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Sagot :
Answer:
First Part
Given that
[tex]Volume = \frac{4}{3} \pi r^{3}[/tex]
We have that
[tex]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}[/tex]
Second Part
Given that
[tex]Volume = \frac{4}{3} \pi r^{3}[/tex]
If the Diameter were reduced by half we have that
[tex]Volume = \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi (\frac{r}{2}) ^{3} = \frac{\frac{4}{3} \pi r^{3}}{8}[/tex]
This shows that the volume would be [tex]\frac{1}{8}[/tex] of its original volume
Step-by-step explanation:
First Part
Gather Information
[tex]Diameter = 18cm[/tex]
[tex]Volume = \frac{4}{3} \pi r^{3}[/tex]
Calculate Radius from Diameter
[tex]Radius = \frac{Diameter}{2} = \frac{18}{2} = 9[/tex]
Use the Radius on the Volume formula
[tex]Volume = \frac{4}{3} \pi r^{3} = \frac{4}{3} \pi 9^{3}[/tex]
Before starting any calculation, we try to simplify everything we can by expanding the exponent and then factoring one of the 9s
[tex]Volume = \frac{4}{3} \pi 9^{3} = \frac{4}{3} \pi 9 * 9 * 9 = \frac{4}{3} \pi 9 * 9 * 3 * 3[/tex]
We can see now that one of the 3s can be already divided by the 3 in the denominator
[tex]Volume = \frac{4}{3} \pi 9 * 9 * 3 * 3 = 4 \pi 9 * 9 * 3[/tex]
Finally, since we can't simplify anymore we just calculate it's volume
[tex]Volume = 4 \pi 9 * 9 * 3 = 12 \pi * 9 * 9 = 12 * 81 \pi = 972 \pi cm^{3}[/tex]
[tex]Volume \approx 3053.63 cm^{3}[/tex]
Second Part
Understanding how the Diameter reduced by half would change the Radius
[tex]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}[/tex]
Understanding how the Radius now changes the Volume
[tex]Volume = \frac{4}{3}\pi r^{3}[/tex]
With the original Diameter, we have that
[tex]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}\\\\[/tex]
If the Diameter were reduced by half, we have that
[tex]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}[/tex]
But we can see that the numerator is exactly the original Volume!
This shows us that the Volume would be [tex]\frac{1}{8}[/tex] of the original Volume if the Diameter were reduced by half.
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