Question Number 221315 by klipto last updated on 30/May/25 $$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{function}}\:\boldsymbol{\mathrm{z}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{analytic}}\:\boldsymbol{\mathrm{within}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{on}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{simple}} \\ $$$$\boldsymbol{\mathrm{closed}}\:\boldsymbol{\mathrm{curve}}\:\boldsymbol{\mathrm{C}},−\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{z}}_{\mathrm{0}} \:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{point}}\:\boldsymbol{\mathrm{within}}\:\boldsymbol{\mathrm{C}} \\ $$$$\boldsymbol{\mathrm{using}}\:\boldsymbol{\mathrm{cauchy}}'\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{integral}}\:\boldsymbol{\mathrm{formula}} \\ $$$$\oint\frac{\boldsymbol{\mathrm{sin}\pi\mathrm{z}}^{\mathrm{2}} +\boldsymbol{\mathrm{cos}\pi\mathrm{z}}^{\mathrm{2}} }{\left(\boldsymbol{\mathrm{x}}−\mathrm{1}\right)\left(\boldsymbol{\mathrm{x}}−\mathrm{2}\right)}\boldsymbol{\mathrm{dz}} \\ $$ Commented by MathematicalUser2357 last…
Question Number 221288 by alcohol last updated on 29/May/25 Answered by mahdipoor last updated on 29/May/25 $$\frac{{x}+\frac{{x}+…}{\mathrm{1}+…}}{\mathrm{1}+\frac{{x}+…}{\mathrm{1}+…}}={A}=\frac{{x}+{A}}{\mathrm{1}+{A}}\:\Rightarrow{A}=\sqrt{{x}} \\ $$$$\int\sqrt{{x}}{dx}=\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}/\mathrm{2}} +{C} \\ $$$$ \\ $$ Terms…
Question Number 221268 by behi834171 last updated on 29/May/25 Commented by behi834171 last updated on 29/May/25 $$\boldsymbol{{CF}}=\sqrt{\mathrm{2}}\:\:,\:\:\boldsymbol{{CD}}=\mathrm{2}\:\:\:,\:\:\:\:\boldsymbol{{CE}}=\mathrm{3} \\ $$$$\boldsymbol{{possible}}\:\boldsymbol{{value}}\left(\boldsymbol{{s}}\right)\:\boldsymbol{{for}}:\boldsymbol{{R}}=? \\ $$$$\boldsymbol{{R}}=\boldsymbol{{radius}}.\:\:\:\boldsymbol{{A}}=\boldsymbol{{center}} \\ $$ Answered by…
Question Number 221270 by SdC355 last updated on 29/May/25 $${p},{q}\in\mathbb{P}\: \\ $$$$\: \\ $$$$\mathrm{Use}\:\mathrm{prime}\:\mathrm{number}\:{p},{q}\:\mathrm{to}\:\mathrm{find}\:\mathrm{all}\:\mathrm{prime}\:\mathrm{number}\: \\ $$$$\mathrm{represented}\:\mathrm{by}\:{p}^{{q}} +{q}^{{p}} \\ $$ Answered by Frix last updated on…
Question Number 221271 by gregori last updated on 29/May/25 $$\:\:{Find}\:{the}\:{remainder}\:{when}\:{x}^{\mathrm{100}} \: \\ $$$$\:\:{is}\:{divided}\:{by}\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right) \\ $$ Answered by A5T last updated on 29/May/25 $$\mathrm{x}^{\mathrm{2}} \equiv\left(−\mathrm{1}−\mathrm{x}\right)\left(\mathrm{mod}\:\:\mathrm{x}^{\mathrm{2}}…
Question Number 221298 by ajfour last updated on 29/May/25 $${Find}\:{the}\:{area}\:{of}\:\bigtriangleup{ABC}. \\ $$$${sides}\:{are}\:\sqrt{\mathrm{20}},\:\sqrt{\mathrm{26}}.\:{and}\:\sqrt{\mathrm{34}}\:. \\ $$ Answered by Ghisom last updated on 29/May/25 $$\mathrm{area}\:\mathrm{of}\:\mathrm{triangle}:\:\mathrm{Heron}'\mathrm{s}\:\mathrm{Formula} \\ $$$$\mathrm{area}=\frac{\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}+{c}−{b}\right)\left({b}+{c}−{a}\right)}}{\mathrm{4}} \\…
Question Number 221245 by fantastic last updated on 28/May/25 $${Which}\:{is}\:{greater}\:\:\:\:\:\:\:\left(\sqrt{\mathrm{7}}+\mathrm{1}\right)\:\:{or}\:\:\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}\right)????? \\ $$ Answered by mr W last updated on 28/May/25 $$\left(\sqrt{\mathrm{7}}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{8}+\mathrm{2}\sqrt{\mathrm{7}} \\ $$$$\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{8}+\mathrm{2}\sqrt{\mathrm{15}}>\mathrm{8}+\mathrm{2}\sqrt{\mathrm{7}}=\left(\sqrt{\mathrm{7}}+\mathrm{1}\right)^{\mathrm{2}}…
Question Number 221247 by fantastic last updated on 28/May/25 Commented by fantastic last updated on 28/May/25 $${what}\:{is}\:{the}\:{yellow}\:{chord}\:{length}? \\ $$ Answered by mehdee7396 last updated on…
Question Number 221238 by fantastic last updated on 28/May/25 $${a}=\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\:{then}\:\left\{\sqrt{{a}}−\frac{\mathrm{1}}{\:\sqrt{{a}}}\right\}=?? \\ $$ Answered by Rasheed.Sindhi last updated on 28/May/25 $${a}=\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\:{then}\:\left\{\sqrt{{a}}−\frac{\mathrm{1}}{\:\sqrt{{a}}}\right\}=?? \\ $$$${a}=\left(\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} +\mathrm{2}\left(\sqrt{\mathrm{2}}\:\right)\left(\sqrt{\mathrm{3}}\:\right) \\…
Question Number 221239 by fantastic last updated on 28/May/25 $$\mathrm{2}^{{x}} =\mathrm{4}^{{y}} =\mathrm{8}^{{z}} \:\:{and}\:\left(\frac{\mathrm{1}}{\mathrm{2}{x}}+\mathrm{1}\frac{\mathrm{1}}{\mathrm{4}{y}}+\frac{\mathrm{1}}{\mathrm{6}{z}}\right)=\frac{\mathrm{24}}{\mathrm{7}}\:\:\: \\ $$$${z}=?? \\ $$ Answered by Rasheed.Sindhi last updated on 28/May/25 $$\mathrm{2}^{{x}}…