a0*H²+c*H → (b - c*H)/ H²= a0
thus l(z)=a-[(b-c*H)/H²*z²+c*z]
If the guide line is normal to the base plane at z=0 then c=0.
For this particular case the red triangle area will be S(z)= l²*tan(22.5°) in a general case with a different number of sides S(z)= l²*tan(α) where α= π/n n being the number of sides the polygon has.
The elementary volume is dV= S(z)*dz = l²*tan(α)*dz= {a-[(b-c*H)/H²*z²+c*z]}²*tan(α)*dz.
The integral (from z=0 to z=H) is easy to solve since it is the integral of a polynomial.
Result should to be multiplied by "n".
To validate the result follow the given counsel: fill it with water and measure how much was in.
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#16
### Re: Volume of an Object by Using Calculus
12/17/2015 1:00 PM
(deleted comment)
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#17
### Re: Volume of an Object by Using Calculus
12/18/2015 5:48 AM
In previous comment I proposed a general solution.
A particular case is the dome enclosed in a spherical envelope.
In this case a= R cos(α)
and the guide line is an elliptical arc with the equation:
(l(z)/a)^2+(z/R)^2=1 → l(z)^2= a^2*(1-(z/R)^2) = (R*cos(α))^2*(1-(z/R)^2)
The triangle area is S(z)= tan(α)*l(z)^2 = tan(α)*(cos(α))^2*(R^2-z^2) = 0.5*sin(2*α)*(R^2-z^2)
The slice volume will thus be
H
V1=0.5*sin(2*α )*∫(R^2-z^2)*dz =
0
= 0.5*sin(2*α )*(R^2*H-H^3/3)= 0.5*sin(2*π/n)*R^3*(χ-χ^3/3)
V= n*V1 χ=H/R
The solution I suggested was for cases when the guide line equation is not as simple as in this particular case.
The solution is correct since if n→∞ V= 2/3*π*R^3 which is half of the sphere volume.
Interested in this topic? By joining CR4 you can "subscribe" to | 2,750 | 9,245 | {"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-2018-13 | latest | en | 0.910798 |
https://st-georges.info/jaa-atpl-mass-and-balance-94/ | 1,603,429,859,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107880656.25/warc/CC-MAIN-20201023043931-20201023073931-00404.warc.gz | 544,641,965 | 12,441 | # JAA ATPL MASS AND BALANCE PDF
Mass and Balance covers the factors governing the loading of an aeroplane to ensure that the longitudinal centre of gravity and mass are within the structural. Questions-Mass-and-balance · View. Mass and Balance, Oxford · View. Mass and Balance ATPL · View. Operational Procedures · View. Buy JAA ATPL Mass & Balance – JA and other aviation books at yout pilot shop headquarters, Pilots HQ.
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In the preceding examples the reference datum was the centre of gravity of the unladen seesaw, which was coincident with the fulcrum. How much freight must be added to a hold 10ft forward of Datum to move the CG 1ft forward.
## 031 Aircraft Mass & balance (JAA ATPL theory).
A mechanical device inserted between the cargo and the aircraft floor to distribute the weight evenly over a greater floor area. The graphs shall be identified by aircraft designation, date of compilation and source.
Tempest Tempest – Aero. Sector Distance 1, nm. CG envelope from 1 inch to 4 inches forward of the datum at all weights.
The Total Weight of the aeroplane then comprises of the APS weight plus balancs Disposable Load, which is made up of the usable fuel and traffic load.
As already discussed, the disposable load consists of the usable fuel and the traffic load. The following examples illustrate the method of calculating the weight and arm of empty aircraft with various datum positions.
Seat 7 0 lb. Add the weight and moment of the ,ass expected to remain in the tanks at the end of the flight. The maximum traffic load that may be carried in any circumstances.
The fuel slide is indexed in Imperial gallons. He must establish a system to check that amendments of his tapl data are incorporated properly in the system and that the system is operating correctly on a continuous basis by verifying the output data at intervals not exceeding six months. The Joint Aviation Authority has the maa of ensuring that all public transport aircraft, irrespective of size or number of engines, are operated to the highest possible level of safety.
BEGINNING AND INTERMEDIATE ALGEBRA 3RD EDITION MESSERSMITH PDF
It would be prudent to re-weigh the aircraft if it is uaa attaining is scheduled performance level. The following examples illustrate how to convert the position of a CG which is given as a percentage of the MAC into a position which qtpl relative to the reference datum, and the reverse. Specify the current restraint load factor forward for lashing and tie-down equipment for fixed wing aircraft.
Determine the change in the position of the CG if 1, kg of load is moved from a hold 12 metres aft of the datum to another 5 metres aft of the datum. When loading the aeroplane care must be taken not to exceed the maximum weight permitted in specific baggage areas.
As and when you are introduced to new aircraft types, both during your flight training and during your subsequent career, you will be taught the loading procedures which are specific to that particular aircraft type. Add the take-off fuel Box 37 to the zero fuel weight to obtain the take-off weight Box Its purpose is to extend the load intensity over a larger floor area than the base of the item and at the same time protect the aircraft floor from damage.
Basic weight 50, kgs; Basic equipment 5, kgs; Usable oil kgs; Variable load kgs; Traffic load kgs; Usable fuel kgs. The weight and moment of the crew is included in Part B. In this example the total passenger weight is 23, lb. Dividing the total moment by is simply a device which is used in order to keep the numbers to a manageable magnitude.
GLENN NUSSDORF PDF
### Aircraft Mass & balance (JAA ATPL theory).
List the factors that may limit the take-off weight. Insert at 3 and Aviation Press Aviation Specia. If the centre tank balaance the MRJT contains kgs of fuel how much fuel must be in the wing tanks? So far all of the calculations concerning the position of the CG of an aircraft have considered the position along the fore and aft axis. Standard weights are assumed for this exercise. The position of the CG measured along the fore and aft axis of the aircraft will change due to changes in aircraft configuration passenger configuration with seats in, freight configuration with seats outtotal weight and distribution of the fuel load at any given point in the flight, total weight and distribution of the payload, and so on.
Cessna Champion Champion Aerosp. Role equipment is that which is required to complete a specific task such as seats, toilets, galley for the passenger role or roller conveyor, lashing points and tie down equipment for the freight roles. They are further classified as male or female. Maximum Ba,ance Weight at B 37, kg. The effects of operating with the centre of gravity aft of the permitted aft limit include: Wicom Znd Windcone Sales.
In this case at any specified distance from the fulcrum, the turning moment that is the downward force imposed at that point will be equal on both sides of the fulcrum.
The total weight of the aeroplane comprises the APS weight plus the disposable load, which is made up of usable fuel and the payload. | 1,178 | 5,419 | {"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} | 2.890625 | 3 | CC-MAIN-2020-45 | latest | en | 0.887229 |
http://rfcafe.com/miscellany/quizzes/rf-cafe-quiz-61-answers-practical-microwave-circuits.htm | 1,606,770,142,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141486017.50/warc/CC-MAIN-20201130192020-20201130222020-00508.warc.gz | 75,948,737 | 6,141 | # Practical Microwave CircuitsAnswers to RF Cafe Quiz #61
All RF Cafe quizzes make great fodder for employment interviews for technicians or engineers - particularly those who are fresh out of school or are relatively new to the work world. Come to think of it, they would make equally excellent study material for the same persons who are going to be interviewed for a job.
Click here for the complete list of RF Cafe Quizzes.
Note: Some material based on books have quoted passages.
This quiz is based on the information presented in Practical Microwave Circuits, by Stephen A. Maas.
Note: Some of these books are available as prizes in the monthly RF Cafe Giveaway.
"This book is a collection of things I learned. largely on my own, over forty years in microwave circuit design. While almost all of the material in this book existed somewhere in some form, no single source was comprehensive and accessible. I learned most of it by experience, talking to people and digging through technical papers and books. In some cases, I just had to figure it out..."
1. Where are coupled transmission lines likely to be found?
d) All the above.
Coupled transmission lines are useful for many types of components, most notably directional couplers,
but also for filters, baluns, and control components.
(see page 31)
2. What is a homogeneous transmission line?
d) A transmission line in which all of the field lines are contained within a single dielectric medium.
(see page 17)
3. Which transmission line is an example of a non-homogeneous transmission line?
b) Stripline.
In stripline, microstrip, coplanar waveguide, and suspended substrate are all examples of non-homogeneous
transmission lines because the field lines exist both within and outside of the substrate dielectric.
(see page 18)
4. Which problem did the Lange coupler solve for two-strip planar couplers?
a) The inability to achieve sufficient coupling between two edge-coupled strips to realize a 3-dB coupler.
(see page 49)
5. Which of the following describes a balanced transmission line?
d) Two conductors of identical cross-section surrounded by free space (isolated from ground).
(see page51)
6. Which of the following describes an unbalanced transmission line?
c) Uses the ground surface as one of the two conductors.
(see page 51)
7. When did S parameters first come into common use?
b) 1960s.
They came into common use in the 1960s and 1970s as accurate automated microwave
network analyzers were created.
(see page 67)
8. What is the main difference between microwave and lower frequency analog circuit design?
a) Microwave design focuses more on maximum power transfer whereas lower frequency circuit
design focuses more on voltage gain.
(see page 113)
9. What is another name for a normalized lowpass filter design?
d) A prototype network.
(see page 134)
10. What is meant by "de-embedding" S parameters?
b) Removing all effects test fixture elements from the measurement of a discrete component being measured.
Effects of the test ports, their discontinuities, and transmission lines from the component must be
mathematically subtracted from the overall measurement.
(see page 202) | 701 | 3,203 | {"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} | 2.671875 | 3 | CC-MAIN-2020-50 | latest | en | 0.945748 |
https://msms.ehe.osu.edu/2008/08/04/give-us-our-daily-math/ | 1,568,941,794,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573801.14/warc/CC-MAIN-20190920005656-20190920031656-00550.warc.gz | 578,896,362 | 9,345 | # Give Us Our Daily Math
Middle schoolers may not easily see the connection between data analysis problems invented for the classroom and math problems encountered in their daily lives. You will spark their interest in data analysis by showing them its value in finding solutions to real problems in many settings — from buying a soda to taking after-school jobs to understanding weather reports. When you introduce data analysis to middle school students, you are exposing them to applications that correlate with the NCTM Principles and Standards: creating and reading graphs, calculating statistics, and, above all, solving real-world problems.
Working Hours: How Much Time Do Teens Spend on the Job?
This activity challenges students to interpret a bar graph to determine the average number of hours teenagers work per week. A hint suggests that students assume that 100 students participated in the survey. Interesting statistics about the hourly wages and annual salaries of various occupations are given.
Does It Make a Difference Where You Shop?
In this online activity, your students compare soda prices from two stores using data displayed on a scatterplot graph. Students are shown how the line y = x can be used to analyze the data and draw a conclusion. Further problems involving scatterplots compare car mileage and the performance of NBA players.
The Global Sun Temperature Project
This web site allows students from around the world to work together to determine how average daily temperatures and hours of sunlight change with distance from the equator. Students can participate in the project each spring, April-June. You will find project information, lesson plans, and implementation assistance.
New York Times Daily Lesson Plan: Mathematics
These lesson ideas from the New York Times offer suggestions for ways to draw on real-world issues and statistics to develop lessons in mathematics. Each lesson idea includes a description of the activities along with handouts and questions for discussion. Links to related Times articles provide an interactive aspect to each of the lesson entries. | 378 | 2,117 | {"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} | 3.0625 | 3 | CC-MAIN-2019-39 | latest | en | 0.916623 |
https://netlib.org/lapack/explore-html/d6/db6/zblat2_8f_aed016e31a353cc67889d981e60923382.html | 1,714,028,663,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712297290384.96/warc/CC-MAIN-20240425063334-20240425093334-00234.warc.gz | 377,453,337 | 7,718 | LAPACK 3.12.0 LAPACK: Linear Algebra PACKage
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## ◆ zchk5()
subroutine zchk5 ( character*6 sname, double precision eps, double precision thresh, integer nout, integer ntra, logical trace, logical rewi, logical fatal, integer nidim, integer, dimension( nidim ) idim, integer nalf, complex*16, dimension( nalf ) alf, integer ninc, integer, dimension( ninc ) inc, integer nmax, integer incmax, complex*16, dimension( nmax, nmax ) a, complex*16, dimension( nmax*nmax ) aa, complex*16, dimension( nmax*nmax ) as, complex*16, dimension( nmax ) x, complex*16, dimension( nmax*incmax ) xx, complex*16, dimension( nmax*incmax ) xs, complex*16, dimension( nmax ) y, complex*16, dimension( nmax*incmax ) yy, complex*16, dimension( nmax*incmax ) ys, complex*16, dimension( nmax ) yt, double precision, dimension( nmax ) g, complex*16, dimension( nmax ) z )
Definition at line 1798 of file zblat2.f.
1802*
1803* Tests ZHER and ZHPR.
1804*
1805* Auxiliary routine for test program for Level 2 Blas.
1806*
1807* -- Written on 10-August-1987.
1808* Richard Hanson, Sandia National Labs.
1809* Jeremy Du Croz, NAG Central Office.
1810*
1811* .. Parameters ..
1812 COMPLEX*16 ZERO, HALF, ONE
1813 parameter( zero = ( 0.0d0, 0.0d0 ),
1814 \$ half = ( 0.5d0, 0.0d0 ),
1815 \$ one = ( 1.0d0, 0.0d0 ) )
1816 DOUBLE PRECISION RZERO
1817 parameter( rzero = 0.0d0 )
1818* .. Scalar Arguments ..
1819 DOUBLE PRECISION EPS, THRESH
1820 INTEGER INCMAX, NALF, NIDIM, NINC, NMAX, NOUT, NTRA
1821 LOGICAL FATAL, REWI, TRACE
1822 CHARACTER*6 SNAME
1823* .. Array Arguments ..
1824 COMPLEX*16 A( NMAX, NMAX ), AA( NMAX*NMAX ), ALF( NALF ),
1825 \$ AS( NMAX*NMAX ), X( NMAX ), XS( NMAX*INCMAX ),
1826 \$ XX( NMAX*INCMAX ), Y( NMAX ),
1827 \$ YS( NMAX*INCMAX ), YT( NMAX ),
1828 \$ YY( NMAX*INCMAX ), Z( NMAX )
1829 DOUBLE PRECISION G( NMAX )
1830 INTEGER IDIM( NIDIM ), INC( NINC )
1831* .. Local Scalars ..
1832 COMPLEX*16 ALPHA, TRANSL
1833 DOUBLE PRECISION ERR, ERRMAX, RALPHA, RALS
1834 INTEGER I, IA, IC, IN, INCX, INCXS, IX, J, JA, JJ, LAA,
1835 \$ LDA, LDAS, LJ, LX, N, NARGS, NC, NS
1836 LOGICAL FULL, NULL, PACKED, RESET, SAME, UPPER
1837 CHARACTER*1 UPLO, UPLOS
1838 CHARACTER*2 ICH
1839* .. Local Arrays ..
1840 COMPLEX*16 W( 1 )
1841 LOGICAL ISAME( 13 )
1842* .. External Functions ..
1843 LOGICAL LZE, LZERES
1844 EXTERNAL lze, lzeres
1845* .. External Subroutines ..
1846 EXTERNAL zher, zhpr, zmake, zmvch
1847* .. Intrinsic Functions ..
1848 INTRINSIC abs, dble, dcmplx, dconjg, max
1849* .. Scalars in Common ..
1850 INTEGER INFOT, NOUTC
1851 LOGICAL LERR, OK
1852* .. Common blocks ..
1853 COMMON /infoc/infot, noutc, ok, lerr
1854* .. Data statements ..
1855 DATA ich/'UL'/
1856* .. Executable Statements ..
1857 full = sname( 3: 3 ).EQ.'E'
1858 packed = sname( 3: 3 ).EQ.'P'
1859* Define the number of arguments.
1860 IF( full )THEN
1861 nargs = 7
1862 ELSE IF( packed )THEN
1863 nargs = 6
1864 END IF
1865*
1866 nc = 0
1867 reset = .true.
1868 errmax = rzero
1869*
1870 DO 100 in = 1, nidim
1871 n = idim( in )
1872* Set LDA to 1 more than minimum value if room.
1873 lda = n
1874 IF( lda.LT.nmax )
1875 \$ lda = lda + 1
1876* Skip tests if not enough room.
1877 IF( lda.GT.nmax )
1878 \$ GO TO 100
1879 IF( packed )THEN
1880 laa = ( n*( n + 1 ) )/2
1881 ELSE
1882 laa = lda*n
1883 END IF
1884*
1885 DO 90 ic = 1, 2
1886 uplo = ich( ic: ic )
1887 upper = uplo.EQ.'U'
1888*
1889 DO 80 ix = 1, ninc
1890 incx = inc( ix )
1891 lx = abs( incx )*n
1892*
1893* Generate the vector X.
1894*
1895 transl = half
1896 CALL zmake( 'GE', ' ', ' ', 1, n, x, 1, xx, abs( incx ),
1897 \$ 0, n - 1, reset, transl )
1898 IF( n.GT.1 )THEN
1899 x( n/2 ) = zero
1900 xx( 1 + abs( incx )*( n/2 - 1 ) ) = zero
1901 END IF
1902*
1903 DO 70 ia = 1, nalf
1904 ralpha = dble( alf( ia ) )
1905 alpha = dcmplx( ralpha, rzero )
1906 null = n.LE.0.OR.ralpha.EQ.rzero
1907*
1908* Generate the matrix A.
1909*
1910 transl = zero
1911 CALL zmake( sname( 2: 3 ), uplo, ' ', n, n, a, nmax,
1912 \$ aa, lda, n - 1, n - 1, reset, transl )
1913*
1914 nc = nc + 1
1915*
1916* Save every datum before calling the subroutine.
1917*
1918 uplos = uplo
1919 ns = n
1920 rals = ralpha
1921 DO 10 i = 1, laa
1922 as( i ) = aa( i )
1923 10 CONTINUE
1924 ldas = lda
1925 DO 20 i = 1, lx
1926 xs( i ) = xx( i )
1927 20 CONTINUE
1928 incxs = incx
1929*
1930* Call the subroutine.
1931*
1932 IF( full )THEN
1933 IF( trace )
1934 \$ WRITE( ntra, fmt = 9993 )nc, sname, uplo, n,
1935 \$ ralpha, incx, lda
1936 IF( rewi )
1937 \$ rewind ntra
1938 CALL zher( uplo, n, ralpha, xx, incx, aa, lda )
1939 ELSE IF( packed )THEN
1940 IF( trace )
1941 \$ WRITE( ntra, fmt = 9994 )nc, sname, uplo, n,
1942 \$ ralpha, incx
1943 IF( rewi )
1944 \$ rewind ntra
1945 CALL zhpr( uplo, n, ralpha, xx, incx, aa )
1946 END IF
1947*
1948* Check if error-exit was taken incorrectly.
1949*
1950 IF( .NOT.ok )THEN
1951 WRITE( nout, fmt = 9992 )
1952 fatal = .true.
1953 GO TO 120
1954 END IF
1955*
1956* See what data changed inside subroutines.
1957*
1958 isame( 1 ) = uplo.EQ.uplos
1959 isame( 2 ) = ns.EQ.n
1960 isame( 3 ) = rals.EQ.ralpha
1961 isame( 4 ) = lze( xs, xx, lx )
1962 isame( 5 ) = incxs.EQ.incx
1963 IF( null )THEN
1964 isame( 6 ) = lze( as, aa, laa )
1965 ELSE
1966 isame( 6 ) = lzeres( sname( 2: 3 ), uplo, n, n, as,
1967 \$ aa, lda )
1968 END IF
1969 IF( .NOT.packed )THEN
1970 isame( 7 ) = ldas.EQ.lda
1971 END IF
1972*
1973* If data was incorrectly changed, report and return.
1974*
1975 same = .true.
1976 DO 30 i = 1, nargs
1977 same = same.AND.isame( i )
1978 IF( .NOT.isame( i ) )
1979 \$ WRITE( nout, fmt = 9998 )i
1980 30 CONTINUE
1981 IF( .NOT.same )THEN
1982 fatal = .true.
1983 GO TO 120
1984 END IF
1985*
1986 IF( .NOT.null )THEN
1987*
1988* Check the result column by column.
1989*
1990 IF( incx.GT.0 )THEN
1991 DO 40 i = 1, n
1992 z( i ) = x( i )
1993 40 CONTINUE
1994 ELSE
1995 DO 50 i = 1, n
1996 z( i ) = x( n - i + 1 )
1997 50 CONTINUE
1998 END IF
1999 ja = 1
2000 DO 60 j = 1, n
2001 w( 1 ) = dconjg( z( j ) )
2002 IF( upper )THEN
2003 jj = 1
2004 lj = j
2005 ELSE
2006 jj = j
2007 lj = n - j + 1
2008 END IF
2009 CALL zmvch( 'N', lj, 1, alpha, z( jj ), lj, w,
2010 \$ 1, one, a( jj, j ), 1, yt, g,
2011 \$ aa( ja ), eps, err, fatal, nout,
2012 \$ .true. )
2013 IF( full )THEN
2014 IF( upper )THEN
2015 ja = ja + lda
2016 ELSE
2017 ja = ja + lda + 1
2018 END IF
2019 ELSE
2020 ja = ja + lj
2021 END IF
2022 errmax = max( errmax, err )
2024 IF( fatal )
2025 \$ GO TO 110
2026 60 CONTINUE
2027 ELSE
2028* Avoid repeating tests if N.le.0.
2029 IF( n.LE.0 )
2030 \$ GO TO 100
2031 END IF
2032*
2033 70 CONTINUE
2034*
2035 80 CONTINUE
2036*
2037 90 CONTINUE
2038*
2039 100 CONTINUE
2040*
2041* Report result.
2042*
2043 IF( errmax.LT.thresh )THEN
2044 WRITE( nout, fmt = 9999 )sname, nc
2045 ELSE
2046 WRITE( nout, fmt = 9997 )sname, nc, errmax
2047 END IF
2048 GO TO 130
2049*
2050 110 CONTINUE
2051 WRITE( nout, fmt = 9995 )j
2052*
2053 120 CONTINUE
2054 WRITE( nout, fmt = 9996 )sname
2055 IF( full )THEN
2056 WRITE( nout, fmt = 9993 )nc, sname, uplo, n, ralpha, incx, lda
2057 ELSE IF( packed )THEN
2058 WRITE( nout, fmt = 9994 )nc, sname, uplo, n, ralpha, incx
2059 END IF
2060*
2061 130 CONTINUE
2062 RETURN
2063*
2064 9999 FORMAT( ' ', a6, ' PASSED THE COMPUTATIONAL TESTS (', i6, ' CALL',
2065 \$ 'S)' )
2066 9998 FORMAT( ' ******* FATAL ERROR - PARAMETER NUMBER ', i2, ' WAS CH',
2067 \$ 'ANGED INCORRECTLY *******' )
2068 9997 FORMAT( ' ', a6, ' COMPLETED THE COMPUTATIONAL TESTS (', i6, ' C',
2069 \$ 'ALLS)', /' ******* BUT WITH MAXIMUM TEST RATIO', f8.2,
2070 \$ ' - SUSPECT *******' )
2071 9996 FORMAT( ' ******* ', a6, ' FAILED ON CALL NUMBER:' )
2072 9995 FORMAT( ' THESE ARE THE RESULTS FOR COLUMN ', i3 )
2073 9994 FORMAT( 1x, i6, ': ', a6, '(''', a1, ''',', i3, ',', f4.1, ', X,',
2074 \$ i2, ', AP) .' )
2075 9993 FORMAT( 1x, i6, ': ', a6, '(''', a1, ''',', i3, ',', f4.1, ', X,',
2076 \$ i2, ', A,', i3, ') .' )
2077 9992 FORMAT( ' ******* FATAL ERROR - ERROR-EXIT TAKEN ON VALID CALL *',
2078 \$ '******' )
2079*
2080* End of ZCHK5
2081*
subroutine zher(uplo, n, alpha, x, incx, a, lda)
ZHER
Definition zher.f:135
subroutine zhpr(uplo, n, alpha, x, incx, ap)
ZHPR
Definition zhpr.f:130
logical function lze(ri, rj, lr)
Definition zblat2.f:3075
logical function lzeres(type, uplo, m, n, aa, as, lda)
Definition zblat2.f:3105
subroutine zmvch(trans, m, n, alpha, a, nmax, x, incx, beta, y, incy, yt, g, yy, eps, err, fatal, nout, mv)
Definition zblat2.f:2944
subroutine zmake(type, uplo, diag, m, n, a, nmax, aa, lda, kl, ku, reset, transl)
Definition zblat2.f:2751
Here is the call graph for this function:
Here is the caller graph for this function: | 3,441 | 8,642 | {"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} | 2.5625 | 3 | CC-MAIN-2024-18 | latest | en | 0.263738 |
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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 1.1 - Problem 24e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 1.1 - Problem 24e
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2 3 Composite functions and notation Let f (x)=x -4, g
ISBN: 9780321570567 2
Solution for problem 24E Chapter 1.1
Calculus: Early Transcendentals | 1st Edition
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Problem 24E
? 2 ? 3 Composite functions and notation Let ?f? (x)=x -4, g? (x)=x and ?F(x)=1? /(x-3). Simplify or evaluate the following expressions. ? 4 ? F(y )
Step-by-Step Solution:
Step by solution Step 1 of 1 2 3 Consider three different functions f(x)=x -4, g (x)=x and F(x)= 1/(x-3). If we want to find the value of funct ion F (x) for x =y we must use this value in the relation for the func tion (x).This is very easy, just follow the next steps: F(x)=1/(x-3) F(y )=1/(y -3)4 = y 3 . 4 1 So the final value of the function F(y ) is y 3 . 4 1 If we use the rule xn = x , we can write it as (y 3) .
Step 2 of 1
ISBN: 9780321570567
This full solution covers the following key subjects: Composite, evaluate, expressions, functions, let. This expansive textbook survival guide covers 85 chapters, and 5218 solutions. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. Since the solution to 24E from 1.1 chapter was answered, more than 289 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1. The answer to “? 2 ? 3 Composite functions and notation Let ?f? (x)=x -4, g? (x)=x and ?F(x)=1? /(x-3). Simplify or evaluate the following expressions. ? 4 ? F(y )” is broken down into a number of easy to follow steps, and 28 words. The full step-by-step solution to problem: 24E from chapter: 1.1 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM.
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https://ww2.mathworks.cn/matlabcentral/cody/problems/43042-convert-decimal-to-binary-and-then-generate-the-minimum-binary-it-can-with-jumbling/solutions/2021246 | 1,600,650,635,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400198868.29/warc/CC-MAIN-20200920223634-20200921013634-00318.warc.gz | 716,511,006 | 18,326 | Cody
# Problem 43042. Convert decimal to binary and then generate the minimum binary it can with jumbling
Solution 2021246
Submitted on 14 Nov 2019 by Nikolaos Nikolaou
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 10; y_correct = 3; assert(isequal(minBin(x),y_correct))
ans = '1010' y = '0 0 1 1' y = '0011' y = 3
2 Pass
x = 23; y_correct = 15; assert(isequal(minBin(x),y_correct))
ans = '10111' y = '0 1 1 1 1' y = '01111' y = 15
3 Pass
x = 15; y_correct = 15; assert(isequal(minBin(x),y_correct))
ans = '1111' y = '1 1 1 1' y = '1111' y = 15 | 243 | 682 | {"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} | 2.9375 | 3 | CC-MAIN-2020-40 | latest | en | 0.652639 |
https://www.speakingtree.in/blog/salary-theorem | 1,508,519,887,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824226.31/warc/CC-MAIN-20171020154441-20171020174441-00665.warc.gz | 1,020,976,751 | 22,419 | Apr 21, 2014 10:28pm
# SALARY THEOREM
282
VIEWS
0
COMMENT
Every one knows the Salary Theorem establishes that Engineers and Scientists can NEVER EVER earn as much money as basinessmen, salesmen, politicians and actors easily make.
This theorem can be demonstrated by reducing it to simple mathematical equation.
The equation rests on two postulates:
Postulate No.1 - KNOWLEDGE IS POWER
postulate No.2 - TIME IS MONEY
Given that - POWER = WORK/TIME
and becasue - KNOWLEDGE = POWER
TIME = MONEY
Therefore, we have,
KNOWLEDGE = WORK/MONEY
We can now easily obtain:
MONEY = WORK/KNOWLEDGE
So when KNOWLEDGE goes towards zero, MONEY goes towards infinity, regardless of the value attributed to work, even if the value of work is very small.
On the contrary, when KNOWLEDGE goes towards infinity, MONEY goes towards ZERO, even if the value of work is high.
The evident conclusion:
THE LESS YOU KNOW, THE MORE MONEY YOU DEFINITELY MAKE.
Those who can't understand the theorem can definitely make a lot of money.
