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
Hi there!
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I believe your answer is:
"Isolate the variable using inverse operations."
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Here’s why:
To solve for a variable, we would have to isolate it on one side.
To isolate it, we would use inverse operations on both sides on the equation until the variable is isolated.
There are no like terms in the given equation.
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Hope this helps you. I apologize if it’s incorrect.
Answer:
B ∩ (A ∪ C)'
Step-by-step explanation:
It's the B part excluding A and C union. Hence,
(A ∪ C)' ∩ B
All work is show in the picture below.
Answer:
14.598
Step-by-step explanation:
When you round that number to the nearest hundredth, you'll get 14.6. Hope this helps.