Question Number 193520 by Tawa11 last updated on 15/Jun/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 193513 by Mastermind last updated on 15/Jun/23 $$\mathrm{Evaluate}\:\mathrm{I}=\underset{\mathrm{s}} {\int}\int\mathrm{x}^{\mathrm{3}} \mathrm{dydz}\:+\:\mathrm{x}^{\mathrm{2}} \mathrm{ydzdx}. \\ $$$$\mathrm{where}\:\mathrm{S}\:\mathrm{is}\:\mathrm{the}\:\mathrm{closed}\:\mathrm{surface}\:\mathrm{consis}− \\ $$$$\mathrm{ting}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cylinder}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} ,\:\mathrm{0}\leqslant\mathrm{z}\leqslant\mathrm{b} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{cylinder}\:\:\mathrm{disks}\:\mathrm{z}=\mathrm{0}\:\mathrm{and}\:\mathrm{z}=\mathrm{b}, \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}}…
Question Number 193512 by SaRahAli last updated on 15/Jun/23 Answered by Subhi last updated on 15/Jun/23 $$\int\frac{\mathrm{1}}{\frac{\mathrm{1}}{{cos}\left({x}\right)}.\frac{{sin}\left({x}\right)}{{cos}\left({x}\right)}}\:\Rrightarrow\:\int\frac{{cos}^{\mathrm{2}} \left({x}\right)}{{sin}\left({x}\right)}.{dx} \\ $$$$\int\frac{\mathrm{1}−{sin}^{\mathrm{2}} \left({x}\right)}{{sin}\left({x}\right)}.{dx}\:\:\looparrowright\:\int{csc}\left({x}\right)\:+\:\int−{sin}\left({x}\right) \\ $$$$\int{csc}\left({x}\right)\:+{cos}\left({x}\right) \\ $$$$\int{csc}\left({x}\right).\frac{{csc}\left({x}\right)+{cot}\left({x}\right)}{{csc}\left({x}\right)+{cot}\left({x}\right)}\:+\:{cos}\left({x}\right)…
Question Number 193511 by Lekhraj last updated on 15/Jun/23 Answered by a.lgnaoui last updated on 17/Jun/23 $$\measuredangle\mathrm{OAE}=\alpha\:\:\:\mathrm{OA}=\mathrm{OB}=\mathrm{OF}=\mathrm{R} \\ $$$$\Rightarrow\measuredangle\mathrm{OAF}=\measuredangle\mathrm{OFA}=\alpha \\ $$$$\Rightarrow\mathrm{cos}\:\alpha=\frac{\mathrm{AF}}{\mathrm{2R}}=\frac{\mathrm{11}}{\mathrm{2R}}\:\Rightarrow\:\:\:\:\:\boldsymbol{\mathrm{R}}=\frac{\mathrm{11}}{\mathrm{2cos}\:\boldsymbol{\alpha}}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Surface}\:\mathrm{triangle}\:\left(\mathrm{AOE}\right)\:\mathrm{est}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{S}}=\frac{\boldsymbol{\mathrm{OA}}×\boldsymbol{\mathrm{AE}}\mathrm{sin}\:\boldsymbol{\alpha}}{\mathrm{2}}=\frac{\mathrm{11}×\mathrm{4sin}\:\boldsymbol{\alpha}}{\mathrm{cos}\:\boldsymbol{\alpha}}…
Question Number 193504 by aba last updated on 15/Jun/23 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}x}^{\mathrm{x}^{\mathrm{x}} } =? \\ $$ Commented by Frix last updated on 15/Jun/23 $$\mathrm{0} \\ $$…
Question Number 193502 by horsebrand11 last updated on 15/Jun/23 $$\:\cancel{\underline{{o}}}\mathrm{olve}\: \\ $$$$\:\:\mathrm{cos}\:\mathrm{2x}\:.\mathrm{tan}\:\left(\frac{\mathrm{7}\pi}{\mathrm{19}}\right)=\mathrm{tan}\:\left(\frac{\mathrm{17}\pi}{\mathrm{23}}\right)+\mathrm{tan}\:\left(\frac{\mathrm{6}\pi}{\mathrm{23}}\right)+\mathrm{tan}\:\left(\frac{\mathrm{12}\pi}{\mathrm{19}}\right) \\ $$ Answered by cortano12 last updated on 15/Jun/23 $$\:\:\mathrm{If}\:\alpha+\beta=\pi\:\mathrm{then}\:\begin{cases}{\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta=\mathrm{0}}\\{\mathrm{tan}\:\alpha\:\mathrm{cot}\:\beta=−\mathrm{1}}\end{cases} \\ $$$$\:\mathrm{so}\:\mathrm{cos}\:\:\mathrm{2x}\:\mathrm{tan}\:\left(\frac{\mathrm{7}\pi}{\mathrm{19}}\right)=\mathrm{0}+\mathrm{tan}\:\left(\frac{\mathrm{12}\pi}{\mathrm{19}}\right) \\…
Question Number 193492 by Socracious last updated on 15/Jun/23 Answered by a.lgnaoui last updated on 15/Jun/23 $$\mathrm{DE}\mid\mid\mathrm{BC}\:\: \\ $$$$\mathrm{tan}\:\mathrm{30}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}=\frac{\boldsymbol{\mathrm{HB}}}{\boldsymbol{\mathrm{HA}}}=\frac{\boldsymbol{\mathrm{b}}}{\mathrm{2}\boldsymbol{\mathrm{x}}}\Rightarrow\:\:\boldsymbol{\mathrm{x}}=\frac{\boldsymbol{\mathrm{b}}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{S}}=\boldsymbol{\mathrm{x}}.\frac{\boldsymbol{\mathrm{b}}}{\mathrm{2}}=\frac{\boldsymbol{\mathrm{b}}^{\mathrm{2}} }{\mathrm{4}}\sqrt{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{Area}}=\frac{\mathrm{b}^{\mathrm{2}} \sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$…
Question Number 193494 by aba last updated on 15/Jun/23 $$\boldsymbol{\mathrm{Let}}\:\boldsymbol{\mathrm{n}}\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{fixed}}\:\boldsymbol{\mathrm{positive}}\:\boldsymbol{\mathrm{integer}}\:\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}} \\ $$$$\mathrm{sin}\left(\frac{\pi}{\mathrm{2n}}\right)+\mathrm{cos}\left(\frac{\boldsymbol{\pi}}{\mathrm{2n}}\right)=\frac{\sqrt{\mathrm{n}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{Then}}\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{n}} \\ $$ Commented by maths_plus last updated on 15/Jun/23 $$\mathrm{cool} \\…
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