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2 years ago

Round 17242 to the nearest ten thousandth

Mathematics

1 answer:

MrRa [10]2 years ago

7 0

Its 20,000. If you look of make a number line use numbers 17,242 - 20,000… <3

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If the function h is defined by h(x)=<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" a

harkovskaia [24]

Given:

The function is:

$h(x)=x^2-3x+5$

To find:

The value of $h(2x+1)$ .

Solution:

We have,

$h(x)=x^2-3x+5$

Putting $x=2x+1$ , we get

$h(2x+1)=(2x+1)^2-3(2x+1)+5$

$h(2x+1)=(2x)^2+2(2x)(1)+(1)^2-3(2x)-3(1)+5$

$h(2x+1)=4x^2+4x+1-6x-3+5$

On combining like terms, we get

$h(2x+1)=4x^2+(4x-6x)+(1-3+5)$

$h(2x+1)=4x^2-2x+3$

Therefore, the required function is $h(2x+1)=4x^2-2x+3$ .

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3 years ago

What is 6.04 in simplest form

Reptile [31]

6 1/25 is the answer

6 0

3 years ago

100 points!!! Which graph represents an exponential function?

drek231 [11]

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2 years ago

Read 2 more answers

(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 :

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