Kayla claims that the absolute value of a number is always greater than the number itself. Which statement about her claim is true? Kayla is correct. Any positive number is greater than a negative, and absolute value of a number is always positive.
Kayla is correct. The distance between 0 and a number is always more than the number.
Kayla is incorrect. The absolute value of a positive number is negative, and negative numbers are less than positive numbers.
Kayla is incorrect. The absolute value of 0 or a positive number is equal to the number.

D. Kayla is incorrect. The absolute value of 0 or a positive number is equal to the number.

Step by step explanation:

Because an absolute number can not be a negative number and is the number for both sides of the number on the number line.

for this pair of functions g(x)=x-2 and f(x)=|x| is (g o f)(a)= |a| -2

step-by-step explanation:

g(x)=x-2

f(x)=|x|

(g o f)(a)=?

(g o f)(x)=g(f(x))

(g o f)(x)=g(|x|)

(g o f)(x)=|x|-2

(g o f)(a)=|a|-2

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.