BC = Before Christ AD = Anno Domini (In the Year of the Lord)
The 4th millennium BC began 4000 BC and ended in 3000 BC. The 4th millennium BC marks the beginning of the bronze age and of writing.
A. BC degree is a bachlor degree of chemistry.
Circa 140 BC
bc is 200 centuries before christ
From the year 1001 to 2000.
* dorian - d ef g a bc d * phrygian -ef g a bc d e* lydian - f g a bc d ef* mixolydian - g a bc d ef g
(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
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
time pass
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 + bc + d
Nagavamsam, BC-"D"
echo "Enter number:" read num r=0 d=0 a=$num while [ $num -gt 10 ] do d=`echo $num % 10 |bc` r=`echo $r \* 10 + $d |bc` num=`echo $num / 10 |bc` done d=`echo $num % 10 |bc` r=`echo $r \* 10 + $d |bc` if [ $a -eq $r ] then echo "given number is polindrome" else echo "given number is not polindrome" fi
Geogina Rockford Livingston! :D
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 :)
The -5.25 is worse than the -4.00. Those numbers give the power of the lens, bc and dia arethe size (curvature and diameter).
bc I d on t k no w