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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."

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How is a plus b over a minus b equals c plus d over c minus 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


What do you do after cross multiplying when comparing fractions with unlike denominators?

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


What is c b stand for in football?

I think C B stand for Cornerback.


How do you derived the sine law?

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.


Can you always find another fraction in between ant two fractions why?

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.

Related Questions

How is a plus b over a minus b equals c plus d over c minus 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


What do you do after cross multiplying when comparing fractions with unlike denominators?

Suppose the two fractions are a/b and c/d ad that b, d &gt; 0. Then cross multiplication gives ad and bc. If ad &gt; bc then a/b &gt; c/d, If ad = bc then a/b = c/d, and If ad &lt; bc then a/b &lt; c/d


What is c b stand for in football?

I think C B stand for Cornerback.


A B C on the tree?

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 " .....


How do you find the missing denominator?

For A/B = C/D, B = AD/C


How do you derived the sine law?

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.


Can you always find another fraction in between ant two fractions why?

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.


What is the ful foam of t b c?

ad


What does airwaybreathingcirculation stand for?

A,b,c


If a over b equals c over d then is it possible that a plus b equals c plus d If not then give Mathematical Reason for that?

a/b=c/d =&gt;ad=bc =&gt;a =bc/d b =ad/c c =ad/b d =bc/a so if a+b=c+d is true =&gt; (bc/d)+(ad/c)=(ad/b)+(bc/a) =&gt; (bc2+ad2)/dc=(da2+cb2)/ab =&gt; ab(bc2+ad2)=dc(da2+cb2) and since ad=bc, =&gt; ab(adc+add) =dc(ada+adc) =&gt; abadc+abadd =dcada + dcadc =&gt; abadc-dcadc =dcada-abadd =&gt; (ab-dc)adc =(dc-ab)add ad cancels out =&gt; (ab-dc)c =(dc-ab)d =&gt; -(dc-ab)c =(dc-ab)d =&gt; -c = d so there's your answer :)


If bd does not equal 0 then a over b plus c over d equals?

If bd &ne; 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


Factor ac plus ad plus bc plus bd?

a(c+d)+b(c+d)=(a+b)(c+d)