what is the square root of 10,000?
100 is square root of 10 ,000
if a number is multiplied by itself then the product obtained is called is square root of the number
if like x × x = x²
the square of an integer is a called a perfect square or perfect second power
the square root of a number is the number which when multiplied by itself given as the product
Based on the definition of integers, ordering them would depend on if they are negative, positive, or have a zero value.
<h3>How are integers ordered?</h3><h3 />
Integers are whole numbers such as 1, 15, and 55. There are no decimals and they do not come in the form of fractions.
Integers can be negative or positive. Positive integers are always higher than negative integers. For instance, 1 is more than -500,000.
If both integers are negative, the larger looking number is considered smaller. For instance, -5 is more than -55. For positive integers, this is the reverse with larger looking numbers being larger.
An example of the correct order of intergers based on this dataset (1, 52, -800, 86, 5, and -4) is:
= -800, -4, 1, 5, 52, 86
Find out more on ordering integers at brainly.com/question/12399107.
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Check if the equation is exact, which happens for ODEs of the form
![M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0](https://tex.z-dn.net/?f=M%28x%2Cy%29%5C%2C%5Cmathrm%20dx%2BN%28x%2Cy%29%5C%2C%5Cmathrm%20dy%3D0)
if
.
We have
![M(x,y)=x^2+y^2+x\implies\dfrac{\partial M}{\partial y}=2y](https://tex.z-dn.net/?f=M%28x%2Cy%29%3Dx%5E2%2By%5E2%2Bx%5Cimplies%5Cdfrac%7B%5Cpartial%20M%7D%7B%5Cpartial%20y%7D%3D2y)
![N(x,y)=xy\implies\dfrac{\partial N}{\partial x}=y](https://tex.z-dn.net/?f=N%28x%2Cy%29%3Dxy%5Cimplies%5Cdfrac%7B%5Cpartial%20N%7D%7B%5Cpartial%20x%7D%3Dy)
so the ODE is not quite exact, but we can find an integrating factor
so that
![\mu(x,y)M(x,y)\,\mathrm dx+\mu(x,y)N(x,y)\,\mathrm dy=0](https://tex.z-dn.net/?f=%5Cmu%28x%2Cy%29M%28x%2Cy%29%5C%2C%5Cmathrm%20dx%2B%5Cmu%28x%2Cy%29N%28x%2Cy%29%5C%2C%5Cmathrm%20dy%3D0)
<em>is</em> exact, which would require
![\dfrac{\partial(\mu M)}{\partial y}=\dfrac{\partial(\mu N)}{\partial x}\implies \dfrac{\partial\mu}{\partial y}M+\mu\dfrac{\partial M}{\partial y}=\dfrac{\partial\mu}{\partial x}N+\mu\dfrac{\partial N}{\partial x}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%28%5Cmu%20M%29%7D%7B%5Cpartial%20y%7D%3D%5Cdfrac%7B%5Cpartial%28%5Cmu%20N%29%7D%7B%5Cpartial%20x%7D%5Cimplies%20%5Cdfrac%7B%5Cpartial%5Cmu%7D%7B%5Cpartial%20y%7DM%2B%5Cmu%5Cdfrac%7B%5Cpartial%20M%7D%7B%5Cpartial%20y%7D%3D%5Cdfrac%7B%5Cpartial%5Cmu%7D%7B%5Cpartial%20x%7DN%2B%5Cmu%5Cdfrac%7B%5Cpartial%20N%7D%7B%5Cpartial%20x%7D)
![\implies\mu\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)=M\dfrac{\partial\mu}{\partial y}-N\dfrac{\partial\mu}{\partial x}](https://tex.z-dn.net/?f=%5Cimplies%5Cmu%5Cleft%28%5Cdfrac%7B%5Cpartial%20N%7D%7B%5Cpartial%20x%7D-%5Cdfrac%7B%5Cpartial%20M%7D%7B%5Cpartial%20y%7D%5Cright%29%3DM%5Cdfrac%7B%5Cpartial%5Cmu%7D%7B%5Cpartial%20y%7D-N%5Cdfrac%7B%5Cpartial%5Cmu%7D%7B%5Cpartial%20x%7D)
Notice that
![\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}=y-2y=-y](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20N%7D%7B%5Cpartial%20x%7D-%5Cdfrac%7B%5Cpartial%20M%7D%7B%5Cpartial%20y%7D%3Dy-2y%3D-y)
is independent of <em>x</em>, and dividing this by
gives an expression independent of <em>y</em>. If we assume
is a function of <em>x</em> alone, then
, and the partial differential equation above gives
![-\mu y=-xy\dfrac{\mathrm d\mu}{\mathrm dx}](https://tex.z-dn.net/?f=-%5Cmu%20y%3D-xy%5Cdfrac%7B%5Cmathrm%20d%5Cmu%7D%7B%5Cmathrm%20dx%7D)
which is separable and we can solve for
easily.
