Atank originally contains 160 gal of fresh water. then water containing 1 4 lb of salt per gallon is poured into the tank at a rate of 4 gal/min, and the mixture is allowed to leave at the same rate. after 8 min the process is stopped, and fresh water is poured into the tank at a rate of 6 gal/min, with the mixture again leaving at the same rate. find the amount of salt in the tank at the end of an additional 8 min.

245.86 lb of salt

Step-by-step explanation:

Let y(t) be the amount of salt in the tank at instant t. Then y'(t) represents the rate of change of the amount of salt in the tank at instant t.

The water entering the tank contains 14 lb of salt per gallon and enters at a speed of 4 gal/min, so every minute 14*4=56 lb of salt are entering the tank the first 8 min.

After 8 minutes, as the volume of tank keeps 160 gal., the concentration of salt is .

Since the water is now leaving the tank at 6 gal/min, the salt is leaving the tank at a speed of .

Notice that the speed is negative because the salt is leaving the tank, so the amount of salt is decreasing.

Now we have the following ordinary differential equation

Solving this differential equation we get a piecewise function defined as follow

where C is a constant.

Since the tank had no salt at the beginning of the process. Y(0)=0, so

For we have

and solving this equation we get

where C is a constant that can be found with the initial condition y(8) = 56*8 = 448

and our solution is

To know what the amount of salt is at the end of the process, we must evaluate y(16), which is

So, at the end of the process the tank contains 245.867 lb of salt.

a) 0.25x times 3. (could be wrong on equation.)

b) 2.25 pound of dark chocolate. (can i eat all the chocolate? plz? anyway, again, could be wrong, but i hope im right. all the chocolate together should be 5.25 pounds)

have a good day!

300pi

step-by-step explanation: