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Dmitry_Shevchenko [17]
3 years ago
14

Jada walks up to a tank of water that can hold up to 10 gallons. When it is active, a drain empties water from the tank at a con

stant rate. When Jada first sees the tank, it contains 8 gallons of water. Three minutes later, the tank contains 5 gallons of water. At what rate is the amount of water in the tank changing? (How many gallons per minute?)
How many more minutes will it take for the tank to drain completely? *
How many minutes before Jada arrived was the water tank completely full? Will give brainliest
Mathematics
1 answer:
mars1129 [50]3 years ago
5 0

Answer:

1 gallon per minute

it will take 5 minutes to completely drain

2 minutes before Jada arrived

Step-by-step explanation:

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Answer: A) y=13e^{-0.1359t}

              B) H = 5.10

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Step-by-step explanation: <u>Exponential</u> <u>Decay</u> <u>function</u> is a model that describes the reducing of an amount by a constant rate over time. Generally, it is written in the form: y(t)=Ce^{rt}

A) C is initial quantity, in this case, the initial concentration of DDT. To determine r, using the data given:

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e^{13r}=0.1708

Using a natural logarithm property called <em>power rule:</em>

13r=ln(0.1708)

r=\frac{ln(0.1708)}{13}

r=-0.1359

The decay function for concentration of DDT through the years is y(t)=13e^{-0.1359t}

B) The value of H is calculated by y=C(0.5)^{\frac{t}{H} }

2.22=13(0.5)^{\frac{13}{H} }

(0.5)^{\frac{13}{H} }=0.1708

Again, using power rule for logarithm:

\frac{13}{H} log(0.5)=log(0.1708)

\frac{13}{H} =\frac{log(0.1708)}{log(0.5)}

\frac{13}{H} =2.55

H = 5.10

Constant H in the half-life formula is H=5.10

C) Using model y(t)=13e^{-0.1359t} to determine concentration of DDT in 1995:

y(24)=13e^{-0.1359.24}

y(24) = 0.5

By 1995, the concentration of DDT is 0.5 ppm, so using this model is possible to reduce such amount and more of DDT.

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