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Question-195287

Question Number 195287 by Mingma last updated on 29/Jul/23 Answered by MM42 last updated on 29/Jul/23 $${CD}={DE}\Rightarrow\angle{C}\mathrm{2}=\angle{E}\:\:\:\&\:\angle{D}\mathrm{1}+\angle{E}=\mathrm{60} \\ $$$$\angle{C}\mathrm{1}=\mathrm{60}−\angle{C}\mathrm{2}=\mathrm{60}−\angle{E}=\angle{D}\mathrm{1} \\ $$$$\frac{{CD}}{{Sin}\mathrm{60}}=\frac{\mathrm{2}}{{SinC}\mathrm{1}}\:\:\&\:\:\frac{{DE}}{{Sin}\mathrm{120}}=\frac{{BE}}{{SinD}\mathrm{1}} \\ $$$$\frac{\mathrm{2}}{{SinC}\mathrm{1}}=\frac{{BE}}{{SinD}\mathrm{1}}\Rightarrow{BE}=\mathrm{2}\:\checkmark \\ $$…

y-sin-x-cos-x-2-4sin-x-2y-cos-x-y-tan-x-

Question Number 195263 by dimentri last updated on 28/Jul/23 $$\:\:\:\:\:\:\begin{cases}{{y}\:\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\:=\:\mathrm{2}}\\{\mathrm{4sin}\:{x}−\mathrm{2}{y}\:\mathrm{cos}\:{x}\:=\:{y}}\end{cases} \\ $$$$\:\:\:\:\:\:\:\mathrm{tan}\:{x}\:=?\: \\ $$ Answered by cortano12 last updated on 28/Jul/23 $$\:\:\:\:\:\:\begin{cases}{\:\cancel{\underline{\underbrace{ }}}}\end{cases} \\ $$…

determinant-

Question Number 195231 by cortano12 last updated on 28/Jul/23 $$\:\:\:\:\:\begin{array}{|c|}{\:\cancel{\underline{\underbrace{ }}}}\\\hline\end{array} \\ $$ Answered by MM42 last updated on 28/Jul/23 $${lim}_{{x}\rightarrow\mathrm{0}} \:\:\sqrt{\frac{\mathrm{1}−{cos}\sqrt{\pi{x}}}{{x}\left(\mathrm{1}+\sqrt{{cos}\sqrt{\pi{x}}}\right)}}\:\: \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\sqrt{\frac{\frac{\mathrm{1}}{\mathrm{2}}\pi{x}}{{x}\left(\mathrm{1}+\sqrt{\left.{cos}\sqrt{\pi{x}}\right)}\right.}}…

prove-that-lim-n-e-n-n-n-n-n-2pi-

Question Number 195259 by mathlove last updated on 28/Jul/23 $${prove}\:{that} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{e}^{{n}} \centerdot\left({n}!\right)}{{n}^{{n}} \:\sqrt{{n}}}=\sqrt{\mathrm{2}\pi} \\ $$ Commented by Frix last updated on 28/Jul/23 $$\mathrm{Easy}\:\mathrm{because}\:\mathrm{for}\:{n}\rightarrow\infty:\:{n}!\rightarrow\frac{{n}^{{n}}…

1-3-i-2-n-x-i-i-2-n-1-i-j-2-n-j-3-i-1-j-1-x-i-i-j-1-n-x-j-i-1-j-1-j-i-i-j-1-n-j-i-n-3-i-1-

Question Number 195227 by York12 last updated on 28/Jul/23 $$ \\ $$$$\alpha_{\mathrm{1}} ^{\mathrm{3}} \left[\frac{\underset{{i}=\mathrm{2}} {\overset{{n}} {\prod}}\left({x}−\alpha_{{i}} \right)}{\underset{{i}=\mathrm{2}} {\overset{{n}} {\prod}}\left(\alpha_{\mathrm{1}} −\alpha_{{i}} \right)}\right]+\underset{{j}=\mathrm{2}} {\overset{{n}} {\sum}}\left(\alpha_{{j}} ^{\mathrm{3}} \left[\frac{\underset{{i}=\mathrm{1}}…

dx-cos-3-x-4sin-xcos-x-

Question Number 195255 by Spillover last updated on 28/Jul/23 $$\int\frac{{dx}}{\mathrm{cos}\:^{\mathrm{3}} {x}\sqrt{\mathrm{4sin}\:{x}\mathrm{cos}\:{x}}} \\ $$ Answered by Frix last updated on 28/Jul/23 $$\left[\mathrm{Using}\:{t}=\sqrt{\mathrm{tan}\:{x}}\right] \\ $$$$=\int\left({t}^{\mathrm{4}} +\mathrm{1}\right){dt}=…=\frac{\mathrm{5}+\mathrm{tan}^{\mathrm{2}} \:{x}}{\mathrm{5}}\sqrt{\mathrm{tan}\:{x}}\:+{C}…

spillover-sin-2-xcos-2-x-sin-5-x-cos-3-xsin-2-x-sin-3-xcos-2-x-cos-5-x-2-dx-

Question Number 195254 by Spillover last updated on 02/Aug/23 $$\int^{\boldsymbol{{spillover}}} \frac{\mathrm{sin}\:^{\mathrm{2}} {x}\mathrm{cos}\:^{\mathrm{2}} {x}}{\left(\mathrm{sin}\:^{\mathrm{5}} {x}+\mathrm{cos}\:^{\mathrm{3}} {x}\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{sin}\:^{\mathrm{3}} {x}\mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{5}} {x}\right)^{\mathrm{2}} }{dx} \\ $$ Answered by Spillover…