Answer:

Explanation:
The volume of fluid in the tank =350 liters
Initial Amount of Salt, A(0)=50 grams
<u>Amount of Salt in the Tank
</u>

Pure water is then pumped into the tank at a rate of 5 L/min. Therefore:
=(concentration of salt in inflow)(input rate of brine)

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

Therefore:

We then solve the resulting differential equation by separation of variables.

Taking the integral of both sides

Recall that when t=0, A(0)=50 grams (our initial condition)

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