Lines m and n are cut by transversal l. Lines m and n are cut by transversal l. On line m where it intersects line l, 4 angles are created. Labeled clockwise, from uppercase left: angle 1, angle 2, angle 4, angle 3. On line n where it intersects line l, 4 angles are created. Labeled clockwise, from uppercase left: angle 5, angle 6, angle 8, angle 7.
Which angle relationships are correct? Check all that apply.

∠1 and ∠8 are alternate exterior angles.
∠4 and ∠6 are same side interior angles.
∠5 and ∠7 are vertical angles.
∠2 and ∠8 are corresponding angles.
∠3 and ∠6 are alternate interior angles.

From the attached diagrams, the correct options are;

∠1 and ∠8 are alternate exterior angles

∠4 and ∠6 are same side interior angles

∠3 and ∠6 are alternate interior angles

Step-by-step explanation:

The given parameters are;

Line m and line n are cut by a transversal l

The angles formed between lines m and l are ∠1, ∠2, ∠4, and ∠3, while the corresponding angles formed between lines n and l are ∠5, ∠6, ∠8, and ∠7

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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