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Author: Tinku Tara

Question-201854

Question Number 201854 by cortano12 last updated on 14/Dec/23 Commented by cortano12 last updated on 14/Dec/23 $$\:\:\begin{cases}{\mathrm{x}^{\mathrm{2}} −\mathrm{3y}^{\mathrm{2}} =\frac{\mathrm{17}}{\mathrm{x}}}\\{\mathrm{3x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} =\frac{\mathrm{23}}{\mathrm{y}}}\end{cases} \\ $$$$\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\:\sqrt[{\mathrm{m}}]{\mathrm{n}}\:,\:\mathrm{m},\mathrm{n}\:\in\mathbb{Z}^{+}…

Question-201848

Question Number 201848 by cortano12 last updated on 14/Dec/23 Answered by dimentri last updated on 14/Dec/23 $$\:\:\mathrm{1820}\:=\:\mathrm{2}^{\mathrm{2}} ×\mathrm{5}×\mathrm{7}×\mathrm{13} \\ $$$$\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{positive}\:\mathrm{factors} \\ $$$$\:\mathrm{from}\:\mathrm{1820}\:=\:\underset{\mathrm{i}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}\begin{pmatrix}{\mathrm{5}}\\{\mathrm{i}}\end{pmatrix}\:=\:\mathrm{2}^{\mathrm{5}} =\:\mathrm{32}…

Question-201873

Question Number 201873 by sonukgindia last updated on 14/Dec/23 Answered by witcher3 last updated on 14/Dec/23 $$\mathrm{Re}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{e}^{\mathrm{i}\theta} }\right)=\frac{\mathrm{1}+\mathrm{e}^{−\mathrm{i}\theta} }{\left(\mathrm{1}+\mathrm{e}^{\mathrm{i}\theta} \right)\left(\mathrm{1}+\mathrm{e}^{−\mathrm{i}\theta} \right)}=\frac{\mathrm{1}+\mathrm{cos}\left(\theta\right)}{\mathrm{2}+\mathrm{2cos}\left(\theta\right)}=\frac{\mathrm{1}}{\mathrm{2}};\forall\theta\in\mathbb{R}−\left\{\left(\mathrm{1}+\mathrm{2k}\right)\pi\right\} \\ $$$$\mathrm{I}==\frac{\varphi^{\mathrm{3}} −\mathrm{1}}{\mathrm{2}} \\…

n-1-1-n-H-n-n-1-

Question Number 201860 by MrGHK last updated on 14/Dec/23 $$\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{n}}+\mathrm{1}}=?? \\ $$ Answered by mnjuly1970 last updated on 14/Dec/23 $$\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty}…