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kirill [66]
3 years ago
13

 Ill Mark Brainliest if you answer correctly. The figure below shows a shaded circular region inside a larger circle:

Mathematics
1 answer:
pogonyaev3 years ago
8 0

First we have to find the areas of both circle using the formula

Area = \pi r^2

For larger circle,

Area=  \pi(5)^2 = 25 \pi

For smaller circle,

Area = \pi (4)^2 = 16 \pi

Required probability

= \frac{25 \pi -16 \pi}{25 \pi} *100 = \frac{9}{25}*100 = 36%

Correct option is the second option .

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Solve the inequality for w
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Find the indefinite integral. (Note: Solve by the simplest method—not all require integration by parts. Use C for the constant o
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Step-by-step explanation:

∫▒〖1st .2nd dx=1st∫▒〖2nd dx〗-∫▒〖(derivative of 1st) dx∫▒〖2nd dx〗〗〗

Let 1st=arctan⁡(x)

And 2nd=1

∫▒〖arctan⁡(x).1 dx=arctan⁡(x) ∫▒〖1 dx〗-∫▒〖(derivative of arctan(x))dx∫▒〖1 dx〗〗〗

As we know that  

derivative of arctan(x)=1/(1+x^2 )

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So  

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-∫▒〖(1/(1+x^2 ))dx.x〗…………Eq1

Let’s solve ∫▒(1/(1+x^2 ))dx by substitution now  

Let 1+x^2=u

du=2xdx

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1/2 ∫▒〖(2/(1+x^2 ))dx.x〗=1/2 ∫▒(2xdx/u)  

1/2 ∫▒(2xdx/u) =1/2 ∫▒(du/u)  

1/2 ∫▒(2xdx/u) =1/2  ln⁡(u)+C

1/2 ∫▒(2xdx/u) =1/2  ln⁡(1+x^2 )+C

Putting values in Eq1 we get

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2  ln⁡(1+x^2 )+C  (required soultion)

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