Answer:
96 i think
Step-by-step explanation:
i did 480/5
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
The distance depends on the time.
So now we need to find the constant of variation or the constant of proportionality.
To do so we must find y(the dependent variable) and x(the independent variable).
To find the constant(k) we must find y/x
Since y/x=k
Then 2.25/.75=k
k=3
1 gal = 3.8 liters....so 65 gallons = 65 * 3.8 = 247 liters
Which is the domain of the function shown in the graph?
A. x ≤ 3
B. -∞ < x < ∞ ( ✔ )
C. x > 0
D. x ≥ 3