(1, -1, 3)
4x + 2y + 3z = 11
x - 2y + z = 6
2x + y + 2z = 7

What is a possible solutions of the x> 3?

The width of a rectangle is 20 units. can the perimeter p of the rectangle be 64 units when its length x is 11 units? no, because p = 2x + 2(20) no, because p ≠ 2x + 2(20) yes, because p ≠ 2x + 2(20) yes, because p = 2x + 2(20)

Set $r$ is a set of rectangles such that (1) only the grid points shown here are used as vertices, (2) all sides are vertical or horizontal and (3) no two rectangles in the set are congruent. if $r$ contains the maximum possible number of rectangles given these conditions, what fraction of the rectangles in set $r$ are squares? express your answer as a common fraction.