Answered

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Wilma tosses a ball into the air. The function h(t) represents the height, in feet, of the ball after t seconds. The difference quotient of h(t) is –32t – 16h + 9. What is the average rate of change in the height of the ball between 1 and 1.5 seconds? –47 ft/s –31 ft/s 31 ft/s 47 ft/s

Sagot :

abidemiokin

Answer:

-32ft/s

Step-by-step explanation:

Given the function of the height expressed as h(t) is –32t² – 16t + 9.

Rate of change in height is the velocity and expressed as;

v(t) = d(h(t))/dt = -64t - 16

at t = 1sec

v(1) =  -64(1)- 16

v(1) = -80ft/s

at t = 1.5secs

v(1.5) =  -64(1.5)- 16

v(1.5) = -96-16

v(1.5) = -112ft/s

The average rate of change in the height of the ball between 1 and 1.5 seconds = -112-(-80) = -112+80 = -32ft/s

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