Question Number 61027 by Gulay last updated on 28/May/19 Commented by Gulay last updated on 28/May/19 $$\mathrm{sir}\:\mathrm{could}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me} \\ $$ Commented by mathsolverby Abdo last updated…
Question Number 192099 by a.lgnaoui last updated on 08/May/23 $$\mathrm{demontrer}\:\:\mathrm{l}\:\mathrm{expression}\:\mathrm{suivante} \\ $$ Commented by a.lgnaoui last updated on 08/May/23 Terms of Service Privacy Policy Contact:…
Question Number 192098 by cortano12 last updated on 08/May/23 Answered by mehdee42 last updated on 08/May/23 $$\left({c}\right)\:\:\frac{\mathrm{6}}{\mathrm{7}}\:\:\checkmark \\ $$ Commented by Skabetix last updated on…
Question Number 192095 by Mastermind last updated on 07/May/23 $$\mathrm{Prove}\:\mathrm{a}\:\mathrm{non}−\mathrm{empty}\:\mathrm{set}\:\mathrm{S}\:\mathrm{of}\:\mathrm{a}\:\mathrm{group} \\ $$$$\mathrm{G}\:\mathrm{wrt}\:\mathrm{binary}\:\mathrm{operation}\:\ast\:\mathrm{is}\:\mathrm{a}\:\mathrm{sub}− \\ $$$$\mathrm{group}\:\mathrm{of}\:\mathrm{G}.\:\mathrm{Iff}\: \\ $$$$\left.\mathrm{1}\right)\:\mathrm{a},\mathrm{b}\:\in\:\mathrm{S}\:\Rightarrow\:\mathrm{a}\ast\mathrm{b}\in\mathrm{S} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{a}\:\in\:\mathrm{S}\:\Rightarrow\:\mathrm{a}^{−\mathrm{1}} \:\in\:\mathrm{S}. \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Hello}…
Question Number 192094 by Mastermind last updated on 08/May/23 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{order}\:\mathrm{of}\:\mathrm{a}\:\mathrm{subgroup} \\ $$$$\mathrm{S}\:\mathrm{of}\:\mathrm{a}\:\mathrm{finite}\:\mathrm{group}\:\mathrm{G},\:\mathrm{always}\:\mathrm{divide} \\ $$$$\mathrm{the}\:\mathrm{order}\:\mathrm{of}\:\mathrm{group}\:\mathrm{G}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 126557 by I want to learn more last updated on 21/Dec/20 Commented by I want to learn more last updated on 21/Dec/20 $$\mathrm{Find}\:\:\mathrm{k}…
Question Number 126552 by mathocean1 last updated on 21/Dec/20 $${u}\:;\:{v}\in\:\mathbb{C}\:{such}\:{that}\:\mid{u}\mid=\mid{v}\mid=\mathrm{1} \\ $$$${and}\:{uv}\neq−\mathrm{1}. \\ $$$${show}\:{that}\:\frac{{u}+{v}}{\mathrm{1}+{uv}}\:\in\:\mathbb{R}. \\ $$ Answered by Olaf last updated on 21/Dec/20 $${u}\:=\:{e}^{{i}\alpha} ,\:{v}\:=\:{e}^{{i}\beta}…
Question Number 126553 by mathocean1 last updated on 21/Dec/20 $${show}\:{thatfor}\:{z}\in\mathbb{C}\: \\ $$$$\mid{z}+\mathrm{1}\mid^{\mathrm{2}} =\mathrm{2}\mid{z}\mid^{\mathrm{2}\:} \Leftrightarrow\mid{z}−\mathrm{1}\mid^{\mathrm{2}} =\mathrm{2}. \\ $$$${This}\:{formula}\:{can}\:{be}\:{used}: \\ $$$$\mid{z}\mid^{\mathrm{2}} ={z}×\overset{−} {{z}} \\ $$ Answered by…
Question Number 192084 by sciencestudentW last updated on 07/May/23 $${f}\left({x}\right)+{x}\centerdot{f}\left(−{x}\right)={x}^{\mathrm{2}} +\mathrm{1} \\ $$$${f}\left(\sqrt{\mathrm{2}}\right)=? \\ $$ Answered by AST last updated on 07/May/23 $${x}^{\mathrm{2}} +\mathrm{1}=−\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)+\mathrm{2}=−\left(\mathrm{1}+{x}−{x}\left(\mathrm{1}+{x}\right)\right)+\mathrm{2} \\…
Question Number 192087 by Mastermind last updated on 07/May/23 $$\mathrm{Let}\:\left\{\mathrm{H}_{\alpha} \right\}\:\in\:\Omega,\:\mathrm{be}\:\mathrm{a}\:\mathrm{family}\:\mathrm{of}\: \\ $$$$\mathrm{subgroup}\:\mathrm{of}\:\mathrm{a}\:\mathrm{group}\:\mathrm{G},\:\mathrm{then}\: \\ $$$$\mathrm{prove}\:\mathrm{that}\:\cap_{\alpha\:\in\:\Omega} \mathrm{H}_{\alpha} . \\ $$$$ \\ $$$$ \\ $$ Commented by…