Answer:
B i worked it out in my head, sorry if i dont get it right
Part A:
Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4
The perimeter of the square is given by 4(x + 4) = 4x + 16
The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12
For the perimeters to be the same
4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2
The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.
Part B:
The area of the square is given by

The area of the rectangle is given by 2(3x + 4) = 6x + 8
For the areas to be the same

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
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Look at the back for answers ( my workbook has answers)
real numbers hope it helps
Okay, so first, the quadrants on a coordinate plane go counter clockwise.
2 | 1
__________|__________
3 | 4
(1/2, -1.8) Would be a little to the right of the y axis, and a little below the x axis.
So, your answer would be D) Quadrant IV