Question Number 209342 by essaad last updated on 07/Jul/24 Answered by Berbere last updated on 07/Jul/24 $${U}_{{n}+\mathrm{1}} ={f}\left({U}_{{n}} \right) \\ $$$${x}\overset{{f}} {\rightarrow}\frac{{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}};{f}\:{increase} \\ $$$${f}\left(\left[\mathrm{0},\mathrm{1}\right]\right)=\left[\mathrm{0},\frac{\mathrm{3}}{\mathrm{4}}\right]\subset\left[\mathrm{0},\mathrm{1}\right]…
Question Number 209166 by Tawa11 last updated on 02/Jul/24 Answered by mr W last updated on 02/Jul/24 $$\left({a}\right) \\ $$$${p}=\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{3}} =\mathrm{0}.\mathrm{42} \\ $$$$\left({b}\right) \\ $$$${p}={C}_{\mathrm{2}}…
Question Number 209128 by Spillover last updated on 02/Jul/24 Answered by A5T last updated on 02/Jul/24 $${The}\:{length}\:{of}\:{the}\:{diagonal}\:{of}\:{the}\:{square}=\mathrm{2}{r} \\ $$$$\Rightarrow{s}^{\mathrm{2}} +{s}^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} \Rightarrow\mathrm{2}{s}^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} \Rightarrow{s}^{\mathrm{2}} =\mathrm{2}{r}^{\mathrm{2}}…
Question Number 209167 by lmcp1203 last updated on 02/Jul/24 $${please}\:{convert}\:\:\mathrm{2531}_{\left(\mathrm{5000}\right)\:} {to}\:\:{base}\:\mathrm{5002}.\:\:{thanks}.\:\: \\ $$ Answered by mr W last updated on 03/Jul/24 $$\mathrm{2531}_{\left(\mathrm{5000}\right)} =\mathrm{2}×\mathrm{5000}^{\mathrm{3}} +\mathrm{5}×\mathrm{5000}^{\mathrm{2}} +\mathrm{3}×\mathrm{5000}+\mathrm{1}…
Question Number 209127 by Spillover last updated on 02/Jul/24 Answered by efronzo1 last updated on 02/Jul/24 $$\:\:\underline{\underbrace{\boldsymbol{{x}}}} \\ $$ Commented by Spillover last updated on…
Question Number 209123 by Spillover last updated on 02/Jul/24 Answered by A5T last updated on 02/Jul/24 $${max}\left(\mathrm{1000}{x}+\mathrm{600}{c}\right)\:{given}\:{c}\geqslant\mathrm{4}{x}\:{and}\:{x}+{c}\leqslant\mathrm{200} \\ $$$$\mathrm{4}{x}−{c}\leqslant\mathrm{0}\:{and}\:{x}+{c}\leqslant\mathrm{200}\Rightarrow{x}\leqslant\mathrm{40} \\ $$$$\Rightarrow{max}\left(\mathrm{1000}{x}+\mathrm{600}{c}\right)=\mathrm{1000}×\mathrm{40}+\mathrm{600}×\mathrm{160} \\ $$$$=\mathrm{136000}. \\ $$…
Question Number 209066 by Tawa11 last updated on 01/Jul/24 Commented by Tawa11 last updated on 01/Jul/24 Answered by A5T last updated on 01/Jul/24 $${cos}\mathrm{48}°=\frac{{XA}}{\mathrm{19}}\Rightarrow{XA}=\mathrm{19}{cos}\mathrm{48}° \\…
Question Number 208980 by lmcp1203 last updated on 30/Jun/24 $${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}} }…