Answer:
1. see attached diagram
2. The dimensions are 75 sqrt(2) by 75 sqrt(2) ft
3. The perimeter is 300 sqrt(2) ft
4. cannot fit 40 machines along the walls
Step-by-step explanation:
We are using a top down view. We have a square building with side length s. The diagonal is 150 ft
Using the Pythagorean theorem
s^2 + s^2 = 150^2
2s^2 = 22500
s^2 =11250
Taking the square root of each side
sqrt(s^2) = sqrt(11250)
s = sqrt(6225*2)
s = 75 sqrt(2)
The dimensions are 75 sqrt(2) by 75 sqrt(2) ft
The perimeter of a square is given by
P = 4s = 4(75 sqrt(2)) =300 sqrt(2) ft
The perimeter is 300 sqrt(2) ft
Assuming the machines are square ( that they are as wide as they are long) 75 sqrt(2) is approximately 106 ft so you can fit 10 along the wall
Putting them along the top and bottom walls = 20 machines
We are only left with 86 ft along the side walls
86/10 = 8 machines
machines * 2 walls = 16
We can fit 20+16 machines = 36 machines not 40
Answer:
Perimeter: 119.11 units length
Step-by-step explanation:
Assuming that EC is the length of one side of the square and one of the legs (l) of the isosceles right triangle, then the hypotenuse (h) of the right triangle is:
h² = 2*l²
h = √(2*22²)
h = 22√2 units length
The perimeter of the figure is the addition of 3 sides of the square, one leg of the triangle and the hypotenuse of the triangle, that is: 4*22 + 22√2 = 22*(4 + √2) ≈ 119.11 units length
Answer:
associative property
Step-by-step explanation:
Your answer is 41.93529593 or 41.9
Answer:
I could be wrong, but I believe that the answer is D, none. This is because there is only proof of one congruent side and angle.
Step-by-step explanation: