I need help on question 4!! geometry unit 5.
1 answer:
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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:
None of these.
Step-by-step explanation:
Let's assume we are trying to figure out if (x-6) is a factor. We got the quotient (x^2+6) and the remainder 13 according to the problem. So we know (x-6) is not a factor because the remainder wasn't zero.
Let's assume we are trying to figure out if (x^2+6) is a factor. The quotient is (x-6) and the remainder is 13 according to the problem. So we know (x^2+6) is not a factor because the remainder wasn't zero.
In order for 13 to be a factor of P, all the terms of P must be divisible by 13. That just means you can reduce it to a form that is not a fraction.
If we look at the first term x^3 and we divide it by 13 we get we cannot reduce it so it is not a fraction so 13 is not a factor of P
None of these is the right option.