Given that ABCD is a parallelogram. What must be true? What may be true? What can’t be true? ∠C is supplementary to both ∠B and ∠D ∠A is supplementary to both ∠B and ∠D AB¯¯¯¯¯¯¯¯⊥AD¯¯¯¯¯¯¯¯ AD¯¯¯¯¯¯¯¯≅BC¯¯¯¯¯¯¯¯ AD¯¯¯¯¯¯¯¯∥BC¯¯¯¯¯¯¯¯ AB¯¯¯¯¯¯¯¯≅AD¯¯¯¯¯¯¯¯ AC¯¯¯¯¯¯¯¯ bisects BD¯¯¯¯¯¯¯¯ AB¯¯¯¯¯¯¯¯∥CD¯¯¯¯¯¯¯¯ ∠A≅∠C AB¯¯¯¯¯¯¯¯≅CD¯¯¯¯¯¯¯¯ ABCD is a square BD¯¯¯¯¯¯¯¯ bisects AC¯¯¯¯¯¯¯¯ ABCD is a kite ABCD is a rhombus m∠A=90° ∠B≅∠D AB¯¯¯¯¯¯¯¯ intersects CD¯¯¯¯¯¯¯¯ AD¯¯¯¯¯¯¯¯ intersects BC¯¯¯¯¯¯¯¯ AC¯¯¯¯¯¯¯¯≅BD¯¯¯¯¯¯¯¯ ABCD is a rectangle ∠A≅∠B English

no i hate broccoli i like nothing

step-by-step explanation:

the graph on the bottom right

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'.

if a = 1,   and b = 2, then the answer is 47.99999999. if you round that to the nearest whole number, the answer is 48.