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If B is between P and Q, then:

P<B<Q

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โˆ™ 2010-03-01 23:03:00
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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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Q: If B is between P and Q?
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Can the product of two rational numbers be irrational?

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.


What does p over q mean in algebra?

P! / q!(p-q)!


What formula is used to find the length between two points in a coordinate plane?

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]


What is 4 times the sum of q and p?

4(p + q), or 4p + 4q


Is not p and q equivalent to not p and not q?

Think of 'not' as being an inverse. Not 1 = 0. Not 0 = 1. Using boolean algebra we can look at your question. 'and' is a test. It wants to know if BOTH P and Q are the same and if they are 1 (true). If they are not the same, or they are both 0, then the result is false or 0. not P and Q is rewritten like so: (P and Q)' = X not P and not Q is rewritten like: P' and Q' = X (the apostrophe is used for not) We will construct a truth table for each and compare the output. If the output is the same, then you have found your equivalency. Otherwise, they are not equivalent. P and Q are the inputs and X is the output. P Q | X P Q | X ------ 0 0 | 1 0 0 | 1 0 1 | 1 0 1 | 0 1 0 | 1 1 0 | 0 1 1 | 0 1 1 | 0 Since the truth tables are not equal, not P and Q is not equivalent to not P and not Q. Perhaps you meant "Is NOT(P AND Q) equivalent to NOT(P) AND NOT(Q)?" NOT(P AND Q) can be thought of intuitively as "Not both P and Q." Which if you think about, you can see that it would be true if something were P but not Q, Q but not P, and neither P nor Q-- so long as they're not both true at the same time. Now, "NOT(P) AND NOT(Q)" is clearly _only_ true when BOTH P and Q are false. So there are cases where NOT(P AND Q) is true but NOT(P) AND NOT(Q) is false (an example would be True(P) and False(Q)). NOT(P AND Q) does have an equivalence however, according to De Morgan's Law. The NOT can be distributed, but in doing so we have to change the "AND" to an "OR". NOT(P AND Q) is equivalent to NOT(P) OR NOT(Q)

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Relationship between diagonals and sides of a parallelogram?

p^2+q^2=2(a^2+b^2) where p,q=diagonals of the parallelogram a,b=sides of the parallelogram


Calculate the number p and q?

p/q form of the number is 0.3 is: (A) (B)


What is the relationship between the roots of quadratic?

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


If p q and q r what is the relationship between the values p and r?

Ifp < q and q < r, what is the relationship between the values p and r? ________________p


Prove that if a and b are rational numbers then a multiplied by b is a rational number?

If a is rational then there exist integers p and q such that a = p/q where q&gt;0. Similarly, b = r/s for some integers r and s (s&gt;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 &gt; 0 means that q*s &gt; 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.


What is the Analogy for d is B as r is to p or T or q or P?

P


What is the geometrical proof of a-b whole square?

Given the graphic capability of this site, you are going to have to use some imagination! &lt;---------a---------&gt; &lt;---a-b---&gt;&lt;--b--&gt; +-----------+-------+ |...............|..........| |.......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


If a divides b and b divides a then a is equal to b?

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 =&gt; b(1-pq)=0 =&gt; pq=1 since b!=0 =&gt; p=q=1 or p=q=-1 =&gt; a=b or a=-b


What is the answer to the analogy d is to B as r is to either p T q or P?

d is to B as r is to P


Is the statement P q valid?

No, it is not valid because there is no operator between P and q.


How do we find an irrational number between two rational numbers?

If a and b are rational, with a &lt; 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 &lt; a + (b-a) [sqrt(2)/3] &lt; 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


Can the product of two rational numbers be irrational?

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.

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