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$$-x^y$$

## What does -(-... mean in this equation?

...The author seems to be assuming that the unary minus has a higher precedence than exponentiation; i.e. that $$-x^y$$ means $$(-x)^y$$ rather than $$-(x^y)$$ . Without any parentheses...
$$- x^y$$

## Why isn't $x^\frac{-1}{2}$ equal to $-\sqrt{x}$

...$$x^{-y}$$ is $$\frac1{x^y}$$ , not $$- x^y$$ . Added: $$\sqrt[-2]{x}$$ is the positive real number $$u$$ such that $$u^{-2}=x$$ , so it's $$u=\frac1{\sqrt x}$$ .
$$(y-x^2)$$

## Exercise in Hartshorne

https://math.stackexchange.com/questions/69015/exercise-in-hartshorne#comment-163203
[email protected] In fact, I don't get it. It is possible that a prime ideal strictly contains another one. Imagine the second example. the map is onto and $$(y-x^2)$$ is prime, but still there is another prime ideal $$(y-x^2,z-x^3)$$ strictly containing it.
$$(y-x^2)$$

## Showing a point is generic

https://math.stackexchange.com/questions/246087/showing-a-point-is-generic
...How does one show that something is a generic point? I feel I may be missing some subtly. Vakil in his notes asks one to show that the prime ideal $$(y-x^2)$$ is a generic point of $$V(y-x^2)$$ ? That is we need to show that the closure of $$(y-x^2)$$ is $$V(y-x^2)$$ . But this seems trivial since $$V(y-x^2)$$ is by definition the smallest closed subset containing $$y-x^2$$ . I am tired, though...
$$y^{2}-x^{3}$$

## Can we classify plane cubics, What are they?

https://math.stackexchange.com/questions/1157292/can-we-classify-plane-cubics%2C-what-are-they%3F
...There are four qualitatively distinct pictures of the plane cubics. What are the polynomials corresponding to them? I know two of them have special names: nodal cubic and cuspidal cubic with polynomials $$y^{2}-x^{3}-x^{2}$$ and $$y^{2}-x^{3}$$ respectively. What are the others?
$$\sqrt{9-x^2-y^2}$$

## Why are the contour lines of a sphere more spread out near the equator, if a sphere is homogeneous?

...I realize that this question may not be a formal as you're used too, but I'd appreciate it if you'd bear with me. So, I have a conceptual question about the contour lines of a sphere (in this case $$\sqrt{9-x^2-y^2}$$ , the top half of a sphere. It is not so hard to follow the actual computations. But the lines are concentrated towards the edges of the sphere along the x - y axis. Since the curvature of the sphere is the same at all points, this seems to imply that the sphere is 'steeper' in that region than t...
$$y^2-x^3-x^2$$

## Show that two quotient rings are not isomorphic

https://math.stackexchange.com/questions/3102311/show-that-two-quotient-rings-are-not-isomorphic#comment-6394972
...Why need the characteristic zero in order to conclude that $$y^2-x^3-x^2$$ is irreducible? This holds over every field (and over every integral domain).
$$z=1-x^2-y^2$$

## the angle of a shifted cone and its equation

https://math.stackexchange.com/questions/1796046/the-angle-of-a-shifted-cone-and-its-equation#comment-3667701
...The equation $$z=1-x^2-y^2$$ does not describe a cone but an upside down paraboloid.
$$-\sqrt{4-x^2-y^2}$$

## Find the volume inside both $x^2+y^2+z^2=4$ and $x^2+y^2=1$.

https://math.stackexchange.com/questions/429943/find-the-volume-inside-both-%24x%5E2%2By%5E2%2Bz%5E2%3D4%24-and-%24x%5E2%2By%5E2%3D1%24.#comment-919248
...Because $$z$$ is from $$-\sqrt{4-x^2-y^2}$$ to $$\sqrt{4-x^2-y^2}$$ .
$$z=64-x^2-y^2$$

## Use Stoke's Theorem to evaluate $∫_CF⋅dr$

https://math.stackexchange.com/questions/2390828/use-stoke%27s-theorem-to-evaluate-%24-%E2%88%AB_cf%E2%8B%85dr-%24#comment-4931469
...I believe you mean $$z=64-x^2-y^2$$ in your first line after 'Answer:'...
$$x=\sqrt{1-y^2-z^2}$$

