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2 years ago

The population of a town increased from 5,600 to 6,300 people. What was the percent of increase?

Mathematics

2 answers:

tatuchka [14]2 years ago

5 0

Answer:

12.5%

Step-by-step explanation:

To find the percent increase, we will need to find the difference between the two populations.

6300 - 5600 = 700 people.

Now, divide this increase by the original population:

$\frac{700}{5600} = 0.125$

Multiply this by 100 to get your percent of increase:

0.125 × 100 = 12.5%

Marat540 [252]2 years ago

4 0

Answer:

13%

Step-by-step explanation:

not for sure but i think

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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 µ(x, y) 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 µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

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

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 F(x, y) = C. 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 x :

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

Differentiate both sides of this with respect to y :

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

Then the general solution to the DE is

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

5 0

2 years ago

G must be which measure for this quadrilateral to be a parallelogram?

PSYCHO15rus [73]

Answer:

Step-by-step explanation:

Because Angle G needs to be the Same as I to be a parallelogram.

Sorry about my bad Explanation. :(

8 0

2 years ago

Read 2 more answers

How are dot plots, bar graphs, and histograms used to analyze data?

cestrela7 [59]

Dot plots, bar graphs, and histograms are visual representations of data. They present data in a way that makes it easier to analyze and compare data. They allow understanding data with ease by visually showing relations, trends and other useful data

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2 years ago

Kobe, Amari, and DJ raised $125.45 for their football team. Kobe collected $24.55 more

vekshin1

Answer:

x+8.95 and x+24.55

Step-by-step explanation:

im not going to do this

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2 years ago

Which of the following coordinates are not correctly named based on the image?

timama [110]

Answer:

Hi, there the answer is (4,2)

Step-by-step explanation:

The reason is that 4 is the y-axis and 2 is the x-axis and

the rule is x is first then y

The correct format for writing this it is supposed to be (2,4).

Therefore, your answer will (4,2)

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2 years ago