Solve the problem. Use 3.14 as the approximate value of $\pi$. Round to the nearest tenth unless instructed otherwise.
-The volume of a ball is $85.33 \pi \mathrm{in}^{3}$. Find the dimensions of a rectangular box that is just large enough to hold the ball.
A) $8 \pi$ in. by $8 \pi$ in. by $8 \pi$ in.
B) $4 \pi$ in. by $4 \pi$ in. by $4 \pi$ in.
C) 4 in. by 4 in. by 4 in.
D) 8 in. by 8 in. by 8 in.

Question

Quizplus · Accepted Answer

The volume of a sphere is given by the formula $V = \frac{4}{3}\pi r^3$. Given the volume is $85.33 \pi \mathrm{in}^3$, we can solve for the radius $r$. Substituting the given volume into the formula and solving for $r$ gives a radius that, when doubled to find the diameter, matches the dimensions of a cube that can contain the sphere. The correct dimensions are those that correspond to the diameter of the sphere, which is 8 inches, making the cube 8 in. by 8 in. by 8 in.

Solve the Problem π\piπ Round to the Nearest Tenth Unless Instructed Otherwise

Solve the Problem $\pi$ Round to the Nearest Tenth Unless Instructed Otherwise