what is the volume of a hemisphere with a radius of 42.1 ft, rounded to the nearest tenth of a cubic foot?

Question

Accepted Answer

# Explanation:
## Step1: Recall the formula for the volume of a sphere
The volume of a sphere is $V_{sphere}=\frac{4}{3}\pi r^{3}$. A hemisphere is half of a sphere, so the volume of a hemisphere $V_{hemisphere}=\frac{2}{3}\pi r^{3}$.
## Step2: Substitute the value of $r = 42.1$ into the formula
$$
\begin{align*}
V&=\frac{2}{3}\pi(42.1)^{3}\
&=\frac{2}{3}\pi(42.1\times42.1\times42.1)\
&=\frac{2}{3}\pi(74509.261)\
\end{align*}
$$
## Step3: Calculate the value
$$
\begin{align*}
V&=\frac{149018.522}{3}\pi\
&\approx\frac{149018.522}{3}\times3.14159\
&= 49672.8407\times3.14159\
&\approx156098.0
\end{align*}
$$

# Answer:
$156098.0$ cubic feet

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Question

Explanation:

Step1: Recall the formula for the volume of a sphere

Step2: Substitute the value of \(r = 42.1\) into the formula

Step3: Calculate the value

Answer: