Question Number 85756 by TawaTawa1 last updated on 24/Mar/20 Commented by TawaTawa1 last updated on 24/Mar/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{red}. \\ $$ Commented by Tony Lin last updated…
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 $$\mathrm{a}=\frac{\mathrm{3}^{\mathrm{x}−\mathrm{1}}…
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Question Number 151276 by peter frank last updated on 19/Aug/21 $$\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}…
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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 $$\mathrm{I}=\int\frac{{e}^{\sqrt{{x}−\mathrm{1}}} }{\:\sqrt{{x}−\mathrm{1}}}{dx},\:{u}=\sqrt{{x}−\mathrm{1}}\Rightarrow{du}=\frac{{dx}}{\mathrm{2}\sqrt{{x}−\mathrm{1}}} \\ $$$$\:\:=\mathrm{2}\int{e}^{{u}}…
Question Number 85729 by jagoll last updated on 24/Mar/20 $$\mathrm{if}\:\mathrm{march}\:\mathrm{24},\:\mathrm{2020}\:\mathrm{is}\:\mathrm{Tuesday}, \\ $$$$\mathrm{then}\:\mathrm{march}\:\mathrm{24},\:\mathrm{2032}\:\mathrm{is}\:\mathrm{the}\:\mathrm{day}\:? \\ $$ Commented by jagoll last updated on 24/Mar/20 $$\mathrm{what}\:\mathrm{the}\:\mathrm{simple}\:\mathrm{method} \\ $$$$\mathrm{for}\:\mathrm{calculate}\:\mathrm{it}? \\…
Question Number 151248 by mathdanisur last updated on 19/Aug/21 Answered by EDWIN88 last updated on 19/Aug/21 $$\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} \\ $$$$\left\{\mathrm{2sin}\:\left(\mathrm{4}{x}−\mathrm{60}°\right)\mathrm{sin}\:\left(\mathrm{6}{x}−\mathrm{60}°\right)\right\}\mathrm{2sin}\:\left(\mathrm{120}°−\mathrm{10}{x}\right)+\mathrm{sin}\:\mathrm{60}°=\mathrm{0} \\ $$$$\left\{\mathrm{cos}\:\mathrm{2}{x}−\mathrm{cos}\:\left(\mathrm{10}{x}−\mathrm{120}°\right)\right\}\mathrm{2sin}\:\left(\mathrm{120}°−\mathrm{10}{x}\right)+\mathrm{sin}\:\mathrm{60}°=\mathrm{0} \\ $$$$\mathrm{2cos}\:\mathrm{2}{x}\:\mathrm{sin}\:\left(\mathrm{120}°−\mathrm{10}{x}\right)−\mathrm{sin}\:\left(\mathrm{240}°−\mathrm{20}{x}\right)+\mathrm{sin}\:\mathrm{60}°=\mathrm{0} \\ $$$$\mathrm{sin}\:\left(\mathrm{120}°−\mathrm{8}{x}\right)−\mathrm{sin}\:\left(\mathrm{12}{x}−\mathrm{120}°\right)+\mathrm{sin}\:\left(\mathrm{60}°−\mathrm{20}{x}\right)+\mathrm{sin}\:\mathrm{60}°=\mathrm{0}…
Question Number 151247 by mathdanisur last updated on 19/Aug/21 $$\frac{\mathrm{1}}{\mathrm{5}}\:+\:\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} \centerdot\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{3}} \centerdot\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{4}} \centerdot\mathrm{4}}\:+\:…\:=\:? \\ $$ Answered by qaz last updated on 19/Aug/21 $$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n5}^{\mathrm{n}}…
Question Number 151241 by bagjagugum123 last updated on 19/Aug/21 Answered by hknkrc46 last updated on 19/Aug/21 $$\left(\mathrm{2}\right)\:\mid\mathrm{2}\boldsymbol{{x}}\:−\:\mathrm{3}\mid\:<\:\mid\boldsymbol{{x}}\:−\:\mathrm{1}\mid \\ $$$$\:\:\:\:\:\:\:\Rightarrow\:\left(\mathrm{2}\boldsymbol{{x}}\:−\:\mathrm{3}\right)^{\mathrm{2}} \:<\:\left(\boldsymbol{{x}}\:−\:\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\Rightarrow\:\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} \:−\:\mathrm{12}\boldsymbol{{x}}\:+\:\mathrm{9}\:<\:\boldsymbol{{x}}^{\mathrm{2}} \:−\:\mathrm{2}\boldsymbol{{x}}\:+\:\mathrm{1} \\…