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) \\ $$$$\:\:\:\:\:={T}\left(−\mathrm{6},−\mathrm{4}\right)=\left(−\mathrm{6}+\mathrm{8},−\mathrm{4}+\mathrm{7}\right)=\left(\mathrm{2},\mathrm{3}\right) \\ $$$$\: \\ $$$$\bullet{T}\left({U}\right)+{T}\left({V}\right)={T}\left(−\mathrm{12},\mathrm{12}\right)+{T}\left(\mathrm{6},−\mathrm{16}\right) \\ $$$$\:\:\:=\left(−\mathrm{12}+\mathrm{8},\mathrm{12}+\mathrm{7}\right)+\left(\mathrm{6}+\mathrm{8},−\mathrm{16}+\mathrm{7}\right)…
Question Number 204313 by serenity last updated on 12/Feb/24 $${lim}\frac{\mathrm{3}×^{\mathrm{2}} −\mathrm{8}×−\mathrm{16}}{\mathrm{2}×^{\mathrm{2}} \mathrm{9}×+\mathrm{4}} \\ $$$$ \\ $$ Answered by Faetmaaa last updated on 27/Feb/24 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}{x}^{\mathrm{2}}…
Question Number 204310 by BaliramKumar last updated on 12/Feb/24 Answered by mr W last updated on 12/Feb/24 Commented by mr W last updated on 12/Feb/24…
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Question Number 204300 by depressiveshrek last updated on 11/Feb/24 $${x}^{\mathrm{2}} \mathrm{log}_{\mathrm{3}} {x}^{\mathrm{2}} −\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)\mathrm{log}_{\mathrm{9}} \left(\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{3log}_{\mathrm{3}} \left(\frac{{x}}{\mathrm{2}{x}+\mathrm{3}}\right) \\ $$ Answered by Rasheed.Sindhi last updated on 11/Feb/24…