Question Number 12040 by Mr Chheang Chantria last updated on 10/Apr/17 $$\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{that}}\:\forall\boldsymbol{{x}},\boldsymbol{{y}}\in\boldsymbol{{R}} \\ $$$$\Rightarrow\:\mathrm{1}+\:\frac{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{xy}}}{\mathrm{3}}\:\geqslant\:\boldsymbol{{x}}+\boldsymbol{{y}} \\ $$ Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 10/Apr/17…
Question Number 143114 by bounhome last updated on 10/Jun/21 $${solve}: \\ $$$$\underset{{z}\rightarrow{i}} {\mathrm{lim}}\frac{\mathrm{3}{z}^{\mathrm{4}} −\mathrm{2}{z}^{\mathrm{3}} +\mathrm{8}{z}^{\mathrm{2}} −\mathrm{2}{z}+\mathrm{5}}{{z}−{i}}=? \\ $$ Answered by liberty last updated on 10/Jun/21…
Question Number 143100 by lapache last updated on 10/Jun/21 $$\int\sqrt{{e}^{{x}} +\mathrm{1}\:}=…..??? \\ $$ Answered by qaz last updated on 10/Jun/21 $$\int\sqrt{\mathrm{e}^{\mathrm{x}} +\mathrm{1}}\mathrm{dx} \\ $$$$=\int\sqrt{\mathrm{u}+\mathrm{1}}\frac{\mathrm{du}}{\mathrm{u}} \\…
Question Number 143097 by ZiYangLee last updated on 10/Jun/21 $$\mathrm{If}\:{z}=\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta,\: \\ $$$$\mathrm{prove}\:\mathrm{that}\:\frac{\mathrm{1}−{z}^{\mathrm{2}} }{\mathrm{1}+{z}^{\mathrm{2}} }=−{i}\mathrm{tan}\:\theta \\ $$ Answered by MJS_new last updated on 10/Jun/21 $$\frac{\mathrm{1}−\left(\mathrm{c}+\mathrm{is}\right)^{\mathrm{2}} }{\mathrm{1}+\left(\mathrm{c}+\mathrm{is}\right)^{\mathrm{2}}…
Question Number 143090 by mathdave last updated on 10/Jun/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 12015 by Nayon last updated on 09/Apr/17 $${prove}\:{that}\:{for}\:{all}\:{x}\:\in{R}, \\ $$$${e}^{{x}} \geqslant{x}^{{e}} \\ $$ Answered by mrW1 last updated on 18/Apr/17 $${at}\:{x}={e},\:{e}^{{x}} ={x}^{{e}} \\…
Question Number 77533 by vishalbhardwaj last updated on 07/Jan/20 $$\mathrm{If}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\:\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:\mathrm{where}\:\mathrm{x}\neq\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{function}\:\mathrm{at}\:\mathrm{x}=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{given}\:\mathrm{that}\:\mathrm{function}\:\mathrm{is}\:\mathrm{continuous}\:\mathrm{at}\:\mathrm{x}=\mathrm{0}\:? \\ $$ Answered by MJS last updated on 07/Jan/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{x}\mathrm{sin}\:\frac{\mathrm{1}}{{x}}\:=…
Question Number 77523 by MJS last updated on 07/Jan/20 $$\mathrm{how}\:\mathrm{many}\:\mathrm{silly}\:\mathrm{questions}\:\mathrm{can}\:\mathrm{a}\:\mathrm{person}\:\mathrm{ask} \\ $$$$\mathrm{within}\:\mathrm{the}\:\mathrm{first}\:\delta\:\mathrm{days}\:\mathrm{of}\:\mathrm{the}\:\mathrm{year}\:\psi\:\mathrm{when} \\ $$$$\mathrm{the}\:\mathrm{number}\:\chi_{\mathrm{0}} \:\mathrm{of}\:\mathrm{his}\:\mathrm{IDs}\notin\mathbb{R}\:\mathrm{in}\:\mathrm{the}\:\mathrm{year}\:\psi−\mathrm{1} \\ $$$$\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\left[\mathrm{ln}\:\left(\frac{\psi}{\mathrm{2}}+\mathrm{3}\mathbb{M}\right)\right]\leqslant\chi_{\mathrm{0}} <\underset{{q}_{\mathrm{0}} ,\:{x}\rightarrow\infty} {\mathrm{lim}}\frac{{q}_{\mathrm{0}} ^{{x}} \sqrt{\mathrm{2}\pi{x}}}{\mathrm{e}^{{q}_{\mathrm{0}} } } \\…
Question Number 77514 by liki last updated on 07/Jan/20 Answered by jagoll last updated on 07/Jan/20 $${L}_{\mathrm{1}} :\:{y}=−\frac{\mathrm{3}}{\mathrm{4}}\left({x}−\mathrm{2}\right)+\left(−\mathrm{1}\right) \\ $$$${L}_{\mathrm{1}} :\:{y}\:=\:−\frac{\mathrm{3}}{\mathrm{4}}{x}+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{a}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${let}\:{poin}\:{C}\left({x},{y}\right)\:,\:{CA}\:{perpendicular} \\ $$$${to}\:{L}_{\mathrm{1}}…
Question Number 143048 by 0731619 last updated on 09/Jun/21 Answered by mindispower last updated on 09/Jun/21 $$=\underset{{n}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\sqrt{\Gamma\left({n}\right)}−\mathrm{1}/\left(\Gamma\left({n}\right)−\mathrm{1}\right)=\frac{\mathrm{1}}{\:\sqrt{\Gamma\left({n}\right)}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Terms of Service…