Question Number 19634 by Tinkutara last updated on 13/Aug/17 $$\mathrm{How}\:\mathrm{many}\:\mathrm{ordered}\:\mathrm{triplets}\:\left({x},\:{y},\:{z}\right)\:\mathrm{of} \\ $$$$\mathrm{positive}\:\mathrm{integer}\:\mathrm{satisfy}\:\mathrm{lcm}\left({x},\:{y}\right)\:=\:\mathrm{72}, \\ $$$$\mathrm{lcm}\left({x},\:{z}\right)\:=\:\mathrm{600}\:\mathrm{and}\:\mathrm{lcm}\left({y},\:{z}\right)\:=\:\mathrm{900}? \\ $$ Answered by Tinkutara last updated on 16/Oct/17 Terms of…
Question Number 148397 by nadovic last updated on 27/Jul/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 82792 by M±th+et£s last updated on 24/Feb/20 $${show}\:{that}\:{if}\:{A}\subset\mathbb{R}^{{m}} \:{and}\:{B}\subset\mathbb{R}^{{n}} \:{are}\: \\ $$$${compact}\:{sets}.\: \\ $$$${then}\:{A}×{B}=\left\{\left({a},{b}\right)\in\mathbb{R}^{{m}+{n}} :{a}\in{A}\:{and}\:{b}\in{B}\right\} \\ $$ Commented by M±th+et£s last updated on…
Question Number 14028 by tawa tawa last updated on 26/May/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 144639 by sdfg last updated on 27/Jun/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 13200 by Joel577 last updated on 16/May/17 $$\sqrt[{\sqrt[{\sqrt[{\sqrt{\mathrm{3}}}]{\mathrm{2}}}]{\mathrm{5}}}]{\mathrm{6}}\:=\:{x} \\ $$$$\mathrm{How}\:\mathrm{to}\:\mathrm{write}\:{x}\:\mathrm{in}\:\mathrm{standard}\:\mathrm{form}? \\ $$ Answered by ajfour last updated on 16/May/17 $${x}=\mathrm{6}^{\left[\frac{\mathrm{1}}{\mathrm{5}^{\left(\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}/\sqrt{\mathrm{3}}} }\right)} }\right]} \:…
Question Number 245 by 123456 last updated on 25/Jan/15 $$\mathrm{A}=\left\{\mathrm{0},\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{8},\mathrm{10}\right\} \\ $$$$\mathrm{B}=\left\{\mathrm{0},\mathrm{1},\mathrm{3},\mathrm{4},\mathrm{6},\mathrm{7},\mathrm{9}\right\} \\ $$$$\mathrm{C}=\left\{\mathrm{1},\mathrm{2},\mathrm{4},\mathrm{5},\mathrm{7},\mathrm{8},\mathrm{10}\right\} \\ $$$$\mid\left(\mathrm{A}\cup\mathrm{B}\right)\cap\left(\mathrm{B}\cup\mathrm{C}\right)\mid \\ $$ Answered by mreddy last updated on 17/Dec/14…
Question Number 242 by 123456 last updated on 25/Jan/15 $$\mathrm{give}\:\mathrm{the}\:\mathrm{sets} \\ $$$$\mathrm{A}=\left\{{x}\in\mathbb{N}:\mathrm{0}\leqslant{x}\leqslant\mathrm{9}\right\} \\ $$$$\mathrm{B}=\left\{\mathrm{0}\right\} \\ $$$$\mathrm{C}=\left\{\mathrm{1}\right\} \\ $$$$\mathrm{D}=\left\{\mathrm{2},\mathrm{3},\mathrm{5},\mathrm{7}\right\} \\ $$$$\mathrm{E}=\left\{\mathrm{4},\mathrm{6},\mathrm{8},\mathrm{9}\right\} \\ $$$$\mathrm{compute} \\ $$$$\mid\mathrm{A}\backslash\mathrm{B}\mid+\mid\mathrm{A}\backslash\mathrm{C}\mid+\mid\mathrm{A}\backslash\mathrm{D}\mid+\mid\mathrm{A}\backslash\mathrm{E}\mid \\…
Question Number 131164 by aklife last updated on 02/Feb/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 12291 by 1kanika# last updated on 18/Apr/17 Answered by 433 last updated on 07/May/17 $$\begin{bmatrix}{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}×\begin{bmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}=\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{−\mathrm{1}}\end{bmatrix}={A} \\ $$$$\begin{bmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}×\begin{bmatrix}{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}=\begin{bmatrix}{−\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}=−{A} \\ $$$$\begin{bmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}×\begin{bmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}=\begin{bmatrix}{−\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{−\mathrm{1}}\end{bmatrix}=−{I}_{\mathrm{2}} \\ $$$$\begin{bmatrix}{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{0}}\end{bmatrix}^{\mathrm{2}} ={I}_{\mathrm{2}} \\…