The volume of a tetrahedron can be solved for by using the equation:

\(\displaystyle V= \frac{a^3}{6\sqrt{2}}\)

where \(\displaystyle a\) is the measurement of the edge of the tetrahedron. 

This problem can be quickly solved by substituting 6 in for \(\displaystyle a\). 

\(\displaystyle V=\frac{6^3}{6\sqrt{2}}=\frac{216}{6\sqrt{2}}=\frac{36}{\sqrt{2}}\)

\(\displaystyle {\color{Blue} V=25.5}\)

The volume of a tetrahedron is \(\displaystyle V=\frac{s^3}{6\sqrt2}\).

This tetrahedron has a side with a length of 8. 

\(\displaystyle V=\frac{8^3}{6\sqrt2}\), which becomes \(\displaystyle \frac{512}{6\sqrt2}\).

You can reduce that answer further, so that it becomes 

\(\displaystyle \frac{256}{3\sqrt2}\). 

The volume of a sphere is given by the equation:

\(\displaystyle V=\frac{4}{3} \pi r^3\)

The problem says that the radius is 15, so plug in 15 for r and simplify.

\(\displaystyle V=\frac{4}{3} \pi r^3=\frac{4}{3} \pi 15^3=4500\pi\)

Write the equation for the volume of a sphere.

\(\displaystyle V=\frac{4}{3}\pi r^3\)

The radius is half the diameter, or \(\displaystyle \frac{3}{2}\).

Substitute the radius into the equation.

\(\displaystyle V=\frac{4}{3}\pi( \frac{3}{2})^3\)

\(\displaystyle V=\frac{4}{3}\pi( \frac{27}{8})\)

Reduce the fractions.

The answer is:  \(\displaystyle \frac{9}{2}\pi\)

Write the formula for the volume of a rectangular prism.

\(\displaystyle A=\textup{ (Length) (Width) (Height)}\)

Substitute the dimensions into the formula.

\(\displaystyle A = \frac{1}{2}\cdot \frac{2}{3}\cdot \frac{5}{6} = \frac{10}{36}\)

Reduce this fraction.

\(\displaystyle \frac{5\times 2}{18\times 2}\)

The answer is:  \(\displaystyle \frac{5}{18}\)

Call \(\displaystyle L\), \(\displaystyle H\), and \(\displaystyle W\) the length, width, and height of the crate. 

The length of the crate is half its height, so 

\(\displaystyle L = \frac{1}{2} H\)

\(\displaystyle 2 \cdot L = 2 \cdot \frac{1}{2} H\)

\(\displaystyle 2 L = H\)

The length of the crate is two-thirds its width, so

\(\displaystyle L = \frac{2}{3} W\)

\(\displaystyle \frac{3} {2} \cdot L = \frac{3} {2} \cdot \frac{2}{3} W\)

\(\displaystyle \frac{3} {2} L = W\)

The dimensions of the crate in terms of \(\displaystyle L\) are \(\displaystyle L\), \(\displaystyle \frac{3} {2} L\), and \(\displaystyle 2L\). The volume is their product:

\(\displaystyle V= LWH\)

Substitute:

\(\displaystyle LWH = V\)

\(\displaystyle L \cdot \frac{3} {2} L \cdot 2L = 4\)

\(\displaystyle 3 L ^{3} = 4\)

Solve for \(\displaystyle L\):

\(\displaystyle 3 L ^{3} = 4\)

\(\displaystyle \frac{3 L ^{3} }{3} = \frac{4 }{3}\)

\(\displaystyle L ^{3} = \frac{4 }{3}\)

\(\displaystyle L =\sqrt[3]{ \frac{4 }{3}}\)

\(\displaystyle L \approx 1.101\) meters.

Since one meter comprises 100 centimeters, multiply by 100 to convert to centimeters:

\(\displaystyle 1.101 \cdot 100 = 110.1\),

which rounds to 110 centimeters.