A tank contains 100 kg of salt and 1000 L of water. A solution of a concentration 0.05 kg of salt per liter enters a tank at the rate 10 L/min. The solution is mixed and drains from the tank at the same rate.

(a) What is the concentration of our solution in the tank initially?
concentration = (kg/L)

(b) Find the amount of salt in the tank after 1 hours.
amount = (kg)

(c) Find the concentration of salt in the solution in the tank as time approaches infinity.
concentration = (kg/L)

Step-by-step explanation:

Initial Quantity of salt Q(0) = 100 kg

Capacity of tank = 1000 L

Inflow of liquid = 10 l /minute and outflow = same rate thus leaving the content of volume 1000 L at any time

Assume after entering constantly mixes and drains out.

Concentration flow = inflow - outflow

= 0.05(10) -

i.e. , Q(0) = 100

Use Q(0) = 5

-45 = A  

So the equation would be

a) Initial concentratin =

b) Use the solution fo rDE substitute t =1

Q(1) = 50-45e^(-0.01) = 5.4477 kg

c) As t becomes large Q = 50

So concentration limit = 50/1000 = 0.05

ddd

step-by-step explanation:

x coordinate can be written by using the formula, x= x1 + 2/5(x2 - x1)

step-by-step explanation:

formula

x = x1 + p/w(x2 - x1)

let the line segment ab with a(x1,y1) and b(x2,y2)

a point(x,y) divide ab in the ratio, 2: 3 the x coordinate can be written as

x = x1 + p/w(x2 - x1)

here p=2 and w = 2+3

therefore,

x coordinate can be written by using the formula, x= x1 + 2/5(x2 - x1)

similarly y coordinate y = y1 + 2/5(y2 - y1)