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.
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.
