5 + 4 • (8 - 6)2 = 5 + 4 • (2)2 = 5 + 4 • 4 = 5 + 16 = 21
C. 200
They earn $2.50 per box of markers, meaning that to make $500 they have to sell 200. [500/2.5=200]
If you're using the app, try seeing this answer through your browser: brainly.com/question/2867785_______________
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:

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which is fine, because those two functions have the same derivative, as the difference between them is a constant:
![\mathsf{\dfrac{1}{2}\,arcsin(x^2)-\left(-\dfrac{1}{2}\,arccos(x^2)\right)}\\\\\\ =\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\ =\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\ =\mathsf{\dfrac{1}{2}\cdot \dfrac{\pi}{2}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7B1%7D%7B2%7D%5C%2Carcsin%28x%5E2%29-%5Cleft%28-%5Cdfrac%7B1%7D%7B2%7D%5C%2Carccos%28x%5E2%29%5Cright%29%7D%5C%5C%5C%5C%5C%5C%0A%3D%5Cmathsf%7B%5Cdfrac%7B1%7D%7B2%7D%5C%2Carcsin%28x%5E2%29%2B%5Cdfrac%7B1%7D%7B2%7D%5C%2Carccos%28x%5E2%29%7D%5C%5C%5C%5C%5C%5C%0A%3D%5Cmathsf%7B%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cleft%5B%5C%2Carcsin%28x%5E2%29%2Barccos%28x%5E2%29%5Cright%5D%7D%5C%5C%5C%5C%5C%5C%0A%3D%5Cmathsf%7B%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cdfrac%7B%5Cpi%7D%7B2%7D%7D)

✔
and that constant does not interfer in the differentiation process, because the derivative of a constant is zero.
I hope this helps. =)
Answer: 16.6
Step-by-step explanation:
a = slope = 0.6
b = y-intercept = -4
c = x intercept = 20/3. (Set y=0 and solve for x)
So, a+b+3c = 0.6 - 4 + 3*20/3 = 16.6
Use the volume of a cylinder equation, V = pi*r^2*h to find the volume of the cylinder. To find the answer to the second one, find the volume of the square and subtract the volume of the cylinder from it.