Answer:
a)-2000
b)-2500
Step-by-step explanation:
Given:
W(t) = 100(t-15)²
applying derivation on both sides
W'(t) = 200(t-15)
->a) At what rate is the water running out at the end of 5 minutes?
Evaluating at t=5
W'(5)= 100(5-15) =>200(-10)
W'(5)=-2000
->b) What is the average rate at which the water flows out during the first 5 minutes?
[tex]\frac{W(5)-W(0)}{5-0} =\frac{100(5-15)^2*100(0-15)^2}{5} =>\frac{10000-22500}{5}[/tex]
=>-2500