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Bogdan [553]
3 years ago
13

Rope on a boat. A rope passing through a capstan on a dock is attached to a boat offshore. The rope is pulled in at a constant r

ate of 3ft/s and the capstan is 5ft vertically above the water. How fast is the boat traveling when it is 10ft from the dock?
Mathematics
1 answer:
katrin2010 [14]3 years ago
8 0
If it is being pulled in at a constant rate of 3ft/sec, then it is travelling 3 ft/sec is it not? knot? lol
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Ben consumes an energy drink that contains caffeine. After consuming the energy drink, the amount of caffeine in Ben's body decr
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Answer:

The 5-hour decay factor for the number of mg of caffeine in Ben's body is of 0.1469.

Step-by-step explanation:

After consuming the energy drink, the amount of caffeine in Ben's body decreases exponentially.

This means that the amount of caffeine after t hours is given by:

A(t) = A(0)e^{-kt}

In which A(0) is the initial amount and k is the decay rate, as a decimal.

The 10-hour decay factor for the number of mg of caffeine in Ben's body is 0.2722.

1 - 0.2722 = 0.7278, thus, A(10) = 0.7278A(0). We use this to find k.

A(t) = A(0)e^{-kt}

0.7278A(0) = A(0)e^{-10k}

e^{-10k} = 0.7278

\ln{e^{-10k}} = \ln{0.7278}

-10k = \ln{0.7278}

k = -\frac{\ln{0.7278}}{10}

k = 0.03177289938&#10;

Then

A(t) = A(0)e^{-0.03177289938t}

What is the 5-hour growth/decay factor for the number of mg of caffeine in Ben's body?

We have to find find A(5), as a function of A(0). So

A(5) = A(0)e^{-0.03177289938*5}

A(5) = 0.8531

The decay factor is:

1 - 0.8531 = 0.1469

The 5-hour decay factor for the number of mg of caffeine in Ben's body is of 0.1469.

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