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3 years ago

Which equation represents the line that passes through (0, 2) and (3, 1)?

Mathematics

1 answer:

Anna11 [10]3 years ago

6 0

<h3>Answer:</h3>

A. y = -1/3x + 2 . . . . . . . . assuming a typo in your problem statement

<h3>Explanation:</h3>

The point (0, 2) is the y-intercept. This tells you the equation will be something of the form ...

... y = mx + 2 . . . . . . matching only answer choices A and C

_____

Knowing this, you can either substitute the values x=3, y=1 into these two equations to see which one works (it is not C), or you can look at the two points and determine the value of the slope.

The slope is ...

... (difference in y)/(difference in x) = (1-2)/(3-0) = -1/3 . . . . . matches choice A

The equation of the line is ...

... y = -1/3x + 2 . . . . . matches answer choice A

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3 years ago

The population of mosquitoes in a certain area increases at a rate proportional to the current pop-ulation, and in the absence o

ollegr [7]

Answer:

The population of mosquitoes in the area at any time <em>t</em> is:

$P(t)=201977.31-1977.31\times 2^{t}$

Step-by-step explanation:

The rate of growth of mosquitoes can be expressed as:

$\frac{dP}{dt}=kP$

$\frac{dP}{P}=k\ dt$

Integrate the above expression as follows:

$\int {\frac{dP}{P}} \, =\int {k\ dt} \, \\\ln|P|=kt+c\\e^{\ln|P|}=e^{kt+c}\\P=Ce^{kt}$

$\Rightarrow P=P_{0}e^{kt}$

It is provided that the population doubles every day.

Compute the value of <em>k</em> as follows:

$2=1\times e^{k\times1}\\2=e^{k}\\k=\ln (2)$

It is also provided that every day 20,000 mosquitoes are eaten.

The rate of growth per week can be expressed as:

$\frac{dP}{dt}=\ln(2)P-14000\\\frac{dP}{dt}-\ln(2)P=14000$

The integrating factor for this is:

$e^{\int {\ln(2)dP}}=e^{\ln(2)\int {dt}}=e^{\ln(2)t}$

Then,

$P(t)\ e^{-\ln(2)t}=\int {e^{-\ln(2)t}}-14000\, dt\\=-14000\int {e^{-\ln(2)t}}\, dt\\=-14000\times \frac{e^{-\ln(2)t}}{-\ln(2)}+C\\P(t)=(e^{-\ln(2)t})\times [-14000\times \frac{e^{-\ln(2)t}}{-\ln(2)}+C]\\=\frac{14000}{\ln(2)}+Ce^{-\ln(2)t}$

The initial population is 200,000.

Compute the value of <em>C</em> as follows:

$P(t)=\frac{140000}{\ln(2)}+Ce^{-\ln(2)t}\\200000=\frac{14000}{\ln(2)}+Ce^{-\ln(2)(0)}\\C=200000-\frac{140000}{\ln(2)}\\C=-1977.31$

Now substitute <em>C</em> in P (t),

$P(t)=\frac{140000}{\ln(2)}+Ce^{\ln(2)t}\\P(t)=201977.31-1977.31\times 2^{t}$

6 0

3 years ago