Mathematics: sketch the region enclosed by y=10x and y=x^2+24. then find the area of the region. how to get this answer?

sketch the region enclosed by y=10x and y=x^2+24. - id102241884461, hannahliebl2000

candy7693

31.12.2021

The area enclosed by given curves is $\frac{4}{3}$ square units.

Step-by-step explanation:

The given equation are

y=10x

y=x^2+24

Draw the graph of both equation.

From the below graph it is clear that the graph intersect each other at (4,40) and (6,60).

The area of region enclosed by curves is

$A=\int_{4}^{6}10xdx-\int_{4}^{6}(x^2+24)dx$

$A=10(\frac{1}{2}[x^2]_{4}^{6})-\frac{1}{3}[x^3]_{4}^{6}-24[x]_{4}^{6}$

$A=5\left(6^2-4^2\right)-\frac{1}{3}\left(6^3-4^3\right)-24\left(6-4\right)$

$A=5[20]-\frac{1}{3}[152]-24[2]$

$A=100-\frac{152}{3}-48$

$A=\frac{4}{3}$

Therefore the area enclosed by given curves is $\frac{4}{3}$ square units.

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phillipselijah2

14.10.2021

8(-10x+3.5y-7)

8×-10x + 8×3.5y + 8×-7

-80x + 28y - 56

therefore

-80x+28y-56 is the correct answer

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