What set of reflections would carry hexagon abcdef onto itself? hexagon abcdef on the coordinate plane with point a at 0, 1, point b at negative 1, 0, point c at negative 2, 1, point d at negative 2, 3, point e at negative 1, 4, and point f at 0, 3. a. x-axis, y=x, x-axis, y=x b. y-axis, x-axis, y-axis c. x-axis, y-axis, y-axis d. y=x, x-axis, y=x, y-axis

When a point is reflected, it must be reflected across a line. The reflection of ABCDEF that carries it onto itself is:

D. y=x, x-axis, y=x, y-axis

Of the given options, (D) is correct.

The proof is as follows:

Using coordinate point A.

We have:

The first reflection in (D) is

This means that:

So, we have:

The next is across the x-axis.

The rule of this reflection is:

So, we have:

The next is

This means that:

So, we have:

The last is across the y-axis.

The rule of this reflection is:

So, we have:

Compare the end point to point A;

We can see that both points are the same i.e. (0,1)

Hence, the set of reflections would carry hexagon ABCDEF onto itself is (d)

Read more about reflections at:

link

You need an even number of reflections to get the original figure, leaving only A and D as potentially viable choices.

The reflection across the x-axis makes the tranformation (x, y) ⇒ (x, -y).
The reflection across the line y=x makse the transformation (x, y) ⇒ (y, x).
The two pairs of transformations of A give
  (x, y) ⇒ (x, -y) ⇒ (-y, x) ⇒ (-y, -x) ⇒ (-x, -y) . . . . . not the original

The reflection across the y-axis makes the transformation (x, y) ⇒ (-x, y).
The two pairs of transformations of D give
  (x, y) ⇒ (y, x) ⇒ (y, -x) ⇒ (-x, y) ⇒ (x, y) . . . . . . the original point

Selection D is appropriate.

42 (the answer to life the universe and everything)

Step-by-step explanation: do more of what makes you happy <3 and be positive!