A tank contains 500 gallons of salt-free water. A brine containing 0.25 lb of salt per gallon runs into the tank at the rate of 1 gal min , and the well-stirred mixture runs out at 1 gal min . In pounds per gallon, what is the concentration of salt in the tank at the end of 10 minutes? (Round your answer to four decimal places.) lbs per gal

0.0198 lbs per gallon

Step-by-step explanation:

amount of salt free water = 500 gallons

salt rate in = 1 gal/min

salt rate out = 1 gal/min

amount o salt in brine = 0.25 lb per gallon

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

salt rate in = 0.25 x 1 = 0.25

salt rate out =

The differential equation is given by

where, A(0) = 0

So, the equation becomes

Here the integrating factor is

The solution of the above differential equation is given by

where, C is the integrating constant.

Put, A(0) = 0

C = - 500

As concentration is defined as

Concentration = Quantity / Volume

Now concentration at t = 10 min

Put, t = 10 min

C (10) = 0.0198 lbs per gallon

Thus, teh concentration after 10 min is 0.0198 lbs per gallon.

ask the math teacher she or he would know lol.

step-by-step explanation:

(5x-1)^2= (5x)^2-2(5x)×1+(1)^2=25x^2-10x+1