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>
The answer is D
Explanation 4 ones is 4
Answer:
60
Step-by-step explanation:
360/6=60
To solve this problem you must apply the proccedure shown below:
1. You have the following information given in the problem above:
- The<span> room has square dimensions and it has been built with two pieces of sheetrock, a smaller one and a larger one.
2. Therefore, let's call
x: the smaller one.
y: the larger one.
3. Then, you have that the lenght of the wall is the sum of the smaller one and the larger one:
x+y
4. So, the area of the room is:
(x+y)(x+y)
(x+y)</span>²
Therefore, the answer is: (x+y)²
If you simplify what's in the bracket, you get 3n
=(3n)+4=32/2
=3n+4= 32/2(here we can either crossmultiply or equate)
3n+4-4=32/2-4(subtract 4 from both sides)
3n=32/2-4
3n=24/2
3n=12
3n=12(divide both sides by the coefficient of the unknown)
3n/3=12/3
n=4