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Category: Arithmetic

xcos-a-ysin-b-1-xsin-ycos-a-2-sin-2-b-2-cos-2-Eliminate-

Question Number 207594 by MATHEMATICSAM last updated on 19/May/24 $$\frac{{x}\mathrm{cos}\theta}{{a}}\:+\:\frac{{y}\mathrm{sin}\theta}{{b}}\:=\:\mathrm{1} \\ $$$${x}\mathrm{sin}\theta\:−\:{y}\mathrm{cos}\theta\:=\:\sqrt{{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta\:+\:{b}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \theta} \\ $$$$\mathrm{Eliminate}\:\theta. \\ $$ Answered by Frix last updated…

The-real-roots-of-the-equation-x-2-6x-c-0-differ-by-2n-where-n-is-a-real-non-zero-Show-that-n-2-9-c-Given-that-the-roots-also-have-opposite-signs-find-the-set-of-possible-values-of-n-

Question Number 207546 by pete last updated on 18/May/24 $$\mathrm{The}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{2}} +\mathrm{6x}+\mathrm{c}=\mathrm{0} \\ $$$$\mathrm{differ}\:\mathrm{by}\:\mathrm{2n},\:\mathrm{where}\:\mathrm{n}\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\mathrm{non}−\mathrm{zero}. \\ $$$$\mathrm{Show}\:\mathrm{that}\:\mathrm{n}^{\mathrm{2}} =\mathrm{9}−\mathrm{c} \\ $$$$\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{also}\:\mathrm{have}\:\mathrm{opposite} \\ $$$$\mathrm{signs},\:\mathrm{find}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{possible}\:\mathrm{values}\:\mathrm{of}\:\mathrm{n} \\ $$ Answered by A5T…

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}…