Answer:
[tex]A(t) = 1600 - e^{-(\frac{t}{250} - 7.33)}[/tex]
Explanation:
A(t) is the amount of salt in the tank at time t, measured in kilograms.
A(0) = 70
[tex]\frac{dA}{dt} =[/tex]rate in - rate out
=0.4* 16 - (A/1000)*4
= [tex]\frac{1600 - A}{250}[/tex] kg/min
[tex]\int\limits {\frac{1}{1600-A} } \, dA = \int\ {\frac{1}{250} } \, dt\\\\ -ln(1600-A) = \frac{t}{250} + C\\\\[/tex]
A(0) = 70
-ln (1600 - 70) = 0/250 + C
-7.33 = C
[tex]-ln(1600-A) = \frac{t}{250} - 7.33\\\\[/tex]
[tex]1600 - A = e^{-(\frac{t}{250} - 7.33)} \\\\A = 1600 - e^{-(\frac{t}{250} - 7.33)}[/tex][tex]1600-A = e^{-(\frac{t}{250} - 7.33)} \\\\ A = 1600 - e^{-(\frac{t}{250} - 7.33)}[/tex]
