Answer:
t = 460.52 min
Step-by-step explanation:
Here is the complete question
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.
Solution
Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.
inflow = 0 (since the incoming water contains no dye)
outflow = concentration × rate of water inflow
Concentration = Quantity/volume = Q/200
outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.
So, Q' = inflow - outflow = 0 - Q/100
Q' = -Q/100 This is our differential equation. We solve it as follows
Q'/Q = -1/100
∫Q'/Q = ∫-1/100
㏑Q = -t/100 + c

when t = 0, Q = 200 L × 1 g/L = 200 g

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

㏑0.01 = -t/100
t = -100㏑0.01
t = 460.52 min
Answer:
The mistake she made was ; She didn't change the sign when subtracting -3/4 from both sides
She wrote 5/4+3/4 instead of 5/4-3/4
Step-by-step explanation:
Correct solution:

Subtract-3/4 from both sides

Simplify

1) Since we have the slope, and one point we can find that linear equation using this point-slope formula. Plugging them we have:

2) Note that we had to add 1 to both sides, to keep the "y "on the left.
3) Therefore, y=-9x+10 is the answer.