Atransformation of adef results in ad'ef.
which transformation maps the pre-image to the image?
the transformation is a dilation.
the transformation is a reflection.
the transformation is a rotation.
the transformation is a translation.

  The transformation is a rotation.

Step-by-step explanation:

Corresponding sides of the image and pre-image are marked congruent, so there is no dilation involved. Dilation changes the lengths of corresponding segments.

The vertex order DEF is clockwise in both the image and pre-image, so there is no reflection involved. (Reflection reverses the vertex order to counterclockwise.)

Segment FD points to the northeast, but segment F'D' points to the west, so a rotation is involved.

With no coordinate axes, we cannot tell if translation is involved or not. (We presume some translation is required, but the primary transformation of interest here is the rotation.)

right

step-by-step explanation:

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

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