## Use Stokes’ Theorem to evaluate $\iint_s \rm curl\ F \cdot dS$.

https://math.stackexchange.com/questions/682870/use-stokes%E2%80%99-theorem-to-evaluate-%24%5Ciint_s-%5Crm-curl%5C-f-%5Ccdot-ds%24.#comment-1433163
...I meant $$x=\sqrt{1-y^2-z^2}$$ , I edited it now.
$$\sqrt{\left(x-x_{o}\right)^{2}+\left(y-y_{o}\right)^{2}}$$

## Proving the set $S=\left\{ \left(x,y\right);ax+by https://math.stackexchange.com/questions/2613163/proving-the-set-%24s%3D%5Cleft%5C%7B-%5Cleft%28x%2Cy%5Cright%29%3Bax%2Bby%3Cc%5Cright%5C%7D-%24-is-open#comment-5396735 [email protected]астонвіллаолофмэллбэрг yes i know it is $$\sqrt{\left(x-x_{o}\right)^{2}+\left(y-y_{o}\right)^{2}}$$ ... $$(x_1-y_1)^2 + (x_2-y_2)^2$$ ##$d(x,y) = \sqrt{ (x_2-x_1)^2 + (y_2-y_1)^2 }.(\mathbb R^2,d)$is a metric space . https://math.stackexchange.com/questions/1513042/%24d%28x%2Cy%29-%3D-%5Csqrt%7B-%28x_2-x_1%29%5E2-%2B-%28y_2-y_1%29%5E2-%7D.%24-%24%28%5Cmathbb-r%5E2%2Cd%29%24-is-a-metric-space-.#comment-3081775 ...Are you sure the question is written correctly? Normally it is $$(x_1-y_1)^2 + (x_2-y_2)^2$$ in the square root. $$(x-y)^2(y-z)^2(z-x)^2$$ ## Let$(x, y, z)\in \Bbb R$, prove that$\big(\frac{2x-y}{x-y}\big)^2+\big(\frac{2y-z}{y-z}\big)^2 +\big(\frac{2z-x}{z-x}\big)^2 \geqslant 5$https://math.stackexchange.com/questions/2419508/let-%24%28x%2C-y%2C-z%29%5Cin-%5Cbbb-r%24%2C-prove-that-%24%5Cbig%28%5Cfrac%7B2x-y%7D%7Bx-y%7D%5Cbig%29%5E2%2B%5Cbig%28%5Cfrac%7B2y-z%7D%7By-z%7D%5Cbig%29%5E2-%2B%5Cbig%28%5Cfrac%7B2z-x%7D%7Bz-x%7D%5Cbig%29%5E2-%5Cgeqslant-5%24#answer-2419568 ...um_{cyc} y^4 z^2 -4 \sum_{cyc} x y z^4 +\\ 14 \sum_{cyc} xy^2 z^3 -10 \sum_{cyc} x y^3 z^2 -3x^2 y^2 z^2 \geq 0 \end{eqnarray*} This can be rearranged to \begin{eqnarray*} (2x-y)^2(y-z)^2(z-x)^2+(x-y)^2(2y-z)^2(z-x)^2+(x-y)^2(y-z)^2(2z-x)^2 \geq 5(x-y)^2(y-z)^2(z-x)^2 \end{eqnarray*} Now divide by $$(x-y)^2(y-z)^2(z-x)^2$$ and we have \begin{eqnarray*} \frac{(2x-y)^2}{(x-y)^2}+\frac{(2y-z)^2}{(y-z)^2}+\frac{(2z-x)^2}{(z-x)^2} \geq 5. \end{eqnarray*} Edit: I worked my way backwards from the result, using the "reduce" algebra package. I then guessed a square that would "consume" some of the terms & got lucky ! (2*x-y)... $$(x-y)^2+(y-z)^4+(z-x)^6$$ ## Factorize$\det\left[\begin{smallmatrix}yz-x^2&zx-y^2&xy-z^2\\zx-y^2&xy-z^2&yz-x^2\\xy-z^2&yz-x^2&zx-y^2\end{smallmatrix}\right]$using factor theorem https://math.stackexchange.com/questions/2675951/factorize-%24%5Cdet%5Cleft%5B%5Cbegin%7Bsmallmatrix%7Dyz-x%5E2%26zx-y%5E2%26xy-z%5E2%5C%5Czx-y%5E2%26xy-z%5E2%26yz-x%5E2%5C%5Cxy-z%5E2%26yz-x%5E2%26zx-y%5E2%5Cend%7Bsmallmatrix%7D%5Cright%5D%24-using-factor-theorem#comment-5526753 ...How do you infer that $$(x-y)^2+(y-z)^2+(z-x)^2$$ is a factor? Couldn't you use the same reasoning to determine that $$(x-y)^2+(y-z)^4+(z-x)^6$$ is a factor? $$(x-y)^2+(y-z)^2+(z-x)^2$$ ## Factorize$\det\left[\begin{smallmatrix}yz-x^2&zx-y^2&xy-z^2\\zx-y^2&xy-z^2&yz-x^2\\xy-z^2&yz-x^2&zx-y^2\end{smallmatrix}\right]$using factor theorem https://math.stackexchange.com/questions/2675951/factorize-%24%5Cdet%5Cleft%5B%5Cbegin%7Bsmallmatrix%7Dyz-x%5E2%26zx-y%5E2%26xy-z%5E2%5C%5Czx-y%5E2%26xy-z%5E2%26yz-x%5E2%5C%5Cxy-z%5E2%26yz-x%5E2%26zx-y%5E2%5Cend%7Bsmallmatrix%7D%5Cright%5D%24-using-factor-theorem#comment-5526767 [email protected] thnx. that seems true. then how do I extract $$(x-y)^2+(y-z)^2+(z-x)^2$$ in the first place ? $$(x-x)^2-(x+y-z)^2+(y-x)^2$$ ## Convert$-2xy+2zx-x^2$into a sum of three squares of expressions of$x,y$and$z$https://math.stackexchange.com/questions/3487950/convert-%24-2xy%2B2zx-x%5E2%24-into-a-sum-of-three-squares-of-expressions-of-%24x%2Cy%24-and-%24z%24#comment-7171876 ... $$(x-x)^2-(x+y-z)^2+(y-x)^2$$ ... $$(x-y)^2 +(y-z)^2+(z-x)^2$$ ## Solving a system of equations$x^2 +y^2 −z(x+y)=2,y^2 +z^2 −x(y+z)=4,z^2 +x^2 −y(z+x)=8\$

