Question Number 203749 by patrice last updated on 27/Jan/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 203748 by patrice last updated on 27/Jan/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 203750 by patrice last updated on 27/Jan/24 Answered by witcher3 last updated on 27/Jan/24 $$\frac{\mathrm{4}\left(\mathrm{k}+\mathrm{2}\right)−\mathrm{k}}{\mathrm{k}\left(\mathrm{k}+\mathrm{2}\right)\mathrm{2}^{\mathrm{k}} }=\frac{\mathrm{4}}{\mathrm{k}.\mathrm{2}^{\mathrm{k}} }−\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{2}\right)\mathrm{2}^{\mathrm{k}} }=\frac{\mathrm{1}}{\mathrm{k}.\mathrm{2}^{\mathrm{k}−\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{k}+\mathrm{2}\right)\mathrm{2}^{\mathrm{k}} }=\mathrm{V}_{\mathrm{k}} −\mathrm{V}_{\mathrm{k}+\mathrm{2}} \\ $$$$\mathrm{s}_{\mathrm{n}}…
Question Number 203736 by Tawa11 last updated on 26/Jan/24 A conductor of length 500cm carried a current of 0.95A when kept in a magnetic field…
Question Number 203701 by Noorzai last updated on 26/Jan/24 Answered by mr W last updated on 26/Jan/24 $$\mathrm{0}<\frac{\mathrm{2}^{{x}} }{{x}!}=\frac{\mathrm{2}×\mathrm{2}×\mathrm{2}×\mathrm{2}×….×\mathrm{2}}{\mathrm{1}×\mathrm{2}×\mathrm{3}×\mathrm{4}×…×{x}}<\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} \\ $$$$\mathrm{0}<\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}^{{x}} }{{x}!}<\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}} =\mathrm{0}…
Question Number 203693 by Numsey last updated on 26/Jan/24 Answered by MM42 last updated on 26/Jan/24 $${I}=\int\sqrt{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}{dx}=\mathrm{2}\sqrt{\mathrm{2}}{sin}\frac{{x}}{\mathrm{2}}\:+{c} \\ $$ Commented by Frix last updated on…
Question Number 203695 by Numsey last updated on 26/Jan/24 Commented by BaliramKumar last updated on 26/Jan/24 $$\mathrm{Step}\:\mathrm{III}\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}\left(\mathrm{27}\right)}\left(\mathrm{1}.\mathrm{5}\right) \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 203568 by Mastermind last updated on 22/Jan/24 $$\mathrm{Reduce}\:\mathrm{a}\:\mathrm{differential}\:\mathrm{equation} \\ $$$$\mathrm{x}^{'''} \:+\:\mathrm{ax}^{''} \:+\:\mathrm{bx}^{'} \:+\:\mathrm{cx}\:=\:\mathrm{0}\:\left(\mathrm{where}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{are}\right. \\ $$$$\left.\mathrm{constants}\right)\:\mathrm{to}\:\mathrm{an}\:\mathrm{equivalent}\:\mathrm{system}\:\mathrm{of}\:\mathrm{first} \\ $$$$\mathrm{order}\:\mathrm{equation}. \\ $$$$\mathrm{x}_{\mathrm{1}} =\:\mathrm{x},\:\mathrm{x}_{\mathrm{2}} \:=\:\mathrm{x}^{'} ,\:\mathrm{x}_{\mathrm{3}} \:=\:\mathrm{x}^{''}…
Question Number 203565 by Mastermind last updated on 22/Jan/24 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{equation}\:\mathrm{simultaneously} \\ $$$$\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{stationary}\:\mathrm{points}: \\ $$$$\mathrm{2xy}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} \:−\:\mathrm{4x}^{\mathrm{3}} \mathrm{y}^{\mathrm{2}} \:−\:\mathrm{2xy}^{\mathrm{4}} \:=\:\mathrm{0}\:—–\left(\mathrm{1}\right) \\ $$$$\mathrm{2x}^{\mathrm{2}} \mathrm{yc}^{\mathrm{2}} \:−\:\mathrm{2x}^{\mathrm{4}} \mathrm{y}\:−\:\mathrm{4x}^{\mathrm{2}} \mathrm{y}^{\mathrm{3}}…
Question Number 203567 by Mastermind last updated on 22/Jan/24 Terms of Service Privacy Policy Contact: info@tinkutara.com