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Category: Algebra

5-2-1-3-5-2-1-3-2014-

Question Number 182188 by mathlove last updated on 05/Dec/22 $$\left(\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}+\mathrm{2}}+\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}−\mathrm{2}}\right)^{\mathrm{2014}} =? \\ $$ Answered by Frix last updated on 05/Dec/22 $$\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}=\varphi\wedge\sqrt[{\mathrm{3}}]{−\mathrm{2}+\sqrt{\mathrm{5}}}=\frac{\mathrm{1}}{\varphi} \\ $$$$\varphi+\frac{\mathrm{1}}{\varphi}=\sqrt{\mathrm{5}} \\ $$$$\left(\sqrt{\mathrm{5}}\right)^{\mathrm{2014}}…

Question-182176

Question Number 182176 by peter frank last updated on 05/Dec/22 Answered by MikeH last updated on 05/Dec/22 $$\mathrm{let}\:\overset{\rightarrow} {\mathrm{w}}\:=\:{x}\mathrm{i}\:+\:\mathrm{6j}\:+\:{y}\:\mathrm{k} \\ $$$$\mathrm{orthogonal}\:\mathrm{to}\:\overset{\rightarrow} {\mathrm{u}}\:\mathrm{and}\:\overset{\rightarrow} {\mathrm{v}}\:\Rightarrow\:\overset{\rightarrow} {\mathrm{w}}\:=\:\overset{\rightarrow} {\mathrm{u}}×\:\overset{\rightarrow}…

Question-116633

Question Number 116633 by zakirullah last updated on 05/Oct/20 Answered by Olaf last updated on 05/Oct/20 $$\mathrm{If}\:\mathrm{B}\:\mathrm{is}\:\mathrm{a}\:\mathrm{inverse}\:\mathrm{matrix}\:\mathrm{of}\:\mathrm{A} \\ $$$$\mathrm{then}\:\mathrm{B}.\mathrm{A}\:=\:\mathrm{A}.\mathrm{B}\:=\:\mathrm{I}\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{C}\:\mathrm{is}\:\mathrm{another}\:\mathrm{inverse}\:\mathrm{matrix}\:\mathrm{of}\:\mathrm{A} \\ $$$$\mathrm{then}\:\mathrm{C}.\mathrm{A}\:=\:\mathrm{A}.\mathrm{C}\:=\:\mathrm{I}\:\left(\mathrm{2}\right)…