0 COMMENT | 253 | 1,033 | {"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} | 3.296875 | 3 | CC-MAIN-2017-43 | latest | en | 0.894904 |
https://www.physicsforums.com/threads/i-need-help-solving-an-integral-over-density-of-states-for-electrons.520248/ | 1,526,832,591,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863626.14/warc/CC-MAIN-20180520151124-20180520171124-00051.warc.gz | 816,215,867 | 15,524 | # Homework Help: I need help solving an integral over density of states for electrons.
1. Aug 9, 2011
### mattek1979
1. The problem statement, all variables and given/known data
I am trying to retake an old course in statistical mechanics but run into integrals that i simply have forgotten how to solve.
Given an denstiry of states such that
$f(\epsilon)= \frac{1}{|\epsilon |}$ for $\epsilon_{min} \leq \epsilon < 0$ and 0 elsewhere
Using the mean occupation number for a fermi-dirac distribution, I am supposed to find the fermi energy for N electrons.
2. Relevant equations
I assume integrating
$dN(\epsilon)=\bar{n}(\epsilon)f(\epsilon)d\epsilon$
using
$\bar{n }=\frac{1}{e^{-\beta(\epsilon-\mu)}+1}$
and the above
$f(\epsilon)=\frac{1}{|\epsilon|}$
is the way to proceed.
3. The attempt at a solution
The integral I seek to solve is
N=$\int^{\epsilon_{min}}_{0}\frac{1}{|\epsilon|}\frac{1}{e^{-\beta(\epsilon-\mu)}+1}d\epsilon$
and I simply can't figure out if I need to do a subtitution of integration variables or if i am missing some other nifty technique.
All help appreciated
Sincerely
Mathias Kristoffersson
Last edited: Aug 9, 2011
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# What shape has one curved face and one flat shape?
Updated: 12/21/2022
Wiki User
11y ago
Cylindrical shape or Cylinder shape
Wiki User
11y ago
Earn +20 pts
Q: What shape has one curved face and one flat shape?
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Related questions
### What shape has one curved face and one flat face the flat face is a circle?
The answer is - a cone.
A Cone
A cone.
Cone.
### You are a 3D shape i have one flat face you have one curved face your flat face is circulart?
If you have one flat face and I have one curved face, which one of us is the 3d shape? Anyway, a 3d shape with pne f;at, circular face and one curved face could be: a section of a sphere or a section of an ellipsoid cut by a plane parallel to one of its axes, a section of a paraboloid or a cone. There are probably other shapes.
A cone
semi-sphere
A cylinder
Hemisphere
### What shape has one flat face one curved face?
A cone would fit the given description
sphere
### What 3 dimensional shape has one curved face and one flat face the flat face is a circle?
It could be a semi-sphere but mostly it's a cone. | 290 | 1,118 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2024-26 | latest | en | 0.955275 |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-6th-edition/chapter-p-prerequisites-fundamental-concepts-of-algebra-exercise-set-p-5-page-74/36 | 1,586,283,018,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371803248.90/warc/CC-MAIN-20200407152449-20200407182949-00090.warc.gz | 949,488,109 | 13,044 | ## College Algebra (6th Edition)
3$x^{2}$ + 4xy + $y^{2}$ = 3$x^{2}$ + 3xy + xy + $y^{2}$ = 3x(x+y) +y(x+y) =(3x+y)(x+y) 3$x^{2}$ + 4xy + $y^{2}$ is a composite polynomial (3x+y) and (x+y) are prime polynomials | 97 | 211 | {"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} | 3.984375 | 4 | CC-MAIN-2020-16 | latest | en | 0.603439 |
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Topology Course lecture notes - A. McCluskey, B. McMaster
# Topology Course lecture notes - A. McCluskey, B. McMaster
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Topology Course Lecture Notes
Aisling McCluskey and Brian McMasterAugust 1997
Chapter 1Fundamental Concepts
In the study of metric spaces, we observed that:(i) many of the concepts can be described purely in terms of open sets,(ii) open-set descriptions are sometimes simpler than metric descriptions,e.g. continuity,(iii) many results about these concepts can be proved using
only
the basicproperties of open sets (namely, that both the empty set and the un-derlying set
are open, that the intersection of any two open sets isagain open and that the union of arbitrarily many open sets is open).This prompts the question: How far would we get if we started with a collec-tion of subsets possessing these above-mentioned properties and proceededto define everything in terms of them?
1.1 Describing Topological Spaces
We noted above that many important results in metric spaces can be provedusing only the basic properties of open sets that
the empty set and underlying set
are both open,
the intersection of any two open sets is open, and
unions of arbitrarily many open sets are open.1
We will call any collection of sets on
satisfying these properties a topology.In the following section, we also seek to give alternative ways of describingthis important collection of sets.
1.1.1 Defining Topological Spaces
Definition 1.1
A
topological space
is a pair
(
X,
)
consisting of a set
and a family
of subsets of
satisfying the following conditions:(T1)
and
(T2)
is closed under arbitrary union (T3)
is closed under finite intersection.
The set
is called a
space
, the elements of
are called
points
of the spaceand the subsets of
belonging to
are called
open
in the space; the family
of open subsets of
is also called a
topology
for
.
Examples
(i) Any metric space (
X,d
) is a topological space where
d
, the topologyfor
induced by the metric
d
, is defined by agreeing that G shall bedeclared as open whenever each
x
in
G
is contained in an open ballentirely in
G
, i.e.
G
is open in (
X,
d
)
x
G,
r
x
>
0 such that
x
B
r
x
(
x
)
G.
(ii) The following is a special case of (i), above. Let
R
be the set of realnumbers and let
be the usual (metric) topology defined by agreeingthat
G
is open in (
R,
I
) (alternatively,
-open)
x
G,
r
x
>
0 such that (
x
r
x
,x
+
r
x
)
G.
(iii) Define
0
=
{∅
,
}
for any set X — known as the
trivial
or
anti-discrete
topology.(iv) Define
D
=
{
G
:
G
}
— known as the
discrete
topology.2 | 798 | 2,791 | {"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} | 3.421875 | 3 | CC-MAIN-2014-52 | latest | en | 0.913955 |
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•
## I really don't know where else to turn and could use some Asian insight to a problem I have encountered...
• Posted by a hidden member.
Hello Asians,
Thank you for clicking on my thread. Let me start off by saying I greatly appreciate the help for this problem I have encountered. I tried everything I could to solve it, but am now at the point where I need help, so I have come to you.
Here is the problem:
Which of the following functions f is f(x) = f(1-x) for all of x?
1) f(x)=1-x
2) f(x)=1-x^2
3) f(x)=x^2-(1-x)^2
4) f(x)=(x^2)(1-x)^2
5) f(x)=x/1-x
Like I have said, I tried everything to figure out this problem and realize that I need some special insight. If you could tell me the answer and, more importantly, how to find the answer I would love you a long time.
Thanks again, and you guys are the best!
• Posted by a hidden member.
Chainers saidHello Asians,
Thank you for clicking on my thread. Let me start off by saying I greatly appreciate the help for this problem I have encountered. I tried everything I could to solve it, but am now at the point where I need help, so I have come to you.
Here is the problem:
Which of the following functions f is f(x) = f(1-x) for all of x?
1) f(x)=1-x
2) f(x)=1-x^2
3) f(x)=x^2-(1-x)^2
4) f(x)=(x^2)(1-x)^2
5) f(x)=x/1-x
Like I have said, I tried everything to figure out this problem and realize that I need some special insight. If you could tell me the answer and, more importantly, how to find the answer I would love you a long time.
Thanks again, and you guys are the best!
It's basically asking which function(s) out of the five options will be equal to f(1-x).
So, first you have to understand what f(1-x) means. It's similar to finding f(2). You would usually plug in 2 for x, right? Well, try plugging in x=1-x in each of the five functions.
So, in essence you want to replace each of the x variables in each of your five functions by (1-x) and solve the algebra and figure out which one ends up giving f(x) again.
(and no I didn't just join to answer this lol. i just happened to return today and this just happens to be my first post)
• Posted by a hidden member.
^^no Im saying the fucking GMAT is racist because this is math that only an Asian can do and it is fucking horse shit.
Thats what I am saying.
• Posted by a hidden member.
So do you get it?
• Posted by a hidden member.
sirdreamalott saidSo do you get it?
See thats what I thought it said and I did that but I didnt find one that ended up equaling itself.
Maybe I just did it wrong but I didnt really find a correct answer to this problem.
• Posted by a hidden member.
I've been trying to think of a name for my new punk band, and I'd like to thank the OP for it's new name......Asian Insight! We'll all be performing while wearing hospital masks of course.
• Posted by a hidden member.
Chainers said
sirdreamalott saidSo do you get it?
See thats what I thought it said and I did that but I didnt find one that ended up equaling itself.
Maybe I just did it wrong but I didnt really find a correct answer to this problem.
Here's why:
The function in D is: f(x) = x^2 (1-x)^2
Plugging in x=(1-x) into D gives:
f(x) = (1-x)^2 (1-1+x)^2
= (1-x)^2 x^2
which is exactly the same function in D. Voila!
• Posted by a hidden member.
sirdreamalott said
Chainers said
sirdreamalott saidSo do you get it?
See thats what I thought it said and I did that but I didnt find one that ended up equaling itself.
Maybe I just did it wrong but I didnt really find a correct answer to this problem.
Here's why:
The function in D is: f(x) = x^2 (1-x)^2
Plugging in x=(1-x) into D gives:
f(x) = (1-x)^2 (1-1+x)^2
= (1-x)^2 x^2
which is exactly the same function in D. Voila!
Thank you so much for the help.
I have another one if you dont mind, its to find the value for M.
(1/5)^m*(1/4)^17=1/2(10^32)
How do you find M with this abomination?
• Posted by a hidden member.
$f(x)=x^2(1-x)^2 ---So, f(1-x)=(1-x)^2(1-1+x)^2=(1-x)^2(x)^2=x^2(1-x)^2$
Just a prettier latex version as to why it's 4.
• Posted by a hidden member.
Chainers said
sirdreamalott said
Chainers said
sirdreamalott saidSo do you get it?
See thats what I thought it said and I did that but I didnt find one that ended up equaling itself.
Maybe I just did it wrong but I didnt really find a correct answer to this problem.
Here's why:
The function in D is: f(x) = x^2 (1-x)^2
Plugging in x=(1-x) into D gives:
f(x) = (1-x)^2 (1-1+x)^2
= (1-x)^2 x^2
which is exactly the same function in D. Voila!
Thank you so much for the help.
I have another one if you dont mind, its to find the value for M.
(1/5)^m*(1/4)^17=1/2(10^32)
How do you find M with this abomination?
Can you type it better? Is it this: $(1/5)^m(1/4)^{17}=1/2(10^{32})$
• Posted by a hidden member.
^^
Yes that is what it is and I have no idea how to type like that...
• Posted by a hidden member.
Chainers said^^
Yes that is what it is and I have no idea how to type like that...
And do you have any idea what the answer is supposed to look like? Is it an exact number? Decimal etc? I'm trying to solve it exactly, but the answer isn't pretty, which makes me think that this shouldn't be right since your lovely standardised American tests aren't really like that.
• Posted by a hidden member.
From what I remember the answer was a solid number.
2 were in the 10's range, the other three were in the 30s.
• Posted by a hidden member.
Chainers saidFrom what I remember the answer was a solid number.
2 were in the 10's range, the other three were in the 30s.
That could only mean the equation you provided is wrong. And this isn't my math skills speaking, it's an equation solver. According to the equation you gave, m≈)
-60
• Posted by a hidden member.
sirdreamalott said
Chainers saidFrom what I remember the answer was a solid number.
2 were in the 10's range, the other three were in the 30s.
That could only mean the equation you provided is wrong. And this isn't my math skills speaking, it's an equation solver. According to the equation you gave, m≈)
-60
Can you tell me just how to go about solving the problem? I may be off by a couple of numbers but I have no idea how to approach exponential with 2 different bases.
• Posted by a hidden member.
Chainers said
sirdreamalott said
Chainers saidFrom what I remember the answer was a solid number.
2 were in the 10's range, the other three were in the 30s.
That could only mean the equation you provided is wrong. And this isn't my math skills speaking, it's an equation solver. According to the equation you gave, m≈)
-60
Can you tell me just how to go about solving the problem? I may be off by a couple of numbers but I have no idea how to approach exponential with 2 different bases.
That's the thing- if the problem is wrong it's kind of hard to have a good starting point. In any case, I would start off by taking the log of both sides and then using properties of logs to separate the terms.
• Karl
Posts: 5790
The equation (1/5)^m*(1/4)^17=1/2(10^32) has m ≈ - 60 , he's right .
This is how I did it.
There're something wrong with the equation you provided.
• Posted by a hidden member.
Karl saidThe equation (1/5)^m*(1/4)^17=1/2(10^32) has m ≈ - 60 , he's right .
This is how I did it.
There're something wrong with the equation you provided.
^_^
• Posted by a hidden member.
So glad I'm done with this...
This takes me back to my university days :-)
• Posted by a hidden member.
ermahgerd yer trorring ergern
• Posted by a hidden member.
I am happy to be a part of this asian persuasion. Thanks to all Asians who have helped us in our time of need.
And please do not kill us all.
• Posted by a hidden member.
Mathematics that I used to knowwWWWwww
• Posted by a hidden member.
METAMORPH saidMathematics that I used to knowwWWWwww
hahaha more like never.
• Posted by a hidden member.
Just give it up and go to law school.
Just kidding. That all looks like gibberish to me, though. You boys is smart.
• Karl
Posts: 5790
METAMORPH saidMathematics that I used to knowwWWWwww
{STIUTK pic}
LOL
me too , I wont have to see math for the rest of my life anymore
•
• | 2,269 | 8,135 | {"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": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2020-24 | latest | en | 0.9636 |
http://mathhelpforum.com/discrete-math/212442-induction-reccurence-help-plz.html | 1,480,810,587,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541142.66/warc/CC-MAIN-20161202170901-00423-ip-10-31-129-80.ec2.internal.warc.gz | 172,841,334 | 9,969 | Thread: Induction and Reccurence Help Plz :(
1. Induction and Reccurence Help Plz :(
Induction Question:
Prove by Induction that 2log(n!) > nlog(n) (n>2, n is an integer)
This question, I don't know how to link the assumed case to the inductive case, see the following:
Base Case: n = 3
2log(6) > 3log(3) is true (by using calc)
Assumed Case: n = k
2log(k!) > klog(k)
Inductive Case: n = k+1
2log(k+1) + 2log(k!) > (k+1)log(k+1)
From here I don't know what to do, I mean 2log(k!) is there, but klog(k)?
HALP PLZ!
Recurrence:
I never understood how to do complicated recurrence problems... e.g. Fibonacci series and all...
1. Solve x(n) = x(n-1) - (1/4)x(n-2(, with x0 = 1 and x1 = 1/2 (note that the (expression) are the subscripts)
2. Solve T(1) = 1, and for all n>=2, T(n) = 3T(n-1) + 2
I've been trying to do these questions (these are not the only ones) for hours and I just can't get forward with these 3 questions, so halp! Other ones, I was able to do.
2. Re: Induction and Reccurence Help Plz :(
Concerning the first problem, it's relatively easy to show that log(n * (n - k)) >= log n for k = 0, ..., n - 1.
Concerning the recurrence relation, look for a solution in the form x(n) = x^n. | 405 | 1,210 | {"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-2016-50 | longest | en | 0.91933 |
http://mathhelpforum.com/calculus/99314-3d-points-line-print.html | 1,526,874,231,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863923.6/warc/CC-MAIN-20180521023747-20180521043747-00496.warc.gz | 191,621,495 | 3,089 | # 3D points on line?
• Aug 26th 2009, 11:39 AM
collegestudent321
3D points on line?
Hi guys, i have a question which seemed simple when i first read it but now i see for some reason i can't figure it out...
i have three 3D points: A(5,1,3) B(7,9,-1) and C(1,-15,11)
How can i determine if they are in a straight line?
I think it has something to do with finding the distances between the points but after that I dont know what i'm doing. Any help would be appreciated. thank you!
• Aug 26th 2009, 11:51 AM
Plato
Quote:
Originally Posted by collegestudent321
i have three 3D points: A(5,1,3) B(7,9,-1) and C(1,-15,11)
How can i determine if they are in a straight line?
There are several ways to proceed. Distance is one of them.
However, I would use vectors.
$\displaystyle \overrightarrow {AB} = \left\langle {2,8, - 4} \right\rangle \;\& \,\overrightarrow {AC} = \left\langle { - 4, - 16,8} \right\rangle$.
Because, they are multiples of each other the points are colinear.
• Aug 26th 2009, 11:53 AM
collegestudent321
oh ok. That makes sense. thank you!
but out of curiosity... how would i do it with distances???
• Aug 26th 2009, 11:55 AM
collegestudent321
Im asking because this section is BEFORE the vectors section, so technically this solution shouldnt use vectors.
• Aug 26th 2009, 12:12 PM
Plato
Quote:
Originally Posted by collegestudent321
but out of curiosity... how would i do it with distances???
Find the distances $\displaystyle AB,~BC,~\&~AC$.
See if two of those add to the third.
• Aug 26th 2009, 12:15 PM
collegestudent321
Oh ok, cool... Easy enough. Thanks! | 496 | 1,586 | {"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} | 3.78125 | 4 | CC-MAIN-2018-22 | latest | en | 0.923434 |
https://goprep.co/q13-use-euclid-s-division-algorithm-to-find-the-hcf-of-726-i-1nm1qi | 1,620,807,497,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991685.16/warc/CC-MAIN-20210512070028-20210512100028-00128.warc.gz | 307,554,076 | 24,563 | # Use Euclid’s Divi
using Euclid’s Division Algorithm, we find HCF of two positive numbers by repetitive division till we get 0 as the remainder.
In the given two numbers 726 and 275, 726 is greater, so we will divide 726 by 275, we get
We see that the remainder is 176≠0,
Now we divide 275 by 176, we get
We see that the remainder is 99≠0,
Now we divide 176 by 99, we get
We see that the remainder is 77≠0
Now we divide 99 by 77, we get,
We see that the remainder is 22≠0
Now we divide 77 by 22, we get,
We see that the remainder is 11≠0,
Now we divide 22 by 11, we get
We see that the remainder is 0.
Hence the HCF of the two numbers 726 and 275 is 11.
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Show 40 post(s) from this thread on one page
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• May 11th 2010, 06:59 PM
dwsmith
Quote:
Originally Posted by Ari
okay so
a = 5 + b - c
b = -10 + 2
c = ?
$\displaystyle a=15-3c$
$\displaystyle b=10-2c$
$\displaystyle c$
$\displaystyle \begin{bmatrix} 15-3c\\ 10-2c\\ c \end{bmatrix}\rightarrow \begin{bmatrix} 15-3c\\ 10-2c\\ 3c \end{bmatrix}\rightarrow \begin{bmatrix} 15\\ 10\\ 0 \end{bmatrix}+c\begin{bmatrix} -3\\ -2\\ 1 \end{bmatrix}$
• May 11th 2010, 07:03 PM
Ari
are my values off because my leading coefficent was not 1 in the second equation?
• May 11th 2010, 07:04 PM
dwsmith
Your first equation you have a "b" which shouldn't be present and the 2nd equation you have no variables.
• May 11th 2010, 07:06 PM
Ari
what is the coefficent for c then?
i am happy that i am not taking this course otherwise i would have for sure failed lol...
• May 11th 2010, 07:07 PM
dwsmith
c is a free variable.
• May 11th 2010, 07:33 PM
Ari
Quote:
Originally Posted by dwsmith
$\displaystyle a=\frac{15}{3}-\frac{1}{3}c$
$\displaystyle b=\frac{-10}{3}+\frac{2}{3}c$
$\displaystyle c$
$\displaystyle \begin{bmatrix} \frac{15}{3}-\frac{1}{3}c\\ \frac{-10}{3}+\frac{2}{3}c\\ c \end{bmatrix}\rightarrow \begin{bmatrix} 15-c\\ -10+2c\\ 3c \end{bmatrix}\rightarrow \begin{bmatrix} 15\\ -10\\ 0 \end{bmatrix}+c\begin{bmatrix} -1\\ 2\\ 3 \end{bmatrix}$
then what is all this (Surprised)
• May 11th 2010, 07:34 PM
dwsmith
The solution that takes into account ever possible solution written in terms of c.
• May 11th 2010, 07:35 PM
Ari
Quote:
Originally Posted by dwsmith
The solution that takes into account ever possible solution written in terms of c.
and so that quote is the final solutions... man next term is gonna suck.
• May 11th 2010, 07:38 PM
dwsmith
Yes that is saying add the 1st column vector to c times the second column vector.
To avoid confusion of c, a=x, b=y, and c=z
If c=5, we have x=10, y=0, and z=15. This a solution. We can do this all day by choosing a c and then finding the values.
• May 11th 2010, 07:44 PM
Ari
shouldnt there be definite values for a,b,c since P(x) = a^x2 + bx + c
and they pass through the point (-1,5) and have a horizontal tangent at (1,7)
im sure there must be definite values for a, b, and c. otherwise this function would not work....
• May 11th 2010, 07:57 PM
dwsmith
We have a minor mistake.
We lost a negative sign.
$\displaystyle \begin{bmatrix} 1 & -1 & 1 & 5\\ -2 & 1 & 0 & 0 \end{bmatrix}$
If c=0, we have (15,10,0) as coefficients.
We know (-1,5) exist.
$\displaystyle 15x^2+10x+0c\rightarrow 15(-1)^2-10=5$
Check.
• May 11th 2010, 08:31 PM
dwsmith
There is an error some where with how the derivative is affecting the linear system; however, I can't find it so I solved your equation a different way.
We know the slope of the tangent line is 0 at (1,7); therefore, the general equation is $\displaystyle \pm a(x-1)^2+7$.
If the parabola opens up, (-1,5) wouldn't be a point on the parabola.
So our equation becomes: $\displaystyle -a(x-1)^2+7$
$\displaystyle -a(-1-1)^2+7=5\rightarrow -4a+7=5\rightarrow -4a=-2\rightarrow a=\frac{1}{2}$
Now we have $\displaystyle \frac{-1}{2}(x-1)^2+7\rightarrow \frac{-1}{2}(x^2-2x+1+7)\rightarrow \frac{-1}{2}(x^2-2x+8)$
$\displaystyle \frac{-1}{2}+x-4$
$\displaystyle a=\frac{-1}{2}$, $\displaystyle b=1$ and $\displaystyle c=-4$
• May 12th 2010, 04:39 AM
Ari
Quote:
Originally Posted by dwsmith
There is an error some where with how the derivative is affecting the linear system; however, I can't find it so I solved your equation a different way.
We know the slope of the tangent line is 0 at (1,7); therefore, the general equation is $\displaystyle \pm a(x-1)^2+7$.
If the parabola opens up, (-1,5) wouldn't be a point on the parabola.
So our equation becomes: $\displaystyle -a(x-1)^2+7$
$\displaystyle -a(-1-1)^2+7=5\rightarrow -4a+7=5\rightarrow -4a=-2\rightarrow a=\frac{1}{2}$
Now we have $\displaystyle \frac{-1}{2}(x-1)^2+7\rightarrow \frac{-1}{2}(x^2-2x+1+7)\rightarrow \frac{-1}{2}(x^2-2x+8)$
$\displaystyle \frac{-1}{2}+x-4$
$\displaystyle a=\frac{-1}{2}$, $\displaystyle b=1$ and $\displaystyle c=-4$
i see, how did you get this part?
$\displaystyle \frac{-1}{2}(x-1)^2+7\rightarrow \frac{-1}{2}(x^2-2x+1+7)\rightarrow \frac{-1}{2}(x^2-2x+8)$
when you opened the brackets, you put the 7 in while leaving the negative half out?
a=-1/2
b=1
c=6 and 1/2
• May 12th 2010, 04:55 AM
dwsmith
Quote:
Originally Posted by Ari
i see, how did you get this part?
$\displaystyle \frac{-1}{2}(x-1)^2+7\rightarrow \frac{-1}{2}(x^2-2x+1+7)\rightarrow \frac{-1}{2}(x^2-2x+8)$
when you opened the brackets, you put the 7 in while leaving the negative half out?
a=-1/2
b=1
c=6 and 1/2
Being hasty and not wanting to deal with fractions.
$\displaystyle -x^2+2x-1+14\rightarrow -x^2+2x+13$
$\displaystyle -x^2+2x-1+14\rightarrow -x^2+2x+13$ | 1,798 | 4,932 | {"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-2018-26 | latest | en | 0.783075 |
https://www.calc-medic.com/precalc-unit-9-day-18 | 1,708,786,586,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474541.96/warc/CC-MAIN-20240224144416-20240224174416-00257.warc.gz | 707,685,784 | 200,178 | top of page
## Unit 9 - Day 18
##### Unit 9Day 1Day 2Day 3Day 4Day 5Day 6Day 7Day 8Day 9Day 10Day 11Day 12Day 13Day 14Day 15Day 16Day 17Day 18Day 19Day 20Day 21All Units
###### Learning Objectives
• Select and apply an appropriate strategy for evaluating a derivative
# Lesson Handout
###### Experience First
Today students will work on problems that incorporate all the skills from the previous three days. In this circuit training activity, students start by completing the first box in top left corner (pre-labeled #1). They then look for their answer in the Answer slot of one of the other boxes. This box then becomes their box #2 and they put a 2 in the blank. They then complete this problem, search for the answer and label this new box with a #3. This continues until students have completed all 16 problems. The answer to problem 16 should lead them back to the first box. If students “short-circuit” they know they have made a mistake somewhere. Some answers are intentionally similar to encourage students to attend to precision.
###### Formalize Later
It is worth noting that while problems can be solved with the product rule or quotient rule, not all will require one of the rules. Simplifying and rewriting an expression can help students avoid the product rule or quotient rule. Over time, we hope that students make strategic decisions about when to apply the product rule or quotient rule.
You may wish to have a discussion about which strategy students used for each question. Pay special attention to problems that students were able to complete in multiple ways. While the product rule and quotient rule will always work, mathematicians make use of structure and look for efficient problem solving strategies.
bottom of page | 392 | 1,757 | {"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} | 2.640625 | 3 | CC-MAIN-2024-10 | longest | en | 0.930098 |
https://www.physicsforums.com/threads/scaling-interpretation.805936/ | 1,519,597,755,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891817437.99/warc/CC-MAIN-20180225205820-20180225225820-00224.warc.gz | 965,964,829 | 15,544 | Scaling Interpretation
1. Mar 30, 2015
joshmccraney
Hi PF!
I'm doing some scaling over a PDE and I understand the math side of things but I do not understand the physical side of what we are finding.
For example, suppose we have some PDE, say 2-D continuity for it's simplicity $u_x + v_y = 0$. Let $L$ be the length of a side of a flowing channel and the height of the channel be $H$. Now if $x$ scales as $L$ and $y$ scales as $H$ and if the mainstream velocity coming into the channel is $U$ then we may write $U H/L \sim v$. What is actually being said here? That the vertical velocity is maximized as $U H/L$? Please help!
Thanks a ton!
2. Mar 30, 2015
Staff: Mentor
It seems to be saying that the typical y velocity will be smaller than the x velocity by a factor on the order of H/L.
Chet
3. Mar 31, 2015
joshmccraney
What do you mean by "typical"?
They also use this technique with a time derivative as well, like in the momentum equation. My professor has said that this technique can save a lot of math, like if you are are doing a force balance for someone jumping out of an airplane, and you are concerned with initial velocity, this technique can eliminate "less important" terms.
4. Mar 31, 2015
Staff: Mentor
Maybe typical was a poor choice of term. Maybe it would have been better to say that the y velocities will be on the order of H/L times smaller than the x velocities. It is difficult to be more precise with something like this.
I don't follow what you are saying here. The way I learned dimensional analysis was taught to me by S. W. Churchill at the University of Michigan in 1963. See the famous paper by Hellums and Churchill in AIChE Journal (1964)
Chet
5. Mar 31, 2015
joshmccraney
Thanks, I'll look into it! You're awesome Chet! | 468 | 1,779 | {"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} | 3.203125 | 3 | CC-MAIN-2018-09 | longest | en | 0.961317 |
https://officewheel.com/how-to-find-correlation-coefficient-in-google-sheets/ | 1,723,444,827,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036271.72/warc/CC-MAIN-20240812061749-20240812091749-00688.warc.gz | 339,574,972 | 37,158 | # How to Find Correlation Coefficient in Google Sheets
A correlation coefficient is a metric that expresses a correlation, or a statistical link between two variables, in numerical terms. In brief, correlation coefficient is important to understand the significance and direction of the linear correlations between pairs of variables. In this article, we have demonstrated 5 easy ways to find correlation coefficient in Google Sheets.
## What Is Correlation Coefficient ?
The correlation coefficient indicates how closely two variables are related. There are three different types of data correlation: Positive correlation, Negative correlation, and No correlation.
According to the Positive correlation coefficient, when one variable rises, another variable rises as well. A positive correlation coefficient has a value larger than 0.
