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
6 is the answer 18 divided by 3 equals to 6 so that’s the answer
? what do you mean. do you have questions that need to be solved?
U + 4/5 = 2 and 1/3
Subtract 4/5 from each side of the equation:
U = (2 and 1/3) - (4/5)
Now it's just problem in plain old adding and subtracting fractions,
just like the ones you've done many times before.
First let's change (2 and 1/3) to a fraction: 2 and 1/3 = 7/3
So you have to find the value of (7/3) - (4/5) .
In order to add or subtract fractions, they need to have a common denominator.
The least common multiple of 3 and 5 is 15, so that's a good choice.
7/3 = 35/15
4/5 = 12/15
Now the problem is: (35/15) - (12/15).
That's 23/15 . . . . . the same thing as <u>1 and 8/15</u> .
That's the value of ' U '. What an ugly number !
