Question Number 222662 by MrGaster last updated on 04/Jul/25 Commented by MrGaster last updated on 04/Jul/25 $$\exists\bigtriangleup{ABC},\angle{B}=\mathrm{90}°,{BA}={AC},{BD}=\mathrm{3}\wedge\angle{CED}=\mathrm{45}°,{CE}=\sqrt{\mathrm{2}}{AE},{CE}=? \\ $$ Commented by mr W last updated…
Question Number 222619 by ahmed2025 last updated on 01/Jul/25 Answered by wewji12 last updated on 02/Jul/25 $$\int\:\:\frac{\mathrm{2}}{\mathrm{2}−\mathrm{3}{x}}\:\:\mathrm{d}{x}=−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\left(\mathrm{2}−\mathrm{3}{x}\right)+{C} \\ $$$$\int_{−\mathrm{1}} ^{\:\:\mathrm{2}} \:\frac{\mathrm{2}}{\mathrm{2}−\mathrm{3}{x}}\:\mathrm{d}{x}=\int_{\:−\mathrm{1}} ^{\frac{\mathrm{2}}{\mathrm{3}}} \frac{\mathrm{2}}{\mathrm{2}−\mathrm{3}{x}}\:\mathrm{d}{x}\:+\int_{\frac{\mathrm{2}}{\mathrm{3}}} ^{\:\mathrm{2}} \:\frac{\mathrm{2}}{\mathrm{2}−\mathrm{3}{x}}\:\mathrm{d}{x}…
Question Number 222599 by MrGaster last updated on 01/Jul/25 $$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} }=\frac{\pi^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}^{\mathrm{2}{n}+\mathrm{2}} \left(\mathrm{2}{n}\right)!}\mid{E}_{\mathrm{2}{m}} \mid \\ $$ Commented by MathematicalUser2357 last updated on…
Question Number 222532 by MrGaster last updated on 29/Jun/25 $$\mathrm{Prove}: \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}{n}} }\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}^{\mathrm{2}} \frac{\mathscr{K}_{{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{192}}{\pi}\mathfrak{I}\mathrm{Li}_{\mathrm{4}} \left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)+\frac{\mathrm{16}}{\pi}\mathfrak{I}\mathrm{Li}_{\mathrm{3}} \left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)\mathrm{ln}\left(\mathrm{2}\right)+\frac{\mathrm{15}\pi^{\mathrm{2}} }{\mathrm{8}}\mathrm{ln}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)^{\mathrm{3}} −\frac{\mathrm{148}}{\pi}\beta\left(\mathrm{4}\right),\mathscr{K}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\ldots+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}} \\ $$…
Question Number 222487 by MathematicalUser2357 last updated on 28/Jun/25 $$'{Delete}\:{all}\:{lines}'\:{function}\:{deletes}\:{all}\:{lines}\:{without}\:{asking}\:{after}\:{deleting}\:{all}\:{lines}\:{for}\:{the}\:{first}\:{time}\:{in}\:{the}\:{equation}\:{editor} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 222462 by fantastic last updated on 27/Jun/25 $${i}^{{i}} =?? \\ $$ Answered by vnm last updated on 27/Jun/25 $$=\left({e}^{{i}\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}{n}\pi\right)} \right)^{{i}} ={e}^{\pi\left(−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}{n}\right)} ,\:{n}\in\mathbb{Z} \\…
Question Number 222441 by fantastic last updated on 27/Jun/25 $$\mathrm{0}^{{i}} \\ $$ Answered by wewji12 last updated on 27/Jun/25 $$\mathrm{0}\:\mathrm{because}\:{f}\left({z}\right)=\mathrm{0}^{{z}} \:\mathrm{for}\:\mathrm{all}\:{z}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$$${f}\left({z}\right)=\mathrm{0}\:\mathrm{const}. \\ $$$${f}\left({z}\right)=\begin{cases}{\:\mathrm{1}\:,\:{z}=\mathrm{0}}\\{\:\mathrm{0}\:,\:{z}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\}}\end{cases}…
Question Number 222466 by klipto last updated on 27/Jun/25 $$\mathrm{find}\:\mathrm{the}\:\mathrm{nth}\:\mathrm{term}.\: \\ $$ Commented by klipto last updated on 27/Jun/25 Commented by Frix last updated on…
Question Number 222424 by wewji12 last updated on 26/Jun/25 $$\int_{\mathrm{2}} ^{\:\infty} \:\:\:\:\frac{\mathrm{d}{z}}{\mathrm{ln}\left({z}\right)}−\underset{{l}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{ln}\left({l}\right)}=?? \\ $$ Answered by MrGaster last updated on 26/Jun/25 $${I}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\int_{{z}}…
Question Number 222425 by klipto last updated on 26/Jun/25 $$\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\infty} \left(\mathrm{4}\boldsymbol{\mathrm{x}}+\sqrt{\mathrm{16}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{\mathrm{x}}}\right) \\ $$$$\boldsymbol{\mathrm{ans}}:\frac{\mathrm{3}}{\mathrm{8}} \\ $$ Answered by Frix last updated on 26/Jun/25 $$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\left(\mathrm{4}{x}+\sqrt{\mathrm{16}{x}^{\mathrm{2}}…