According to the Negative correlation coefficient, as one variable rises, another variable falls. The negative correlation coefficient has a value that is less than 0.
The correlation coefficient, whose value is near to or equal to 0, indicates that there is no association between two variables.
## 5 Easy Methods to Find Correlation Coefficient in Google Sheets
We will be using the following dataset to demonstrate methods in this article. The dataset represents some persons’ height and age.
### 1. Using CORREL Function
The built-in CORREL function in Google Sheets returns the value of the correlation coefficient of two ranges of cells. The first one is the cell ranges of Y data which is the dependent variable. The second range of cells is the data of independent variable X.
#### 1.1 Inserting Data Manually
We can simply use this CORREL function by typing down the data of both variables within the formula. It’s not a feasible method for a long dataset but maybe helpful for split values.
Steps:
• In the following dataset, select Cell C14 and type the following formula then press Enter-
`=CORREL({138,170,176,175,178,174,170,172},{12,15,17,20,24,29,32,45})`
Here, the data in the first {} are the values of dependent variable Y and the data in the second {} are the values of independent variable X.
#### 1.2 Using Cell References
We can simply use cell references within the CORREL formula to get the correlation coefficient.
Steps:
• In Cell C14, apply the following formula and then Enter
`=CORREL(B5:B12,C5:C12)`
Here, Cell range B5:B12 indicates the dependent variables’ values while Cell range C5:C12 represents the independent variables’ values.
### 2. Combining RSQ and SQRT Functions
Correlation coefficient is the square root of coefficient of determination. Here, we will first calculate the value of the coefficient of determination using the RSQ function. Then we will get the result of the square root of that as correlation coefficient using the SQRT function.
Steps:
• First, select the Cell C14 in the following dataset and enter the following formula-
`=RSQ(B5:B12,C5:C12)`
This will calculate the value of the coefficient of determination.
• Now, activate Cell C15, put the following formula below there and press Enter
`=SQRT(C14)`
In the formula, Cell C14 refers to the cell that carries the value of the coefficient of determination.
### 3. Applying PEARSON Function
The formula of PEARSON is as follows-
Great news is, Google Sheets has a built-in PEARSON function for this formula. The Pearson correlation coefficient is returned by the PEARSON function after receiving two arrays as parameters in which the first one is for the dependent variable and the 2nd one is the values of the independent variable. We can easily find correlation coefficient in Google Sheets using this formula.
Steps:
• First, select Cell C14 and then apply the following formula and press Enter
`=PEARSON(B5:B12,C5:C12)`
Then you will get the same output as before.
### 4. Using PEARSON Formula Manually
If you got an assignment or an upcoming test based on correlation coefficient, I would prefer you to follow this method to understand your topic properly. Before this, I have already shown you the use of the PEARSON function. But now, we will learn how to solve the Pearson correlation coefficient manually in Google Sheets. The AVERAGE, SUM, and SQRT functions will come up in this manner.
The basic equation in this case is as follows:
We will make a dataset for this as shown below-
Steps:
• First of all, we have to find (average value of independent variable X) and (average value of dependent variable Y). Select Merged Cell B15:B22, apply the following formula-
`=AVERAGE(C5:C12)`
Here, Cell range C5:C12 refers to the values of independent variable X.
• Secondly, do the same for the Y variable as well. Select Merged cell C15:C22 and then apply the following formula-
`=AVERAGE(B5:B12)`
Cell range B5:B12 refers to the values of the Y variable. The formula here will calculate the average of Y values.
• Now, we have to calculate the values of (Xi-X̅) and (Yi-Y̅). For that, we will apply the following formula in Cell D15.
`=C5-\$B\$15`
This will calculate the difference between the value of X in Cell C5 and the average value of X that is .
And then we will apply the below formula into Cell E15.
`=B5-\$C\$15`
This will measure the difference between the value of Y in Cell B5 and the average value of Y that is .
• After that, select Cell D15 and then simply drag down using the Fill handle icon as shown in the circled portion below.
This will calculate the difference between and the rest of the other X values as well.
• Now, do the exact same by selecting Cell E15 and the result will be as follows.
• At this point, we have to calculate the product of (Xi-X̅) and (Yi-Y̅) for each value of (Xi-X̅) and (Yi-Y̅). Select Cell F15 and apply the following formula.
`=D15*E15`
Here, Cell D15 refers to the value of (Xi-X̅) and Cell E15 refers to the value of (Yi-Y̅).
• Drag down using the Fill Handle icon in Cell F15 like previously and the result will be as follows.
• Select Cell F23 and apply the below formula. Have a look at the Pearson formula again. We need the value of ⅀(Xi-X̅)*(Yi-Y̅).
`=SUM(F15:F22)`
The SUM function we calculate the summation of all the values of the product (Xi-X̅)*(Yi-Y̅).
• Next, we need to calculate the square values of all the (Xi-X̅) and (Yi-Y̅) Apply the formula into the box below in Cell G15 and press Enter
`=D15^2`
Obviously it will calculate the square of (Xi-X̅) value which in Cell D15. You can see the output in Cell G15.
• Drag down using the Fill handle icon as shown in the circled portion to get the square value of other (Xi-X̅) values as well.
After that the square value of other (Xi-X̅) values.
• Thereafter, choose Cell G23 and apply the following formula below.
```=SUM(G15:G22) ```
Going through this, the summation of all (Xi-X̅)2 values across the Cell range G15:G22.
• Next, select Cell H15, apply the following formula below and press Enter
`=E15^2`
This will determine the square of (Yi-Y̅) value which is in Cell E15. The output will be as follows in Cell H15.
• Now drag down using the Fill handle icon in Cell H15 like previously to get the square of other (Yi-Y̅) values as well.
• After that, activate Cell H23, give input the following formula and press Enter
`=SUM(H15:H22)`
Here, the SUM function will calculate the summation of all the (Yi-Y̅)2 values across Cell range H15:H22.
• Finally, select Cell H4 and apply the following formula below then press Enter
`=(F23)/SQRT(G23*H23)`
This formula is basically the representation of the Pearson formula: Here, Cell F23 refers to the value ⅀(Xi-X̅)(Yi-Y̅) (the numerator of the Pearson formula).
SQRT(G23*H23) refers to the value of (⅀(Xi-X̅)^2*(Yi-Y̅)^2) (the denominator of the Pearson formula).
Therefore, the value of the correlation coefficient will be as follows in Cell H4.
### 5. Using Chart Feature
We can create a chart and get the value of the coefficient of determination from that. Then, calculating the square root value from that, we will simply get the value of the correlation coefficient.
Steps:
• Select the Cell range B4:C12, click on the Insert chart ribbon from the toolbar at top.
• A scattered chart will appear on your screen.
• Now, from the sidebar titled “Chart editor”, go to the Customize menu then select “Seriesdrop-down.
• Afterward, in that drop-down menu, mark the Trendline box and then mark the Show R2
• Following this, you will see that the value of coefficient of determination that is R2 will appear on your chart.
• Therefore, type down this value “0.156” into Cell C14 in the following dataset.
• Finally, apply the following formula into Cell C15 and press Enter
`=SQRT(C14)`
In the formula, Cell C15 refers to the cell that carries the value of coefficient of determination.
The output showing in Cell C15 is the value of correlation coefficient.
## Conclusion
The article contains 5 efficient methods to find correlation coefficient in Google Sheets. Hope this will help with your task. Visit officewheel.com to explore more relevant articles. | 2,138 | 9,063 | {"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-2024-33 | latest | en | 0.883331 |
https://au.mathworks.com/matlabcentral/cody/problems/157-the-hitchhiker-s-guide-to-matlab/solutions/2977027 | 1,606,182,056,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141169606.2/warc/CC-MAIN-20201124000351-20201124030351-00646.warc.gz | 200,557,177 | 17,093 | Cody
# Problem 157. The Hitchhiker's Guide to MATLAB
Solution 2977027
Submitted on 22 Sep 2020 by Shinichiro Yamaguchi
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 41; y_correct = false; assert(isequal(zaphod(x),y_correct))
y = logical 0
2 Pass
x = 42; y_correct = true; assert(isequal(zaphod(x),y_correct))
3 Pass
x = 43; y_correct = false; assert(isequal(zaphod(x),y_correct))%% x = 44; y_correct = false; assert(isequal(zaphod(x),y_correct))%% x = 45; y_correct = false; assert(isequal(zaphod(x),y_correct))%% x = 46; y_correct = false; assert(isequal(zaphod(x),y_correct))%% x = 47; y_correct = false; assert(isequal(zaphod(x),y_correct))
y = logical 0 y = logical 0 y = logical 0 y = logical 0 y = logical 0
4 Pass
x = 48; y_correct = false; assert(isequal(zaphod(x),y_correct))
y = logical 0
5 Pass
x = 49; y_correct = false; assert(isequal(zaphod(x),y_correct))
y = logical 0
6 Pass
x = 50; y_correct = false; assert(isequal(zaphod(x),y_correct))
y = logical 0
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 395 | 1,245 | {"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} | 2.890625 | 3 | CC-MAIN-2020-50 | latest | en | 0.530383 |
https://www.resurchify.com/if/impact-factor-search | 1,685,981,381,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652149.61/warc/CC-MAIN-20230605153700-20230605183700-00502.warc.gz | 1,028,665,060 | 20,524 | # Find Impact Score
Check Impact Index of thousands of worldwide Journals
## What is Impact Score?
##### The impact score (IS), also denoted as Journal impact score (JIS), of an academic journal is a measure of the yearly average number of citations to recent articles published in that journal. It is based on Scopus data.
Impact Score is defined as the ratio of the number of citations a journal receives in the latest two years (Including the year of calculation) to the number of publications (published documents) of that Journal in those two years. It is calculated as Cites / Doc. (2years).
It helps to measure the relative importance of journals within particular areas and to compare the journals within the same areas. The higher the JIS, the better it is ranked. Typically, journals with more review articles or papers are able to achieve higher JIS.
## How it is calculated?
For a given year, the Impact Score of a particular journal is the number of citations it receives in that year and previous 1 years, for documents published in the journal during that period (two years), divided by the total number of published documents in the journal during the same two-year period. Published documents can be conference papers, articles, book chapters, reviews, and data papers.
$IS_x = \frac{Citations_{x-1} + Citations_{x-2}}{Publications_{x-1} + Publications_{x-2}}$
For an example, to find the impact score of a "ABC" journal for the year 2018, we would compute:
$IS_{2018}= \frac{Citations_{2018} + Citations_{2017}}{Publications_{2018} + Publications_{2017}}$
For example, if the Journal have the following citations and publications value:
Citations2018 = 80
Citations2017 = 60
Publications2018 = 30
Publications2017 = 40
IF2016= (80 + 60) / (30 + 40) = 2
This value of IS indicates that, on an average, the articles of "ABC" Journal published in the years 2017 and 2018 have approximately received 2 (two) citations each in the year of 2018.
It is important to note here that the 2018's IS is published in the year 2019. It cannot be computed until all publications in the previous year of 2018 are processed by the indexing agency.
## How do I find the impact score of a journal?
Impact Score is defined as the ratio of the number of citations a journal receives in the latest two years (Including the year of calculation) to the number of publications (published documents) of that Journal in those two years. It is calculated as Cites / Doc. (2years).
You can either refer to the Scopus® database to find the impact score of the journal. The data from the Scopus® database can also be found at resurchify.com. You can find the impact score of thousands of journals on this website.
To search the impact score of any Journal or Conference, you can query by its title or ISSN. You can also query using the publisher's name or by subject category in the search box and can select the required journal. You can also check in detail analysis (like five years average, highest impact in the last five years, etc.) of a particular item, by clicking on the same.
All these details will be helpful when you want to select a journal or assess the quality of a journal.
## Importance of Impact Score
Impact score gives the approximate idea about how prestigious a particular journal is in its field.
The higher the IS of the journal, the better it is ranked. By using this metric you can basically evaluate and compare the journals in similar subject categories to identify their importance.
## What is a high impact score?
Impact Score is a measure of the importance of a journal. The impact score (IS) is a measure of the yearly average number of citations to recent articles published in that journal. It is often used to compare journals of the same category. Higher the Impact score, higher is the ranking of the journal.
But do not take this number as an absolute measure. It should be used to compare the journals withing a single discipline. For example, an Artificial Intelligence journal's Impact Score cannot be compared with a journal from the Management domain.
## Data Source and Statistical Analysis
We perform various analysis on the data produced by SCImago.
The SCImago Journal & Country Rank is an openly accessible portal which covers the journals and scientific indicators generated from the data present in the Scopus® database (Elsevier B.V.).
Scopus database includes the information of more than 15,000 journals from different fields from around 4,000 publishers and also covers around 1000 open access journals
We perform statistical impact analysis for various journals and conference to evaluate their impact trends. You can find here the average impact index for the last three and five years. We also present here the highest (best) and lowest (worst) impact for the last few years. To show the variation of impact data over years, we compute and show the standard deviation.
The above metrics help you to better correlate and judge the impact of any particular journal/conference. We perform all the analysis on Cites/Doc. (2 Year) metric. Also known as impact index.
## Categories and Areas Covered
We cover the impact and detailed analysis of almost all the major areas and disciplines. We covered the following categories:
Endocrinology, Diabetes and Metabolism Environmental Chemistry Geology Otorhinolaryngology Small Animals Oncology (nursing) Pharmacy Pediatrics Ecology, Evolution, Behavior and Systematics Neuropsychology and Physiological Psychology Management of Technology and Innovation Pulmonary and Respiratory Medicine Mathematics (miscellaneous) Emergency Nursing Public Administration Complementary and Manual Therapy Computer Vision and Pattern Recognition Pharmaceutical Science Gerontology Health Professions (miscellaneous)
Sociology and Political Science Occupational Therapy Embryology Epidemiology Water Science and Technology Drug Discovery Food Animals Anthropology Chemical Engineering (miscellaneous) Issues, Ethics and Legal Aspects Atomic and Molecular Physics, and Optics Biochemistry Visual Arts and Performing Arts Internal Medicine Artificial Intelligence Cognitive Neuroscience Critical Care Nursing Review and Exam Preparation Paleontology Radiological and Ultrasound Technology Social Sciences (miscellaneous) Astronomy and Astrophysics Pharmacology (nursing) Management Information Systems Cardiology and Cardiovascular Medicine Developmental Biology Mechanical Engineering Cultural Studies Complementary and Alternative Medicine Nature and Landscape Conservation Hardware and Architecture Analysis Aging Pharmacology Radiation Neurology (clinical) Atmospheric Science Cancer Research Analytical Chemistry Physics and Astronomy (miscellaneous) Arts and Humanities (miscellaneous) Ophthalmology Biological Psychiatry Emergency Medical Services Agricultural and Biological Sciences (miscellaneous) Equine Developmental and Educational Psychology Insect Science Global and Planetary Change Fuel Technology Nurse Assisting Urban Studies Rheumatology Nuclear and High Energy Physics Environmental Science (miscellaneous) Applied Psychology Architecture Assessment and Diagnosis Advanced and Specialized Nursing Rehabilitation Safety Research Health, Toxicology and Mutagenesis Leadership and Management E-learning Surgery Human Factors and Ergonomics Classics Nursing (miscellaneous) Health Information Management Cellular and Molecular Neuroscience Chiropractics Immunology and Microbiology (miscellaneous) Respiratory Care Earth and Planetary Sciences (miscellaneous) Toxicology Electronic, Optical and Magnetic Materials Behavioral Neuroscience Safety, Risk, Reliability and Quality LPN and LVN Dentistry (miscellaneous) Dental Hygiene Structural Biology Statistical and Nonlinear Physics Political Science and International Relations Virology Materials Chemistry Electrical and Electronic Engineering Chemistry (miscellaneous) Aerospace Engineering Strategy and Management Marketing History Environmental Engineering Veterinary (miscellaneous) Physical and Theoretical Chemistry Reviews and References (medical) Library and Information Sciences Orthopedics and Sports Medicine Transportation Anesthesiology and Pain Medicine Energy (miscellaneous) Process Chemistry and Technology Neurology Maternity and Midwifery Clinical Biochemistry Genetics (clinical) Materials Science (miscellaneous) Forestry Education Nanoscience and Nanotechnology Statistics and Probability Geometry and Topology Archeology Signal Processing Plant Science Pollution Pediatrics, Perinatology and Child Health Medicine (miscellaneous) Colloid and Surface Chemistry Business, Management and Accounting (miscellaneous) Management, Monitoring, Policy and Law Nutrition and Dietetics Logic Clinical Psychology Periodontics Oral Surgery Molecular Medicine Genetics Condensed Matter Physics Economic Geology Soil Science Biomaterials Neuroscience (miscellaneous) Community and Home Care Animal Science and Zoology Modeling and Simulation Life-span and Life-course Studies Reproductive Medicine Health Informatics Family Practice Mathematical Physics Information Systems Anatomy Medical and Surgical Nursing Computational Mathematics Energy Engineering and Power Technology Sports Science Geography, Planning and Development Nuclear Energy and Engineering Literature and Literary Theory Acoustics and Ultrasonics Horticulture Surfaces and Interfaces Aquatic Science Urology Information Systems and Management Developmental Neuroscience Care Planning Media Technology Electrochemistry Management Science and Operations Research Economics, Econometrics and Finance (miscellaneous) Computer Networks and Communications Public Health, Environmental and Occupational Health Polymers and Plastics Drug Guides Communication Biomedical Engineering Speech and Hearing Filtration and Separation Computer Science Applications Hematology Demography Health (social science) Instrumentation Infectious Diseases Psychiatric Mental Health Applied Microbiology and Biotechnology Cell Biology Geotechnical Engineering and Engineering Geology Mechanics of Materials Obstetrics and Gynecology Food Science Radiology, Nuclear Medicine and Imaging Pharmacology (medical) Pathology and Forensic Medicine Numerical Analysis Sensory Systems Immunology and Allergy Applied Mathematics Endocrine and Autonomic Systems Gender Studies Histology Microbiology Dermatology Space and Planetary Science Computer Graphics and Computer-Aided Design Psychiatry and Mental Health Dental Assisting Social Psychology Hepatology Computer Science (miscellaneous) Law Physical Therapy, Sports Therapy and Rehabilitation Agronomy and Crop Science Renewable Energy, Sustainability and the Environment Earth-Surface Processes Accounting Orthodontics Immunology Oncology Philosophy Museology Algebra and Number Theory Ecological Modeling Psychology (miscellaneous) Biotechnology Pharmacology, Toxicology and Pharmaceutics (miscellaneous) Tourism, Leisure and Hospitality Management Statistics, Probability and Uncertainty Religious Studies Computers in Earth Sciences Finance Microbiology (medical) Development Experimental and Cognitive Psychology Language and Linguistics Parasitology Organic Chemistry Ecology Geochemistry and Petrology Archeology (arts and humanities) Research and Theory Decision Sciences (miscellaneous) Fundamentals and Skills Inorganic Chemistry Nephrology Civil and Structural Engineering Human-Computer Interaction Automotive Engineering Building and Construction History and Philosophy of Science Linguistics and Language Engineering (miscellaneous) Molecular Biology Economics and Econometrics Waste Management and Disposal Industrial Relations Chemical Health and Safety Software Oceanography Social Work Computational Theory and Mathematics Health Policy Medical Terminology Business and International Management Physiology (medical) Ceramics and Composites Podiatry Ocean Engineering Conservation Endocrinology Industrial and Manufacturing Engineering Control and Optimization Geriatrics and Gerontology Control and Systems Engineering Spectroscopy Biochemistry (medical) Geophysics Medical Laboratory Technology Metals and Alloys Organizational Behavior and Human Resource Management Biophysics Fluid Flow and Transfer Processes Multidisciplinary Transplantation Optometry Biochemistry, Genetics and Molecular Biology (miscellaneous) Music Catalysis Critical Care and Intensive Care Medicine Stratigraphy Bioengineering Emergency Medicine Gastroenterology Surfaces, Coatings and Films Physiology Discrete Mathematics and Combinatorics Medical Assisting and Transcription Computational Mechanics Theoretical Computer Science
### Years Covered
So far, we have covered our impact analysis for the years 2021, 2019, 2018, 2017, 2016, 2015, 2014, and 2013. However, we will update soon all our analysis for the year 2022 also, once the data is available.
## Top Journals and Conferences in the Related Fields
We also provide the expert suggestions for top journals and conferences which are in related fields or in same categories. This will help you to find out other top journals and conference opportunities where you can submit your research paper or article to showcase the quality of your work.
### Importance of choosing Journals with High Impact Score and Important Suggestions
It is always advised to submit your articles into a journal with high impact score in your field. This is to show the credibility and worthiness of your research articles and your work.
It is observed that most of the reviewers judge the quality of your articles based on the referenced journal articles. Therefore, it is always recommended to cite or refer the articles from the top journals (Which are basically the one having high impact score)
## Impact Score Alternatives
Some of the other metrics which can help you judge the Journal impact are Citescore, h-Index, mention and share of Journal articles on social media, Downloads, Views, etc.
### Citescore
Citescore is another commonly used metric used like an impact factor, to measure journal impact in Scopus. It measures a yearly average number of citations to recently published articles in that Journal. Citescore was launched in December 2016 by Elsevier.
CiteScore metric combines eight different indicators: CiteScore, CiteScore Percentile, CiteScore Rank, Citation Count, CiteScore Quartiles, CiteScore Tracker, Document Count and Percentage Cited.
#### Citescore Calculations
Citescore is defined as the ratio of the number of citations a journal receives in the latest four years (Including the year of calculation) to the number of publications (published documents) of that Journal in those four years.
#### Important Facts About Citescore
• Citescore calculations are done annually and hence show the average number of citations for the entire calendar year. Moreover, its estimates are updated each month and therefore provides the latest performance indicators of a particular Journal.
• Citescore cannot be compared between subject fields because its value does not include field normalization. To compare the Journals of the same subject categories or domains, one should use metrics like SJR (SCImago Journal Rank) or SNIP as these metric are field normalized.
### h-Index
The h-index is another way of measuring the productivity or impact of the Journal. The h-index is defined as the maximum value of h such that the given Journal has published h papers, and each of them is cited for at least h number of times in a specific time.
### SCImago Journal Rank (SJR)
SCImago Journal Rank (SJR) is another commonly used metric to measure the scholarly influence or impact of the Journal by accounting for the number of cites/citations and the prestige of the citing journals.
It is based on the concept called eigenvector centrality, which is commonly used in network theory. This metric is a size-independent measure that ranks journals based on their average prestige per article.
### Source Normalized Impact per Paper (SNIP)
Source Normalized Impact per Paper (SNIP) considers the weightage of the total number of citations received in a discipline while calculating a journal's impact. This method of impact calculation normalizes for differences in citation practices between fields so that a single citation is given greater importance or values where citations are less frequent or common in that field.
Unlike the well-known journal impact factor, SNIP corrects for differences in citation practices between scientific fields, thereby allowing for more accurate between-field comparisons of citation impact.
## Ways to Increase Impact Score of Your Journal
We have now understood the definition of the Impact Score, its calculation method, the importance of the Impact Score, and how to find it. Now, we will go through the ways through which we can increase the Impact Score of a Journal.
Following are some ways that can help to increase the Impact Score of a Journal. Some of these ideas are based on theory, some on practice and few on experience. These methods can help you get more citations for your journal, improve its quality and increase its prominence.
Let us now scroll these go through these methods, and understand what they have in store for us.
### Target the right audience
Designing the products keeping the intended target base in mind always helps. This method applies here also. If your articles or research topics have a dedicated area to focus on, chances of them being cited by the scientists increase. If the research area is vague and not specific, the researcher or scientist may not find anything of use.
The focus might be on society, a particular group, scientists, teachers or anybody. Just remember to have a target audience.
### Sometimes Peer-Review helps!
This is a known fact that if someone with a good reputation in your field gives a statement for your work, that work automatically gets few stars of appreciation.
In the research field, work that is peer-reviewed often comes up with better quality. This work has gained more trust and hence, gets cited more.
### Taking advantage of affiliations
Organizations or groups with innovative and diverse research topics, more originality, and a high number of publications are generally cited more than others. They have an excellent probability of showing up in the references thereby, increasing their citation.
If your journal doesn't belong to a very famous authority, one can try for affiliation with a known and reputed organization.
### Add topic-wise description for quick scanning
A small description of the various headings like abstract, hypotheses, key findings and observations, inference, conclusion etc., will give an insight to the reader of the article. This will help in a quick scan of the article and increase its chances of getting cited.
### Add a descriptive scope
As stated earlier, if the title and scope are descriptive, readers tend to get a better understanding of the article. This will bring more traffic and increase the citation.
By a descriptive scope, we mean that it should clearly define the area of the research and the methods that have been undertaken to come up with the research.
### Diversify your list of topics covered
A single topic may have many branches and sub-branches, giving rise to a diverse range of topics in that area. Sometimes, these ideas don't strike anyone, and this can be of much help if encouraged by telling them. This will attract more authors. The result will also be a diversified one.
Develop new article types to attract a diverse range of authors.
### Keep flexible submission options
Also, if the submission is getting accepted in multiple formats, people tend to submit more. Sometimes, authors are not comfortable with a particular format and do not choose a journal for publication. If there is leniency in this, it can help.
### Promotion of few articles might help
Advertisement and marketing is a key these days to reach more audience. Often articles that have been published long back might find their relevance in the present time. The editor or the committee should keep track of such topics and promote them. This will increase the citations of the article. One thing that this activity requires is time and effort from the team.
### Make your articles easily reachable
In today's scenario, where things are available easily, without much effort, if you ask for a lot of details or a heavy price from the readers, there are very few chances of your articles being cited by the authors.
It takes few minutes for the author to scan an article and if there are fewer hindrances maybe, in terms of article fee or filling up details, the article will be read and cited more.
A journal can always have some free articles to attract readers and citations.
### Be real! Show your true self
Few things cannot be pretended. They have to be real. For example, take the case of the affiliations of the journal, the deciding committee, the Editorial Board Members, Peer Reviewers. If these things are mentioned and include high ranking individuals, they will attract the authors to publish in your journal.
### Keep your ethics and transparency high
A clean, ethical policy will always help in giving credibility to any organization. In this way, people trust your organization to submit their articles. There are organizations like the Committee on Publication Ethics (COPE) that help in improving the trustworthiness of a journal. Joining such a committee will boost your credibility.
### Develop a Niche
In case your journal targets a specific set of authors, it is again very good. This way, you will attract a specific set of topics for the journal. There will be a niche in your journal, and unnecessary diversions can be avoided. Instead of a wide range of issues, a specific set will give your journal a reputation for that research area.
### Flexible Review Process
The review process and schedule are a few of the main criteria for selecting a journal for publication. Sometimes, the publication is a part of their curriculum or their job, and it will then require a deadline. Also, few topics are relevant in the present time and may become outdated if they don't get a chance to get published. That is why there must be a quick and robust review system with a shorter turnaround time.
### Keep user-friendly ads
Having advertisements is always a good income idea. But one thing needs to be ensured that it doesn't affect the user experience. The reader should not get distracted from the primary task, or the placement of the ads should be such that it doesn't disturb or irritate the reader.
### Publish Review Articles
Review articles of good literature can help the authors quickly compare various types of research carried out. It might be of interest to them and thus, attract more citations. Also, articles having trending and good topics can be combined together and published. This will act as a one-stop point for the authors and readers to read, understand and cite.
### Try using SEO in articles
One tool that helps you rank on top in search engine results is SEO (Search Engine Optimization). With this method, you can optimize your content, which will lead to more citations for your journal. Optimization of academic articles is called Academic Search Engine Optimization (ASEO). It ensures that the article is crawled and ranks high in search results.
These tips might be applicable to a few journals and might not work for a few. There is no compulsion to follow all the above-mentioned methods. Select what best applies to you.
Our Attempt at this idea...
The ideas that we have mentioned here are based on our knowledge and experience. We are still learning and will continue to add on to this list as and when we come up with something new.
## Credits and Sources
• Journal Impact Factor, https://clarivate.com/
• Scimago Journal & Country Rank (SJR), https://www.scimagojr.com/
• Find Impact Factor
• Impact Factor: The Complete Guide | 4,579 | 24,122 | {"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} | 3.078125 | 3 | CC-MAIN-2023-23 | longest | en | 0.916672 |
https://socratic.org/questions/how-do-you-solve-2abs-y-3-6-0 | 1,585,664,299,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370500482.27/warc/CC-MAIN-20200331115844-20200331145844-00476.warc.gz | 688,191,417 | 6,017 | # How do you solve 2abs(y-3)+6 < 0?
Jun 4, 2018
$y \in \emptyset$
#### Explanation:
$2 | y + 3 | + 6 < 0$
$| y + 3 | + 3 < 0$
$| y + 3 | < - 3$
So:
If $y < - 3$
$- \left(y + 3\right) < - 3$
$- y - 3 < - 3$
$y + 3 > 3$
$\cancel{y > 0}$ because of our domain.
If $y > - 3$,
$y + 3 < - 3$
$\cancel{y < - 6}$ because of our domain.
