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Author: Tinku Tara

Question-201411

Question Number 201411 by MrGHK last updated on 05/Dec/23 Commented by mr W last updated on 06/Dec/23 $${Mr}\:{Laplac}:\:{please}\:{post}\:{your}\:{answer} \\ $$$${in}\:{the}\:{thread}\:{corresponding}\:{to}\:{the} \\ $$$${question}!\:{otherwise}\:{it}'{s}\:{not}\:{clear} \\ $$$${what}\:{your}\:{post}\:{is}\:{for}. \\…

Question-201308

Question Number 201308 by Mingma last updated on 04/Dec/23 Commented by aleks041103 last updated on 04/Dec/23 $$−{Q}\Leftrightarrow−{Q} \\ $$$${Q}\Leftrightarrow−\left(−{Q}\right) \\ $$$${P}\Rightarrow{Q}\Leftrightarrow−\left(−{Q}\right) \\ $$$$\therefore{P}\Rightarrow−\left({Q}\right) \\ $$…

Question-201329

Question Number 201329 by MrGHK last updated on 04/Dec/23 Answered by witcher3 last updated on 04/Dec/23 $$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{2n}+\mathrm{1}}\underset{\mathrm{m}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{m}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{m}+\mathrm{2n}+\mathrm{1}} \mathrm{dx} \\…

Question-201322

Question Number 201322 by cherokeesay last updated on 04/Dec/23 Answered by AST last updated on 05/Dec/23 $${Let}\:\angle{EDC}=\beta\Rightarrow\angle{ABD}=\mathrm{2}\beta−\mathrm{90} \\ $$$${EC}^{\mathrm{2}} =\mathrm{2}−\mathrm{2}{cos}\beta…\left({i}\right) \\ $$$${BC}^{\mathrm{2}} ={BD}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{BD}\boldsymbol{{D}}{Ccos}\left(\mathrm{2}\beta\right)…\left({ii}\right) \\…