Question

an igloo can be modeled as a hemisphere. its radius measures 3.2 m. find its volume in cubic meters. round your answer to the nearest tenth if necessary.

1. a can of soda can be modeled as a right cylinder. jackson measures its radius as 2.4 cm and volume as 186 cubic centimeters. find the height of the can in centimeters. round your answer to the nearest tenth if necessary.
2. the volume of a tree stump can be modeled by considering it as a right cylinder. adrian measures its height as 2.5 ft and its radius as 32 in. find the volume of the stump in cubic inches. round your answer to the nearest tenth if necessary.
3. a can of soda can be modeled as a right cylinder. jordan measures its radius as 2.1 cm and volume as 116 cubic centimeters. find the height of the can in centimeters. round your answer to the nearest tenth if necessary.

Explanation:

Step1: <Formula for volume of hemisphere>

The volume formula for a hemisphere is \(V=\frac{2}{3}\pi r^{3}\). Given \(r = 3.2\) m.

Step2: <Substitute the value of r into the formula>

Substitute \(r = 3.2\) into \(V=\frac{2}{3}\pi r^{3}\), we get \(V=\frac{2}{3}\times\pi\times(3.2)^{3}\).
First calculate \((3.2)^{3}=3.2\times3.2\times3.2 = 32.768\).
Then \(V=\frac{2}{3}\times\pi\times32.768\).
\(V=\frac{65.536\pi}{3}\approx\frac{65.536\times3.14}{3}\).
\(65.536\times3.14 = 205.78304\).
\(V=\frac{205.78304}{3}\approx68.6\) \(m^{3}\).

Step1: <Formula for volume of a right - cylinder>

The volume formula for a right - cylinder is \(V=\pi r^{2}h\). For problem 7, first convert height from feet to inches. Since \(1\) ft \(= 12\) in, \(h = 2.5\) ft \(=2.5\times12=30\) in, \(r = 32\) in.

Step2: <Substitute the values of r and h into the formula>

Substitute into \(V=\pi r^{2}h\), \(V=\pi\times(32)^{2}\times30\).
\((32)^{2}=1024\), then \(V=\pi\times1024\times30=30720\pi\approx30720\times3.14 = 96460.8\) \(in^{3}\).

For problem 8:

Step1: <Rearrange the volume formula for height>

From \(V=\pi r^{2}h\), we can solve for \(h\) as \(h=\frac{V}{\pi r^{2}}\). Given \(V = 116\) \(cm^{3}\), \(r = 2.1\) cm.

Step2: <Substitute the values of V and r into the formula for h>

\(h=\frac{116}{\pi\times(2.1)^{2}}\).
\((2.1)^{2}=4.41\), then \(h=\frac{116}{4.41\pi}\approx\frac{116}{4.41\times3.14}\).
\(4.41\times3.14 = 13.8474\), \(h=\frac{116}{13.8474}\approx8.4\) cm.

For problem 9:

Step1: <Rearrange the volume formula for height>

From \(V=\pi r^{2}h\), \(h=\frac{V}{\pi r^{2}}\). Given \(V = 186\) \(cm^{3}\), \(r = 2.4\) cm.

Step2: <Substitute the values of V and r into the formula for h>

\(h=\frac{186}{\pi\times(2.4)^{2}}\).
\((2.4)^{2}=5.76\), then \(h=\frac{186}{5.76\pi}\approx\frac{186}{5.76\times3.14}\).
\(5.76\times3.14 = 18.0864\), \(h=\frac{186}{18.0864}\approx10.3\) cm.

Answer:

• For the igloo: \(68.6\) \(m^{3}\)
• For problem 7: \(96460.8\) \(in^{3}\)
• For problem 8: \(8.4\) cm
• For problem 9: \(10.3\) cm