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

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

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

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

For this page on Distributed Proofreaders:

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

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

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

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

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

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

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

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

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

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

For this page:

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

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

$\{\begin{array}{c}\frac{x+2y}{3x-y}=1,\\ \frac{4y-x}{3+x-2y}=2\frac{1}{2}.\end{array}$

$\{\begin{array}{c}\frac{x+2y}{x-2}=-5\frac{2}{3},\\ \frac{2y-4x}{3-y}=-6.\end{array}$

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

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

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

[1]

$\begin{array}{c}{x}^3+2x^2+3x\\ \underset{\_}{3x^2-2x+1}\\ 3x^5+6x^4+9x^3\\ -2x^4-4x^3-6x^2\\ \underset{\_}{x^3+2x^2+3x}\\ 3x^5+4x^4+6x^3-4x^2+3x\end{array}$

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