Question Number 214247 by hardmath last updated on 02/Dec/24 $$\mathrm{If}\:\:\:\mathrm{2a}\:=\:\mathrm{1}\:−\:\mathrm{2}\sqrt{\mathrm{a}} \\ $$$$\mathrm{Find}\:\:\:\frac{\mathrm{2a}^{\mathrm{2}} \:+\:\sqrt{\mathrm{a}}}{\mathrm{2a}}\:=\:? \\ $$ Answered by A5T last updated on 02/Dec/24 $$\mathrm{2}{a}=\mathrm{1}−\mathrm{2}\sqrt{{a}}\Rightarrow\begin{cases}{\sqrt{{a}}=\frac{\mathrm{2}{a}−\mathrm{1}}{−\mathrm{2}}}\\{\left(\mathrm{2}{a}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}{a}\Rightarrow\mathrm{4}{a}^{\mathrm{2}} +\mathrm{1}=\mathrm{8}{a}}\end{cases}…
Question Number 214191 by golsendro last updated on 01/Dec/24 $$\:\:\:\:\:\:\begin{cases}{\mathrm{y}^{\mathrm{3}} =\:\mathrm{x}^{\mathrm{x}+\mathrm{y}} }\\{\mathrm{y}^{\mathrm{x}+\mathrm{y}} \:=\:\mathrm{x}^{\mathrm{6}} \mathrm{y}^{\mathrm{3}} }\end{cases} \\ $$$$\:\:\:\:\cancel{\underline{ }} \\ $$ Answered by TonyCWX08 last updated…
Question Number 214186 by hardmath last updated on 30/Nov/24 $$\mathrm{If}\:\:\:\mathrm{x}\:+\:\frac{\mathrm{1}}{\mathrm{x}}\:=\:\mathrm{1}\:\:\:\mathrm{find}\:\:\:\mathrm{x}^{\mathrm{61}} \:+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{61}} }+\:\mathrm{4}\:\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 30/Nov/24 $${x}+\frac{\mathrm{1}}{{x}}=\mathrm{1};\:{x}^{\mathrm{61}} +\frac{\mathrm{1}}{{x}^{\mathrm{61}} }+\mathrm{4}=? \\…
Question Number 214144 by a.lgnaoui last updated on 29/Nov/24 $$\mathrm{calculer}\:\:\mathrm{n}\:\:\:\: \\ $$ Commented by a.lgnaoui last updated on 29/Nov/24 Commented by issac last updated on…
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Question Number 214122 by a.lgnaoui last updated on 28/Nov/24 $$\mathrm{Etude}\:\mathrm{d}\:\mathrm{une}\:\boldsymbol{\mathrm{suite}} \\ $$ Commented by a.lgnaoui last updated on 28/Nov/24 Terms of Service Privacy Policy Contact:…
Question Number 214098 by issac last updated on 28/Nov/24 $$\mathrm{Let}\:{F}\:\mathrm{be}\:\:\mathrm{Field}\:\mathrm{of}\:\mathrm{characteristic}\:\mathrm{0} \\ $$$${L}_{{i}} \:\left({i}=\mathrm{1},\mathrm{2}\right)\:\mathrm{be}\:\mathrm{two}\:\mathrm{algebraic}\:\mathrm{extension} \\ $$$$\mathrm{of}\:{F}\:,\:\mathrm{and}\:{L}_{\mathrm{1}} {L}_{\mathrm{2}} \:\mathrm{be}\:\mathrm{a}\:\mathrm{field}\:\mathrm{in}\:\bar {{F}}\: \\ $$$$\left(\mathrm{where}\:\bar {{F}}\:\:\mathrm{is}\:\mathrm{the}\:\mathrm{algebraic}\:\mathrm{closure}\:\:\mathrm{of}\:{F}\right) \\ $$$$\mathrm{defined}\:\mathrm{by}\:\left\{{l}_{\mathrm{1}} {l}_{\mathrm{2}} \mid{l}_{{i}}…
Question Number 214061 by a.lgnaoui last updated on 25/Nov/24 $$\mathrm{determiner}\:\mathrm{l}\:\mathrm{equation}\:\mathrm{de}\:\mathrm{la}\:\mathrm{parabole} \\ $$$$\mathrm{represente}\:\mathrm{par}\:\mathrm{la}\:\mathrm{courbe}\:\mathrm{ci}−\mathrm{dessous} \\ $$$$\mathrm{avec}\:\boldsymbol{\mathrm{y}}=−\mathrm{3} \\ $$ Commented by a.lgnaoui last updated on 25/Nov/24 Terms of…
Question Number 214059 by RoseAli last updated on 25/Nov/24 Answered by a.lgnaoui last updated on 25/Nov/24 $$\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{5}\right)\:=\mathrm{0}\:\:\:\: \\ $$$$\: \\ $$$$\begin{cases}{\mathrm{x}_{\mathrm{1}} \:\:\:\:=\left\{\frac{\mathrm{10}}{\mathrm{10}}\:\:,\frac{−\mathrm{10}}{−\mathrm{10}}\right\}}\\{\mathrm{x}_{\mathrm{2}} \:\:\:=\left\{\frac{\mathrm{50}}{\mathrm{10}}\:\:\:\:,\frac{−\mathrm{50}}{−\mathrm{10}}\:\:\:\right\}\:\:}\end{cases} \\ $$$$\boldsymbol{\mathrm{S}}=\left\{\frac{−\mathrm{50}}{−\mathrm{10}}\:;\:\:\frac{−\mathrm{10}}{−\mathrm{10}}\:\:;\:\frac{+\mathrm{10}}{+\mathrm{10}}\:;\:\frac{+\mathrm{50}}{+\mathrm{10}}\right\}…
Question Number 214051 by issac last updated on 25/Nov/24 $$\mathrm{evaluate}\:\int_{\mathrm{0}} ^{\:\pi} \:{e}^{\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u}…\: \\ $$$$\mathrm{i}\:\mathrm{use}\:\mathrm{Feynman}'\mathrm{s}\:\mathrm{trick}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{integral} \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \:{e}^{\mathrm{sin}^{\mathrm{2}} \left({u}\right)} \mathrm{d}{u}={I} \\ $$$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\:\pi}…