Procedure:
1) Integrate the function, from t =0 to t = 60 minutues to obtain the number of liters pumped out in the entire interval, and
2) Substract the result from the initial content of the tank (1000 liters).
Hands on:
Integral of (6 - 6e^-0.13t) dt ]from t =0 to t = 60 min =
= 6t + 6 e^-0.13t / 0.13 = 6t + 46.1538 e^-0.13t ] from t =0 to t = 60 min =
6*60 + 46.1538 e^(-0.13*60) - 0 - 46.1538 = 360 + 0.01891 - 46.1538 = 313.865 liters
2) 1000 liters - 313.865 liters = 613.135 liters
Answer: 613.135 liters
-4g is what I got if ur evaluating
If it’s something about slope I got -4
9= h/9
Mutiply both sides by 9
9*9= h/9*9
Cross out 9 and 9 , divide by 9 and then becomes h
9*9= 81
Answer: h= 81
15.69 because 49.57-2.50 divided by 3 =15.69
