Construct a pentagon/admin

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Revision as of 15:13, 25 June 2006 by Bear (talk | contribs) (Math: add eq 24)
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01 <math> \frac\mbox{NO + ND}\mbox{NO} = \frac\mbox{NO}\mbox{ND}</math> 02 <math> \frac{\frac{s}{t} + \frac{t}{t}}{\frac{s}{t}} = \frac{s}{t} </math> 03 <math> \frac{\varphi + 1}{\varphi} = \varphi </math> 04 <math> \frac{s+t}{s} = \frac{s}{t} </math> 05 <math> \varphi + 1 = \varphi^2 </math> 06 <math>\varphi^2 - \varphi - 1 = 0</math> 07 <math>{ax}^2 - {bx} - {c} = 0</math> 08 <math>x = \frac{-b \pm \sqrt {b^2-4ac}}{2a} </math> 09 <math>x = \frac{-(-1) \pm \sqrt {(-1)^2-4(1)(-1)}}{2(1)} </math> 10 <math>\varphi = {1 \pm \sqrt{5} \over 2}</math> 11 <math>\varphi = {1 + \sqrt{5} \over 2} 12 <math> \frac{\varphi}{\varphi} + \frac{1}{\varphi} = \varphi </math> 13 <math> 1 + \frac{1}{\varphi} = \varphi </math> 14 <math> \frac{1}{\varphi} = \varphi - 1 </math> 15 <math> \mbox{DA}' = {g} = \frac{1}{\varphi} = \varphi - 1</math> 16 <math>\mbox{MA}' = \sqrt {\left(\frac{1}{2}\right)^2 + (1)^2} </math> 17 <math>\mbox{MA}' = \sqrt {\frac{1}{4} + 1} = \sqrt {\frac{5}{4}} = {\sqrt {5} \over \sqrt {4}}</math> 18 <math>\mbox{MA}' = {\sqrt {5} \over {2}}</math> 19 <math>\mbox{PA}' = {\sqrt {5} \over {2}} + \frac{1}{2} = {1 + \sqrt{5} \over 2}</math> 20 <math>\mbox{QA}' = {\sqrt {5} \over {2}} - \frac{1}{2}</math> 21 <math>\mbox{QA}' = {\sqrt {5} \over {2}} + \left(\frac{1}{2} - \frac{1}{2}\right) - \frac{1}{2}</math> 22 <math>\mbox{QA}' = {1+\sqrt {5} \over {2}} - \frac{1}{2} - \frac{1}{2}</math> 23 <math>\mbox{QA}' = {1+\sqrt {5} \over {2}} - 1</math> 24 <math>\mbox{AB} = \sqrt {{5-\sqrt {5} \over {2}}}</math>