Category: Arithmetic

Question Number 209166 by Tawa11 last updated on 02/Jul/24 Answered by mr W last updated on 02/Jul/24 $$\left(a\right)$$$$p={\left(\frac{3}{4}\right)}^3=0.42$$$$\left(b\right)$$$${p}={C}_{\mathrm{2}}…

Question Number 209167 by lmcp1203 last updated on 02/Jul/24 $$pleaseconvert{2531}_{\left(5000\right)}tobase5002.thanks.$$ 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 209123 by Spillover last updated on 02/Jul/24 Answered by A5T last updated on 02/Jul/24 $$max(1000x+600c)givenc⩾4xandx+c⩽200$$$$4x-c⩽0andx+c⩽200\Rightarrow x⩽40$$$$\Rightarrow max(1000x+600c)=1000\times 40+600\times 160$$$$=136000.$$…

Question Number 208980 by lmcp1203 last updated on 30/Jun/24 $$please.find{2}^{{11001}^{666}}mod23thanks.$$ 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 208445 by alcohol last updated on 16/Jun/24 $$u_{n+1}=u_n-u_n^3,u_0\in ]0,1[$$$${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}} }…