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Find the volume of a sphere with a diameter of 3.5 ft. Use 3.14 for pi. Round your answer to the nearest hundredth, if necessar...Asked by liquidlew
Find the volume of a sphere with a diameter of 3.5 ft. Use 3.14 for pi. Round your answer to the nearest hundredth, if necessary.(
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Answered by
GPT-4o mini
To find the volume of a sphere, you can use the formula:
\[
V = \frac{4}{3} \pi r^3
\]
where \( r \) is the radius of the sphere.
First, we need to find the radius from the diameter. The radius is half the diameter:
\[
r = \frac{d}{2} = \frac{3.5 \, \text{ft}}{2} = 1.75 \, \text{ft}
\]
Now we can substitute the radius into the volume formula:
\[
V = \frac{4}{3} \times 3.14 \times (1.75)^3
\]
Next, we calculate \( (1.75)^3 \):
\[
(1.75)^3 = 1.75 \times 1.75 \times 1.75 = 5.359375
\]
Now substituting this value back into the volume formula:
\[
V = \frac{4}{3} \times 3.14 \times 5.359375
\]
Calculating \( \frac{4}{3} \times 3.14 \):
\[
\frac{4}{3} \times 3.14 \approx 4.18666667
\]
Now multiply this value by \( 5.359375 \):
\[
V \approx 4.18666667 \times 5.359375 \approx 22.435
\]
Rounding to the nearest hundredth:
\[
V \approx 22.44 \, \text{ft}^3
\]
Therefore, the volume of the sphere is approximately \( \boxed{22.44} \, \text{ft}^3 \).
\[
V = \frac{4}{3} \pi r^3
\]
where \( r \) is the radius of the sphere.
First, we need to find the radius from the diameter. The radius is half the diameter:
\[
r = \frac{d}{2} = \frac{3.5 \, \text{ft}}{2} = 1.75 \, \text{ft}
\]
Now we can substitute the radius into the volume formula:
\[
V = \frac{4}{3} \times 3.14 \times (1.75)^3
\]
Next, we calculate \( (1.75)^3 \):
\[
(1.75)^3 = 1.75 \times 1.75 \times 1.75 = 5.359375
\]
Now substituting this value back into the volume formula:
\[
V = \frac{4}{3} \times 3.14 \times 5.359375
\]
Calculating \( \frac{4}{3} \times 3.14 \):
\[
\frac{4}{3} \times 3.14 \approx 4.18666667
\]
Now multiply this value by \( 5.359375 \):
\[
V \approx 4.18666667 \times 5.359375 \approx 22.435
\]
Rounding to the nearest hundredth:
\[
V \approx 22.44 \, \text{ft}^3
\]
Therefore, the volume of the sphere is approximately \( \boxed{22.44} \, \text{ft}^3 \).
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