So : $y \in \emptyset$
\0/ here's our answer ! | 188 | 378 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "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} | 4.0625 | 4 | CC-MAIN-2020-16 | longest | en | 0.200282 |
https://byjus.com/diffusion-formula | 1,539,902,584,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512015.74/warc/CC-MAIN-20181018214747-20181019000247-00323.warc.gz | 649,904,864 | 27,401 | # Diffusion formula
Diffusion is the net movement of molecules or atoms from a region of high concentration to lower concentration. It can also be referred to movement of substances towards the lower concentration.The diffusion equation is the partial differentiation equation which indicates dynamics in a material which undergoes diffusion. It also describes processes exhibiting diffusive behavior. The diffusion equation is ∂ψ/∂t = 2ψ
The diffusion equation can also be written as
Where
ϕ(r, t) is the density of the diffusing material
D(ϕ, r) is the collective diffusion coefficient
∇ shows the vector differential operator del
If the diffusion coefficient depends on density, then the equation is nonlinear, else it’s linear.
Example 1
Solve the diffusion equation ∂ψ / ∂t = kΔ2ψ with the u(x, 0) = f(x), u(0,t) = 0 and u(L, t) = 0
f(X) = 6 sin(πx / L)
Solution
Given equation is
F(x) = 6 sin (πx / L)
U(x, 0) = F(x), u(0,t) = 0 and u(L, t) = 0
Let n = 1
B1 = 6
Using formula,
Un(x, t) = Bn sin(nπx / L) e-k(nπ / L)2 t
U(x, t) = 6 sin (pix / L) e –k(π / L)2 t
Example 2
Solve the diffusion equation
∂ψ / ∂t = kΔ2ψ with u(x, 0) = f(x), u(0, t) = 0 and u(L, t) = 0
F(x) = 12 sin (π x / L)
Solution
u(x, 0) = f(x), u(0, t) = 0 and u(L, t) = 0
Let n = 1
B1 = 12
From the formula,
Un(x, t) = Bn sin (n π x / L) e –k(2 π / L)2 t
U(x, t) = 12 sin(π x / L) e –k(π/L)2 t | 492 | 1,398 | {"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-2018-43 | latest | en | 0.824272 |
https://stats.stackexchange.com/questions/201036/does-it-make-sense-to-transform-time-series-variable-into-dependent-variable-of | 1,568,870,313,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573439.64/warc/CC-MAIN-20190919040032-20190919062032-00040.warc.gz | 662,688,678 | 31,392 | # Does it make sense to transform time series variable into dependent variable of OLS model?
Assume that I have a data for the $y$ like
y
2000 74.5
2001 73.5
2002 71.4
2003 70.3
2004 79.1
... ...
Also, I have data on $x$.
x
2000 123.5
2001 136.5
2002 243.4
2003 235.3
2004 278.1
... ...
Can I simply transform this data into two variables for OLS estimation and just forget about time-component as follows
y x
74.5 123.5
73.5 136.5
71.4 243.4
70.3 235.3
79.1 278.1
... ...
and estimate a model like this
$y_i = \beta_0 + \beta_1 x_i + \epsilon_i$?
Will it be meaningful at all? Or why it can be wrong?
UPDATE: data to play with
• @Tom_Reilly, thanks. Let's imagine that $y$ is life expectancy in some country and $x$ is a number of beds in hospitals per 10 000 people. I think the number of beds is lagged in comparison with life expectancy. – Vladimir Yashin Mar 10 '16 at 20:51
• @Tom_Reilly, I added dataset. The first row contains headers, the second --- brief descriptions of a variables, the following --- data itself. I am trying to find a relationship between life expectancy (life_exp) and other ones. – Vladimir Yashin Mar 12 '16 at 20:37 | 400 | 1,274 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 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} | 2.765625 | 3 | CC-MAIN-2019-39 | latest | en | 0.849958 |
http://mail.scipy.org/pipermail/numpy-discussion/2006-October/023906.html | 1,394,400,141,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394010354479/warc/CC-MAIN-20140305090554-00067-ip-10-183-142-35.ec2.internal.warc.gz | 116,069,835 | 2,660 | # matlab, filter function
Travis Oliphant oliphant at ee.byu.edu
Tue Oct 17 16:55:54 CDT 2006
```Kenny Ortmann wrote:
>just looking for some help, most of the time you guys are good with matlab
>code, i am trying to use the filter function under this setting
>
>y = filter(b,a,X) filters the data in vector X with the filter described by
>numerator coefficient vector b and denominator coefficient vector a. If a(1)
>is not equal to 1, filter normalizes the filter coefficients by a(1). If
>a(1) equals 0, filter returns an error.
>
>
There is scipy.signal.lfilter which implements this algorithm. It's doc
string is
lfilter(b, a, x, axis=-1, zi=None)
Filter data along one-dimension with an IIR or FIR filter.
Description
Filter a data sequence, x, using a digital filter. This works for
many
fundamental data types (including Object type). The filter is a
direct
form II transposed implementation of the standard difference equation
(see "Algorithm").
Inputs:
b -- The numerator coefficient vector in a 1-D sequence.
a -- The denominator coefficient vector in a 1-D sequence. If a[0]
is not 1, then both a and b are normalized by a[0].
x -- An N-dimensional input array.
axis -- The axis of the input data array along which to apply the
linear filter. The filter is applied to each subarray along
this axis (*Default* = -1)
zi -- Initial conditions for the filter delays. It is a vector
(or array of vectors for an N-dimensional input) of length
max(len(a),len(b)). If zi=None or is not given then initial
Outputs: (y, {zf})
y -- The output of the digital filter.
zf -- If zi is None, this is not returned, otherwise, zf holds the
final filter delay values.
Algorithm:
The filter function is implemented as a direct II transposed
structure.
This means that the filter implements
y[n] = b[0]*x[n] + b[1]*x[n-1] + ... + b[nb]*x[n-nb]
- a[1]*y[n-1] + ... + a[na]*y[n-na]
using the following difference equations:
y[m] = b[0]*x[m] + z[0,m-1]
z[0,m] = b[1]*x[m] + z[1,m-1] - a[1]*y[m]
...
z[n-3,m] = b[n-2]*x[m] + z[n-2,m-1] - a[n-2]*y[m]
z[n-2,m] = b[n-1]*x[m] - a[n-1]*y[m]
where m is the output sample number and n=max(len(a),len(b)) is the
model order.
The rational transfer function describing this filter in the
z-transform domain is
-1 -nb
b[0] + b[1]z + ... + b[nb] z
Y(z) = ---------------------------------- X(z)
-1 -na
a[0] + a[1]z + ... + a[na] z
-------------------------------------------------------------------------
Using Tomcat but need to do more? Need to support web services, security?
Get stuff done quickly with pre-integrated technology to make your job easier | 755 | 2,636 | {"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} | 3.140625 | 3 | CC-MAIN-2014-10 | longest | en | 0.810461 |
https://www.khanacademy.org/math/arithmetic/arith-decimals/arith-review-decimals-to-fractions/v/converting-decimals-to-fractions-1-ex-3 | 1,618,567,773,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038056325.1/warc/CC-MAIN-20210416100222-20210416130222-00372.warc.gz | 962,445,013 | 33,831 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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Main content
# Rewriting decimals as fractions: 0.36
CCSS.Math:
## Video transcript
let's see if we can write 0.36 as a fir action as a fraction and there are several ways of doing it the way I like to do it is to say well 0.36 this is the same thing as 36 hundredths or one way to think about it is this is in the hundreds place hundreds hundreds place this is in the tens place or you could view this as 30 hundredths you could view this as three tenths or thirty hundreds so we could say that this is the same thing as thirty six hundredths or this is equal to thirty six over one hundred we've already expressed it as a fraction but now we could actually simplify it because both 36 and one-hundred have some common factors they are both divisible by well looks like they're both divisible by four so if we could divide the numerator by 4 and the denominator by 4 we get we're doing the same thing to both so we're not changing the value of the fraction 36 divided by 4 is 9 and then 100 divided by 4 is 25 and now these two characters don't seem to share any other common factors and so we've written it we've written it in simplified form and we're done | 320 | 1,384 | {"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} | 3.296875 | 3 | CC-MAIN-2021-17 | latest | en | 0.974241 |
http://centraledesmaths.uregina.ca/QandQ/topics/two%20equations | 1,679,538,068,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944606.5/warc/CC-MAIN-20230323003026-20230323033026-00629.warc.gz | 12,835,102 | 5,400 | Centrale des maths - centraledesmaths.uregina.ca
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Two equations with fractions 2020-10-22 mia pose la question :I am supposed to look for x and y. my teacher didn't explain very well. please help : (( 2x - y = 5 x/4 + y/3= 2Penny Nom lui répond. Are those two equations the same? 2018-06-05 Francisco pose la question :Solve for y: 2(x+3)=y x=5 Solve for y: (x+3)2=y x=5 Are those two equations the same?Penny Nom lui répond. Algebra 2017-05-28 Kerim pose la question :I need to solve this equation. y=at^2 - 2at x =2a√t Express y in terms of x and a Give in the form y=(x^p/ma^3) - (x^q/na) where p,q,m and n are integersPenny Nom lui répond. Two equations 2016-09-17 Nilesh pose la question :How to solve (x-y)^1/2 + 3(x-y) = 30 ; xy + 3(x-y) = 11?Penny Nom lui répond. Two equations 2015-02-16 nigel pose la question :2x+1/2y=1 6x-3/2y=21Penny Nom lui répond. Two equations in x and y 2014-09-25 seyilogo pose la question :solve y=2x - 3 and (4x - 2y) / x + y = 1 simultaneouslyHarley Weston lui répond. Two equations in two variables 2014-09-18 Susan pose la question :(28x + 36y) - [20000 - .75(28x + 36y) + 60000] = 5000 x + y = 10000 solving two equations involving variablesPenny Nom lui répond. Substitution 2010-02-14 Izayah pose la question :what is 3 ways to find the meeting point of y=2x+4 and 3x+y=7 by making 2 table and using a graphStephen La Rocque lui répond. Solving two equations, one with a square root 2009-11-23 kacie pose la question :y = square root of x+3 x-4y = -7 im having trouble with this problem...i have to find where they intersect.Harley Weston lui répond. Two equations in two unknowns 2009-09-18 Citizen pose la question :x+-3y=7 -x+4y=7Penny Nom lui répond. The method of substitution 2009-09-17 laura pose la question :ok! i really need help with this question plz help!! x= - 4y +5 x+2y = 7Penny Nom lui répond. Two equations 2009-05-31 tony pose la question :y = 1.2x - 32.7, y = -0.7x + 46.15 can you please solve this for me? i am not able to figure it out... thank you tonyPenny Nom lui répond. Two equations 2009-05-18 Bob pose la question :solve the system: x/2 - 2y= -8 3x + 2/3y= -10Stephen La Rocque lui répond. Two equations in two unknowns 2008-08-01 James pose la question :Question from james, a student: 3x+2y=9;24x-2y=18 My math teacher says that it is (2,-4) Can someone please explain how to figure this out by graphing?Victoria West lui répond. Two equations in x and y 2008-02-01 leslie pose la question :6x+8y=40 4x+5y=25Penny Nom lui répond. Dividing money unequally 2007-02-06 Linda pose la question :I have a total of \$13,826.48. It is to be divided by 15 kids - 2 of the kids only get half of what the other 13 do - how much do the 13 get and how much does the 2 get? ThanksHaley Ess and Steve La Rocque lui répond. Two equations in two unknowns 2003-02-06 Patrick pose la question :What are all ordered pairs of real numbers (x, y) for which: yx2 - 7x + 12 = 1 and x + y = 6? Penny Nom lui répond. Simultaneous equations 2001-12-17 Matthew pose la question :4x + y = 4 2x - 3y = 5 what is x and y Penny Nom lui répond. Two equations in two unknowns 2001-12-04 Courtney pose la question :y = 3x + 2 y = 4x - 5 solve for xPenny Nom lui répond.
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` < < < < < < < < < < `
## Camomile Game -- who is the winner?
Time limit = 5 seconds
The Camomile Game has following rules. Camomile is a flower with N petals.
Two playes make their moves in turn. For a move a player can pull out any petal or two petals growing side by side.
The player who can't make move looses.
It's easy to find out that the second player has winning strategy for N > 2.
We have M camomiles each with one petal pulled out. Your program should determine who has winning strategy in this game.
Input The number of camomiles M, 1 ≤ M ≤ 10 folowed by M lines with numbers of petals on camomiles Ni ( Ni < 200 ).
Output Line with "`First wins.`" or "`Second wins.`".
Input#1```1 10 ``` Output#1```First wins. ```
Input#2```2 10 10 ``` Output#2```Second wins. ```
Input#3```3 6 7 8 ``` Output#3```Second wins. ```
Author:
Voroztsov Artem
<PREV | Problem set | Search related messages | NEXT>
© acm.mipt DevGroupThe page was generated in 210ms | 303 | 1,009 | {"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} | 2.578125 | 3 | CC-MAIN-2020-16 | latest | en | 0.900078 |
https://www.thestudentroom.co.uk/showthread.php?t=2300981 | 1,529,564,043,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864039.24/warc/CC-MAIN-20180621055646-20180621075646-00019.warc.gz | 945,731,937 | 46,115 | You are Here: Home >< Maths
# Implicit equations help watch
1. I'm stuck on question 4
Posted from TSR Mobile
Attached Images
2. And this is as far as I've got
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Attached Images
3. Oh that's horrible!
Really!
You have separated the variables correctly but the next line is just drivel. Try substituting a couple of numbers and you will agree.
Integrate the y term by recognising a logarithmic integral and the x term by making a substitution for the denominator or by expressing it in divided out form.
4. (Original post by SDavis123)
And this is as far as I've got
Posted from TSR Mobile
Line 1 = good
Line 2 = rubbish
Ditto the y side being incorrect
5. (Original post by Mr M)
Oh that's horrible!
Really!
You have separated the variables correctly but the next line is just drivel. Try substituting a couple of numbers and you will agree.
Integrate the y term by recognising a logarithmic integral and the x term by making a substitution for the denominator or by expressing it in divided out form.
Ah yes I see what you mean so take the 2 out of the y bit to make the top differential of the bottom so you'll get ln(y^2 +1) ?
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7. (Original post by SDavis123)
Ah yes I see what you mean so take the 2 out of the y bit to make the top differential of the bottom so you'll get ln(y^2 +1) ?
Posted from TSR Mobile
Sort of.
You need to deal with the 2 you just threw away.
8. Ah yh sorry, (1/2)(ln(y^2 + 1))
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9. (Original post by SDavis123)
Ah yh sorry, (1/2)(ln(y^2 + 1))
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Yes.
10. (Original post by Mr M)
Yes.
I'm not sure on the x part but I said
x/(x-1)=x(x-1)^-1
And when I integrate that I get
((-x^2)/4).(x-1) that right?
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11. (Original post by SDavis123)
I'm not sure on the x part but I said
x/(x-1)=x(x-1)^-1
And when I integrate that I get
((-x^2)/4).(x-1) that right?
Posted from TSR Mobile
You seem to be making up your own integration rules now.
which I hope you can integrate.
12. (Original post by Mr M)
You seem to be making up your own integration rules now.
which I hope you can integrate.
Ermmm ok haha but how did you get
x/(x-1)=(x-1+1)/(x-1) ?
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13. (Original post by SDavis123)
Ermmm ok haha but how did you get
x/(x-1)=(x-1+1)/(x-1) ?
Posted from TSR Mobile
Because
14. (Original post by SDavis123)
Ermmm ok haha but how did you get
x/(x-1)=(x-1+1)/(x-1) ?
Posted from TSR Mobile
Because x-1+1 = x ???
Edit: SNAP!
15. (Original post by davros)
Because x-1+1 = x ???
Snap!
16. (Original post by Mr M)
Because
Ah yh obviously can't believe I didn't see that haha
Posted from TSR Mobile
17. (Original post by Mr M)
Snap!
I'm trying to write maths while watching the football. I should stop now before I write something stupid!
18. (Original post by davros)
I'm trying to write maths while watching the football. I should stop now before I write something stupid!
Football? And there was me thinking mathematicians were clever
And what's the next move after
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19. (Original post by SDavis123)
Football? And there was me thinking mathematicians were clever
And what's the next move after
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Integrate it - come on this is getting a bit painful!
20. (Original post by Mr M)
Integrate it - come on this is getting a bit painful!
Yh I did, did you see the picture?
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sir can you tell how to do the mahesh and ramesh question using unit law, i reached to part where we get common value for both but how take it with respect to savings as both the saving are different.
Dhruval Sharma
5 months ago
in 9:40 mins . 1/4 of his income means 1/4 of X. then where has 1/4 gone.
Sumant Raj Garg
10 months ago
his income is taken as 4x so when 1/4 is multiplied by income which is 4x it simply becomes x.
sir in last question 7-3=4 then savings of b should be multiple of 4 ,but answer is not multiple of 4???
brother a and b had some savings in ratio 4:5. they decided to buy a gift for their sister. sharing the cost in the ratio 3:4. after they bought, A spent two third of his amount, while B is left with 145. then, the value of the gift is?
Jyoti Meena
2 years ago
how to apply unit method in this question?
A
Ayush
a year ago
4x , 5x is their savings and spend in 3y, 4y then 5x-4y=145 and 4x*2/3 =3y
PS
sir how can we assume 1/4 as x? I assumed the same in another question but I didn't get the answer..
1. RATIO AND PROPORTION Class -5 (Income and Expenditure) Part-1
2. If income of Ram and Shyam are in the ratio 5:6 and their expenditure are in the ratio 7:9. If both of them saves Rs3000. Find income of Shyam. a. Rs8000 b. Rs6000 C. Rs12000 d. Rs10000
3. If income of Ram and Shyam are in the ratio 5:6 and their expenditure are in the ratio 7:9. If both of them saves Rs3000. Find income of Shyam. a. Rs8000 b. Rs6000 C. Rs12000 d. Rs10000 Salwe 30D
4. If income of Ram and Shyam are in the ratio 5:6 and their expenditure are in the ratio 7:9. If both of them saves Rs3000. Find income of Shyam. a. Rs8000 b. Rs6000 C. Rs12000 d. Rs10000 43 42 9 Bors 3om
5. Ratio of income of Ramesh and Mahesh is 3 4 and ratio of their expenditure is 5 8. If Ramesh saves Rs2500 and Mahesh saves Rs2000. Find sum of income of Mahesh and Ramesh a. Rs17000 b. Rs14000 c. Rs17500 d. Rs21000
6. Ratio of income of Ramesh and Mahesh is 3:4 and ratio of their expenditure is 5 8. If Ramesh saves Rs2500 and Mahesh saves Rs2000. Find sum of income of Mahesh and Ramesh. a. Rs17000 b. Rs14000 c. Rs17500 d. Rs21000 3 C. Save Sovivn
7. Ratio of income of Ramesh and Mahesh is 3:4 and ratio of their expenditure is 5 8. If Ramesh saves Rs2500 and Mahesh saves Rs2000. Find sum of income of Mahesh and Ramesh. a. Rs17000 b. Rs14000 c. Rs17500 d. Rs21000 CO (9 3 C. Sovivn
8. Weekly incomes of two persons are in the ratio of 7:3 and their weekly expenses are in the ratio of 5:2. If each of them saves Rs 300 per week, then the weekly income of the first person is a. Rs 7500 b. Rs 4500 c. Rs 6300 d. Rs 5400
9. Weekly incomes of two persons are in the ratio of 7:3 and their weekly expenses are in the ratio of 5:2. If each of them saves Rs 300 per week, then the weekly income of the first person is a. Rs 7500 b. Rs 4500 c. Rs 6300 d. Rs 5400 A B A B 2
10. The ratio of the incomes of A, B and C is 4:5:6 and the ratio of their expenditure is 9:8:7. If savings of A is one-fourth of his income, then what is the ratio of their savings ? a, 7 : 3 : 11 b.11 : 7:3 c. 3: 11:7 d.3:7:11 | 1,168 | 3,518 | {"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-2020-34 | latest | en | 0.962405 |
http://webwork.maa.org/viewvc/system/trunk/pg/macros/contextLimitedPolynomial.pl?sortby=author&view=diff&pathrev=5991&r1=5050&r2=5051 | 1,369,467,378,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705749262/warc/CC-MAIN-20130516120229-00075-ip-10-60-113-184.ec2.internal.warc.gz | 293,499,001 | 3,213 | [system] / trunk / pg / macros / contextLimitedPolynomial.pl Repository: Repository Listing bbplugincoursesdistsnplrochestersystemwww
# Diff of /trunk/pg/macros/contextLimitedPolynomial.pl
Revision 5050 Revision 5051
19# each degree can be included in the polynomial: 19# each degree can be included in the polynomial:
20# 20#
21# Context("LimitedPolynomial")->flags->set(singlePowers=>1); 21# Context("LimitedPolynomial")->flags->set(singlePowers=>1);
22# 22#
23 23
24=cut
25
24# 26#
25# Handle common checking for BOPs 27# Handle common checking for BOPs
26# 28#
27
28=cut
29
30package LimitedPolynomial::BOP; 29package LimitedPolynomial::BOP;
31 30
32# 31#
33# Do original check and then if the operands are numbers, its OK. 32# Do original check and then if the operands are numbers, its OK.
34# Otherwise, do an operator-specific check for if the polynomial is OK. 33# Otherwise, do an operator-specific check for if the polynomial is OK.
59 LimitedPolynomial::markPowers(\$l); 58 LimitedPolynomial::markPowers(\$l);
60 LimitedPolynomial::markPowers(\$r); 59 LimitedPolynomial::markPowers(\$r);
61 my \$exponents = \$self->{exponents} = \$r->{exponents}; 60 my \$exponents = \$self->{exponents} = \$r->{exponents};
62 delete \$r->{exponents}; delete \$r->{powers}; 61 delete \$r->{exponents}; delete \$r->{powers};
63 if (\$l->{exponents}) { 62 if (\$l->{exponents}) {
64 my \$single = \$self->{equation}{context}->flag('singlePowers'); 63 my \$single = \$self->context->flag('singlePowers');
65 foreach my \$i (0..scalar(@{\$exponents})-1) { 64 foreach my \$i (0..scalar(@{\$exponents})-1) {
66 \$self->Error("A variable can appear only once in each term of a polynomial") 65 \$self->Error("A variable can appear only once in each term of a polynomial")
67 if \$exponents->[\$i] && \$l->{exponents}[\$i] && \$single; 66 if \$exponents->[\$i] && \$l->{exponents}[\$i] && \$single;
68 \$exponents->[\$i] += \$l->{exponents}[\$i]; 67 \$exponents->[\$i] += \$l->{exponents}[\$i];
69 } 68 }
78# and record the new power list 77# and record the new power list
79# 78#
80sub checkPowers { 79sub checkPowers {
81 my \$self = shift; 80 my \$self = shift;
82 my (\$l,\$r) = (\$self->{lop},\$self->{rop}); 81 my (\$l,\$r) = (\$self->{lop},\$self->{rop});
83 my \$single = \$self->{equation}{context}->flag('singlePowers'); 82 my \$single = \$self->context->flag('singlePowers');
84 LimitedPolynomial::markPowers(\$l); 83 LimitedPolynomial::markPowers(\$l);
85 LimitedPolynomial::markPowers(\$r); 84 LimitedPolynomial::markPowers(\$r);
86 \$self->{isPoly} = 1; 85 \$self->{isPoly} = 1;
87 \$self->{powers} = \$l->{powers} || {}; delete \$l->{powers}; 86 \$self->{powers} = \$l->{powers} || {}; delete \$l->{powers};
88 return 1 unless \$r->{powers}; 87 return 1 unless \$r->{powers};
316# 315#
317# Now build the new context that calls the 316# Now build the new context that calls the
318# above classes rather than the usual ones 317# above classes rather than the usual ones
319# 318#
320 319
321\$context{LimitedPolynomial} = Context("Numeric"); 320\$context{LimitedPolynomial} = Parser::Context->getCopy(undef,"Numeric");
322\$context{LimitedPolynomial}->operators->set( 321\$context{LimitedPolynomial}->operators->set(
323 '+' => {class => 'LimitedPolynomial::BOP::add'}, 322 '+' => {class => 'LimitedPolynomial::BOP::add'},
324 '-' => {class => 'LimitedPolynomial::BOP::subtract'}, 323 '-' => {class => 'LimitedPolynomial::BOP::subtract'},
325 '*' => {class => 'LimitedPolynomial::BOP::multiply'}, 324 '*' => {class => 'LimitedPolynomial::BOP::multiply'},
326 '* ' => {class => 'LimitedPolynomial::BOP::multiply'}, 325 '* ' => {class => 'LimitedPolynomial::BOP::multiply'},
Legend:
Removed from v.5050 changed lines Added in v.5051 | 1,139 | 3,736 | {"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} | 2.765625 | 3 | CC-MAIN-2013-20 | latest | en | 0.345143 |
https://net-informations.com/python/iq/lambda.htm | 1,685,419,512,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645089.3/warc/CC-MAIN-20230530032334-20230530062334-00739.warc.gz | 467,247,257 | 5,697 | # What is lambda in Python?
Lambda, the 11th letter of the Greek alphabet , is the symbol for wavelength . Lambda comes from the Lambda Calculus and refers to anonymous functions in programming. In Python, Lambda is an expression . Lambda's body is a single expression, not a block of statements. Because it is limited to an expression, a lambda is less general than a def you can only squeeze so much logic into a lambda body without using statements such as if. This is not exactly the same as lambda in functional programming languages, but it is a very powerful concept that's well integrated into Python and is often used in conjunction with typical functional concepts like map() , filter() and reduce() . Moreover, Lambda can be used wherever function objects are required.
The general syntax of a lambda function is quite simple:
lambda argument_list: expression
The argument list consists of a comma separated list of arguments and the expression is an arithmetic expression using these arguments.
Here's an example. You can build a function in the normal way, using def, like this:
def square_root(x): return math.sqrt(x)
Using lambda:
square_root = lambda x: math.sqrt(x)
example
square_root = lambda x: x*x print(square_root(2))
Lambda allows you to write quick throw away functions without naming them. It also provides a nice way to write closures also.
example
Find the sum of two numbers using lambda
output
30
Lambda in Conditional expressions:
result = lambda x: "Bigger than 100" if x > 100 else "Smaller than 100"
print(result(99))
output
Smaller than 100
### Map example using lambda
my_list = [1, 2, 3, 4, 5, 6] squared = map(lambda x: x**2, my_list) print(list(squared))
output
[1, 4, 9, 16, 25, 36]
### Filter example using lambda
my_list = [1, 3,5, 7, 9, 11, 13, 15] new_list = list(filter(lambda x: (x%3 == 0) , my_list)) print(new_list)
output
[3, 9, 15]
### Reduce example using lambda
from functools import reduce result = reduce((lambda x, y: x * y), [1, 2, 3, 4,5]) print(result)
output
120
### Calculating the sum of the numbers from 1 to 1000 using lambda
from functools import reduce total = reduce(lambda x, y: x+y, range(1,1001)) print(total)
output
500500 | 559 | 2,216 | {"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} | 2.921875 | 3 | CC-MAIN-2023-23 | longest | en | 0.782075 |
https://programmer.ink/think/machine-learning-algorithm-series-logistic-regression-algorithm.html | 1,643,172,950,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304915.53/warc/CC-MAIN-20220126041016-20220126071016-00086.warc.gz | 532,767,017 | 7,880 | # Machine learning algorithm series - Logistic Regression Algorithm
Posted by phpzone on Tue, 11 Jan 2022 07:50:23 +0100
# 1, Introduction
continue with a log probability regression algorithm (I), which introduces two methods to optimize the cost function of log probability regression - Gradient descent method and Newton's method. However, when using some third-party machine learning libraries, it will be found that the above two methods are not simply used directly, but some optimized versions or algorithm variants are used. For example, the optional solvers in scikit learn described earlier are shown in the following table:
solver / solveralgorithm
sagaRandom average gradient descent acceleration method (SAGA)
lbfgsL-BFGS algorithm (limited memory Broyden – Fletcher – Goldfarb – Shanno/L-BFGS)
let's introduce these algorithms one by one. Why don't the direct gradient descent method and Newton method in the general third-party library? What are the defects of these two original algorithms? Due to the author's limited ability, the following algorithm only gives the iterative formula, and the source of the iterative formula cannot be deduced in detail here. Interested readers can refer to the proof in the corresponding paper.
the original gradient descent algorithm, also known as Batch gradient descent/BGD, takes the whole data set as input to calculate the gradient.
$$w=w-\eta \nabla_{w} \operatorname{Cost}(w)$$
the main disadvantage of this algorithm is that it uses the whole data set. When the data set is large, it may be abnormally time-consuming to calculate the gradient.
each iteration update only randomly processes one data, not the whole data set.
$$w=w-\eta \nabla_{w} \operatorname{Cost}\left(w, X_{i}, y_{i}\right) \quad i \in[1, N]$$
because the algorithm is a random data point, the cost function does not decline all the time, but fluctuates up and down. Adjusting the step size makes the result of the cost function decline as a whole, so the convergence rate does not decline as fast as the batch gradient.
Mini batch gradient descent / MBGD
the small batch gradient descent method combines the above two algorithms. When calculating the gradient, it uses neither the whole data set nor one data randomly selected at a time, but part of the data to update at a time.
$$w=w-\eta \nabla_{w} \operatorname{Cost}\left(w, X_{(i: i+k)}, y_{(i: i+k)}\right) \quad i \in[1, N]$$
stochastic average gradient descent method is the optimization of stochastic gradient descent method. Due to the randomness of SGD, its convergence speed is slow. SAG records the gradient of the last position so that more information can be seen.
\begin{aligned} w_{k+1} &=w_{k}-\frac{\eta}{N} \sum_{i=1}^{N}\left(d_{i}\right)_{k} \\ d_{k} &=\left\{\begin{array}{ll} \nabla_{w} \operatorname{Cost}\left(w_{k}, X_{i}, y_{i}\right) & i=i_{k} \\ d_{k-1} & i \neq i_{k} \end{array}\right. \end{aligned}
variance reduction stochastic gradient descent method is another optimization of stochastic gradient descent method. Because the convergence problem of SGD is that the variance of gradient is assumed to have an upper bound of constant, SVRG makes the convergence process more stable by reducing this variance.
$$w_{k+1}=w_{k}-\eta\left(\nabla_{w} \operatorname{Cost}\left(w_{k}, X_{i}, y_{i}\right)-\nabla_{w} \operatorname{Cost}\left(\hat{w}, X_{i}, y_{i}\right)+\frac{1}{N} \sum_{j=1}^{N} \nabla_{w} \operatorname{Cost}\left(\hat{w}, X_{j}, y_{j}\right)\right)$$
Variation of random mean gradient descent method3(SAGA)
SAGA is the optimization of random average gradient descent method, which combines the variance reduction random gradient descent method.
$$w_{k+1}=w_{k}-\eta\left(\nabla_{w} \operatorname{Cost}\left(w_{k}, X_{i}, y_{i}\right)-\nabla_{w} \operatorname{Cost}\left(w_{k-1}, X_{i}, y_{i}\right)+\frac{1}{N} \sum_{j=1}^{N} \nabla_{w} \operatorname{Cost}\left(w, X_{j}, y_{j}\right)\right)$$
# 3, Newton method
Newton Method
the original version of Newton's method takes the whole data set as input to calculate the gradient and Hesse matrix, and then calculate the falling direction
$$w_{k+1}=w_{k}-\eta\left(H^{-1} \nabla_{w} \operatorname{Cost}\left(w_{k}\right)\right)$$
the main disadvantage of this algorithm is that it needs to find the Hesse matrix and its inverse matrix. When the dimension of x is too many, the process of finding the Hesse matrix will be extremely difficult.
DFP method4(Davidon–Fletcher–Powell)
DFP method is a quasi Newton method. The algorithm is as follows:
\begin{aligned} g_{k} &=\nabla_{w} \operatorname{Cost}\left(w_{k}\right) \\ d_{k} &=-D_{k} g_{k} \\ s_{k} &=\eta d_{k} \\ w_{k+1} &=w_{k}+s_{k} \\ g_{k+1} &=\nabla_{w} \operatorname{Cost}\left(w_{k+1}\right) \\ y_{k} &=g_{k+1}-g_{k} \\ D_{k+1} &=D_{k}+\frac{s_{k} s_{k}^{T}}{s_{k}^{T} y_{k}}-\frac{D_{k} y_{k} y_{k}^{T} D_{K}}{y_{k}^{T} D_{k} y_{k}} \end{aligned}
it can be seen that DFP no longer directly solves the Hessian matrix, but obtains the approximate value through iteration, where D is the approximation of the inverse matrix of the Hessian matrix.
BFGS method5(Broyden–Fletcher–Goldfarb–Shanno)
The BFGS method as like as two peas Newton methods is basically the same as the DFP method.
\begin{aligned} g_{k} &=\nabla_{w} \operatorname{Cost}\left(w_{k}\right) \\ d_{k} &=-D_{k} g_{k} \\ s_{k} &=\eta d_{k} \\ w_{k+1} &=w_{k}+s_{k+1} \\ g_{k+1} &=\nabla_{w} \operatorname{Cost}\left(w_{k+1}\right) \\ y_{k} &=g_{k+1}-g_{k} \\ D_{k+1} &=\left(I-\frac{s_{k} y_{k}^{T}}{y_{k}^{T} s_{k}}\right) D_{k}\left(I-\frac{y_{k} s_{k}^{T}}{y_{k}^{T} s_{k}}\right)+\frac{s_{k} s_{k}^{T}}{y_{k}^{T} s_{k}} \end{aligned}
it can be seen that the only difference between the BFGS method is the approximation method for the Hesse matrix, where D is the approximation of the inverse matrix of the Hesse matrix.
L-BFGS method6(Limited-memory Broyden–Fletcher–Goldfarb–Shanno)
because the BFGS method needs to store an approximate Hessian matrix, when the dimension of x is too large, the memory occupied by the Hessian matrix will be abnormally large. The L-BFGS method is another approximation of the BFGS method. The algorithm is as follows:
\begin{aligned} g_{k} &=\nabla_{w} \operatorname{Cost}\left(w_{k}\right) \\ d_{k} &=-\operatorname{calcDirection}\left(s_{k-m: k-1}, y_{k-m: k-1}, \rho_{k-m: k-1}, g_{k}\right) \\ s_{k} &=\eta d_{k} \\ w_{k+1} &=w_{k}+s_{k} \\ g_{k+1} &=\nabla_{w} \operatorname{Cost}\left(w_{k+1}\right) \\ y_{k} &=g_{k+1}-g_{k} \\ \rho_{k} &=\frac{1}{y_{k}^{T} s_{k}} \end{aligned}
it can be seen that the L-BFGS method no longer directly saves the approximate Hessian matrix, but directly calculates it through a group of vectors when it is to be used, so as to save memory. The method of calculating the direction can refer to the implementation in the following code.
Newton conjugate gradient / Newton CG
Newton conjugate gradient method is the optimization of Newton method. The algorithm is as follows:
\begin{aligned} g_{k} &=\nabla_{w} \operatorname{Cost}\left(w_{k}\right) \\ H_{k} &=\nabla_{w}^{2} \operatorname{Cost}\left(w_{k}\right) \\ \Delta w &=c g\left(g_{k}, H_{k}\right) \\ w_{k+1} &=w_{k}-\Delta w \end{aligned}
it can be seen that Newton's Conjugate Gradient method no longer solves the inverse matrix of Hesse matrix, but directly solves it through the Conjugate Gradient method Δ w. The Conjugate Gradient method is recommended in the references article 7. The principle and application of the algorithm are introduced in detail.
# 4, Code implementation
Use Python to implement the logarithmic probability regression algorithm (random gradient descent method):
import numpy as np
def logisticRegressionSGD(X, y, max_iter=100, tol=1e-4, step=1e-1):
w = np.zeros(X.shape[1])
xy = np.c_[X.reshape(X.shape[0], -1), y.reshape(X.shape[0], 1)]
for it in range(max_iter):
s = step / (np.sqrt(it + 1))
np.random.shuffle(xy)
X_new, y_new = xy[:, :-1], xy[:, -1:].ravel()
for i in range(0, X.shape[0]):
d = dcost(X_new[i], y_new[i], w)
if (np.linalg.norm(d) <= tol):
return w
w = w - s * d
return w
Use Python to implement the logarithmic probability regression algorithm (batch random gradient descent method):
import numpy as np
def logisticRegressionMBGD(X, y, batch_size=50, max_iter=100, tol=1e-4, step=1e-1):
w = np.zeros(X.shape[1])
xy = np.c_[X.reshape(X.shape[0], -1), y.reshape(X.shape[0], 1)]
for it in range(max_iter):
s = step / (np.sqrt(it + 1))
np.random.shuffle(xy)
for start in range(0, X.shape[0], batch_size):
stop = start + batch_size
X_batch, y_batch = xy[start:stop, :-1], xy[start:stop, -1:].ravel()
d = dcost(X_batch, y_batch, w)
if (np.linalg.norm(p_avg) <= tol):
return w
w = w - s * d
return w
Use Python to implement the logarithmic probability regression algorithm (random average gradient descent method):
import numpy as np
def logisticRegressionSAG(X, y, max_iter=100, tol=1e-4, step=1e-1):
w = np.zeros(X.shape[1])
p = np.zeros(X.shape[1])
d_prev = np.zeros(X.shape)
for it in range(max_iter):
s = step / (np.sqrt(it + 1))
for it in range(X.shape[0]):
i = np.random.randint(0, X.shape[0])
d = dcost(X[i], y[i], w)
p = p - d_prev[i] + d
d_prev[i] = d
p_avg = p / X.shape[0]
if (np.linalg.norm(p_avg) <= tol):
return w
w = w - s * p_avg
return w
Use Python to implement the logarithmic probability regression algorithm (variance reduction random gradient descent method):
import numpy as np
def logisticRegressionSVRG(X, y, max_iter=100, m = 100, tol=1e-4, step=1e-1):
w = np.zeros(X.shape[1])
for it in range(max_iter):
s = step / (np.sqrt(it + 1))
g = np.zeros(X.shape[1])
for i in range(X.shape[0]):
g = g + dcost(X[i], y[i], w)
g = g / X.shape[0]
tempw = w
for it in range(m):
i = np.random.randint(0, X.shape[0])
d_tempw = dcost(X[i], y[i], tempw)
d_w = dcost(X[i], y[i], w)
d = d_tempw - d_w + g
if (np.linalg.norm(d) <= tol):
break
tempw = tempw - s * d
w = tempw
return w
Implement logarithm probability regression algorithm (SAGA) using Python:
import numpy as np
def logisticRegressionSAGA(X, y, max_iter=100, tol=1e-4, step=1e-1):
w = np.zeros(X.shape[1])
p = np.zeros(X.shape[1])
d_prev = np.zeros(X.shape)
for i in range(X.shape[0]):
d_prev[i] = dcost(X[i], y[i], w)
for it in range(max_iter):
s = step / (np.sqrt(it + 1))
for it in range(X.shape[0]):
i = np.random.randint(0, X.shape[0])
d = dcost(X[i], y[i], w)
p = d - d_prev[i] + np.mean(d_prev, axis=0)
d_prev[i] = d
if (np.linalg.norm(p) <= tol):
return w
w = w - s * p
return w
Implement the logarithmic probability regression algorithm (DFP) using Python:
import numpy as np
def logisticRegressionDPF(X, y, max_iter=100, tol=1e-4):
w = np.zeros(X.shape[1])
D_k = np.eye(X.shape[1])
g_k = dcost(X, y, w)
for it in range(max_iter):
d_k = -D_k.dot(g_k)
s = lineSearch(X, y, w, d_k, 0, 10)
s_k = s * d_k
w = w + s_k
g_k_1 = dcost(X, y, w)
if (np.linalg.norm(g_k_1) <= tol):
return w
y_k = (g_k_1 - g_k).reshape(-1, 1)
s_k = s_k.reshape(-1, 1)
D_k = D_k + s_k.dot(s_k.T) / s_k.T.dot(y_k) - D_k.dot(y_k).dot(y_k.T).dot(D_k) / y_k.T.dot(D_k).dot(y_k)
g_k = g_k_1
return w
Implement the logarithmic probability regression algorithm (BFGS) using Python:
import numpy as np
def logisticRegressionBFGS(X, y, max_iter=100, tol=1e-4):
w = np.zeros(X.shape[1])
D_k = np.eye(X.shape[1])
g_k = dcost(X, y, w)
for it in range(max_iter):
d_k = -D_k.dot(g_k)
s = lineSearch(X, y, w, d_k, 0, 10)
s_k = s * d_k
w = w + s_k
g_k_1 = dcost(X, y, w)
if (np.linalg.norm(g_k_1) <= tol):
return w
y_k = (g_k_1 - g_k).reshape(-1, 1)
s_k = s_k.reshape(-1, 1)
a = s_k.dot(y_k.T)
b = y_k.T.dot(s_k)
c = s_k.dot(s_k.T)
D_k = (np.eye(X.shape[1]) - a / b).dot(D_k).dot((np.eye(X.shape[1]) - a.T / b)) + c / b
g_k = g_k_1
return w
Implement the logarithmic probability regression algorithm (L-BFGS) using Python:
import numpy as np
def calcDirection(ss, ys, rhos, g_k, m, k):
delta = 0
L = k
q = g_k.reshape(-1, 1)
if k > m:
delta = k - m
L = m
alphas = np.zeros(L)
for i in range(L - 1, -1, -1):
j = i + delta
alpha = rhos[j].dot(ss[j].T).dot(q)
alphas[i] = alpha
q = q - alpha * ys[j]
r = np.eye(g_k.shape[0]).dot(q)
for i in range(0, L):
j = i + delta
beta = rhos[j].dot(ys[j].T).dot(r)
r = r + (alphas[i] - beta) * ss[j]
return -r.ravel()
def logisticRegressionLBFGS(X, y, m=100, max_iter=100, tol=1e-4):
w = np.zeros(X.shape[1])
g_k = dcost(X, y, w)
d_k = -np.eye(X.shape[1]).dot(g_k)
ss = []
ys = []
rhos = []
for it in range(max_iter):
d_k = calcDirection(ss, ys, rhos, g_k, m, it)
s = lineSearch(X, y, w, d_k, 0, 1)
s_k = s * d_k
w = w + s_k
g_k_1 = dcost(X, y, w)
if (np.linalg.norm(g_k_1) <= tol):
return w
y_k = (g_k_1 - g_k).reshape(-1, 1)
s_k = s_k.reshape(-1, 1)
ss.append(s_k)
ys.append(y_k)
rhos.append(1 / (y_k.T.dot(s_k)))
g_k = g_k_1
return w
Implement the logarithmic probability regression algorithm (Newton conjugate gradient method) using Python:
import numpy as np
def cg(H, g, max_iter=100, tol=1e-4):
"""
H * deltaw = g
"""
deltaw = np.zeros(g.shape[0])
i = 0
r = g
d = r
delta = np.dot(r, r)
delta_0 = delta
while i < max_iter:
q = H.dot(d)
alpha = delta / (np.dot(d, q))
deltaw = deltaw + alpha * d
r = r - alpha * q
delta_prev = delta
delta = np.dot(r, r)
if delta <= tol * tol * delta_0:
break
beta = delta / delta_prev
d = r + beta * d
i = i + 1
return deltaw
def logisticRegressionNewtonCG(X, y, max_iter=100, tol=1e-4, step = 1.0):
"""
args:
X - Training data set
y - Target tag value
max_iter - Maximum number of iterations
tol - Tolerance value of variation
return:
w - weight coefficient
"""
# Initialize w to zero vector
w = np.zeros(X.shape[1])
# Start iteration
for it in range(max_iter):
d = dcost(X, y, w)
# When the gradient is small enough, end the iteration
if np.linalg.norm(d) <= tol:
break
# Calculate Hesse matrix
H = ddcost(X, y, w)
# Calculated by conjugate gradient method Δ w
deltaw = cg(H, d)
w = w - step * deltaw
return w
# 5, Third party library implementation
scikit-learn 8 realize logarithmic probability regression (random average gradient descent method):
from sklearn.linear_model import LogisticRegression
# Initialize log probability regressor, no regularization
reg = LogisticRegression(penalty="none", solver="sag")
# Fitting linear model
reg.fit(X, y)
# weight coefficient
w = reg.coef_
# intercept
b = reg.intercept_
scikit-learn 8 implement log probability regression (SAGA):
from sklearn.linear_model import LogisticRegression
# Initialize log probability regressor, no regularization
reg = LogisticRegression(penalty="none", solver="saga")
# Fitting linear model
reg.fit(X, y)
# weight coefficient
w = reg.coef_
# intercept
b = reg.intercept_
scikit-learn 8 log probability regression (L-BFGS):
from sklearn.linear_model import LogisticRegression
# Initialize log probability regressor, no regularization
reg = LogisticRegression(penalty="none", solver="lbfgs")
# Fitting linear model
reg.fit(X, y)
# weight coefficient
w = reg.coef_
# intercept
b = reg.intercept_
scikit-learn 8 realize logarithmic probability regression (Newton conjugate gradient method):
from sklearn.linear_model import LogisticRegression
# Initialize log probability regressor, no regularization
reg = LogisticRegression(penalty="none", solver="newton-cg")
# Fitting linear model
reg.fit(X, y)
# weight coefficient
w = reg.coef_
# intercept
b = reg.intercept_ | 4,861 | 15,305 | {"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": 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} | 3.390625 | 3 | CC-MAIN-2022-05 | latest | en | 0.889503 |
https://arriofeedbspos.web.app/265.html | 1,621,121,957,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991488.53/warc/CC-MAIN-20210515223209-20210516013209-00528.warc.gz | 130,136,100 | 5,292 | # Tangent of a circle worksheet pdf
These exercises are curated for students of grade 4 through high school. Jul 23, 2015 designed for the new gcse specification, this worksheet allows students to practise sketching circles and finding equations of tangents. A radius drawn to a tangent at the point of tangency is perpendicular to the tangent. Its cross section matches the function sinx from 0 to.
Use the pythagorean theorem to determine missing sides of right triangles learn the definitions of the sine, cosine, and tangent ratios of a right triangle set up proportions using sin, cos, tan to determine missing sides of right triangles use inverse trig functions to determine missing angles of a right triangle. The circle theorem circle theorem revision which tells us that a tangent is perpendicular to the radius. It will always form a right angle 90 with the radius. A line that intersects a circle at two points is an of the circle. The graph above shows a circle centred at the origin, o. Displaying all worksheets related to tangent and secant lines for circles. Students are given multiple situations and types of problems to find missing measures of the circles. Measure the arc intercepted ab by the chord by measuring the central angle.
Url on the angles and arcs formed by a tangent and a chord. If px, y are the coordinates of any point on the unit circle, and is the angle of rotation from the xaxis to point p, then tan. Tangent segment a touches a circle at one of the segments endpoints and lies in the line that is tangent to the circle at that point. The normal is a straight line which is perpendicular to the tangent.
Tangents to circles worksheet pdf october 3, 2019 july 9, 2019 some of the worksheets below are tangents to circles worksheet in pdf, tangents to circles. By the time your class has completed this extensive series, they will undoubtedly be experts at finding the area of a circle. Secant and tangent angles find the measure of the arc or angle indicated. Oct 03, 2019 tangents to circles worksheet pdf october 3, 2019 july 9, 2019 some of the worksheets below are tangents to circles worksheet in pdf, tangents to circles. Draw a random chord through your circle with endpoints a and b. Feel free to make copies of this worksheet for the sole purpose of use in your own classroom. Assume that lines which appear tangent are tangent.
Opposite angles in a cyclic quadrilateral sum to 180. The ime pqr is a tangent to a circle with centre o. Draw a circle on the half sheet and make a dot at the center. The fact that the gradient of a line is the negative reciprocal of the line perpendicular to it. Ap calculus ab worksheet 19 tangent and normal lines power rule learn. The tangent is a straight line which just touches the curve at a given point. Finding the trig values of points on unit circle examples.
Area of circles worksheet 1 here is a nine problem worksheet that will allow your students to practice calculating the area of a circle. Geometry circle theorems secant and tangent angles puzzle. The point p lies on the circle and has coordinates 20, 15. Each exercise provides a drawing of the circle as well as the length of the radius. Communicating about circles identifying special segments and lines, identifying common tangents, examples, exercises, once you find your worksheet s, you can either click on the popout icon or download button. In the figure at right, point e is the center and m. Equation of a tangent to a circle teaching resources. An of a circle is a line that intersects the circle in exactly one point. Equation of tangent line to circle worksheet tessshebaylo. Worksheets are 11 tangents to circles, arcs and angles formed by secants and tangents from a, kuta geometry circle tangents, assignment, work 1, assignment, 11 secant tangent and tangent tangent angles, parts of circle moderate s1. But it is sometimes useful to work in coordinates and this requires us to know the standard equation of a circle, how to interpret that equation and how to.
The is the point at which a tangent intersects a circle. In the diagram shown below, say whether ef is tangent to the circle with center at d. Circle graphs and tangents worksheets questions revision mme. The of a circle is a fixed point in space that is an equal distance from every point on the circle. The tangent at a point on a circle is at right angles to this radius. The wire is 48 cm long and is a tangent to the circle at points a and b. Displaying all worksheets related to tangent to a circle. A tangent is a line that just skims the surface of a circle. Tangent segments to the same circle from the same external point are congruent. Arcs and angles formed by secants and tangents from a. Gradient of straight line graphs revision if any of this feels unfamiliar to you.
Derivative slope of the tangent line at that points xcoordinate example. The point at witch a tangent line intersects the circle to witch it is tangent is the point of tangency. F worksheet by kuta software llc find the measure of the arc or angle indicated. Communicating about circles identifying special segments and lines, identifying common tangents, examples, exercises. A n angle whose vertex is on a circle and whose sides contain chords of the circle is an.
Tangent to a circle is perpendicular to the radius drawn to the point of tangency. Sheet matches the video exactly, including real exam question on the topic. Tangents to circles date period river dell regional. Equations of tangents to circles teaching resources. Feel free to make copies of this worksheet for the sole purpose. A radius is obtained by joining the centre and the point of tangency. A howto worksheet with qr code link to a video tutorial. From the same external point, the tangent segments to a circle are equal. Use your results from exercise 1 to make a conjecture about the lengths of tangent segments that have a common endpoint. Tangent to a curve over a giant snowdrift has formed on the quad. Infinite geometry tangent and secant angles and segments. Assume that lines which appear to be tangent are tangent. Oct 04, 2019 some of the worksheets below are segments in circles worksheet in pdf, line and segment relationships in the circle, geometry notes circles.
Tangent line of a circle can be determined once the tangent point or the slope of the line is known. Tangent lines in circles theorems practice foldable. Tangent segments from an exterior point to a circle are congruent. E r 2m ia td he1 cwmist 9h u vijn7fuizn 5i4tge0 vgzevo nmce etzr9yy.
Differentiate the terms relating to a circle, once you find your worksheet s, you can either click on the popout icon or download button to print or download your desired. The circle is a familiar shape and it has a host of geometric properties that can be proved using the traditional euclidean format. The exclusive pages contain a lot of pdf worksheets in finding area, circumference, arc length, and area of sector. A circle centered at the origin with a radius of 1. Angle at centre is twice angle at circumference 4 angle abc 92 reason. There are two main theorems that deal with tangents. Circle segment revision how to calculate the gradient of a line between two points.
The unit circle sec, cot 2tt 900 tt 3tt 2 2700 positive. The angle opq is 340 not drawn accurately calculate the size of angle tqr. Algebra2trig chapter 9 packet polk school district. This page contains circle worksheets based on identifying parts of a circle and finding radius or diameter. The point at which a tangent line intersects the circle to which it is tangent is the point of tangency.
Create a tangent line from the chords endpoints b in one direction. Step 2 draw tangents draw lines ab and cb so that they intersectp only ata and c,respectively. A line that intersects a circle at two points is a n of the circle. The exterior of a circle consists of the points that are outside the circle. Gradients of straight line revision the circle theorem circle theorem revision which tells us that a tangent is perpendicular to the radius. Worksheets are 11 secant tangent and tangent tangent angles, infinite geometry, arcs and angles formed by secants and tangents from a, circles 7 secant and tangent angles, topics include arcs angles chords secants tangents, assignment, circlecheat, find the measure of the arc or. Enjoy redistribution in any other form is prohibited. An angle whose vertex is on a circle and whose sides contain chords of the circle is an. In a circle, or in congruent circles, congruent central angles intercept congruent arcs. T is a point on the circumference of the circle such that pot is a straight line. Circles tangent and secant lines in circles colorbynumber worksheet this colorbynumber worksheet covers the concepts of tangent and secant lines in circles.
Secant and tangent angles 1 70 2 60 3 114 4 105 5 53 6 39 7 55 8 80. S secants and tangents oh my teaching geometry circle. Worksheets are 11 tangents to circles, arcs and angles formed by secants and tangents from a, kuta geometry circle tangents, assignment, work 1, assignment, 11 secant tangent. Y worksheet by kuta software llc find the area of each. Solved calculus bc worksheet on implicit diffeiation. Line qr is a tangent to the circle passing through the point p. Fillin the blank notes on the properties of tangents in circles for your students notebooks. Differentiate the terms relating to a circle, once you find your worksheets, you can either click on the popout icon or download button to print or download your desired worksheets. O tangent ratio classwork worksheet find the value of each trigonometric ratio. Tangents to circles date period river dell regional school. Angle between tangent and radius is 90 3 angle abc 67. Some of the worksheets below are segments in circles worksheet in pdf, line and segment relationships in the circle, geometry notes circles. A tangent line of a circle will always be perpendicular to the radius of that circle. Arcs and angles formed by a tangent and a chord url on the angles and arcs formed by a tangent and a chord.
These notes are lesson 1 in my 7 lesson circle unit notes for high school geometry. In a circle, or in congruent circles, congruent chords intercept congruent arcs. District programs, activities, and practices shall be free from discrimination based on race, color, ancestry, national origin, ethnic group identification, age, religion, marital or parental status, physical or mental disability, sex, sexual orientation, gender, gender identity or expression, or genetic information. In each circle, c is the center and ab is tangent to the circle at point b. Designed for the new gcse specification, this worksheet allows students to practise sketching circles and finding equations of tangents. These notes get folded in half to fit nicely in a spiral or composition book. Tangents and normals mctytannorm20091 this unit explains how di. Some of the worksheets below are tangents to circles worksheet in pdf, tangents to circles. In example 1, c is the center, so the circle is called c. Tangent and secant lines for circles worksheets lesson. B k qm5agd6ei aw6i zt uhl 3ihnafqibn niethey 5g 0eko moebtlr xy6.
1355 765 425 851 1125 50 1407 92 932 638 1111 16 1000 1252 1481 1592 523 1061 284 838 679 1336 12 1039 1260 1460 826 425 1149 7 946 | 2,514 | 11,373 | {"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-2021-21 | latest | en | 0.886176 |
https://www.opengl.org/discussion_boards/archive/index.php/t-181274.html | 1,534,817,621,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221217909.77/warc/CC-MAIN-20180821014427-20180821034427-00061.warc.gz | 928,441,482 | 6,015 | PDA
View Full Version : rotate object in 3d using camera
marito
03-19-2013, 10:58 AM
Hello! I am trying to make a camera rotate around an object that is already rotating. I want to move the camera using keyboard. Each time I push a button I change one of the three angles which show where the camera is. The camera rotates bythe Euler angles. I don't understand what I am doing wrong. Please help, I have searched and tried many things, but nothing works...
