Question Number 196832 by Frix last updated on 01/Sep/23 $$\int{x}\mathrm{e}^{\frac{\mathrm{1}}{\mathrm{2}{x}}} {dx}=? \\ $$ Commented by mokys last updated on 02/Sep/23 $${u}\:=\:\frac{\mathrm{1}}{\mathrm{2}{x}}\:\rightarrow\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}{u}}\:\rightarrow\:{dx}\:=\:−\:\frac{{du}}{\mathrm{2}{u}^{\mathrm{2}} } \\ $$$$ \\…
Question Number 196828 by ERLY last updated on 01/Sep/23 Answered by Skabetix last updated on 01/Sep/23 $$\left.\mathrm{2}.{a}\right)\:{U}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{{U}_{{n}} } \\ $$$${Comme}\:{U}_{{n}} >\mathrm{0}\rightarrow{U}_{{n}} \geqslant\frac{\mathrm{1}}{{U}_{{n}} }\rightarrow{U}_{{n}+\mathrm{1}} \leqslant\frac{\mathrm{1}}{\mathrm{2}}{U}_{{n}}…
Question Number 196829 by cortano12 last updated on 01/Sep/23 Answered by universe last updated on 01/Sep/23 $${by}\:{apollonius}\:{theorem} \\ $$$${BC}^{\mathrm{2}} \:+\:{PC}^{\mathrm{2}\:} \:=\:\mathrm{2}\left({QC}^{\mathrm{2}} +{PQ}^{\mathrm{2}} \right)\:\:\:…..\left(\mathrm{1}\right) \\ $$$${AC}^{\mathrm{2}}…
Question Number 196816 by ERLY last updated on 01/Sep/23 Answered by MM42 last updated on 01/Sep/23 $$\left.{a}\right){S}=\frac{\mathrm{1}}{\mathrm{2}}+{cost}+{cos}\mathrm{2}{t}+…+{cosnt} \\ $$$$\mathrm{2}{sin}\frac{{t}}{\mathrm{2}}{S}={sin}\frac{{t}}{\mathrm{2}}+{sin}\frac{\mathrm{3}{t}}{\mathrm{2}}−{sin}\frac{{t}}{\mathrm{2}}+{sin}\frac{\mathrm{5}{t}}{\mathrm{2}}−{sin}\frac{\mathrm{3}{t}}{\mathrm{2}}+…+{sin}\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}{t}−{sin}\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}}{t} \\ $$$${S}=\frac{{sin}\left(\frac{\mathrm{2}{n}+\mathrm{1}}{\mathrm{2}}\right){t}}{\mathrm{2}{sin}\frac{{t}}{\mathrm{2}}}\:\:\checkmark \\ $$$$\left.{b}\right){S}={sint}+{sin}\mathrm{3}{t}+…+{sin}\left(\mathrm{2}{n}+\mathrm{1}\right){t} \\ $$$$\mathrm{2}{sintS}=\mathrm{1}−{cos}\mathrm{2}{t}+{cos}\mathrm{2}{t}−{cos}\mathrm{4}{t}+{cos}\mathrm{4}{t}−{cos}\mathrm{6}{t}+…+{cos}\mathrm{2}\left({n}+\mathrm{1}\right){t}−{cos}\mathrm{2}{nt}…
Question Number 196817 by universe last updated on 01/Sep/23 Answered by witcher3 last updated on 04/Sep/23 $$\mathrm{are}\:\mathrm{You}\:\mathrm{sur}\:\mathrm{of}\:\mathrm{the}\:\mathrm{expression}? \\ $$ Commented by universe last updated on…
Question Number 196815 by mathlove last updated on 01/Sep/23 $$!\mathrm{6}×\frac{\sqrt[{\mathrm{4}}]{!\mathrm{5}+\mathrm{9}!!!!!+\mathrm{7}!!!−\mathrm{16}}}{!\mathrm{10}}=? \\ $$ Answered by Tokugami last updated on 01/Sep/23 $$\mathrm{9}!!!!!=\mathrm{9}×\left(\mathrm{9}−\mathrm{5}\right)=\mathrm{36} \\ $$$$\mathrm{7}!!!=\mathrm{7}×\left(\mathrm{7}−\mathrm{3}\right)×\left(\mathrm{7}−\mathrm{6}\right)=\mathrm{28} \\ $$$$!{n}=\left[\frac{{n}!}{{e}}\right] \\…
Question Number 196806 by cortano12 last updated on 01/Sep/23 $$\:\:\:\:\:\cancel{\underline{\underbrace{ }}} \\ $$ Answered by dimentri last updated on 01/Sep/23 $$\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{1}/\mathrm{6}} \left(\mathrm{2}−\mathrm{2}{x}\right)^{\mathrm{1}/\mathrm{6}} −{x}}{\mathrm{2}{x}\:−{x}\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{1}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{1}}+\:\sqrt[{\mathrm{3}}]{\left({x}^{\mathrm{3}}…
Question Number 196766 by mr W last updated on 31/Aug/23 Commented by mr W last updated on 31/Aug/23 $${find}\:{the}\:{minimum}\:{and}\:{maximum} \\ $$$${of}\:{m}+{n}\:{with}\:{P}\:{on}\:{the}\:{circle}. \\ $$ Commented by…
Question Number 196760 by tri26112004 last updated on 31/Aug/23 $${f}:\:{R}\rightarrow{R} \\ $$$${f}\left({f}\left({x}+{y}\right)\right)={f}\left({x}\right)+{f}\left({y}\right) \\ $$$${Find}\:{f}\left({x}\right)=¿ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 196793 by MrGHK last updated on 31/Aug/23 Answered by Mathspace last updated on 02/Sep/23 $${B}\left({x},{y}\right)=\frac{\Gamma\left({x}\right)\Gamma\left({y}\right)}{\Gamma\left({x}+{y}\right)} \\ $$$$\Rightarrow\frac{\partial{B}}{\partial{x}}\left({x},{y}\right)=\Gamma\left({y}\right).\frac{\Gamma^{'} \left({x}\right)\Gamma\left({x}+{y}\right)−\Gamma\left({x}\right)\Gamma^{'} \left({x}+{y}\right)}{\Gamma^{\mathrm{2}} \left({x}+{y}\right)} \\ $$$$=\Gamma\left({y}\right)\frac{\Psi\left({x}\right)\Gamma\left({x}\right)\Gamma\left({x}+{y}\right)−\Gamma\left({x}\right)\Psi\left({x}+{y}\right)\Gamma\left({x}+{y}\right)}{\Gamma^{\mathrm{2}} \left({x}+{y}\right)}…