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Evaluate the indefinite integral:
Make a trigonometric substitution:
so the integral (i) becomes
Now, substitute back for t = arcsin(x²), and you finally get the result:
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You could also make
x² = cos t
and you would get this expression for the integral:
✔
which is fine, because those two functions have the same derivative, as the difference between them is a constant:
✔
and that constant does not interfer in the differentiation process, because the derivative of a constant is zero.
I hope this helps. =)
Perform the indicated multiplication first: -15 + 5q + 4 = 5q - 11
Note that 5q appears on both sides of this equation. Cancelling, we get :
-11 = -11. This is always true. Thus, -5(3-q) +4=5q-11 is true for all q.
Your question is Incomplete if you read the policies you will see this isn't accepted
Well, first off, you see how all the answers say a fraction, then pi? that means that you don't calculate pi into your answer.
But lets see. The radius is 4. V=(4/3)pi*r^2
V=(4/3)pi*4^2
V=(4/3)pi*16
V=(4/3)*16*pi
V=(4/3)*(16/1)*pi
V=(4*16)/(3*1)*pi
V=64/3*pi
so the answer is (B) 64/3pi units cubed.