A tank contains 120 liters of fluid in which 20 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 3 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.

A(t) = 120 - 100 e^(-t/40)

Step-by-step explanation:

Since The Fluid is been pumped both in and out,  there will always be a  120 Liters  of fluid in a  tank

The amount of salt at time t in tank is= A(t) (in grams)

the amount of salt in a tank at time t is given as = A(t) grams

Level of concentrations of salt is = A/120 g/L

The brine quantity  that is pumped in the tank =3L (of salt concentration = 1 g/L)

The salt pumped in: 3L x 1g/L = 3g rams

The level of  Amount of brine that is  pumped out at time t: 3L (of  salt concentration = A/120 g/L)

The quantity of salt pumped out at time t: 3L x A/120 g/L = A/40 g

Therefore, dA/dt = 3 - A/40

40 dA/dt = 120 - A

40/(120 - A) dA = dt

-40 ln|A-120| = t + C₀

ln|A-120| = -t/40 + C₁ ,  where C₁ = -C₀/40

A - 120 = Ce^(-t/40) .. where C = e^C₁

A = 120 + C e^(-t/40)

Recall that, the initial brine contains 20 grams of salt

so,

A(0) = 20

120 + C e^0 = 20

C = -100 or 100

Then

A(t) = 120 - 100 e^(-t/40)

6hryt%4578855

i think it is c, not positive though

step-by-step explanation: