Question Number 107080 by Khalmohmmad last updated on 08/Aug/20 Answered by abdomsup last updated on 08/Aug/20 $${A}_{{n}} =\frac{{n}^{{n}} \left\{\frac{{n}}{{n}^{{n}} }+\frac{{n}^{\mathrm{2}} }{{n}^{{n}} }+….+\mathrm{1}\right\}}{{n}^{{n}} \left\{\frac{\mathrm{1}}{{n}^{{n}} }\:+\left(\frac{\mathrm{2}}{{n}}\right)^{{n}\:} +\left(\frac{\mathrm{3}}{{n}^{}…
Question Number 107088 by mathdave last updated on 08/Aug/20 Commented by mohammad17 last updated on 08/Aug/20 $$\int_{−\mathrm{1}} ^{\:\mathrm{1}} \frac{\mathrm{1}−{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx}=\left({sin}^{−\mathrm{1}} {x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)_{−\mathrm{1}} ^{\:\mathrm{1}} =\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}=\pi \\…
Question Number 41536 by mondodotto@gmail.com last updated on 09/Aug/18 $$\boldsymbol{\mathrm{if}}\:\boldsymbol{{y}}=\sqrt{\frac{\mathrm{1}+\boldsymbol{\mathrm{sin}{x}}}{\mathrm{1}−\boldsymbol{\mathrm{sin}{x}}}\:}\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}} \\ $$$$\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}=\frac{\mathrm{1}}{\mathrm{1}−\boldsymbol{\mathrm{sin}{x}}} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18 $${sinx}=\frac{\mathrm{2}{tan}\frac{{x}}{\mathrm{2}}}{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}} \\ $$$${y}=\sqrt{\frac{\mathrm{1}+\frac{\mathrm{2}{tan}\frac{{x}}{\mathrm{2}}}{\mathrm{1}+{tan}^{\mathrm{2}}…
Question Number 41534 by mondodotto@gmail.com last updated on 09/Aug/18 $$\boldsymbol{\mathrm{Three}}\:\boldsymbol{\mathrm{chidren}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{playing}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{game}} \\ $$$$\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{claping}}\:\boldsymbol{\mathrm{hands}},\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{first}}\:\boldsymbol{\mathrm{child}}\:\boldsymbol{\mathrm{claping}}\:\boldsymbol{\mathrm{hands}} \\ $$$$\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{every}}\:\boldsymbol{\mathrm{after}}\:\mathrm{1}\boldsymbol{\mathrm{sec}},\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{second}}\: \\ $$$$\boldsymbol{\mathrm{child}}\:\boldsymbol{\mathrm{clap}}\:\boldsymbol{\mathrm{hands}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{every}}\:\boldsymbol{\mathrm{after}}\:\mathrm{10}\boldsymbol{\mathrm{sec}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{third}}\:\boldsymbol{\mathrm{child}}\:\boldsymbol{\mathrm{claps}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{every}}\:\boldsymbol{\mathrm{after}}\:\mathrm{5}\boldsymbol{\mathrm{sec}}.\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{long}}\:\boldsymbol{\mathrm{do}}\:\boldsymbol{\mathrm{all}} \\ $$$$\boldsymbol{\mathrm{three}}\:\boldsymbol{\mathrm{children}}\:\boldsymbol{\mathrm{will}}\:\boldsymbol{\mathrm{clap}}\:\boldsymbol{\mathrm{their}}\:\boldsymbol{\mathrm{hands}}\:\boldsymbol{\mathrm{together}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{same}}\:\boldsymbol{\mathrm{time}}? \\ $$ Answered by alex041103…
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Question Number 107069 by mathdave last updated on 08/Aug/20 Answered by bemath last updated on 09/Aug/20 $$\:\:\:\:\:\:\:@{bemath}@ \\ $$$$\left(\sqrt{\mathrm{cos}\:{x}}\:+\sqrt{\mathrm{sin}\:{x}}\right)^{\mathrm{5}} =\left(\sqrt{\mathrm{cos}\:{x}}\left(\mathrm{1}+\sqrt{\mathrm{tan}\:{x}}\right)^{\mathrm{5}} \right. \\ $$$$=\:\mathrm{cos}\:^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{cos}\:{x}}\:\left(\mathrm{1}+\sqrt{\mathrm{tan}\:{x}}\right)^{\mathrm{5}} \\…
Question Number 107064 by mathocean1 last updated on 08/Aug/20 $${Solve}\:{in}\:\mathbb{R} \\ $$$${x}+\sqrt{−{x}^{\mathrm{2}} +{x}+\mathrm{6}}=\sqrt{\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{6}}−{x}} \\ $$ Answered by Her_Majesty last updated on 09/Aug/20 $${defined}\:{for}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{3} \\…
Question Number 107065 by mathocean1 last updated on 08/Aug/20 $${Solve}\:{in}\:\mathbb{R}^{\mathrm{3}} \\ $$$$\begin{cases}{{x}^{\mathrm{2}} +\mathrm{2}{xy}+{y}^{\mathrm{2}} −\mathrm{4}{z}^{\mathrm{2}} =\mathrm{0}}\\{\mathrm{3}{x}−\mathrm{2}{y}+{z}=\mathrm{3}}\end{cases} \\ $$$$\:\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{3}{z}^{\mathrm{2}} −\mathrm{4}{xy}+{yz}+{x}+\mathrm{2}{y}−\mathrm{3}{z}+\mathrm{7}=\mathrm{0} \\ $$ Answered by Her_Majesty…
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