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The height of an object off the ground, h (in feet) t seconds after it is launched into the air is given by

h(t) = −16t2 + 128t, 0 ≤ t ≤ 8.

Find the average rate of change of h over the interval

[4, 8].

Respuesta :

aabdulsamir

Answer: -64

Step-by-step explanation:

h(t) = -16t² + 128t

dh/dt = -32t + 128

At t = 4,

dh/dt= -16(4)² + 128 = 0

At t = 4, dh/dt = 0

At t = 8,

dh/dt = -16(8)² + 128 = -128

At t = 8, dh/dt = -128

The rate of change of h is constant, the average rate change is

(0 - 128) / 2 = -64

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