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enot [183]
2 years ago
12

On last week's climb, each climber carried a pack that was 28% of his or her body weight. One climber weighed 78 pounds. What wa

s the weight of his pack?
Mathematics
1 answer:
11Alexandr11 [23.1K]2 years ago
8 0
<h3><u>Solution</u> : </h3>

weight of bag :

  • 28   \% \:  \: of \:  \:  \: climber's \:  \: weight

  • \dfrac{28}{100}  \times 78

  • \dfrac{2184}{100}

  • 21.84 \:  \:  \: pounds
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The angle of depression from a helicopter to a landing pad is 65 if the horizantal distance from the helicopter to the landing p
MrRissso [65]

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During optimal conditions, the rate of change of the population of a certain organism is proportional to the population at time
Lana71 [14]

Answer:

The population is of 500 after 10.22 hours.

Step-by-step explanation:

The rate of change of the population of a certain organism is proportional to the population at time t, in hours.

This means that the population can be modeled by the following differential equation:

\frac{dP}{dt} = Pr

In which r is the growth rate.

Solving by separation of variables, then integrating both sides, we have that:

\frac{dP}{P} = r dt

\int \frac{dP}{P} = \int r dt

\ln{P} = rt + K

Applying the exponential to both sides:

P(t) = Ke^{rt}

In which K is the initial population.

At time t = 0 hours, the population is 300.

This means that K = 300. So

P(t) = 300e^{rt}

At time t = 24 hours, the population is 1000.

This means that P(24) = 1000. We use this to find the growth rate. So

P(t) = 300e^{rt}

1000 = 300e^{24r}

e^{24r} = \frac{1000}{300}

e^{24r} = \frac{10}{3}

\ln{e^{24r}} = \ln{\frac{10}{3}}

24r = \ln{\frac{10}{3}}

r = \frac{\ln{\frac{10}{3}}}{24}

r = 0.05

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P(t) = 300e^{0.05t}

At what time t is the population 500?

This is t for which P(t) = 500. So

P(t) = 300e^{0.05t}

500 = 300e^{0.05t}

e^{0.05t} = \frac{500}{300}

e^{0.05t} = \frac{5}{3}

\ln{e^{0.05t}} = \ln{\frac{5}{3}}

0.05t = \ln{\frac{5}{3}}

t = \frac{\ln{\frac{5}{3}}}{0.05}

t = 10.22

The population is of 500 after 10.22 hours.

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