Suppose that the available coins are in the denominations that are powers of cc, i. e., the denominations are c^0, c^1, \ldots, c^kc 0 ,c 1 ,…,c k for some integers c > 1c>1 and k \ge 1k≥1. Show that the greedy algorithm always yields an optimal solution.

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'.

14.3

step-by-step explanation:

add all the numbers together.

9 + 12 + 15 + 16 + 17 + 17 = 86

now, divide 86 by how many numbers we have: 6.

86 / 6 = 14.3