Question Number 208980 by lmcp1203 last updated on 30/Jun/24 $${please}\:.\:\:\:\:\:{find}\:\:\mathrm{2}^{\mathrm{11001}^{\mathrm{666}} } {mod}\:\mathrm{23}\:\:\:\:\:\:\:\:{thanks}. \\ $$ Answered by A5T last updated on 30/Jun/24 $$\mathrm{2}^{\mathrm{11001}^{\mathrm{666}} } {mod}\left(\mathrm{23}\right)\equiv\mathrm{2}^{\mathrm{11001}^{\mathrm{666}} \left[{mod}\:\phi\left(\mathrm{23}\right)\right]}…
Question Number 208876 by Tawa11 last updated on 26/Jun/24 The 𝚌a𝚕𝚎𝚗𝚍𝚊𝚛 𝚘𝚏 𝚝𝚑𝚎 𝚢𝚎𝚊𝚛 2024 𝚒𝚜 𝚝𝚑𝚎 𝚜𝚊𝚖𝚎 𝚏𝚘𝚛 𝙰.2044 𝙱.2032 𝙲.2040 𝙳.2036 Commented by…
Question Number 208445 by alcohol last updated on 16/Jun/24 $$\left.{u}_{{n}+\mathrm{1}} \:=\:{u}_{{n}} −{u}_{{n}} ^{\mathrm{3}} ,\:{u}_{\mathrm{0}} \in\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$$${v}_{{n}} \:=\:\frac{\mathrm{1}}{{u}_{{n}+\mathrm{1}} ^{\mathrm{2}} }−\frac{\mathrm{1}}{{u}_{{n}} ^{\mathrm{2}} }\:=\:{f}\left({u}_{{n}} ^{\mathrm{2}} \right)\:;\:{f}\left({x}\right)\:=\:\frac{\mathrm{2}−{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }…
Question Number 208217 by mr W last updated on 07/Jun/24 $$\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} +\mathrm{13}^{\mathrm{2}} +\mathrm{21}^{\mathrm{2}} =? \\ $$ Answered by A5T last updated…
Question Number 207979 by Tawa11 last updated on 01/Jun/24 generate nth term for the sequence: 1, 1, 1, 2, 3, 5, 9, 18, 35, 75…
Question Number 207731 by efronzo1 last updated on 24/May/24 $$\:\:\:\downharpoonleft\underline{\:} \\ $$ Answered by A5T last updated on 24/May/24 Commented by A5T last updated on…
Question Number 207594 by MATHEMATICSAM last updated on 19/May/24 $$\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…
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…
Question Number 207416 by efronzo1 last updated on 14/May/24 $$\:\:\: \\ $$ Answered by mr W last updated on 14/May/24 $$\frac{{U}_{\mathrm{2}} }{{q}}+{U}_{\mathrm{2}} +{qU}_{\mathrm{2}} =\mathrm{9}\left(\frac{{q}}{{U}_{\mathrm{2}} }+\frac{\mathrm{1}}{{U}_{\mathrm{2}}…
Question Number 207194 by sniper237 last updated on 09/May/24 $${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}}…