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} \\ $$$${x}={y}={z}\left({prime}\:{of}\:\mathrm{4}{m}+\mathrm{3}\:{type}\right) \\ $$$$\frac{{z}^{\mathrm{2}{k}} +{z}^{\mathrm{2}{k}} }{{z}^{{k}} +{z}^{{k}} }=\frac{\mathrm{2}{z}^{\mathrm{2}{k}}…
Question Number 151858 by mathdanisur last updated on 23/Aug/21 Answered by Olaf_Thorendsen last updated on 23/Aug/21 $${f}\left({x}\right)\:=\:\mathrm{sin}{x}+\frac{\mathrm{sin3}{x}}{\mathrm{3}}+\frac{\mathrm{sin5}{x}}{\mathrm{5}}+\mathrm{sin}\frac{\mathrm{7}{x}}{\mathrm{7}} \\ $$$${f}'\left({x}\right)\:=\:\mathrm{cos}{x}+\mathrm{cos3}{x}+\mathrm{cos5}{x}+\mathrm{cos7}{x} \\ $$$${f}'\left({x}\right)\:=\:\mathrm{Re}\left({e}^{{ix}} +{e}^{\mathrm{3}{ix}} +{e}^{\mathrm{5}{ix}} +{e}^{\mathrm{7}{ix}} \right)…
Question Number 151841 by mathdanisur last updated on 23/Aug/21 $$\mathrm{The}\:\mathrm{volue}\:\mathrm{of}\:\mathrm{the}\:\mathrm{limit}:\: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}^{−\boldsymbol{\mathrm{n}}^{\mathrm{2}} } }{\underset{\boldsymbol{\mathrm{k}}=\boldsymbol{\mathrm{n}}+\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{2}^{−\boldsymbol{\mathrm{k}}^{\mathrm{2}} } }\:\:\:;\:\:\:\left(\mathrm{a}\right)\mathrm{0}\:\:\left(\mathrm{b}\right)\mathrm{some}\:\mathrm{c}\in\left(\mathrm{0};\mathrm{1}\right)\:\:\left(\mathrm{c}\right)\mathrm{1} \\ $$ Terms of Service Privacy…
Question Number 86298 by 21042009 last updated on 28/Mar/20 $${let}\:{x}^{{x}^{{x}^{\iddots} } } =\mathrm{2} \\ $$$${x}^{\mathrm{2}} =\mathrm{2} \\ $$$${x}=\pm\sqrt{\mathrm{2}} \\ $$$${then}\:{let}\:{x}^{{x}^{{x}^{\iddots} } } =\mathrm{4} \\ $$$${x}^{\mathrm{4}}…
Question Number 151826 by mathdanisur last updated on 23/Aug/21 $$\mathrm{if}\:\:\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2}\right)\left(\mathrm{y}^{\mathrm{2}} +\mathrm{2}\right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{2}\right)\:\geqslant\:\mathrm{9}\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com