Answer:
Hexagon AGHBGHCGHDGHEGHFGH coincides with hexagon ABCDEF when GH passes through the midpoints of opposite sides; that is, it is a perpendicular bisector of the two sides. HexagonAGHBGHCGHDGHEGHFGH also coincides with hexagon ABCDEF when the line of reflection joins a pair of vertices opposite one another on the hexagon. There are three perpendicular bisectors and three pairs of opposite vertices. In all, there are six lines of reflection that will map the hexagon back onto itself.
Step-by-step explanation:
There are two cases to consider.
A) The removed square is in an odd-numbered column (and row). In this case, the board is divided by that column and row into parts with an even number of columns, which can always be tiled by dominos, and the column the square is in, which has an even number of remaining squares that can also be tiled by dominos.
B) The removed square is in an even-numbered column (and row). In this case, the top row to the left of that column (including that column) can be tiled by dominos, as can the bottom row to the right of that column (including that column). The remaining untiled sections of the board have even numbers of rows, so can be tiled by dominos.
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Perhaps the shorter answer is that in an odd-sized board, the corner squares are the ones that there is one of in excess. Cutting out one that is of that color leaves an even number of squares, and equal numbers of each color. Such a board seems like it <em>ought</em> to be able to be tiled by dominos, but the above shows there is actually an algorithm for doing so.
Answer:
$37.50
Step-by-step explanation:
$125 times30%=$37.50
Answer:
C = (2,2)
Step-by-step explanation:
B = (10 ; 2)
M = (6 ; 2)
C = (x ; y )
|___________|___________|
B (10;2) M (6;2) C ( x; y)
So:
dBM = dMC
√[(2-2)^2 + (6-10)^2] = √[(y-2)^2 + (x - 6)^2]
(2-2)^2 - (6-10)^2 = (y-2)^2 + (x - 6)^2
0 + (-4)^2 = (y-2)^2 + (x - 6)^2
16 = (y-2)^2 + (x - 6)^2
16 - (x - 6)^2 = (y-2)^2
Also:
2*dBM = dBC
2*√[(2-2)^2 + (6-10)^2] = √[(y-2)^2 + (x - 10)^2]
4*[(0)^2 + (-4)^2] = (y-2)^2 + (x - 10)^2
4*(16) = (y-2)^2 + (x - 10)^2
64 = (y-2)^2 + (x - 10)^2
64 = 16 - (x - 6)^2 + (x - 10)^2
48 = (x - 10)^2 - (x - 6)^2
48 = x^2 - 20*x + 100 - x^2 + 12*x - 36
48 = - 20*x + 100 + 12*x - 36
8*x = 16
x = 2
Thus:
16 - (x - 6)^2 = (y-2)^2
16 - (2 - 6)^2 = (y-2)^2
16 - (-4)^2 = (y-2)^2
16 - 16 = (y-2)^2
0 = (y-2)^2
0 = y - 2
2 = y
⇒ C = (2,2)