Answer:
[-1,5]
Step-by-step explanation:
all reals between -1 and 5
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
It’s graph 2 because it is. There is a pause for the taking the bath and then the emptiest mhm so it can’t be graph 3
Answer:
x = 2 x = -6
Step-by-step explanation:
2 (x+2) ^2-4=28
Add 4 to each side
2 (x+2) ^2-4+4=28+4
2 (x+2) ^2= 32
Divide by 2
2/2 (x+2) ^2=32/2
(x+2)^2 = 16
Take the square root of each side
sqrt((x+2)^2) =±sqrt( 16)
x+2 = ±4
Subtract 2 from each side
x+2-2 = -2±4
x = -2±4
x = -2+4 and x = -2-4
x = 2 x = -6