Menu Close

Category: Arithmetic

Let-x-cos-pi-9-Show-that-8x-3-6x-1-0-Deduce-x-is-not-rational-

Question Number 207194 by sniper237 last updated on 09/May/24 $${Let}\:\:{x}\:=\:{cos}\frac{\pi}{\mathrm{9}}\: \\ $$$${Show}\:{that}\:\mathrm{8}{x}^{\mathrm{3}} −\mathrm{6}{x}−\mathrm{1}=\mathrm{0} \\ $$$${Deduce}\:{x}\:{is}\:{not}\:\:{rational}\: \\ $$ Answered by Berbere last updated on 09/May/24 $${cos}\left(\mathrm{3}.\frac{\pi}{\mathrm{9}}\right)={cos}\left(\frac{\pi}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{2}}…

Question-206919

Question Number 206919 by BaliramKumar last updated on 30/Apr/24 Answered by A5T last updated on 30/Apr/24 $${MA}=\mathrm{2}{x};\:{MC}=\mathrm{3}{x}\Rightarrow{AC}=\mathrm{5}{x} \\ $$$$\Rightarrow{AB}+{BC}=\mathrm{4}+\mathrm{3}{x}+\mathrm{2}{x}=\mathrm{5}{x}+\mathrm{4} \\ $$$$\left(\mathrm{2}+\mathrm{3}{x}\right)^{\mathrm{2}} +\left(\mathrm{2}+\mathrm{2}{x}\right)^{\mathrm{2}} =\mathrm{25}{x}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}{x}^{\mathrm{2}}…

Question-206805

Question Number 206805 by BaliramKumar last updated on 26/Apr/24 Answered by A5T last updated on 26/Apr/24 $$\left(\mathrm{225}\right)^{\mathrm{40}} =\mathrm{10}^{{x}} \Rightarrow{x}=\mathrm{40}{log}\mathrm{225}\approx\mathrm{94}.\mathrm{087}<\mathrm{95} \\ $$$$\Rightarrow\mathrm{10}^{\mathrm{94}} <\left(\mathrm{225}\right)^{\mathrm{40}} <\mathrm{10}^{\mathrm{95}} \Rightarrow\mathrm{225}^{\mathrm{40}} \:{has}\:\mathrm{95}\:{digits}…

Find-the-missing-number-determinant-72-24-6-96-16-12-108-18-A-12-B-16-C-18-D-20-Please-help-

Question Number 206709 by Nimnim111118 last updated on 23/Apr/24 $$\:\mathrm{Find}\:\mathrm{the}\:\mathrm{missing}\:\mathrm{number} \\ $$$$\:\:\:\:\:\:\:\:\:\:\begin{array}{|c|c|c|}{\:\mathrm{72}}&\hline{\mathrm{24}}&\hline{\:\:\mathrm{6}}\\{\:\mathrm{96}}&\hline{\mathrm{16}}&\hline{\mathrm{12}}\\{\mathrm{108}}&\hline{\:?}&\hline{\mathrm{18}}\\\hline\end{array} \\ $$$$\mathrm{A}.\mathrm{12}\:\:\:\:\:\:\mathrm{B}.\mathrm{16}\:\:\:\:\mathrm{C}.\mathrm{18}\:\:\:\:\:\:\mathrm{D}.\mathrm{20} \\ $$$$\mathrm{Please}\:\mathrm{help}… \\ $$ Commented by Frix last updated on 24/Apr/24…