Question Number 197345 by sonukgindia last updated on 14/Sep/23 Answered by HeferH last updated on 14/Sep/23 Commented by HeferH last updated on 14/Sep/23 $$\:{T}\:=\:\mathrm{4}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}}…
Question Number 197344 by MathedUp last updated on 14/Sep/23 $$\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\centerdot\frac{\mathrm{d}{x}^{\boldsymbol{\lambda}} }{\mathrm{d}{t}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}^{\boldsymbol{\lambda\alpha}} \left(\partial_{\boldsymbol{\mu}} ^{\:} \mathrm{g}_{\boldsymbol{\alpha\nu}} +\partial_{\boldsymbol{\nu}} ^{\:} \mathrm{g}_{\boldsymbol{\alpha\mu}} −\partial_{\boldsymbol{\alpha}} ^{\:} \mathrm{g}_{\boldsymbol{\mu\nu}} \right)\frac{\mathrm{d}{x}^{\boldsymbol{\mu}} }{\mathrm{d}{t}}\centerdot\frac{\mathrm{d}{x}^{\boldsymbol{\nu}} }{\mathrm{d}{t}}=\mathrm{0} \\ $$…
Question Number 197362 by sonukgindia last updated on 14/Sep/23 Answered by MM42 last updated on 15/Sep/23 $$\frac{{e}^{{x}} −\mathrm{1}}{{e}^{{x}} +\mathrm{1}}={u}\Rightarrow{e}^{{x}} {u}+{u}={e}^{{x}} −\mathrm{1}\Rightarrow{e}^{{x}} \left({u}−\mathrm{1}\right)=−\mathrm{1}−{u} \\ $$$${e}^{{x}} =\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\Rightarrow{f}\left({x}\right)=\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)^{\mathrm{2}}…
Question Number 197310 by sonukgindia last updated on 13/Sep/23 Answered by Frix last updated on 13/Sep/23 $$\left(\mathrm{1}\right)\:{z}^{\mathrm{2}} −\mathrm{7}{z}+\mathrm{11}=\mathrm{1} \\ $$$$\Rightarrow\:{z}=\mathrm{2}\vee{z}=\mathrm{5} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{ln}\:\left({z}^{\mathrm{2}} −\mathrm{3}{z}+\mathrm{e}^{\mathrm{2}} \right)\:=\mathrm{0} \\…
Question Number 197336 by Huy last updated on 13/Sep/23 $${lim}_{{x}\rightarrow+\infty} \left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{cos}{x}\right)=? \\ $$ Answered by TheHoneyCat last updated on 13/Sep/23 $$\mathrm{This}\:\mathrm{limit}\:\mathrm{is}\:\mathrm{undefined} \\ $$$$\left(\mathrm{because}\:\mathrm{lim}\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{0}\:\mathrm{and}\:\mathrm{lim}\:\mathrm{cos}{x}\:\mathrm{is}\:\mathrm{undefined}\right)…
Question Number 197338 by yusufkhabibulloh last updated on 13/Sep/23 $$\int\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$ Answered by TheHoneyCat last updated on 13/Sep/23 $$=\int\frac{{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{x} \\…
Question Number 197335 by sonukgindia last updated on 13/Sep/23 Answered by witcher3 last updated on 13/Sep/23 $$\mathrm{x}\rightarrow\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\mathrm{I}=\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\sqrt{\mathrm{3}}} \frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{3}}\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} +\mathrm{2}}\right)\mathrm{dx}…
Question Number 197325 by sonukgindia last updated on 13/Sep/23 Answered by witcher3 last updated on 13/Sep/23 $$\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{if}\:\mathrm{x}\in\mathbb{R}\:\mid\mathrm{f}^{\mathrm{2}} \left(\mathrm{x}\right)=\mathrm{x}\Rightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{x}\in\left\{\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}},\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right\} \\…
Question Number 197320 by tri26112004 last updated on 13/Sep/23 $$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}\left({k}+\mathrm{1}\right)} \\ $$ Answered by Erico last updated on 13/Sep/23 $$\forall\mathrm{k}\in\left[\mathrm{1},\mathrm{n}\right]\:\:\:\:\:\left(−\mathrm{1}\right)^{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)} =\mathrm{1} \\ $$$$\Rightarrow\:\underset{\mathrm{k}=\mathrm{1}}…
Question Number 197272 by Erico last updated on 12/Sep/23 $$\mathrm{How}\:\mathrm{to}\:\mathrm{calculate}\:\mathrm{this}\:\mathrm{integral} \\ $$$$\underset{\:\mathrm{0}} {\int}^{\:\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{ln}\left(\mathrm{1}+{sint}\right)}{{sint}}{dt} \\ $$ Answered by Mathspace last updated on 12/Sep/23 $${tan}\left(\frac{{t}}{\mathrm{2}}\right)={x}\:\Rightarrow \\…