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jok3333 [9.3K]
2 years ago
9

The equation P = 16,250 + 75n can be used to determine the population, P, of a city after n years. Which statements are true? Se

lect ALL that apply. The initial population was 16,250 people. The initial population was 16,250 people. The population after 10 years was 16,325 people. The population after 10 years was 16,325 people. The population increases by 75 people each year. The population increases by 75 people each year. The population increases by 16,250 people each year. The population increases by 16,250 people each year. The population increases by 16,250 people every 75 years. The population increases by 16,250 people every 75 years.
Mathematics
1 answer:
I am Lyosha [343]2 years ago
4 0

Answer:

15000+1250x

Step-by-step explanation:

Since the initial population is 15,000, we want to know how many people are going to Jonesville each year.  Assuming that in 2001 and in 2002, each year adds 1,250 people and that it is constant.  The equation is 15000+1250x in which 15,000 represents the initial number and 1250 is the amount of people added each year followed by x which is the number in years.

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142.6 divided by 2.3
AveGali [126]

Answer:

62

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3 years ago
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You have $1,500 to spend over a five month semester to cover your living expenses. In the first month, you spend $150 at restaur
alekssr [168]

Answer:

$250

Step-by-step explanation:

1st month you spent $300. And the flight back is $200. 300+200=500. 1500-500=1000. Yoy have 1000 dollars to spend in 4 months. 1000/4=250, so each month you can spend $250.

6 0
3 years ago
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(x^2y+e^x)dx-x^2dy=0
klio [65]

It looks like the differential equation is

\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0

Check for exactness:

\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x

As is, the DE is not exact, so let's try to find an integrating factor <em>µ(x, y)</em> such that

\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0

*is* exact. If this modified DE is exact, then

\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}

We have

\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu

Notice that if we let <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}

The modified DE,

\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0

is now exact:

\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}

So we look for a solution of the form <em>F(x, y)</em> = <em>C</em>. This solution is such that

\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}

Integrate both sides of the first condition with respect to <em>x</em> :

F(x,y) = -e^{-x}y - \dfrac1x + g(y)

Differentiate both sides of this with respect to <em>y</em> :

\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C

Then the general solution to the DE is

F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}

5 0
3 years ago
If you vertically stretch the exponential function f(x)=2^x by a factor of 3, what is the equation of the new function
AleksAgata [21]

Answer:

y=3*2^{x}

Step-by-step explanation:

A vertical stretch of a function means the output values have changed by a factor of 3 or multiplication by 3. Recall, an exponential function has the basic form

y=ab^{2}.

If our equation is f(x)=2^x, then a=1. To stretch it vertically by a factor of 3, we multiply a by 3. So 1(3)=3. The value of a now becomes 3.

y=3*2^{x}


7 0
3 years ago
The number of students who picked each subject is shown.
Naddik [55]

Answer: 90% of students did NOT choose history.

Step-by-step explanation:

Art - 32.5%

Spanish - 7.5%

ELA - 10%

Math - 15%

Science - 25%

When added altogether each subject equals 90%, leaving the other 10 percent history. 90% is our answer since that is the number of people that voted for subjects other than history.

Hope This Helped, Have A Great Day!

5 0
3 years ago
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