Solve for x. Assume that lines which appear tangent are tangent.

Presumably, x is standing in for the question mark.

The pair of vertical angles (DFE and CFV) shown here both have measure 125°. This angle measure is equal to the average of the measures of the arcs intercepted by the chords CE and DV.

So we have

angle DFE = 1/2 (arc CV + arc DE)

125° = 1/2 (x + 200°)

250° = x + 200°

x = 50°

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

answer: 12

step-by-step explanation:

because yes