The pool can be seen as a cylinder with depth (or height) 2.5 meters and a base with diameter 18 meters - and radius half this, or 9 meters. 

The bottom of the pool - the base of the cylinder - is a circle with radius 9 meters, so its area is

\(\displaystyle B = \pi r ^{2} = 3.14 \cdot 9 \cdot 9 = 254.34\) square meters.

Its side - the lateral face of the cylinder - has area

\(\displaystyle LA = 2 \pi r h = 2 \cdot 3.14 \cdot 18 \cdot 2.5 = 141.3\) square meters.

Their sum - the total area to be painted - is \(\displaystyle 254.34 + 141.3 = 395.64\) square feet. Since one can of paint covers 40 square meters, divide:

\(\displaystyle 395.64 \div 40 \approx 9.891\)

Nine cans of paint and part of a tenth will be required, so the correct response is ten.

Six inches is equal to 0.5 feet, so the radius of the interior of the tank is 

\(\displaystyle 24-0.5= 23.5\) feet.

The surface area of the interior of the tank can be calculated using the formula

\(\displaystyle SA = 4 \pi r^{2}\)

\(\displaystyle SA \approx 4 \cdot 3.14 \cdot 23.5 ^{2} \approx 6,936\),

which rounds to 6,900 square feet.

The base is a circle with radius \(\displaystyle r = 4\), and its area can be calculated using the area formula for a circle:

\(\displaystyle B= \pi r^{2} = \pi \times 4^{2} \approx 3.14 \times 16 = 50.24\) square meters.

To find the lateral area, we need the slant height of the cone. This can be found by way of the Pythagorean Theorem. Treating the height \(\displaystyle h = 10\) and the radius \(\displaystyle r = 4\) as the legs and slant height \(\displaystyle l\) as the hypotenuse, calculate:

\(\displaystyle l^{2} = h^{2} + r^{2}\)

\(\displaystyle l^{2} = 10^{2} + 4^{2} = 100 + 16 = 116\)

\(\displaystyle l = \sqrt{116} \approx 10.77\) meters.

The formula for the lateral area can be applied now:

\(\displaystyle LA = \pi r l \approx 3.14 \cdot 4 \cdot 10.77 \approx 135.34\)

Add the base and the lateral area to obtain the total surface area:

\(\displaystyle SA = B + LA \approx 50.24 + 135.34 \approx 185.58\).

This rounds to 186 square meters.

Three inches is equal to 0.25 feet, so the height of the interior of the tank is 

\(\displaystyle 40 - 2 \times 0.25 = 39.5\) feet.

The radius of the interior of the tank is 

\(\displaystyle 15 - 0.25 = 14.75\) feet.

The surface area of the interior of the tank can be determined by using this formula:

\(\displaystyle SA = 2 \pi r (r + h)\)

\(\displaystyle SA = 2 \cdot 3.14 \cdot 14.75 (14.75 + 39.5)\)

\(\displaystyle SA \approx 2 \cdot 3.14 \cdot 14.75 \cdot 54.25\approx 5,025\),

which rounds to 5,000 square feet.

The surface area of a cone with radius \(\displaystyle r\) and slant height \(\displaystyle l\) is calculated using the formula \(\displaystyle S = \pi r^{2} + \pi r l\).

Substitute 35 for \(\displaystyle r\) and 100 for \(\displaystyle l\) to find the surface area in square meters:

\(\displaystyle S = \pi \cdot 35^{2} + \pi \cdot 35 \cdot 100\)

\(\displaystyle \approx 3.14 \cdot 1,225 + 3.14\cdot 35 \cdot 100\)

\(\displaystyle \approx 3,848 + 10,996\)

\(\displaystyle \approx 14,844\) square meters.

The paint covers 40 square meters per gallon, so the city needs

\(\displaystyle 14,844 \div 40 =371.1\) gallons of paint. 

Since the city buys the paint in multiples of 25 gallons, it will need to buy the next-highest multiple of 25, or 375 gallons.

The area of an equilateral triangle is given by the formula

\(\displaystyle A = \frac{s^{2}\sqrt{3}}{4}\).

Since there are eight equilateral triangles that comprise the surface of the octahedron, the total surface area is 

\(\displaystyle SA = 8 A = 8 \cdot \frac{s^{2}\sqrt{3}}{4} = 2 s^{2}\sqrt{3}\).

Substitute \(\displaystyle s = 3\):

\(\displaystyle SA =2 \cdot 3^{2}\sqrt{3} = 18 \sqrt{3}\) square inches.

The area of an equilateral triangle is given by the formula

\(\displaystyle A = \frac{s^{2}\sqrt{3}}{4}\).

Since there are twenty equilateral triangles that comprise the surface of the icosahedron, the total surface area is 

\(\displaystyle SA = 20 A = 20 \cdot \frac{s^{2}\sqrt{3}}{4} = 5 s^{2}\sqrt{3}\).

Substitute \(\displaystyle s = 2\):

 \(\displaystyle SA = 5 \cdot 2^{2}\sqrt{3} = 5 \cdot 4 \sqrt{3} = 20 \sqrt{3}\) square inches.

The area of an equilateral triangle is given by the formula

\(\displaystyle A = \frac{s^{2}\sqrt{3}}{4}\).

Since there are four equilateral triangles that comprise the surface of the tetrahedron, the total surface area is 

\(\displaystyle SA = 4 A = 4 \cdot \frac{s^{2}\sqrt{3}}{4} = s^{2}\sqrt{3}\).

Substitute \(\displaystyle s = 5\):

\(\displaystyle SA = 5^{2}\sqrt{3} = 25 \sqrt{3}\) square inches.

To find the surface area of a cube, we will use the following formula:

\(\displaystyle SA = 6 \cdot l \cdot w\)

where l is the length, and w is the width of the cube.

Now, we know the height of the cube is 9cm. Because it is a cube, all lengths, widths, and heights are the same. Therefore, the length and the width are also 9cm.

Knowing this, we can substitute into the formula. We get

\(\displaystyle SA = 6 \cdot 9\text{cm} \cdot 9\text{cm}\)

\(\displaystyle SA = 6 \cdot 81\text{cm}^2\)

\(\displaystyle SA = 486\text{cm}^2\)

To find the surface area of a sphere, we will use the following formula:

\(\displaystyle SA = 4 \cdot \pi \cdot r^2\)

where r is the radius of the sphere.

Now, we know the radius of the sphere is 7in. 

So, we can substitute into the formula. We get

\(\displaystyle SA = 4 \cdot \pi \cdot (7\text{in})^2\)

\(\displaystyle SA = 4 \cdot \pi \cdot 49\text{in}^2\)

\(\displaystyle SA = 196\text{in}^2 \cdot \pi\)

\(\displaystyle SA = 196\pi \text{ in}^2\)