Assuming you are referring to the abbreviations that typically follow years, A.D. stands for anno domini, which means "in the year of our lord" in Latin, while B.C. stands for "before Christ."
(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
I think C B stand for Cornerback.
Consider any triangle ABC, and let AD be the altitude from A on to BC. Then sin(B) = AD/AB so that AD = AB*sin(B) and sin(C) = AD/AC so that AD = AC*sin(C) Therefore AB*sin(B) = AC*sin(C) or c*sin(B) = b*sin(C) where the lower case letter represents the side opposite the angle with the upper case name. Divide both sides by bc to give sin(B)/b = sin(C)/c. Similarly, using the altitude from B you can show that sin(A)/a = sin(C)/c. Combining with the previous result, sin(A)/a = sin(B)/b = sin(C)/c.
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)/(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
I think C B stand for Cornerback.
A B C on the tree . A stand for A B stand for Bee C stand for see means " A bee see on the tree " .....
For A/B = C/D, B = AD/C
Consider any triangle ABC, and let AD be the altitude from A on to BC. Then sin(B) = AD/AB so that AD = AB*sin(B) and sin(C) = AD/AC so that AD = AC*sin(C) Therefore AB*sin(B) = AC*sin(C) or c*sin(B) = b*sin(C) where the lower case letter represents the side opposite the angle with the upper case name. Divide both sides by bc to give sin(B)/b = sin(C)/c. Similarly, using the altitude from B you can show that sin(A)/a = sin(C)/c. Combining with the previous result, sin(A)/a = sin(B)/b = sin(C)/c.
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.
ad
A,b,c
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 :)
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(c+d)+b(c+d)=(a+b)(c+d)