A tank contains 1000L of pure water. Brine that contains 0.04kg of salt per liter enters the tank at a rate of 5L/min. Also, brine that contains 0.06kg of salt per liter enters the tank at a rate of 10L/min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15L/min. Answer the following questions. 1. How much salt is in the tank after t minutes

s(t) = 160/3  ( 1 - e^(-3t / 200) )

Step-by-step explanation:

volume of pure water in tank  = 1000 L

Brine contains 0.04kg of salt/L

Inflow rate of Brine containing 0.04kg of salt/L = 5L/min

Brine containing 0.06 kg of salt/L

Inflow rate of Brine containing 0.06 kg of salt/L = 10L/min

Solution is thoroughly mixed and drains from tank at 15L/min

a) Determine the amount of salt is in the tank after t minutes

rate of salt entering = 0.2 + 0.6 = 0.8 kg/min

rate of salt leaving = s/1000 * 15

amount of salt at time (t) = s(t)

initial condition s( 0 ) = 0

ds/dt = 0.8 - 15s/1000 = 0.8 - 3s/200

200 ds/dt = ( 160 - 3s )

-200/3  In ( 160 - 3s ) = t + c

Given that ;  t = 0 , s = 0    

c =  - 200/3  In ( 160 )

∴  -200/3  In ( 160 - 3s ) = t - 200/3  In ( 160 )

- 200/3  [ In ( 60 - 3s ) - In ( 160 ) ] = t

therefore:  

In ( 160 - 3s / 160 ) = -3t/200

= ( 160 - 3s / 160 ) = e ^ (-3t/200 )

Hence amount of salt in tank after t minutes

s(t) = 160/3  ( 1 - e^(-3t / 200) )

18a

step-by-step explanation:

simply because it is. it's the highest common factor

-1   1/6.   you dont subtract the denominators

step-by-step explanation: