Question Number 216493 by issac last updated on 09/Feb/25 $$\mathrm{Res}_{{z}={c}} \left\{{f}\left({z}\right)\right\}=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\:\oint_{\:\mathrm{C}} \:{f}\left({z}\right)\mathrm{d}{z} \\ $$$$\mathrm{Res}_{{z}=\mathrm{1}} \left\{\frac{{z}^{\mathrm{21}} +{z}^{\mathrm{2}} +{z}+\mathrm{1}}{\left({z}−\mathrm{1}\right)^{\mathrm{3}} }\right\}=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\:\oint_{\:{C}} \:\frac{{z}^{\mathrm{21}} +{z}^{\mathrm{2}} +{z}+\mathrm{1}}{\left({z}−\mathrm{1}\right)^{\mathrm{3}} }\mathrm{d}{z} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\:\oint_{\:{C}} \:\:\frac{\frac{{z}^{\mathrm{21}}…
Question Number 216489 by manxsol last updated on 09/Feb/25 $${find}\:\:{residuo} \\ $$$$\:\:\:\:\:\:\:\:\frac{{x}^{\mathrm{21}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$ Commented by issac last updated on 09/Feb/25 $$\mathrm{Q216493}…
Question Number 216491 by Jubr last updated on 09/Feb/25 Answered by MrGaster last updated on 09/Feb/25 $$\left(\mathrm{1}\right): \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\int_{\mathrm{0}} ^{{x}} \left(\mathrm{1}−{t}^{\mathrm{2}} +\frac{{t}^{\mathrm{4}} }{\mathrm{2}!}−\frac{{t}^{\mathrm{6}} }{\mathrm{3}!}+\ldots\right){dt}}{\:\sqrt{\mathrm{1}−{e}^{−{x}^{\mathrm{2}}…
Question Number 216513 by CrispyXYZ last updated on 09/Feb/25 $$\mathrm{Prove}\:\mathrm{or}\:\mathrm{disprove}\:\mathrm{that}: \\ $$$$\mathrm{If}\:{p}=\sqrt{\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{3}^{{k}} }\:\left({n}>\mathrm{0}\right)\:\mathrm{is}\:\mathrm{an}\:\mathrm{integer},\:\mathrm{then}\:{p}\:\mathrm{is}\:\mathrm{prime}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 216515 by Jubr last updated on 09/Feb/25 Answered by Jubr last updated on 09/Feb/25 Commented by Jubr last updated on 09/Feb/25 $${Find}\:{the}\:{surface}\:{area}\:{of}\:{the}\:{figure}. \\…
Question Number 216477 by Jubr last updated on 08/Feb/25 Answered by A5T last updated on 08/Feb/25 $$\left.\mathrm{2}\right)\:\sqrt{\mathrm{1}−\sqrt{−\mathrm{3}}}=\mathrm{u}\:;\:\sqrt{\mathrm{1}+\sqrt{−\mathrm{3}}}=\mathrm{v} \\ $$$$\Rightarrow\mathrm{u}^{\mathrm{2}} +\mathrm{v}^{\mathrm{2}} =\mathrm{2}\:;\:\mathrm{uv}=\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\sqrt{−\mathrm{3}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{4}}=\mathrm{2} \\ $$$$\Rightarrow\left(\mathrm{u}+\mathrm{v}\right)^{\mathrm{2}}…
Question Number 216445 by Tawa11 last updated on 08/Feb/25 $$\mathrm{Prove}\:\mathrm{that}\:\:\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:\:=\:\:\sqrt{\pi} \\ $$ Commented by Mathstar last updated on 08/Feb/25 $$\Gamma\left(\mathrm{x}\right)\Gamma\left(\mathrm{1}−\mathrm{x}\right)=\frac{\pi}{\mathrm{sin}\left(\pi\mathrm{x}\right)} \\ $$$$\mathrm{Let}\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\pi\:\Rightarrow\:\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\:\sqrt{\pi}…
Question Number 216478 by Tawa11 last updated on 08/Feb/25 Commented by mr W last updated on 09/Feb/25 $${the}\:{vertex}\:{of}\:{the}\:{cone}\:{may}\:{touch}\: \\ $$$${any}\:{point}\:{of}\:{one}\:{end}\:{of}\:{the}\:{cylinder}, \\ $$$${we}\:{always}\:{have} \\ $$$${V}_{{S}} =\mathrm{2}{V}…
Question Number 216437 by issac last updated on 08/Feb/25 Answered by issac last updated on 08/Feb/25 $$\mathrm{Q}. \\ $$$$\mathrm{Maxwells}\:\mathrm{equations}\:\mathrm{for}\:\mathrm{electric}\:\mathrm{and} \\ $$$$\mathrm{magntic}\:\mathrm{fields}\:\mathrm{are}\:\mathrm{as}\:\mathrm{follows}. \\ $$$$\: \\ $$$$\mathrm{Choose}\:\mathrm{one}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be}\:\mathrm{described}\:\mathrm{by}\:\mathrm{the}…
Question Number 216471 by Ismoiljon_008 last updated on 08/Feb/25 Commented by Ismoiljon_008 last updated on 08/Feb/25 $$\:\:\:{help},\:{please} \\ $$ Commented by mr W last updated…