Atank contains 30 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) = 2 + sin(t/4) lb/gal. a(t) = incorrect: your answer is incorrect. lb without actually graphing, conjecture what the solution curve of the ivp should look like. then use a graphing utility to plot the graph of the solution on the interval [0, 300].

At any time (min), the volume of solution in the tank is

If is the amount of salt in the tank at any time , then the solution has a concentration of .

The net rate of change of the amount of salt in the solution, , is the difference between the amount flowing in and the amount getting pumped out:

Dropping the units and simplifying, we get the linear ODE

Multiplying both sides by allows us to identify the left side as a derivative of a product:

Integrate and divide both sides by to get

The tanks starts off with 30 lb of salt, so and we can solve for to get a particular solution of

x equals 1

step-by-step explanation:

f(4)=f(×)

x=1

the answer is (-8,-5)

step-by-step explanation:

the endpoint to midpoint is half the length. the length is (-2,-1) so when you are at the midpoint just add it to it to find your answer -6+(-2)= -8 and -4+(-1)=-5 so (-8,-5)