A tank contains 300 liters of fluid in which 10 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 5 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.

Step-by-step explanation:

The volume of fluid in the tank = 300 Liters

Initial amount of Salt in the tank, A(0)=10 grams

Change in the Amount of Salt in the Tank

Rate In =(concentration of salt in inflow)(input rate of brine)

Rate Out =(concentration of salt in outflow)(output rate of brine)

Therefore:

We solve the resulting linear differential equation for A(t)

Taking the integral of both sides

Recall that when t=0, A(t)=10 grams (our initial condition)

Therefore, the number A(t) of grams of salt in the tank at time t is:

the correct answer is 5 6 and 7

the correct answer is 1: 2

step-by-step explanation:

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