#include <stdio.h>
#include <stdlib.h>
#include <glut.h>
#include <math.h>
#define pi 3.14159265
typedef GLfloat point3d[3];
point3d p0={-1,-1,-13}, p1={1,-1,-13}, p2={-1,1,-13}, p3={1,1,-13};
point3d p4={-1,-1,-15}, p5={1,-1,-15}, p6={-1,1,-15}, p7={1,1,-15};
point3d p20={1,1,-13}, p21={1.5,1,-13}, p22={1,1.5,-13}, p23={1.5,1.5,-13};
point3d p24={1,1,-14}, p25={1.5,1,-14}, p26={1,1.5,-14}, p27={1.5,1.5,-14};
int delay=100, angle; //delay in msec
float xea=0., yea=0., zea=0., xe, ye, ze, a=0., b=0., c=0., move=20.;
void InitWindow()
{
GLfloat ambientLight[] = {0.3, 0.3, 0.3, 1};//ambient grey light
GLfloat diffuseLight[] = {0.7, 0.7, 0.7, 1};//diffuse white light
GLfloat lightPos[] = {1, 1, -12, 1};
//ENABLE SPECULAR LIGHT
//GLfloat specularLight[] = {0.8, 0.8, 0.8, 1};
//GLfloat specref[] = {1, 1, 1, 1};//full reflectance
//GLfloat spotDir[]={0, 0, -14};
//glLightfv(GL_LIGHT0,GL_SPECULAR,specularLight);
//glLightfv(GL_LIGHT0,GL_SPOT_DIRECTION, spotDir);
//glLightf(GL_LIGHT0,GL_SPOT_CUTOFF,15);
//glMaterialfv(GL_FRONT_AND_BACK, GL_SPECULAR,specref);
//glMateriali(GL_FRONT,GL_SHININESS,128);
glLightfv(GL_LIGHT0,GL_AMBIENT,ambientLight);
glLightfv(GL_LIGHT0,GL_DIFFUSE,diffuseLight);
glLightfv(GL_LIGHT0, GL_POSITION, lightPos);
glEnable(GL_LIGHTING);
glEnable(GL_LIGHT0);
glEnable(GL_DEPTH_TEST);
glEnable(GL_COLOR_MATERIAL);
glColorMaterial(GL_FRONT,GL_AMBIENT_AND_DIFFUSE);
glClearColor(0.,0.,40./255.,0); //blue sky
glMatrixMode(GL_PROJECTION);
glFrustum(-2,2,-2,2,10,15000);
// glOrtho(-4,5,-3,5,-13,15000);
glMatrixMode(GL_MODELVIEW);
}
void Stars()
{
float d,e;
srand(1);
glEnable(GL_POINT_SMOOTH);
glColor3f(1.,1.,0.);
glPointSize(1);
glBegin(GL_POINTS);
for (int i=1;i<=1000;i++){
e=200*rand()/RAND_MAX-100;
d=200*rand()/RAND_MAX-100;
glVertex3i(e,d,-300);
}
glEnd();
glPointSize(4);
glBegin(GL_POINTS);
for (int i=1;i<=100;i++){
e=200*rand()/RAND_MAX-100;
d=200*rand()/RAND_MAX-100;
glVertex3i(e,d,-300);
}
glEnd();
glPointSize(8);
glBegin(GL_POINTS);
for (int i=1;i<=10;i++){
e=200*rand()/RAND_MAX-100;
d=200*rand()/RAND_MAX-100;
glVertex3i(e,d,-300);
}
glEnd();
}
void Cubes()
{
glColor3f(0.3,0.3,0.9);//blue
glVertex3fv(p0); glVertex3fv(p1); glVertex3fv(p3); glVertex3fv(p2); //front
glColor3f(0.1,0.6,0.9);
glVertex3fv(p1); glVertex3fv(p5); glVertex3fv(p7); glVertex3fv(p3); //right
glColor3f(0,0.3,0.9);
glVertex3fv(p5); glVertex3fv(p4); glVertex3fv(p6); glVertex3fv(p7); //back
glColor3f(0.3,0,0.9);
glVertex3fv(p4); glVertex3fv(p0); glVertex3fv(p2); glVertex3fv(p6); //left
glColor3f(0.3,0.3,0);
glVertex3fv(p2); glVertex3fv(p3); glVertex3fv(p7); glVertex3fv(p6); //top
glColor3f(1,0.3,1);
glVertex3fv(p1); glVertex3fv(p0); glVertex3fv(p4); glVertex3fv(p5); //bottom
glEnd();
glColor3f(0.3,0.3,0.9);//blue
glVertex3fv(p20); glVertex3fv(p21); glVertex3fv(p23); glVertex3fv(p22); //front
glColor3f(0.1,0.6,0.9);
glVertex3fv(p21); glVertex3fv(p25); glVertex3fv(p27); glVertex3fv(p23); //right
glColor3f(0,0.3,0.9);
glVertex3fv(p25); glVertex3fv(p24); glVertex3fv(p26); glVertex3fv(p27); //back
glColor3f(0.3,0,0.9);
glVertex3fv(p24); glVertex3fv(p20); glVertex3fv(p22); glVertex3fv(p26); //left
glColor3f(0.3,0.3,0);
glVertex3fv(p22); glVertex3fv(p23); glVertex3fv(p27); glVertex3fv(p26); //top
glColor3f(1,0.3,1);
glVertex3fv(p21); glVertex3fv(p20); glVertex3fv(p24); glVertex3fv(p25); //bottom
glEnd();
}
void Display(void)
{
glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
glPushMatrix();
Stars();
glPopMatrix();
xe=(cos(c*pi/180.)*cos(b*pi/180.)*cos(a*pi/180.)-sin(c*pi/180.)*sin(a*pi/180.))*xea+(cos(c*pi/180.)*cos(b*pi/180.)*sin(a*pi/180.)+sin(c*pi/180.)*cos(a*pi/180.))*yea-cos(c*pi/180.)*sin(b*pi/180.)*zea;
ye=-(sin(c*pi/180.)*cos(b*pi/180.)*cos(a*pi/180.)-cos(c*pi/180.)*sin(a*pi/180.))*xea+(-sin(c*pi/180.)*cos(b*pi/180.)*sin(a*pi/180.)+cos(c*pi/180.)*cos(a*pi/180.))*yea+sin(c*pi/180.)*sin(b*pi/180.)*zea;
ze=sin(b*pi/180.)*cos(a*pi/180.)*xea+sin(b*pi/180.)*sin(a*pi/180.)*yea+cos(b*pi/180.)*zea;
gluLookAt(xe,ye,ze,0,0,-14,0,1,0);// (0,1,0) the angle of the camera,(xe,ye,ze) the camera,(0,0,-14) the focus point
xea=xe;
yea=ye;
zea=ze;
glTranslatef(0,0,-14);
glRotatef(angle,0,1,0);
glTranslatef(0,0,14);
Cubes();
glutSwapBuffers();
}
void Camera(unsigned char key, int x, int y)
{
switch(key){
case 'q': a+=move; break;
case 'w': a-=move; break;
case 'a': b+=move; break;
case 's': b-=move; break;
case 'z': c+=move; break;
case 'x': c-=move; break;
case '0': a=0.; b=0.; c=0.; //go to the initial state
}
if(key==27) exit(0); //turn off camera
glutPostRedisplay();
}
void Rotate(int n) // the glutTimerFunc
{
n++;
angle+=5;
glutPostRedisplay();
glutTimerFunc(delay,Rotate,n);
}
void main(int argc, char **argv)
{
glutInit(&argc, argv);
glutInitDisplayMode(GLUT_DOUBLE|GLUT_RGB|GLUT_DEPT H);
glutInitWindowPosition(200,100);
glutInitWindowSize(400,400);
glutCreateWindow("3D-cubes rotating in space and hit by light");
InitWindow();
glutDisplayFunc(Display);
glutTimerFunc(delay,Rotate,0);
glutKeyboardFunc(Camera);
glutMainLoop();
} | 2,106 | 5,459 | {"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} | 2.90625 | 3 | CC-MAIN-2018-34 | latest | en | 0.486873 |
https://responsivedesigntool.com/qa/does-2-tablespoons-equal-20-ml.html | 1,618,377,106,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038076819.36/warc/CC-MAIN-20210414034544-20210414064544-00019.warc.gz | 589,002,074 | 8,327 | # Does 2 Tablespoons Equal 20 Ml?
## Does 2 tablespoons equal 10 mL?
Official Answer 10mL equals two teaspoons (2tsp).
A tablespoon is three times bigger than a teaspoon and three teaspoons equal one tablespoon (1Tbsp or 1Tb).
One tablespoon also equals 15mL..
## Is 5 ml half a teaspoon?
Also, remember that 1 level teaspoon equals 5 mL and that ½ a teaspoon equals 2.5 mL.
## How many mL is a tablespoon of soy sauce?
Volumecentimeter³236.59metric dessertspoon23.66metric tablespoon15.77metric teaspoon47.32milliliter236.5914 more rows
## Is a tablespoon 15 or 20 mL?
Traditional definitions In nutrition labeling in the US and the UK, a tablespoon is defined as 15 mL (0.51 US fl oz). A metric tablespoon is exactly equal to 15 mL (0.51 US fl oz).
## What is 20 ml in teaspoons or tablespoons?
How Many Teaspoons is 20ml? – 20 ml is equal to 4.06 teaspoons. 20 ml to tsp converter to calculate how many teaspoons is 20ml.
## How do I measure 5 ml?
1 mL = 1 cc.2.5 mL = 1/2 teaspoon.5 mL = 1 teaspoon.15 mL = 1 tablespoon.3 teaspoons = 1 tablespoon.
## How much is a tablespoon?
A tablespoon is a unit of measure equal to 1/16 cup, 3 teaspoons, or 1/2 fluid ounce in the USA. It is either approximately or (in some countries) exactly equal to 15 mL. “Tablespoon” may be abbreviated as T (note: uppercase letter), tbl, tbs or tbsp.
## What is 4 mL in tablespoons?
0.27054 Milliliters = 0.2705 Tablespoons Volume Calculator Conversions.
## What is 15mL in spoons?
15ml equals to 3.04 teaspoons or there are 3.04 tsp in 15 milliliters.
## How many cups is 30 ml of water?
Volume (liquid)1 tablespoon15 ml2 tablespoon or 1 fluid ounce30 ml1/4 cup or 2 fluid ounces59 ml1/3 cup79 ml14 more rows
## Does 30 mL equal 2 tablespoons?
30 Milliliters = 2.0288 Tablespoons.
## How do I measure 20 ml?
A teaspoon, such for making coffee or tea, holds 5 ml. A smaller desert spoon holds 15 ml, the larger ones hold 20 mls. Cap-fuls usually vary, but if you plan on using one, you could measure out 20 mls once by using spoon-measure, and then use that.
## What is 2.5ml in teaspoons?
measurement of medicines1/4 teaspoon1.25 ml1/2 teaspoon2.5 ml3/4 teaspoon3.75 ml1 teaspoon5 ml1-1/2 teaspoon7.5 ml1 more row
## How many mils is a teaspoon?
4.92892159 milliliters1 US teaspoon (tsp) is equal to 4.92892159 milliliters (mL).
## Is 5mL the same as 5mg?
mL is a measurement of volume, while mg is a measurement of mass. … For instance, if you have a solution that contains 10mg of medication in 5mL and you want to give 5mg of medication, then “give over have” or 5/10 or one half, so half of 5mL is 2.5 mL. 2.5 mL of the solution will contain 5mg of the medication.
## What is 30 ml in spoons?
Convert 30 Milliliters to TablespoonsmLtbsp30.002.028830.012.029530.022.030230.032.030996 more rows
## How much is 30 ml in a glass?
Overview: Glass & Plastic Container Size Conversion ChartContainer SizeDramMilliliter10 ml2.711015 ml4.061530 ml8.12301 oz.8~ 309 more rows
## How much is 20 mL in a spoon?
Convert 20 Milliliters to TablespoonsmLtbsp20.001.352620.011.353220.021.353920.031.354696 more rows | 950 | 3,123 | {"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} | 2.78125 | 3 | CC-MAIN-2021-17 | latest | en | 0.844474 |
http://roommeimei.info/cyzu/work-out-exchange-rate-752.php | 1,563,806,999,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528037.92/warc/CC-MAIN-20190722133851-20190722155851-00479.warc.gz | 136,210,666 | 7,073 | Select Page
# Work out exchange rate
Some call this profit a South African Rands. You discover that 1 US single value or volume values Product and Quotient Rules - you in your local currency. Include your email address to fee or commission. How to convert naira into tutorial please. You would divide the US price by the exchange rate to get the cost to you can find through Google or by calling the Department. Read Question Reply to All.
## How to convert currencies in Microsoft Excel?
Kutools for Excel 's Currency exchange services that use companies you have to multiply by travelers exchange their home country's in the denominatorin exchange rates from the reliable. For example, If you have pounds in the numeratorby spending one unit of than 50 currencies in your money when they arrive in the foreign currency to buy. I have an international hotel smallest unit of value in ended up working it out. Most major airports have currency that converts both directions - places, you might round it to pesos; my formulas are "at war" with one another. Pip -A pip is the I was searching for but a bid-ask spread. The "rate" is what you get of the foreign currency quick way to convert more your local currency, or what Excel by using the latest published in The Journal of minutes before meals. .
Major currencies in the world today take on the physical out the exchange rate than exchange rate, divide 20, by Try now Kutools for Excel: of a person's currency is on a Google search, or. Currency exchange rates fluctuate frequently. Buying foreign currency from a use a calculator to work the selling ask price, which to turn on a PC, on a person, but most up and then search your favourites or a search engine. For instance, say you want South African Rands. The Product and Quotient Rules to convert US dollars to. The approaches described here are great, but they only work fill handle over the range unit of another currency. The forex is able to US dollars you would need to save at the current are equal in value. If not, there is a currency exchange at almost every major airport that will. How to convert naira into.
1. Customized Currency Exchange Rate
The euro is different in that it's given in terms for this math problem. If you select Only commentall of the converted of domestic currency per unit of foreign currency; s is. If you want to convert much money you'll need in Dollars, you would divide the and start with the foreign. Or, if you know how Sri Lankan Rupees into Australian the other country, work backwards number of rupees you have. Rea Oct 17 '17 at What is the general formula. I need to use the to assert that garcinia cambogia were split into two groups meal 2 and 4. Brings up the Edit menu of a forex quote:.
1. How Exchange Rates Work
Exchange rates are a common sight for both travelers to international investors. While exchange rate quotes are relatively easy to find these days, reading and making calculations based on them can be a little more challenging for those that aren't familiar with the techniques. Foreign Exchange Charges Calculator. Find out how much you pay in currency exchange fees. This calculator will reveal your provider’s hidden charges. A great companion to .
1. Working out the exchange rate on an calculator
Highlight the range that you. Currency forwards contracts and future parity, forward exchange rate is your jobs easier. There are many websites [1] because one unit of currency calculated using the following formula:. The Product and Quotient Rules more than handy functions, makes. This market is a necessity decentralized, over-the-counter market for the the currency risk. Using the covered interest rate to convert US dollars to. Real world currency exchanges with possible spreads, while sellers want the highest spreads. To view the entire catalogue modern revival of hunting for third most effective brand I've weight loss results in daily.
1. With Live Exchange Rate
What is the current value. Or some reference data sources. For instance, say you want of Sierra Leonean Leones in. Marginal point, but you need to assume an efficient market stand for the same as otherwise transitivity does not hold interest rate; and I f is foreign interest rate. Just being a pedant: Where, fs and n ie, no arbitrage opportunitiesinflation rates in the domestic and foreign countries. They are sometimes called the to calculate exchange rates. The current technology behind cryptocurrencies of Treasury [3] or other governmental agencies [4] to get transactions across a peer-to-peer network A pip is sometimes called. | 945 | 4,600 | {"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} | 2.9375 | 3 | CC-MAIN-2019-30 | latest | en | 0.935099 |
https://www.physicsforums.com/threads/demonstrate-that-a-matrix-that-has-a-null-row-is-not-invertible.285969/ | 1,721,448,497,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514981.25/warc/CC-MAIN-20240720021925-20240720051925-00758.warc.gz | 810,128,151 | 16,486 | # Demonstrate that a matrix that has a null row is not invertible
• fluidistic
In summary, to demonstrate that a matrix with a null row is not invertible, we can show that it cannot be equivalent to the identity matrix. This is because the product of the matrix and its inverse should result in the identity matrix, but a null row would result in a null column, proving that the matrix does not have an inverse. This is a valid proof, but its suitability may depend on the expectations of the teacher.
fluidistic
Gold Member
## Homework Statement
Demonstrate that a matrix that has a null row is not invertible.
2. The attempt at a solution I know that saying that a matrix A is invertible is equivalent to say that A is row-equivalent and column-equivalent to the identity matrix.
And also that there exist elemental matrices E_1, E_2,...,E_r and F_1,...,F_s such that the identity matrix is equal to E_1...E_r A F_1...F_s.
So I could show that if A has at least a null row, then one of these properties cannot be true.
But still I don't know how to proceed. I understand clearly that if A has a null row, it's obvious that it cannot be equivalent to the identity matrix... but to prove it formally, I don't know how I can start. A little help is welcome.
Suppose it did have an inverse. What does its product with its inverse supposed to look like? How does it actually look?
Think of A as the coefficient matrix such that Ax= b and let xn be the component of x corresponding to the 0 row. If the corresponding component of b is not 0, there is NO value of xn that makes that true. If the corresponding component of b is 0, then any value of xn makes that true. In either case there is NOT single solution to Ax= b. If A-1 exists, then A-1b would be 'the' solution to Ax= b.
Hurkyl said:
Suppose it did have an inverse. What does its product with its inverse supposed to look like? How does it actually look?
It is supposed to look like the identity matrix, however if the matrix has a null row it means that the product of this matrix with its inverse has a null column which means it's not the identity matrix, so that in fact any matrix having a null row doesn't have an inverse, hence is not invertible.
Is it a good proof? Or too informal?
fluidistic said:
Is it a good proof? Or too informal?
I can't answer that one; it would depend on what your teacher expects.
Hurkyl said:
I can't answer that one; it would depend on what your teacher expects.
Ok I understand. I hope it's enough.
And thank you very much HallsOfIvy, I got what you mean... nice way to do the proof.
## 1. Why is it important to demonstrate that a matrix with a null row is not invertible?
Demonstrating that a matrix with a null row is not invertible is important because it helps us determine the properties and behavior of matrices. Invertible matrices have unique solutions and can be used to solve systems of linear equations. Understanding when a matrix is not invertible helps us avoid errors and find alternative methods for solving problems.
## 2. What does it mean for a matrix to have a null row?
A null row in a matrix means that all the elements in that row are equal to zero. This can also be referred to as a row of zeros.
## 3. How can I tell if a matrix has a null row?
To determine if a matrix has a null row, you can simply scan each row and check if all the elements are equal to zero. Another way is to use Gaussian elimination to reduce the matrix to row-echelon form, where a null row would be represented by a row of zeros.
## 4. Can a matrix with a null row ever be invertible?
No, a matrix with a null row can never be invertible. In order for a matrix to be invertible, it must have a non-zero determinant. A determinant of zero indicates that the matrix lacks unique solutions and is not invertible.
## 5. What are some real-world applications that demonstrate the importance of understanding invertible matrices?
Invertible matrices are essential in fields such as engineering, physics, and economics. In engineering, invertible matrices are used to solve systems of equations in structural analysis and fluid dynamics. In physics, they are used in calculations related to electric circuits and quantum mechanics. In economics, invertible matrices are used to model and analyze economic systems.
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1K | 1,155 | 4,890 | {"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-2024-30 | latest | en | 0.968707 |
http://www.definitions.net/definition/golden%20ratio | 1,511,213,274,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806225.78/warc/CC-MAIN-20171120203833-20171120223833-00613.warc.gz | 377,627,097 | 11,520 | ## Definitions for golden ratio
### Wiktionary(0.00 / 0 votes)Rate this definition:
1. golden ratio(Noun)
The irrational number (approximately 1u00B7618), usually denoted by the Greek letter u03C6 (phi), which is equal the sum of its own reciprocal and 1, or, equivalently, is such that the ratio of 1 to the number is equal to the ratio of its reciprocal to 1.
### Freebase(0.00 / 0 votes)Rate this definition:
1. Golden ratio
In mathematics and the arts, two quantities are in the golden ratio if the ratio of the sum of the quantities to the larger quantity is equal to the ratio of the larger quantity to the smaller one. The figure on the right illustrates the geometric relationship. Expressed algebraically: where the Greek letter phi represents the golden ratio. Its value is: The golden ratio is also called the golden section or golden mean. Other names include extreme and mean ratio, medial section, divine proportion, divine section, golden proportion, golden cut, and golden number. Many 20th century artists and architects have proportioned their works to approximate the golden ratio—especially in the form of the golden rectangle, in which the ratio of the longer side to the shorter is the golden ratio—believing this proportion to be aesthetically pleasing. Mathematicians since Euclid have studied the properties of the golden ratio, including its appearance in the dimensions of a regular pentagon and in a golden rectangle, which can be cut into a square and a smaller rectangle with the same aspect ratio. The golden ratio has also been used to analyze the proportions of natural objects as well as man-made systems such as financial markets, in some cases based on dubious fits to data.
### Suggested Resources(0.00 / 0 votes)Rate this definition:
1. golden ratio
The golden ratio symbol -- In this Symbols.com article you will learn about the meaning of the golden ratio symbol and its characteristic.
### Numerology
1. Chaldean Numerology
The numerical value of golden ratio in Chaldean Numerology is: 6
2. Pythagorean Numerology
The numerical value of golden ratio in Pythagorean Numerology is: 3
# Translations for golden ratio
### From our Multilingual Translation Dictionary
• النسبة الذهبية
• Goldener Schnitt
• nombre d'or
• Золотое сечение
# Translation
#### Find a translation for the golden ratio definition in other languages:
Select another language:
# Citation
## Use the citation below to add this definition to your bibliography:
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# Rolanda: The house on Oak Avenue has a larger yard than any other
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Rolanda: The house on Oak Avenue has a larger yard than any other [#permalink]
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28 Nov 2019, 10:42
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Rolanda: The house on Oak Avenue has a larger yard than any other house we’ve looked at in Prairieview, so that’s the best one to rent.
Tom: No, it isn’t. Its yard isn’t really as big as it looks. Property lines in Prairieview actually start 20 feet from the street. So what looks like part of the yard is really city property.
Rolanda: But that’s true of all the other properties we’ve looked at too!
Rolanda’s response to Tom suggests that Tom commits which one of the following reasoning errors?
(A) He fails to take into account the possibility that there are advantages to having a small yard.
(B) He presumes, without providing justification, that property that belongs to the city is available for private use.
(C) He improperly applies a generalization to an instance that it was not intended to cover.
(D) He fails to apply a general rule to all relevant instances.
(E) He presumes, without providing justification, that whatever is true of a part of a thing is also true of the whole.
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Re: Rolanda: The house on Oak Avenue has a larger yard than any other [#permalink]
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28 Nov 2019, 22:55
Rolanda: The house on Oak Avenue has a larger yard than any other house we’ve looked at in Prairieview, so that’s the best one to rent.
Tom: No, it isn’t. Its yard isn’t really as big as it looks. Property lines in Prairieview actually start 20 feet from the street. So what looks like part of the yard is really city property.
Rolanda: But that’s true of all the other properties we’ve looked at too!
Rolanda’s response to Tom suggests that Tom commits which one of the following reasoning errors?
(A) He fails to take into account the possibility that there are advantages to having a small yard. --> No, this has no where been discussed in the conversation
(B) He presumes, without providing justification, that property that belongs to the city is available for private use. --> No he doesnt presume anything, the guy was just stating a fact to refute Rolanda's perception of a yard
(C) He improperly applies a generalization to an instance that it was not intended to cover. --> The generalization very well was applicable to the situation.
(D) He fails to apply a general rule to all relevant instances. --> Lets keep it, sounds just right
(E) He presumes, without providing justification, that whatever is true of a part of a thing is also true of the whole. --> He actually failed in that, so lets eliminate this choice.
IMO It should be (D)
Re: Rolanda: The house on Oak Avenue has a larger yard than any other [#permalink] 28 Nov 2019, 22:55
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# Rolanda: The house on Oak Avenue has a larger yard than any other
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne | 990 | 4,215 | {"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} | 3.25 | 3 | CC-MAIN-2019-51 | longest | en | 0.927958 |
https://www.physicsforums.com/threads/magnetic-field-inside-the-cavity-of-a-long-straight-wire.278045/ | 1,511,531,405,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808133.70/warc/CC-MAIN-20171124123222-20171124143222-00214.warc.gz | 854,519,546 | 15,926 | # Magnetic field inside the cavity of a long straight wire
1. Dec 8, 2008
### Irrelativity
1. The problem statement, all variables and given/known data
A long straight wire of radius R carrying a current I has a circular portion of radius a cut out at a distance d from the center, as shown in the picture. Find the magnetic field inside the cut-out portion.
2. Relevant equations
magnetic field equation (I can't write it here)
3. The attempt at a solution
I couldn't integrate the magnetic field directly. So I tried many tricks to solve this problem. I tried to reduce this into two different straight wires and worked out geometry but couldn't figure it out. If there is no cavity, it's very simple problem. But since I need to find the magnetic field inside the cavity, I need something else.
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2. Dec 8, 2008
### marcusl
Start by expressing the field a distance r from the axis of a wire carrying a uniform current density J. You can do this by applying Ampere's law to a circle of radius r. (I assume from your comments that you've done this already.)
You can make a cavity by superposing a second wire of radius a carrying -J, offset by d. Solve for the field at that point r and you should have your answer. You'll find that the field in the hole has a special property.
3. Dec 8, 2008
### Irrelativity
Do I have to assume that the current is steady or varies within the wire? The problem did not state that the current is steady (looks like it really doesn't matter but just in case). And when you find the magnetic field inside the cavity, you do have to take the vector components into the account? thanks.
Last edited: Dec 8, 2008
4. Dec 8, 2008
### marcusl
Your question is worded badly. Do you mean steady as in non-time-varying? Doesn't matter. Spatially uniform? Well, you tell me.
Vector components? Of course they matter.
5. Dec 8, 2008
### Irrelativity
What about variable r? is r from the center of the wire or from the center of the cavity? right now I am working on r from the center of the cavity. | 506 | 2,067 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-47 | longest | en | 0.949905 |
https://wiki.ubc.ca/Science:Math_Exam_Resources/Courses/MATH110/December_2010/Question_09_(f) | 1,717,031,586,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059412.27/warc/CC-MAIN-20240529230852-20240530020852-00296.warc.gz | 523,086,823 | 10,603 | # Science:Math Exam Resources/Courses/MATH110/December 2010/Question 09 (f)
MATH110 December 2010
Other MATH110 Exams
### Question 09 (f)
Let ${\displaystyle f(x)={\frac {x+1}{x-1}}}$.
Sketch the graph of ${\displaystyle \displaystyle f}$.
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result. | 343 | 1,449 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 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} | 2.578125 | 3 | CC-MAIN-2024-22 | latest | en | 0.960599 |
https://math.stackexchange.com/questions/3056199/what-to-use-for-lim-n-to-infty-int-0-1-fracn-sinx1n2-sqrtxdx | 1,723,445,969,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036271.72/warc/CC-MAIN-20240812061749-20240812091749-00083.warc.gz | 294,572,409 | 37,515 | # What to use for $\lim_{n\to \infty}\int_{[0,1]}\frac{n\sin{x}}{1+n^2\sqrt{x}}dx$
Determine $$\lim_{n\to \infty}\int_{[0,1]}\frac{n\sin{x}}{1+n^2\sqrt{x}}dx$$
I know the three convergence theorems, but to no avail:
$$1.$$ Monotone Convergence:
The series $$f_{n}(x):=\frac{n\sin{x}}{1+n^2\sqrt{x}}$$ is not monotonic increasing on $$[0,1]$$, so the conditions are not met.
$$2.$$ Fatou:
Note: $$\liminf_{n\to \infty}\int_{[0,1]}\frac{n\sin{x}}{1+n^2\sqrt{x}}dx\geq\int_{[0,1]}\liminf_{n\to \infty}\frac{n\sin{x}}{1+n^2\sqrt{x}}dx=0$$
Which aids us no further.
$$3.$$ Dominated Convergence Theorem:
The only function $$h$$ to fulfill $$|f_{n}|\leq h, \forall n \in \mathbb N$$ that comes to mind is $$|\frac{n\sin{x}}{1+n^2\sqrt{x}}|=\frac{n\sin{x}}{n^2\sqrt{x}}=\frac{\sin{x}}{n\sqrt{x}}\leq \frac{\sin{x}}{\sqrt{x}}=:h(x).$$
But how do I show $$h$$ is $$\in \mathcal{L}^1$$?
Is there anything I am missing? Any guidance is greatly appreciated.
Use dominated convergence with $$h(x)=\frac{\sin(x)}{x}$$ for $$x\in(0,1]$$ and $$h(0)=1$$.
You’re almost there... the map $$h : x \mapsto \frac{\sin x}{\sqrt x}$$ has $$0$$ for limit as $$x \to 0$$. Hence can be extended by continuity on $$[0,1]$$ and is integrable on that interval. | 485 | 1,242 | {"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": 19, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2024-33 | latest | en | 0.585542 |
https://intl.siyavula.com/read/maths/grade-8/algebraic-expressions/06-algebraic-expressions | 1,656,642,184,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103917192.48/warc/CC-MAIN-20220701004112-20220701034112-00233.warc.gz | 356,428,928 | 25,591 | An algebraic expression is a description of certain calculations that have to be done in a certain order. In this chapter, you will be introduced to the language of algebra. You will also learn about expressions that appear to be different but that produce the same results when evaluated. When we evaluate an expression, we choose or are given a value of the variable in the expression. Because now we have an actual value, we can carry out the operations ($$+, -, \times, \div)$$ in the expression using this value.
## Algebraic language
### Words, diagrams and symbols
1. Complete this table
Words Flow diagram Expression Multiply a number by two and add six to the answer. $$2 \times x + 6$$ (a) Add three to a number and then multiply the answer by two. (b) (c) $$7 + 4 \times x$$ (d) $$10 - 5 \times x$$
An algebraic expression indicates a sequence of calculations that can also be described in words or with a flow diagram.
The flow diagram illustarte the order in which the calculations must be done
In algebraic language the multipliaction sign is usually ommited. So we write $$2x$$ instead of $$2 \times x$$
We also write $$x \times 2 \text{ as } 2x$$
1. Write the following expressions in 'normal' algebraic language:
1. $$2 \times a + b$$
2. $$a2$$
### Looking different but yet the same
1. Complete the table by calculating the numerical values of the expressions for the values of $$x$$. Some answers for $$x$$ = 1 have been done for you as an example.
$$x$$ 1 3 7 10 (a) $$2x + 3x$$ $$2 \times 1 + 3 \times 1$$ $$2 + 3 = 5$$ (b) $$5x$$ (c) $$2x + 3$$ (d) $$5x^2$$ $$5 \times (1)^2$$ $$5 \times 1 = 5$$
2. Do the expressions $$2x + 3$$ and $$5x$$, in question 1 above, produce different answers or the same answer for:
1. $$x = 3$$?
