A ball thrown in the air vertically from ground level with initial velocity 37 m/s has height h(t) = 37t − 4.9t2 m at time t (in seconds). Find the average height over the time interval extending from the ball's release to its return to ground level. (Round your answer to two decimal places.)

46.57

Step-by-step explanation:

Given that:

h(t) = 37t - 4.9t²

When the ball reaches the ground, then the height is zero;

So, we equate h(t) to zero and solve for t

37t - 4.9t² = 0

t = 0, 7.55

Let the Initial time be  a = t = 0

The time the ball reaches the ground be b = 7.55

So, the average height can now be calculated as:

Average height =

=

=

=

= 46.57091665

≅ 46.57  to (2 decimal places)

answer: a) (0,0), (1,1), (2,8), and (3,27).

step-by-step explanation:

0^3=0

1^3=1

2^3=8

3^3=27

in effect, the exponential equation for this would me y=3^x. each x-coordinate cubed is giving the y coordinate, so a is your answer : )