Question Number 82937 by niroj last updated on 26/Feb/20 $$\:\: \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{between}\:\mathrm{the}\:\mathrm{curves} \\ $$$$\:\mathrm{y}=\:\mathrm{log}\:\mathrm{x}\:\:\mathrm{and}\:\mathrm{y}=\:\left(\mathrm{log}\:\mathrm{x}\right)^{\mathrm{2}} . \\ $$ Commented by jagoll last updated on 26/Feb/20 $$\mathrm{area}\:=\:\int\underset{\mathrm{1}}…
Question Number 17397 by tawa tawa last updated on 05/Jul/17 Answered by student last updated on 05/Jul/17 $$\boldsymbol{\mathrm{i}}.\:\mathrm{C}_{\mathrm{3}} \mathrm{H}_{\mathrm{8}} \:+\:\mathrm{5O}_{\mathrm{2}} \:\rightarrow\:\mathrm{3CO}_{\mathrm{2}} \:+\:\mathrm{4H}_{\mathrm{2}} \mathrm{O} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{answer}\:\mathrm{for}\:\boldsymbol{\mathrm{ii}}\:\mathrm{should}\:\mathrm{be}\:\mathrm{the}…
Question Number 148467 by bramlexs22 last updated on 28/Jul/21 Answered by puissant last updated on 28/Jul/21 $$\mathrm{x}=\mathrm{tan}\left(\mathrm{t}\right)\Rightarrow\mathrm{dx}=\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\mathrm{t}\right)\mathrm{dt} \\ $$$$\mathrm{0}\leqslant\mathrm{x}\leqslant\infty\:\Rightarrow\:\mathrm{0}\leqslant\mathrm{t}\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{arctan}\left(\mathrm{tan}\left(\mathrm{t}\right)\right)}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\mathrm{t}\right)}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}}…
Question Number 17393 by tawa tawa last updated on 05/Jul/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{modulus}\:\mathrm{of}\:\:\mathrm{z}\:=\:\mathrm{6}\:+\:\mathrm{8i} \\ $$ Answered by ajfour last updated on 05/Jul/17 $$\mid\mathrm{z}\mid=\sqrt{\left(\mathrm{6}\right)^{\mathrm{2}} +\left(\mathrm{8}\right)^{\mathrm{2}} }\:=\mathrm{10}\:. \\ $$…
Question Number 148466 by Ar Brandon last updated on 28/Jul/21 $$\mathrm{xdx}+\mathrm{ydy}=\mathrm{xdy}−\mathrm{ydx} \\ $$ Answered by bramlexs22 last updated on 28/Jul/21 $$\left({x}+{y}\right){dx}=\left({x}−{y}\right){dy} \\ $$$$\:\frac{{dy}}{{dx}}=\:\frac{{x}+{y}}{{x}−{y}} \\ $$$$\:{let}\:{y}={ux}\:\Rightarrow\frac{{dy}}{{dx}}\:=\:{u}\:+\:{x}\:\frac{{du}}{{dx}}…
Question Number 82929 by VBash last updated on 26/Feb/20 $${if}\:\mathrm{2}{B}+{A}=\mathrm{45}° \\ $$$${show}\:{that}; \\ $$$${tan}\:{B}=\:\frac{\mathrm{1}−\mathrm{2}{tanA}−{tan}^{\mathrm{2}} {A}}{\mathrm{1}+\mathrm{2}{tanA}−{tan}^{\mathrm{2}} {A}} \\ $$ Answered by jagoll last updated on 26/Feb/20…
Question Number 17392 by tawa tawa last updated on 05/Jul/17 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{cube}\:\mathrm{root}\:\mathrm{of}\:\:\mathrm{z}\:=\:−\:\mathrm{1} \\ $$ Answered by mrW1 last updated on 05/Jul/17 $$−\mathrm{1} \\ $$$$−\omega=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$−\omega^{\mathrm{2}}…
Question Number 17391 by tawa tawa last updated on 05/Jul/17 $$\mathrm{write}\:\:\:\mathrm{z}\:=\:\left(\mathrm{2}\:+\:\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{i}\right)^{\mathrm{3}} \:\:\mathrm{in}\:\mathrm{polar}\:\mathrm{form}. \\ $$ Answered by ajfour last updated on 05/Jul/17 $$\mathrm{z}=\mathrm{4}^{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{3}} =\mathrm{64}\left(\mathrm{e}^{\boldsymbol{\mathrm{i}}\pi/\mathrm{3}} \right)^{\mathrm{3}}…
Question Number 82924 by jagoll last updated on 26/Feb/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{tan}\:\mathrm{x}}−\sqrt{\mathrm{sin}\:\mathrm{x}}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{x}}} \\ $$ Commented by jagoll last updated on 26/Feb/20 $$\mathrm{the}\:\mathrm{ans}\::\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Commented…
Question Number 17386 by ajfour last updated on 05/Jul/17 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{log}\:_{\mathrm{2}} \mathrm{x}\:\mathrm{log}\:_{\mathrm{3}} \mathrm{x}\:\mathrm{log}\:_{\mathrm{5}} \mathrm{x}=\mathrm{log}\:_{\mathrm{2}} \mathrm{x}\:\mathrm{log}\:_{\mathrm{3}} \mathrm{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{log}\:_{\mathrm{3}} \mathrm{x}\:\mathrm{log}\:_{\mathrm{5}} \mathrm{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{log}\:_{\mathrm{5}} \mathrm{x}\:\mathrm{log}\:_{\mathrm{2}} \mathrm{x}\:.…