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}} \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\frac{\left(\sqrt{\mathrm{2}}{cosx}-\mathrm{1}\right)^{'} }{\left({tanx}-\mathrm{1}\right)^{'} }=\frac{\sqrt{\mathrm{2}}\left(-{sinx}\right)}{{sec}^{\mathrm{2}} {x}} \\…
Question Number 109606 by mathdave last updated on 24/Aug/20 Commented by malwan last updated on 24/Aug/20 $${What}\:{do}\:{you}\:{do}\:{sir}\:? \\ $$$${You}\:{post}\:{the}\:{solution}\:{as}\:{question} \\ $$$${and}\:??\:{where}\:{is}\:{the}\:{final}\:{result}?? \\ $$ Commented by…
Question Number 44068 by peter frank last updated on 21/Sep/18 Answered by kunal1234523 last updated on 21/Sep/18 $${join}\:{OB}\:{and}\:{OD}\:={R} \\ $$$${l}^{\mathrm{2}} +{x}^{\mathrm{2}} ={R}^{\mathrm{2}} \:\:\:\:…..\left(\mathrm{1}\right) \\ $$$${h}^{\mathrm{2}}…
Question Number 109605 by mathdave last updated on 24/Aug/20 Commented by Rasheed.Sindhi last updated on 24/Aug/20 $${mathdave}\:{sir}, \\ $$$$\mathcal{T}{he}\:{space}\:{in}\:{which}\:{you}\:{have} \\ $$$${written}\:{solution}\:\left({of}\:{an}\:{unknown}\right. \\ $$$$\left.{question}\right)\:{is}\:{specified}\:{for} \\ $$$$\boldsymbol{{asking}}\:\boldsymbol{{question}}.{Please}\:{don}'{t}…
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