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:
14.9 min
Step-by-step explanation:
r=rate
r25=5*7
r25=35
r=35/25
r=1.4
substitution
r160=t*15
t=time
r=1.4
1.4*160=t*15
224=t*15/15
224/15=t
14.9 min
11 school days are left.
In the attached picture of a calendar, I marked the days the person has left. There are 11.
Answer:
22.5
Step-by-step explanation:
y=kx
20=k15
k=1•33
30=1•33x
x=22•5
To solve, simply simplify each fraction
(16/32)/(16/16) = 1/2
(12/24)/(12/12) = 1/2
True, the equation given to us is true.
hope this helps
