Question Number 161815 by HongKing last updated on 22/Dec/21 Answered by Lordose last updated on 23/Dec/21 $$\Omega\left(\mathrm{x}\right)\:=\:\Gamma\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−\frac{\mathrm{x}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−\left(\frac{\mathrm{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\mathrm{sin}\left(\pi\mathrm{x}\right) \\ $$$$\Omega\left(\mathrm{x}\right)\:=\:\boldsymbol{\pi}\mathrm{csc}\left(\frac{\boldsymbol{\pi}\mathrm{x}}{\mathrm{2}}\right)\centerdot\boldsymbol{\pi}\mathrm{csc}\left(\frac{\boldsymbol{\pi}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{2}}\right)\mathrm{sin}\left(\pi\mathrm{x}\right) \\ $$$$\Omega\left(\mathrm{x}\right)\:=\:\mathrm{2}\boldsymbol{\pi}^{\mathrm{2}} \\ $$$$\mathrm{x}^{\mathrm{2}} \:−\:\frac{\mathrm{4x}}{\Omega\left(\mathrm{x}\right)}\:+\:\frac{\mathrm{1}}{\boldsymbol{\pi}^{\mathrm{4}} }\:=\:\mathrm{0}…
Question Number 30739 by abdo imad last updated on 25/Feb/18 $${let}\:\left({u}_{{n}} \right)\:/\:{u}_{\mathrm{1}} =\mathrm{1}−{i}\:{and}\:\:\forall{p}\in\left\{\mathrm{2},\mathrm{3},…{n}\right\}\:{u}_{{p}} ={u}_{{p}−\mathrm{1}} {j}\:{with} \\ $$$${j}={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$$\left.\mathrm{1}\right){verify}\:{that}\:{u}_{\mathrm{1}} \:+{u}_{\mathrm{2}} \:+{u}_{\mathrm{3}} =\mathrm{0} \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:\forall{p}\in\:\left\{\mathrm{4},\mathrm{5},…,{n}\right\}\:\:{u}_{{p}} ={u}_{{p}−\mathrm{3}}…
Question Number 161800 by mathlove last updated on 22/Dec/21 $$\frac{\mathrm{1}^{\mathrm{2}} \centerdot\mathrm{2}!+\mathrm{2}^{\mathrm{2}} \centerdot\mathrm{3}!+\mathrm{3}^{\mathrm{2}} \centerdot\mathrm{4}!+\centerdot\centerdot\centerdot+{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)!−\mathrm{2}}{\left({n}+\mathrm{1}\right)!}=\mathrm{108} \\ $$$${n}=? \\ $$ Commented by Rasheed.Sindhi last updated on 22/Dec/21…
Question Number 96244 by Don08q last updated on 31/May/20 $$ \\ $$$$\:\:\mathrm{The}\:\mathrm{line}\:{y}\:=\:{mx}\:\:\mathrm{meets}\:\mathrm{the}\:\mathrm{parabola} \\ $$$$\:\:{y}\:=\:\left({x}\:−\:{a}\right)\left({b}\:−\:{x}\right)\:\mathrm{tangentially}\:\mathrm{where} \\ $$$$\:\:\mathrm{0}\:<\:{a}\:<\:{b}.\:\mathrm{Show}\:\mathrm{that}\:{m}\:=\:\left(\sqrt{{b}}\:−\:\sqrt{{a}}\right)^{\mathrm{2}} \\ $$$$ \\ $$ Commented by bobhans last updated…
Question Number 161764 by Tawa11 last updated on 22/Dec/21 Commented by Tawa11 last updated on 22/Dec/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{yellow}\:\mathrm{circle}. \\ $$ Commented by cortano last updated on…
Question Number 161745 by HongKing last updated on 21/Dec/21 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{3}\:\sqrt{\mathrm{e}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left[\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{n}!}\:\right]\mathrm{k2}^{-\boldsymbol{\mathrm{k}}} \\ $$ Commented by mr W last updated on…
Question Number 161744 by HongKing last updated on 21/Dec/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\frac{\mathrm{x}}{\mathrm{y}}\:+\:\frac{\mathrm{5}}{\mathrm{x}}\:+\:\frac{\mathrm{y}\:-\:\mathrm{5}}{\mathrm{5}}\:=\:\frac{\mathrm{y}\:+\:\mathrm{x}}{\mathrm{y}\:+\:\mathrm{5}}\:+\:\frac{\mathrm{5}\:+\:\mathrm{y}}{\mathrm{5}\:+\:\mathrm{x}} \\ $$ Commented by Rasheed.Sindhi last updated on 22/Dec/21 $${Not}\:{unique}\:{solution}.{Because} \\ $$$${only}\:\boldsymbol{{one}}\:{equation}\:{is}\:{given}\:{in}\:\boldsymbol{{two}} \\…
Question Number 161733 by Tawa11 last updated on 21/Dec/21 Answered by FongXD last updated on 21/Dec/21 $$\:\:\:\mathrm{let}\:\mathrm{O}\:\mathrm{be}\:\mathrm{the}\:\mathrm{point}\:\mathrm{where}\:\mathrm{the}\:\mathrm{2} \:\mathrm{intersect} \\ $$$$\:\:\:\mathrm{let}\:\mathrm{T}\in\left(\mathrm{QR}\right),\:\mathrm{where}\:\left(\mathrm{PT}\right)\bot\left(\mathrm{QR}\right) \\ $$$$\bullet\:\mathrm{in}\:\mathrm{the}\:\mathrm{right}\:\mathrm{triangle}\:\mathrm{PQT} \\ $$$$\mathrm{we}\:\mathrm{have}:\:\mathrm{PQ}^{\mathrm{2}} =\mathrm{PT}^{\mathrm{2}}…
Question Number 96189 by bemath last updated on 30/May/20 $$\mathrm{find}\:\mathrm{a}\:\mathrm{common}\:\mathrm{roots}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{quadratic}\:\mathrm{eq} \\ $$$$\mathrm{24x}^{\mathrm{2}} +\left(\mathrm{p}+\mathrm{4}\right)\mathrm{x}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{6x}^{\mathrm{2}} +\mathrm{11x}+\mathrm{p}+\mathrm{2}=\mathrm{0} \\ $$ Answered by john santu last…
Question Number 161704 by HongKing last updated on 21/Dec/21 $$\mathrm{let}\:\:\mathrm{n}\in\mathbb{N}\:\:\mathrm{fixed}\:,\:\mathrm{solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers} \\ $$$$\left[\mathrm{x}\right]\left\{\mathrm{x}\right\}=\mathrm{nx} \\ $$ Commented by mr W last updated on 21/Dec/21 $${two}\:{solutions}: \\ $$$${x}=−\frac{\mathrm{1}}{{n}+\mathrm{1}},\:\mathrm{0}…