Start by finding the volume of one water balloon.

Recall how to find the volume of a sphere:

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

Plug in the given value for the radius.

\(\displaystyle \text{Volume}=\frac{4}{3}\pi (3.25)^3=143.79\)

Now, since one water balloon will require \(\displaystyle 143.79 in^3\) of water, divide the total volume of water by this value to find how many balloons can be filled.

\(\displaystyle \text{Number of balloons filled}=\frac{5000}{143.79}=34.77\)

Since the question asks for the number of complete balloons that can be filled, we will have to round down to the nearest whole number, \(\displaystyle 34\).

Consider a tube which is 3 ft wide and 18 ft long.

Find the volume of the largest sphere which could fit within the tube described above. 

To find the volume of a sphere, we simply need its radius

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

Now, the largest sphere which will fit within the tube will need to have a radius equal to the tube. Therefore, we can say our radius must be half the diameter, making it 1.5 ft.

Next, plug 1.5 ft into our formula to find our Volume

\(\displaystyle V=\frac{4}{3}\pi (1.5ft)^3=4.5\pi ft^3\)

So, our answer is:

\(\displaystyle V= 4.5\pi ft^3\)

This problem is deceptively simple. In order to solve for the volume of a sphere, all you need is the formula: \(\displaystyle V=\frac{4}{3}\pi r^3\), where r is the radius. 

This problem has provided additional information alongside the pertinent information. We don't need to know what the circumference is if we have been provided with the radius. 

Therefore, this problem can be quickly solved for by substituting \(\displaystyle x\) for \(\displaystyle r\) in the volume formula. 

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

therefore, the volume of this sphere is \(\displaystyle \frac{4}{3}\pi x^3\)

This problem is deceptively simple. In order to solve for the volume of a sphere, all you need is the formula: \(\displaystyle V=\frac{4}{3}\pi r^3\), where r is the radius. 

This problem has provided us with the diameter, so we just need to do a little bit of work to solve for the radius. What is the relationship between radius and diameter? The diameter is twice the radius. Or in math speak: \(\displaystyle d = 2r\). This means we can solve for our radius by taking half of the diameter. Therefore, the radius will be \(\displaystyle 18.\) 

Now that we have r, we can substitute in the value for r and solve for the volume!

\(\displaystyle V=\frac{4}{3}\pi 18^3\)

\(\displaystyle V=\frac{4}{3}\pi \times 5832\)

\(\displaystyle V=\frac{5832 \times 4}{3}\pi\)

\(\displaystyle V=7776\pi\)

This problem is very easy to solve as long as you have the volume formula for a sphere handy. 

The formula is \(\displaystyle V=\frac{4}{3}\pi r^3\), where r is the radius. 

The problem provides us with V, the volume. If we substitute in the volume, the only unknown in the problem is r. This is exactly what we want. 

\(\displaystyle 284\pi=\frac{4}{3}\pi r^3\)

The goal is to get r by itself. We can begin this process by dividing by \(\displaystyle \pi\).

\(\displaystyle \frac{284\pi}{{\color{Red} \pi}}=\frac{\frac{4}{3}{\color{Red} \pi} r^3}{{\color{Red} \pi}}\)

\(\displaystyle 284=\frac{4}{3} r^3\)
Now we may multiply both sides of the equation by \(\displaystyle \frac{3}{4}\) to remove the fraction from the right side of the equals sign. This allows us to get closer to solving for r. 

\(\displaystyle 284 \times \frac{3}{4}=\frac{4}{3} r^3 \times \frac{3}{4}\)

\(\displaystyle \frac{284 \times 3}{4}= r^3\)

\(\displaystyle 213= r^3\)

Now we need to take the cubed root of each side and we will have solved for the radius!

\(\displaystyle 5.97= r\)

One cubic meter of water is equal to \(\displaystyle 100^{3} = 1,000,000\) cubic centimeters; one-fourth of a cubic meter is 250,000 centimeters. The volume of the aquarium is 

\(\displaystyle 60 \times 60 \times 120 = 432,000\) cubic centimeters.

Therefore, after having one-fourth of a cubic meter of water poured in, there is room left for 

\(\displaystyle 432,000 - 250,000 = 182,000\) cubic centimeters of water.

The aquarium is a rectangular prism with dimensions 9 meters by 9 meters by 10.5 meters, each of which can be converted to centimeters by multiplying by 100. So the dimensions are 900 cm x 900 cm x 1,050 cm, and the volume is the product of the three, or

\(\displaystyle 900 \times 900 \times 1,050 = 850,500,000 \textup{ cm}^{3}\)

or

\(\displaystyle 850,500,000 \textup{ cm}^{3} \div 1,000 = 850,500 \textup{ L}\)

80% of this is

\(\displaystyle 850,500 \textup{ L } \times 80 \% = 680,400\textup{ L}\)

Each pipe fills the aquarium at 200 L per minute, so the eight pipes working together fill it at a rate of 1,600 L per minute. Divide, and the aquarium is filled at 80% capacity in

\(\displaystyle 680,400\textup{ L } \div 1,600 \textup{ L \ min}= 425.25\textup{ min}\)

In hours, this is

\(\displaystyle 425.25 \div 60 \approx 7.1\) 

so 7 hours is the correct response.

The volume of the rectangular solid can be written as:

\(\displaystyle V= (LW)H = BH\)

The length times width constitutes the base of the rectangular solid, and is given in the question.

Substitute the known dimensions into the formula.

\(\displaystyle V= 20\textup{ ft}^2(8 \textup{ ft}) = 160 \textup{ ft}^3\)

The answer is:  \(\displaystyle 160 \textup{ ft}^3\)

Convert all the dimensions into feet.

\(\displaystyle 6 \textup{ inches} = \frac{1}{2} \textup{ ft}\)

\(\displaystyle 8 \textup{ inches} = \frac{8}{12} \textup{ ft}=\frac{2}{3} \textup{ ft}\)

Write the formula for the volume of the block.

\(\displaystyle V= LWH\)

Substitute the dimensions and solve for volume.

\(\displaystyle V= (\frac{1}{2} \textup{ ft})(\frac{2}{3} \textup{ ft})(1\textup{ ft}) = \frac{1}{3}\textup{ ft}^3\)

The answer is:  \(\displaystyle \frac{1}{3}\textup{ ft}^3\)

Write the formula for the volume of a rectangular prism.

\(\displaystyle A=LWH\)

Substitute the dimensions into the formula.

\(\displaystyle A=6(\frac{1}{2})(\frac{1}{3}) = 6(\frac{1}{6}) = 1\)

The answer is:  \(\displaystyle 1\)