Question Number 218119 by MrGaster last updated on 30/Mar/25 $$\begin{vmatrix}{\varepsilon}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{1}}&{\varepsilon}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\varepsilon}&{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\varepsilon}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\varepsilon}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}&{\varepsilon}\end{vmatrix}=? \\ $$$$ \\ $$ Answered by SdC355 last updated on 30/Mar/25 $$\mathrm{Holy}\:\mathrm{shit}….\mathrm{what}\:\mathrm{is}\:\mathrm{that}\:\mathrm{Lol} \\ $$$$\mathrm{But}\:\mathrm{that}\:\mathrm{det}\left\{\mathrm{A}\right\}\:\mathrm{is}\:\varepsilon^{\mathrm{6}} −\mathrm{6}\varepsilon^{\mathrm{4}}…
Question Number 218128 by mr W last updated on 30/Mar/25 $$\begin{array}{|c|}{?}&\hline{?}&\hline{?}&\hline{?}\\\hline\end{array}×\begin{array}{|c|}{?}\\\hline\end{array}=\mathrm{8044}\begin{array}{|c|}{?}\\\hline\end{array} \\ $$ Answered by A5T last updated on 30/Mar/25 $$\frac{\mathrm{8044}?}{\mathrm{8}}>\frac{\mathrm{80000}}{\mathrm{8}}=\mathrm{10000}\: \\ $$$$\Rightarrow\:\mathrm{8044}?\:\mathrm{should}\:\mathrm{be}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{9} \\ $$$$\Rightarrow\mathrm{9}\:\mid\:\mathrm{8}+\mathrm{0}+\mathrm{4}+\mathrm{4}+?\Rightarrow\:?=\mathrm{2}…
Question Number 218103 by ArshadS last updated on 29/Mar/25 $${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\:,\:{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=? \\ $$ Answered by Rasheed.Sindhi last updated on 29/Mar/25 $${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\:,\:{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=? \\…
Question Number 218099 by Rasheed.Sindhi last updated on 29/Mar/25 $${x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0}\:,\:{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}=? \\ $$ Answered by Rasheed.Sindhi last updated on 29/Mar/25 $$\mathrm{Another}\:\mathrm{way} \\ $$$$\left({x}^{\mathrm{2}}…
Question Number 218093 by ArshadS last updated on 29/Mar/25 $${Solve}\:{for}\:{x} \\ $$$$\sqrt{\mathrm{2}{x}+\mathrm{3}}\:−\sqrt{{x}−\mathrm{2}}\:=\sqrt{{x}+\mathrm{2}}\: \\ $$ Answered by vnm last updated on 29/Mar/25 $$\mathrm{2}{x}+\mathrm{3}=\left(\sqrt{{x}+\mathrm{2}}+\sqrt{{x}−\mathrm{2}}\right)^{\mathrm{2}} = \\ $$$${x}+\mathrm{2}+{x}−\mathrm{2}+\mathrm{2}\sqrt{\left({x}+\mathrm{2}\right)\left({x}−\mathrm{2}\right)}…
Question Number 218088 by ArshadS last updated on 29/Mar/25 $${Solve} \\ $$$$\sqrt{{x}+\mathrm{5}}\:+\sqrt{{x}−\mathrm{3}}\:=\mathrm{4} \\ $$ Answered by A5T last updated on 29/Mar/25 $$\sqrt{\mathrm{x}+\mathrm{5}}=\mathrm{u};\:\sqrt{\mathrm{x}−\mathrm{3}}=\mathrm{v} \\ $$$$\mathrm{u}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}}…
Question Number 218047 by dscm last updated on 27/Mar/25 $${Determine}\:{x}: \\ $$$$\sqrt{{x}+\mathrm{3}}\:+\:\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:=\mathrm{2} \\ $$ Answered by Rasheed.Sindhi last updated on 27/Mar/25 $$\sqrt{{x}+\mathrm{3}}\:+\:\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:=\mathrm{2} \\ $$$$\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:={y}\Rightarrow{x}={y}^{\mathrm{3}} −\mathrm{1}…
Question Number 218057 by ArshadS last updated on 27/Mar/25 $${Solve} \\ $$$$\left({x}−\mathrm{3}\right)^{\mathrm{5}/\mathrm{3}} −\mathrm{6}\left({x}−\mathrm{3}\right)^{\mathrm{2}/\mathrm{3}} −\mathrm{8}=\mathrm{0} \\ $$ Answered by Frix last updated on 27/Mar/25 $$\left({x}−\mathrm{3}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \left({x}−\mathrm{9}\right)=\mathrm{8}…
Question Number 218055 by ArshadS last updated on 27/Mar/25 Commented by ArshadS last updated on 27/Mar/25 $${Solve} \\ $$ Answered by Frix last updated on…
Question Number 218036 by dscm last updated on 26/Mar/25 $${Solve} \\ $$$$\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{xy}\:=\:\mathrm{19}\: \\ $$$$\:\mathrm{x}\:+\:\mathrm{y}=\:\mathrm{5} \\ $$ Answered by Hanuda354 last updated on 26/Mar/25…