Question Number 191966 by Shlock last updated on 04/May/23 Answered by mr W last updated on 04/May/23 $${y}={x}^{{x}^{…} } ={x}^{{y}} \\ $$$${y}={e}^{{y}\mathrm{ln}\:{x}} \\ $$$$\left(−{y}\mathrm{ln}\:{x}\right){e}^{−{y}\mathrm{ln}\:{x}} =−\mathrm{ln}\:{x}…
Question Number 191961 by Shlock last updated on 04/May/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 191960 by Rupesh123 last updated on 04/May/23 Answered by a.lgnaoui last updated on 05/May/23 $$\:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)−\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}\right)\right)=\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)−\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}\right)+\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}^{\mathrm{2}} \right)\right. \\ $$$$\:\:\Rightarrow\boldsymbol{\mathrm{f}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)−\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{y}}\right)\right)=\boldsymbol{\mathrm{f}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\right)−\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{f}}^{−\mathrm{1}} \left(\boldsymbol{\mathrm{y}}\right)+\boldsymbol{\mathrm{f}}^{−\mathrm{1}}…
Question Number 126427 by Study last updated on 20/Dec/20 $${li}\underset{{x}\rightarrow−\mathrm{3}} {{m}}\frac{\mid{x}+\mathrm{3}\mid}{{x}+\mathrm{3}}\:\:\:\:\:\:{show}\:{right}\:{and}\:{left}\:{side} \\ $$$${limit}?? \\ $$ Answered by liberty last updated on 20/Dec/20 $$\underset{{x}\rightarrow−\mathrm{3}^{−} } {\mathrm{lim}}\left[\:\frac{\mid{x}+\mathrm{3}\mid}{{x}+\mathrm{3}}\:\right]\:=\:\underset{{x}\rightarrow−\mathrm{3}^{−}…
Question Number 191962 by ajfour last updated on 04/May/23 Answered by mr W last updated on 04/May/23 $$\left[\sqrt{\left({a}+{x}\right)^{\mathrm{2}} −\left({a}−{x}\right)^{\mathrm{2}} }−\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} }\right]^{\mathrm{2}} =\left({b}+{x}\right)^{\mathrm{2}} −\left(\mathrm{2}{a}−{b}−{x}\right)^{\mathrm{2}} \\…
Question Number 60888 by ajfour last updated on 26/May/19 Commented by ajfour last updated on 27/May/19 $${If}\:{length}\:\boldsymbol{{l}}\:{encloses}\:{maximum} \\ $$$${such}\:{area},\:{find}\:{the}\:{area}. \\ $$$$\left({a}\right)\:{if}\:{l}\:{be}\:{circular} \\ $$$$\left({b}\right)\:{if}\:{shape}\:{of}\:{l}\:{not}\:{be}\:{given} \\ $$$$\left({someone}\:{please}\:{consider}\right.…
Question Number 191958 by Rupesh123 last updated on 04/May/23 Commented by Rupesh123 last updated on 04/May/23 x=? Commented by mr W last updated on 04/May/23…
Question Number 60881 by aliesam last updated on 26/May/19 $$\underset{−\pi} {\overset{\pi} {\int}}{sin}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)\:{dx} \\ $$ Commented by MJS last updated on 26/May/19 $$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{this},\:\mathrm{not}\:\mathrm{even} \\ $$$$\mathrm{approximate}.\:\mathrm{it}'\mathrm{s}\:\mathrm{undefined}\:\mathrm{at}\:{x}=\pm\mathrm{1}\:\mathrm{and}…
Question Number 126414 by sdfg last updated on 20/Dec/20 Answered by physicstutes last updated on 20/Dec/20 $$\:\mathrm{Q}_{\mathrm{2}} .\:\mathrm{normally}\:\mathrm{we}\:\mathrm{should}\:\mathrm{be}\:\mathrm{computing}\:\mid\boldsymbol{\mathrm{A}}\:×\:\boldsymbol{\mathrm{B}}\mid\: \\ $$$$\:\mathrm{the}\:\mathrm{unit}\:\mathrm{vector}\:\boldsymbol{\mathrm{n}}\:\mathrm{could}\:\mathrm{be}\:\mathrm{any}\:\mathrm{unit}\:\mathrm{vector}\:\mathrm{they}\:\mathrm{are}\:\mathrm{millions}\:\mathrm{of}\:\mathrm{them}\:\mathrm{so} \\ $$$$\mathrm{i}\:\mathrm{guess}\:\mathrm{we}\:\mathrm{can}\:\mathrm{not}\:\mathrm{get}\:\mathrm{a}\:\mathrm{specific}\:\boldsymbol{\mathrm{A}}×\:\boldsymbol{\mathrm{B}}\:\mathrm{vector}\:\mathrm{but}\:\mathrm{we}\:\mathrm{can}\:\mathrm{get}\:\mid\boldsymbol{\mathrm{A}}×\boldsymbol{\mathrm{B}}\mid\:\mathrm{thus}: \\ $$$$\:\mid\boldsymbol{\mathrm{A}}×\boldsymbol{\mathrm{B}}\mid\:=\:\mid\boldsymbol{\mathrm{A}}\mid\mid\boldsymbol{\mathrm{B}}\mid\:\mathrm{sin}\:\theta\:\mid\:\overset{\wedge} {\boldsymbol{\mathrm{n}}}\mid…
Question Number 191950 by Shrinava last updated on 04/May/23 $$\sqrt{\mathrm{a}}\:=\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{a}}\:\:\:\mathrm{find}:\:\:\:\mathrm{a}^{\mathrm{2}} −\mathrm{a}−\sqrt{\mathrm{a}}\:=\:? \\ $$ Answered by AST last updated on 04/May/23 $$\frac{\mathrm{1}}{\:\sqrt{{a}}−\mathrm{1}}=\frac{\sqrt{{a}}+\mathrm{1}}{{a}−\mathrm{1}}={a}\Rightarrow{a}^{\mathrm{2}} −{a}−\sqrt{{a}}=\mathrm{1} \\ $$ Terms…