Question Number 151907 by mathdanisur last updated on 24/Aug/21 $$\mathrm{If}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{convex}\:\mathrm{quadrilateral} \\ $$$$\mathrm{is}\:\mathrm{2}\boldsymbol{\mathrm{k}}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{its}\:\mathrm{diagonals} \\ $$$$\mathrm{is}\:\mathrm{4}\boldsymbol{\mathrm{k}}^{\mathrm{2}} ,\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\:\mathrm{this}\:\mathrm{quadrilateral} \\ $$$$\mathrm{is}\:\mathrm{an}\:\mathrm{orthodiagonal}\:\mathrm{one}. \\ $$ Commented by mr W last…
Question Number 151905 by mathdanisur last updated on 24/Aug/21 Answered by mindispower last updated on 24/Aug/21 $$\mathrm{8}{x}^{{x}} {y}^{{y}} {z}^{{z}} \mathrm{2}^{{x}+{y}+{z}} =\mathrm{2}\left(\mathrm{2}{x}\right)^{{x}} .\mathrm{2}\left(\mathrm{2}{y}\right)^{{y}} .\left(\mathrm{2}{z}\right)^{{z}_{} } \\…
Question Number 20823 by ANTARES_VY last updated on 04/Sep/17 Answered by ajfour last updated on 04/Sep/17 $$\mathrm{2}{y}+\mathrm{5}\left(\frac{{x}−\mathrm{9}}{\mathrm{2}}\right)=\mathrm{5} \\ $$$$\Rightarrow\:\:\:\:\:\:\:\:\:\:\mathrm{4}{y}+\mathrm{5}{x}=\mathrm{55}\:\:\:\:…\left({i}\right) \\ $$$$\left.\:\:\:{and}\:\:\:\mathrm{4}{x}−{y}=\mathrm{11}\:\:\:\:\right]×\mathrm{4} \\ $$$$\:\:\:\:\:\:\:{so}\:\:\:\:\:\begin{cases}{\mathrm{16}{x}−\mathrm{4}{y}\:=\mathrm{44}}\\{\mathrm{4}{y}+\mathrm{5}{x}\:=\mathrm{55}}\end{cases} \\ $$$$\Rightarrow\:\:\:\:\:\:\:\:\mathrm{21}{x}=\mathrm{99}\:\:\:\:…
Question Number 151889 by mathdanisur last updated on 23/Aug/21 $$\mathrm{Compare}: \\ $$$$\sqrt[{\mathrm{2020}}]{\left(\mathrm{2020}!\right)^{\mathrm{3}} }\:\:\:\mathrm{and}\:\:\:\mathrm{505}\centerdot\mathrm{2021}^{\mathrm{2}} \\ $$ Answered by MJS_new last updated on 24/Aug/21 $$\forall{n}\in\mathbb{N}\mid{n}>\mathrm{1}:\left({n}!\right)^{\frac{\mathrm{3}}{{n}}} <\frac{{n}}{\mathrm{4}}×\left({n}+\mathrm{1}\right)^{\mathrm{2}} \\…
Question Number 151890 by mathdanisur last updated on 23/Aug/21 Answered by mnjuly1970 last updated on 24/Aug/21 $$\:\:\:{x}^{\:\mathrm{2}} +{y}^{\:\mathrm{2}} +{z}^{\:\mathrm{2}} \:+\left({x}^{\:\mathrm{2}} +\:\frac{\mathrm{1}}{{x}^{\:\mathrm{2}} }\right)+\left({y}^{\:\mathrm{2}} +\frac{\mathrm{1}}{{y}^{\:\mathrm{2}} }\:\right) \\…
Question Number 151884 by mathdanisur last updated on 23/Aug/21 $$\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{x}-\mathrm{1}\right)\left(\mathrm{x}-\mathrm{2}\right)…\left(\mathrm{x}-\mathrm{2021}\right) \\ $$$$\mathrm{f}\:^{'} \left(\mathrm{2021}\right)\:=\:? \\ $$ Answered by mr W last updated on 24/Aug/21 $$\mathrm{ln}\:{y}=\mathrm{ln}\:\left({x}−\mathrm{1}\right)+\mathrm{ln}\:\left({x}−\mathrm{2}\right)+…+\mathrm{ln}\:\left({x}−\mathrm{2021}\right) \\…
Question Number 151874 by miktun last updated on 23/Aug/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 20800 by Tinkutara last updated on 03/Sep/17 $$\mathrm{If}\:\mathrm{Re}\left(\frac{{z}\:+\:\mathrm{4}}{\mathrm{2}{z}\:−\:\mathrm{1}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}},\:\mathrm{then}\:{z}\:\mathrm{is}\:\mathrm{represented} \\ $$$$\mathrm{by}\:\mathrm{a}\:\mathrm{point}\:\mathrm{lying}\:\mathrm{on} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{A}\:\mathrm{circle} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{An}\:\mathrm{ellipse} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{A}\:\mathrm{straight}\:\mathrm{line} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{No}\:\mathrm{real}\:\mathrm{locus} \\ $$ Commented by ajfour…
Question Number 20799 by Tinkutara last updated on 03/Sep/17 $$\mathrm{If}\:{z}^{\mathrm{2}} \:+\:{z}\mid{z}\mid\:+\:\mid{z}\mid^{\mathrm{2}} \:=\:\mathrm{0},\:\mathrm{then}\:\mathrm{locus}\:\mathrm{of}\:{z}\:\mathrm{is} \\ $$ Answered by ajfour last updated on 03/Sep/17 $${let}\:{z}={re}^{{i}\theta} \\ $$$$\Rightarrow\:\:{r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\theta}…
Question Number 151849 by mathdanisur last updated on 23/Aug/21 Commented by Rasheed.Sindhi last updated on 23/Aug/21 $${Special}\:{general}\:{case} \\ $$$${x}={y}={z}\left({prime}\:{of}\:\mathrm{4}{m}+\mathrm{3}\:{type}\right) \\ $$$$\frac{{z}^{\mathrm{2}{k}} +{z}^{\mathrm{2}{k}} }{{z}^{{k}} +{z}^{{k}} }=\frac{\mathrm{2}{z}^{\mathrm{2}{k}}…