![-\mu=-x\dfrac{\mathrm d\mu}{\mathrm dx}](https://tex.z-dn.net/?f=-%5Cmu%3D-x%5Cdfrac%7B%5Cmathrm%20d%5Cmu%7D%7B%5Cmathrm%20dx%7D)
![\dfrac{\mathrm d\mu}\mu=\dfrac{\mathrm dx}x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%5Cmu%7D%5Cmu%3D%5Cdfrac%7B%5Cmathrm%20dx%7Dx)
![\ln|\mu|=\ln|x|](https://tex.z-dn.net/?f=%5Cln%7C%5Cmu%7C%3D%5Cln%7Cx%7C)
![\implies \mu=x](https://tex.z-dn.net/?f=%5Cimplies%20%5Cmu%3Dx)
So, multiply the original ODE by <em>x</em> on both sides:
![(x^3+xy^2+x^2)\,\mathrm dx+x^2y\,\mathrm dy=0](https://tex.z-dn.net/?f=%28x%5E3%2Bxy%5E2%2Bx%5E2%29%5C%2C%5Cmathrm%20dx%2Bx%5E2y%5C%2C%5Cmathrm%20dy%3D0)
Now
![\dfrac{\partial(x^3+xy^2+x^2)}{\partial y}=2xy](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%28x%5E3%2Bxy%5E2%2Bx%5E2%29%7D%7B%5Cpartial%20y%7D%3D2xy)
![\dfrac{\partial(x^2y)}{\partial x}=2xy](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%28x%5E2y%29%7D%7B%5Cpartial%20x%7D%3D2xy)
so the modified ODE is exact.
Now we look for a solution of the form
, with differential
![\mathrm dF=\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0](https://tex.z-dn.net/?f=%5Cmathrm%20dF%3D%5Cdfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20x%7D%5C%2C%5Cmathrm%20dx%2B%5Cdfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y%7D%5C%2C%5Cmathrm%20dy%3D0)
The solution <em>F</em> satisfies
![\dfrac{\partial F}{\partial x}=x^3+xy^2+x^2](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20x%7D%3Dx%5E3%2Bxy%5E2%2Bx%5E2)
![\dfrac{\partial F}{\partial y}=x^2y](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y%7D%3Dx%5E2y)
Integrating both sides of the first equation with respect to <em>x</em> gives
![F(x,y)=\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+f(y)](https://tex.z-dn.net/?f=F%28x%2Cy%29%3D%5Cdfrac%7Bx%5E4%7D4%2B%5Cdfrac%7Bx%5E2y%5E2%7D2%2B%5Cdfrac%7Bx%5E3%7D3%2Bf%28y%29)
Differentiating both sides with respect to <em>y</em> gives
![\dfrac{\partial F}{\partial y}=x^2y+\dfrac{\mathrm df}{\mathrm dy}=x^2y](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y%7D%3Dx%5E2y%2B%5Cdfrac%7B%5Cmathrm%20df%7D%7B%5Cmathrm%20dy%7D%3Dx%5E2y)
![\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C](https://tex.z-dn.net/?f=%5Cimplies%5Cdfrac%7B%5Cmathrm%20df%7D%7B%5Cmathrm%20dy%7D%3D0%5Cimplies%20f%28y%29%3DC)
So the solution to the ODE is
![F(x,y)=C\iff \dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+C=C](https://tex.z-dn.net/?f=F%28x%2Cy%29%3DC%5Ciff%20%5Cdfrac%7Bx%5E4%7D4%2B%5Cdfrac%7Bx%5E2y%5E2%7D2%2B%5Cdfrac%7Bx%5E3%7D3%2BC%3DC)
![\implies\boxed{\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3=C}](https://tex.z-dn.net/?f=%5Cimplies%5Cboxed%7B%5Cdfrac%7Bx%5E4%7D4%2B%5Cdfrac%7Bx%5E2y%5E2%7D2%2B%5Cdfrac%7Bx%5E3%7D3%3DC%7D)
It would use a closed circle and the ray will move to the left.
Answer:
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Step-by-step explanation: