Answer:
Therefore the concentration of salt in the incoming brine is 1.73 g/L.
Step-by-step explanation:
Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.
Let the concentration of salt be a gram/L
Let the amount salt in the tank at any time t be Q(t).

Incoming rate = (a g/L)×(1 L/min)
=a g/min
The concentration of salt in the tank at any time t is =
g/L
Outgoing rate =



Integrating both sides

[ where c arbitrary constant]
Initial condition when t= 20 , Q(t)= 15 gram


Therefore ,
.......(1)
In the starting time t=0 and Q(t)=0
Putting t=0 and Q(t)=0 in equation (1) we get









Therefore the concentration of salt in the incoming brine is 1.73 g/L
Answer:
x + y = 8
20x + 110y = 340
Step-by-step explanation:
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
Answer:
The order of ordered pairs of a function and its inverse reverse. The graph of the inverse of this function passes through the points (2,-1), (6,0), (-4,3).
Explanation:
I just took the test.
I think it is because they are both different differences