Question Number 192340 by Mastermind last updated on 15/May/23 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{order}\:\mathrm{of}\:\mathrm{any}\:\mathrm{permuta}− \\ $$$$\mathrm{tion}\:\theta\:\mathrm{is}\:\mathrm{the}\:\mathrm{least}\:\mathrm{common}\:\mathrm{multiple} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{its}\:\mathrm{disjoint}\:\mathrm{cycles}. \\ $$$$ \\ $$$$\:\mathrm{hi} \\ $$ Answered by aleks041103 last updated…
Question Number 61270 by necx1 last updated on 31/May/19 Answered by Askash last updated on 31/May/19 $$\mathrm{24} \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 192343 by cortano12 last updated on 15/May/23 $$\:\mathrm{Find}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\:\:\:\:\:\:\:\mathrm{y}=\frac{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{2}\:\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}\: \\ $$ Answered by manxsol last updated on 15/May/23 $$\:\:{x}\neq{k}\pi\frac{\pi}{\mathrm{2}}\:\:\:\:{y}>\mathrm{0} \\…
Question Number 61269 by alphaprime last updated on 31/May/19 $${Let}\:{p}\left({x}\right)\:{be}\:{a}\:{quadratic}\:{polynomial}\:{such} \\ $$$${that}\:{for}\:{distinct}\:\alpha\:{and}\:\beta\:, \\ $$$${p}\left(\alpha\right)\:=\:\alpha\:{and}\:{p}\left(\beta\right)\:=\beta \\ $$$${prove}\:{that}\:\alpha\:{and}\:\beta\:{are}\:{roots}\:{of}\:\:{p}\left[{p}\left({x}\right)\right]−{x}=\mathrm{0}\: \\ $$$${Find}\:{the}\:{remaining}\:{roots}\:. \\ $$ Answered by ajfour last updated…
Question Number 192342 by Mastermind last updated on 15/May/23 $$\left.\mathrm{1}\right)\:\mathrm{Compute}\:\mathrm{in}\:\mathrm{S}_{\mathrm{a}} \:,\:\mathrm{a}^{−\mathrm{1}} \mathrm{ba}\:\:\mathrm{where}\: \\ $$$$\mathrm{a}=\left(\mathrm{1}\:\mathrm{2}\right)\left(\mathrm{1}\:\mathrm{3}\:\mathrm{5}\right),\:\mathrm{b}=\left(\mathrm{1}\:\mathrm{5}\:\mathrm{7}\:\mathrm{1}\right) \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Given}\:\mathrm{permutation}\:\alpha\:=\:\left(\mathrm{1}\:\mathrm{2}\right)\left(\mathrm{3}\:\mathrm{4}\right), \\ $$$$\beta\:=\:\left(\mathrm{1}\:\mathrm{3}\right)\left(\mathrm{5}\:\mathrm{6}\right).\:\mathrm{Find}\:\mathrm{a}\:\mathrm{permutation} \\ $$$$\mathrm{x}\in\mathrm{S}_{\mathrm{6}} \:\exists\alpha\mathrm{x}\:=\:\beta. \\ $$$$…
Question Number 61268 by alphaprime last updated on 31/May/19 $${Let}\:{a},{b},{c},{d},{e}\:\geqslant\:−\mathrm{1}\:{and}\:{a}+{b}+{c}+{d}+{e}=\mathrm{5} \\ $$$${Find}\:{the}\:{maximum}\:{and}\:{minimum} \\ $$$${value}\:{of}\:{S}\:=\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{d}\right)\left({d}+{e}\right)\left({e}+{a}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 61267 by behi83417@gmail.com last updated on 31/May/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 192336 by Mastermind last updated on 14/May/23 Answered by Rajpurohith last updated on 27/May/23 $${Since}\:\boldsymbol{{S}}\:{is}\:{bounded}\:,{inf}\left(\boldsymbol{{S}}\right)\:{and}\:{sup}\left(\boldsymbol{{S}}\right)\:{exist}\:{and}\:\lambda\in\mathbb{R}. \\ $$$$\left({a}\right)\forall{s}\in\boldsymbol{{S}}\:,\:{inf}\left({S}\right)\leqslant{s} \\ $$$$\Rightarrow\:\:\forall{s}\in\boldsymbol{{S}}\:,\:{inf}\left({S}\right)+\lambda\leqslant{s}+\lambda \\ $$$$\Rightarrow{inf}\left(\boldsymbol{{S}}\right)+\lambda\:{is}\:{a}\:{lower}\:{bound}\:{of}\:\boldsymbol{{S}}+\lambda. \\ $$$${if}\:\:{inf}\left(\boldsymbol{{S}}\right)+\lambda<{t}\:\left({be}\:{a}\:{lower}\:{bound}\:{of}\:\boldsymbol{{S}}+\lambda\right)…
Question Number 126803 by john_santu last updated on 24/Dec/20 $$\:\:{B}\left(\frac{\mathrm{7}}{\mathrm{3}},\frac{\mathrm{2}}{\mathrm{3}}\right)\:=? \\ $$$${B}\:=\:{betha}\:{function}\: \\ $$ Answered by Dwaipayan Shikari last updated on 24/Dec/20 $${B}\left(\frac{\mathrm{7}}{\mathrm{3}},\frac{\mathrm{2}}{\mathrm{3}}\right)=\frac{\Gamma\left(\frac{\mathrm{7}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{\Gamma\left(\mathrm{3}\right)}=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right).\frac{\mathrm{4}}{\mathrm{3}}.\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{2}!}=\frac{\mathrm{2}}{\mathrm{9}}.\frac{\pi}{{sin}\frac{\pi}{\mathrm{3}}}=\frac{\mathrm{4}\pi}{\mathrm{9}\sqrt{\mathrm{3}}} \\ $$…
Question Number 192339 by Mastermind last updated on 15/May/23 $$\mathrm{Express}\:\mathrm{as}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{disjoint}\: \\ $$$$\mathrm{cycle}\:\mathrm{the}\:\mathrm{permutation} \\ $$$$\left.\mathrm{a}\right)\:\theta\left(\mathrm{1}\right)=\mathrm{4}\:\:\theta\left(\mathrm{2}\right)=\mathrm{6}\:\:\theta\left(\mathrm{1}\right)=\mathrm{5}\:\:\theta\left(\mathrm{4}\right)=\mathrm{1} \\ $$$$\theta\left(\mathrm{5}\right)=\mathrm{3}\:\:\theta\left(\mathrm{6}\right)=\mathrm{2} \\ $$$$ \\ $$$$\left.\mathrm{b}\right)\:\left(\mathrm{1}\:\mathrm{6}\:\mathrm{3}\right)\left(\mathrm{1}\:\mathrm{3}\:\mathrm{5}\:\mathrm{7}\right)\left(\mathrm{6}\:\mathrm{7}\right)\left(\mathrm{1}\:\mathrm{2}\:\mathrm{3}\:\mathrm{4}\:\mathrm{5}\right) \\ $$$$ \\ $$$$\left.\mathrm{c}\right)\:\left(\mathrm{1}\:\mathrm{2}\:\mathrm{3}\:\mathrm{4}\:\mathrm{5}\right)\left(\mathrm{6}\:\mathrm{7}\right)\left(\mathrm{1}\:\mathrm{3}\:\mathrm{5}\:\mathrm{7}\right) \\…