Question Number 199738 by sonukgindia last updated on 08/Nov/23 Answered by AST last updated on 08/Nov/23 $$\frac{{sin}\left({x}\right)}{{DB}}=\frac{{sin}\left(\alpha\right)}{{BC}};\frac{{sin}\left(\mathrm{2}{x}\right)}{{BE}={DB}}=\frac{{sinCEB}={sin}\left(\mathrm{90}+{x}\right)}{{BC}} \\ $$$$\Rightarrow\frac{{DB}}{{BC}}=\frac{{sin}\left({x}\right)}{{sin}\left(\alpha\right)}=\frac{{sin}\left(\mathrm{2}{x}\right)}{{sin}\left(\mathrm{90}+{x}\right)} \\ $$$$\Rightarrow{sin}\left(\alpha\right)=\frac{{sin}\left({x}\right)\left[{sin}\mathrm{90}{cosx}\right]}{\mathrm{2}{sinxcosx}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\alpha=\mathrm{30}° \\ $$ Commented by…
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Question Number 199707 by SANOGO last updated on 08/Nov/23 $${study}\:{the}\:{convergence} \\ $$$$\underset{{n}\geqslant{o}} {\overset{} {\sum}}{sin}\left(\pi\sqrt{\mathrm{4}{n}^{\mathrm{2}} +\mathrm{2}\:\:\:\:}\right. \\ $$ Answered by witcher3 last updated on 09/Nov/23 $$\sqrt{\mathrm{4n}^{\mathrm{2}}…
Question Number 199732 by Rupesh123 last updated on 08/Nov/23 Answered by Rasheed.Sindhi last updated on 08/Nov/23 $${y}=\frac{{x}^{\mathrm{2}} +\mathrm{2}}{{x}+\mathrm{3}}=\frac{{x}^{\mathrm{2}} −\mathrm{9}+\mathrm{9}+\mathrm{2}}{{x}+\mathrm{3}} \\ $$$$\:\:\:=\frac{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{3}\right)+\mathrm{11}}{{x}+\mathrm{3}} \\ $$$$=\left({x}−\mathrm{3}\right)+\frac{\mathrm{11}}{{x}+\mathrm{3}}\in\mathbb{Z}\Rightarrow\left({x}+\mathrm{3}\right)\mid\mathrm{11} \\ $$$${x}+\mathrm{3}=\pm\mathrm{1},\pm\mathrm{11}…
Question Number 199733 by Rupesh123 last updated on 08/Nov/23 Answered by mr W last updated on 08/Nov/23 $${x}^{\mathrm{6}} −{x}^{\mathrm{4}} +\mathrm{1}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} \right)^{\mathrm{3}} −\left({x}^{\mathrm{2}} \right)^{\mathrm{2}}…
Question Number 199728 by Mingma last updated on 08/Nov/23 Answered by AST last updated on 08/Nov/23 $$\frac{{sin}\mathrm{120}°}{{BC}}=\frac{{sin}\mathrm{20}°}{{AB}}\Rightarrow{AB}=\frac{\mathrm{2}\sqrt{\mathrm{3}}{BCsin}\mathrm{20}°}{\mathrm{3}} \\ $$$$\frac{{sin}\mathrm{20}°}{{AB}}=\frac{{sin}\mathrm{40}°}{{AC}}\Rightarrow{AC}=\mathrm{2}{ABcos}\mathrm{20}°=\frac{\mathrm{2}\sqrt{\mathrm{3}}{BCsin}\mathrm{40}}{\mathrm{3}} \\ $$$$\frac{{sin}\left({x}\right)}{{DC}}=\frac{{sinADC}}{{AC}}=\frac{{sin}\left(\mathrm{140}−{x}\right)}{{AC}}\Rightarrow\frac{{AC}}{{DC}}=\frac{{sin}\left(\mathrm{140}−{x}\right)}{{sin}\left({x}\right)} \\ $$$$=\frac{{sin}\mathrm{40}}{{sin}\mathrm{20}}=\mathrm{2}{cos}\mathrm{20}=\frac{{sin}\mathrm{140}{cosx}−{sin}\left({x}\right){cos}\left(\mathrm{140}\right)}{{sinx}} \\ $$$$={sin}\mathrm{40}{tan}\left({x}\right)−{cos}\left(\mathrm{90}+\mathrm{50}\right)={sin}\mathrm{40}{tanx}+{sin}\mathrm{50}…
Question Number 199729 by Mingma last updated on 08/Nov/23 Answered by mathfreak01 last updated on 08/Nov/23 $${N}\:=\:\underset{{n}\:=\:\mathrm{1}} {\overset{\mathrm{450}} {\sum}}\left(\mathrm{10}^{{n}} \:−\:\mathrm{1}\right) \\ $$$${N}\:=\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{450}} {\sum}}\mathrm{10}^{{n}} \:−\:\mathrm{450}…
Question Number 199762 by hardmath last updated on 08/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 199730 by Rupesh123 last updated on 08/Nov/23 Answered by mr W last updated on 10/Nov/23 Commented by mr W last updated on 10/Nov/23…
Question Number 199731 by Rupesh123 last updated on 08/Nov/23 Answered by des_ last updated on 08/Nov/23 $$\left({x}\:+\:\frac{\mathrm{1}}{{x}}\:\right)^{\mathrm{3}} =\:\mathrm{3}\sqrt{\mathrm{3}}\:\Rightarrow \\ $$$$\Rightarrow\:{x}^{\mathrm{3}} \:+\:\mathrm{3}\left({x}\:+\:\frac{\mathrm{1}}{{x}}\:\right)\:+\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:\mathrm{3}\sqrt{\mathrm{3}}\:\:\Rightarrow \\ $$$$\Rightarrow\:{x}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{3}}…