Question
What is the radius of a sphere with a volume of 3675, m, cubed, comma3675 m
3
, to the nearest tenth of a meter?
3
, to the nearest tenth of a meter?
Answers
GPT-4o mini
To find the radius of a sphere given its volume, we can use the formula for the volume of a sphere:
\[
V = \frac{4}{3} \pi r^3
\]
where \( V \) is the volume and \( r \) is the radius. We can rearrange this formula to solve for \( r \):
\[
r^3 = \frac{3V}{4\pi}
\]
Now, substituting the volume \( V = 3675 \, m^3 \):
\[
r^3 = \frac{3 \times 3675}{4\pi}
\]
Now we calculate \( r^3 \):
1. Calculate \( 3 \times 3675 = 11025 \).
2. Now, find \( 4\pi \). Using \( \pi \approx 3.14159 \):
\[
4\pi \approx 4 \times 3.14159 \approx 12.56636
\]
3. Divide \( 11025 \) by \( 12.56636 \):
\[
r^3 \approx \frac{11025}{12.56636} \approx 879.027
\]
Now we take the cube root to find \( r \):
\[
r \approx \sqrt[3]{879.027} \approx 9.54
\]
Finally, rounding the radius to the nearest tenth of a meter:
\[
r \approx 9.5 \, m
\]
Thus, the radius of the sphere, to the nearest tenth of a meter, is **9.5 m**.
\[
V = \frac{4}{3} \pi r^3
\]
where \( V \) is the volume and \( r \) is the radius. We can rearrange this formula to solve for \( r \):
\[
r^3 = \frac{3V}{4\pi}
\]
Now, substituting the volume \( V = 3675 \, m^3 \):
\[
r^3 = \frac{3 \times 3675}{4\pi}
\]
Now we calculate \( r^3 \):
1. Calculate \( 3 \times 3675 = 11025 \).
2. Now, find \( 4\pi \). Using \( \pi \approx 3.14159 \):
\[
4\pi \approx 4 \times 3.14159 \approx 12.56636
\]
3. Divide \( 11025 \) by \( 12.56636 \):
\[
r^3 \approx \frac{11025}{12.56636} \approx 879.027
\]
Now we take the cube root to find \( r \):
\[
r \approx \sqrt[3]{879.027} \approx 9.54
\]
Finally, rounding the radius to the nearest tenth of a meter:
\[
r \approx 9.5 \, m
\]
Thus, the radius of the sphere, to the nearest tenth of a meter, is **9.5 m**.