Thread:Talk:LiquidThreads testing/Verhunzt/reply (2): Difference between revisions

Latest revision as of 19:49, 11 May 2010

Pythagoras: $a^{2} + b^{2} = c^{2}$

Relativistischer Pythagoras: $\sqrt{1 - \frac{a^{2} + b^{2}}{c^{2}}} = 0$

:-)

Revision as of 19:49, 11 May 2010 view source 92.229.245.171 (talk) Reply to Verhunzt ← Older edit	Latest revision as of 19:49, 11 May 2010 view source 92.229.245.171 (talk) Reply to Verhunzt
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