Question Number 201873 by sonukgindia last updated on 14/Dec/23 Answered by witcher3 last updated on 14/Dec/23 $$\mathrm{Re}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{e}^{\mathrm{i}\theta} }\right)=\frac{\mathrm{1}+\mathrm{e}^{−\mathrm{i}\theta} }{\left(\mathrm{1}+\mathrm{e}^{\mathrm{i}\theta} \right)\left(\mathrm{1}+\mathrm{e}^{−\mathrm{i}\theta} \right)}=\frac{\mathrm{1}+\mathrm{cos}\left(\theta\right)}{\mathrm{2}+\mathrm{2cos}\left(\theta\right)}=\frac{\mathrm{1}}{\mathrm{2}};\forall\theta\in\mathbb{R}−\left\{\left(\mathrm{1}+\mathrm{2k}\right)\pi\right\} \\ $$$$\mathrm{I}==\frac{\varphi^{\mathrm{3}} −\mathrm{1}}{\mathrm{2}} \\…
Question Number 201839 by depressiveshrek last updated on 14/Dec/23 Commented by depressiveshrek last updated on 14/Dec/23 $${The}\:{last}\:{one}\:{means}: \\ $$$$\underset{{i}=\mathrm{0}} {\overset{{k}} {\sum}}{i}^{\mathrm{2}} \begin{pmatrix}{{k}}\\{{i}}\end{pmatrix}={k}\left({k}+\mathrm{1}\right)\mathrm{2}^{{k}−\mathrm{2}} \\ $$ Terms…
Question Number 201864 by ajfour last updated on 14/Dec/23 $$\left({ct}^{\mathrm{2}} −\frac{\mathrm{1}}{{ct}^{\mathrm{2}} }+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} \left({t}^{\mathrm{4}} +\frac{\mathrm{1}}{{t}^{\mathrm{4}} }\right) \\ $$$${Find}\:\:{t}={f}\left({c}\right). \\ $$ Commented by Frix last updated…
Question Number 201860 by MrGHK last updated on 14/Dec/23 $$\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{n}}+\mathrm{1}}=?? \\ $$ Answered by mnjuly1970 last updated on 14/Dec/23 $$\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty}…
Question Number 201857 by cortano12 last updated on 14/Dec/23 $$\:\:\mathrm{If}\:{a},{b},{c},{d},{e}\:\mathrm{are}\:\mathrm{thr}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$$\:\:\mathrm{2x}^{\mathrm{5}} −\mathrm{3x}^{\mathrm{3}} +\mathrm{2x}−\mathrm{7}=\mathrm{0}\:,\:\mathrm{find}\:\mathrm{the}\: \\ $$$$\:\mathrm{value}\:\mathrm{of}\:\underset{\mathrm{cyc}} {\prod}\left({a}^{\mathrm{3}} −\mathrm{1}\right)\: \\ $$ Commented by Frix last updated…
Question Number 201859 by York12 last updated on 14/Dec/23 $$\mathrm{If}\:{xyz}\:\in\mathbb{R}^{+} \:,\:{xyz}=\mathrm{1}\:,\:\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}\:\mathrm{holds}: \\ $$$$\frac{{x}}{\mathrm{2}{x}^{\mathrm{5}} +{x}+\mathrm{4}}+\frac{{y}}{\mathrm{2}{y}^{\mathrm{5}} +{y}+\mathrm{4}}+\frac{{z}}{\mathrm{2}{z}^{\mathrm{5}} +{z}+\mathrm{4}}\geqslant\frac{\mathrm{3}}{\mathrm{7}}. \\ $$$$\boldsymbol{\mathrm{Solution}}\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{advice}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{get}}\:\boldsymbol{\mathrm{better}} \\ $$$$\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{inequalities}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{which}}\:\boldsymbol{\mathrm{book}}\:\boldsymbol{\mathrm{would}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{recommend}}. \\ $$$$\boldsymbol{\mathrm{Thanks}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{advance}}! \\ $$$$\: \\…
Question Number 201820 by cortano12 last updated on 13/Dec/23 $$\:\:\:\frac{\mid\mathrm{3x}+\mathrm{1}\mid−\mid\mathrm{x}+\mathrm{2}\mid}{\mathrm{3}−\mid\mathrm{2x}\mid}\:\geqslant\:\mathrm{0}\: \\ $$$$\:\:\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{set}. \\ $$ Answered by dimentri last updated on 13/Dec/23 $$\:\:\frac{\left(\mathrm{3x}+\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{3}^{\mathrm{2}} −\left(\mathrm{2x}\right)^{\mathrm{2}}…
Question Number 201837 by MrGHK last updated on 13/Dec/23 $${y}'''−{y}''+{y}'={sec}\left({t}\right),−\frac{\pi}{\mathrm{2}}<{t}<\frac{\pi}{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 201822 by MathedUp last updated on 13/Dec/23 $$\mathrm{Do}\:\mathrm{Not}\:\mathrm{Use}\:\mathrm{sin}\left(\theta\right)\sim\theta\:\left(\theta\:\:\mathrm{is}\:\mathrm{small}\:\mathrm{Enough}\right) \\ $$$$\ddot {\theta}+\frac{\mathrm{g}}{\ell}\mathrm{sin}\left(\theta\right)=\mathrm{0} \\ $$$${y}''\left({t}\right)+\frac{\mathrm{g}}{\ell}\:\mathrm{sin}\left({y}\left({t}\right)\right)=\mathrm{0} \\ $$$${y}''\left({t}\right){y}'\left({t}\right)+\frac{\mathrm{g}}{\ell}\mathrm{sin}\left({y}\left({t}\right)\right){y}'\left({t}\right)=\mathrm{0} \\ $$$${y}'\left({t}\right){y}''\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\left({y}'\left({t}\right)\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{g}}{\ell}\mathrm{sin}\left({y}\left({t}\right)\right){y}'\left({t}\right)=−\frac{\mathrm{g}}{\ell}\centerdot\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\mathrm{cos}\left({y}\left({t}\right)\right) \\ $$$$\therefore\:\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\left[\frac{\mathrm{1}}{\mathrm{2}}\left({y}'\left({t}\right)\right)^{\mathrm{2}} −\frac{\mathrm{g}}{\ell}\mathrm{cos}\left({y}\left({t}\right)\right)\right]=\mathrm{0} \\…
Question Number 201832 by sonukgindia last updated on 13/Dec/23 Answered by aleks041103 last updated on 13/Dec/23 $${x}={e}^{−{t}} \:\Rightarrow\:{dx}=−{e}^{−{t}} {dt} \\ $$$$\Rightarrow{I}=\mathrm{32}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\mathrm{5}} \left(−{t}\right)}{−{t}}{e}^{−{t}} {dt}=\mathrm{32}\int_{\mathrm{0}}…