The volume of a cone is given by:

\(\displaystyle Volume=\frac{1}{3}\pi r^2h\)

where \(\displaystyle r\)is the radius of the circular base, and \(\displaystyle h\) is the height; the perpendicular distance from the base to the vertex. Substitute the known values in the formula:

\(\displaystyle Volume=\frac{1}{3}\pi r^2h\Rightarrow 8\pi=\frac{1}{3}\pi\times (2t)^2\times 3t\)

\(\displaystyle \Rightarrow 8\pi=4\pi t^3\Rightarrow t^3=2\Rightarrow t=\sqrt[3]{2}\)

First, divide the diameter in half to find the radius.

\(\displaystyle \frac{d}{2}=r\)

\(\displaystyle \frac{12}{2}=6\:m\)

Now, use the formula to find the volume of the cone.

\(\displaystyle \text{Volume}=\frac{1}{3}\pi r^2 h\)

\(\displaystyle \text{Volume}=\frac{1}{3}\pi \times 6^2\times4=\frac{1}{3}\pi \times 36\times4=\frac{144}{3}\pi=48\pi\:m^3\)

The volume of a cone is given by the formula:

\(\displaystyle \text{Volume}=\frac{1}{3}\pi r^2h\)

Now, plug in the values of the radius and height to find the volume of the given cone.

\(\displaystyle \text{Volume}=\frac{1}{3}\pi\times3^2\times12=\frac{1}{3}\pi\times9\times12=\frac{108}{3}\pi=36\pi\:in^3\)

The volume of a cube is the cube of its sidelength, which is also the sidelength of each square face. This sidelength is the square root of the area 64:

\(\displaystyle \sqrt{64}=8\) inches.

Multiply this by 2.5 to get the sidelength in centimeters:

\(\displaystyle 8\) \(\displaystyle \times 2.5\) \(\displaystyle = 20\) centimeters.

The cube of this is 

\(\displaystyle 20^{3}\)  \(\displaystyle =8000\) cubic centimeters

A cube has all edges the same length. The volume of a cube is found by multiplying the length of any edge by itself twice. As a formula:

\(\displaystyle Volume=s^3\) where \(\displaystyle s\) is the length of any edge of the cube.

The Surface Area of a cube can be calculated as \(\displaystyle 6s^2\).

So we get:

Volume \(\displaystyle =s^3=5^3=125 in^3\)

Surface area\(\displaystyle =6s^2=6\times 5^2=6\times 25=150 in^2\)

First we need to determine how many of the small cubes of gouda would fit along one dimension of the large cheese block. One edge of the large block is 24 inches, so 16 smaller cubes \(\displaystyle (24\div 1.5)\) would fit along the edge. Now we simply cube this one dimension to see how many cubes fit within the whole cube. \(\displaystyle 16^{3}=4096\).

Let \(\displaystyle s\) be the length of one edge of the cube. By the three-dimensional extension of the Pythagorean Theorem, 

\(\displaystyle s^{2}+s^{2}+s^{2}=N^{2}\)

\(\displaystyle 3s^{2}=N^{2}\)

\(\displaystyle s^{2}=\frac{N^{2}}{3}\)

\(\displaystyle s =\sqrt{\frac{N^{2}}{3}} = \frac{N}{\sqrt{3}} = \frac{N\sqrt{3}}{3}\)

Cube this sidelength to get the volume:

\(\displaystyle s^{3} = \left ( \frac{N\sqrt{3}}{3} \right )^{3} = \frac{N^{3} \left ( \sqrt{3}\right ) ^{3} }{3^{3} } = \frac{N^{3} \left (3 \sqrt{3}\right ) }{27 }= \frac{N^{3} \sqrt{3} }{9 }\)

Let \(\displaystyle s\) be the length of one edge of the cube. Since its surface area is \(\displaystyle N\), one face has one-sixth of this area, or \(\displaystyle \frac{N}{6}\). Therefore, 

\(\displaystyle s^{2} = \frac{N}{6}\), 

and

\(\displaystyle s =\sqrt{ \frac{N}{6}}= \frac{ \sqrt{ N}}{\sqrt{ 6}}= \frac{ \sqrt{ 6N}}{6}\)

Cube this sidelength to get the volume:

\(\displaystyle s ^{3}= \left ( \frac{ \sqrt{ 6N}}{6} \right )^{3} = \frac{\left ( \sqrt{ 6N} \right )^{3}}{6^{3}}= \frac{ 6N\sqrt{ 6N} }{216}= \frac{N\sqrt{ 6N} }{36}\)

Since a diagonal of a square face of the cube is 10, each side of each square has length \(\displaystyle \frac{\sqrt{2}}{2}\) times this, or \(\displaystyle 10 \cdot \frac{\sqrt{2}}{2} = 5\sqrt{2}\).

Cube this to get the volume of the cube:

\(\displaystyle \left (5\sqrt{2} \right )^{3}= 5^{3} \left ( \sqrt{2} \right )^{3} = 125 \left (2 \sqrt{2} \right ) = 250 \sqrt{2}\).

A perfect cube has square faces; if a face has area 1.44 square meters, then each side of each face measures the square root of this, or 1.2 meters. The volume of the tank is the cube of this, or

\(\displaystyle 1.2^{3} = 1.728\) cubic meters.

Its capacity in liters is \(\displaystyle 1.728 \times 1,000 = 1,728\) liters.

90% of this is 

\(\displaystyle 1,728 \times 0.9 = 1,555.2\) liters. 

This rounds to 1,600 liters, the correct response.