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Sedaia [141]
2 years ago
13

H(x) = ax^2 +1 Find the value of a when h(-2) = 21.

Mathematics
1 answer:
blagie [28]2 years ago
4 0

Answer:

a = 5

Step-by-step explanation:

Solving h(-2) = 21:

h(-2) = 21 \\ a(-2)^2 +1 = 21 \\ a(4) +1 = 21 \\ 4a +1 = 21 \\ 4a +1 -1 = 21 -1 \\ 4a = 20 \\ \frac{4a}{4} = \frac{20}{4} \\ a = 5

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Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\

And the second one would be

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If you rotate the first region around the "y" axis you get that

{\displaystyle A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51 }

And if you rotate the second region around the "y" axis you get that

{\displaystyle A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188 }

And the sum would be  2.51+4.188 = 6.698

If you revolve just the outer curve you get

If you rotate the first  region around the x axis you get that

{\displaystyle A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708 }

And if you rotate the second region around the x axis you get that

{\displaystyle A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472 }

And the sum would be 1.5708+1.0472 = 2.618

7 0
3 years ago
Quadrilateral ABCD has vertices A(-3, 4), B(1, 3), C(3, 6), and D(1, 6). Match each set of vertices of quadrilateral EFGH with t
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Answer:

See explanation

Step-by-step explanation:

First plot points A, B, C and D on the coordinate plane and connect them to get quadrilateral ABCD. Then plot the vertices E, F, G, H of the quadrilateral EFGH in each case.

For each of these cases there are attached diagrams

1 case - reflection acraoss the x-axis

2 case - translation 5 units down

3 case - reflection across the y-axis

4 case - translation 7 units right

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slamgirl [31]

Answer:

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Step-by-step explanation:

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