Question Number 109734 by mathdave last updated on 25/Aug/20 $${solve}\:{the}\:{following}\:{integral} \\ $$$$\left.\mathrm{1}\right)\int_{\mathrm{3}} ^{\mathrm{7}} \mathrm{4}\sqrt{\left({x}−\mathrm{3}\right)\left(\mathrm{7}−{x}\right)}{dx} \\ $$$$\left.\mathrm{2}\right)\int_{\mathrm{0}} ^{\infty} \frac{{x}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }{dx} \\ $$$$\left.\mathrm{3}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left[\mathrm{ln}\left(\mathrm{1}−\mathrm{tan}{x}\right)\right]^{\mathrm{2}} {dx}=\frac{\pi}{\mathrm{2}}\mathrm{ln2}−\mathrm{2}{G}…
Question Number 44194 by peter frank last updated on 23/Sep/18 Commented by maxmathsup by imad last updated on 23/Sep/18 $$\left.{ii}\right)\:{fog}\left({x}\right)={f}\left({g}\left({x}\right)\right)=\mid{x}−\mathrm{2}\mid^{\mathrm{2}} −\mathrm{1}=\left({x}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1}={x}^{\mathrm{2}} \:−\mathrm{4}{x}+\mathrm{4}−\mathrm{1}={x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3} \\…
Question Number 175264 by Beginner last updated on 25/Aug/22 Answered by Rasheed.Sindhi last updated on 25/Aug/22 $${S}_{{n}} =\mathrm{5}{n}^{\mathrm{2}} −\mathrm{2}{n}={n}\left(\mathrm{5}{n}−\mathrm{2}\right) \\ $$$${U}_{{r}} ={S}_{{r}} −{S}_{{r}−\mathrm{1}} =\left(\mathrm{5}{r}^{\mathrm{2}} −\mathrm{2}{r}\right)−\left(\:\mathrm{5}\left({r}−\mathrm{1}\right)^{\mathrm{2}}…
Question Number 175251 by MathsFan last updated on 24/Aug/22 $$\boldsymbol{\mathrm{Given}}\:\boldsymbol{\mathrm{that}}\: \\ $$$$\:\mathrm{4}\left(\boldsymbol{\mathrm{x}}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{9}\left(\boldsymbol{\mathrm{y}}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{27} \\ $$$$\:\boldsymbol{\mathrm{graph}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{ellipse}} \\ $$ Answered by MJS_new last updated on 26/Aug/22…
Question Number 175240 by Best1 last updated on 24/Aug/22 Commented by Best1 last updated on 25/Aug/22 $${how} \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 44161 by MJS last updated on 22/Sep/18 $$\mathrm{LOL}!\:\mathrm{found}\:\mathrm{this}\:\mathrm{on}\:\mathrm{the}\:\mathrm{web}: \\ $$$$\mathrm{1}=\sqrt{\mathrm{1}}=\sqrt{\left(−\mathrm{1}\right)\left(−\mathrm{1}\right)}=\sqrt{−\mathrm{1}}\sqrt{−\mathrm{1}}=\mathrm{i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$\mathrm{each}\:\mathrm{step}\:\mathrm{seems}\:\mathrm{right},\:\mathrm{so}\:\mathrm{where}'\mathrm{s}\:\mathrm{the}\:\mathrm{mistake}? \\ $$ Answered by rahul 19 last updated on 22/Sep/18…
Question Number 175211 by MathsFan last updated on 23/Aug/22 $$\boldsymbol{\mathrm{Assume}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{sequence}}\:\boldsymbol{\mathrm{terms}}\:\boldsymbol{\mathrm{tend}} \\ $$$$\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{constant}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{{u}},\:\boldsymbol{\mathrm{so}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{as}} \\ $$$$\boldsymbol{\mathrm{n}}\rightarrow\infty,\:\boldsymbol{{u}}_{\boldsymbol{{n}}−\mathrm{1}} \rightarrow\boldsymbol{{u}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{{u}}_{\boldsymbol{{n}}} \rightarrow\boldsymbol{{u}}. \\ $$$$\:\left(\boldsymbol{\mathrm{i}}\right)\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\:\boldsymbol{{u}}^{\mathrm{2}} +\boldsymbol{{u}}−\mathrm{1}=\mathrm{0} \\ $$$$\:\left(\boldsymbol{\mathrm{ii}}\right)\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+…..}}}}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$ Answered by…
Question Number 175193 by Best1 last updated on 22/Aug/22 Answered by Rasheed.Sindhi last updated on 22/Aug/22 $$ \\ $$$${A}=\begin{bmatrix}{\mathrm{2}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{3}}&{\mathrm{3}}&{\mathrm{2}}\end{bmatrix},\:{A}^{{T}} =\begin{bmatrix}{\mathrm{2}}&{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{2}}\end{bmatrix} \\ $$$${AA}^{{T}} =\begin{bmatrix}{\mathrm{4}+\mathrm{1}+\mathrm{0}}&{\mathrm{2}+\mathrm{2}+\mathrm{0}}&{\mathrm{6}+\mathrm{3}+\mathrm{0}}\\{\mathrm{2}+\mathrm{2}+\mathrm{0}}&{\mathrm{1}+\mathrm{4}+\mathrm{1}}&{\mathrm{3}+\mathrm{6}+\mathrm{2}}\\{\mathrm{6}+\mathrm{3}+\mathrm{0}}&{\mathrm{3}+\mathrm{6}+\mathrm{2}}&{\mathrm{9}+\mathrm{9}+\mathrm{4}}\end{bmatrix}\: \\ $$$${AA}^{{T}}…
Question Number 109640 by hhryhrry2 last updated on 24/Aug/20 Commented by kaivan.ahmadi last updated on 24/Aug/20 $${a}. \\ $$$$\underset{\mathrm{2}} {\overset{\mathrm{6}} {\sum}}\left({g}^{\mathrm{2}} +\mathrm{2}{g}\right)−\underset{\mathrm{2}} {\overset{\mathrm{6}} {\sum}}\left(\mathrm{3}\left({g}+\mathrm{3}\right)−\mathrm{5}\right)=\underset{\mathrm{2}} {\overset{\mathrm{6}}…
Question Number 175160 by thean last updated on 21/Aug/22 Commented by Alisajadrajaee last updated on 21/Aug/22 $$\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\left(\frac{\sqrt{\mathrm{2}}{cosx}-\mathrm{1}}{{tanx}-\mathrm{1}}\right)=\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\frac{\left(\sqrt{\mathrm{2}}{cosx}-\mathrm{1}\right)^{'} }{\left({tanx}-\mathrm{1}\right)^{'} }=\frac{\sqrt{\mathrm{2}}\left(-{sinx}\right)}{{sec}^{\mathrm{2}} {x}} \\…