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:
ST=10
Step-by-step explanation:
Answer:
The home would be worth $249000 during the year of 2012.
Step-by-step explanation:
The price of the home in t years after 2004 can be modeled by the following equation:

In which P(0) is the price of the house in 2004 and r is the growth rate.
Since 2003 median home prices in Midvale, UT have been growing exponentially at roughly 4.7 % per year.
This means that 
$172000 in 2004
This means that 
What year would the home be worth $ 249000 ?
t years after 2004.
t is found when P(t) = 249000. So







2004 + 8.05 = 2012
The home would be worth $249000 during the year of 2012.
S/4-6.8=-9.8
S/4= -9.8+6.8
S/4= -3
S=-3*4
S= -12
Hello,
y=k*(x-2)(x-4)
and is passing throught (3,1)
==>1=k*(3-2)(3-4)==>k=-1
y=-(x-2)(x-4) is an answer