Area of a rectangle = W X L
Area of a rectangle = 5 X 2
Area of a rectangle = 10
Area of triangle 1= 1/2 X B X H
Area of triangle 1= 1/2 X 2 X 2
Area of triangle 1= 1/2 X 4
Area of triangle 1= 2
Area of triangle 2= 1/2 X B X H
Area of triangle 2= 1/2 X 7 X 4
Area of triangle 2= 1/2 X 28
Area of triangle 2= 14
Area of a rectangle + Area of triangle 1 + Area of triangle 2=
10 + 2 + 14 = 36
If you had 1 model and it equaled 100. You then would have 32 to equal 3200, because 32 X 100 = 3200. The answer would be 32 models to get you 3200.
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
9*13+n
I think that is correct, try and split up the question.
9*
A sum of a number AND 13
So that means that 9 is multiplying by the SUM (adding) of 13 and a number (n)
