Answer:
1. Find the difference between the areas.
<u>Area of the small rectangle</u>: 
<u>Area of the big rectangle</u>: 
The difference is: 

2.
You can solve this question just by looking at the graph.
a) The height is 4 meters.

To find the height of the bleachers, we should consider the moment before the shoot, when the distance is equal to 0.


The height is 4 meters.
b) 9 meters.
For 



b) The ball travels 4 meters.
But to calculate it, it is when 

Using the quadratic formula:




It will give us to solutions, once it is a quadratic equation, but we are talking about a positive distance.

3.
In this question, we have to find the area of the cylinder and the sphere.
From the information given, we have
a = 5mm and d = 5mm, therefore the radius is 2.5 mm.
The volume of a cylinder:




The volume of the sphere:


The volume of the capsule is approximately 