Answer:
Total peanut=324
Step-by-step explanation:
Let the peanut=x
Philip took 1/3 of x
Remaining peanut=x-1/3x
=3x-x/3
=2/3x
Joy took 1/4 of the remaining leaving 3/4
Remaining=(3/4)(2x/3)
=6x/12
=1/2x
Brent took 1/2 of those, leaving 1/2
=(1/2)*(1/2x)
=1/4x
Preston took 10
Remaining
1/4x - 10 = 71
1/4x = 81
x=81÷1/4
=81*4/1
x = 324
Total peanut =324
Philip=1/3x
=1/3*324
=108
Remaining 324-108
=216
Joy=1/4 of 216
=54
Remaining
324-108-54=162
Brent=1/2 of 162
=81
Preston=10
Total taken=Philip+joy+Brent+Preston
=108+54+81+10
=253
Total remaining=Total peanut - total taken
=324-253
=71
Mark: 52 meatball sandwiches
Julie: 64 meatball sandwiches
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:
5) √(3² + 5²) = √34
6) √(4² - 3²) = √7
7) the visible answers all say exactly the same thing in different ways and are all correct.
Step-by-step explanation: