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July 23, 2014

July 23, 2014

Posted by **Melissa** on Monday, September 13, 2010 at 7:21pm.

- Calculus -
**Melissa**, Monday, September 13, 2010 at 7:42pmignore the "d=". Sorry!

- Calculus -
**Reiny**, Monday, September 13, 2010 at 7:45pmthe instantaneous rate of change is given by

F'(x) which is 10s + 5

which, when s=1, becomes F'(1) = 15

for the "average" rate of change from 0 to d

F(0) = 4

F(d) = 5d^2 + 5d + 4

average rate of change = (5d^2 + 5d + 4 - 4)/(d-0)

so solve for d in

(5d^2 + 5d)/d = 15

d^2 + d = 3d

d^2 - 2d = 0

d(d-2) = 0

d = 0 or d = 2 ,but you want d > 0,

so d = 2

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