Menu Close

Category: Algebra

Question-116780

Question Number 116780 by ajfour last updated on 06/Oct/20 Commented by ajfour last updated on 06/Oct/20 $${The}\:{curve}\:{in}\:{black}\:{is}\:\boldsymbol{{y}}=\boldsymbol{{x}}^{\mathrm{3}} −\boldsymbol{{x}}−\boldsymbol{{c}} \\ $$$$\:\:\:\left({c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\right)\:.\:{Find}\:{the}\:{radius}\:{of}\:{circle}. \\ $$ Answered by mr…

Question-116759

Question Number 116759 by bemath last updated on 06/Oct/20 Answered by bobhans last updated on 06/Oct/20 $$=\underset{\mathrm{n}=\mathrm{0}} {\overset{\mathrm{21}} {\prod}}\left(\mathrm{2030}^{\mathrm{2}} −\left(\mathrm{2040}−\mathrm{n}\right)^{\mathrm{2}} \right) \\ $$$$=\underset{\mathrm{n}=\mathrm{0}} {\overset{\mathrm{21}} {\prod}}\left(\mathrm{2030}+\left(\mathrm{2040}−\mathrm{n}\right)\left(\mathrm{2030}−\left(\mathrm{2040}−\mathrm{n}\right)\right)\right.…

1-a-37-1-b-37-1-7a-8b-19ab-0-mod-37-a-and-b-natural-nambers-a-1-b-1-a-2-b-2-a-n-b-n-n-

Question Number 182272 by SEKRET last updated on 06/Dec/22 $$\:\:\:\:\mathrm{1}\leqslant\boldsymbol{\mathrm{a}}\leqslant\mathrm{37} \\ $$$$\:\:\:\:\mathrm{1}\leqslant\boldsymbol{\mathrm{b}}\leqslant\mathrm{37} \\ $$$$\:\:\:\mathrm{1}+\mathrm{7a}+\mathrm{8}\boldsymbol{\mathrm{b}}\:+\mathrm{19}\boldsymbol{\mathrm{ab}}\:\:=\:\mathrm{0}\:\boldsymbol{\mathrm{mod}}\left(\mathrm{37}\right)\: \\ $$$$\:\:\boldsymbol{\mathrm{a}}\:\:\:\boldsymbol{\mathrm{and}}\:\:\boldsymbol{\mathrm{b}}\:\:\boldsymbol{\mathrm{natural}}\:\:\boldsymbol{\mathrm{nambers}} \\ $$$$\:\:\:\:\:\left(\boldsymbol{\mathrm{a}}_{\mathrm{1}} ;\boldsymbol{\mathrm{b}}_{\mathrm{1}} \right)\:\left(\boldsymbol{\mathrm{a}}_{\mathrm{2}} ;\boldsymbol{\mathrm{b}}_{\mathrm{2}} \right)……\left(\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} ;\boldsymbol{\mathrm{b}}_{\boldsymbol{\mathrm{n}}} \right) \\…

Find-9-4-5-9-4-5-

Question Number 182247 by Shrinava last updated on 06/Dec/22 $$\mathrm{Find}: \\ $$$$\sqrt{\mathrm{9}\:+\:\mathrm{4}\:\sqrt{\mathrm{5}}}\:−\:\sqrt{\mathrm{9}\:−\:\mathrm{4}\:\sqrt{\mathrm{5}}}\:=\:? \\ $$ Answered by mr W last updated on 06/Dec/22 $$\mathrm{9}\pm\mathrm{4}\sqrt{\mathrm{5}}=\left(\sqrt{\mathrm{5}}\right)^{\mathrm{2}} \pm\mathrm{4}\sqrt{\mathrm{5}}+\mathrm{2}^{\mathrm{2}} =\left(\sqrt{\mathrm{5}}\pm\mathrm{2}\right)^{\mathrm{2}}…

It-is-given-a-family-of-open-interval-set-U-r-r-Q-of-R-that-satifies-condition-r-Q-r-U-r-Prove-that-there-exists-a-family-set-U-r-r-Q-which-not-cover-R-or-gt-0-r-Q-U-r-

Question Number 182241 by Matica last updated on 06/Dec/22 $$\:\mathrm{It}\:\mathrm{is}\:\mathrm{given}\:\mathrm{a}\:\mathrm{family}\:\mathrm{of}\:\mathrm{open}\:\mathrm{interval}\:\mathrm{set}\:\left({U}_{{r}} \right)_{{r}\in\mathbb{Q}} \:\mathrm{of}\:\mathbb{R} \\ $$$$\mathrm{that}\:\mathrm{satifies}\:\mathrm{condition}\:\forall{r}\in\mathbb{Q},\:{r}\in{U}_{{r}\:} . \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{there}\:\mathrm{exists}\:\mathrm{a}\:\mathrm{family}\:\mathrm{set}\:\left({U}_{{r}} \right)_{{r}\in\mathbb{Q}} \mathrm{which}\:\mathrm{not}\:\mathrm{cover}\:\mathbb{R}\: \\ $$$$\mathrm{or}\:\forall\varepsilon>\mathrm{0},\:\:\lambda\left(\underset{{r}\in\mathbb{Q}} {\cup}\:{U}_{{r}} \:\right)\leqslant\:\varepsilon\:. \\ $$…