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Let-the-region-bounded-by-the-curve-y-x-1-x-and-the-x-axis-be-R-The-line-y-mx-divides-R-into-two-parts-find-the-value-of-1-m-3-

Question Number 161461 by ZiYangLee last updated on 18/Dec/21 $$\mathrm{Let}\:\mathrm{the}\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{curve}\: \\ $$$${y}={x}\left(\mathrm{1}−{x}\right)\:\mathrm{and}\:\mathrm{the}\:{x}-\mathrm{axis}\:\mathrm{be}\:\mathrm{R}. \\ $$$$\mathrm{The}\:\mathrm{line}\:{y}={mx}\:\mathrm{divides}\:\mathrm{R}\:\mathrm{into}\:\mathrm{two}\:\mathrm{parts}, \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\left(\mathrm{1}−{m}\right)^{\mathrm{3}} . \\ $$ Commented by cortano last updated on…

Let-f-x-x-x-1-Find-lim-x-0-f-x-f-0-x-0-and-lim-x-0-f-x-f-0-x-0-Hence-determine-whether-f-x-is-differentiable-at-x-0-

Question Number 161463 by ZiYangLee last updated on 18/Dec/21 $$\mathrm{Let}\:{f}\left({x}\right)=\:{x}+\mid{x}\mid−\mathrm{1}.\: \\ $$$$\mathrm{Find}\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{{f}\left({x}\right)−{f}\left(\mathrm{0}\right)}{{x}−\mathrm{0}}\:\mathrm{and}\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\frac{{f}\left({x}\right)−{f}\left(\mathrm{0}\right)}{{x}−\mathrm{0}}. \\ $$$$\mathrm{Hence},\:\mathrm{determine}\:\mathrm{whether}\:{f}\left({x}\right)\:\mathrm{is}\: \\ $$$$\mathrm{differentiable}\:\mathrm{at}\:{x}=\mathrm{0}. \\ $$ Answered by TheSupreme…

Find-the-particular-solution-to-the-differential-equation-2y-5y-2y-0-subject-to-the-initial-conditions-y-0-2y-y-0-1-

Question Number 161462 by ZiYangLee last updated on 18/Dec/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{particular}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{the}\:\mathrm{differential} \\ $$$$\mathrm{equation}\:\mathrm{2}{y}''+\mathrm{5}{y}'+\mathrm{2}{y}=\mathrm{0}\:\mathrm{subject}\:\mathrm{to}\:\mathrm{the}\:\mathrm{initial} \\ $$$$\mathrm{conditions}\:\:{y}\left(\mathrm{0}\right)=\mathrm{2}{y}\:,\:\:{y}'\left(\mathrm{0}\right)=\mathrm{1}\:. \\ $$ Answered by TheSupreme last updated on 18/Dec/21 $$\mathrm{2}\lambda^{\mathrm{2}} +\mathrm{5}\lambda+\mathrm{2}=\mathrm{0}…

Question-30350

Question Number 30350 by mondodotto@gmail.com last updated on 21/Feb/18 Answered by Rasheed.Sindhi last updated on 21/Feb/18 $${w}:\mathrm{width}\:,\:{l}:\mathrm{length} \\ $$$$\mathrm{2}{w}={l}+\mathrm{3} \\ $$$$\mathrm{2}{l}+\mathrm{2}{w}=\mathrm{36} \\ $$$$\mathrm{2}{l}+\left({l}+\mathrm{3}\right)=\mathrm{36} \\ $$$$\mathrm{3}{l}=\mathrm{36}−\mathrm{3}=\mathrm{33}…

Question-30340

Question Number 30340 by mondodotto@gmail.com last updated on 20/Feb/18 Answered by mrW2 last updated on 20/Feb/18 $$\mathrm{4}^{{x}} =\mathrm{16}{x} \\ $$$${e}^{{x}\mathrm{ln}\:\mathrm{4}} =\mathrm{16}{x} \\ $$$${xe}^{−{x}\mathrm{ln}\:\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{16}} \\…

A-B-B-C-C-D-D-

Question Number 95860 by omere hum aFreeN last updated on 28/May/20 $$\overset{−} {{A}}\centerdot\left({B}+\overset{−} {{B}}\right)\centerdot\left({C}+\overset{−} {{C}}\right)\centerdot\left({D}+\overset{−} {{D}}\right) \\ $$ Answered by omere hum aFreeN last updated…