A large tank is filled to capacity with 400 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into the tank at a rate of 4 gal/min. The well-mixed solution is pumped out at the same rate. Find the number A(t) of pounds of salt in the tank at time t.

A(t) =  800*[1 - 1/e^(t/100 )]

Step-by-step explanation:

Suppose A (t) is the amount of salt in the tank at time t.

Thus:

A '(t) = rate of salt inflow - rate of salt outflow

would be:

dA / dt = 2 * 4 - 4 * (A / 400)

dA / dt = (800 - A) / 100

separating variables we are left with:

dA / (800 - A) = dt / 100

We integrate on each side and we have:

- ln | 800 - A | = t / 100 + c

Now, at A (0) = 0, therefore:

- ln | 800 - 0 | = 0/100 + c, we solve for c:

c = - ln | 800 |

we replace:

- ln | 800 - A | = t / 100 - ln | 800 |

ln | 800 - A | = ln | 800 | - t / 100

we apply e

800 - A = 800 - e ^ (t / 100)

800 - A = 800 / e ^ (t / 100)

A = 800 - 800 / e ^ (t / 100)

A = 800 * [1 - 1 / e ^ (t / 100)]

Thus,

A (t) = 800 * [1 - 1 / e ^ (t / 100)]

a - base

b - exponent

so,

answer: 193

explanation:

200-7 = 193