For a finite sequence of nonzero numbers, the number of variations in sign is defined as the number of pairs of consecutive terms of the sequence for which the product of the two consecutive terms is negative. what is the number of variations in sign for the sequence 1, –3, 2, 5, –4, –6 ? one

the number of variations in sign = 3

Step-by-step explanation:

Let the finite sequence 1, –3, 2, 5, –4, –6 , the only pairs of consecutive terms for which the product of the two consecutive terms is negative are :

{1, –3} ; {–3, 2}; {5, –4} then

the number of variations in sign = 3

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this

the correct answer is 5 6 and 7