Question Number 196352 by SANOGO last updated on 23/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 196355 by MATHEMATICSAM last updated on 23/Aug/23 $$\mathrm{log}_{\mathrm{5}} \sqrt{\mathrm{5}\sqrt{\mathrm{5}\sqrt{\mathrm{5}\sqrt{\mathrm{5}…..}}}\:}=\:? \\ $$ Commented by bbbbbbbb last updated on 23/Aug/23 $$\mathrm{log}_{\mathrm{5}} \sqrt{\mathrm{5}\sqrt{\mathrm{5}\sqrt{\mathrm{5}\sqrt{\mathrm{5}…..}}}\:}=\:? \\ $$$$\sqrt{\mathrm{x}\sqrt{\mathrm{x}\sqrt{\mathrm{x}\sqrt{\mathrm{x}…..}}}}=\mathrm{x} \\…
Question Number 196286 by galivan last updated on 22/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 196309 by sonukgindia last updated on 22/Aug/23 Commented by Frix last updated on 22/Aug/23 $$\sqrt{\mathrm{65}}\:\mathrm{at}\:{x}=−\frac{\mathrm{13}}{\mathrm{11}} \\ $$ Answered by AST last updated on…
Question Number 196311 by sonukgindia last updated on 22/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 196337 by SaRahAli last updated on 22/Aug/23 Answered by MM42 last updated on 22/Aug/23 $${lim}_{{x}\rightarrow\infty} \:\frac{{ln}\left(\mathrm{1}+\frac{{a}}{{x}}\right)}{\frac{\mathrm{1}}{{x}}}\:\overset{\frac{\mathrm{1}}{{x}}=\:{u}} {=}{lim}_{{u}\rightarrow\mathrm{0}} \:\frac{{ln}\left(\mathrm{1}+{au}\right)}{{u}} \\ $$$${hop}={lim}_{{u}\rightarrow\mathrm{0}} \:\frac{\frac{{a}}{\mathrm{1}+{au}}}{\mathrm{1}}\:={a}\:\checkmark \\ $$…
Question Number 196304 by cortano12 last updated on 22/Aug/23 Answered by a.lgnaoui last updated on 24/Aug/23 $$\:\boldsymbol{\mathrm{S}}=\mathrm{shaded}\:\mathrm{Area} \\ $$$$\boldsymbol{\mathrm{S}}\mathrm{1}=\mathrm{Arc}\left(\mathrm{AMC}\right)\:\:\:\boldsymbol{\mathrm{S}}\mathrm{2}=\mathrm{Arc}\left(\mathrm{OBD}\right) \\ $$$$\:\boldsymbol{\mathrm{S}}=\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{ABCD}}\right)−\boldsymbol{\mathrm{S}}\mathrm{1}+\boldsymbol{\mathrm{S}}\mathrm{2} \\ $$$$\bullet\boldsymbol{\mathrm{Calcul}}\:\boldsymbol{\mathrm{de}}\:\boldsymbol{\mathrm{S}}\left(\mathrm{ABCD}\right) \\ $$$$\:\:\mathrm{OBsin}\:\mathrm{30}=\mathrm{OAsin}\:\mathrm{45}\Rightarrow\:\:\boldsymbol{\mathrm{OA}}=\mathrm{5}\sqrt{\mathrm{2}}\:…
Question Number 196300 by hardmath last updated on 22/Aug/23 $$\mathrm{If}\:\rightarrow\:\mathrm{n}\:\in\:\mathbb{N}\:\:\:\:\:\mathrm{and}\:\:\:\:\:\mathrm{n}\:\geqslant\:\mathrm{2} \\ $$$$\mathrm{Then}\:\rightarrow\:\mathrm{tan}\:\left(\frac{\mathrm{1}}{\mathrm{n}\:−\:\mathrm{1}}\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{2}} {\overset{\boldsymbol{\mathrm{n}}} {\sum}}\:\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{k}}\right)\:<\:\frac{\mathrm{2}}{\mathrm{5}}\:+\:\frac{\boldsymbol{\gamma}}{\mathrm{n}\:−\:\mathrm{1}} \\ $$ Answered by sniper237 last updated on 22/Aug/23 $${n}=\mathrm{2}\Rightarrow\mathrm{0},\mathrm{5}={tan}\left({arctan}\left(\mathrm{1}/\mathrm{2}\right)\right)\overset{?} {<}\mathrm{2}/\mathrm{5}\:=\mathrm{0},\mathrm{4}\:…
Question Number 196303 by sonukgindia last updated on 22/Aug/23 Answered by sniper237 last updated on 22/Aug/23 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:{Im}\left(\frac{{e}^{{i}\pi/\mathrm{4}} }{\mathrm{2}}\right)^{{n}} ={Im}\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({e}^{{i}\pi/\mathrm{4}} /\mathrm{2}\right)^{{n}} \right)…
Question Number 196302 by cortano12 last updated on 22/Aug/23 Answered by Rasheed.Sindhi last updated on 22/Aug/23 $$\mathrm{60}+\left(\mathrm{180}−{a}\right)+\left(\mathrm{180}−{b}\right)+\left(\mathrm{180}−{c}\right)=\mathrm{360} \\ $$$$\mathrm{60}+\mathrm{180}×\mathrm{3}−\mathrm{360}={a}+{b}+{c} \\ $$$${a}+{b}+{c}=\mathrm{240} \\ $$ Commented by…