Question Number 221350 by MrGaster last updated on 31/May/25 $$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\:\pi{x}}{\Gamma\left(\mathrm{2}+{x}\right)\Gamma\left(\mathrm{2}−{x}\right)}{dx} \\ $$$$ \\ $$$$ \\ $$ Commented by Ghisom last updated on 31/May/25…
Question Number 221347 by RoseAli last updated on 31/May/25 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{4}−{x}^{\mathrm{2}} }{{x}−\mathrm{2}} \\ $$ Answered by AntonCWX8 last updated on 31/May/25 $$\frac{\mathrm{4}−{x}^{\mathrm{2}} }{{x}−\mathrm{2}} \\ $$$$=\frac{\left(\mathrm{2}+{x}\right)\left(\mathrm{2}−{x}\right)}{{x}−\mathrm{2}}…
Question Number 221321 by dr1001sa last updated on 30/May/25 $${if}\:\mathrm{0}<{x}<{y}<{e}^{\mathrm{2}} \:{then} \\ $$$${y}^{\sqrt{{x}}} +{x}^{\mathrm{2}} +\mathrm{6}{xy}+\mathrm{18}{y}^{\mathrm{2}} +\frac{\mathrm{8}}{{x}}+\frac{\mathrm{16}}{\mathrm{9}{x}^{\mathrm{2}} {y}^{\mathrm{2}} }>\mathrm{2}+{x}^{\sqrt{{y}}} \\ $$ Terms of Service Privacy Policy…
Question Number 221322 by Frix last updated on 30/May/25 $$\mathrm{Solve}\:\mathrm{for}\:{x} \\ $$$${x}^{\frac{\mathrm{1}}{{a}}} +\sqrt{{x}^{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}} }={x}^{\frac{\mathrm{1}}{{b}}} \\ $$ Answered by fantastic last updated on 30/May/25 $${or}\:{x}^{\frac{\mathrm{1}}{{a}}} +\sqrt{{x}^{\frac{{a}+{b}}{{ab}}}…
Question Number 221306 by Gbenga last updated on 30/May/25 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{csch}^{\mathrm{2}} \left(\pi{n}\right)}{{n}^{\mathrm{2}} } \\ $$ Answered by SdC355 last updated on 30/May/25 $$\underset{{l}=\mathrm{1}} {\overset{\infty}…
Question Number 221332 by fantastic last updated on 30/May/25 Answered by mr W last updated on 30/May/25 $${AB}={OB}=\sqrt{\mathrm{3}}\:{OD} \\ $$$$\frac{\Delta{DOE}}{\Delta{ABC}}=\left(\frac{{OD}}{{AB}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\Delta{DOE}=\frac{\mathrm{12}}{\mathrm{3}}=\mathrm{4}\:{sq}.\:{units} \\…
Question Number 221315 by klipto last updated on 30/May/25 $$\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…