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
Part a), solve this system of equations:
x + y = 5
x - y = 2.25
2x = 7.25
x = 3.625
y = 1.375
part b), solve this system of equations:
x + y = 5
x = 3y
y = 1.25
x = 3.75
(12 x 2) - 5 because twice as many means to multiply by 2. fewer means to subtract.
Answer:
The amount in the account after six years is $2,288.98
Step-by-step explanation:
In this question, we are asked to calculate the amount that will be in an account that has a principal that is compounded quarterly.
To calculate this amount, we use the formula below
A = P(1+r/n)^nt
Where P is the amount deposited which is $1,750
r is the rate which is 4.5% = 4.5/100 = 0.045
t is the number of years which is 6 years
n is the number of times per year, the interest is compounded which is 4(quarterly means every 3 months)
we plug these values into the equation
A = 1750( 1 + 0.045/4)^(4 * 6)
A = 1750( 1 + 0.01125)^24
A = 1750( 1.01125)^24
A = 2,288.98
The amount in the account after 6 years is $2,288.98