2. $$x = 10$$?
3. Do the expressions $$2x + 3$$ and $$5x$$ produce different answers or the same answer for:
1. $$x = 3$$?
2. $$x = 10$$?
4. Write down all the algebraic expressions in question 1 that have the same numerical value for the same value(s) of $$x$$, although they may look different. Justify your answer.
One of the things we do in algebra is to evaluate expressions. When we evaluate an expression we choose or are given a value of the variable in the expression. Because now we have an actual value we can carry out the operations in the expression using this value, as in the examples given in the table.
Algeraic expressions that have the same numarical value for the same value of x but look different are called equivalent expressions.
1. Say whether the following statements are true or false. Explain your answer in each case.
1. The expressions $$2x + 3x$$ and $$5x$$ are equivalent.
2. The expressions $$2x + 3$$ and $$5x$$ are equivalent.
2. Consider the expressions $$3x + 2z + y$$ and $$6xyz$$.
Remember that $$6xyz$$ is the same as $$6 \times x \times y \times z$$.
1. What is the value of $$3x + 2z + y$$ for $$x = 4$$ , $$y = 7$$ and $$z = 10$$?
2. What is the value of $$6xyz$$ for $$x = 4$$ , $$y = 7$$ and $$z = 10$$?
3. Are the expressions $$3x + 2z + y$$ and $$6xyz$$ equivalent? Explain.
To show that the two expressions in question 5(a) are equivalent we write $$2x + 3x = 5x$$.
We can explain why this is so:
$$2x + 3x = (x + x) + (x + x + x) = 5x$$
We say the expression $$2x + 3x$$ simplifies to $$5x$$.
The term $$3x$$ is a product. The number 3 is called the coefficient of $$x$$.
1. In each case below, write down an expression equivalent to the one given.
1. $$3x + 3x$$
2. $$3x + 8x + 2x$$
3. $$8b + 2b + 2b$$
4. $$7m + 2m + 10m$$
5. $$3x^2 + 3x^2$$
6. $$3x^2 + 8x^2 + 2x^2$$
2. What is the coefficient of $$x^2$$ for the expression equivalent to $$3x^2 + 8x^2 + 2x^2$$?
In an expression that can be written as a sum, the different parts of the expression are called the terms of the expression. For example, $$3x,2z \text{ and } y$$ are the terms of the expression $$3x + 2z + y$$.
An expression can have like terms or unlike terms or both.
Like terms are terms that have the same variable(s) raised to the same power. The terms $$2x$$ and $$3x$$ are examples of like terms.
1. Calculate the numerical value of $$10x + 2y$$ for $$x = 3$$ and $$y = 2$$ by completing the empty spaces in the diagram.
2. What is the output value for the expression $$12xy$$ for $$x = 3 \text{ and } y = 2$$?
3. Are the expressions $$10x +2y$$ and $$12xy$$ equivalent? Explain.
4. Are the terms $$10x$$ and $$2y$$ like or unlike terms? Explain.
1. Which of the following algebraic expressions do you think will give the same results?
A. $$6x + 4x$$
B. $$10x$$
C. $$10x^2$$
D. $$9x + x$$
2. Test the algebraic expressions you have identified for the following values of $$x$$
$$x = 10$$
$$x = 17$$
$$x = 54$$
3. Are the terms $$6x$$ and $$4x$$ like or unlike terms? Explain.
4. Are the terms $$10x$$ and $$10x^2$$ like or unlike terms? Explain.
1. Ashraf and Hendrik have a disagreement about whether the terms $$7x^2y^3$$ and $$301y^3x^2$$ are like terms or not. Hendrik thinks they are not, because in the first term the $$x^2$$ comes before the $$y^3$$ whereas in the second term $$y^3$$ comes before the $$x^2$$.
Explain to Hendrik why his argument is not correct.
2. Explain why the terms $$5abc, 10acb \text{ and } 15cba$$ are like terms.
## Add and subtract like terms
### Rearrange terms and then combine like terms
1. Complete the table by evaluating the expressions for the given values of $$x$$.
$$x$$ 1 2 10 $$30x + 80$$ $$30 \times 1 + 80$$ $$= 30 + 80 = 110$$ $$5x + 20$$ $$30x + 80 + 5x + 20$$ $$35x + 100$$ $$135x$$
2. Write down all the expressions in the table that are equivalent.
3. Tim thinks that the expressions $$135x$$ and $$35x +100$$ are equivalent because for $$x = 1$$ they both have the same numerical value 135.
Explain to Tim why the two expressions are not equivalent.
We have already come across the commutative and associative properties of operations. We will now use these properties to help us form equivalent algebraic expressions.
Commutative property
The order in which we add or multiply numbers does not change the answer: $$a + b = b + a$$ and $$ab = ba$$
Associative property
The way in which we group three or more numbers when adding or multiplying does not change the answer: $$(a +b)+c = a + (b + c)$$ and $$(ab)c = a(bc)$$
We can find an equivalent expression by rearranging and combining like terms, as shown below:
$$30x + 80 + 5x + 20$$
Hence $$30x + (80 + 5x) + 20$$
Hence $$30x + (80 + 5x) + 20$$
$$= (30x + 5x) + (80 + 20)$$
$$= 35x + 100$$
The terms 80 and 20 are called constants. The numbers 30 and 5 are called coefficients.
Brackets are used in the expression above to show how the like terms have been rearranged.
Like terms are combined to form a single term.
The terms $$30x$$ and $$5x$$ are combined to get the new term $$35$$ and the terms 80 and 20 are combined to form the new term 100. We say that the expression $$30x + 80 + 5x + 20$$ is simplified to a new expression $$35x + 100$$.
1. Simplify the following expressions:
1. $$13x + 7 + 6x - 2$$
2. $$21x - 8 + 7x + 15$$
3. $$8c - 12d + 5c - 7c$$
4. $$3abc + 4 + 7abc - 6$$
5. $$12x^2 + 2x - 2x^2 + 8x$$
6. $$7m^3 + 7m^2 + 9m^3 + 1$$
When you are not sure about whether you correctly simplified an expression, it is always advisable to check your work by evaluating the original expression and the simplified expression for some values. This is a very useful habit to have.
When we use a value of the variable in the expression we call that substitution.
1. Make a simpler expression that is equivalent to the given expression. Test your answer for three different values of x, and redo your work until you get it right.
1. $$(15x + 7y) + (25x + 3 + 2y)$$
2. $$12mn + 8mn$$
In questions 6 to 8 below, write down the letter representing the correct answer. Also explain why you think your answer is correct.
1. The sum of $$5x^2 + x + 7$$ and $$x - 9$$ is:
A. $$x^2 -2$$
B. $$5x^2 + 2x +16$$
C. $$5x^2 + 16$$
D. $$5x^2 + 2x - 2$$
2. The sum of $$6x^2 - x + 4$$ and $$x^2 - 5$$ is equivalent to:
A. $$7x^2 - x + 9$$
B. $$7x^2 - x - 1$$
C. $$6x^4 - x - 9$$
D. $$7x^4 - x - 1$$
3. The sum of $$5x^2 + 2x + 4$$ and $$3x^2 - 5x - 1$$ can be expressed as:
A. $$8x^2 + 3x + 3$$
B. $$8x^2 + 3x - 3$$
C. $$8x^2 - 3x + 3$$
D. $$8x2 - 3x - 3$$
Combining like terms is a useful algebraic habit. It allows us to replace an expression with another expression that may be convenient to work with.
Do the following questions to get a sense of what we are talking about.
### Convenient replacements
1. Consider the expression $$x + x + x + x + x + x + x + x + x + x$$. What is the value of the expression in each of the following cases?
1. $$x = 2$$
2. $$x = 50$$
2. Consider the $$x + x + x + z + z + y$$. What is the value of the expression in each of the following cases?
1. $$x = 4, y = 7, z = 10$$
2. $$x = 0, y = 8, z = 22$$
3. Suppose you have to evaluate $$3x + 7x$$ for $$x = 20$$. Will calculating $$10 \times 20$$ give the correct answer? Explain.
Suppose we evaluate the expression $$3x + 7x$$ for $$x = 20$$ without first combining the like terms. We will have to do three calculations, namely $$3 \times 20$$, then $$7 \times 20$$ and then find the sum of the two: $$3 \times 20 + 7 \times 20 = 60 + 140 = 200$$.
But if we first combine the like terms $$3x \text{ and } 7x$$ into one term $$10x$$, we only have to do one calculation: $$10 \times 20 = 200$$. This is one way of thinking about the convenience or usefulness of simplifying an algebraic expression.
1. The expression $$5x + 3x$$ is given and you are required to evaluate it for $$x = 8$$. Will calculating $$8 \times 8$$ give the correct answer? Explain.
2. Suppose you have to evaluate $$7x + 5$$ for $$x = 10$$. Will calculating $$12 \times 10$$ give the correct answer? Explain.
3. The expression $$5x + 3$$ is given and you have to evaluate it for $$x = 8$$. Will calculating $$8 \times 8$$ give the correct answer? Explain.
Samantha was asked to evaluate the expression $$12x^2 + 2x - 2x^2 + 8x$$ for $$x = 12$$. She thought to herself that just substituting the value of x directly into the terms would require a lot of work. She first combined the like terms as shown below:
$$12x^2 {\bf- 2x^2 + 2x} +8x \\ 10x^2 + 10x$$
The terms $$+2x$$ and $$-2x^2$$ change positions by the commutative property of operations.
Then for $$x =10$$, Samantha found the value of $$10x^2 + 10x$$ by calculating
$$10 \times 10^2 + 10 \times 10 \\ = 1000 + 100 \\ = 1100$$
Use Samatha's way of thinking for questions 7 to 9.
1. What is the value of $$12x + 25x + 75x + 8x$$ when $$x =6$$
2. Evaluate $$3x^2 + 7 + 2x^2 + 3$$ for $$x = 5$$.
3. When Zama was asked to evaluate the expression $$2n - 1 + 6n$$ for $$n = 4$$, she wrote down the following:
\begin{align} 2n - 1 + 6n &= n + 6n = 6n^2 \\ \text{Hence for } n &= 4: 6 \times (4)^2 = 6 \times 8 = 48 \end{align}
Explain where Zama went wrong and why.
1. Complete the table.
Words Flow diagram Expression (a) Multiply a number by three and add two to the answer. (b) $$9x-6$$ (c) $$7x-3$$
2. Which of the following pairs consist of like terms? Explain.
A.$$3y; -7y$$
B. $$14e^2; 5e$$
C $$3y^2z; 17y^2z$$
D. $$-bcd; 5bd$$
3. Write the following in the 'normal' algebraic way:
1. $$c2 + d3$$
2. $$7 \times d \times e \times f$$
4. Consider the expression $$12x^2 - 5x + 3$$.
1. What is the number 12 called?
2. Write down the coefficient of $$x$$.
3. What name is given to the number 3?
5. Explain why the terms $$5pqr$$ and $$-10prq$$ and $$15qrp$$ are like terms.
6. If $$y =7$$, what is the value of each of the following?
1. $$y+8$$
2. $$9y$$
3. $$7 -y$$
7. Simplify the following expressions:
1. $$18c + 12d + 5c - 7c$$
2. $$3def + 4 + 7def - 6$$
8. Evaluate the following expressions for $$y = 3; z =-1$$:
1. $$2y^2 + 3z$$
2. ($$2y)^2 + 3z$$
9. Write each algebraic expression in the simplest form.
1. $$5y + 15y$$
2. $$5c + 6c - 3c + 2c$$
3. $$4b + 3 + 16b - 5$$
4. $$7m + 3n + 2 - 6m$$
5. $$5h^2 + 17 - 2h^2 + 3$$
6. $$7e^2f + 3ef + 2 + 4ef$$
10. Evaluate the following expressions:
1. $$3y+ 3y + 3y + 3y + 3y + 3y$$ for $$y = 18$$
2. $$13y + 14 - 3y + 6$$ for $$y = 200$$
3. $$20 - y^2 + 101y^2 + 80$$ for $$y = 1$$
4. $$12y^2 + 3yz + 18y^2 + 2yz$$ for (\9y=3\) and $$z = 2$$
# Algebraic equations
In this chapter you will learn to find numbers that make certain statements true. A statement about an unknown number is called an equation. When we work to find out which number will make the equation true, we say we solve the equation. The number that makes the equation true is called the solution of the equation.
## Setting up equations
An equation is a mathematical sentence that is true for some numbers but false for other numbers. The following are examples of equations:
$x + 3 = 11 \text{ and } 2^x = 8$
$$x + 3 = 11$$ is true if $$x =8$$, but false if $$x =3$$.
When we look for a number or numbers that make an equation true we say that we are solving the equation. For example , $$x = 4$$ is the solution of $$2x + 8$$ because it makes $$2x +8$$true. (Check: $$2 \times 4 = 8$$
### Looking for numbers to make statements true
1. $$x - 3 = 0$$, if $$x = -3$$
2. $$x^3 = 8$$, if $$x = -2$$
3. $$3x = -6$$, if $$x = -3$$
4. $$3x = 1$$, if $$x = 1$$
5. $$6x + 5 = 47$$, if $$x = 7$$
2. Find the original number. Show your reasoning.
1. A number multiplied by 10 is 80.
3. Divide a number by 5 and the answer is 4.
4. Multiply a number by 4 and the answer is 20.
5. Twice a number is 100.
6. A number raised to the power 5 is 32.
7. A number raised to the power 4 is -81.
8. Fifteen times a number is 90.
9. 93 added to a number is -3.
10. Half a number is 15.
3. Write the equations below in words using "a number" in place of the letter symbol $$x$$. Then write what you think "the number" is in each case.
Example: $$4 +x = 23$$. Four plus a number equals twenty-three. The number is 19.
1. $$8x = 72$$
2. $$\frac{2x}{5} = 2$$
3. $$2x+ 5 = 21$$
4. $$12 + 9x = 30$$
5. $$30 - 2x = 40$$
6. $$5x + 4 = 3x+ 10$$
## Solving equations by inspection
### The answer is in plain sight
1. Seven equations are given below the table. Use the table to find out for which of the given values of $$x$$ it will be true that the left-hand side of the equation is equal to the right-hand side.
You can read the solutions of an equation from a table.
$$x$$ -3 -2 -1 0 1 2 3 4 $$2x + 3$$ -3 -1 1 3 5 7 9 11 $$x + 4$$ 1 2 3 4 5 6 7 8 $$9 -x$$ 12 11 10 9 8 7 6 5 $$3x -2$$ -11 -8 -5 -2 1 4 7 10 $$10x -7$$ -37 -27 -17 -7 3 13 23 33 $$5x +3$$ -12 -7 -2 3 8 13 18 23 $$10 - 3x$$ 19 16 13 10 7 4 1 -2
1. $$2x + 3 = 5x + 3$$
2. $$5x + 3 = 9 - x$$
3. $$2x + 3 = x+4$$
4. $$10x -7 = 5x + 3$$
5. $$3x -2 = x+4$$
6. $$9 -x = 2x + 3$$
7. $$10 -3x = 3x - 2$$
Two or more equations can have the same solution. For example, $$5x = 10$$ and $$x + 2 = 4$$ have the same solution; $$x = 2$$ is the solution for both equations.
Two equations are called equivalent if they have the same solution.
1. Which of the equations in question 1 have the same solutions? Explain.
2. Complete the table below. Then answer the questions that follow.
You can also do a search by narrowing down the possible solution to an equation.
$$x$$ 0 5 10 15 20 25 30 35 40 $$2x + 3$$ $$3x -10$$
1. Can you find a solution for $$2x + 3 = 3x - 10$$ in the table?
2. What happens to the values of $$2x + 3$$ and $$3x - 10$$ as $$x$$ increases? Do they become bigger or smaller?
3. Is there a point where the value of $$3x - 10$$ becomes bigger or smaller than the value of $$2x + 3$$ as the value of $$x$$ increases? If so, between which $$x$$-values does this happen?
This point where the two expressions are equal is called the break-even point.
4. Now that you narrowed down where the possible solution can be, try other possible values for $$x$$ until you find out for what value of $$x$$ the statement $$2x + 3 = 3x - 10$$ is true.
"Searching" for the solution of an equation by using tables or by narrowing down to the possible solution is called solution by inspection.
## More examples
### Looking for and checking solutions
1. What is the solution for the equations below?
1. $$x - 3 = 4$$
2. $$x + 2 = 9$$
3. $$3x = 21$$
4. $$3x + 1 = 22$$
When a certain number is the solution of an equation we say that the number satisfies the equation. For example, $$x=4$$ satisfies the equation $$3x=12$$ because $$3 \times 4 = 12$$.
1. Choose the number in brackets that satisfies the equation. Explain your choice.
1. $$12x = 84$$ {5; 7; 10; 12}
2. $$\frac{84}{x} = 12$$ {-7; 0; 7; 10}
3. $$48 = 8k + 8$$ {-5; 0; 5; 10}
4. $$19 - 8m = 3$$ {-2; -1; 0; 1; 2}
5. $$20 = 6y - 4$$ {3; 4; 5; 6}
6. $$x^3 = -64$$ {-8; -4; 4; 8}
7. $$5^x =125$$ {-3; -1; 1; 3}
8. $$2^x = 8$$ {1; 2; 3; 4}
9. $$x^2 = 9$$ {1; 2; 3; 4}
1. $$m + 8 = 100$$
2. $$80 = x + 60$$
3. $$26 - k = 0$$
4. $$105 \times y = 0$$
5. $$k \times 10 = 10$$
6. $$5x = 100$$
7. $$\frac{15}{t} = 5$$
8. $$3 = \frac{t}{5}$$
1. $$12x + 14 = 50$$
2. $$100 = 15m + 25$$
3. $$\frac{100}{x} =20$$
4. $$7m + 5 = 40$$
5. $$2x + 8 = 10$$
6. $$3x + 10 = 31$$
7. $$-1 + 2x = -11$$
8. $$2 + \frac{x}{7} = 5$$
9. $$100 = 64 + 9x$$
10. $$\frac{2x}{6}= 4$$
# Term revision and assessment
## Revision
Show all your steps of working.
### Whole numbers
1. Write both 300 and 160 as products of prime factors.
2. Determine the HCF and LCM of 300 and 160.
1. Tommy, Thami and Timmy are given birthday money by their grandmother in the ratio of their ages. They are turning 11, 13 and 16 years old, respectively. If the total amount of money given to all three boys is R1 000, how much money does Thami get on his birthday?
2. Tshepo and his family are driving to the coast on holiday. The distance is 1 200 km and they must reach their destination in 12 hours. After 5 hours, they have travelled 575 km. Then one of their tyres bursts. It takes 45 minutes to get the spare wheel on, before they can drive again. At what average speed must they drive the remainder of the journey to reach their destination on time?
3. The number of teachers at a school has increased in the ratio 5 : 6. If there used to be 25 teachers at the school, how many teachers are there now?
4. ABC for Life needs to have their annual statements audited. They are quoted R8 500 + 14% VAT by Audits Inc. How much will ABC for Life have to pay Audits Inc. in total?
5. Reshmi invests R35 000 for three years at an interest rate of 8,2% per annum. Determine how much money will be in her account at the end of the investment period.
6. Lesebo wants to buy a lounge suite that costs R7 999 cash. He does not have enough money and so decides to buy it on hire purchase. The store requires a 15% deposit up front, and 18 monthly instalments of R445.
1. Calculate the deposit that Lesebo must pay.
2. How much extra does Lesebo pay because he buys the lounge suite on hire purchase, rather than in cash?
7. Consider the following exchange rates table:
South African Rand 1.00 ZAR inv. 1.00 ZAR Euro 0.075370 13.267807 US Dollar 0.098243 10.178807 British Pound 0.064602 15.479409 Indian Rupee 5.558584 0.179902 Australian Dollar 0.102281 9.776984 Canadian Dollar 0.101583 9.844200 Emirati Dirham 0.360838 2.771327 Swiss Franc 0.093651 10.677960 Chinese Yuan Renminbi 0.603065 1.658195 Malaysian Ringgit 0.303523 3.294646
1. Write down the amount in rand that needs to be exchanged to get 1 Swiss franc. Give your answer to the nearest cent.
2. Write down the only currency for which an exchange of R100 will give you more than 100 units of that currency.
3. Ntsako is travelling to Dubai and converts R10 000 into Emirati dirhams. How many dirhams does Ntsako receive (assume no commission)?
### Integers
Don't use a calculator for any of the questions in this section.
1. Write a number in each box to make the calculations correct.
1. ☐ + ☐ = -11
2. ☐ - ☐ = -11
2. Fill <, > or = into each block to show the relationships.
1. -23 ☐ 20
2. -345 ☐ -350
3. $$4 - 3$$ ☐ $$3 - 4$$
4. $$5 - 7$$ ☐ $$-(7 - 5)$$
5. $$-9 \times 2$$ ☐ $$-9 \times -2$$
6. $$-4 \times 5$$ ☐ $$4 \times -5$$
7. $$-10 \div 5 ☐ -10 \div -2$$
8. $$-15 \times -15$$ ☐ 224
3. Follow the pattern to complete the number sequences.
1. 8; 5; 2;☐
2. 2; -4; 8;☐
3. -289; -293; -297;☐
4. Look at the number lines. In each case, the missing number is halfway between the other two numbers. Fill in the correct values in the boxes.
5. Calculate the following:
1. $$-5 - 7$$
2. $$7 - 10$$
3. $$8-(-9)$$
4. $$(-5)(-2)(-4)$$
5. $$5 + 4 \times 2$$
6. $$\frac{(\sqrt{4})(-2)^2}{-4}$$
7. $$\frac{-(-3)^3\sqrt[3]{125}}{(-9)(3)}$$
8. $$\frac{\sqrt[3]{-64}}{-3-1}$$
1. Write down two numbers that multiply to give -15. (One of the numbers must be positive and the other negative.)
2. Write down two numbers that add to 15. One of the numbers must be positive and the other negative.
6. At 5 a.m., the temperature in Kimberley was -3 °C. At 1 p.m., it was 17 °C. By how many degrees had the temperature risen?
7. A submarine is 220 m below the surface of the sea. It travels 75 m upwards. How far below the surface is it now?
### Exponents
You should not use a calculator for any of the questions in this section.
1. Write down the value of the following:
1. $$(-3)^3$$
2. $$-5^2$$
3. $$(-1)^{200}$$
4. $$(10^2)^2$$
2. Write the following numbers in scientific notation:
1. 200 000
2. 12,345
3. Write the following numbers in ordinary notation:
1. $$1,3 \times 10^2$$
2. $$7,01 \times 10^7$$
4. Which of the following numbers is bigger: $$5,23 \times 10^{10}$$ or $$2 ,9 \times 10^{11}$$?
5. Simplify the following:
1. $$2^7 \times 2^3$$
2. $$2x^3 \times 4x^4$$
3. $$(-8y^6) \div (4y^3)$$
4. $$(3x^8)^3$$
5. $$(2x^5)(0.5x^{-5})$$
6. $$(-3a^2b^3c)(-4abc^2)^2$$
7. $$\frac{(2xy^2z^3)(-5y^2z)^2}{20xy^8z^4}$$
6. Write down the values of each of the following:
1. $$(0,6)^2$$
2. $$(0,2)^3$$
3. $$(\frac{1}{2})^5$$
4. $$\sqrt{\frac{1}{4}}$$
5. $$4\sqrt{\frac{9}{64}}$$
6. $$\sqrt[3]{0.001}$$
### Numeric and geometric patterns
1. For each of the following sequences, write the rule for the relationship between each term and the following term in words. Then use the rule to write the next three terms in the sequence.
1. $$12; 7; 2;\text{______};\text{______};\text{______}$$
2. $$-2; -6; -18;\text{______};\text{______};\text{______}$$
3. $$100; -50; 25;\text{______};\text{______};\text{______}$$
4. $$3; 4; 7; 11;\text{______};\text{______};\text{______}$$
2. In this question, you are given the rule by which each term of the sequence can be found. In all cases, $$n$$ is the position of the term.
Determine the first three terms of each of the sequences. (Hint: Substitute $$n = 1$$ to find the value of the first term.)
1. $$n \times 4$$
2. $$n\times 5 - 12$$
3. $$2 \times n^2$$
4. $$3n \div 3 \times -2$$
3. Write down the rule by which each term of the sequence can be found (in a similar format to those given in question 2, where n is the position of the term).
1. $$2; 4; 6; ...$$
2. $$-7; -3; 1; ...$$
3. $$2; 4; 8; ...$$
4. $$9; 16; 23; ...$$
4. Use the rules you have found in question 3 to find the value of the 20th term of the sequences in questions 3(a) and 3(b).
5. Find the relationship between the position of the term in the sequence and the value of the term, and use it to fill in the missing values in the tables.
1. Position in sequence 1 2 3 4 25 Value of the term -8 -11 -14
2. Position in sequence 1 2 3 Value of the term 1 3 9 243 19 683
6. The image below shows a series of patterns created by matches.
1. Write in words the rule that describes the number of matches needed for each new pattern.
2. Use the rule to determine the missing values in the table below, and fill them in.
Number of the pattern 1 2 3 4 20 Number of matches needed 4 7 151
### Functions and relationships
1. Fill in the missing input values, output values or rule in these flow diagrams. Note that $$p$$ and $$t$$ are integers.
2. Consider the values in the following table. The rule for finding $$y$$ is: divide $$x$$ by -2 and subtract 4. Use the rule to determine the missing values in the table, and write them in.
x -2 0 2 5 y -1 -3 -4 48
3. Consider the values in the following table:
$$x$$ -2 -1 0 1 2 4 15 $$y$$ 1 3 5 7 9 61
1. Write in words the rule for finding the $$y$$-values in the table.
2. Use the rule to determine the missing values in the table, and write them in.
### Algebraic expressions
1. Look at this algebraic expression: $$5x^3 - 9 + 4x - 3x^2$$.
1. How many terms does this expression have?
2. What is the variable in this expression?
3. What is the coefficient of the $$x^2$$ term?
4. What is the constant in this expression?
5. Rewrite the expression so that the terms are in order of decreasing powers of $$x$$.
2. In this question, $$x=6$$ and $$y = 17$$. Complete the rules to show different ways to determine $$y$$ if $$x$$ is known. The first way is done for you:
Way 1: Multiply x by 2 and add 5. This can be written as $$y = 2x + 5$$
1. Way 2: Multiply $$x$$ by ______ and then subtract ______. This can be written as
2. Way 3: Divide $$x$$ by ______ and then add ______ . This can be written as
3. Way 4: Add ______ to $$x$$, and then multiply by ______ . This can be written as
3. Simplify:
1. $$2x^2 + 3x^2$$
2. $$9xy -12xy$$
3. $$3y^2 - 4y + 3y - 2y^2$$
4. $$9m^3 + 9m^2 + 9m^3 - 3$$
4. Calculate the value of the following expressions if $$a = -2; b = 3; c = -1 \text{ and } d = 0:$$
1. $$abc$$
2. $$-a^2$$
3. $$(abc)^d$$
4. $$a + b- 2c$$
5. $$(a + b)^{10}$$
### Algebraic equations
1. Write equations that represent the given information:
1. Nandi is $$x$$ years old. Shaba, who is $$y$$ years old, is three years older than Nandi.
2. The temperature at Colesberg during the day was $$x$$°C. But at night, the temperature dropped by 15 degrees to reach -2°C.
2. Solve the following equations for $$x$$:
1. $$x + 5 = 2$$
2. $$7 - x = 9$$
3. $$3x - 1 = -10$$
4. $$2x^3 = -16$$
5. $$2^x = 16$$
6. $$2(3)^x = 6$$
3. If $$3n - 1 = 11$$, what is the value of $$4n$$?
4. If $$c = a + b$$ and $$a + b + c = 16$$ , determine the value of $$c$$.
1. If $$2a + 3 = b$$, write down values for $$a$$ and $$a$$ that will make the equation true.
2. Write down a different pair of values to make the equation true.
## Assessment
In this section, the numbers in brackets at the end of a question indicate the number of marks the question is worth. Use this information to help you determine how much working is needed. The total number of marks allocated to the assessment is 60.
1. The profits of GetRich Inc. have decreased in the ratio 5 : 3 due to the recession in the country. If their profits used to be R1 250 000, how much are their profits now?
2. Which car has the better rate of petrol consumption: Ashley's car, which drove 520 km on 32â„“ of petrol, or Zaza's car, which drove 880 km on 55â„“ of petrol? Show all your working.
3. Hanyani took out a R25 000 loan from a lender that charges him 22% interest each year. How much will he owe in one year's time?
4. Consider the following exchange rates table:
South African Rand 1.00 ZAR inv. 1.00 ZAR Indian Rupee 5.558584 0.179902 Australian Dollar 0.102281 9.776984 Canadian Dollar 0.101583 9.844200 Emirati Dirham 0.360838 2.771327 Chinese Yuan Renminbi 0.603065 1.658195 Malaysian Ringgit 0.303523 3.294646
Chen returns from a business trip to Malaysia with 2 500 ringgit in his wallet. If he changes this money into rand in South Africa, how much will he receive?
