Question Number 161464 by ZiYangLee last updated on 18/Dec/21 $$\mathrm{Given}\:\mathrm{that}\:\mathrm{in}\:\Delta\mathrm{ABC}, \\ $$$$\left(\mathrm{sin}\:\mathrm{A}+\mathrm{sin}\:\mathrm{B}\right):\left(\mathrm{sin}\:\mathrm{B}+\mathrm{sin}\:\mathrm{C}\right):\left(\mathrm{sin}\:\mathrm{C}+\mathrm{sin}\:\mathrm{A}\right)=\:\mathrm{6}:\:\mathrm{4}:\:\mathrm{5} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{A}. \\ $$ Commented by cortano last updated on 18/Dec/21 $$\:\begin{cases}{\mathrm{sin}\:{A}+\mathrm{sin}\:{B}=\mathrm{6}{k}}\\{\mathrm{sin}\:{B}+\mathrm{sin}\:{C}=\mathrm{4}{k}}\\{\mathrm{sin}\:{C}+\mathrm{sin}\:{A}=\mathrm{5}{k}}\end{cases} \\…
Question Number 30390 by mondodotto@gmail.com last updated on 21/Feb/18 Commented by abdo imad last updated on 21/Feb/18 $${we}\:{have}\:{x}^{\mathrm{2}} \:+\mathrm{4}{x}\:+\mathrm{3}=\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1}\:\:{tbe}\:{ch}\:.{x}+\mathrm{2}={cht}\:{give} \\ $$$${I}=\:\int\:\sqrt{{ch}^{\mathrm{2}} {t}\:−\mathrm{1}}\:{shtdt}\:=\:\int\:{sh}^{\mathrm{2}} {t}\:{dt}\:=\:\int\frac{\mathrm{1}−{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt} \\…
Question Number 161461 by ZiYangLee last updated on 18/Dec/21 $$\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…
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…
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 $$\mathrm{2}\lambda^{\mathrm{2}} +\mathrm{5}\lambda+\mathrm{2}=\mathrm{0}…
Question Number 30373 by Tinkutara last updated on 21/Feb/18 Commented by Tinkutara last updated on 21/Feb/18 What is the output of this code segment? Commented by mrW2 last updated on 22/Feb/18 $$\mathrm{1}\:\:\mathrm{1}…
Question Number 30355 by mondodotto@gmail.com last updated on 21/Feb/18 Answered by $@ty@m last updated on 21/Feb/18 $$\mathrm{120} \\ $$ Commented by mondodotto@gmail.com last updated on…
Question Number 30350 by mondodotto@gmail.com last updated on 21/Feb/18 Answered by Rasheed.Sindhi last updated on 21/Feb/18 $${w}:\mathrm{width}\:,\:{l}:\mathrm{length} \\ $$$$\mathrm{2}{w}={l}+\mathrm{3} \\ $$$$\mathrm{2}{l}+\mathrm{2}{w}=\mathrm{36} \\ $$$$\mathrm{2}{l}+\left({l}+\mathrm{3}\right)=\mathrm{36} \\ $$$$\mathrm{3}{l}=\mathrm{36}−\mathrm{3}=\mathrm{33}…
Question Number 30340 by mondodotto@gmail.com last updated on 20/Feb/18 Answered by mrW2 last updated on 20/Feb/18 $$\mathrm{4}^{{x}} =\mathrm{16}{x} \\ $$$${e}^{{x}\mathrm{ln}\:\mathrm{4}} =\mathrm{16}{x} \\ $$$${xe}^{−{x}\mathrm{ln}\:\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{16}} \\…
Question Number 161410 by saly last updated on 17/Dec/21 Commented by saly last updated on 17/Dec/21 $$\:\:{Thank}\:{you} \\ $$ Commented by saly last updated on…