Square efgh has vertices: e = (-1,1), f = (-1,2), g = (-2,1), h = (-2, 2). in which quadrant will the image of vertex e' be if the square is rotated 270° clockwise about the origin o?

If we were to imagine a single point (1,1) which is in quadrant 1, and we rotate it 90o then we get a negative x. Another 90o then both x and y is negative. another 90 (270) and we get a negative y.

thus:

(e)(f)(g)(h) = (-1,-1)(-1,-2)(-2,-1)(-2,-2)

38 dollars is your answer.