Question Number 205559 by mnjuly1970 last updated on 24/Mar/24 $$ \\ $$$$\:\:\:{Question}.\:\left({math}\:{analysis}\right) \\ $$$$\:\:\left({X}\:,{d}\:\right)\:{is}\:{a}\:{metric}\:{space}\:{and} \\ $$$$\:\:\left({p}_{{n}} \right)_{{n}=\mathrm{1}} ^{\infty} \:{is}\:{a}\:{sequence}\:{in}\:{X}. \\ $$$$\:\:\:\left({p}_{{n}} \right)_{{n}=\mathrm{1}} ^{\:\infty} {is}\:{cauchy}\:{if}\:{and}\:\:{only}\:{if} \\…
Question Number 205534 by hardmath last updated on 23/Mar/24 $$\mathrm{Find}: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\mathrm{n}\:\mathrm{x}^{\boldsymbol{\mathrm{n}}} \:\mathrm{e}^{\boldsymbol{\mathrm{x}}^{\mathrm{2}} } \:\mathrm{dx}\:=\:? \\ $$ Answered by Mathspace last updated…
Question Number 205528 by hardmath last updated on 23/Mar/24 $$\mathrm{Let}\:\:\:\forall\mathrm{x}\:\in\:\mathrm{A}\:\rightarrow\:\mathrm{x}\:\in\:\mathbb{R} \\ $$$$\mathrm{And}\:\:\:\mathrm{card}\left(\mathrm{A}\right)\:>\:\mathrm{card}\:\mathrm{N} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{card}\left(\mathrm{A}'\right)\:>\:\mathrm{card}\:\mathrm{N} \\ $$ Answered by Berbere last updated on 24/Mar/24…
Question Number 205530 by cherokeesay last updated on 23/Mar/24 Answered by mr W last updated on 24/Mar/24 $${P}\left({a}\:\mathrm{cos}\:\theta,\:{b}\:\mathrm{sin}\:\theta\right) \\ $$$$\mathrm{tan}\:\varphi=−\frac{{dy}}{{dx}}=\frac{{b}}{{a}\:\mathrm{tan}\:\theta} \\ $$$${r}={a}\:\mathrm{cos}\:\theta−{r}\:\mathrm{sin}\:\varphi\:\Rightarrow{r}\:\mathrm{sin}\:\varphi={a}\:\mathrm{cos}\:\theta−{r} \\ $$$${r}={b}\:\mathrm{sin}\:\theta−{r}\:\mathrm{cos}\:\varphi\:\Rightarrow{r}\:\mathrm{cos}\:\varphi={b}\:\mathrm{sin}\:\theta−{r} \\…
Question Number 205527 by MATHEMATICSAM last updated on 23/Mar/24 $$\mathrm{If}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{are}\:\mathrm{one} \\ $$$$\mathrm{another}'\mathrm{s}\:\mathrm{cube}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$$\left({b}^{\mathrm{2}} \:−\:\mathrm{2}{ac}\right)^{\mathrm{2}} \:=\:{ac}\left({a}\:+\:{c}\right)^{\mathrm{2}} . \\ $$ Answered by A5T last updated…
Question Number 205516 by MathedUp last updated on 23/Mar/24 $$\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\boldsymbol{\zeta}\left(\mathrm{2}{h}\right)−\mathrm{1}}{{h}}\:=\:…..? \\ $$$$\underset{{h}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(\boldsymbol{\zeta}\left(\mathrm{2}{h}+\mathrm{1}\right)−\mathrm{1}\right)=……? \\ $$$$\mathrm{pls}\:\mathrm{help}\:\mathrm{me} \\ $$ Terms of Service Privacy Policy…
Question Number 205517 by BaliramKumar last updated on 23/Mar/24 Answered by mr W last updated on 23/Mar/24 $${AB}=\sqrt{\left(\mathrm{5}−\mathrm{4}\right)^{\mathrm{2}} +\left(\mathrm{7}−\mathrm{3}\right)^{\mathrm{2}} }=\sqrt{\mathrm{17}} \\ $$$${say}\:{D}\:{divides}\:{AB}\:{externally}\:{and} \\ $$$${C}\:{divides}\:{AB}\:{internally}. \\…
Question Number 205514 by aba last updated on 23/Mar/24 $$\mathrm{Quelle}\:\mathrm{est}\:\mathrm{la}\:\mathrm{decomposition}\:\mathrm{en}\:\mathrm{cycles} \\ $$$$\mathrm{a}\:\mathrm{support}\:\mathrm{disjoints}\:\mathrm{de}\:\mathrm{c}^{\mathrm{k}} \:,\:\mathrm{ou}\:\mathrm{c}=\left(\mathrm{1}\:\mathrm{2}\:\mathrm{3}\:…\:\mathrm{n}\right)\:? \\ $$ Commented by TheHoneyCat last updated on 01/Apr/24 On parle de permutations? et si oui... Vous êtes sur que la permutation en question c'est (1,...n) (à savoir l'identité ?) Commented by…
Question Number 205515 by aba last updated on 23/Mar/24 $$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{decomposition}\:\mathrm{into}\:\mathrm{cycles} \\ $$$$\mathrm{with}\:\mathrm{disjoints}\:\mathrm{support}\:\mathrm{of}\:\mathrm{c}^{\mathrm{k}} ,\:\mathrm{where}\:\mathrm{c}=\left(\mathrm{123}…\mathrm{n}\right)\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 205506 by Lindemann last updated on 23/Mar/24 $$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{4}} }{dx} \\ $$ Answered by sniper237 last updated on 23/Mar/24 $$\overset{{u}={x}^{\mathrm{4}} } {=}\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}}…