Question Number 118647 by cantor last updated on 18/Oct/20 Answered by mathmax by abdo last updated on 18/Oct/20 $$\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{z}^{\mathrm{2}} −\mathrm{2z}}{\left(\mathrm{z}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} +\mathrm{4}\right)}\:\:\:−\mathrm{1}\:\mathrm{is}\:\mathrm{double}\:\mathrm{poles}\:\mathrm{for}\:\mathrm{f}\:\mathrm{so} \\ $$$$\mathrm{Res}\left(\mathrm{f},−\mathrm{1}\right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow−\mathrm{1}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}+\mathrm{1}\right)^{\mathrm{2}}…
Question Number 118627 by mohammad17 last updated on 18/Oct/20 $${find}\:{the}\:{infmum}\left({s}\right)\:{if}\:{S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\:}\: \\ $$$${help}\:{me}\:{sir} \\ $$ Answered by TANMAY PANACEA last updated on 18/Oct/20…
Question Number 184159 by Davidtim last updated on 03/Jan/23 $${prove}\:{that}\:\mathrm{0}^{\mathrm{0}} =\mathrm{1} \\ $$ Commented by Frix last updated on 03/Jan/23 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{moon}\:\mathrm{is}\:\mathrm{a}\:\mathrm{star}. \\ $$ Answered by…
Question Number 184144 by paul2222 last updated on 03/Jan/23 $$\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\boldsymbol{\sum}}}\underset{\boldsymbol{\mathrm{m}}=\mathrm{1}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}+\boldsymbol{\mathrm{m}}} }{\boldsymbol{\mathrm{nm}}\left(\boldsymbol{\mathrm{n}}+\boldsymbol{\mathrm{m}}\right)} \\ $$ Answered by SEKRET last updated on 04/Jan/23 $$\:\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}}…
Question Number 53055 by Otchere Abdullai last updated on 16/Jan/19 $${The}\:{area}\:{of}\:\bigtriangleup{XYZ}\:{is}\:\mathrm{50}{cm}^{\mathrm{2}} .\:{L}\: \\ $$$${is}\:{the}\:{midpoint}\:{of}\:{XZ}\:{andKY}\:\parallel{LM}. \\ $$$${find}\:{the}\:{area}\:{of}\:\:\bigtriangleup{ZLY}. \\ $$$${help}\:{please}\:{sir} \\ $$ Commented by mr W last…
Question Number 118585 by Tinku Tara last updated on 18/Oct/20 $${v}\mathrm{2}.\mathrm{227}\:\mathrm{reverts}\:\mathrm{app}\:\mathrm{first}\:\mathrm{screen}\:\mathrm{to} \\ $$$$\mathrm{two}\:\mathrm{parts},\:\mathrm{saved}\:\mathrm{and}\:\mathrm{forum}. \\ $$$$\mathrm{To}\:\mathrm{start}\:\mathrm{in}\:\mathrm{forum}\:\mathrm{by}\:\mathrm{default}\:\mathrm{change} \\ $$$$\mathrm{settings}\:\mathrm{and}\:\mathrm{Enable}\:\mathrm{start}\:\mathrm{in}\:\mathrm{forum}. \\ $$$$\mathrm{Non}\:\mathrm{English}\:\mathrm{users}\:\mathrm{were}\:\mathrm{having} \\ $$$$\mathrm{difficulty}\:\mathrm{in}\:\mathrm{locating}\:\mathrm{offline}\:\mathrm{editor}. \\ $$$$\mathrm{Work}\:\mathrm{is}\:\mathrm{in}\:\mathrm{progress}\:\mathrm{to}\:\mathrm{translate} \\ $$$$\mathrm{in}-\mathrm{app}\:\mathrm{menu}\:\mathrm{to}\:\mathrm{different}\:\mathrm{languages}.…
Question Number 53033 by gopikrishnan005@gmail.com last updated on 16/Jan/19 $$\mathrm{in}\:\mathrm{the}\:\mathrm{law}\:\mathrm{of}\:\mathrm{mean}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\theta\:\mathrm{satisfies}\:\mathrm{the}\:\mathrm{condition}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 184103 by paul2222 last updated on 02/Jan/23 $$\boldsymbol{\mathrm{evaluate}}\:\overset{\frac{\boldsymbol{\pi}}{\mathrm{2}}} {\int}_{\mathrm{0}} \boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)} \boldsymbol{\mathrm{dx}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 184102 by paul2222 last updated on 02/Jan/23 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{ln}}^{\mathrm{4}} \left(\mathrm{1}−\boldsymbol{\mathrm{x}}\right)}{\:\sqrt{\mathrm{1}−\boldsymbol{\mathrm{x}}}}\boldsymbol{\mathrm{dx}} \\ $$ Answered by ARUNG_Brandon_MBU last updated on 03/Jan/23 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{4}}…
Question Number 184093 by Ib last updated on 02/Jan/23 Answered by manxsol last updated on 03/Jan/23 $$\left(\mathrm{1}+\frac{{t}}{\mathrm{100}}\right)^{\mathrm{10}} =\mathrm{3} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\left(\frac{\mathrm{100}}{{t}}\right)}\right)^{\mathrm{10}} =\mathrm{3} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\left(\frac{\mathrm{100}}{{t}}\right)}\right)^{\mathrm{10}×\frac{\mathrm{10}}{{t}}} =\mathrm{3}^{\frac{\mathrm{10}}{{t}}} \\…