A tank contains 100 gallons of water. Five gallons of brine per minute flow into the tank, each gallon of brine containing 1 pound of salt. Five gallons of brine flow out of the tank per minute. Assume that the tank is kept well stirred. A) find a differential equation for the number of lbs of salt in the tank (call it A)
B) assume the tank initially contains 10 lbs of salt, solve this differential equation.
C) how much salt in the tank after 10 minutes?
D) at what time will there be 199 lbs of salt in the tank?

A) (dA/dt) = 5 - 0.05A

B) A(t) = 100 - 20e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾

C) At t = 10 mins, A = 45.42 g

D) When A = 199 lb, t = 62.1 minute

Step-by-step explanation:

First of, we take the overall balance for the system,

Let V = volume of solution in the tank at any time = 100 gallons (constant because flowrate in and out is the same)

Rate of flow into the tank = Fᵢ = 5 gallons/min

Rate of flow out of the tank = F = 5 gallons/min

A) Component balance for the concentration.

Let the initial amount of salt in the tank be A₀

The rate of flow of salt coming into the tank be 1 lb/gallon × 5 gallons/min = 5 lb/min

Amount of salt in the tank, at any time = A

Rate of flow of salt out of the tank = (A × 5 m³/min)/V = (5A/V) g/min

But V = 100 gallons

Rate of flow of salt out of the tank = 0.05A lb/min

The balance,

Rate of Change of the amount of salt in the tank = (rate of flow of salt into the tank) - (rate of flow of salt out of the tank)

(dA/dt) = 5 - 0.05A

B) dA/(5 - 0.05A) = dt

∫ dA/(5 - 0.05A) = ∫ dt

Integrating the left hand side from A₀ to A and the right hand side from 0 to t

- 20 In [(5 - 0.05A)/(5 - 0.05A₀)] = t

In (5 - 0.05A) - In (5 - 0.05A₀) = - 0.05t

In (5 - 0.05A) = In (5 - 0.05A₀) - 0.05t

A₀ = 10 lb

In (5 - 0.05A) = (In 4.5) - 0.05t

In (5 - 0.05A) = 1.504 - 0.05t

(5 - 0.05A) = e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾

A(t) = 100 - 20e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾

C) When t = 10 min

A = 100 - 20e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾

1.504 - 0.05 (10) = 1.004

A = 100 - 20e⁽¹•⁰⁰⁴⁾

A = 100 - 54.58

A = 45.42 g

D) When A = 199

A = 100 - 20e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾

199 = 100 - 20e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾

20e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾ = 99

e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾ = 4.95

-(1.504 - 0.05t) = In 4.95 = 1.599

0.05t = 1.599 + 1.504 = 3.103

t = 62.1 minute

f . i can go whenever i feel like it bc it is confidential meaning only i would know how much i deposited besides the bank and they wouldnt care bc when you make a deposit you are using their services which means they are using there competition and gaining more customers

and if they dont well then thats less customers for them in a less profitable way

plz mark brainliest

your answer is

a.

5

step-by-step explanation: