A tank contains 40 lb of salt dissolved in 500 gallons of water. A brine solution is pumped into the tank at a rate of 5 gal/min; it mixes with the solution there, and then the mixture is pumped out at a rate of 5 gal/min. Determine A(t), the amount of salt in the tank at time t, if the concentration of salt in the inflow is variable and given by cin(t)

Step-by-step explanation:

Volume of water in the Tank =500 gallons

Let A(t) be the amount of salt in the tank at time t.

Initially, the tank contains 40 lbs of salt, therefore:

A(0)=40 lbs

Rate of change of 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 then solve the resulting differential equation by separation of variables.

Taking the integral of both sides

Recall that when t=0, A(t)=40 lbs (our initial condition)

-2 is the right

let x be the money she spent so that she can earn the free t-shirt.

since, she already spent $112.49

⇒

also, as per the statement: jennifer must spend $250 or more in order to qualify for a free t-shirt at the store.

then,

we have an inequality i.,e

subtract 112.49 from both sides we get;

therefore, $137. 51 more money does jennifer need to spend so that she can earn the free t-shirt