Question Number 204739 by hardmath last updated on 26/Feb/24 Answered by mr W last updated on 26/Feb/24 $${x}^{\mathrm{4}} +\mathrm{2}{x}=−\mathrm{4} \\ $$$$\Rightarrow{x}^{\mathrm{3}} +\mathrm{2}=−\frac{\mathrm{4}}{{x}} \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{3}}…
Question Number 204702 by mnjuly1970 last updated on 25/Feb/24 $$ \\ $$$$\:\:\:\mathrm{prove}\:\mathrm{that}\:: \\ $$$$\:\:\:\:\:\:\:\mathrm{cl}\left(\mathbb{Q}×\mathbb{Q}\:\right)\overset{?} {=}\:\mathbb{R}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{note}:\:\:\:\left({X}\:,{d}\:\right)\:{is}\:{a}\:{metric}\:{space} \\ $$$$\:\:\:\:\:\:\:\:\:\:,\:\:\:{A}\:\subseteq\:{X}\::\:\:\:\:\:{x}\in\:\overset{\:\:−} {{A}}=\mathrm{cl}\left({A}\right)\:\Leftrightarrow\:\forall\:{r}\:>\mathrm{0}\:,\:{B}_{{r}} \:\left({x}\right)\:\cap\:{A}\:\neq\:\phi \\ $$ Answered by…
Question Number 204701 by Abdullahrussell last updated on 25/Feb/24 Answered by A5T last updated on 25/Feb/24 $${x}^{\mathrm{3}} ={x}−\mathrm{2}\Rightarrow{x}^{\mathrm{4}} ={x}^{\mathrm{2}} −\mathrm{2}{x}\Rightarrow{x}^{\mathrm{5}} ={x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} =−\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{2} \\…
Question Number 204664 by cortano12 last updated on 25/Feb/24 $$\:\:\mathrm{8}+\sqrt{\mathrm{8}^{\mathrm{2}} +\sqrt{\mathrm{8}^{\mathrm{4}} +\sqrt{\mathrm{8}^{\mathrm{8}} +\sqrt{\mathrm{8}^{\mathrm{16}} +\sqrt{…}}}}}\:=\:?\: \\ $$ Answered by Frix last updated on 25/Feb/24 $$=\mathrm{8}+\mathrm{8}\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+…}}}} \\…
Question Number 204666 by Ghisom last updated on 25/Feb/24 $$\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{closed}\:\mathrm{curve}: \\ $$$${f}\left(\theta\right)=\left(\mathrm{e}^{\mathrm{i}\theta} \right)^{\left(\mathrm{e}^{\mathrm{i}\theta} \right)} =\mathrm{e}^{−\theta\mathrm{sin}\:\theta} \mathrm{e}^{\mathrm{i}\theta\mathrm{cos}\:\theta} ;\:−\pi<\theta\leqslant\pi \\ $$$${f}:\:\begin{cases}{{x}\left(\theta\right)=\mathrm{e}^{−\theta\mathrm{sin}\:\theta} \mathrm{cos}\:\left(\theta\mathrm{cos}\:\theta\right)}\\{{y}\left(\theta\right)=\mathrm{e}^{−\theta\mathrm{sin}\:\theta} \mathrm{sin}\:\left(\theta\mathrm{cos}\:\theta\right)}\end{cases} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{area} \\ $$…
Question Number 204647 by Engr_Jidda last updated on 24/Feb/24 Answered by Rasheed.Sindhi last updated on 24/Feb/24 $$\left(\frac{{x}}{\mathrm{2}}\right)^{\frac{{x}}{\mathrm{2}}−\mathrm{1}} =\mathrm{3}^{\mathrm{2}} \\ $$$$\Leftarrow\frac{{x}}{\mathrm{2}}=\mathrm{3}\:\wedge\:\frac{{x}}{\mathrm{2}}−\mathrm{1}=\mathrm{2}\Rightarrow{x}=\mathrm{6} \\ $$ Commented by Engr_Jidda…
Question Number 204658 by hardmath last updated on 24/Feb/24 $$\mathrm{If}\:\:\:\mathrm{a}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{9}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{3}}\:+\:\mathrm{1} \\ $$$$\mathrm{Find}\:\:\:\left(\frac{\mathrm{4}\:−\:\mathrm{a}}{\mathrm{a}}\right)^{\mathrm{6}} =\:? \\ $$ Answered by Rasheed.Sindhi last updated on 24/Feb/24 $$\:\mathrm{a}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{9}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{3}}\:+\:\mathrm{1};\:\left(\frac{\mathrm{4}\:−\:\mathrm{a}}{\mathrm{a}}\right)^{\mathrm{6}} =\:? \\…
Question Number 204632 by mr W last updated on 23/Feb/24 Answered by witcher3 last updated on 23/Feb/24 $$\left.\mathrm{x}>\mathrm{1};\mathrm{x}=\frac{\mathrm{1}}{\mathrm{sin}\left(\mathrm{t}\right)};\mathrm{t}\in\right]\mathrm{0},\frac{\pi}{\mathrm{2}}\left[\right. \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{sin}\left(\mathrm{t}\right)}+\frac{\frac{\mathrm{1}}{\mathrm{sin}\left(\mathrm{t}\right)}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \left(\mathrm{t}\right)}−\mathrm{1}}}=\frac{\mathrm{1}}{\mathrm{sin}\left(\mathrm{t}\right)}+\frac{\mathrm{1}}{\mathrm{cos}\left(\mathrm{t}\right)}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{cauchy}\:\mathrm{shwartz}\:\left(\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{sin}\left(\mathrm{t}\right)}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{cos}\left(\mathrm{t}\right)}}\right)^{\mathrm{2}} \right)\left(\left(\sqrt{\mathrm{sin}\left(\mathrm{t}\right)}\right)^{\mathrm{2}}…
Question Number 204621 by hardmath last updated on 23/Feb/24 $$\mathrm{a}\:,\:\mathrm{b}\:,\:\mathrm{c}\:\in\:\mathbb{R}^{+} \\ $$$$\mathrm{If}\:\:\:\sqrt{\mathrm{a}}\:+\:\sqrt{\mathrm{b}}\:+\:\sqrt{\mathrm{c}}\:=\:\mathrm{1} \\ $$$$\mathrm{Prove}\:\mathrm{that}:\:\:\:\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$ Answered by A5T last updated on 23/Feb/24 $$\frac{{a}+{b}+{c}}{\mathrm{3}}\geqslant\left(\frac{\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{9}}\Rightarrow{a}+{b}+{c}\geqslant\frac{\mathrm{1}}{\mathrm{3}}…
Question Number 204610 by mnjuly1970 last updated on 23/Feb/24 $$ \\ $$$$\:\:\:{If}\:,\:\:\:{f}\left({x}\right)\:=\:\begin{cases}{\:\mathrm{2}^{\mathrm{2}{x}} −\:{log}_{\mathrm{3}} \:\left(\:{x}+\mathrm{3}\:\right)\:\:\:\:;\:\:\:{x}\:\geqslant\mathrm{5}}\\{\:{f}\:\left(\mathrm{1}+\:{x}\:\right)\:\:−\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\:;\:\:{x}\:<\:\mathrm{5}}\end{cases}\:\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\:\:{f}\:\left(\mathrm{0}\:\right)=\:? \\ $$$$ \\ $$ Answered by Rasheed.Sindhi…