Question

reflect △efg over the y - axis, then that image over the x - axis. label the coordinates of e. rotate △cde 90° clockwise about the origin. label the coordinates of c.

Explanation:

Step1: Recall reflection rules

The rule for reflecting a point $(x,y)$ over the $y -$axis is $(-x,y)$. Let the coordinates of point $E$ be $(x_1,y_1)$. After reflecting $\triangle EFG$ over the $y -$axis, the coordinates of the new - point $E'$ of $E$ is $(-x_1,y_1)$.

Step2: Apply second - reflection rule

The rule for reflecting a point $(x,y)$ over the $x -$axis is $(x, - y)$. So, after reflecting the image (where $E'$ has coordinates $(-x_1,y_1)$) over the $x -$axis, the coordinates of $E''$ is $(-x_1,-y_1)$.
Assume the coordinates of point $E$ are $(-3,1)$.
After reflecting over the $y -$axis: $E'=(3,1)$
After reflecting $E'$ over the $x -$axis: $E''=(3, - 1)$

Answer:

If the original coordinates of $E$ are $(x,y)$, the coordinates of $E''$ are $(-x,-y)$. For example, if $E=(-3,1)$, then $E''=(3, - 1)$