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algol13
3 years ago
9

Evaluate5− t/3 when t=12 ​

Mathematics
2 answers:
Novosadov [1.4K]3 years ago
8 0

Answer:

1

Step-by-step explanation:

5 -  \frac{t}{3} =  \frac{5}{1} - \frac{12}{3} = 5 - 4 = 1

or

5 -  \frac{t}{3} =  \frac{5}{1} -  \frac{12}{3} =  \frac{15 - 12}{3} =  \frac{3}{3} = 1

Hope this helps ;) ❤❤❤

Nikitich [7]3 years ago
6 0
Answer would be 1:) I hope this helps you!
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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
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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

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

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Applying the exponential to both sides:

P(t) = Ke^{rt}

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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}

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So

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}

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\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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3 years ago
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Answer:

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