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:

answer: y = -2.5x - 15

              (-6,0)

step-by-step explanation:

could u provide us some answer choices or hmm idk maybe a table for us to anser with ? ?