Question Number 62281 by behi83417@gmail.com last updated on 19/Jun/19 $$\begin{cases}{\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\boldsymbol{\mathrm{y}}^{\mathrm{3}} =\mathrm{3}\boldsymbol{\mathrm{xy}}}\\{\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\boldsymbol{\mathrm{y}}^{\mathrm{4}} =\mathrm{4}\boldsymbol{\mathrm{xy}}}\end{cases}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\neq\mathrm{0}\right] \\ $$ Answered by mr W last updated on 19/Jun/19 $${let}\:{u}={x}+{y},\:{v}={xy}…
Question Number 127814 by dactor last updated on 02/Jan/21 Answered by mnjuly1970 last updated on 02/Jan/21 $${solution} \\ $$$$\mathrm{2}{cos}^{\mathrm{2}} \left({x}\right)−\mathrm{1}−{cos}\left({x}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{cos}\left({x}\right)\left(\mathrm{2}{cos}\left({x}\right)−\mathrm{1}\right)=\mathrm{0} \\ $$$${cos}\left({x}\right)=\mathrm{0}\Rightarrow\:{x}=\mathrm{2}{k}\pi\:\:;\:{k}\in\mathbb{Z} \\…
Question Number 127815 by Algoritm last updated on 02/Jan/21 Commented by Algoritm last updated on 02/Jan/21 $$\mathrm{prove}\:\mathrm{the}\:\mathrm{equation} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 127813 by dactor last updated on 02/Jan/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 62276 by behi83417@gmail.com last updated on 18/Jun/19 $$\begin{cases}{\frac{\sqrt{\boldsymbol{\mathrm{x}}}}{\boldsymbol{\mathrm{a}}}+\frac{\sqrt{\boldsymbol{\mathrm{y}}}}{\boldsymbol{\mathrm{b}}}=\mathrm{1}}\\{\frac{\sqrt{\boldsymbol{\mathrm{a}}}}{\boldsymbol{\mathrm{x}}}+\frac{\sqrt{\boldsymbol{\mathrm{b}}}}{\boldsymbol{\mathrm{y}}}=\mathrm{1}}\end{cases}\:\:\:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\in\mathrm{R}^{+} \\ $$ Answered by mr W last updated on 19/Jun/19 $${let}\:{X}=\sqrt{{x}},\:{Y}=\sqrt{{y}},\:{A}=\sqrt{{a}},\:{B}=\sqrt{{b}} \\ $$$$\frac{{X}}{{A}^{\mathrm{2}} }+\frac{{Y}}{{B}^{\mathrm{2}} }=\mathrm{1}…
Question Number 62275 by behi83417@gmail.com last updated on 18/Jun/19 $$\begin{cases}{\boldsymbol{\mathrm{a}}\sqrt{\boldsymbol{\mathrm{x}}}+\boldsymbol{\mathrm{b}}\sqrt{\boldsymbol{\mathrm{y}}}=\mathrm{2}\sqrt{\boldsymbol{\mathrm{ab}}}}\\{\boldsymbol{\mathrm{x}}\sqrt{\boldsymbol{\mathrm{a}}}+\boldsymbol{\mathrm{y}}\sqrt{\boldsymbol{\mathrm{b}}}=\mathrm{2}\sqrt{\boldsymbol{\mathrm{ab}}}}\end{cases}\:\:\:\:\:\:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\in\mathrm{R}^{+} \\ $$ Answered by mr W last updated on 19/Jun/19 $${let}\:{X}=\sqrt{{x}},{Y}=\sqrt{{y}},{A}=\sqrt{{a}},{B}=\sqrt{{b}} \\ $$$${A}^{\mathrm{2}} {X}+{B}^{\mathrm{2}} {Y}=\mathrm{2}{AB}…
Question Number 127811 by mr W last updated on 02/Jan/21 Commented by mr W last updated on 02/Jan/21 $${a}\:{paraboloid}\:{bowl}\:{has}\:{a}\:{depth}\:{of}\:{H} \\ $$$${and}\:{an}\:{opening}\:{with}\:{diameter}\:{L}. \\ $$$${a}\:{small}\:{ball}\:{of}\:{mass}\:{m}\:{is}\:{released} \\ $$$${from}\:{rest}\:{at}\:{the}\:{rim}\:{of}\:{the}\:{bowl}.…
Question Number 62274 by aliesam last updated on 18/Jun/19 $$\underset{\mathrm{0}} {\overset{\infty} {\int}}{e}^{−{x}^{\mathrm{2}} } \:{dx} \\ $$ Commented by maxmathsup by imad last updated on 19/Jun/19…
Question Number 62273 by alphaprime last updated on 18/Jun/19 $${Mr}.\:{Rasheed}.{Sindhi}\: \\ $$$${I}\:{sense}\:{you}'{re}\:{much}\:{engaged}\:{in}\:{making}\: \\ $$$${olympiad}\:{contents}\:{these}\:{days}\:,\:{I}\:{wish}\: \\ $$$${that}\:{you}\:{join}\:{my}\:{workspace}\:{concerning}\:{that}\:{same}. \\ $$ Commented by Rasheed.Sindhi last updated on 20/Jun/19…
Question Number 62266 by aliesam last updated on 18/Jun/19 $$\int\frac{\mathrm{2}{sin}\left({x}\right)+\mathrm{3}{cos}\left({x}\right)}{\mathrm{3}{sin}\left({x}\right)+\mathrm{4}{cos}\left({x}\right)}{dx} \\ $$ Commented by maxmathsup by imad last updated on 19/Jun/19 $${let}\:{A}\:=\int\:\:\:\frac{\mathrm{2}{sinx}\:+\mathrm{3}{cosx}}{\mathrm{3}{sinx}\:+\mathrm{4}{cosx}}\:{dx}\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${A}\:=\int\:\:\:\frac{\frac{\mathrm{4}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+\mathrm{3}\frac{\mathrm{1}−{t}^{\mathrm{2}}…