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.
</span>
- 16:10
- 24:15
- 40:25
8:5 is just a simplier version of these other ratios
Answer:
-11x=-66
x=6
Explanation:
-11x-12=-78
1) Add 12 to both sides.
-11x=-78+12
-11x=-66
2) Divide both sides by -11.
x=-66/-11
x=6
Hope this helps! Please leave a thanks and brainliest answer if you can :)
No 8 is the sum (not factor)of the factors 2 and 4