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2 years ago

A city's population is represented by the function P=25,000(1.0095)t , where t is time in years.

Mathematics

2 answers:

dedylja [7]2 years ago

6 0

Daily growth rate: 0.003%

P = 25,000 [1.0095^(1/365)]^(t / 365)

Sophie [7]2 years ago

6 0

Answer:

The approximate daily growth rate is 0.0026%

The function can be rewritten as P = 25,000*[1.0095^(1/365)]^t, where now t is in days

Step-by-step explanation:

Given the formula:

P = 25,000*(1.0095)^t

25,000 indicates the initial population, t is the time elapsed in years, and P is the population after t years. After 1 year the population will be:

P = 25,000*(1.0095)

which is equivalent to:

P = 25,000*100.95%

that represents an increment of 0.95 % in a year. Given that the year has 365 days, then this represent a daily growth of 0.95/365 = 0.0026%

Dividing t by 365 in the original expression, so that, time is expressed in days, we get:

P = 25,000*(1.0095)^(t/365)

Which can be rewritten as:

P = 25,000*[1.0095^(1/365)]^t

P = 25,000*(1.000026)^t

That represents a daily growth rate of 0.0026% and now t is in days

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Read 2 more answers

Find the dimensions of the box with volume 1728 cm3 that has minimal surface area. (Let x, y, and z be the dimensions of the box

Gala2k [10]

Answer:

The dimensions of the box are 12 cm , 12 cm , 12 cm

Step-by-step explanation:

Let x , y and z be the dimensions of box

Volume of box =xyz=1728

$z=\frac{1728}{xy}$

Surface area of box = $2xy+2yz+2xz=2xy+2y(\frac{1728}{xy})+2x(\frac{1728}{xy})$

Let $f(x,y)=2xy+2(\frac{1728}{x})+2(\frac{1728}{y})$

To get minimal surface area

$\frac{\partial f}{\partial x}=0$ and $\frac{\partial f}{\partial y}=0$

$\frac{\partial(2xy+2(\frac{1728}{x})+2(\frac{1728}{y}))}{\partial x}=0$

$2y-2(\frac{1728}{x^2})=0$

$y=\frac{1728}{x^2}$ ----1

$\frac{\partial(2xy+2(\frac{1728}{x})+2(\frac{1728}{y}))}{\partial y}=0$

$2x-2(\frac{1728}{y^2})=0\\x=\frac{1728}{y^2} \\y^2=\frac{1728}{x}$

Using 1

$(\frac{1728}{x^2} )^2=\frac{1728}{x}$

x=0 and $x^3=1728$

Side can never be 0

So, $x^3=1728$

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1 year ago

Solve for x. Round to the nearest tenth, if necessary.

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1 year ago

Investigate the following harvesting model both qualitatively and analytically. If a constant number h of fish are harvested fro

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Answer:

a. The population does not become extinct in finite time.

Step-by-step explanation:

The model for the population of the fishery is

$dP/dt = P(a-bP)-h, P(0) = P_0$

If we rearrange and replace the constants we have:

$\frac{dP}{P(7-P)-49/4} =dt\\\\-4 (\frac{dP}{4(P-7)P+49}) =dt\\\\-4 \frac{dP}{(2P-7)^2} =dt\\\\-4 \int\frac{dP}{(2P-7)^2} =\int dt\\\\-4(-\frac{1}{2(2P-7)})=t+C\\\\\frac{2}{2P-7}=t+C\\\\ t=0 \,\,\, P(0)=P_0\\\\\frac{2}{2P_0-7}=0+C\\\\C=\frac{2}{2P_0-7}$

Now we can calculate if the population become 0 in any finite time

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To be a finite time, t>0

$t=-\frac{2}{7}-\frac{2}{2P_0-7}=0\\\\-\frac{2}{2P_0-7}=\frac{2}{7}\\\\7-2P_0=7\\\\P_0=0$

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Because P0 is a positive constant, we can say that the population does not become extint in finite time.

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