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3 years ago

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1 answer:

iren2701 [21]3 years ago

8 0

Are these the actual figures?

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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

$P(t) = 300e^{rt}$

$1000 = 300e^{24r}$

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

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

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

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

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

$r = 0.05$

So

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

At what time t is the population 500?

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

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

$500 = 300e^{0.05t}$

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

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

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

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

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

$t = 10.22$

The population is of 500 after 10.22 hours.

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3 years ago

Find the zeros of polynomial function and solve polynomials equations.<br> f(x)=x^2-x-90

victus00 [196]

The zeros are:

10 and -9

(ten and negative-nine)

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3 years ago

What is 2.333333 in fraction form

iragen [17]

$x=2.(3)\ \ \ /\cdot10\\\\10x=23.(3)\\\\10x-x=23.(3)-2.(3)\\\\9x=21\ \ \ /:9\\\\x=\frac{21}{9}\\\\x=\frac{7}{3}\\\\x=2\frac{1}{3}$

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Cara has $25 to buy dry pet food and treats for the animal shelter. A pound of dog food cost $2 and treats are $1 apiece. If she

nordsb [41]

<span>If dog food costs $2 a pound and Cara buys 9 pounds, her expenditure on dog food is 9 x 2. This makes $18. If she has a total budget of $25 and spends $18 on dog food, she has $7 left over for treats (25 - 18 = 7). At $1 each, she can buy up to 7 treats.</span>

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Jenny paid $17 for 2 large boxes of popcorn and 3 large soft drinks. Justin paid $10 for 1 large box of popcorn and 2 large soft

wolverine [178]

2p + 3d = 17

1p + 2d = 10

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3

D is the correct answer

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3 years ago