A tank contains 80 kg of salt and 1000 L of water. A solution of a concentration 0.04 kg of salt per liter enters a tank at the rate 8 L/min. The solution is mixed and drains from the tank at the same rate. Let y be the number of kg of salt in the tank after t minutes. Write the differential equation for this situation

Let A(t) denote the amount of salt (in kg) in the tank at time t.

At the start, there are 80 kg of salt in the tank, so A(0) = 80.

Solution flows into the tank at a rate of 8 L/min at a concentration of 0.04 kg/L, so that salt flows in at a rate of

(8 L/min) * (0.04 kg/L) = 0.32 kg/min

Solution flows out at the same rate, but its concentration depends on the amount of salt in the tank. The concentration of the solution is the proportion of salt in the liquid to the total volume of the liquid. Solution flows in and out at 8 L/min, so the volume of liquid (1000 L) stays the same. A(t) is the amount of salt in the tank, so the concentration is A(t)/1000 kg/L. Hence salt flows out at a rate of

(8 L/min) * (A(t)/1000 kg/L) = 0.008 A(t) kg/min

The net rate at which salt flows through the system is then given by the differential equation,

dA(t)/dt = 0.32 - 0.008 A(t)

(Don't forget to include the initial condition)

answer:   n ≤ -3 or n < -5

step-by-step explanation:

-5 + 2n ≥ -11

+5             +5

2n ≥ -6

2/2 ≤ -6/2

n ≤ -3

-6 - 10n > 44

+6           +6

-10n > 50

-10/-10 50/-10

n < -5

what are the answear choices to the question.

step-by-step explanation: