What is the radius of hemisphere with a volume of 6043 in^3, to the nearest tenth of an inch?
3 answers
The volume of a hemisphere with radius $r$ is $V=\frac{2}{3}\pi r^3=\frac{1}{2}\pi r^3$. Thus $\frac{1}{2}\pi r^3=6043$, so $r^3=\frac{2\cdot 6043}{\pi}\approx 3832$, and $r\approx \boxed{15.3}$ inches to the nearest tenth of an inch.
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I apologize for the error. Let's go through the solution again.
The volume of a hemisphere with radius $r$ is given by $V=\frac{2}{3}\pi r^3$.
Substituting the given volume of $6043$ in$^3$, we have $\frac{2}{3}\pi r^3 = 6043$.
Solving for $r^3$, we get $r^3 = \frac{3\cdot 6043}{2\pi} \approx 2893.7$.
Taking the cube root of both sides, we find $r \approx \sqrt[3]{2893.7} \approx 14.4$.
Rounding to the nearest tenth of an inch, the radius of the hemisphere is $\boxed{14.4}$ inches.
The volume of a hemisphere with radius $r$ is given by $V=\frac{2}{3}\pi r^3$.
Substituting the given volume of $6043$ in$^3$, we have $\frac{2}{3}\pi r^3 = 6043$.
Solving for $r^3$, we get $r^3 = \frac{3\cdot 6043}{2\pi} \approx 2893.7$.
Taking the cube root of both sides, we find $r \approx \sqrt[3]{2893.7} \approx 14.4$.
Rounding to the nearest tenth of an inch, the radius of the hemisphere is $\boxed{14.4}$ inches.
1 answer