Which statement accurately describes how to reflect point a (−2, 1) over the x‐axis? construct a line from a perpendicular to the x-axis, determine the distance from a to the x-axis along this perpendicular line, find a new point on the other side of the x-axis that is equidistant from the x-axis. construct a line from a parallel to the x-axis, determine the distance from a to the x-axis along this parallel line, find a new point on the other side of the x-axis that is equidistant from the x-axis. construct a line from a perpendicular to the y-axis, determine the distance from a to the y-axis along this perpendicular line, find a new point on the other side of the y-axis that is equidistant from the y-axis. to reflect a over the x-axis you must find the intersection of a line perpendicular to the x-axis through point a, that will act as a bisector. this option is incorrect.#^ construct a line from a parallel to the y-axis, determine the distance from a to the y-axis along this parallel line, find a new point on the other side of the y-axis that is equidistant from the y-axis.

Construct a line from A perpendicular to the x-axis, determine the distance from A to the x-axis along this perpendicular line, find a new point on the other side of the x-axis that is equidistant from the x-axis

Step-by-step explanation:

we know that

The rule of the reflection of a point across the x-axis is equal to

(x,y) > (x,-y)

That means > The reflected point will be located equidistant from the x-axis

so

- Construct a line from A perpendicular to the x-axis

- Determine the distance from A to the x-axis along this perpendicular line

- Find a new point on the other side of the x-axis that is equidistant from the x-axis

therefore

A(-2,1) > A'(-2,-1)

The reflection point is A'

The distance of point A to the x-axis is equal to 1 unit

The distance from point A' to the x-axis is also 1 unit

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0