Question Number 195165 by Nimnim111118 last updated on 25/Jul/23 $$\mathrm{If}\:\overset{\rightarrow} {\mathrm{r}_{\mathrm{1}} }=\left(\mathrm{sin}\theta,\mathrm{cos}\theta,\theta\right),\:\overset{\rightarrow} {\mathrm{r}_{\mathrm{2}} }=\left(\mathrm{cos}\theta,−\mathrm{sin}\theta,−\mathrm{3}\right)\:\mathrm{and} \\ $$$$\:\overset{\rightarrow} {\mathrm{r}_{\mathrm{3}} }=\left(\mathrm{2},\mathrm{3},−\mathrm{1}\right),\:\mathrm{find}\:\frac{\mathrm{d}}{\mathrm{d}\theta}\left\{\overset{\rightarrow} {\mathrm{r}_{\mathrm{1}} }×\left(\overset{\rightarrow} {\mathrm{r}_{\mathrm{2}} }×\overset{\rightarrow} {\mathrm{r}_{\mathrm{3}} }\right)\right\}\:\mathrm{at}\:\theta=\mathrm{0} \\…
Question Number 195135 by 073 last updated on 25/Jul/23 Answered by som(math1967) last updated on 25/Jul/23 $$\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} −\mathrm{3}{x}.\frac{\mathrm{1}}{{x}}\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$=\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{3}\sqrt{\mathrm{3}}=\mathrm{0} \\ $$$$\:\frac{{x}^{\mathrm{6}} +\mathrm{1}}{{x}^{\mathrm{3}}…
Question Number 195129 by a.lgnaoui last updated on 25/Jul/23 $$\mathrm{Calculer}\:\mathrm{la}\:\mathrm{valeur}\:\mathrm{de}\:\mathrm{la}\:\mathrm{serie}\:\mathrm{suivante}: \\ $$$$\boldsymbol{\mathrm{S}}=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{5}}{\mathrm{8}}+\frac{\mathrm{7}}{\mathrm{32}}+\frac{\mathrm{9}}{\mathrm{128}}+….. \\ $$ Answered by MM42 last updated on 25/Jul/23 $$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }{S}=\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{5}} }+\frac{\mathrm{7}}{\mathrm{2}^{\mathrm{7}}…
Question Number 195157 by Erico last updated on 25/Jul/23 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{{x}^{\mathrm{3}} }{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\:\frac{{x}}{{y}}\right)}+\frac{{y}^{\mathrm{3}} }{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}{arctan}\:\frac{{y}}{{x}}\right)}=\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$ Answered by Frix last updated…
Question Number 195124 by Shlock last updated on 25/Jul/23 Answered by witcher3 last updated on 25/Jul/23 $$\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\leqslant\sqrt{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{y}+\mathrm{1}\right)} \\ $$$$\Leftrightarrow\mathrm{x}+\mathrm{y}+\mathrm{2}\sqrt{\mathrm{xy}}\leqslant\mathrm{xy}+\mathrm{x}+\mathrm{y}+\mathrm{1}\Leftrightarrow\mathrm{xy}+\mathrm{1}\geqslant\mathrm{2}\sqrt{\mathrm{xy}},\mathrm{AM}−\mathrm{GM} \\ $$$$\Rightarrow\forall\left(\mathrm{x},\mathrm{y}\right)\in\mathbb{R}_{+} \sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}\leqslant\sqrt{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{y}+\mathrm{1}\right)} \\ $$$$\Rightarrow\forall\left(\mathrm{a},\mathrm{b}\right)\in\left[\mathrm{1},\infty\left[^{\mathrm{2}} \right.\right.…
Question Number 195126 by Erico last updated on 25/Jul/23 $$\mathrm{Soit}\:{f}_{{n}} \left({x}\right)=\mathrm{2}^{{n}+\mathrm{1}} \left[\frac{\frac{\mathrm{1}}{\mathrm{2}^{{n}} }{cotan}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)−{cotanx}}{{sin}\left(\frac{{x}}{\mathrm{2}^{{n}} }\right)}\right] \\ $$$${Calculer}\:\underset{{x}\rightarrow\mathrm{0}} {{lim}f}_{{n}} \left({x}\right)\:{et}\:\underset{{n}\rightarrow+\infty} {{lim}}\:\frac{{f}_{{n}} \left({x}\right)}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{2}} } \\ $$ Answered…
Question Number 195121 by mathlove last updated on 25/Jul/23 $${x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}^{\mathrm{8}} +\mathrm{2}{x}^{\mathrm{7}} −\mathrm{47}{x}=? \\ $$ Answered by qaz last updated on 25/Jul/23 $${x}^{\mathrm{8}}…
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Question Number 195154 by cortano12 last updated on 25/Jul/23 $$\:\:\:\:\underset{{x}\rightarrow\mathrm{2}\pi} {\mathrm{lim}}\:\left(\frac{\mathrm{tan}\:\left(\pi\:\mathrm{cos}\:{x}\right)}{{x}^{\mathrm{2}} \left({x}−\mathrm{5}\pi\right)+\mathrm{4}\pi^{\mathrm{2}} \left(\mathrm{2}{x}−\pi\right)}\right)=? \\ $$$$ \\ $$ Answered by dimentri last updated on 25/Jul/23 $$\:\:\:\underbrace{\Subset}…
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