Question

A can is in the shape of a cylinder. The can has a volume of 450 cubic inches and a diameter of 4 inches. To the nearest tenth of an inch, what is the height of the can?(1 point)
Responses

143.2 in
143.2 in

11.8 in
11.8 in

9 in
9 in

35.8 in

Answer 1

To calculate the height of the can, we'll use the formula for the volume of a cylinder, which is given by:

\[
V = \pi r^2 h
\]

Where:
- \( V \) is the volume,
- \( r \) is the radius,
- \( h \) is the height.

1. First, let's find the radius of the can. The diameter is given as 4 inches, so the radius \( r \) is half of the diameter:

\[
r = \frac{d}{2} = \frac{4}{2} = 2 \text{ inches}
\]

2. Next, we can substitute the known values into the volume formula. We know the volume \( V = 450 \) cubic inches, and we have calculated \( r = 2 \) inches:

\[
450 = \pi (2)^2 h
\]

3. Now, calculate \( (2)^2 \):

\[
(2)^2 = 4
\]

4. Substitute this back into the equation:

\[
450 = \pi (4) h
\]

5. Now, divide both sides by \( 4\pi \) to solve for \( h \):

\[
h = \frac{450}{4\pi}
\]

6. Calculate \( 4\pi \):

\[
4\pi \approx 12.5664
\]

7. Substitute this value back in:

\[
h = \frac{450}{12.5664} \approx 35.8 \text{ inches}
\]

Therefore, the height of the can, to the nearest tenth of an inch, is approximately \( 35.8 \) inches.

So the correct response is:

**35.8 in**