Question Number 213846 by issac last updated on 18/Nov/24 $$\mathrm{path}\:\mathscr{C}\:\mathrm{is}\:\mathrm{closed} \\ $$$${f}\:\mathrm{is}\:\mathrm{regular}\:\mathrm{function}\:\mathrm{in}\:\mathrm{Path}\:\mathscr{C} \\ $$$${f}\:\mathrm{is}\:\mathrm{have}\:\mathrm{zero}\:\mathrm{point}\:\mathrm{and}\:\mathrm{poles}\:\mathrm{in}\:\mathscr{C} \\ $$$$\mathrm{show}\:\mathrm{that}\: \\ $$$$\oint_{\:\mathscr{C}} \:\frac{{f}\left({z}\right)}{{f}'\left({z}\right)}\:\mathrm{d}{z}=\mathrm{2}\pi\boldsymbol{{i}}\left({Z}−{P}\right) \\ $$$$\mathrm{and}\:\mathrm{if}\:\mathrm{poles}\:\mathrm{are}\:\mathrm{not}\:\mathrm{exist} \\ $$$$\mathrm{show}\:\mathrm{that} \\ $$$$\oint_{\:\mathscr{C}\:}…
Question Number 213841 by mnjuly1970 last updated on 18/Nov/24 $$ \\ $$$$\:\:{Find}\:{the}\:{vertical}\:{asymptots} \\ $$$$\: \\ $$$$\:\:{of}\:\:,\:\:\:{f}\left({x}\right)=\:\mathrm{tan}\left(\frac{\:\pi}{\mathrm{2}{x}\:+\:\mathrm{2}}\:\right)\:\:{in}\: \\ $$$$\: \\ $$$$\:\:\:\:\:\left[\:\mathrm{0}\:\:,\:\:\:\mathrm{4}\:\right] \\ $$$$\:−−−−−−−−−−−−− \\ $$$$ \\…
Question Number 213821 by hardmath last updated on 17/Nov/24 $$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{sinx}}{\mathrm{x}}\right)^{\frac{\mathrm{sinx}}{\mathrm{x}\:−\:\mathrm{sinx}}} \:\:=\:\:? \\ $$ Answered by mehdee7396 last updated on 17/Nov/24 $${lim}_{{x}\rightarrow\mathrm{0}} \left(\frac{{sinx}}{{x}}−\mathrm{1}\right)\frac{{sinx}}{{x}−{sinx}} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}}…
Question Number 213817 by essaad last updated on 17/Nov/24 Commented by essaad last updated on 17/Nov/24 numenclature stp Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 213818 by ajfour last updated on 17/Nov/24 Commented by ajfour last updated on 17/Nov/24 $${Find}\:{R}\:{in}\:{terms}\:{smaller}\:{radii}\:{a},\:{b}. \\ $$ Answered by mr W last updated…
Question Number 213802 by efronzo1 last updated on 17/Nov/24 Answered by A5T last updated on 17/Nov/24 $${t}_{\mathrm{1}} ={k}−\mathrm{1};{t}_{\mathrm{2}} ={k} \\ $$$${t}_{\mathrm{3}} =\frac{\mathrm{5}{k}+\mathrm{1}}{\mathrm{25}{k}−\mathrm{25}};\:\:\:\:{t}_{\mathrm{4}} =\frac{\frac{\mathrm{10}{k}−\mathrm{4}}{\mathrm{5}{k}−\mathrm{5}}}{\mathrm{25}{k}}=\frac{\mathrm{10}{k}−\mathrm{4}}{\mathrm{125}{k}\left({k}−\mathrm{1}\right)} \\ $$$${t}_{\mathrm{5}}…
Question Number 213803 by muallimRiyoziyot last updated on 17/Nov/24 Commented by Frix last updated on 17/Nov/24 $$\mathrm{There}\:\mathrm{is}\:\mathrm{a}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{complex}\:\mathrm{solutions}\:\mathrm{but}\:\mathrm{the} \\ $$$$\mathrm{exact}\:\mathrm{form}\:\mathrm{is}\:\mathrm{not}\:\mathrm{useable}. \\ $$$${x}\approx\mathrm{1}.\mathrm{32848492}\pm.\mathrm{570204126i} \\ $$ Commented by…
Question Number 213796 by efronzo1 last updated on 17/Nov/24 Answered by A5T last updated on 17/Nov/24 $${x}_{\mathrm{0}} ={k}\Rightarrow{x}_{\mathrm{1}} =\frac{\mathrm{1}+{k}}{\mathrm{1}−{k}}\Rightarrow{x}_{\mathrm{2}} =\frac{−\mathrm{1}}{{k}}\Rightarrow{x}_{\mathrm{3}} =\frac{{k}−\mathrm{1}}{\mathrm{1}+{k}}\Rightarrow{x}_{\mathrm{4}} =\frac{\mathrm{2}{k}}{\mathrm{2}}={k} \\ $$$$\Rightarrow{x}_{\mathrm{4}{n}} ={k}=\mathrm{2022}…
Question Number 213790 by issac last updated on 16/Nov/24 $$\mathrm{So}\:\mathrm{Weird}…… \\ $$$$\int_{\mathrm{0}} ^{\:\infty} {J}_{\nu} \left({t}\right){e}^{−{st}} \mathrm{d}{t}=\frac{\left({s}+\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{s}^{\mathrm{2}} +\mathrm{1}}}\: \\ $$$${J}_{−\nu} \left({t}\right)=\left(−\mathrm{1}\right)^{\nu} {J}_{\nu} \left({t}\right)\:\: \\…
Question Number 213791 by hardmath last updated on 16/Nov/24 $$\mathrm{If}\:\:\:\mathrm{x}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{x}}\:−\:\frac{\mathrm{4}}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}}}\:\:=\:\:\mathrm{10} \\ $$$$\mathrm{Find}\:\:\:\sqrt[{\mathrm{3}}]{\mathrm{x}}\:−\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}}}\:\:+\:\:\mathrm{3}\:\:=\:\:? \\ $$ Commented by muallimRiyoziyot last updated on 19/Nov/24 $${x}−\mathrm{8}=\sqrt[{\mathrm{3}}]{{x}}+\mathrm{2}+\frac{\mathrm{4}}{\:\sqrt[{\mathrm{3}}]{{x}}} \\ $$$$\left(\sqrt[{\mathrm{3}}]{{x}}−\mathrm{2}\right)\left(\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }+\mathrm{2}\sqrt[{\mathrm{3}}]{{x}}+\mathrm{4}\right)=\frac{\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}}…