Question Number 204733 by hardmath last updated on 26/Feb/24 $$\mathrm{Find}:\:\:\:\frac{\mathrm{59}^{\mathrm{2}} \:+\:\mathrm{48}^{\mathrm{2}} \:+\:\mathrm{41}^{\mathrm{2}} \:−\:\mathrm{30}^{\mathrm{2}} }{\mathrm{68}^{\mathrm{2}} \:+\:\mathrm{52}^{\mathrm{2}} \:+\:\mathrm{32}^{\mathrm{2}} \:−\:\mathrm{48}^{\mathrm{2}} }\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated…
Question Number 204712 by cortano12 last updated on 26/Feb/24 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{real}\:\mathrm{solution}\: \\ $$$$\:\:\:\:\sqrt{\mathrm{3x}^{\mathrm{2}} +\mathrm{x}−\mathrm{1}}\:+\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{3}}\:=\: \\ $$$$\:\:\sqrt{\mathrm{3x}^{\mathrm{2}} +\mathrm{3x}+\mathrm{5}}\:+\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{3}}\: \\ $$ Commented by Rasheed.Sindhi last updated…
Question Number 204742 by York12 last updated on 26/Feb/24 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:{x} \\ $$ Commented by York12 last updated on 26/Feb/24 Commented by TonyCWX08 last updated on…
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}} =\:? \\…