Question Number 222280 by klipto last updated on 21/Jun/25 $$\boldsymbol{\mathrm{y}}=\mathrm{3}\boldsymbol{\mathrm{x}}^{\mathrm{2024}} −\mathrm{18}\boldsymbol{\mathrm{x}}^{\mathrm{2020}} +\mathrm{5}\boldsymbol{\mathrm{x}}^{\mathrm{47}} −\mathrm{8} \\ $$$$\boldsymbol{\mathrm{find}}\:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2025}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2025}} } \\ $$ Commented by mr W last updated…
Question Number 222193 by MrGaster last updated on 20/Jun/25 $$\mathrm{Prove}: \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)}=\mathrm{2}\left(\mathrm{cos}\left(\frac{\pi}{\mathrm{8}}\right)−\mathrm{sin}\left(\frac{\pi}{\mathrm{8}}\right)\right) \\ $$$$\frac{\mathrm{2}}{\mathrm{sin}\left(\frac{\mathrm{8}\pi}{\mathrm{9}}\right)}−\frac{\mathrm{1}}{\mathrm{sin}\left(\frac{\mathrm{5}\pi}{\mathrm{9}}\right)}=\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{4sin}\left(\frac{\pi}{\mathrm{9}}\right) \\ $$$$\frac{\mathrm{2}}{\mathrm{sin}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)}+\frac{\mathrm{1}}{\mathrm{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)}=\mathrm{4}\left(\mathrm{sin}\left(\frac{\pi}{\mathrm{7}}\right)+\mathrm{cos}\left(\frac{\pi}{\mathrm{14}}\right)\right) \\ $$ Answered by som(math1967) last updated on 20/Jun/25…
Question Number 222191 by wewji12 last updated on 20/Jun/25 $$\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\:\int_{\:{V}^{\:\mathrm{3}} } \rho_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)\mathrm{d}{V}=−\oint_{\:\partial{V}} \:\boldsymbol{\mathrm{J}}_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)\centerdot\mathrm{d}\boldsymbol{\mathrm{a}}+\int_{\:{V}^{\:\mathrm{3}} } \:{S}_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)\mathrm{d}{V} \\ $$$$\int_{\:{V}^{\:\mathrm{3}} } \:\frac{\partial\rho_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)}{\partial{t}}\:\mathrm{dV}=−\int_{{V}^{\:\mathrm{3}} } \overset{\rightarrow}…
Question Number 222142 by fantastic last updated on 19/Jun/25 $${a}=\mathrm{3}\sqrt{\mathrm{2}}\:,{b}=\frac{\mathrm{1}}{\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{6}}} \sqrt{\mathrm{6}}}\:{and}\:{x},{y}\epsilon\mathbb{R}\:{such}\:{that} \\ $$$$\mathrm{3}{x}\:+\mathrm{2}{y}=\mathrm{log}\:_{{a}} \left(\mathrm{18}\right)^{\frac{\mathrm{5}}{\mathrm{4}}} \\ $$$$\mathrm{2}{x}−{y}=\mathrm{log}\:_{{b}} \left(\sqrt{\mathrm{1080}}\right) \\ $$$${then}\:{find}\:{the}\:{value}\:{of}\:\: \\ $$$$\mathrm{4}{x}+\mathrm{5}{y} \\ $$ Answered by…
Question Number 222151 by Ismoiljon_008 last updated on 19/Jun/25 Commented by Ismoiljon_008 last updated on 19/Jun/25 $$\:\:\:{help},\:{please} \\ $$$$ \\ $$ Commented by Rasheed.Sindhi last…
