No.Suppose a and b are two rational numbers.Then they can be written as follows: a = p/q, b = r/s where p, q, r and s are integers and q, s >0.Then a*b = (p*r)/(q*s).Using the properties of integers, p*r and q*s are integers and q*s is non-zero. So a*b can be expressed as a ratio of two integers and so the product is rational.
It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]
There are six letters between M and T. They are n, o, p, q, r, and s. The midway point is in the middle between P and Q, so there is no one letter midway between M and T. If you must have an answer, it would be both P and Q.
P! / q!(p-q)!
4(p + q), or 4p + 4q
p^2+q^2=2(a^2+b^2) where p,q=diagonals of the parallelogram a,b=sides of the parallelogram
Suppose the roots a quadratic, in the form ax2 + bx + c = 0, are p and q. Then p + q = -b/a and pq = c/a
p/q form of the number is 0.3 is: (A) (B)
If a is rational then there exist integers p and q such that a = p/q where q>0. Similarly, b = r/s for some integers r and s (s>0) Then a*b = p/q * r/s = (p*r)/(q*s) Now, since p, q r and s are integers, p*r and q*s are integers. Also, q and s > 0 means that q*s > 0 Thus a*b can be expressed as x/y where p and r are integers implies that x = p*r is an integer q and s are positive integers implies that y = q*s is a positive integer. That is, a*b is rational.
Given the graphic capability of this site, you are going to have to use some imagination! <---------a---------> <---a-b---><--b--> +-----------+-------+ |...............|..........| |.......P......|....Q...| |...............|..........| +-----------+-------+ |.......R......|....S....| |...............|..........| +-----------+-------+ In the above graphic, P, S and the whole figure are meant to be squares. The total area is P+Q+R+S = a2 P = (a-b)2 Q = b*(a-b) = (a-b)*b = a*b - b2 R = (a-b)*b = a*b = a*b - b2 and S = b2 Now, P = {P+Q+R+S} - Q - R - S = a2 - ab + b2 - ab + b2 - b2 = a2 - 2ab + b2
Ifp < q and q < r, what is the relationship between the values p and r? ________________p
how to un b o q n s p e
If a divides b and b divides a then either a is equal to b or a is equal to -b. Additional note: if a divides b, there exist a p such that ap=b. and if b divides a, there exist a q such that a=bq. then ap=(bq)p=b => b(1-pq)=0 => pq=1 since b!=0 => p=q=1 or p=q=-1 => a=b or a=-b
If a and b are rational, with a < b, then a + (b-a) [sqrt(2)/ 2] is an irrational number between a and b. This number is between a and b because sqrt(2)/2 is less than one and positive, so that a < a + (b-a) [sqrt(2)/3] < a + (b-a) [1] = b. To prove that a + (b-a) [sqrt(2)/2] is not rational, suppose that a + (b-a) [sqrt(2)/2] = p/q where p and q are integers. Then, sqrt(2) = ( p/q -a ) 2/(b-a) which is rational since the rationals are a field, closed under arithmetical operation, but sqrt(2) not rational
P
d is to B as r is to P
No.Suppose a and b are two rational numbers.Then they can be written as follows: a = p/q, b = r/s where p, q, r and s are integers and q, s >0.Then a*b = (p*r)/(q*s).Using the properties of integers, p*r and q*s are integers and q*s is non-zero. So a*b can be expressed as a ratio of two integers and so the product is rational.