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Question Number 196946 by Khalmohmmad last updated on 04/Sep/23 Answered by trisetyo last updated on 04/Sep/23 $$\mathrm{3}^{{x}} +\mathrm{2}^{{x}} =\mathrm{5} \\ $$$$\mathrm{3}^{{x}} =−\left(\mathrm{2}^{{x}} −\mathrm{5}\right)\vee\mathrm{2}^{{x}} =−\left(\mathrm{3}^{{x}} −\mathrm{5}\right)…

Question Number 196914 by SANOGO last updated on 03/Sep/23 Answered by witcher3 last updated on 03/Sep/23 $$−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}>−\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\mathrm{I}_{\mathrm{n}} =\left[−\frac{\mathrm{1}}{\mathrm{n}},\mathrm{1}\right] \\ $$$$\mathrm{I}_{\mathrm{n}+\mathrm{1}} \:\:\subseteq\mathrm{I}_{\mathrm{n}} \:\:\mathrm{suite}\:\mathrm{decroissante}\:\mathrm{minore} \\…

Question Number 196913 by ERLY last updated on 02/Sep/23 $${soit}\:\left\{_{{r}_{{n}+\mathrm{1}} ={r}_{{n}} /\left(\mathrm{2}+{r}_{{n}} ^{\mathrm{2}} \right)} ^{{r}_{\mathrm{0}} =\mathrm{1}} \right. \\ $$$${demontrer}\:{sans}\:{recurrence}\:{que}\:{r}_{{n}} >\mathrm{0} \\ $$$${demontrer}\:{par}\:{recurrence}\:{que}\:{r}_{{n}+\mathrm{1}} \leq\frac{\mathrm{1}}{\mathrm{2}}{r}_{{n}} \\ $$$${demontrer}\:{sans}\:{recurrence}\:{que}\:{r}_{{n}}…

Question Number 196848 by sonukgindia last updated on 01/Sep/23 Commented by Frix last updated on 01/Sep/23 $$\mathrm{The}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{42}\:\left(\mathrm{or}\:\mathrm{else}\:\mathrm{prove}\:\mathrm{it}'\mathrm{s}\:\boldsymbol{{not}}\:\mathrm{42}\right) \\ $$ Answered by AST last updated on…

Question Number 196850 by sonukgindia last updated on 01/Sep/23 Terms of Service Privacy Policy Contact: info@tinkutara.com