Question Number 136656 by 0731619177 last updated on 24/Mar/21 Answered by Ñï= last updated on 24/Mar/21 $${y}={x}^{{x}^{{x}^{…} } } ={x}^{{y}} \\ $$$${y}'=\left({e}^{{ylnx}} \right)'={x}^{{y}} \left({y}'{lnx}+\frac{{y}}{{x}}\right)={x}^{{y}} {y}'{lnx}+{yx}^{{y}−\mathrm{1}}…
Question Number 136643 by Ñï= last updated on 24/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{sin}^{−\mathrm{1}} \sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}+{x}^{\mathrm{2}} }}{dx}=\frac{\pi}{\mathrm{4}}{ln}\mathrm{3} \\ $$ Answered by Ñï= last updated on 24/Mar/21 $${I}=\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 136635 by mohammad17 last updated on 24/Mar/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 5546 by FilupSmith last updated on 19/May/16 $$\mathrm{I}\:\mathrm{have}\:\mathrm{two}\:\mathrm{circles}\:\mathrm{side}\:\mathrm{by}\:\mathrm{side}. \\ $$$$\mathrm{Circle}\:\mathrm{1}\:\mathrm{has}\:\mathrm{radius}\:{r}. \\ $$$$\mathrm{Circle}\:\mathrm{2}\:\mathrm{has}\:\mathrm{radius}\:\frac{\mathrm{1}}{\mathrm{3}}{r}. \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{I}\:\mathrm{roll}\:\mathrm{circle}\:\mathrm{2}\:\mathrm{around}\:\mathrm{circle}\:\mathrm{1}\:\mathrm{until} \\ $$$$\mathrm{it}\:\mathrm{reaches}\:\mathrm{the}\:\mathrm{begining}, \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{times}\:\mathrm{will}\:\mathrm{it}\:\mathrm{roll}? \\ $$ Commented…
Question Number 136596 by 0731619177 last updated on 23/Mar/21 Answered by MJS_new last updated on 24/Mar/21 $$\mathrm{answer}\:\mathrm{is}\: \\ $$$$\frac{\gamma}{\mathrm{1}−\gamma} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{x}^{{x}^{{x}…} } −{x}\Gamma\left({x}\right)}{{x}\Gamma\left({x}\right)^{{x}\Gamma\left({x}\right)} −\mathrm{1}}\:=…
Question Number 136582 by physicstutes last updated on 23/Mar/21 $$\mathrm{Use}\:\mathrm{De}'\mathrm{Moivre}'\mathrm{s}\:\mathrm{theorem}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\:\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{cos}\:{rx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{find}\:\mathrm{a}\:\mathrm{value}\left(\mathrm{or}\:\mathrm{expression}\right)\:\mathrm{for}\:\underset{{r}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{sin}\:{rx} \\ $$$$\mathrm{assume}\:\mathrm{that}\:\mathrm{this}\:\mathrm{two}\:\mathrm{series}\:\mathrm{were}\:\mathrm{convergent}. \\ $$ Answered by Ar Brandon last…
Question Number 136559 by mohammad17 last updated on 23/Mar/21 $${show}\:{that}\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{{n}−\mathrm{4}}{{n}^{\mathrm{2}} −\mathrm{2}{n}+\mathrm{1}}\:{is}\:{converge}\:{or}\:{diverge}\: \\ $$$$ \\ $$$${by}\:{ussing}\:{any}\:{test} \\ $$ Commented by mr W last updated…
Question Number 136551 by Tinku Tara last updated on 23/Mar/21 $$\mathrm{Google}\:\mathrm{released}\:\mathrm{an}\:\mathrm{update}\:\mathrm{to}\:\mathrm{android} \\ $$$$\mathrm{webview}\:\mathrm{which}\:\mathrm{is}\:\mathrm{causing}\:\mathrm{multiple} \\ $$$$\mathrm{android}\:\mathrm{app}\:\mathrm{to}\:\mathrm{crash}.\:\mathrm{Please} \\ $$$$\mathrm{search}\:\mathrm{internet}\:\mathrm{for}\:\mathrm{workaround} \\ $$$$\mathrm{for}\:\mathrm{your}\:\mathrm{device}. \\ $$$$\boldsymbol{\mathrm{This}}\:\boldsymbol{\mathrm{problem}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{not}}\:\boldsymbol{\mathrm{caused}}\:\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{Tinkutara}} \\ $$$$\boldsymbol{\mathrm{Equation}}\:\boldsymbol{\mathrm{Editor}}. \\ $$$$\mathrm{Please}\:\mathrm{also}\:\mathrm{note}\:\mathrm{no}\:\mathrm{android}\:\mathrm{app}…
Question Number 136541 by mr W last updated on 23/Mar/21 $${to}\:{Tinku}\:\:{Tara}\:{sir}: \\ $$$${since}\:{update}\:{to}\:{version}\:\mathrm{2}.\mathrm{262}\:{i}'{m} \\ $$$${encountering}\:{a}\:{big}\:{problem}: \\ $$$${after}\:{some}\:{time}\:{of}\:{use}\:{the}\:{app}\: \\ $$$${crashes},\:{then}\:{i}\:{can}'{t}\:{start}\:{the}\:{app} \\ $$$${any}\:{more},\:{that}\:{means}\:{after}\:{displaying} \\ $$$${the}\:{welcome}\:{screen}\:{the}\:{app}\:{crashes}. \\ $$$${besides}\:{some}\:{other}\:{apps},\:{e}.{g}.\:{google},…
Question Number 136537 by Jamshidbek last updated on 23/Mar/21 Answered by Olaf last updated on 23/Mar/21 $${a}+\frac{\mathrm{1}}{{a}}\:=\:−\mathrm{1} \\ $$$${a}^{\mathrm{2}} +{a}+\mathrm{1}\:=\:\mathrm{0} \\ $$$${a}\:=\:\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\:{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \mathrm{or}\:{e}^{{i}\frac{\mathrm{4}\pi}{\mathrm{3}}} \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+\mathrm{8}+\mathrm{9}+\mathrm{1}+\mathrm{0}+\mathrm{1}+\mathrm{1}…