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