#### Answer

a) $20 < 2x +16 < 30$
b) $ 2 < x < 7$, please see the first graph.
c) $2/3 < x < 10$, please see the second graph.

#### Work Step by Step

a) Two of the sides of the rectangle have lengths of (x+2) and 6. Since these sides are not parallel, the other two sides have sides of the same length.
$20< (x+2) + (x+2) + 6 + 6 < 30$
$20 < 2x +16 < 30$
b)
$20 < 2x +16 < 30$
$20 -16 < 2x + 16 - 16 < 30-16$
$4 < 2x < 14$
$4/2 < 2x/2 < 14/2$
$2 < x < 7$
c)
possible pairs of triangle legs: (6, x+4), (6, 2x), (2x,x+4)
$6 + x+4 > 2x$
$10 + x > 2x$
$10 + x - x > 2x - x $
$10 > x
$6 + 2x > x + 4$
$6 + 2x - x - 4 > x + 4 - x - 4$
$2 + x > 0$
$2 + x - 2 > 0 - 2$
$x > -2 $
$x+4 + 2x > 6$
$3x + 4 > 6$
$3x + 4 - 4 > 6 - 4 $
$3x > 2 $
$3x/3 > 2/3$
$x > 2/3$
If we let x = 11, we would not have a triangle as the lengths would be 6, 15, and 22 (and 6 + 15 < 22). Thus, the solution set is 2/3 < x < 10.