Answer:
44.452
Step-by-step explanation:
Using the Pythagorean theorem you can use:
51^2 = 25^2 + w^2
2601 = 625 +w^2
w^2 = 1976
w = 44.452
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>
I believe it should be 36.00$
Since the court is 9 meters wide we would have to multiply that by 4 because each meter is 4$
The sum of 15 and r.
Sum is the term used in verbal addition problems.
This is an expression therefore, you do not state what it equals. We no this because there is no equals sign.
Answer:
option d. is the correct answer