Which system of linear inequalities is graphed? {y> 2x+1x+y< −2 {y≥2x+1x+y≤−2 {y< 2x+1x+y> −2 {y≤2x+1x+y≥−2

y>2x+1; x+y<−2

Step-by-step explanation:

Since, dotted line represents the inequality with sign '<' or '>',

Thus, y≥2x+1; x+y≤−2  and y≤2x+1 ; x+y≥−2  can not be the system of equations,

Now, by the given graph,

The one of the related equations passes through (0, -2) and (-2, 0)

So, the equation of the related equation,

∵ Shaded region of this line does not contain the origin,

Thus, the inequality would be,

Also, other of the related equations passes through (0, 1) and (-1, -1)

So, the equation of the related equation,

∵ Shaded region of this line does not contain the origin,

Thus, the inequality would be,

y > 2x + 1

Hence, the required system would be,

y>2x+1; x+y<−2

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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