https://math.stackexchange.com/questions/1900201/solving-a-system-of-equations-%24x%5E2-%2By%5E2-%E2%88%92z%28x%2By%29%3D2%2Cy%5E2-%2Bz%5E2-%E2%88%92x%28y%2Bz%29%3D4%2Cz%5E2-%2Bx%5E2-%E2%88%92y%28z%2Bx%29%3D8%24#comment-3898909
...I loved the trick that gives $$(x-y)^2 +(y-z)^2+(z-x)^2$$ (+1)...
$$(x+y+1-x^2-y^2)/(1-x^2-y^2)$$

## Calculating the surface area of sphere above a plane

https://math.stackexchange.com/questions/153472/calculating-the-surface-area-of-sphere-above-a-plane#comment-353749
...ok I get a very messy formula for the cross product: sqrt( $$(x+y+1-x^2-y^2)/(1-x^2-y^2)$$ )... I take that this is right and I set x^2+y^2=r^2? the limits would then be r=0 to r=sqrt(3/4) and \theta from 0 to 2pi? how did you "guess" the answer so quickly when this is so complicated thought?
$$\sqrt{\frac{a^2-x^2-y^2+x^2+y^2}{a^2-x^2-y^2}}=\frac{a}{\sqrt{a^2-x^2-y^2}}$$

## Surface area of a sphere of radius a using double integral area formula

...Just add all the fractions in there. You will have $$\sqrt{\frac{a^2-x^2-y^2+x^2+y^2}{a^2-x^2-y^2}}=\frac{a}{\sqrt{a^2-x^2-y^2}}$$ ...