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>
So each dimention is 32 by 8 by 10
so the surface area=2 times(8 times 10)+(2 times (32 times 10))+(2 times (8 times 32))=2 times 80+2 times 320+2 times 256=1312
if halved
just divide each by 2
(2 times 4 times 5)+(2 times 16 times 5)+(2 times 4 times 16)=(2 times 20)+(2 times 80)+(2 times 64)=40+160+128=328
the new surface area is 328 mm^2
Can the numbers be repeted!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
<u>The original three-digit number is 417</u>
Step-by-step explanation:
Let's find out the solution to this problem, this way:
x = the two digits that are not 7
Original number = 10x+7
The value of the shifted number = 700 + x
Difference between the shifted number and the original number = 324
Therefore, we have:
324 = (700 + x) - (10x + 7)
324 = 700 + x - 10x - 7
9x = 693 - 324 (Like terms)
9x = 369
x = 369/9
x = 41
<u>The original three-digit number is 417</u>
If i were you, i would add one to 7x and fifteen. this gives you x^2+8x+16, a perfect square of x+4.