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:
Substitute each of the values into the expression.
1. 7(7)-3(4)+8-4
49-12+8-4
41
2. 6(5)-10/5
30-2
28
3. 3(1)^2+2^3
3(1)+8
11
4. 5(3)^2
5(9)
45
:)
So solve for x
|x-72.5|=4
the || makes the inside positive so it could be positivie and negative so
x-72.5=4 and x-72.5=-4
solve for x in both situations
x-72.5=4
add 72.5 to both sides
x=76.5
x-72.5=-4
add 72.5 to both sides
x=68.6
the minimum it can be set to is 68.5 and the max is 76.5
The value of t is 3/2 or 1.5
<u>Step-by-step explanation</u>:
The given expression is 17/20 equals t plus -13/20.
<u>Changing it into an equation form :</u>
⇒ t + (-13/20) = 17/20
Keeping t alone on one side and constant terms to the another side,
t = (17/20) + (13/20)
Since the denominators are same, it is easy to simplify those fractions.
⇒ t = (17+13)/20
⇒ t = 30/20
⇒ t = 3/2
⇒ t = 1.5
The solution is t= 3/2 or t= 1.5
Answer:
D
Step-by-step explanation:
10/5=2
20/10=2
40/20=2