Question Number 10995 by Nadium last updated on 06/Mar/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{3}>\left(\mathrm{log}_{\mathrm{2}} \mathrm{3}\right)^{\mathrm{2}} >\mathrm{2}. \\ $$ Commented by FilupS last updated on 06/Mar/17 $$\mathrm{for}\:\:{l}=\mathrm{log}_{{n}} {x} \\ $$$$\mathrm{if}\:{n}>\mathrm{1}\:\mathrm{and}\:{x}\geqslant\mathrm{1},\:{l}\geqslant\mathrm{0}…
Question Number 76531 by benjo 1/2 santuyy last updated on 28/Dec/19 $${what}\:{is}\:{a}\:{and}\:{b}\:{such}\:{that}\:{a}+{b}\:=\:{a}×{b}\: \\ $$$${and}\:{a}+{b}\:={a}/{b}\:? \\ $$ Commented by MJS last updated on 28/Dec/19 $$\left(\mathrm{1}\right)\:{a}+{b}={ab}\:\Rightarrow\:{a}=\frac{{b}}{{b}−\mathrm{1}}\:\Rightarrow\:{b}\neq\mathrm{1} \\…
Question Number 10994 by geovane10math last updated on 06/Mar/17 $$\Gamma\left(\mathrm{2407}\right)\:=\:\left(\mathrm{2406}\right)!\:=\:\int_{\mathrm{0}} ^{\infty} {e}^{−{x}} {x}^{\mathrm{2406}} \:\mathrm{dx} \\ $$$$\mathrm{How}\:\mathrm{evaluate}\:\int{e}^{−{x}} {x}^{\mathrm{2406}} \:\mathrm{dx}\:??? \\ $$ Commented by FilupS last updated…
Question Number 142061 by iloveisrael last updated on 26/May/21 $$\sqrt{\mathrm{5}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}+\mathrm{2}{xy}−\mathrm{4}{xz}+\mathrm{10}}\:+ \\ $$$$\mid\mathrm{2}{x}−{y}−\mathrm{13}\mid\:=\:\mathrm{3}\: \\ $$ Commented by MJS_new last updated on 26/May/21 $$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{question}?…
Question Number 142060 by iloveisrael last updated on 26/May/21 Answered by Ar Brandon last updated on 26/May/21 $$\mathrm{I}=\int\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}} \mathrm{sec}\left(\mathrm{1}+\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}} \right)\mathrm{tan}\left(\mathrm{1}+\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}} \right)\mathrm{dx} \\ $$$${u}=\mathrm{1}+\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{x}}} \Rightarrow\mathrm{d}{u}=−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}}…
Question Number 76524 by john santu last updated on 28/Dec/19 $${find}\:{vector}\:{unit}\:{perpendicular}\: \\ $$$${to}\:{vector}\:\overset{−} {{a}}=\left(\mathrm{1},\mathrm{2},\mathrm{3}\right)\:{and}\:\overset{−} {{b}}=\left(−\mathrm{1},\mathrm{0},\mathrm{2}\right) \\ $$ Answered by MJS last updated on 28/Dec/19 $$\begin{vmatrix}{\mathrm{1}}&{−\mathrm{1}}&{{u}_{{x}}…
Question Number 10989 by ketto last updated on 05/Mar/17 $${find}\:{the}\:{image}\:{of}\:{the}\:{point}\left(\mathrm{5},\mathrm{2}\right)\: \\ $$$${under}\:{a}\:{rotation}\:{of}\:\mathrm{90}°\:{clockwise} \\ $$ Answered by ajfour last updated on 05/Mar/17 Commented by ajfour last…
Question Number 76522 by john santu last updated on 28/Dec/19 $${sir}\:{what}'{s}\:{difference}\:\frac{{d}}{{da}}\:{with}\:\frac{\partial}{\partial{a}}\:?\: \\ $$ Commented by mr W last updated on 28/Dec/19 $${if}\:{the}\:{function}\:{has}\:{only}\:{one} \\ $$$${variable},\:{then}\:\frac{{df}\left({x}\right)}{{dx}}. \\…
Question Number 10987 by Joel576 last updated on 05/Mar/17 $$\mathrm{If}\: \\ $$$$\bullet\:{f}\left(\mathrm{2}{x}\:+\:\mathrm{1}\right)\:+\:{g}\left(\mathrm{3}\:−\:{x}\right)\:=\:{x} \\ $$$$\bullet\:{f}\left(\frac{\mathrm{3}{x}\:+\:\mathrm{5}}{\:{x}\:+\:\mathrm{1}}\right)\:+\:\mathrm{2}{g}\left(\frac{\mathrm{2}{x}\:+\:\mathrm{1}}{{x}\:+\:\mathrm{1}}\right)\:=\:\frac{{x}}{{x}\:+\:\mathrm{1}} \\ $$$$\mathrm{for}\:\mathrm{every}\:{x}\:\in\:\mathbb{R},\:\:{x}\:\neq\:−\mathrm{1} \\ $$$$\mathrm{Find}\:{f}\left({x}\right)\:!! \\ $$ Answered by ajfour last updated…
Question Number 142052 by Rankut last updated on 26/May/21 $$\boldsymbol{\mathrm{Given}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{fog}}\left(\boldsymbol{\mathrm{x}}\right)=\frac{\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{1}}{\boldsymbol{\mathrm{x}}}\:\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{g}}\left(\boldsymbol{\mathrm{x}}\right)=\mathrm{5}\boldsymbol{\mathrm{x}}+\mathrm{2}, \\ $$$$\boldsymbol{\mathrm{Find}}\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right). \\ $$$$ \\ $$ Answered by iloveisrael last updated on 26/May/21 $$\Rightarrow{f}\left(\mathrm{5}{x}+\mathrm{2}\right)=\frac{\mathrm{2}{x}−\mathrm{1}}{{x}} \\…