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zloy xaker [14]
3 years ago
8

Select the correct answer

Mathematics
1 answer:
CaHeK987 [17]3 years ago
5 0

Answer:

u7t54r3d3

Step-by-step explanation:

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Wilbert needs a student loan to finish school. He needs $4200 and has two options.
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A) 4596.96

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Step-by-step explanation:

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3 years ago
PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! IM DESPARETE
amm1812
Total surface area of the present=
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3 0
3 years ago
Let R be the region in the first quadrant bounded by the graphs of y =x^2 and y=2x, as shown in the figure above. The region R i
Kobotan [32]

Answer:

The area between the two functions is approximately 1.333 units.

Step-by-step explanation:

If I understand your question correctly, you're looking for the area surrounded by the the line y = 2x and the parabola y = x², (as shown in the attached image).

To do this, we just need to take the integral of y = x², and subtract that from the area under y = 2x, within that range.

First we need to find where they intersect:

2x = x²

2 = x

So they intersect at (2, 4) and (0, 0)

Now we simply need to take the integrals of each, subtracting the parabola from the line (as the parabola will have lower values in that range):

a = \int\limits^2_0 {2x} \, dx - \int\limits^2_0 {x^2} \, dx\\\\a = x^2\left \{ {{x=2} \atop {x=0}} \right - \frac{x^3}{3}\left \{ {{x=2} \atop {x=0}} \right\\\\a = (2^2 - 0^2) - (\frac{2^3}{3} - \frac{0^3}{3})\\\\a = 2^2 - \frac{2^3}{3}\\a = 4 - 8/3\\a \approx 1.333

So the correct answer is C, the area between the two functions is 4/3 units.

3 0
3 years ago
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