Question Number 193676 by lapache last updated on 18/Jun/23 $${Solve} \\ $$$${arcsin}\left(\mathrm{2}{x}\right)+{arcsin}\left({x}\sqrt{\mathrm{3}}\right)={arcsin}\left({x}\right) \\ $$ Answered by Frix last updated on 18/Jun/23 $$\mathrm{sin}^{−\mathrm{1}} \:{t}\:\in\mathbb{R}\:\Leftrightarrow\:−\mathrm{1}\leqslant{t}\leqslant\mathrm{1} \\ $$$${f}\left({x}\right)=\mathrm{sin}^{−\mathrm{1}}…
Question Number 193733 by Tawa11 last updated on 18/Jun/23 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{ordinary}\:\mathrm{differential}\:\mathrm{equation} \\ $$$$\mathrm{satisfy}\:\mathrm{by}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}\:\:=\:\:\mathrm{x}^{\mathrm{n}} \left(\mathrm{A}\:\:+\:\:\mathrm{Blogx}\right) \\ $$ Answered by Rajpurohith last updated on 19/Jun/23 $${y}\:'={x}^{{n}}…
Question Number 193734 by Mastermind last updated on 18/Jun/23 $$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} \left(\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{2}}\mathrm{dx} \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$ Answered by witcher3…
Question Number 193596 by Mingma last updated on 17/Jun/23 Answered by cortano12 last updated on 17/Jun/23 $$\left(\mathrm{1}\right)\:\gamma=\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{e}^{\mathrm{x}} −\mathrm{2}\alpha\mathrm{x}−\beta}{\mathrm{2x}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\:\:\beta=\mathrm{1}\: \\ $$$$\:\left(\mathrm{3}\right)\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 193656 by York12 last updated on 17/Jun/23 $${please}\:{can}\:{someone}\:{solves}\:{this}\: \\ $$ Commented by York12 last updated on 17/Jun/23 Commented by mahdipoor last updated on…
Question Number 193595 by Rupesh123 last updated on 17/Jun/23 Answered by MM42 last updated on 17/Jun/23 $${sin}^{\mathrm{2}} \alpha+{sin}^{\mathrm{2}} \beta−{sin}\alpha{cos}\beta−{sim}\beta{cos}\alpha=\mathrm{0} \\ $$$$\Rightarrow{sin}\alpha\left({sin}\alpha−{cos}\beta\right)+{sin}\beta\left({sin}\beta−{cos}\alpha\right)=\mathrm{0} \\ $$$${sin}\alpha\left(\mathrm{2}{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right){cos}\left(\frac{\alpha−\beta}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right)+{sin}\beta\left(\mathrm{2}{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right){cos}\left(\frac{\beta−\alpha}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right)=\mathrm{0} \\ $$$$\Rightarrow{sin}\left(\frac{\alpha+\beta}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right)\left({sin}\alpha{cos}\left(\frac{\alpha−\beta}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)+{sin}\beta{cos}\left(\frac{\beta−\alpha}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\right)=\mathrm{0}…
Question Number 193651 by a.lgnaoui last updated on 17/Jun/23 $$\mathrm{determiner}\:\mathrm{rayon}\:\boldsymbol{\mathrm{R}} \\ $$ Commented by a.lgnaoui last updated on 17/Jun/23 Terms of Service Privacy Policy Contact:…
Question Number 193647 by Rupesh123 last updated on 17/Jun/23 Answered by MM42 last updated on 17/Jun/23 $${both}\:{numbers}\:{a}\:,\:{b}\:{can}\:{not}\:{be}\:{odd}. \\ $$$${so}\:{a}=\mathrm{2}\:{or}\:{b}=\mathrm{2} \\ $$$${if}\:\:{a}=\mathrm{2}\:\&\:{b}=\mathrm{3}\:\Rightarrow{p}=\mathrm{2}^{\mathrm{3}} +\mathrm{7}×\mathrm{3}^{\mathrm{2}} =\mathrm{71}\:\checkmark \\ $$$${if}\:\:{a}=\mathrm{3}\:\&\:{b}=\mathrm{2}\Rightarrow\:{p}=\mathrm{3}^{\mathrm{2}}…
Question Number 193646 by Shlock last updated on 17/Jun/23 Answered by som(math1967) last updated on 17/Jun/23 $$\:\left({cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} −\mathrm{2}{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x} \\ $$$$=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{sinxcosx}\right)^{\mathrm{2}} \\…
Question Number 193635 by mr W last updated on 17/Jun/23 Commented by mr W last updated on 17/Jun/23 $${find}\:{the}\:{largest}\:{circle}\:{and}\:{the}\:{largest} \\ $$$${square}\:{which}\:{you}\:{can}\:{completely} \\ $$$${cover}\:{with}\:{three}\:{circular}\:{plates}\:{with} \\ $$$${radius}\:\mathrm{1}\:{respectively}.…