This is your answer, hope this helps :D !
Answer:
Step-by-step explanation:
27x - 45 - 4x + 9
23x - 36
That shi look hard I ain’t gone lie
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
Answer:
Increase?
Step-by-step explanation:
Because think about it let's use 1/2 as an example.
1/2 equals .5 now lets add 2 to both the numerator and denominator
3/4 is the result and it also equals .75 which is larger than .5 so that's why the answer is increase.
and u can keep on going on with the method
4/5=.8