Question Number 204372 by mnjuly1970 last updated on 14/Feb/24 $$ \\ $$$$\:\:{If}\:,\:\:\:\:{f}\::\:\left[\:\mathrm{0}\:,\:{b}\right]\:\overset{{continuous}} {\rightarrow}\:\mathbb{R}\: \\ $$$$\:\:\:\:\:\:\:\:,\:\:\:\:{g}\::\:\mathbb{R}\:\underset{{b}−{periodic}} {\overset{{continuous}} {\rightarrow}}\:\mathbb{R} \\ $$$$\:\:\:\:\:\:\Rightarrow\:\:{lim}_{{n}\rightarrow\infty} \:\int_{\mathrm{0}} ^{\:{b}} {f}\left({x}\right){g}\left({nx}\right){dx}\overset{?} {=}\frac{\mathrm{1}}{{b}}\:\int_{\mathrm{0}} ^{\:{b}} {f}\left({x}\right){dx}\:.\int_{\mathrm{0}}…
Question Number 204373 by esmaeil last updated on 14/Feb/24 $$\mathrm{3}^{{x}} −\mathrm{2}^{{x}} =\mathrm{65} \\ $$$${x}=? \\ $$ Answered by Frix last updated on 14/Feb/24 $$\mathrm{Obviously}\:{x}=\mathrm{4} \\…
Question Number 204374 by mr W last updated on 14/Feb/24 Commented by mr W last updated on 15/Feb/24 $${The}\:{radius}\:{of}\:{the}\:{big}\:{sphere}\:\left({say}\:\right. \\ $$$$\left.{the}\:{Earth}\right)\:{is}\:{R}\:{and}\:{the}\:{radius}\:{of} \\ $$$${the}\:{small}\:{sphere}\:\left({say}\:{theMoon}\right)\:{is}\:{r}. \\ $$$${find}\:{the}\:{maximum}\:{area}\:{of}\:{the}…
Question Number 204334 by universe last updated on 13/Feb/24 Answered by Frix last updated on 13/Feb/24 $${n}\rightarrow\infty\:\Rightarrow\:{a}_{{n}} =\sqrt{\mathrm{2}{a}_{{n}} }\:\Rightarrow\:{a}_{{n}} =\mathrm{2} \\ $$ Answered by MM42…
Question Number 204344 by SEKRET last updated on 13/Feb/24 $$ \\ $$$$\: \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\boldsymbol{\mathrm{ctg}}^{\mathrm{6}} \left(\frac{\pi}{\mathrm{9}}\right)−\mathrm{9}\centerdot\boldsymbol{\mathrm{ctg}}^{\mathrm{4}} \left(\frac{\pi}{\mathrm{9}}\right)+\mathrm{11}\centerdot\boldsymbol{\mathrm{ctg}}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{9}}\right)=?\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$…
Question Number 204328 by MASANJAJJ last updated on 13/Feb/24 Commented by MASANJAJJ last updated on 13/Feb/24 $$ \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 204329 by mr W last updated on 13/Feb/24 $${solve}\:\frac{\mathrm{1}}{\left[{x}\right]}+\frac{\mathrm{1}}{\left[\mathrm{2}{x}\right]}=\left\{{x}\right\}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$ Answered by AST last updated on 13/Feb/24 $${x}\:{cannot}\:{be}\:{negative},{otherwise},{L}.{H}.{S}\:{and}\:{R}.{H}.{S} \\ $$$${will}\:{have}\:{opposite}\:{signs}.\:{x}\:{cannot}\:{also}\:{grow} \\ $$$${arbitrarily}\:{large},{otherwise}\:\frac{\mathrm{1}}{\left[{x}\right]}+\frac{\mathrm{1}}{\left[\mathrm{2}{x}\right]}\ll\frac{\mathrm{1}}{\mathrm{3}}…
Question Number 204330 by universe last updated on 13/Feb/24 Answered by AST last updated on 13/Feb/24 $$\mathrm{3}{a}+\mathrm{2}{b}+\mathrm{15}=\mathrm{0}\:\:\:\:\wedge\:\:\:\mathrm{6}{a}+\mathrm{2}{b}\leqslant\mathrm{0}\Rightarrow{b}\leqslant−\mathrm{3}{a} \\ $$$$\Rightarrow{b}=\frac{−\mathrm{15}−\mathrm{3}{a}}{\mathrm{2}}\leqslant−\mathrm{3}{a}\Rightarrow−\mathrm{15}−\mathrm{3}{a}\leqslant−\mathrm{6}{a}\Rightarrow\mathrm{3}{a}\leqslant\mathrm{15} \\ $$$$\Rightarrow{a}\leqslant\mathrm{5}\Rightarrow{max}\left({a}\mid{a}\in\mathbb{Z}\right)=\mathrm{5} \\ $$ Commented by…
Question Number 204337 by SANOGO last updated on 13/Feb/24 Answered by witcher3 last updated on 13/Feb/24 $$\left(\mathrm{3}\right)\Rightarrow\left(\mathrm{2}\right) \\ $$$$\mathrm{soitU}\:\mathrm{un}\:\mathrm{ouvert}\:\mathrm{de}\:\mathrm{E}\: \\ $$$$\exists\:\mathrm{existe}\:\mathrm{V}\:\mathrm{un}\:\mathrm{ouvert}\:\mathrm{de}\:\mathrm{F} \\ $$$$\mathrm{t}\:\mathrm{elle}\:\mathrm{Que}\:\mathrm{U}=\mathrm{f}^{−} \left(\mathrm{V}\right);\mathrm{car}\:\mathrm{f}\:\mathrm{et}\:\mathrm{bijective}\:\mathrm{donc}\:\mathrm{f}^{−} \\…
Question Number 204318 by MASANJAJJ last updated on 12/Feb/24 Answered by Rasheed.Sindhi last updated on 12/Feb/24 $$\bullet{T}\left({U}+{V}\right)={T}\left(\:\left(−\mathrm{12},\mathrm{12}\right)+\left(\mathrm{6},−\mathrm{16}\right)\:\right) \\ $$$$\:\:\:\:\:={T}\left(−\mathrm{6},−\mathrm{4}\right)=\left(−\mathrm{6}+\mathrm{8},−\mathrm{4}+\mathrm{7}\right)=\left(\mathrm{2},\mathrm{3}\right) \\ $$$$\: \\ $$$$\bullet{T}\left({U}\right)+{T}\left({V}\right)={T}\left(−\mathrm{12},\mathrm{12}\right)+{T}\left(\mathrm{6},−\mathrm{16}\right) \\ $$$$\:\:\:=\left(−\mathrm{12}+\mathrm{8},\mathrm{12}+\mathrm{7}\right)+\left(\mathrm{6}+\mathrm{8},−\mathrm{16}+\mathrm{7}\right)…