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Category: Algebra

Question-215052

Question Number 215052 by Abdullahrussell last updated on 27/Dec/24 Answered by mr W last updated on 27/Dec/24 $${f}\left(\mathrm{1}\right)=\mathrm{1}\:\Rightarrow{f}\left({x}\right)=\left({x}−\mathrm{1}\right){g}\left({x}\right)+\mathrm{1} \\ $$$${f}\left(\mathrm{2}\right)=\left(\mathrm{2}−\mathrm{1}\right){g}\left(\mathrm{2}\right)+\mathrm{1}=\mathrm{4}\:\Rightarrow{g}\left(\mathrm{2}\right)=\mathrm{3}\:\Rightarrow{g}\left({x}\right)=\left({x}−\mathrm{2}\right){h}\left({x}\right)+\mathrm{3} \\ $$$$\Rightarrow{f}\left({x}\right)=\left({x}−\mathrm{1}\right)\left[\left({x}−\mathrm{2}\right){h}\left({x}\right)+\mathrm{3}\right]+\mathrm{1} \\ $$$${f}\left(\mathrm{3}\right)=\left(\mathrm{3}−\mathrm{1}\right)\left[\left(\mathrm{3}−\mathrm{2}\right){h}\left(\mathrm{3}\right)+\mathrm{3}\right]+\mathrm{1}=\mathrm{3}\:\Rightarrow{h}\left(\mathrm{3}\right)=−\mathrm{2}\:\Rightarrow{h}\left({x}\right)=\left({x}−\mathrm{3}\right){k}\left({x}\right)−\mathrm{2} \\…

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Question Number 215072 by hardmath last updated on 27/Dec/24 $$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{2000x}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{Roots}:\:\:\boldsymbol{\mathrm{a}}\:\:\mathrm{and}\:\:\boldsymbol{\mathrm{b}} \\ $$$$\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{2008x}\:−\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{Roots}:\:\:\boldsymbol{\mathrm{c}}\:\:\mathrm{and}\:\:\boldsymbol{\mathrm{d}} \\ $$$$\mathrm{Find}:\:\:\left(\mathrm{a}+\mathrm{c}\right)\left(\mathrm{b}+\mathrm{d}\right)\left(\mathrm{a}−\mathrm{d}\right)\left(\mathrm{b}−\mathrm{c}\right)\:=\:? \\ $$ Commented by TonyCWX08…

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Question Number 215005 by Abdullahrussell last updated on 25/Dec/24 $$\:{a},{b},{c},{d}\in{R}\:{such}\:{that}, \\ $$$$\:\left({a}+{b}\right)\left({c}+{d}\right)=\mathrm{2} \\ $$$$\:\left({a}+{c}\right)\left({b}+{d}\right)=\mathrm{3} \\ $$$$\:\left({a}+{d}\right)\left({b}+{c}\right)=\mathrm{4}\: \\ $$$$\:{find}:\:\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)_{{minimum}.} \\ $$ Answered…

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Question Number 214977 by Abdullahrussell last updated on 25/Dec/24 $$\:{Solve}: \\ $$$$\:\frac{\left(\mathrm{3}{x}−\mathrm{5}\right)^{\mathrm{5}} −\left(\mathrm{2}{x}−\mathrm{3}\right)^{\mathrm{5}} −\left({x}−\mathrm{2}\right)^{\mathrm{5}} }{\left(\mathrm{3}{x}−\mathrm{5}\right)^{\mathrm{3}} −\left(\mathrm{2}{x}−\mathrm{3}\right)^{\mathrm{3}} −\left({x}−\mathrm{2}\right)^{\mathrm{3}} }=\mathrm{65} \\ $$ Answered by MrGaster last updated…

Question-214916

Question Number 214916 by Spillover last updated on 23/Dec/24 Answered by A5T last updated on 23/Dec/24 $$\mathrm{2017}^{\mathrm{2017}^{\mathrm{2017}} } \equiv\mathrm{1}^{\mathrm{2017}^{\mathrm{2017}} } =\mathrm{1}\left({mod}\:\mathrm{16}\right) \\ $$$$\mathrm{2017}^{\mathrm{2017}^{\mathrm{2017}} } \equiv\mathrm{142}^{\mathrm{2017}^{\mathrm{2017}}…

Question-214888

Question Number 214888 by Emmanuel07 last updated on 22/Dec/24 Commented by mr W last updated on 23/Dec/24 $${i}\:{got} \\ $$$${a}_{{n}} =\mathrm{cot}\:\left\{\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{2}−\frac{\mathrm{3}\pi}{\mathrm{8}}\right)+\frac{\pi}{\mathrm{8}}\right]\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} −\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{tan}^{−\mathrm{1}} \mathrm{2}−\frac{\mathrm{3}\pi}{\mathrm{8}}\right)−\frac{\pi}{\mathrm{8}}\right]\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{n}} \right\}…

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Question Number 214805 by universe last updated on 20/Dec/24 $$\:\:\mathrm{let}\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\:=\:\mathrm{x}^{\mathrm{2}} −\mathrm{2xy}+\mathrm{y}^{\mathrm{2}} −\mathrm{y}^{\mathrm{3}} +\mathrm{x}^{\mathrm{5}} \:+\mathrm{show}\:\mathrm{that} \\ $$$$\:\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{has}\:\mathrm{neither}\:\mathrm{a}\:\mathrm{maximum}\:\mathrm{nor}\:\mathrm{a}\: \\ $$$$\:\:\mathrm{minimum}\:\mathrm{at}\:\left(\mathrm{0},\mathrm{0}\right) \\ $$ Answered by TonyCWX08 last updated…

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Question Number 214779 by MrGaster last updated on 19/Dec/24 $$ \\ $$$$\mathrm{seek}\:\mathrm{the}\:\mathrm{solutions}\:\mathrm{to}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\sqrt{\mathrm{3}}+\mathrm{2}\:\mathrm{sin}\:\mathrm{2}{x}}{\:\sqrt{\mathrm{3}}+\mathrm{2}\:\mathrm{sin}\:{x}}=\sqrt{\mathrm{3}}\mathrm{sin}\:{x}+\frac{\mathrm{cos2}{x}}{\mathrm{2}\:\mathrm{cos}\:{x}} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{interval}\:\:\mathrm{between}\:\mathrm{0}\:\mathrm{and}\frac{\pi}{\mathrm{2}}. \\ $$$$ \\ $$ Terms of Service Privacy Policy…