Question Number 196107 by maths_plus last updated on 18/Aug/23 $$ \\ $$$$\mathrm{help},\:\mathrm{please}\:! \\ $$$$\underset{{x}\:\rightarrow\:\frac{\boldsymbol{\pi}}{\mathrm{4}}} {{lim}}\:\frac{{cos}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\:−{x}\right)−{tan}\:{x}}{\mathrm{1}−{sin}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}+{x}\right)}\:=\:???\: \\ $$ Answered by MM42 last updated on 21/Aug/23 $${hop}\rightarrow{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{4}}}…
Question Number 196103 by MrGHK last updated on 18/Aug/23 $$\frac{\boldsymbol{\mathrm{d}}^{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{dx}}^{\boldsymbol{\mathrm{n}}} }\left[\boldsymbol{\Gamma}\left(\boldsymbol{\mathrm{x}}\right)\right]\:\:\boldsymbol{\mathrm{Hence}}\:\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{is}}\:\frac{\boldsymbol{\mathrm{d}}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\boldsymbol{\mathrm{dx}}^{\frac{\mathrm{1}}{\mathrm{2}}} }\left[\boldsymbol{\Gamma}\left(\boldsymbol{\mathrm{x}}\right)\right] \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 196098 by sonukgindia last updated on 18/Aug/23 Answered by mr W last updated on 18/Aug/23 $${x}^{\frac{\mathrm{1}}{\mathrm{2}^{{x}^{\mathrm{4}} } }} ={x}^{\frac{\mathrm{1}}{\mathrm{4}^{{x}^{\mathrm{2}} } }} \\ $$$$\Rightarrow{x}=\mathrm{0}…
Question Number 196063 by Rodier97 last updated on 17/Aug/23 $$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{calcul}\:\:\:\:\int_{\mathrm{0}} ^{+\infty} \:\frac{\sqrt{\mathrm{u}}\:.\mathrm{arctan}\left(\mathrm{u}\right)}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\:\mathrm{du} \\ $$$$ \\ $$$$…
Question Number 196062 by qaz last updated on 17/Aug/23 $$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\left[{gcd}\left({i},{j}\right)=\mathrm{1}\right]=?\:\:\:\:\:\:\:\:\:,\left[{D}\right]=\begin{cases}{\mathrm{1},\:\:\:{D}\:{is}\:{ture}.}\\{\mathrm{0},\:\:\:\:{D}\:{is}\:{false}.\:\:\:}\end{cases} \\ $$ Commented by mr W last updated on 17/Aug/23 $${i}\:{don}'{t}\:{think}\:{we}\:{can}\:{get}\:{it}\:{in}\:{a}\:{closed}…
Question Number 196056 by mr W last updated on 17/Aug/23 $${solve} \\ $$$$\sqrt{\mathrm{2}{x}+\mathrm{3}}−\sqrt{\mathrm{3}{x}+\mathrm{8}}=\sqrt{\mathrm{3}{x}+\mathrm{4}}−\sqrt{\mathrm{2}{x}+\mathrm{7}} \\ $$ Answered by cortano12 last updated on 17/Aug/23 $$\:\sqrt{\mathrm{u}}\:−\sqrt{\mathrm{v}+\mathrm{4}}\:=\sqrt{\mathrm{v}}−\sqrt{\mathrm{u}+\mathrm{4}} \\ $$$$\:\:\sqrt{\mathrm{u}}−\sqrt{\mathrm{v}}\:=\:\sqrt{\mathrm{v}+\mathrm{4}}−\sqrt{\mathrm{u}+\mathrm{4}}…
Question Number 196085 by sonukgindia last updated on 17/Aug/23 Answered by MM42 last updated on 17/Aug/23 $$\left(\mathrm{1}^{\mathrm{17}} +\mathrm{2023}^{\mathrm{17}} \right)+\left(\mathrm{2}^{\mathrm{17}} +\mathrm{2022}^{\mathrm{17}} \right)+…+\mathrm{1012}^{\mathrm{17}} = \\ $$$$\left(\mathrm{1}+\mathrm{2023}\right){k}_{\mathrm{1}} +\left(\mathrm{2}+\mathrm{2022}\right){k}_{\mathrm{2}}…