Question Number 206795 by BaliramKumar last updated on 25/Apr/24 Commented by BaliramKumar last updated on 25/Apr/24 $$\mathrm{can}\:\mathrm{be}\:\:\left(\mathrm{a}\right)\:\mathrm{right}??? \\ $$ Commented by A5T last updated on…
Question Number 206804 by mr W last updated on 25/Apr/24 Answered by mr W last updated on 26/Apr/24 Commented by mr W last updated on…
Question Number 206788 by SANOGO last updated on 25/Apr/24 Answered by A5T last updated on 25/Apr/24 $$\mathrm{1}.\:\overline {\left(\frac{{x}}{{y}}\right)}=\frac{\overset{−} {{x}}}{\overset{−} {{y}}}\Rightarrow\overset{} {\left(\frac{\mathrm{1}}{{z}}\right)}=\frac{\overset{−} {\mathrm{1}}}{\overset{−} {{z}}}=\frac{\mathrm{1}}{\overset{−} {{z}}} \\…
Question Number 206789 by BaliramKumar last updated on 25/Apr/24 Answered by A5T last updated on 25/Apr/24 $$\sqrt[{{n}}]{{a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}} }={P} \\ $$$$\sqrt[{\mathrm{2}{n}}]{{a}_{{n}+\mathrm{1}} {a}_{{n}+\mathrm{2}} …{a}_{\mathrm{3}{n}} }={Q}…
Question Number 206791 by akolade last updated on 25/Apr/24 $$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−\sqrt{\mathrm{x}}}.\mathrm{ln}^{\mathrm{2}} \mathrm{x}\:\mathrm{dx} \\ $$ Commented by Frix last updated on 25/Apr/24 $$\mathrm{Question}\:\mathrm{206754} \\ $$…
Question Number 206787 by Amidip last updated on 25/Apr/24 Commented by mr W last updated on 25/Apr/24 $${you}\:{don}'{t}\:{tell}\:{what}\:{you}\:{want}\:{to}\:{know}? \\ $$ Terms of Service Privacy Policy…
Question Number 206764 by mustafazaheen last updated on 24/Apr/24 Answered by A5T last updated on 24/Apr/24 $${f}\left({g}\left(−\mathrm{3}\right)\right)={f}\left(\sqrt{−\mathrm{3}}\right)={f}\left(\sqrt{\mathrm{3}}{i}\right)=\left(\sqrt{\mathrm{3}}{i}\right)^{\mathrm{2}} =−\mathrm{3} \\ $$ Commented by JDamian last updated…
Question Number 206779 by peter frank last updated on 24/Apr/24 Answered by Spillover last updated on 13/Jul/24 Answered by Spillover last updated on 13/Jul/24 Terms…
Question Number 206773 by MaruMaru last updated on 24/Apr/24 $$\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx}= \\ $$$${I}^{\mathrm{2}} =\int\int_{\:\boldsymbol{\mathcal{D}}} \:\frac{{e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \left({y}^{\mathrm{2}}…
Question Number 206750 by mr W last updated on 23/Apr/24 Answered by HeferH24 last updated on 23/Apr/24 $$\:{By}\:{similar}\:{triangles}: \\ $$$$\:\frac{{d}}{{c}}\:=\:\frac{{x}}{{b}}\:;\:\:\frac{{d}}{{c}}\:=\:\frac{{a}}{{x}}\: \\ $$$$\:\frac{{a}}{{x}}\:=\:\frac{{x}}{{b}}\:\Leftrightarrow\:{x}^{\mathrm{2}} \:=\:{ab}\:\Leftrightarrow\:{x}=\sqrt{{ab}} \\ $$…