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A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt per gallon of water is pumped into the tank at the rate of 5 liters per minute, and the well-stirred mixture is pumped out at the same rate. Let A (t) represent the amount of salt in the tank at time t. The amount of salt in the tank at time t is:

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opudodennis

Answer:

[tex]A(t)=600-300e^t^/^3^0^0[/tex]

Step-by-step explanation:

let [tex]A(t)[/tex] represent the amount of salt in the tank at time [tex]t.[/tex] Then[tex]A(t)=300 \ at \ t=0[/tex] and

[tex]\frac{dA}{dt}=rate \ in- \ rate \ out\\\frac{dA}{dt}=2-\frac{A}{300}\\\\\frac{dA}{dt}+\frac{A}{300}=2\\\\e^t^/^3^0^0 \frac{dA}{dt}+\frac{e^t^/^3^0^0}{300}A=2e^t^/^3^0^0\\\\e^t^/^3^0^0A=\int2e^t^/^3^0^0dt=600e^t^/^3^0^0+C\\\\A(t)=600+Ce^t^/^3^0^0[/tex]#To find [tex]C[/tex] we use  [tex]A=300, \ when \ t=0[/tex]

Hence:

[tex]300=600+Ce^t^/^3^0^0\\C=-300[/tex]

#The amount of salt in tank at time time is expressed as:

[tex]A(t)=600-300e^t^/^3^0^0[/tex]