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
The triangle are congruent.
The one is 5 times longer on the base therefore the height is 5 times higher,
5x2 = 10m
Answer:
units
Step-by-step explanation:
I had this question on flocabulary.
Think: (distance traveled by faster car) - (distance traveled by slower car) = 35 mi.
Using the formula distance = rate times time, we get for the faster car:
(55 mph)(t)
and for the slower car we get (50 mph)(t), so that
(55 mph)(t) - (50 mph)(t) = 35 mi. This can be simplified as follows:
(5 mph)(t) = 35 mi. Dividing both sides by 5 mph, we get t = (35 mi) / (5 mph).
Then t = 7 hours. The two cars will be 35 miles apart, going in the same direction, after 7 hours.