Initially, 50 pounds of salt are dissolved in a large tank holding 300 gallons of water. A brine solution with a concentration of 2 pounds per gallon is pumped into the tank at a rate of 3 gallons per minute, and the well-stirred solution is then pumped out at the same rate. Write a differential equation to model this problem, and use it to determine a function describing the amount of salt in the tank at any given time. How much salt will be in the tank if this process is allowed to continue for an infinite amount of time

dA / dt = 6 - A (t) / 100

600

Step-by-step explanation:

We have to:

dA / dt = In - Out

Input = input salt concentration * input rate of brine

replacing

Inlet = 2 lb / gal * 3 gal / min = 6 lb / min

there is no build up because the inlet flow equals the outlet, so there are 300 gallons in the tank at the start

Output: salt concentration at the output * output rate of brine

replacing

Output: A (t) / 300 * 3 gal / min = A (t) / 100

Thus:

dA / dt = 6 - A (t) / 100

this would be the differential equation.

now, when time tends to infinity, we would be left with:

100 * dA / dt = 600 - A (t)

When t tends to infinity, A (t) tends to 0, therefore, the amount of salt would be 600.

5.5

step-by-step explanation:

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.