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Author: Tinku Tara

Question-202679

Question Number 202679 by Mingma last updated on 31/Dec/23 Answered by esmaeil last updated on 31/Dec/23 $$\curvearrowright\rightarrow<_{{hex}} =\mathrm{120}\rightarrow \\ $$$${tan}\mathrm{30}=\frac{{h}}{\frac{{a}}{\mathrm{2}}}\rightarrow{h}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{6}} \\ $$$$\mathrm{2}\left(\frac{{a}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{6}}×\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{43}=\left({S}_{{blue}} +{S}_{{pink}} \right)\rightarrow \\…

The-sum-of-all-the-divisors-of-24-is-60-Find-the-sum-of-all-the-divisors-of-24-79-

Question Number 202672 by BaliramKumar last updated on 31/Dec/23 $$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{divisors}\:\mathrm{of}\:\mathrm{24}\:\mathrm{is}\:\mathrm{60}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{divisors}\:\mathrm{of}\:\mathrm{24}×\mathrm{79}. \\ $$ Answered by a.lgnaoui last updated on 31/Dec/23 $$\mathrm{60} \\ $$ Commented…

x-1-x-1-1-x-1-1-x-

Question Number 202638 by depressiveshrek last updated on 31/Dec/23 $$\sqrt{{x}−\frac{\mathrm{1}}{{x}}}−\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}=\mathrm{1}−\frac{\mathrm{1}}{{x}} \\ $$ Commented by AST last updated on 31/Dec/23 $${x}=\mathrm{1}.{Suppose}\:{x}\neq\mathrm{1},{let}\:{a}=\sqrt{{x}−\frac{\mathrm{1}}{{x}}};{b}=\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}} \\ $$$${a}−{b}=\mathrm{1}−\frac{\mathrm{1}}{{x}};\left({a}+{b}\right)^{\mathrm{2}} =\mathrm{1}+{x} \\ $$$${a}^{\mathrm{2}}…

determine-le-reste-de-la-division-eucludienne-de-2023-2019-par-13-

Question Number 202698 by Bambamamoudou last updated on 31/Dec/23 $${determine}\:{le}\:{reste}\:{de}\:{la}\:{division}\:{eucludienne}\:{de}\:\mathrm{2023}^{\mathrm{2019}} {par}\:\mathrm{13} \\ $$ Answered by Rasheed.Sindhi last updated on 01/Jan/24 $$\because\:{gcd}\left(\mathrm{2023},\mathrm{13}\right)=\mathrm{1} \\ $$$$\therefore\:\:\:\:\mathrm{2023}^{\phi\left(\mathrm{13}\right)} \equiv\mathrm{1}\left({mod}\:\mathrm{13}\right) \\…