User:Espilceranul

Long list of unnecessary things about me that can be ignored if you want to

File:Lucas brawl.gifThis user mains Lucas in Super Smash Bros. Brawl.

File:Pkstarstorm.gifThis user has used PK Starstorm more than 100 times. Run away!

This user enjoys playing Big 2.

This user enjoys playing poker.

This user enjoys playing Sudoku.

This user plays bridge.

This user plays Hearts.

This user contributes using Mozilla Firefox.

This user thinks an explorer can't be better than a fox!

This user loves chocolate by
HERSHEY'S

This user eats cookies.

This user loves to eat fries, or chips.

This user loves to eat pizza.

This user does not eat fish. Ever.

This user prefers pepperoni pizza.

This user loves cake.

This user is a chocoholic.

This user has been known to eat a hot dog on occasion.

This user loves bacon!

This user thinks you can't trade with animals (but you can offer a home).

This user prefers Mozilla Firefox.

This user thinks polar coordinates are pretty.

${\frac {?}{0}}$

This user can divide by zero.

a²+b²=c²This user is an intermediate mathematician.

e^πi+1=0This user is a mathie.

$\textstyle {\frac {\partial }{\partial x}}\,\;$

This user only partially understands mathematics.

This user loves getting tangled up in Knot theory.

This user enjoys thinking in Four Dimensions.

This user is interested in polyhedra.

This user knows that math is $\int e^{xy}\,$.

This user uses Windows Live Messenger and his/her email address is none of your business.

This user uses µTorrent.

GThis user will use Google before asking dumb questions.

This user uses Google as a primary search engine.

${\text{2009 EOY Bonus Question 17}}\,$

${\text{Make p the subject of the formula.}}\,$

$z^{b}p^{25c}={\sqrt {2z^{3b}\cdot p^{25c}+z^{2a}-z^{4b}}}\,$

$z^{2b}p^{50c}=2z^{3}b{p^{2}5c}+z^{2a}-z^{4b}\,$

$z^{2b}p^{50c}-2z^{3b}p^{25c}+z^{4b}=z^{2a}\,$

$z^{2b}(p^{50c}-2z^{b}p^{25c}+z^{2b})=z^{2a}\,$

$z^{2b}(p^{25c}-z^{b})^{2}=z^{2a}\,$

$p^{25c}-z^{b}=\pm {\frac {z^{a}}{z^{b}}}\,$

$\therefore p={\sqrt[{25c}]{\frac {z^{2b}\pm z^{a}}{z^{b}}}}\,$

${\text{they said assume c is an odd number so no plus minus on the last one}}\,$