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sammy [17]
2 years ago
6

A large mixing tank initially contains 1000 gallons of water in which 40 pounds of salt have been dissolved. Another brine solut

ion is pumped into the tank at the rate of 5 gallons per minute, and the resulting mixture is pumped out at the same rate. The concentration of the incoming brine solution is 3 pounds of salt per gallon. If img represents the amount of salt in the tank at time t, the correct differential equation for A is:______.
A) dA/dT=3--.005A
B) dA/dT=5--.05A
C) dA/dT=15--.005A
D) dA/dT=3--.05A
E) dA/dT=15+.05A
Mathematics
1 answer:
sergey [27]2 years ago
8 0

Answer:

If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt =  15 - 0.005A

Option C) dA/dt =  15 - 0.005A is the correction Answer

Step-by-step explanation:

Given the data in the question;

If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is?

dA/dt = rate in - rate out

first we determine the rate in and rate out;

rate in = 3pound/gallon × 5gallons/min = 15 pound/min

rate out = A pounds/1000gallons × 5gallons/min  = 5Ag/1000pounds/min

= 0.005A pounds/min

so we substitute

dA/dt = rate in - rate out

dA/dt =  15 - 0.005A

Therefore, If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt =  15 - 0.005A

Option C) dA/dt =  15 - 0.005A is the correction Answer

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