A tank contains 400 liters of fluid in which 15 grams of salt aredissolved. Brinecontaining 3 grams of salt per liter is then pumped into the tank at the rate of 8liters per minute, and the well mixed solution is then pumpedout at the slowerrate of 2 liters per minute. Write a differential equation, with initial condition, that models the amount of saltA(t)in the tank at timet. Extra Credit (6 points)

x´  + 2*x  =  24  

Initial condition is  t =  0   x =  15/400 grs/lt    

Step-by-step explanation:

15 grams of salt in 400 liters  means  15/400  or  0,0375 grams per liter

The concentration in the tank at any moment could be written as:

Δ(x)(t)  =  concentration (pumped in )* rate of pumping in *δt -  concentration (pumped out )* rate of pumping out*δt

Δ(x)(t)  = 3*8*δt  -  x(s) *2*δt

Dividing by δt on both sides of the equation we get:

Δ(x)(t)/ δt  =  24  -  2*x(s)

We should remember that the first member of the equation when δt⇒0 is the derivative of x with respect to time then:

D(x)/dt  =  24  -  2*x

or

x´  + 2*x  =  24      

That is a first-order differential linear equation.

Initial condition is  t =  0   x =  15/400

step-by-step explanation:

here's what happens:

1) the graph of f   = x^2 is shifted 2 units to the right.

2) the resulting graph is reflected in the x-axis, so that the parabola now opens down.

3) this final result is transformed up by 5 units.

4.08%

step-by-step explanation: