Question Number 119057 by mathocean1 last updated on 21/Oct/20 $${show}\:{that} \\ $$$$\mid{x}+{y}\mid\leqslant\mid{x}\mid+\mid{y}\mid \\ $$$$ \\ $$ Answered by Bird last updated on 21/Oct/20 $$\left(\mid{x}\mid+\mid{y}\mid\right)^{\mathrm{2}} −\mid{x}+{y}\mid^{\mathrm{2}}…
Question Number 119048 by Hassen_Timol last updated on 21/Oct/20 $$\mathrm{Show}\:\mathrm{that}\:: \\ $$$$ \\ $$$$\:\:\:\:\lfloor{x}\rfloor\:=\:\lfloor\frac{\lfloor{nx}\rfloor}{{n}}\rfloor\:\:\:\:\mathrm{where}\:{n}\in\mathbb{N}^{\ast} \:\:\:{x}\in\mathbb{R} \\ $$$$ \\ $$ Answered by mr W last updated…
Question Number 53500 by Otchere Abdullai last updated on 22/Jan/19 $$\mathrm{1}.\:{If}\:\:\sqrt{\left({x}+\mathrm{2}\right)^{\mathrm{2}\:} +{y}^{\mathrm{2}} }+\sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }=\mathrm{6} \\ $$$${show}\:{that}\:{when}\:{the}\:{equation}\:{is}\: \\ $$$${simplified},\:{it}\:{can}\:{be}\:{express}\:{as} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{9}}+\frac{{y}^{\mathrm{2}} }{\mathrm{5}}=\mathrm{1} \\ $$$$\mathrm{2}.\:{find}\:{the}\:{value}\:{of}\:{n}\:{such}\:{that}\:{the}…
Question Number 119032 by Algoritm last updated on 21/Oct/20 Answered by 1549442205PVT last updated on 22/Oct/20 $$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+…−\frac{\mathrm{1}}{\mathrm{2n}}=\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}+…+\frac{\mathrm{1}}{\mathrm{2n}} \\ $$$$\mathrm{Put}\:\mathrm{S}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}+…−\frac{\mathrm{1}}{\mathrm{2n}} \\ $$$$\Rightarrow\:\mathrm{S}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{\mathrm{2n}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2n}} \\ $$$$−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{\mathrm{2n}}\right) \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+…+\frac{\mathrm{1}}{\mathrm{2n}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2n}}…
Question Number 119035 by ZiYangLee last updated on 21/Oct/20 $$\mathrm{Among}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{less}\:\mathrm{than}\:\mathrm{1200}, \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{of}\:\mathrm{them}\:\mathrm{are}\:\mathrm{relatively}\:\mathrm{prime} \\ $$$$\mathrm{to}\:\mathrm{60}? \\ $$ Answered by floor(10²Eta[1]) last updated on 21/Oct/20 $$\mathrm{60}=\mathrm{2}^{\mathrm{2}} .\mathrm{3}.\mathrm{5}…
Question Number 119008 by zakirullah last updated on 21/Oct/20 Commented by zakirullah last updated on 21/Oct/20 $$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{initial}}\:\boldsymbol{\mathrm{point}}\:\boldsymbol{\mathrm{p}}\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{the}}\:\: \\ $$$$\boldsymbol{\mathrm{terminal}}\:\boldsymbol{\mathrm{point}}\:\boldsymbol{\mathrm{Q}}\:\boldsymbol{{whichever}}\:\boldsymbol{{is}}\:\boldsymbol{{missing}}? \\ $$$$\boldsymbol{{i}}.\:\:\boldsymbol{{P}}\overset{\rightarrow} {\boldsymbol{{Q}}}\:=\:\left[−\mathrm{2},\mathrm{3}\right],\:\boldsymbol{{P}}\left(\mathrm{1},−\mathrm{2}\right) \\ $$$$\boldsymbol{{ii}}.\:\boldsymbol{{P}}\overset{\rightarrow} {\boldsymbol{{Q}}}\:=\left[\mathrm{4},−\mathrm{5}\right],\:\boldsymbol{{Q}}\left(−\mathrm{1},\mathrm{1}\right)…
Question Number 53455 by hassentimol last updated on 22/Jan/19 $$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{derivative} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{function}\:\mathrm{involving}\:\mathrm{complex} \\ $$$$\mathrm{numbers}\:?\:{Thanks} \\ $$ Commented by ajfour last updated on 22/Jan/19 $${study}\:{from}\:{Erwin}\:{Kreizig}'{s} \\…
Question Number 118990 by Algoritm last updated on 21/Oct/20 Answered by mr W last updated on 22/Oct/20 $${x}\neq\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{5}} −\mathrm{1}}{{x}−\mathrm{1}}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{5}} =\mathrm{1} \\…
Question Number 118984 by 675480065 last updated on 21/Oct/20 $$\mathrm{z}=\mathrm{x}+\mathrm{iy} \\ $$$$\mathrm{and}\:\mathrm{z}^{\mathrm{5}} =\mathrm{1} \\ $$$$\mathrm{Show}\:\mathrm{that}\:\mathrm{4x}\left(\mathrm{y}^{\mathrm{4}} −\mathrm{x}^{\mathrm{4}} \right)=\mathrm{1} \\ $$$$\mathrm{Pls}\:\mathrm{help} \\ $$ Answered by Dwaipayan Shikari…
Question Number 184505 by SEKRET last updated on 07/Jan/23 $$\:\:\:\frac{\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)^{\mathrm{5}} −\boldsymbol{\mathrm{a}}^{\mathrm{5}} −\boldsymbol{\mathrm{b}}^{\mathrm{5}} −\boldsymbol{\mathrm{c}}^{\mathrm{5}} }{\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)^{\mathrm{3}} −\boldsymbol{\mathrm{a}}^{\mathrm{3}} −\boldsymbol{\mathrm{b}}^{\mathrm{3}} −\boldsymbol{\mathrm{c}}^{\mathrm{3}} }\:=\:? \\ $$$$\:\mathrm{is}\:\mathrm{there}\:{an}\:\:{easier}\:\:{way}. \\ $$ Answered by Frix…