Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 liter per minute. The water is mixed well and drained at 1 liter per minute. In 20 minutes there are 15 grams of salt in the tank. What is the concentration of salt in the incoming brine?

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt  be a gram/L

Let the amount salt in the tank at any time t be Q(t).

Incoming rate = (a g/L)×(1 L/min)

                       =a g/min

The concentration of salt in the tank at any time t is =  g/L

Outgoing rate =

Integrating both sides

       [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

Therefore ,

.......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0  in equation (1) we get

Therefore the concentration of salt in the incoming brine is 1.73 g/L

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632

false

step-by-step explanation:

rulers and compasses arent required in order to bisect an angle because when bisecting angles, you're making a "half circle" and considering in order to bisect an angle it wouldn't be straight, you wouldn't need either of the two