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amm1812
3 years ago
6

Zachary is solving the following equation

Mathematics
2 answers:
Helga [31]3 years ago
8 0
The answer is C.

The first step in a math problem with distribution involved is to take care of the parentheses (PEMDAS). This means you have to distribute 3/4 to the (x - 8).

I hope this helped!
Elena-2011 [213]3 years ago
5 0

Answer:

C. Distribute 3/4 over (x - 8)

Step-by-step explanation:

There are multiple ways to begin solving this problem.

If we want to cancel the 3/4 without using the distributive property, we would divide both sides by 3/4.  Dividing by 4/3 will not cancel it.

When dividing by fractions, we multiply by the reciprocal; this would be multiplying both sides by 4/3.  Multiplying by 3/4 will not cancel it.

Using the distributive property, we multiply (x-8) by the number in front of parentheses, 3/4.  We cannot distribute 4/3, since it is not part of the equation.

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Answer:

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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  

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∫▒〖1 dx〗=x

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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3 years ago
Read 2 more answers
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Answer:

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Step-by-step explanation:

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