Question Number 222121 by wewji12 last updated on 18/Jun/25 $$\int\:\mathrm{acos}\left(\frac{\mathrm{cos}\left(\varrho\right)}{\mathrm{1}+\mathrm{2cos}\left(\varrho\right)}\right)\:\mathrm{d}\varrho \\ $$ Answered by Nicholas666 last updated on 19/Jun/25 $$\:\:\:\:\int{cos}^{−\mathrm{1}} \:\left(\frac{{cos}\:{x}}{\mathrm{1}\:+\mathrm{2}\:{cos}\:{x}}\right)\:{dx}\:=\:\int\:\frac{{cos}\:{x}}{{cox}\left(\mathrm{1}+\mathrm{2}\:{cos}\:{x}\right)}\:{dx}\:=\int\:\frac{\mathrm{1}\:}{\mathrm{1}\:+\mathrm{2}\:{cos}\:{x}}\:{dx} \\ $$$$\:\:\:\mathrm{let};\:\:\:{t}\:=\:{tan}\:\left(\frac{{x}}{\mathrm{2}}\right)\:\Rightarrow\:{cos}\:{x}\:=\:\frac{\mathrm{1}\:−{t}^{\mathrm{2}} }{\mathrm{1}\:+\:{t}^{\mathrm{2}} }\:\mathrm{and}\:{dx}=\frac{\mathrm{2}}{\mathrm{1}\:+{t}^{\mathrm{2}}…
Question Number 222117 by klipto last updated on 18/Jun/25 $$\boldsymbol{\mathrm{problem}}\:\mathrm{1}.\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{positive}}\:\boldsymbol{\mathrm{integer}} \\ $$$$\boldsymbol{\mathrm{m}},\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{triples}}\left(\boldsymbol{\mathrm{n}},\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{positive}}\:\boldsymbol{\mathrm{integers}},\boldsymbol{\mathrm{with}} \\ $$$$\boldsymbol{\mathrm{n}}\:\boldsymbol{\mathrm{relatively}}\:\boldsymbol{\mathrm{prime}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{m}},\boldsymbol{\mathrm{which}}\:\boldsymbol{\mathrm{satisfy}} \\ $$$$\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} \right)^{\boldsymbol{\mathrm{m}}} =\left(\boldsymbol{\mathrm{xy}}\right)^{\boldsymbol{\mathrm{n}}} \\ $$$$\boldsymbol{\mathrm{hint}}:\boldsymbol{\mathrm{utilize}}\:\boldsymbol{\mathrm{AM\&GM}},\boldsymbol{\mathrm{diophantine}}\:\boldsymbol{\mathrm{eqn}} \\ $$$$\boldsymbol{\mathrm{KLIPTO}}−\boldsymbol{\mathrm{QUANTA}}−\boldsymbol{\mathrm{OOZY}} \\ $$…
Question Number 222095 by wewji12 last updated on 17/Jun/25 $$\mathrm{could}\:\:\mathrm{I}\:\mathrm{consider}\:\:{Y}_{\nu} \left({z}\right)=\mathrm{cot}\left(\nu\pi\right){J}_{\nu} \left({z}\right)−\mathrm{csc}\left(\nu\pi\right){J}_{−\nu} \left({z}\right) \\ $$$$\mathrm{as}\:\infty−\infty\:\mathrm{form}\:\mathrm{limit}\:\mathrm{when}\:\nu\in\mathbb{Z} \\ $$$$\mathrm{and}\:\mathrm{How}\:\mathrm{can}\:\mathrm{i}\:\mathrm{calculate} \\ $$$${Y}_{\nu} \left({z}\right)=\mathrm{cot}\left(\nu\pi\right){J}_{\nu} \left({z}\right)−\mathrm{csc}\left(\nu\pi\right){J}_{−\nu} \left({z}\right)…?? \\ $$$$\underset{\alpha\rightarrow\nu} {\mathrm{lim}}\:\frac{\mathrm{cot}^{\mathrm{2}}…
Question Number 222057 by wewji12 last updated on 16/Jun/25 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}^{\mathrm{1}} ×\mathrm{2}^{\mathrm{2}} ×\mathrm{3}^{\mathrm{3}} ……×{n}^{{n}} }{{n}^{\frac{\mathrm{1}}{\mathrm{2}}{n}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{n}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{12}}} ×{e}^{−\frac{\mathrm{1}}{\mathrm{4}}{n}^{\mathrm{2}} } }=??? \\ $$$$\mathrm{Help}…. \\ $$$$\mathrm{i}\:\mathrm{can}'\mathrm{t}\:\mathrm{Solve}\:\mathrm{that}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{a}_{{n}}…
Question Number 222044 by wewji12 last updated on 15/Jun/25 $$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{evaluate} \\ $$$$\int_{−\infty} ^{\:\:\infty} \:\:\frac{\mathrm{sin}\left({z}+\mathrm{1}\right)}{\left({z}+\mathrm{1}\right)\left({z}^{\mathrm{2}} +\mathrm{1}\right)}\:\mathrm{d}{z} \\ $$ Answered by vnm last updated on 15/Jun/25 $$…