User:Jimregan/Trail of breadcrumbs

I've joked about this twice already, but I really need it. Just using bookmarks doesn't hack it for me, so here goes.

I've just noticed Mozilla keywords... http://www.wikipedia.org/w/wiki.phtml?search=%s&go=Go works well for those links on the mailing list.

Lists

 * List of lists - I know it redirects, I just prefer that name
 * List of record labels
 * List of albums
 * List of record producers

Meta

 * bans

Albums

 * WikiProject Albums - If I'm gonna be doing album articles...
 * Tuf-Kat's comments here too.

Misc.

 * Image use policy
 * Boilerplate text
 * MediaWiki custom messages

TeX
$$\it{m}/(1-\beta^2)^3_2$$

$$\it{m}/(1-\beta^2)\frac{1}{2}$$

$$\ _2^3e^2/\it{a}$$

For this page on Distributed Proofreaders:

$$\left(\frac{a}{b}\right)^2 = \frac{a}{b} \times \frac{a}{b} = \frac{a^2}{b^2}$$

$$\left(\frac{m}{n} + \frac{1}{x}\right)\left(\frac{m}{n} - \frac{1}{x}\right) = \frac{m^2}{n^2} - \frac{1}{x^2}$$

$$\left(\frac{x}{y} - 3\right)\left(\frac{x}{y} - 6\right) = \frac{x^2}{y^2} - \frac{9x}{y} + 18$$

$$\frac{b^4}{a^4} - \frac{y^4}{x^4} = \left(\frac{b^2}{a^2} + \frac{y^2}{x^2}\right) \left(\frac{b}{a} + \frac{y}{x}\right) \left( \frac{b}{a} - \frac{y}{x}\right)$$

$$x^4 + x^2 + \frac{1}{4} = \left(x^2 + \frac{1}{2}\right)^2$$

$$\left(\frac{2x^2}{ab^3}\right)^2$$

$$\left(\frac{a^2b^3}{4y}\right)^3$$

$$\left(\frac{x(a-b)}{3a^2b}\right)^4$$

$$\left(- \frac{5x^2y(a+b^2)^2}{2ab^3(x^2-y)^3}\right)^3$$

$$\left(- \frac{3am^4(2a+3b)^3}{4x^2y^2(m-n)^2}\right)^3$$

For this page:

$$\left\{ \begin{matrix}\frac{3x}{5} - \frac{2y}{7} = 35, \\ x + 2y = -63.\end{matrix}\right.$$

$$\left\{ \begin{matrix}x - \frac{3y}{5} = 6, \\ \frac{2x}{3} + 7y = 189.\end{matrix}\right.$$

$$\left\{ \begin{matrix}{x + 2y \over 3x - y} = 1, \\ {4y - x \over 3 + x - 2y} = 2\frac{1}{2}.\end{matrix}\right.$$

$$\left\{ \begin{matrix}{x + 2y \over x - 2} = -5\frac{2}{3}, \\ {2y - 4x \over 3 - y} = -6.\end{matrix}\right.$$

$$\left\{ \begin{matrix}y - \frac{2y + x}{3} = \frac{2x + y}{4} - 8\frac{3}{4}, \\ \frac{3x + y}{2} - \frac{y}{3} = \frac{109}{10} + \frac{4y - x}{5}.\end{matrix}\right.$$

$$\left\{ \begin{matrix}\frac{3x - 19}{2} + 4 = \frac{3y + x}{3} + \frac{5x - 3}{2}, \\ \frac{4x + 5y}{16} + \frac{2x + y}{2} = \frac{9x - 7}{8} + \frac{3y + 9}{4}.\end{matrix}\right.$$

$$\left\{ \begin{matrix}\frac{1}{5}(3x - 2y) + \frac{1}{3}(5x - 3y) = x, \\ \frac{4x - 3y}{2} + \frac{2}{3}x - y = 1 + y.\end{matrix}\right.$$



$$\begin{matrix}x^3 + 2x^2 + 3x \\ \underline{3x^2 - 2x + 1} \\ 3x^5 + 6x^4 + 9x^3 \\ \qquad\qquad\quad -2x^4 - 4x^3 -6x^2 \\ \underline{\qquad\qquad\qquad\qquad\qquad x^3 + 2x^2 +3x} \\ 3x^5 + 4x^4 +6x^3 - 4x^2 + 3x\end{matrix}$$

$$\frac{5}{6}a^4 - \frac{1}{5}a^3b - \frac{1}{3}a^2b^2 \mbox{by} \frac{6}{5}ab^2.$$