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

1 answer

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:

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

  1. 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:

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

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

\[ (2)^2 = 4 \]

  1. Substitute this back into the equation:

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

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

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

  1. Calculate \( 4\pi \):

\[ 4\pi \approx 12.5664 \]

  1. 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