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eimsori [14]
3 years ago
9

During summer, the height of the water in a pool decreased by 2 inches each week due to evaporation. What is the change in the h

eight of the water over an 8 week period??
Mathematics
1 answer:
Alecsey [184]3 years ago
8 0

Answer:

16in or 1ft 4in

Step-by-step explanation:

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The marginal price per pound (in dollars) at which a coffee store is willing to supply x pounds of Jamaican Blue Mountain coffee
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Answer:

Thus, the coffee shop is willing to supply 6 pounds per week at a price of $4 per pound.

Step-by-step explanation:

We are given the following information in the question:

The marginal price per pound (in dollars) is given by:

p'(x) = \displaystyle\frac{208}{(x+7)^2}

where x is the supply in pounds.

P(x) = \displaystyle\int p'(x)~dx =\displaystyle\int\displaystyle\frac{208}{(x+7)^2}~dx\\\\P(x) = \frac{-208}{(x+7)} + c\\\\\text{where c is the constant of integration.}

The coffee shop is willing to supply 9 pounds per week at a price of $7 per pound.

Thus, we are given that

P(9) = 7

Putting the values, we get,

P(x) = \displaystyle\frac{-208}{(x+7)} + c\\\\P(9) = 7\\\\\displaystyle\frac{-208}{(9+7)} + c = 7\\\\c = 7 + \frac{208}{16} = 20

P(x) = \displaystyle\frac{-208}{(x+7)} + 20

Now, we have to find how many pounds it would be willing to supply at a price of $4 per pound.

P(x) = 4

P(x) = \displaystyle\frac{-208}{(x+7)} + 20 = 4\\\\\frac{-208}{x+7} = -16\\\\x + 7 = 13\\x = 6

Thus, the coffee shop is willing to supply 6 pounds per week at a price of $4 per pound.

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3 years ago
6. A tank contains 100-gallon of pure water. At time t = 0, a solution containing 2 lb of salt per gallon flows into the tank at
SOVA2 [1]

Answer:

The answer is below

Step-by-step explanation:

Let Q represent the amount of salt in the tank at time t.

\frac{dQ}{dt}= flow\ in - flow\ out\\ \\flow\ in=3\ gal/min*2\ lb/gal=6\ lb/min\\\\net\ gain\ in\ tank\ volume=3-4=-1, henceflow\ out= \frac{4Q}{100-t} \\\\\frac{dQ}{dt}= 6-\frac{4Q}{100-t} \\\\\frac{dQ}{dt}+ \frac{4Q}{100-t}=6\\\\The \ integrating\ factor\ is:\\\\IF=e^{\int\limits {\frac{4}{100-t} } \, dt }=e^{-4\int\limits {\frac{-1}{100-t}}=e^{-4ln(100-t)}=(100-t)^{-4}}\\\\Multiplying\ through  \ by\ IF: \\\\(100-t)^{-4}\frac{dQ}{dt}+ (100-t)^{-4}\frac{4Q}{100-t}=6(100-t)^{-4}\\\\

Integrating:\\\\A(100-t)^{-4}=-2(100-t)^{-3}+c\\\\A=-2(100-t)+\frac{c}{(100-t)^{-4}} \\\\at, t=0,A=0\\\\0=-2(100-0)+\frac{c}{(100-0)^{-4}}\\\\c=0.02\\\\A=-2(100-t)+\frac{0.02}{(100-t)^{-4}}

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