1. C
2. 3, because I used a proportion table
4a = 36
to solve for a, divide both sides by 4 to get a by itself.
<u>4a</u> = <u>36
</u> 4 4
<u>
</u>a = 9
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
Try and find the common denominator which is 6 since 2×3 is 6 then once u get ur denominator u have to find the numerator which would be 4/6+3/6 And that would equal 7/6 and if you would u would keep it like that but if u would have to change it into a mixed fraction it would be 1 and 1/6
Hoped this helped