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
Answer:
288 is your answer. 8 is your answer.
Step-by-step explanation:
If you want to know how many logs all together, all you do is multiply.
48 x 6 = 288
288 is your answer.
Or division it is
48/6 = 8
8 is your answer.
Class F=36.6666666667%
class E=33.3333333333%
class H=41.6666666667%
class G=<span>32%</span>
(5 radical 2)^2 =
5^2 × (radical 2)^2 =
25 × 2 = 50
(s÷t) - r is the correct expression.
