Part D Now examine the sum of a rational number, y, and an irrational number, x. The rational number y can be written as y =
where a and b are integers and b +0. Leave the irrational number x as x because it can't be written as the ratio of two
integers.
Let's look at a proof by contradiction. In other words, we're trying to show that x + y is equal to a rational number instead
of an irrational number. Let the sum equal where mand n are integers and n0. The process for rewriting the sum for
x is shown
Statement
Reason
o
substitution
x +
subtraction property of equality
n
() () - (A) ()
Create common denominators
X
bn
bran
bi
Simplify
х
Based on what we established about the classification of x and using the closure of integers, what does the equation tell
you about the type of number x must be for the sum to be rational? What conclusion can I now make about the result
of adding a rational and an irrational number?
0 11:54

step-by-step explanation:

through just a quick inspection, one can see that the graph of this cubing function starts in quadrant iii and ends in quadrant i.   when x = 0, y = 3, so the y-intercept is (0, 3).

if graphs are part of the problem statement, share them.   .

jwjwkwkwkkkdjebd b jdjejdndj he jenej

don't report but were is the digram

step-by-step explanation:

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.