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>
Answer and Step-by-step explanation:
The actual answer is 348.
4 rectangles with area 60 each, which, when added together, equals 240.
1 square with area 36.
Each triangle face is 18 each, so 72.
<u>Add those together and the answer is 348 </u>
<u> :).</u>
<u></u>
<u></u>
<u><em>#teamtrees #PAW (Plant And Water)</em></u>
Answer:
10 < 18 + 7
10 < 25
True yyes
cuz it is above 10
Step-by-step explanation:
(3x-4y)^2=9x^2-24x+16y^2
(3x-2)(9x^2+6x+4)=27x^3-8