Given that line segments are taken to line segments of the same length during rigid transformations, which transformation maps the line segment ab onto itself? a) rotation counterclockwise of 90° → x-axis reflection → rotation counterclockwise of 270° → x-axis reflection b) rotation counterclockwise of 90° → y-axis reflection → rotation counterclockwise of 270° → y-axis reflection c) rotation counterclockwise of 90° → x-axis reflection → rotation counterclockwise of 270° → y-axis reflection d) rotation counterclockwise of 180° → x-axis reflection → rotation counterclockwise of 270° → y-axis reflection

Correct choice is C

Step-by-step explanation:

Points A and B have coordinates (1,-2) and (4,3), respectively.

1. Rotation counterclockwise of 90° about the origin has a rule:

(x,y)→(-y,x).

Then the image of point A is point A'(2,1) and the image of point B is point B'(-3,4).

2. Refection about the x-axis has a rule:

(x,y)→(x,-y).

Then the image of point A' is point A''(2,-1) and the image of point B' is point B''(-3,-4).

3. Rotation counterclockwise of 270° about the origin has a rule:

(x,y)→(y,-x).

Then the image of point A'' is point A'''(-1,-2) and the image of point B'' is point B'''(-4,3).

4.  Refection about the y-axis has a rule:

(x,y)→(-x,y).

Then the image of point A''' is point A(1,-2) and the image of point B''' is point B(4,3).

the correct answer is a.) bc< mn 5 stars if this you : )

step-by-step explanation:

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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