Abeach ball rolls off a cliff and onto the beach. the height, in feet, of the beach ball can be modeled by the function h(t)=64–16t2
h
(
t
)
=
64
–
16
t
2
, where t
t
represents time, in seconds.

what is the average rate of change in the height, in feet per second, during the first 1.25
1.25
seconds that the beach ball is in the air?

The average rate of change between 0 seconds and 1.25 seconds is -20 meters.

Step-by-step explanation:

The height of the ball can be described by the equation

h(t) = 64m - (16m/s^2)t^2

Where we want to find the average rate of change in the first 1.25 seconds that the ball is in the air, we can do this by finding the slope that conects the points h(0s) and h(1.25s).

h(0s) = 64m

h(1.25s) = 64m - (16m/s^2)*(1.25s)^2 = 39m

now, remember that the slope between two points can be written as:

s = (y2 - y1)/(x2 - x1)

where y2 = h(x2) and y1 = h(x1)

so we have that the average rate of change:

s = (39m - 64m)/(1.25 - 0) = -20m

Then we have that the average rate of change betwen 0 seconds and 1.25 seconds is -20m

the nearest power is 1.651.

means 7482 to the 1.651th power is almost equal to 2489287

answer: what do you mean