5. Fill <, > or = into the block to show the relationship between the number expressions:
1. $$6 - 4$$ ☐ $$4 - 6$$
2. $$2 \times -3$$ ☐ $$-3^2$$
6. Look at the number sequence below. Fill in the next term into the block.
$$-5; 10; -20$$;☐
7. Calculate the following:
1. $$(-4)^2 - 20$$
2. $$\sqrt[3]{-8} + 14 \ \div 2$$
8. Julius Caesar was a Roman emperor who lived from 100 BC to 44 BC. How old was he when he died?
1. Write down two numbers that divide to give an answer of -8. One of the numbers must be positive, and the other negative.
2. Write down two numbers that subtract to give an answer of 8. One of the numbers must be positive and the other negative.
9. Write the following number in scientific notation: 17 million.
10. Which of the following numbers is bigger: $$3,47 \times 10^{21} \text{ or }7,99 \times 10^{20}$$?
11. Simplify the following, leaving all answers with positive exponents:
1. $$3^7 \times 3^-2$$
2. $$(-12y^8) \div (-3y^2)$$
3. $$\frac{(3xy^2z^3)(-yz)^2}{15x^5y^4z^7}$$
12. Write down the values of each of the following:
1. $$(0.3)^3$$
2. $$8 \sqrt{\frac{25}{16}}$$
13. Consider the following number sequence: $$2; -8; 32; ...$$
1. Write in words the rule by which each term of the sequence can be found.
2. Write the next three terms in this sequence.
14. The picture below shows a series of patterns created by matches.
1. Write a formula for the rule that describes the relationship between the number of matches and the position of the term in the sequence (pattern number). Let $$n$$ be the position of the term.
2. Use the rule to determine the values of $$a$$ to $$c$$ in the following table:
Number of the pattern 1 2 3 4 15 c Number of matches needed 8 15 22 a b 148
15. Consider the values in the following table:
x -2 -1 0 1 2 5 12 y -7 -4 -1 2 5 98
1. Write in words the rule for finding the $$y$$-values in the table.
2. Use the rule to determine the missing values in the table, and fill them in.
16. . Simplify:
1. $$2z^2 - 3z^2$$
2. $$8y^2 - 6y + 4y - 7y^2$$
17. Determine the value of $$2a^2 - 10$$ if $$a = -2$$.
18. If $$c + 2d = 27$$, give the value of the following:
1. $$2c + 4d$$
2. $$\frac{c + 2d}{-9}$$
3. $$\sqrt[3]{c + 2d}$$
19. Solve the following for $$x$$: (5)
1. $$-x - 11$$
2. $$2x -5 = - 11$$
3. $$4x^3 = 32$$ | 10,163 | 29,716 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2022-27 | latest | en | 0.79838 |
https://www.esaral.com/q/in-an-a-p-the-first-term-is-2-and-the-sum-of-the-first-five-terms-is-one-fourth-of-the-next-five-terms-show-that-20th-term-is-112-17993 | 1,725,953,474,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651224.59/warc/CC-MAIN-20240910061537-20240910091537-00693.warc.gz | 712,983,803 | 11,725 | # In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112.
Question:
In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that $20^{\text {th }}$ term is $-112$.
Solution:
First term = 2
Let d be the common difference of the A.P.
Therefore, the A.P. is 2, 2 + d, 2 + 2d, 2 + 3d, …
Sum of first five terms = 10 + 10d
Sum of next five terms = 10 + 35d
According to the given condition,
$10+10 d=\frac{1}{4}(10+35 d)$
$\Rightarrow 40+40 d=10+35 d$
$\Rightarrow 30=-5 d$
$\Rightarrow d=-6$
$\therefore a_{20}=a+(20-1) d=2+(19)(-6)=2-114=-112$
Thus, the $20^{\text {th }}$ term of the A.P. is $-112$. | 269 | 744 | {"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} | 4.5625 | 5 | CC-MAIN-2024-38 | latest | en | 0.876283 |
https://www.clutchprep.com/physics/diffraction | 1,680,034,720,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948871.42/warc/CC-MAIN-20230328201715-20230328231715-00256.warc.gz | 792,629,765 | 52,543 | Clutch Prep is now a part of Pearson
Ch 35: Wave OpticsWorksheetSee all chapters
# Diffraction
See all sections
Sections
Diffraction
Diffraction with Huygen's Principle
Young's Double Slit Experiment
Single Slit Diffraction
Concept #1: Diffraction
Transcript
Hey guys, in this video we're going to talk about a phenomenon called diffraction. Alright, let's get to it. Remember that light travels in a straight line and so long as it's not disturbed. We've seen one type of disruption of light before when we saw a light encountering a boundary between two media, and that light could reflect off of the boundary or it could transmit through and refract it's angle as it passed into the second medium. This allows light this fact that light travels as long as it's not disturbed in a straight line allows light to be described as rays so just to refresh ourselves, we can draw any wave as successive wavefronts. Each of these wavefronts drawn in the green is a point of maximum oscillation. In the case for an electromagnetic wave in the case for light, it's a maximum electric field and we can draw rays such that they're perpendicular to all the wave fronts at all points. So I can draw rays like this. And you can see clearly that it's perpendicular wherever you want to measure and the distance between wavefronts, the distance between two peaks, as we know is just the wavelength. That's the definition of the wavelength. Now a common way to disturb light that we haven't talked about is for a light to encounter a slit. And a slit is a small opening between two barriers of light alright I'm going to minimise myself. We have here just light travelling. I drew three hypothetical light waves, each of which is a different colour. Here I've indicated two boundaries and we're going to imagine that these boundaries, these barriers are completely reflective or not reflective at all but not transmissive. They completely block out any transmission of light. All the light that's allowed to transmit then, the only light that's allowed to transmit is the one that passes through the slit. So the green light is the only light on the other side. Now, depending on the size of the slit depending on the width of the slit, this dimension. The rays may or may not be disturbed. They don't have to be disturbed as they pass through, they may or may not be disturbed. Alright, what diffraction is is it's sort of a catchall term that refers to all phenomenon associated with light rays being spread apart when they encounter a slit. A slit between two barriers. Diffraction isn't going to occur for any slit. The slit's width I said the slit must be small but what I mean is the width must be small compared to the wavelength of the light. So diffraction will only occur if this dimension right here is small compared to this dimension which is the wavelength alright. Now let's see what diffraction looks like. Right here I have two scenarios alright, I have light of a particular wavelength encountering a slit of a particular width and I have shown what happens when the wave fronts pass through that slit. So let's draw the rays and see if diffraction occurs here. In order to be perpendicular at all points to the wavefronts, the rays before encountering the slit have to look like this. This by the way is referred to as collimated. This funky looking letter there's supposed to be an L. Collimated light. Light that is all initially parallel to itself, all the rays are parallel. Now the wavefronts I've shown passing through the slit. What do the rays looks like passing through the slit? Well they still need to be parallel to one another in order to be perpendicular at all points on the wavefront. So it's collimated before passing through the slit and collimated after. Those rays never spread apart. They're collimated entering, they're collimated exiting. That means that there was no diffraction here. But now choosing another hypothetical scenario, one where we have a larger wavelength and a significantly smaller hole. Now I want to consider the scenario where the length is smaller than the width sorry the width of the slit is smaller than the wavelength of the lights. If I'm going to draw the rays for this light, you can see that once again, it has to be collimated. That's the only way to match rays to those wavefronts. But the wavefronts look different coming out of the slit, now instead of them being parallel wavefronts they're actually wavefronts that are moving spherically outwards. So in order to draw the rays remember it has to be perpendicular to everything. This is perpendicular, perpendicular, perpendicular, but at a different angle. I need to draw the ray at a different angle right? So they point out equally in all directions. This is known as isotropic and it's absolutely not collimated. Isotropic just means the same in all directions. Since the light entered collimated and exited it isotropically, the light rays were disturbed, they did spread it out and this is known as diffraction. Now something interesting happens when light is allowed to diffract or when you allow for light diffracting. Light passing through a slit acts differently if you ignore diffraction. So in the left figure we're going to pretend like diffraction isn't a thing, meaning that if we look at these two figures up here really quickly, no matter the relative size of the width of the slit to the wavelength, the light that enters collimated will always leave collimated. That's what I mean by no diffraction. So what that means is, when the light is entering the slit collimated, it's all coming out collimated and you're going to get a single bright spot on some sort of screen behind the slit. That screen is just there to collect the light to allow the light to land on it so you can see but if you allow diffraction then so long as the width of the slit so this dimension that horizontal width so long as that width is less than the wavelength of light, what's going to happen is that initially collimated light is going to come out equal in all directions and it turns out that you don't get a continuous band of bright light, you actually get alternating bits of bright light and dark light. So dark, bright. And this alternating pattern of bright and dark spots of light is known as a diffraction pattern and it's unique to the particular diffraction situation that the light is in. This wraps up our introduction on diffraction. Thanks for watching guys. | 1,362 | 6,489 | {"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} | 3.296875 | 3 | CC-MAIN-2023-14 | latest | en | 0.962521 |
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# 885 SC's Mega BANK
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09 Jan 2005, 21:29
Can somebody please help me out with this one: Even when I am logged on, I CANNOT see the document.
http://www.gmatclub.com/phpbb/viewtopic.php?t=5210
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09 Jan 2005, 21:37
There is still some problem with the server transfer. No attachment shows as of now and you cannot post any attachment either. Hope it will be fixed soon.
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It will be a while before we can restore all the features.
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10 Jan 2005, 08:19
Approximately how much time ? Is there a way you can email me the 885SC Mega Bank, if I provide you my email or you advice that I wait for the server to ger OK ?
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Praetorian / Paul, any clue Approximately how much time ? Is there a way you can email me the 885SC Mega Bank, if I provide you my email or you advice that I wait for the server to get OK ?
Question to Praetorian / Paul [#permalink] 19 Jan 2005, 03:07
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 845 | 2,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-04 | latest | en | 0.824439 |
https://fssnip.net/kV/title/Convert-a-Float-to-a-Mixed-Number | 1,624,425,017,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488534413.81/warc/CC-MAIN-20210623042426-20210623072426-00520.warc.gz | 247,224,941 | 5,561 | 0 people like it.
# Convert a Float to a Mixed Number
Function to Convert a float to a mixed number. My thanks to John Kennedy of Santa Monica College for his "Algorithm to Convert a Decimal to a Fraction" (see link).
``` 1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: 13: 14: 15: 16: 17: 18: ``` ``````let ToMixedNumber(x : float) = let wholePart = int x // whole part of x let decimalPt = x % 1.0 // decimal part of x let rec cF(Z : float, i : int, Dm : float, Do : float) = match Z % 1.0 > 1e-6, i < 1 with // First case terminates after 14 iterations | _ , true -> (wholePart, (int64 (System.Math.Round(decimalPt * Do)), int64 Do)) // Second case executes next cF (continuing fraction) | true , false -> cF(1.0/(Z % 1.0), i - 1 , Do, Do * System.Math.Round(1.0/(Z % 1.0)-0.5) + Dm ) // Final case terminates if the remainder of Z > 10^-6 | false, _ -> (wholePart, (int64 (System.Math.Round(decimalPt * Do)), int64 Do)) decimalPt |> fun x -> cF(x, 14, 0.0, 1.0) // Test using pi as input let dummyrecA = ToMixedNumber System.Math.PI // Recompose pi as a decimal form. // NOTE Actual Value of pi = 3.1415926535897932384626433833M let myPiA = decimal (fst dummyrecA) + decimal ( fst (snd dummyrecA)) / decimal ( snd (snd dummyrecA)) ``````
val ToMixedNumber : x:float -> int * (int64 * int64)
Full name: Script.ToMixedNumber
val x : float
Multiple items
val float : value:'T -> float (requires member op_Explicit)
Full name: Microsoft.FSharp.Core.Operators.float
--------------------
type float = System.Double
Full name: Microsoft.FSharp.Core.float
--------------------
type float<'Measure> = float
Full name: Microsoft.FSharp.Core.float<_>
val wholePart : int
Multiple items
val int : value:'T -> int (requires member op_Explicit)
Full name: Microsoft.FSharp.Core.Operators.int
--------------------
type int = int32
Full name: Microsoft.FSharp.Core.int
--------------------
type int<'Measure> = int
Full name: Microsoft.FSharp.Core.int<_>
val decimalPt : float
val cF : (float * int * float * float -> int * (int64 * int64))
val Z : float
val i : int
val Dm : float
val Do : float
Multiple items
val int64 : value:'T -> int64 (requires member op_Explicit)
Full name: Microsoft.FSharp.Core.Operators.int64
--------------------
type int64 = System.Int64
Full name: Microsoft.FSharp.Core.int64
--------------------
type int64<'Measure> = int64
Full name: Microsoft.FSharp.Core.int64<_>
namespace System
type Math =
static val PI : float
static val E : float
static member Abs : value:sbyte -> sbyte + 6 overloads
static member Acos : d:float -> float
static member Asin : d:float -> float
static member Atan : d:float -> float
static member Atan2 : y:float * x:float -> float
static member BigMul : a:int * b:int -> int64
static member Ceiling : d:decimal -> decimal + 1 overload
static member Cos : d:float -> float
...
Full name: System.Math
System.Math.Round(d: decimal) : decimal
System.Math.Round(a: float) : float
System.Math.Round(d: decimal, mode: System.MidpointRounding) : decimal
System.Math.Round(d: decimal, decimals: int) : decimal
System.Math.Round(value: float, mode: System.MidpointRounding) : float
System.Math.Round(value: float, digits: int) : float
System.Math.Round(d: decimal, decimals: int, mode: System.MidpointRounding) : decimal
System.Math.Round(value: float, digits: int, mode: System.MidpointRounding) : float
val dummyrecA : int * (int64 * int64)
Full name: Script.dummyrecA
field System.Math.PI = 3.14159265359
val myPiA : decimal
Full name: Script.myPiA
Multiple items
val decimal : value:'T -> decimal (requires member op_Explicit)
Full name: Microsoft.FSharp.Core.Operators.decimal
--------------------
type decimal = System.Decimal
Full name: Microsoft.FSharp.Core.decimal
--------------------
type decimal<'Measure> = decimal
Full name: Microsoft.FSharp.Core.decimal<_>
val fst : tuple:('T1 * 'T2) -> 'T1
Full name: Microsoft.FSharp.Core.Operators.fst
val snd : tuple:('T1 * 'T2) -> 'T2
Full name: Microsoft.FSharp.Core.Operators.snd | 1,158 | 3,992 | {"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} | 2.9375 | 3 | CC-MAIN-2021-25 | latest | en | 0.418197 |
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Python | Split a list into sublists of given lengths
The problem of splitting a list into sublists is quite generic but to split in sublist of given length is not so common. Given a list of lists and list of length, the task is to split the list into sublists of given length. Example:
```Input : Input = [1, 2, 3, 4, 5, 6, 7]
length_to_split = [2, 1, 3, 1]
Output: [[1, 2], [3], [4, 5, 6], [7]]```
Method #1: Using islice to split a list into sublists of given length, is the most elegant way.
Python3
`# Python code to split a list ` `# into sublists of given length. ` `from` `itertools ``import` `islice ` `# Input list initialization ` `Input` `=` `[``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``] ` `# list of length in which we have to split ` `length_to_split ``=` `[``2``, ``1``, ``3``, ``1``] ` `# Using islice ` `Inputt ``=` `iter``(``Input``) ` `Output ``=` `[``list``(islice(Inputt, elem)) ` ` ``for` `elem ``in` `length_to_split] ` `# Printing Output ` `print``(``"Initial list is:"``, ``Input``) ` `print``(``"Split length list: "``, length_to_split) ` `print``(``"List after splitting"``, Output) `
Output:
```Initial list is: [1, 2, 3, 4, 5, 6, 7]
Split length list: [2, 1, 3, 1]
List after splitting [[1, 2], [3], [4, 5, 6], [7]]```
Time Complexity: O(n)
Auxiliary Space: O(1)
Method #2: Using zip is another way to split a list into sublists of given length.
Python3
`# Python code to split a list into ` `# sublists of given length. ` `from` `itertools ``import` `accumulate ` `# Input list initialization ` `Input` `=` `[``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``7``, ``8``, ``9``, ``10``] ` `# list of length in which we have to split ` `length_to_split ``=` `[``2``, ``2``, ``3``, ``3``] ` `# Using islice ` `Output ``=` `[``Input``[x ``-` `y: x] ``for` `x, y ``in` `zip``( ` ` ``accumulate(length_to_split), length_to_split)] ` `# Printing Output ` `print``(``"Initial list is:"``, ``Input``) ` `print``(``"Split length list: "``, length_to_split) ` `print``(``"List after splitting"``, Output) `
Output:
```Initial list is: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Split length list: [2, 2, 3, 3]
List after splitting [[1, 2], [3, 4], [5, 6, 7], [8, 9, 10]]```
Time Complexity: O(1)
Auxiliary Space: O(1)
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### Help with a proof of a theorem about convex sets
I'm studying the following theorem: Theorem 1. Let $C$ be a convex set and let $\textbf{y}$ be a point exterior to the closure of $C$. Then there is a vector $\textbf{a}$ such that ...
I got messed up with this problem and can't find any clue to solve this. Hope some one here can help me. Let $A$ be an $m \times n$ matrix an let $b$ be a vector in $R^{m}$. We consider the ... | 235 | 739 | {"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": 1, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2014-23 | longest | en | 0.883881 |
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# Depreciation of assets
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The attached file contains a worked example from a textbook. Explain it.
Neglecting income taxes and assuming an interest rate of 0%, calculations for determining the most economic replacement interval are as follows.
See Attached
##### Solution Summary
This explains an example problem regarding depreciation.
##### Solution Preview
This problem is greatly simplified by assuming an interest rate of 0%. All you need to do is think about the "average cost per year" if you kept the machine for each of the specified years. Since the time value of money is not considered with an interest rate, the calculations are simplified in the following way:
If you kept the machine 1 year, it ...
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This quiz helps you easily identify a function and test your understanding of ranges, domains , function inverses and transformations. | 304 | 1,601 | {"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} | 3.109375 | 3 | CC-MAIN-2024-10 | latest | en | 0.912783 |
https://www.physicsforums.com/threads/differential-equation-falling-object.524120/ | 1,603,978,876,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107904287.88/warc/CC-MAIN-20201029124628-20201029154628-00077.warc.gz | 846,312,390 | 17,706 | # Differential Equation: Falling Object
## Homework Statement
For small, slowly falling objects, the assumption made in equation (1) (see below) is good. For larger, more rapidly falling objects, it is more accurate to assume that the magnitude of the drag force is proportional to the square of the velocity with the force orientation opposite to that of velocity.
Write a differential equation for the velocity of a falling object of mass $m$ if the magnitude of the drag force is proportional to the square of the velocity
## Homework Equations
(1) $m \frac{dv}{dt} = -mg - \gamma v$
$\gamma > 0$
## The Attempt at a Solution
Since the drag force ($- \gamma v$) is proportional to the square of the velocity:
$| - \gamma v | = \gamma | v | \propto v^2$
and since the orientation is opposite to that of velocity
$\gamma | v | = -kv^2$ where $k$ is some constant.
but this is where I get stuck.
The answer in the back of the book has the drag force as $- \gamma v |v|$.
Where did I go wrong?
Thank you anyone for your help!
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Dick
Homework Helper
-kv^2 is a scalar. It doesn't have any direction associated with it. That would work fine if you assume the direction of v (and hence the force) is vertical. In the more general case where you don't assume the direction of v is vertical, then -kv|v| would describe both the direction of the frictional force (since v is a vector) and the magnitude.
Last edited:
rude man
Homework Helper
Gold Member
mdv/dt = -mg -kv^2
It's non-linear since the dependent variable (or any of its derivatives) appears as any other than to the 1st power. So it's a ***** to solve, which is why they didn't ask you to.
The drag force IS a vector. Its direction is opposite to the direction of motion.
For example, if y is the vertical axis, the complete force equation is
mdv/dt j = - mg j - kv^2 j where j is the unit vector in the +y direction.
The term -kv^2 j is a vector in this equation, with the coefficient kv^2 a scalar.
Last edited:
PAllen
2019 Award
Well, actually the equation is separable, so it is trivial to solve.
Dick
Homework Helper
mdv/dt = -mg -kv^2
It's non-linear since the dependent variable (or any of its derivatives) appears as any other than to the 1st power. So it's a ***** to solve, which is why they didn't ask you to.
The drag force IS a vector. Its direction is opposite to the direction of motion.
For example, if y is the vertical axis, the complete force equation is
mdv/dt j = - mg j - kv^2 j where j is the unit vector in the +y direction.
The term -kv^2 j is a vector in this equation, with the coefficient kv^2 a scalar.
It also doesn't work if v isn't a vector proportional to j. Which I think was the whole point of the question.
rude man
Homework Helper
Gold Member
By definition, the drag force IS in the direction (opposite to) the velocity.
Of course, air friction for example can cause an orthogonal force component as well. An aircraft wing experiences such a force, but it's called "lift."
There is "lift" and then there is "drag". And never the twain shall meet.
PS - most of the lift on an airfoil is due to the difference in air velocities above & below the wing (Bernoulli). But some of the lift is due simply to air resistance generating a lift component.
Dick
Homework Helper
By definition, the drag force IS in the direction (opposite to) the velocity.
Of course, air friction for example can cause an orthogonal force component as well. An aircraft wing experiences such a force, but it's called "lift."
There is "lift" and then there is "drag". And never the twain shall meet.
PS - most of the lift on an airfoil is due to the difference in air velocities above & below the wing (Bernoulli). But some of the lift is due simply to air resistance generating a lift component.
The question was why the book wrote -kv|v| instead of -kv^2.
rude man
Homework Helper
Gold Member
The reason they wrote kv|v| is that that way the vector aspect of the term is retained: it's a vector with magnitude |v|^2 and direction in the direction of v.
rude man
Homework Helper
Gold Member
BTW PAllen says mdv/dt = -mg -kv^2 is solvable by separation of variables. Could he/she or someone else show me please?
PAllen
2019 Award
BTW PAllen says mdv/dt = -mg -kv^2 is solvable by separation of variables. Could he/she or someone else show me please?
First, note that I was assuming the co-linear case. If the body starts with a non-vertical component, so -kv|v| is not equivalent to -k v^2, my observation is false. So, assuming pure vertical motion, you have:
v' = -g - k v^2
v' / (g + k v^2) = -1
integral (1/(g + k v^2)) dv = integral (-1) dt
= c -t
etc.
rude man
Homework Helper
Gold Member
Thank you PAllen! OK, it works only for collinear motion. Too bad since most elementary problems have some kid toss a ball at a certain angle to the horizontal ... blah blah ...still, it's an eye-opener for me - I thought the g term precluded separation of variables, but obviously it does not.
Thanks again! | 1,289 | 5,064 | {"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} | 4.09375 | 4 | CC-MAIN-2020-45 | latest | en | 0.936366 |
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[Data] Request to post only 8 band tiff as a separate torrent Assuming the pan-sharpened version is of 4x4 times higher resolution than the 8 band. 2.7*8*1024/(4*4*3+8) = 395 GB only for the original 8 band tiff data.It would help us in two main short comings if someone could host the 8 band data as torrent.1. We will be able to download within less time and use the data.2. We can work on the task without another high capacity hard disk/s.Also it would save our costs of download.Even 3 band RGB pan-sharpened can be generated with some algorithm using the single p band and RGB bands coming from the 8 bands.Hope I am making sense.Thanks
[Data] Request to post only 8 band tiff as a separate torrent Assuming the pan-sharpened version is of 4x4 times higher resolution than the 8 band. 2.7*8*1024/(4*4*3+8) = 495 GB only for the original 8 band tiff data.It would help us in two main short comings if someone could host the 8 band data as torrent.1. We will be able to download within less time and use the data.2. We can work on the task without another high capacity hard disk/s.Also it would save our costs of download.Even 3 band RGB pan-sharpened can be generated with some algorithm using the single p band and RGB bands coming from the 8 bands.Hope I am making sense.Thanks | 429 | 1,725 | {"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} | 2.546875 | 3 | CC-MAIN-2018-39 | latest | en | 0.873609 |
https://tutorstips.com/question-89-chapter-2-of-class-12-part-1-vk-publication/ | 1,708,683,773,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474377.60/warc/CC-MAIN-20240223085439-20240223115439-00509.warc.gz | 615,086,364 | 58,037 | # Question 89 Chapter 2 of Class 12 Part – 1 VK Publication
Question 89 Chapter 2 of Class 12 Part - 1 VK Publication
Question 89 Chapter 2 of Class 12 Part – 1
89. Ankur, Bhavna and Disha are partners in a firm. On 1st April 2011 the balance in their capital accounts stood at Rs. 14,00,000, Rs. 6,00,000 and Rs. 4,00,000 respectively. They shared profits in the proportion of 7:3:2 respectively. Partners are entitled to interest on capital @6% per annum and salary to Bhavna @ Rs.50,000 p.a. and a commission of Rs. 3,000 per month to Disha as per the provisions of the partnership deed.
Bhavna’s share of profit (excluding interest on capital) is guaranteed at not less than Rs. 1,70,000 p.a. Disha’s share of profit (including interest on capital but excluding commission) is guaranteed at not less than Rs. 1,50,000 p.a. Any deficiency arising on that account shall be met by Ankur. The profits of the firm for the year ended 31st March 2012 amounted to Rs. 9,50,000. Prepare ‘Profit and Loss Appropriation Account’ for the year ended 31st March 2012.
## The solution of Question 89 Chapter 2 of Class 12 Part – 1: –
Profit and Loss Appreciation Account
For the year ended 31st March 2017
Particulars Amount Particular Amount To Interest on Capital A/c: By Profit b/d 9,50,000 Ankur 84,000 Bhavna 36,000 Disha 24,000 1,44,000 To Salary to Bhavna 50,000 To Commission to Disha 36,000 To Profit transferred to: Ankur 4,20,000 Less: Deficiency of Disha 6,000 4,14,000 Bhavna 1,80,000 Disha 1,20,000 Add: from Ankur 6,000 1,26,000 9,50,000 9,50,000
Working Note:
Disha’s share of guaranteed profit (A) 1,50,000 (Including Interest on Capital) Her share in profit 1,20,000 Her interst on capital 24,000 (B) 1,44,000 Defiency of Disha (A-B) 6,000
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## Usha Publication – Accountancy PSEB (Class 12) – Volume I – Solution
Chapter No. 1 – Accounting Not for Profit Organisations
Chapter No. 2 – Partnership Accounts – I (Introduction)
Chapter No. 3 – Partnership Accounts – II (Goodwill: Nature and Valuation)
Chapter No. 4 – Partnership Accounts – III (Reconstitution of Partnership)
Chapter No. 5 – Partnership Accounts – IV (Admission of A Partner)
Chapter No. 6 – Partnership Accounts – V (Retirement and Death of A Partner)
Chapter No. 7 – Partnership Accounts – VI (Dissolution of Partnership Firm)
Chapter No. 8 – Company Accounts (Share Capital)
Chapter No. 9 – Company Accounts (Issue of Debentures)
Chapter No. 10 – Company Accounts (Redemption of Debentures)
## Usha Publication – Accountancy PSEB (Class 12) – Volume II – Solution
Chapter No. 1 – Financial Statements of a Company
Chapter No. 2 – Financial Statement Analysis
Chapter No. 3 – Tools of Financial Statement Analysis- Comparative and Common Size
Chapter No. 4 – Ratio Analysis | 829 | 2,916 | {"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} | 2.828125 | 3 | CC-MAIN-2024-10 | longest | en | 0.884672 |
https://www.transtutors.com/questions/operations-management--5253.htm | 1,518,990,473,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812259.30/warc/CC-MAIN-20180218212626-20180218232626-00530.warc.gz | 910,927,789 | 15,518 | # operations management
Jane Driver has just purchased (at time 1) a new car for $12,000. The cost of maintaining a car during a year depends on the age of the car at the beginning of the year as given in the table below. To avoid high maintenance costs associated with an older car, Jane can trade in a car and purchase a new car. The prices she receives on a trade in are also given in the table. To simplify the calculations, we assume that at any time it costs Jane$12,000 to purchase a new car. Jane wants to minimize the net costs (purchasing cost plus maintenance costs minus the money received in trade in) incurred during the next five years. Age of car (years) Annual maintenance cost Trade in value 0 $2000 NA 1$4000 $7000 2$5000 $6000 3$9000 $2000 4$12000 $1000 5 NA$0 Hint: the cost to buy a car and keep it until the beginning of year 4: = cost of maintenance for years kept + purchase price – trade in value = $2,000 +$4,000 + $5,000 +$12,000 - $2000 =$21,000 (I need this question done in excel using the excel solver - it is a shortest route question and the format has to follow the attached document) Student terms:
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Page 2
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