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Question-197635

Question Number 197635 by sonukgindia last updated on 25/Sep/23 Commented by mr W last updated on 26/Sep/23 $${it}'{s}\:{strange}!!! \\ $$$${even}\:{it}\:{is}\:{not}\:{said}\:{what}\:{the}\:{question} \\ $$$${is},\:{people}\:{gave}\:\mathrm{2}\:{likes}\:{to}\:{the}\:{question}.\: \\ $$$${what}\:{do}\:{these}\:{people}\:{like}? \\…

Question-197667

Question Number 197667 by SANOGO last updated on 25/Sep/23 Answered by Frix last updated on 26/Sep/23 $${X},\:{Y}\in\mathbb{R}^{\mathrm{3}} ,\:{r}\in\mathbb{R} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:{N}\left({X}\right)=\mathrm{0}\:\Rightarrow\:{X}=\mathrm{0}\:\checkmark \\ $$$$\:\:\:\:\:{X}\neq\mathrm{0}\:\Leftrightarrow\:{x}_{\mathrm{1}} \neq\mathrm{0}\vee{x}_{\mathrm{2}}…

Question-197622

Question Number 197622 by sonukgindia last updated on 24/Sep/23 Answered by a.lgnaoui last updated on 26/Sep/23 $$\boldsymbol{\mathrm{Calcul}}\:\boldsymbol{\mathrm{de}}\:\:\:\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{BCDE}}\right) \\ $$$$ \\ $$$$\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{BCDE}}\right)=\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{AMN}}\right)−\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{DEMN}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{ABC}}\right) \\ $$$$…

find-n-1-k-n-

Question Number 197581 by mokys last updated on 22/Sep/23 $${find}\:\underset{{n}=\mathrm{1}} {\overset{{k}} {\sum}}\sqrt{{n}}\:? \\ $$ Answered by salvatore last updated on 07/Jan/24 $$\left.{w}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{24}}\:\:\:{w}\left({m}\right)=\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}}\\{\:{m}}\end{pmatrix}\:\frac{{B}_{{m}+\mathrm{1}} }{{m}+\mathrm{1}}\:−\left(−\mathrm{1}\right)^{{m}} \underset{{h}=\mathrm{1}} {\overset{{m}−\mathrm{1}}…

Question-197524

Question Number 197524 by sonukgindia last updated on 20/Sep/23 Answered by som(math1967) last updated on 20/Sep/23 $$\mathrm{2arc}{tan}\left(\frac{{x}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right) \\ $$$$=\mathrm{arc}{tan}\left\{\frac{\frac{\mathrm{2}{x}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} }}\right\} \\…