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hammer [34]
2 years ago
11

The volume of a sphere whose diameter is 18 centimeters is _ cubic centimeters. If it’s diameter we’re reduced by half, it’s vol

ume would be _ of its original volume
Mathematics
1 answer:
kaheart [24]2 years ago
3 0
<h2>Answer:</h2>

<u>First Part</u>

Given that

Volume = \frac{4}{3} \pi r^{3}

We have that

Volume = \frac{4}{3} \pi r^{3} =  \frac{4}{3} \pi (\frac{Diameter}{2})^{3} =  \frac{4}{3} \pi 9^{3} = 972\pi cm^{3} \approx 3053.63 cm^{3}

<u>Second Part</u>

Given that

Volume = \frac{4}{3} \pi r^{3}

If the Diameter were reduced by half we have that

Volume = \frac{4}{3} \pi r^{3} =  \frac{4}{3} \pi (\frac{r}{2}) ^{3} = \frac{\frac{4}{3} \pi r^{3}}{8}

This shows that the volume would be \frac{1}{8} of its original volume

<h2>Step-by-step explanation:</h2>

<u>First Part</u>

Gather Information

Diameter = 18cm

Volume = \frac{4}{3} \pi r^{3}

Calculate Radius from Diameter

Radius = \frac{Diameter}{2} = \frac{18}{2} = 9

Use the Radius on the Volume formula

Volume = \frac{4}{3} \pi r^{3} =  \frac{4}{3} \pi 9^{3}

Before starting any calculation, we try to simplify everything we can by expanding the exponent and then factoring one of the 9s

Volume = \frac{4}{3} \pi 9^{3} = \frac{4}{3} \pi 9 * 9 * 9 = \frac{4}{3} \pi 9 * 9 * 3 * 3

We can see now that one of the 3s can be already divided by the 3 in the denominator

Volume = \frac{4}{3} \pi 9 * 9 * 3 * 3 = 4 \pi 9 * 9 * 3

Finally, since we can't simplify anymore we just calculate it's volume

Volume = 4 \pi 9 * 9 * 3 = 12 \pi * 9 * 9 = 12 * 81 \pi = 972 \pi cm^{3}

Volume \approx 3053.63 cm^{3}

<u>Second Part</u>

Understanding how the Diameter reduced by half would change the Radius

Radius =\frac{Diameter}{2}\\\\If \\\\Diameter = \frac{Diameter}{2}\\\\Then\\\\Radius = \frac{\frac{Diameter}{2} }{2} = \frac{\frac{Diameter}{2}}{\frac{2}{1}} = \frac{Diameter}{2} * \frac{1}{2} = \frac{Diameter}{4}

Understanding how the Radius now changes the Volume

Volume = \frac{4}{3}\pi r^{3}

With the original Diameter, we have that

Volume = \frac{4}{3}\pi (\frac{Diameter}{2}) ^{3} = \frac{4}{3}\pi \frac{Diameter^{3}}{2^{3}}\\\\ = \frac{4}{3}\pi \frac{Diameter^{3}}{2 * 2 * 2} = \frac{4}{3}\pi \frac{Diameter^{3}}{8}\\\\

If the Diameter were reduced by half, we have that

Volume = \frac{4}{3}\pi (\frac{Diameter}{4}) ^{3} = \frac{4}{3}\pi \frac{Diameter^{3}}{4^{3}}\\\\ = \frac{4}{3}\pi \frac{Diameter^{3}}{4 * 4 * 4} = \frac{4}{3}\pi \frac{Diameter^{3}}{4 * 2 * 2 * 4} = \frac{4}{3}\pi \frac{Diameter^{3}}{8 * 8} = \frac{\frac{4}{3}\pi\frac{Diameter^{3}}{8}}{8}

But we can see that the numerator is exactly the original Volume!

This shows us that the Volume would be  \frac{1}{8} of the original Volume if the Diameter were reduced by half.

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