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Question-226173

Question Number 226173 by sonukgindia last updated on 22/Nov/25 Answered by A5T last updated on 22/Nov/25 $$\mathrm{x}^{\mathrm{2013}} +\left(\mathrm{2014}−\mathrm{x}\right)^{\mathrm{2013}} \:\equiv\:\mathrm{x}^{\mathrm{2013}} +\left(−\mathrm{x}\right)^{\mathrm{2013}} \equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{2014}\right) \\ $$$$\mathrm{1}^{\mathrm{2013}} +\mathrm{2}^{\mathrm{2013}} +\mathrm{3}^{\mathrm{2013}}…

Question-226194

Question Number 226194 by fantastic2 last updated on 22/Nov/25 Answered by gregori last updated on 22/Nov/25 $$\:\:\measuredangle\: =\:\mathrm{60}°\: \\ $$$$\:\:\measuredangle\: =\:\mathrm{60}°\: \\ $$$$\:\:\measuredangle\: =\:\mathrm{360}°−\left(\mathrm{2}×\mathrm{120}°+\mathrm{60}°\right)\:=\:\mathrm{60}°\: \\…

Q226015-

Question Number 226134 by fantastic2 last updated on 20/Nov/25 $${Q}\mathrm{226015} \\ $$ Answered by fantastic2 last updated on 20/Nov/25 $${f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)={x}\:…{i} \\ $$$${put}\:{x}=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\…

S-u-v-r-2-v-sin-u-sin-2piv-rv-cos-u-r-2-v-sin-u-cos-2piv-r-2v-2-D-0-r-lt-pi-u-pi-0-v-pi-2-D-det-S-u-S-v-dV-

Question Number 226101 by Lara2440 last updated on 19/Nov/25 $$\overset{\rightarrow} {\boldsymbol{\mathcal{S}}}\left({u},{v}\right)=\begin{cases}{{r}\centerdot\left(\mathrm{2}+{v}\centerdot\mathrm{sin}\left({u}\right)\right)\mathrm{sin}\left(\mathrm{2}\pi{v}\right)}\\{{rv}\centerdot\mathrm{cos}\left({u}\right)}\\{{r}\centerdot\left(\mathrm{2}+{v}\centerdot\mathrm{sin}\left({u}\right)\right)\mathrm{cos}\left(\mathrm{2}\pi{v}\right)+{r}\centerdot\left(\mathrm{2}{v}−\mathrm{2}\right)}\end{cases} \\ $$$$\mathcal{D}=\left(\mathrm{0}\leq{r}<\infty\:,\:−\pi\leq{u}\leq\pi\:,\:\mathrm{0}\leq{v}\leq\frac{\pi}{\mathrm{2}}\:\right) \\ $$$$\int\int_{\:\mathcal{D}} \:\mathrm{det}\:\overset{\rightarrow} {\boldsymbol{\mathcal{S}}}_{{u}} ×\overset{\rightarrow} {\boldsymbol{\mathcal{S}}}_{{v}} \:\mathrm{d}{V}=?? \\ $$ Terms of Service…