. A 600-gallon tank is lled with 300 gal of pure water. A spigot isopened and a salt solution containing 1 lb of salt per gallon begins owinginto the tank at a rate of 3gal/min. Simultaneously, a drain is opened atthe bottom of the tank allowing the mixture to leave the tank at a rate of1gal/min. What will be the salt content in the tank at the precise momentthat the volume of solution in the tank reaches the tank's capacity?

387.87 lb

Step-by-step explanation:

Rate of flow of salt solution, = (1 lb/gal * 3 gal/min) - (lb/gal * 1 gal/min)

= 3 -

+ = 3

The linear differential equation with Q = , R = 3

I.F. = e =

y * I.F. =

y =

At y = 0, t = o, c =

y =

y = 2(t + 150) -

Net rate of solution accumulation = (3 - 1) gal/min = 2 gal/min

300 gallon was already there when t = 0.

Time taken to fill the gallon = 300 gal / 2 min/gal = 150 min

Salt content at t=150 min is given by

y = 2(150+150) -

y = 387.87 lb

y = - 6

step-by-step explanation:

given the point (- 2, y) lies on the graph then it is a solution of it's equation.

substitute x = - 2 into the equation for value of y

y = 3(- 2) = 3 × - 2 = - 6

the point on the graph is (- 2, 6)

the answer is b.)

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