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The radius of a fully inflated soccer ball is 4.5 inches. After using the ball for a game, the match referee, Jason, measured the ball's radius and found it had decreased to 4.4 inches.

Ignoring the thickness of the materials the soccer ball is made from, find the volume of the ball before and after the match, and find the difference in the volumes before and after the match. Use [tex][tex]$\pi=\frac{22}{7}$[/tex][/tex].

The approximate volume of the ball before the match is [tex]\quad \text{(drop-down)} \quad[/tex] cubic inches.
The approximate volume of the ball after the match is [tex]\quad \text{(drop-down)} \quad[/tex] cubic inches.
The approximate change in the ball's volume before and after the match is [tex]\quad \text{(drop-down)} \quad[/tex] cubic inches.



Answer :

Let's solve this question step by step.

1. Calculate the volume of the ball before the match:
- The formula for the volume [tex]\( V \)[/tex] of a sphere is given by [tex]\( V = \frac{4}{3} \pi r^3 \)[/tex], where [tex]\( r \)[/tex] is the radius.
- Here, the radius before the match is 4.5 inches.
- Using [tex]\( \pi = \frac{22}{7} \)[/tex], substitute in the values:
[tex]\[ V_{\text{before}} = \frac{4}{3} \times \frac{22}{7} \times (4.5)^3 \][/tex]
- This evaluates to approximately [tex]\( 381.857 \)[/tex] cubic inches.

2. Calculate the volume of the ball after the match:
- The radius after the match is 4.4 inches.
- Again, using [tex]\( \pi = \frac{22}{7} \)[/tex], the volume is:
[tex]\[ V_{\text{after}} = \frac{4}{3} \times \frac{22}{7} \times (4.4)^3 \][/tex]
- This evaluates to approximately [tex]\( 356.962 \)[/tex] cubic inches.

3. Calculate the difference in volume:
- The difference in volume [tex]\( \Delta V \)[/tex] is the volume before minus the volume after:
[tex]\[ \Delta V = V_{\text{before}} - V_{\text{after}} \][/tex]
- Substituting in the values calculated:
[tex]\[ \Delta V = 381.857 - 356.962 \][/tex]
- This evaluates to approximately [tex]\( 24.896 \)[/tex] cubic inches.

So, the volume of the ball before the match is approximately [tex]\( \boxed{381.857} \)[/tex] cubic inches. The volume of the ball after the match is approximately [tex]\( \boxed{356.962} \)[/tex] cubic inches. The change in volume is approximately [tex]\( \boxed{24.896} \)[/tex] cubic inches.