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:
8 weeks
Step-by-step explanation:
So first you find out how much money he makes per week, by doing 7.5 x 20 you will find that. Now you need to find how many weeks it will take, find this by dividing 1200 by the amount he makes per week and you will end up with 8.
Answer:
3v-v² = v(3-v)
Step-by-step explanation:
Use distributive property on the right side of the equality.
Answer:
For the tickets sold collum, your going to do 1,2,3,4,5,6,7 and for the total revenue, your going to do 34.00 68.00 102.00 136.00 170.00 204.00 238.00
Step-by-step explanation:
Then for ordered pairs, you´re going to do 1, 34.00 2, 68.00 3, 102.00 and so on. Then you graph it, oh and K= 34.00