Water is pumped out of a holding tank at a rate of 6-6e^-0.13t liters/minute, where t is in minutes since the pump is started. if the holding tank contains 1000 liters of water when the pump is started, how much water does it hold one hour later?

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

613.135 liters

 

48 degrees

Step-by-step explanation:

If you take away the line separating angles 1 and 5, you will get a 90 degree angle. Because angles 3 and 5 are corresponding angles, you will subtract 90-42. This gives you your answer of 48 degrees

Have a good day!