Q: The sum of p and q is the of the middle term of a trinomial?

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NO kidding, he doesn't have a middle name :P

P(x=n1,y=n2) = (n!/n1!*n2!*(n-n1-n2)) * p1^n1*p2^n2*(1-p1-p2) where n1,n2=0,1,2,....n n1+n2<=n

it depends on the power of the leading coefficient, and that is not always a great indication because polynomials can have non real numbers. A factor of a polynomial is where the function crosses the x axis. If the trinomial will not factor into real numbers, then there are not any real zeros but there are still factors. Think of this one x^2+6x+14. this will not factor into real numbers, but complex solutions. But these complex solutions are factors, so the rule still holds. If the trinomial is a cubic, or at a odd power, then its a odd function, and can have one real solution. If the trinomial is squared, or any even power, its a even function and can have two real solutions. With the graph you can determine it this way: if p(x) is a polynomial function of degree n, then the graph has at most n-1 turning points. If the graph of a function P has n-1 turning points, then the degree of p(x) is at least n.

10/21 First, observe that for the sum of the three balls to be odd, either one is odd and two are even, or all three are odd. Since the sum of two odd numbers is even, the sum of two even number is even, and the sum of an even number and an odd number is odd. P(exactly one is odd)=(5/9)*(4/8)*(3/7)*3=180/504=5/14 P(all three are odd)=(5/9)*(4/8)*(3/7)=60/504=5/42 P(sum is odd)=5/14+5/42=15/42+5/42=20/42=10/21, which is approximately .48

Yes,if your name is abbreviated into your first name that starts with p and your middle name that starts with j

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Suppose the trinomial is x2 + Bx + C You need to find a factor pair of C whose sum is B. If the factors are p and q (that is, pq = C and p+q = B), then the trinomial can be factorised as (x + p)*(x + q).

It is 1 if the two are the only factors.