10 Consider the graphs of fx) = x® and of g(x) = 3. Are
f(x)
the composite functions commutative? Why or why
not?
5
g(x)
-5
g(x)
5
-10
10
They are commutative because f(g(1)) = g(f(1)).
They are commutative because the composite
functions both equal x.
They are not commutative because the domains of
f(x) and g(x) are different.
O They are not commutative because the graphs
intersect each other.
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step-by-step explanation:

i'm pretty sure it would take 20 hours because it's half the men so you multiply the number of hours it takes by 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.

and?

step-by-step explanation:

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