Answer:
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
=(new-old)/old *100%
=(90-75)/75 *100%
=20%
<span> hope it helps</span>
Answer:
90 - 9 = 81°
Step-by-step explanation:
It depends how many decimal points you are moving. If it's 1 decimal point then the point would move to the right and the answer would be 3246.50if it was two decimal places then it moves two to the right and the answer would 32465.0. But you don't need to put the .0 at the end so it will be 32465
Answer:d
Step-by-step explanation: