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ki77a [65]
2 years ago
6

Find the perimeter of a square with area 64 cm/2

Mathematics
2 answers:
finlep [7]2 years ago
8 0
The answer is 32cm :))))))
disa [49]2 years ago
5 0

Answer:

Perimeter = 32 feet.

Step-by-step explanation:

Given  :  area 64 cm².

To find : find the perimeter of a square.

Solution : We have given

Area of square =  64 cm².

Area of square = side * sides = 64

side² = 64.

Taking square root both sides.

Side = \sqrt{64}.

Side =  \sqrt{8* 8}.

Side = 8 feet.

Perimeter= 4 * side.

Plug the value

Perimeter = 4 * 8

Perimeter = 32 feet.

Therefore, Perimeter = 32 feet.

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Answer:

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

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)  

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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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Answer:

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Answer:

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Step-by-step explanation:

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