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
32% of 60
32% × 60
0.32 × 60
19.2
Answer:
32 and 40
Step-by-step explanation:
Let's call the smaller number x. If the smaller number is x, the equation for this problem is 2x + 8 = 72
Equation:
2x + 8 = 72
2x = 64
x = 32
So now we know the smaller number is 32, so now add 8 and we get 40.
Therefore the 2 numbers are 32 and 40
Answer:
a. V = (20-x)
b . 1185.185
Step-by-step explanation:
Given that:
- The height: 20 - x (in )
- Let x be the length of a side of the base of the box (x>0)
a. Write a polynomial function in factored form modeling the volume V of the box.
As we know that, this is a rectangular box has a square base so the Volume of it is:
V = h *
<=> V = (20-x)
b. What is the maximum possible volume of the box?
To maximum the volume of it, we need to use first derivative of the volume.
<=> dV / Dx = -3
+ 40x
Let dV / Dx = 0, we have:
-3
+ 40x = 0
<=> x = 40/3
=>the height h = 20/3
So the maximum possible volume of the box is:
V = 20/3 * 40/3 *40/3
= 1185.185