Question

find the value of x.

1. v = 64 in³
2. v = 147π cm³
3. v = 56 m³
4. multiple choice a right cone has a height of 6 feet and a volume of 32π cubic feet. what is its radius?

a. 2 ft
b. 3 ft
c. 4 ft
d. 5 ft
find the volume of the solid. the prisms, pyramids, and cones are right. round your answer to the hundredths place.
11.
12.

Explanation:

Step1: Recall volume formula for a rectangular - based pyramid

The volume formula for a rectangular - based pyramid is $V=\frac{1}{3}lwh$, where $l$ is the length, $w$ is the width, and $h$ is the height. Given $V = 64$ in³, $l = 8$ in, and $w = 4$ in. Substitute these values into the formula: $64=\frac{1}{3}\times8\times4\times x$.

Step2: Simplify the right - hand side of the equation

First, calculate $\frac{1}{3}\times8\times4=\frac{32}{3}$. So the equation becomes $64=\frac{32}{3}x$.

Step3: Solve for $x$

Multiply both sides of the equation by $\frac{3}{32}$ to isolate $x$. $x = 64\times\frac{3}{32}$. $64\times\frac{3}{32}=\frac{192}{32}=6$ in.

Step1: Recall the volume formula for a cone

The volume formula for a cone is $V=\frac{1}{3}\pi r^{2}h$. Given $V = 32\pi$ cubic feet and $h = 6$ feet. Substitute these values into the formula: $32\pi=\frac{1}{3}\pi r^{2}\times6$.

Step2: Simplify the right - hand side of the equation

First, calculate $\frac{1}{3}\times6 = 2$. So the equation becomes $32\pi=2\pi r^{2}$.

Step3: Solve for $r^{2}$

Divide both sides of the equation by $2\pi$. $\frac{32\pi}{2\pi}=r^{2}$, which simplifies to $r^{2}=16$.

Step4: Solve for $r$

Take the square root of both sides. Since $r>0$ (radius is a non - negative quantity), $r = 4$ ft.

Answer:

$x = 6$ in

For the cone problem (question 10):