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

• meta:bans

Albums

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

Misc.

• Wikipedia:Image use policy
• Wikipedia:Boilerplate text
• Wikipedia:MediaWiki custom messages

TeX

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

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

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

For this page on Distributed Proofreaders:

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

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

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

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

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

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

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

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

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

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

For this page:

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

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

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

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

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

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

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

[1]

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

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