Could be either. That is the whole purpose of designating the BC or AD, so that you know which of the two it is. There is a difference of 828 years between the two. However, it is typical that if neither is designated, that the assumption would be that it is AD.
AD stands for "anno Domini" BC stands for "before Christ"
There are 30 years from 20 BC to 10 AD. This is true because it goes 20 to 0 to 10, adding up to 30.
Medeval weapons are most likley to be used as nights in the AD period not BC like in the atient age! -Colin202
Add the two year values together and subtract 1, to allow for the fact that there was no year zero. So from 1 BC to 1 AD is 1 year. 1 + 1 - 1 = 1. From 10 BC to 40 AD is 49. 10 + 40 - 1 = 49.
The difference between the fractions a/b and c/d = abs[(ad - bc)/bd]
The difference of two rational numbers is rational. Let the two rational numbers be a/b and c/d, where a, b, c, and d are integers. Any rational number can be represented this way. Their difference is a/b-c/d = ad/bd-cb/bd = (ad-cb)/bd. Products and differences of integers are always integers. This means that ad-cb is an integer, and so is bd. Thus, (ad-cb)/bd is a rational number (since it is the ratio of two integers). This is equivalent to the difference of the original two rational numbers.
A rational number is one that can be expressed as a/b The sum of two rational numbers is: a/b + c/d =ad/bd + bc/bd =(ad+bc)/bd =e/f which is rational The difference of two rational numbers is: a/b - c/d =ab/bd - bc/bd =(ab-bc)/bd =e/f which is rational The product of two rational numbers is: (a/b)(c/d) =ac/bd =e/f which is rational
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
according to sum of fractions: a/b + c/d = (ad+bc)/bd hence for example: a=c=1 , b=d=2: (ad+bc)/bd = (2+2)/2x2 = 4/4 = 1 sagy
Because it can be derived from a large number of more basic principles applied to numbers. Since a/b and c/d are defined, then b and d are non-zero. Then a/b = ad/bd since multiplying the numerator and denominator of a fraction by a non-zero number leaves it unchanged. Similarly, c/d = bc/bd [These steps implicitly assume the commutative property of multiplication ie there is no difference between premultiplication and postmultiplication] Another property of numbers: If x = p and y = q then x+y = p+q [equals added to equals result in equals]. So a/b + c+d = ad/bd + bc/bd and finally, using the distributive property of multiplication [by 1/bd] over addition, = (ad + bc)/bd Maybe you wish you hadn't asked!
Infinitely many ways!Suppose you have the fraction 2/d.Pick any pair of integers a and b where b � 0.Then 2b-ad is and integer, as is bd so that (2b - ad)/bd is a fraction.Consider the fractions a/b and (2b - ad)/bdTheir sum isa/b + (2b-ad)/bd = ad/bd + (2b-ad)/bd = 2b/bd = 2/d - as required.Since a and b were chosen arbitrarily, there are infinitely many possible answers to the question.
The USA is big, while BD is small.
1960 years
Ad is higher than ac
Yes. Since they are rational numbers, let's call the first one a/b and the second one c/d where a,b,c, and d are integers. Now we can subtract by finding a common denominator. Let's use bd. So we have ad/bd-cb/bc= (ad-bc)/CD which is rational since we know ad and bc are integers being the product of integers and CD is also an integers. Call ad-bd=P and call CD=Q where P and Q are integers. We now see the difference is of two rationals is rational.
For any two fractions, a/b and c/d, where b and d are non-zero, a/b - c/d = (ad - bc)/bd