Step-by-step explanation:
Let the height above which the ball is released be H
This problem can be tackled using geometric progression.
The nth term of a Geometric progression is given by the above, where n is the term index, a is the first term and the sum for such a progression up to the Nth term is
To find the total distance travel one has to sum over up to n=3. But there is little subtle point here. For the first bounce ( n=1 ), the ball has only travel H and not 2H. For subsequent bounces ( n=2,3,4,5...... ), the distance travel is 2×(3/4)n×H
a=2H..........r=3/4
However we have to subtract H because up to the first bounce, the ball only travel H instead of 2H
Therefore the total distance travel up to the Nth bounce is
For N=3 one obtains
D=3.625H
Circumference =

When you plug that in you should have

Solve that and you have your answer(: I would recommend rounding to the nearest hundreth
A and B are the answers.
If you need proof, just ask in the comments, and I’ll get back to you.
I hope this helps!
Answer:
The answer is "
The volume is greater than 50 cubic units
"
Step-by-step explanation:
In this question, the value of the units and figure is missing that's why its solution can be defined as follows:
Let's the unit value =50 and please find the figure in the attachment file.
The given figure shows its box is not filled with 50 cubes. These were ten empty places where cubes could be placed.
This figure indicates that only the box size will be four units per five units per three units, so they realize it will be in...

That's why the volume value exceeds 50 cubic units.