Menu Close

Category: Arithmetic

The-speeds-of-Daniel-and-Robert-are-in-the-ratio-of-3-4-In-a-race-of-300m-Daniel-has-a-start-of-90m-Daniel-won-by-

Question Number 18214 by chux last updated on 16/Jul/17 $$\mathrm{The}\:\mathrm{speeds}\:\mathrm{of}\:\mathrm{Daniel}\:\mathrm{and}\:\mathrm{Robert} \\ $$$$\mathrm{are}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{3}:\mathrm{4}.\mathrm{In}\:\mathrm{a}\:\mathrm{race}\:\mathrm{of} \\ $$$$\mathrm{300m}\:\mathrm{Daniel}\:\mathrm{has}\:\mathrm{a}\:\mathrm{start}\:\mathrm{of}\:\mathrm{90m}.\: \\ $$$$\mathrm{Daniel}\:\mathrm{won}\:\mathrm{by}? \\ $$ Answered by ajfour last updated on 17/Jul/17…

The-greatest-number-that-will-divide-82-111-and-140-leaving-the-same-remainder-in-each-case-is-

Question Number 18213 by chux last updated on 16/Jul/17 $$\mathrm{The}\:\mathrm{greatest}\:\mathrm{number}\:\mathrm{that}\:\mathrm{will} \\ $$$$\mathrm{divide}\:\mathrm{82}\:,\mathrm{111}\:\mathrm{and}\:\mathrm{140}\:\mathrm{leaving}\:\mathrm{the}\: \\ $$$$\mathrm{same}\:\mathrm{remainder}\:\mathrm{in}\:\mathrm{each}\:\mathrm{case}\:\mathrm{is}…….. \\ $$ Answered by Tinkutara last updated on 17/Jul/17 $$\mathrm{82}\:=\:{xa}\:+\:{r} \\…

n-1-30-n-2-1-A-n-1-15-2n-2-30n-224-B-n-1-15-2n-2-30n-225-C-n-1-15-2n-2-30n-226-D-n-1-15-2n-2-30n-227-E-n-

Question Number 17767 by Joel577 last updated on 10/Jul/17 $$\underset{{n}\:=\:\mathrm{1}} {\overset{\mathrm{30}} {\sum}}\left({n}^{\mathrm{2}} \:+\:\mathrm{1}\right)\:=\: \\ $$$$\left(\mathrm{A}\right)\:\underset{{n}\:=\:\mathrm{1}} {\overset{\mathrm{15}} {\sum}}\left(\mathrm{2}{n}^{\mathrm{2}} \:+\:\mathrm{30}{n}\:+\:\mathrm{224}\right) \\ $$$$\left(\mathrm{B}\right)\:\underset{{n}\:=\:\mathrm{1}} {\overset{\mathrm{15}} {\sum}}\left(\mathrm{2}{n}^{\mathrm{2}} \:+\:\mathrm{30}{n}\:+\:\mathrm{225}\right) \\ $$$$\left(\mathrm{C}\right)\:\underset{{n}\:=\:\mathrm{1}}…

Question-148780

Question Number 148780 by Jonathanwaweh last updated on 31/Jul/21 Answered by Kamel last updated on 31/Jul/21 $$ \\ $$$$\Omega=\underset{{n}=\mathrm{2}} {\overset{+\infty} {\prod}}{e}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)^{{n}^{\mathrm{2}} } ={e}^{\underset{{n}=\mathrm{2}} {\overset{+\infty}…