The year 610 was 1,400 years ago as of 2010.
813 years
1776 AD means the year 1776 after christ was born, this year is 2011 AD. Forty years before Jesus was born it was 40 BC. BC: Before Christ AD: After Death
It was 1100-750 b.c.
Russia was consolidated from scattered duchies in 14th Century AD, so the birthday of the term "Russia" referring to a single big State is in AD.
Suppose the two fractions are a/b and c/d ad that b, d > 0. Then cross multiplication gives ad and bc. If ad > bc then a/b > c/d, If ad = bc then a/b = c/d, and If ad < bc then a/b < c/d
AD stands for Anno Domini
a
(a + b)/(a - b) = (c + d)/(c - d) cross multiply(a + b)(c - d) = (a - b)(c + d)ac - ad + bc - bd = ac + ad - bc - bd-ad + bc = -bc + ad-ad - ad = - bc - bc-2ad = -2bcad = bc that is the product of the means equals the product of the extremesa/b = b/c
Domini
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Yes. The simple answer is that rational fractions are infinitely dense. A longer proof follows:Suppose you have two fractions a/b and c/d where a, b, c and d are integers and b, d are positive integers.Without loss of generality, assume a/b < c/d.The inequality implies that ad < bc so that bc-ad>0 . . . . . . . . . . . . . . . . . . . (I)Consider (ad + bc)/(2bd)Then (ad+bc)/2bd - a/b = (ad+bc)/2bd - 2ad/2bd = (bc-ad)/2bdBy definition, b and d are positive so bd is positive and by result (I), the numerator is positive.That is to say, (ad+bc)/2bd - a/b > 0 or (ad+bc)/2bd > a/b.Similarly, by considering c/d - (ad+bc)/2bd is can be shown that c/d > (ad+bc)/2bd.Combining these results,a/b < (ad+bc)/2bd < c/d.
a(b + d).
If bd ≠ 0, then a/b + c/d (the common denominator is bd) = (a x d)/(b x d) + (c x b)/(d x b) = ad/bd + cb/db = ad/bd + cb/bd = (ad + cb)/ bd
a/b=c/d =>ad=bc =>a =bc/d b =ad/c c =ad/b d =bc/a so if a+b=c+d is true => (bc/d)+(ad/c)=(ad/b)+(bc/a) => (bc2+ad2)/dc=(da2+cb2)/ab => ab(bc2+ad2)=dc(da2+cb2) and since ad=bc, => ab(adc+add) =dc(ada+adc) => abadc+abadd =dcada + dcadc => abadc-dcadc =dcada-abadd => (ab-dc)adc =(dc-ab)add ad cancels out => (ab-dc)c =(dc-ab)d => -(dc-ab)c =(dc-ab)d => -c = d so there's your answer :)
PITBULL
Anno Domini (AD) .