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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B. Samuel will pay will 10 dollars less if he buys the boots from store A.
Eq1) 2r+2s=50
eq2) 2r-s=17
solve for s in equation2 (eq2)
-s=17-2r
s=-17+2r
Substitute s into equation1 (eq1)
2r+2(-17+2r)=50
2r-34+4r=50
6r-34=50
6r=50+34
6r=84
r=14
Substitute into either equation and solve for s
2(14)-s=17
28-s=17
-s=17-28
-s=-11
s=11
Answer:
−4304
Step-by-step explanation:
1. The given determinant is :

We need to find its determinant . It can be solved as follows :

So, the value of determinant is equal to −4304.
The final sale price would be $662.35