Which sequences of transformations performed on shape I show its congruency to shape II? a reflection across the x-axis, followed by a 90° clockwise rotation about the origin, and then a translation 5 units lefo a reflection across the x-axis, followed by a 90° counterclockwise rotation about the origin, and then a translation 5 units right a reflection across the y-axis, followed by a 90. clockwise rotation about the origin and then a translation 5 units left a reflection across the x-axis, followed by a 90° counterdockwise rotation about the origin, and then a translation 5 units left a reflection across the y-axis, followed by a 90° counterclockwise rotation about the origin, and then a translation 5 units left

11

step-by-step explanation:

32.50-7.50=25

225/25=11

idk

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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