A third degree polynomial equation with rational coefficients that has roots -4 and 2+i

1 answer:

8 0

If one of the roots is 2+i, another must be 2-i. We can express this polynomial as (x+4)(x-2-i)(x-2+i). This expands to $x^3-11x+20$ .

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Read 2 more answers

Line A: y=-1/2x+3

sashaice [31]

A) For the equation $y=- \frac{1}{2}x+3$ , the slope is $- \frac{1}{2}$ .
Slope-intercept form is y = mx + b, where m is slope and b is the y-intercept. This means that if $m=- \frac{1}{2}$ , the slope is $- \frac{1}{2}$ .

B) Since this equation is in the same form, you just find what m is equal to.
Since $m=2$ , that's the slope.

For the solution, set $y=- \frac{1}{2}x+3$ equal to $y=2x-4$ , so it would be put together like this:
$y=[tex]- \frac{1}{2}x+3=2x-4\\ -\frac{1}{2}=2x-7\\ 1\frac{1}{2}x=-7\\ x=-4.66667$
So your answer is -4.667.

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