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EleoNora [17]
3 years ago
13

Question is on the picture !

Mathematics
1 answer:
Nataly [62]3 years ago
4 0

Answer: 122.5

======================================================

Explanation:

First thing to do is to find the perimeter of figure B. Add up all the sides and we get: 5+9+9+12 = 14+21 = 35.

The scale factor 7:2 means that if the perimeter of figure A was 7, then the perimeter of figure B would be 2. Or it could be 14 for A and 4 for B. And so on. The idea is that the two perimeters scale up or down together. This allows us to set up the proportion below in which we can solve for x

(perimeter of A)/(perimeter of B) = 7/2

x/35 = 7/2

x*2 = 35*7 .... cross multiply

2x = 245

2x/2 = 245/2 .... divide both sides by 2

x = 122.5

The perimeter of figure A is 122.5

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Butoxors [25]
\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx

Setting x=5\sin\theta, you have \mathrm dx=5\cos\theta\,\mathrm d\theta. Then the integral becomes

\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta
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\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta

Now, \sqrt{x^2}=|x| in general. But since we want our substitution x=5\sin\theta to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means \theta=\sin^{-1}\dfrac x5, which implies that \left|\dfrac x5\right|\le1, or equivalently that |\theta|\le\dfrac\pi2. Over this domain, \cos\theta\ge0, so \sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta.

Long story short, this allows us to go from

\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta

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\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta
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Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta

Then integrate term-by-term to get

\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)
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Now undo the substitution to get the antiderivative back in terms of x.

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and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C
4 0
3 years ago
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Umnica [9.8K]
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