Question Number 191934 by Rupesh123 last updated on 04/May/23 Answered by mehdee42 last updated on 04/May/23 $${C}_{\:\mathrm{6}} ^{\:\mathrm{3}} \:×\mathrm{3}!×\mathrm{5}!=\mathrm{14400} \\ $$ Commented by mehdee42 last…
Question Number 191930 by a.lgnaoui last updated on 05/May/23 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{inside}\:\mathrm{Area}\:\mathrm{of} \\ $$$$\left(\mathrm{ABCDEF}\right)? \\ $$$$\mathrm{Such}\:\mathrm{that}:\:\measuredangle\mathrm{AOB}=\mathrm{120}\:\:\:\measuredangle\mathrm{ANB}=\mathrm{60};°\mathrm{R}=\mathrm{ON} \\ $$$$\:\left(\mathrm{OA}=\mathrm{OB}=\mathrm{32cm}\right)\:\mathrm{ArcAE}=\mathrm{ArcBF}\left(\mathrm{r}=\mathrm{12cm}\right) \\ $$$$\mathrm{BASE}\:\mathrm{is}\:\mathrm{circulare} \\ $$$$\left({Aider}\:{le}\:{tailleur}\:{a}\:{savoir}\:{la}\:{surface}\right. \\ $$$${du}\:{tissu}\:{necessaire}\:{pour}\:{couvrir}\: \\ $$$$\left.\:{l}\:'\mathrm{e}{space}\:{indique}\:{dans}\:{la}\:{figure}?\right) \\…
Question Number 60856 by Tony Lin last updated on 26/May/19 $${if}\:\:\mathrm{0}<{x}<\mathrm{1},\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\frac{\left.{x}^{{x}^{{x}^{.^{.^{.^{{x}} } } } } } \right\}{n}}{\left(\left({x}^{{x}} \right)^{{x}} \right)^{\left.{x}…\right\}{n}} }=? \\ $$$$\left(?\:{can}\:{be}\:{expressed}\:{by}\:{x}\right) \\ $$…
Question Number 126391 by mathocean1 last updated on 20/Dec/20 $$ \\ $$$${show}\:{that}\: \\ $$$${m}^{\mathrm{2}} +{m}\:{and}\:\:\mathrm{2}{m}+\mathrm{1}\:{are}\:{prime}\:{betwen} \\ $$$${them}. \\ $$ Answered by akornes last updated on…
Question Number 60854 by Askash last updated on 26/May/19 $$\left({x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)/\left({x}−\mathrm{1}\right) \\ $$$$ \\ $$$$ \\ $$ Commented by Askash last updated on 26/May/19…
Question Number 191927 by a.lgnaoui last updated on 04/May/23 $$\boldsymbol{\mathrm{W}}\mathrm{hat}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{laterale}\:\mathrm{Shaded} \\ $$$$\:\mathrm{Area}\:\:\left(\boldsymbol{\mathrm{couronne}}\:\boldsymbol{\mathrm{circulaire}}\:\:\boldsymbol{\mathrm{coloree}}?\right) \\ $$$$\:\:\boldsymbol{\mathrm{R}}=\mathrm{10}\boldsymbol{\mathrm{cm}}\:\:\:\mathrm{20}\boldsymbol{\mathrm{cm}}\leqslant\:\boldsymbol{\mathrm{h}}\leqslant\mathrm{24}\boldsymbol{\mathrm{cm}}\:\:\measuredangle\mathrm{COD}=\mathrm{45}° \\ $$$$\:\: \\ $$ Commented by a.lgnaoui last updated on 04/May/23…
Question Number 60853 by Tony Lin last updated on 26/May/19 $$\sqrt{\mathrm{5}−\mathrm{12}{i}}+\sqrt{\mathrm{5}+\mathrm{12}{i}}=? \\ $$ Answered by tanmay last updated on 26/May/19 $$\sqrt{\mathrm{9}−\mathrm{4}−\mathrm{12}{i}}\:+\sqrt{\mathrm{9}−\mathrm{4}+\mathrm{12}{i}}\: \\ $$$$\sqrt{\left(\mathrm{3}−\mathrm{2}{i}\right)^{\mathrm{2}} }\:+\sqrt{\left(\mathrm{3}+\mathrm{2}{i}\right)^{\mathrm{2}} }\:…
Question Number 191926 by Rupesh123 last updated on 04/May/23 Answered by AST last updated on 04/May/23 $$\left[{x}\right]+\left\{{x}\right\}={x} \\ $$$$\left[{x}\right].\left[{y}\right]=\left[{x}\right]+\left[{y}\right]+\left\{{x}\right\}+\left\{{y}\right\}<\left[{x}\right]+\left[{y}\right]+\mathrm{2} \\ $$$$\Rightarrow{Case}\:{I}:\left[{x}\right].\left[{y}\right]=\left[{x}\right]+\left[{y}\right]\:{or}\:{II}:\left[{x}\right].\left[{y}\right]=\left[{x}\right]+\left[{y}\right]+\mathrm{1} \\ $$$${I}\Rightarrow\left[{x}\right]\left(\left[{y}\right]−\mathrm{1}\right)=\left[{y}\right]\Rightarrow\left[{x}\right]=\frac{\left[{y}\right]}{\left[{y}\right]−\mathrm{1}}=\mathrm{1}+\frac{\mathrm{1}}{\left[{y}\right]−\mathrm{1}} \\ $$$$\Rightarrow\left(\left[{y}\right]−\mathrm{1}\right)\mid\mathrm{1}\Rightarrow\left[{y}\right]−\mathrm{1}=\mathrm{1}\:{or}\:\left[{y}\right]−\mathrm{1}=−\mathrm{1}…
Question Number 126389 by shaker last updated on 20/Dec/20 Answered by liberty last updated on 20/Dec/20 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\:\left[\frac{\mathrm{3cos}\:\mathrm{3}{x}}{\mathrm{sin}\:\mathrm{3}{x}}\right]}{\:\left[\frac{\mathrm{2cos}\:\mathrm{2}{x}}{\mathrm{sin}\:\mathrm{2}{x}}\right]}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{3cos}\:\mathrm{3}{x}}{\mathrm{sin}\:\mathrm{3}{x}}×\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2cos}\:\mathrm{2}{x}} \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}}×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{sin}\:\mathrm{3}{x}}\:=\:\mathrm{1} \\ $$ Answered…
Question Number 191921 by Rupesh123 last updated on 03/May/23 Answered by AST last updated on 03/May/23 $$\int\left(\mathrm{3}{x}+\mathrm{5}\right){dx}=\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{5}{x}+{c} \\ $$$$\int_{−\mathrm{4}} ^{\mathrm{2}} \left(\mathrm{3}{x}+\mathrm{5}\right){dx}=\left(\mathrm{16}+{c}\right)−\left(\mathrm{4}+{c}\right)=\mathrm{12} \\ $$$$\Rightarrow\left(\mathrm{4}+{log}_{\mathrm{3}} {x}\right)\left({log}_{\mathrm{3}}…