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Question-71149

Question Number 71149 by naka3546 last updated on 12/Oct/19 Commented by MJS last updated on 17/Oct/19 $$\mathrm{if}\:\mathrm{the}\:\mathrm{hexagon}\:\mathrm{is}\:\mathrm{regular}\:\mathrm{it}'\mathrm{s}\:\mathrm{symmetric} \\ $$$${BE}\:\mathrm{is}\:\mathrm{the}\:\mathrm{symmetry}\:\mathrm{axis} \\ $$$${M}\in\left({BE}\right) \\ $$$${A}={C}'\:\wedge\:{F}={D}' \\ $$$$\Rightarrow\:\left({AF}\right)=\left({CD}\right)'\:\mathrm{and}\:\mathrm{similar}\:\mathrm{for}\:\mathrm{all}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{lines}…

2x-x-1-16-

Question Number 71141 by Mr. K last updated on 12/Oct/19 $$\left(\mathrm{2}{x}\right)^{{x}} =\frac{\mathrm{1}}{\mathrm{16}} \\ $$ Commented by mr W last updated on 12/Oct/19 $$\left(\mathrm{2}{x}\right)^{{x}} =\sqrt{\left(\mathrm{2}{x}\right)^{\mathrm{2}{x}} }=\sqrt{{t}^{{t}}…

Question-136656

Question Number 136656 by 0731619177 last updated on 24/Mar/21 Answered by Ñï= last updated on 24/Mar/21 $${y}={x}^{{x}^{{x}^{…} } } ={x}^{{y}} \\ $$$${y}'=\left({e}^{{ylnx}} \right)'={x}^{{y}} \left({y}'{lnx}+\frac{{y}}{{x}}\right)={x}^{{y}} {y}'{lnx}+{yx}^{{y}−\mathrm{1}}…

I-0-1-sin-1-x-1-x-x-2-dx-pi-4-ln3-

Question Number 136643 by Ñï= last updated on 24/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{sin}^{−\mathrm{1}} \sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}+{x}^{\mathrm{2}} }}{dx}=\frac{\pi}{\mathrm{4}}{ln}\mathrm{3} \\ $$ Answered by Ñï= last updated on 24/Mar/21 $${I}=\int_{\mathrm{0}} ^{\mathrm{1}}…

I-have-two-circles-side-by-side-Circle-1-has-radius-r-Circle-2-has-radius-1-3-r-If-I-roll-circle-2-around-circle-1-until-it-reaches-the-begining-how-many-times-will-it-roll-

Question Number 5546 by FilupSmith last updated on 19/May/16 $$\mathrm{I}\:\mathrm{have}\:\mathrm{two}\:\mathrm{circles}\:\mathrm{side}\:\mathrm{by}\:\mathrm{side}. \\ $$$$\mathrm{Circle}\:\mathrm{1}\:\mathrm{has}\:\mathrm{radius}\:{r}. \\ $$$$\mathrm{Circle}\:\mathrm{2}\:\mathrm{has}\:\mathrm{radius}\:\frac{\mathrm{1}}{\mathrm{3}}{r}. \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{I}\:\mathrm{roll}\:\mathrm{circle}\:\mathrm{2}\:\mathrm{around}\:\mathrm{circle}\:\mathrm{1}\:\mathrm{until} \\ $$$$\mathrm{it}\:\mathrm{reaches}\:\mathrm{the}\:\mathrm{begining}, \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{times}\:\mathrm{will}\:\mathrm{it}\:\mathrm{roll}? \\ $$ Commented…

Question-136596

Question Number 136596 by 0731619177 last updated on 23/Mar/21 Answered by MJS_new last updated on 24/Mar/21 $$\mathrm{answer}\:\mathrm{is}\: \\ $$$$\frac{\gamma}{\mathrm{1}−\gamma} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{x}^{{x}^{{x}…} } −{x}\Gamma\left({x}\right)}{{x}\Gamma\left({x}\right)^{{x}\Gamma\left({x}\right)} −\mathrm{1}}\:=…