7*24 = 168 quarts of water per day goes into barrel.
25*48 = 1200 quarts of waste water picked up
1200÷168 =7.142857143
Answer:
b ≈ 9.5, c ≈ 14.7
Step-by-step explanation:
Using the Sine rule in Δ ABC, that is
=
, substitute values
=
( cross- multiply )
b × sin23° = 7 × sin32° ( divide both sides by sin23° )
b =
≈ 9.5 ( to the nearest tenth )
Also
= 
=
( cross- multiply )
c × sin23° = 7 × sin125° ( divide both sides by sin23° )
c =
≈ 14.7 ( to the nearest tenth )
-8w+3z
w=2.2
z=-9.1
-8(2.2)+3(-9.1)
-17.6-27.3
-44.9
The answer is -44.9
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:
85.6cm
Step-by-step explanation:
Area of a triangle =1/2*b*h
Where base=19.5cm while the height is 8 4/5 = 8.8
Area = 1/2 * 19.5cm * 8.8cm
Area = 171.6cm/2
Area=85.6cm