If your in school your teacher taught it: where’s your notes ? haha jk the right answer is B.) bc it’s going over 2 up 3.
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
9+3=12
30+20=50
300+100=400
Then add.
12+50+400=462
<span>So 339+123=462</span>
Step-by-step explanation:
y = 6x-11
substitute y into 2x+3y=7
2x+3(6x-11)=7
2x + 18x-33 = 7
20 x = 7+33
x = 2
y?
y= 6(2)-11
y = 12-11 = 1
so, x = 2 and y = 1