Category: Algebra

Question Number 151300 by mathdanisur last updated on 19/Aug/21 $$\mathrm{5}\:\centerdot\:\mathrm{6},\mathrm{02}\centerdot\mathrm{10}^{\mathrm{23}} \:=\:?\:\left(\mathrm{solution}\right) \\ $$$$\left.\mathrm{a}\left.\right)\mathrm{3},\mathrm{01}\centerdot\mathrm{10}^{\mathrm{24}} \:\:\:\mathrm{b}\right)\mathrm{3},\mathrm{01}\centerdot\mathrm{10}^{\mathrm{22}} \\ $$ Answered by puissant last updated on 19/Aug/21 $$\mathrm{5}×\mathrm{6},\mathrm{02}.\mathrm{10}^{\mathrm{23}} =\mathrm{3},\mathrm{1}.\mathrm{10}^{\mathrm{24}}…

Question Number 151284 by mathdanisur last updated on 19/Aug/21 $$\mathrm{if}\:\:\frac{\mathrm{a}}{\mathrm{3}^{\boldsymbol{\mathrm{x}}-\mathrm{1}} }\:=\:\frac{\mathrm{b}}{\mathrm{3}^{\boldsymbol{\mathrm{y}}+\mathrm{2}} }\:=\:\frac{\mathrm{c}}{\mathrm{3}^{\boldsymbol{\mathrm{z}}-\mathrm{1}} }\:=\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\mathrm{find}\:\:\mathrm{abc}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 19/Aug/21 $$\mathrm{a}=\frac{\mathrm{3}^{\mathrm{x}−\mathrm{1}}…

Question Number 151276 by peter frank last updated on 19/Aug/21 $$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{cos}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{x}\right)\mathrm{dx} \\ $$ Commented by tabata last updated on 19/Aug/21 $$=\int_{\mathrm{0}} ^{\:\mathrm{2}\pi}…

Question Number 151278 by mathdanisur last updated on 19/Aug/21 $$\int\:\frac{\boldsymbol{\mathrm{e}}^{\sqrt{\boldsymbol{\mathrm{x}}\:-\:\mathrm{1}}} }{\:\sqrt{\boldsymbol{\mathrm{x}}\:-\:\mathrm{1}}}\:\mathrm{dx}\:=\:? \\ $$ Answered by Ar Brandon last updated on 19/Aug/21 $$\mathrm{I}=\int\frac{{e}^{\sqrt{{x}−\mathrm{1}}} }{\:\sqrt{{x}−\mathrm{1}}}{dx},\:{u}=\sqrt{{x}−\mathrm{1}}\Rightarrow{du}=\frac{{dx}}{\mathrm{2}\sqrt{{x}−\mathrm{1}}} \\ $$$$\:\:=\mathrm{2}\int{e}^{{u}}…

Question Number 151248 by mathdanisur last updated on 19/Aug/21 Answered by EDWIN88 last updated on 19/Aug/21 $$\mathrm{4sin}\:\left(\mathrm{4}{x}−\mathrm{60}°\right)\mathrm{sin}\:\left(\mathrm{6}{x}−\mathrm{60}°\right)\mathrm{sin}\left(\mathrm{480}°−\mathrm{10}{x}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{0} \\ $$$$\left\{\mathrm{2sin}\:\left(\mathrm{4}{x}−\mathrm{60}°\right)\mathrm{sin}\:\left(\mathrm{6}{x}−\mathrm{60}°\right)\right\}\mathrm{2sin}\:\left(\mathrm{120}°−\mathrm{10}{x}\right)+\mathrm{sin}\:\mathrm{60}°=\mathrm{0} \\ $$$$\left\{\mathrm{cos}\:\mathrm{2}{x}−\mathrm{cos}\:\left(\mathrm{10}{x}−\mathrm{120}°\right)\right\}\mathrm{2sin}\:\left(\mathrm{120}°−\mathrm{10}{x}\right)+\mathrm{sin}\:\mathrm{60}°=\mathrm{0} \\ $$$$\mathrm{2cos}\:\mathrm{2}{x}\:\mathrm{sin}\:\left(\mathrm{120}°−\mathrm{10}{x}\right)−\mathrm{sin}\:\left(\mathrm{240}°−\mathrm{20}{x}\right)+\mathrm{sin}\:\mathrm{60}°=\mathrm{0} \\ $$$$\mathrm{sin}\:\left(\mathrm{120}°−\mathrm{8}{x}\right)−\mathrm{sin}\:\left(\mathrm{12}{x}−\mathrm{120}°\right)+\mathrm{sin}\:\left(\mathrm{60}°−\mathrm{20}{x}\right)+\mathrm{sin}\:\mathrm{60}°=\mathrm{0}…