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Vitek1552 [10]
3 years ago
14

Between 11 P.M. and 8:36 A.M.​, the water level in a swimming pool decreased by 8 25 in. Assuming that the water level decreased

at a constant​ rate, how much did the water level drop each​ hour?
Mathematics
1 answer:
Eva8 [605]3 years ago
6 0

Answer:

0.8594 in/hr

Step-by-step explanation:

Decreased in the water level is 8.25 in.

Time 11 P.M. and 8:36 A.M.  is 9 hr 36 min = 9.6 hr.

The rate of decreasing water level is constant.

So, \text{Rate}=\frac{\text{Level decreased}}{\text{Time taken}}

\Rightarrow \text{Rate}=\frac{8.25}{9.6}=0.8594 in/hr.

Hence, the water level drop at the rate of 0.8594 in/hr.

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A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int
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Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

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By expanding the previous equation:

c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)

The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:

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Since there is no accumulation within the tank, expression is simplified to this:

c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}

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c(t) = \frac{c_{0}}{f_{out}} \cdot ({1-e^{-\frac{f_{out}}{V_{tank}}\cdot t }})

The salt concentration after 8 minutes is:

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The instantaneous amount of salt in the tank is:

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Answer:

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