Question Number 197950 by York12 last updated on 05/Oct/23 $${Let}\:{x},{y},{z}>\mathrm{0}\:,\:{x}+{y}+{z}=\mathrm{3}\:{Prove}\:{That}\:: \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}}}+\frac{\mathrm{1}}{\:\sqrt{{z}^{\mathrm{2}} +\mathrm{2}{z}}}+\sqrt{\mathrm{3}}\left(\frac{\mathrm{1}}{{y}+\mathrm{2}}−\frac{{y}}{\mathrm{9}}\right)+\frac{\sqrt[{\mathrm{3}}]{\sqrt{{x}}+\sqrt{{y}}+\sqrt{{z}}+\mathrm{24}}}{\:\sqrt{\mathrm{3}}}\geqslant\frac{\mathrm{17}}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 197951 by universe last updated on 05/Oct/23 $$\:\underset{{n}=\mathrm{1}\:} {\overset{\infty} {\sum}}\frac{{n}}{{n}^{\mathrm{4}} +{n}^{\mathrm{2}} +\mathrm{1}}\:−\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{n}^{\mathrm{2}} }{{n}^{\mathrm{8}} +{n}^{\mathrm{4}} +\mathrm{1}}\:=\:? \\ $$ Terms of Service Privacy…
Question Number 197944 by liuxinnan last updated on 05/Oct/23 $${tan}\frac{\pi}{\mathrm{12}}=\frac{{sin}\alpha−{sin}\frac{\pi}{\mathrm{12}}}{{cos}\alpha+{cos}\frac{\pi}{\mathrm{12}}} \\ $$$$\alpha=? \\ $$ Answered by MM42 last updated on 05/Oct/23 $${sin}\frac{\pi}{\mathrm{12}}×{cos}\alpha+{sin}\frac{\pi}{\mathrm{12}}×{cos}\frac{\pi}{\mathrm{12}}={sin}\alpha×{cos}\frac{\pi}{\mathrm{12}}−{sin}\frac{\pi}{\mathrm{12}}×{cos}\frac{\pi}{\mathrm{12}} \\ $$$$\Rightarrow{sin}\left(\alpha−\frac{\pi}{\mathrm{12}}\right)=\mathrm{2}{sin}\frac{\pi}{\mathrm{12}}×{cos}\frac{\pi}{\mathrm{12}}={sin}\frac{\pi}{\mathrm{6}} \\…
Question Number 197947 by mnjuly1970 last updated on 05/Oct/23 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{calculate}\ldots \\ $$$$\:\:\mathrm{L}\:=\:\mathrm{lim}\:_{\mathrm{n}\rightarrow\infty} \sqrt[{{n}}]{\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)\ldots\:\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)}\:=\:?\:\:\:\:\:\:\:\: \\ $$$$\:\: \\ $$ Answered by MM42 last updated on…
Question Number 197937 by mathlove last updated on 05/Oct/23 $$\sqrt{\mathrm{3}}\:{sin}^{\mathrm{2}} \theta\centerdot{tan}\beta+{cos}^{\mathrm{2}} \beta=? \\ $$ Answered by Frix last updated on 05/Oct/23 $$\mathrm{It}\:\mathrm{is}\:\mathrm{what}\:\mathrm{it}\:\mathrm{is},\:\mathrm{not}\:\mathrm{getting}\:\mathrm{better}\:\mathrm{any}\:\mathrm{way}. \\ $$$${p}=\mathrm{tan}\:\theta\:\wedge{q}=\mathrm{tan}\:\beta\:\rightarrow\:\frac{\sqrt{\mathrm{3}}{p}^{\mathrm{2}} {q}}{{p}^{\mathrm{2}}…
Question Number 197916 by sonukgindia last updated on 04/Oct/23 Answered by MM42 last updated on 04/Oct/23 Commented by MM42 last updated on 05/Oct/23 $${let}\:\:{OB}={r}\:\:\&\:{CD}={x} \\…
Question Number 197917 by sonukgindia last updated on 04/Oct/23 Commented by Frix last updated on 04/Oct/23 $$\mathrm{I}\:\mathrm{have}\:\mathrm{more}\:\mathrm{joy}\:\mathrm{with}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{one},\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{one} \\ $$$$\mathrm{is}\:\mathrm{boring}. \\ $$ Answered by…
Question Number 197935 by Frix last updated on 04/Oct/23 $$\mathrm{Show}\:\mathrm{that} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}\right)!}\:=\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{27}} \\ $$ Answered by witcher3 last updated on 05/Oct/23 $$\underset{\mathrm{n}\geqslant\mathrm{1}}…
Question Number 197919 by universe last updated on 04/Oct/23 $$\:\:\:\:\:\:\mathrm{I}_{{m}} \:\:\:\:\:=\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\lfloor\mathrm{2}^{{m}} {x}\rfloor}{\mathrm{3}^{{m}} }\:\underset{{n}={m}+\mathrm{1}} {\overset{\infty} {\sum}}\frac{\lfloor\mathrm{2}^{{n}} {x}\rfloor}{\mathrm{3}^{{n}} }\right){dx} \\ $$$$\:\:\:\:\:\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\:\:\:\:\:\:\mathrm{I}\:=\:\:\:\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{I}_{{m}}…
Question Number 197914 by a.lgnaoui last updated on 04/Oct/23 $$\boldsymbol{\mathrm{Determiner}}\:\boldsymbol{\mathrm{la}}\:\boldsymbol{\mathrm{surface}}\:\boldsymbol{\mathrm{hachuree}} \\ $$$$\left(\boldsymbol{\mathrm{voir}}\:\:\boldsymbol{\mathrm{figure}}\right) \\ $$$$\:\boldsymbol{\mathrm{BC}}=\mathrm{10}\boldsymbol{\mathrm{cm}}\:\:\:\:\:\measuredangle\boldsymbol{\mathrm{B}}=\mathrm{45}°\:\:\:\:\:\:\measuredangle\boldsymbol{\mathrm{C}}=\mathrm{30}° \\ $$ Commented by a.lgnaoui last updated on 04/Oct/23 Commented by…