Menu Close

Author: Tinku Tara

Given-that-are-the-three-roots-of-the-equation-x-59-3-2x-64-3-3x-123-3-find-the-value-of-

Question Number 137652 by ZiYangLee last updated on 05/Apr/21 $$\mathrm{Given}\:\mathrm{that}\:\alpha,\beta,\gamma\:\mathrm{are}\:\mathrm{the}\:\mathrm{three}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$$\mathrm{the}\:\mathrm{equation}\:\left({x}−\mathrm{59}\right)^{\mathrm{3}} +\left(\mathrm{2}{x}−\mathrm{64}\right)^{\mathrm{3}} =\left(\mathrm{3}{x}−\mathrm{123}\right)^{\mathrm{3}} , \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\alpha+\beta+\gamma. \\ $$ Answered by mitica last updated on…

Question-72112

Question Number 72112 by A8;15: last updated on 24/Oct/19 Answered by mind is power last updated on 24/Oct/19 $$\mathrm{x}=\frac{\mathrm{1}}{\mathrm{t}}\Rightarrow\mathrm{dx}=\frac{−\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} } \\ $$$$\frac{\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{899999999}} }}{\frac{\left(\mathrm{1}+\mathrm{t}^{\mathrm{9}} \right)^{\mathrm{1000000001}} }{\mathrm{t}^{\mathrm{900}\:\mathrm{000009}}…

Question-137650

Question Number 137650 by bemath last updated on 05/Apr/21 Answered by bemath last updated on 05/Apr/21 $${By}\:{Langrange}\:{multiplier} \\ $$$${f}\left({x},{y},\lambda\right)=\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\lambda\left(\mathrm{6}{x}^{\mathrm{2}} +\mathrm{2}{xy}+\mathrm{6}{y}^{\mathrm{2}} −\mathrm{9}\right) \\ $$$$\frac{\partial{f}}{\partial{x}}\:=\:\mathrm{2}{x}+\lambda\left(\mathrm{12}{x}+\mathrm{2}{y}\right)=\mathrm{0}…

Question-72111

Question Number 72111 by A8;15: last updated on 24/Oct/19 Commented by mathmax by abdo last updated on 24/Oct/19 $${let}\:{I}=\int_{\mathrm{0}} ^{\frac{{e}}{\pi}} \:\frac{{arctan}\left(\frac{\pi{x}}{{e}}\right)}{\pi{x}\:+{e}}{dx}\:{changement}\:\frac{\pi{x}}{{e}}\:={t}\:{give} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{arctant}}{\pi\left(\frac{{et}}{\pi}\right)+{e}}\frac{{e}}{\pi}{dt}\:=\frac{{e}}{\pi}\int_{\mathrm{0}}…

Question-72108

Question Number 72108 by ajfour last updated on 24/Oct/19 Answered by Tanmay chaudhury last updated on 24/Oct/19 $${Torque}\:{due}\:{to}\:{mg}\:\:\:\Gamma=\left(\frac{{l}}{\mathrm{2}}−{a}\right){mgsin}\left(\frac{\pi}{\mathrm{2}}+\alpha\right) \\ $$$$\left(\frac{{l}}{\mathrm{2}}−{a}\right){mgcos}\alpha={I}.\beta \\ $$$${calculation}\:{of}\:{I}=\frac{{ml}^{\mathrm{2}} }{\mathrm{12}}+{m}\left(\frac{{l}}{\mathrm{2}}−{a}\right)^{\mathrm{2}} \\ $$$$\theta=\mathrm{0}.{t}+\frac{\mathrm{1}}{\mathrm{2}}\beta{t}^{\mathrm{2}}…