Question Number 84231 by Rio Michael last updated on 10/Mar/20 $$\mathrm{A}\:\mathrm{compound}\:\mathrm{pendulum}\:\mathrm{oscillates}\:\mathrm{though}\:\mathrm{a} \\ $$$$\mathrm{small}\:\mathrm{angle}\:\theta\:\mathrm{about}\:\mathrm{its}\:\mathrm{equilibrium}\:\mathrm{position} \\ $$$$\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\:\mathrm{10}{a}\left(\frac{{d}\theta}{{dt}}\right)^{\mathrm{2}} \:=\:\mathrm{4}{g}\:\mathrm{cos}\:\theta\:,\:{a}\:>\mathrm{0}\:.\:\mathrm{its}\:\mathrm{period}\:\mathrm{is}\: \\ $$$$\left[\mathrm{A}\right]\:\mathrm{2}\pi\sqrt{\frac{\mathrm{5}{a}}{\mathrm{4}{g}}\:\:}\:\:\:\left[\mathrm{B}\right]\:\frac{\mathrm{2}\pi}{\mathrm{5}}\sqrt{\frac{{a}}{{g}}}\:\:\left[\mathrm{C}\right]\:\mathrm{2}\pi\sqrt{\frac{\mathrm{2}{g}}{\mathrm{5}{a}}\:}\:\:\left[\mathrm{D}\right]\:\mathrm{2}\pi\sqrt{\frac{\mathrm{5}{a}}{{g}}}\: \\ $$ Answered by TANMAY…
Question Number 18693 by Arnab Maiti last updated on 27/Jul/17 $$\mathrm{prove}\:\mathrm{that}\:\mathrm{2tan}^{−\mathrm{1}} \left[\mathrm{tan}\frac{\alpha}{\mathrm{2}}\mathrm{tan}\left(\frac{\pi}{\mathrm{4}}−\frac{\beta}{\mathrm{2}}\right)\right] \\ $$$$=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\alpha\:\mathrm{cos}\beta}{\mathrm{cos}\alpha\:+\mathrm{sin}\beta}\right) \\ $$ Answered by 951172235v last updated on 01/Feb/19 $$\mathrm{tan}\:\frac{\Theta}{\mathrm{2}\:}\:=\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{tan}\:\left(\frac{\overset{−}…
Question Number 149766 by mathdanisur last updated on 07/Aug/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{1}}\:=\:\frac{\mathrm{10}}{\mathrm{3}} \\ $$ Commented by amin96 last updated on 07/Aug/21 $${x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}=\mathrm{y}\:\:\:\:\Rightarrow\:\:{y}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{10}}{\mathrm{3}}…
Question Number 149757 by iloveisrael last updated on 07/Aug/21 Commented by amin96 last updated on 07/Aug/21 $${x}\frac{{dy}}{{dx}}+\mathrm{2}{y}=\frac{{sinx}}{{x}}\:\:\:\:\:\:\frac{{dy}}{{dx}}=\frac{\mathrm{sin}\:{x}−\mathrm{2}{xy}}{{x}^{\mathrm{2}} } \\ $$$$\left(\mathrm{sin}\:{x}−\mathrm{2}{xy}\right){dx}+\left(−{x}^{\mathrm{2}} \right){dy}=\mathrm{0} \\ $$$$\frac{\partial{M}}{\partial{y}}=−\mathrm{2}{x}\:\:\:\:\:\:\:\frac{\partial{N}}{\partial{x}}=−\mathrm{2}{x}\:\:\:\:\:\:{M}_{{y}} ={N}_{{x}} \:\:\:…
Question Number 84220 by Rio Michael last updated on 10/Mar/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{2}}{\mathrm{3cosh2}{x}\:+\mathrm{2}} \\ $$ Commented by mr W last updated on 10/Mar/20 $$\mathrm{cosh}\:\mathrm{2}{x}\:\geqslant\mathrm{1} \\ $$$$\mathrm{3}\:\mathrm{cosh}\:\mathrm{2}{x}+\mathrm{2}\:\geqslant\mathrm{3}×\mathrm{1}+\mathrm{2}=\mathrm{5} \\…
Question Number 84218 by Rio Michael last updated on 10/Mar/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{the}\:\mathrm{planes} \\ $$$$\:\mathrm{2}\boldsymbol{{x}}−\boldsymbol{{y}}−\boldsymbol{{z}}\:=\:\mathrm{24}\:\mathrm{and}\:\mathrm{2}\boldsymbol{{x}}−\boldsymbol{{y}}−\boldsymbol{{z}}\:=\:\mathrm{36} \\ $$ Commented by M±th+et£s last updated on 10/Mar/20 $$\mathrm{2}{x}−{y}−{z}=\mathrm{24}\Rightarrow\Rightarrow\mathrm{2}{x}−{y}−{z}−\mathrm{24}=\mathrm{0} \\ $$$$\mathrm{2}{x}−{y}−{z}=\mathrm{36}\Rightarrow\Rightarrow\mathrm{2}{x}−{y}−{z}−\mathrm{36}=\mathrm{0}…
Question Number 84219 by Rio Michael last updated on 10/Mar/20 $$\int_{−\mathrm{1}} ^{\mathrm{1}} {e}^{\mid{x}\mid} {dx}\:=? \\ $$ Answered by TANMAY PANACEA last updated on 10/Mar/20 $$\int_{−\mathrm{1}}…
Question Number 18681 by Joel577 last updated on 27/Jul/17 $${y}\:=\:\mid\mathrm{sin}\:{x}\mid\:+\:\mathrm{2} \\ $$$${y}\:=\:\mid{x}\mid\:+\:\mathrm{2}\:−\pi \\ $$$$−\pi\:\leqslant\:{x}\:\leqslant\:\pi \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{that}\:\mathrm{have}\:\mathrm{created} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{above} \\ $$ Answered by ajfour last updated…
Question Number 84215 by mathocean1 last updated on 10/Mar/20 $${hi} \\ $$$${show}\:{that}\:{the}\:{following}\:{sequence} \\ $$$${is}\:{limited}: \\ $$$${U}_{{n}} =\frac{\mathrm{3}{n}+\mathrm{2}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$${precise}\:{the}\:{upper}\:{and}\:{lower}. \\ $$ Answered by Rio Michael…
Question Number 18678 by Joel577 last updated on 27/Jul/17 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{4}^{{x}\:+\:\mathrm{1}} \:+\:\mathrm{2}^{{x}\:+\mathrm{1}} \:−\:\mathrm{3}^{{x}\:+\:\mathrm{1}} }{\mathrm{4}^{{x}\:−\:\mathrm{1}} \:+\:\mathrm{2}^{{x}\:−\:\mathrm{1}\:} +\:\mathrm{3}^{{x}\:+\:\mathrm{1}} \:} \\ $$ Answered by 433 last updated on…