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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

Practice Test

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Exercise 10 Page 875

Use the formula for the circumference of a circle to find its radius.

About 29.19 cubic inches

A regulation softball can be modeled by the following sphere with a circumference of C=12 inches.

Let r denote its radius. We know that the circumference of the ball is 12 inches. We are asked to find its volume. First, let's find r using the formula for the circumference of a circle.

C=2π r

C= 12

12=2π r

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Solve for r

Rearrange equation

2π r=12

.LHS /2π.=.RHS /2π.

r=12/2π

a/b=.a /2./.b /2.

r=6/π

Use a calculator

r=1.909859...

Round to 2 decimal place(s)

r≈ 1.91

Therefore, the radius of the sphere is about r=1.91 inches. Now, let's use the formula for the volume of a sphere.

\textcolor{darkorange}{V_\text{ball}}=\dfrac{4}{3}\pi r^3

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Substitute 1.91 for r and evaluate

r= 1.91

\textcolor{darkorange}{V_\text{ball}}=\dfrac{4}{3}\pi({\color{#0000FF}{1.91}})^2

Calculate power and product

\textcolor{darkorange}{V_\text{ball}}=\dfrac{4}{3}\cdot 6.967871\pi

MoveRightFacToNum

a/c* b = a* b/c

\textcolor{darkorange}{V_\text{ball}}=\dfrac{4\cdot 6.967871\pi}{3}

Multiply

\textcolor{darkorange}{V_\text{ball}}=\dfrac{27.871484\pi}{3}

Use a calculator

\textcolor{darkorange}{V_\text{ball}}=29.186949\ldots

Round to 2 decimal place(s)

\textcolor{darkorange}{V_\text{ball}}\approx \textcolor{darkorange}{29.19}

We find that the volume of the softball is about 29.19 cubic inches.