The distance from the edge of the fountain to the centre is approximately 23.87 inches.
The water fountain forms a circle. The distance around the water fountain is the circumference of the circle formed.
Therefore,
circumference = 2πr
150 = 2πr
The distance from the edge of the fountain to the centre is the radius of the circle formed. Therefore,
75 = πr
r = 75 / 3.14159
r = 23.8732616287
r = 23.87 inches
The distance from the edge of the fountain to the centre is approximately 23.87 inches.
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Answer:
B. 45 tens + 623thousandths
Answer:
Connect the intersections of the diameters and the circle with a segment
Step-by-step explanation:
Assuming the construction creates two orthogonal diameters, their ends will be the vertices of the inscribed square. Connecting them with segments in order around the circle will create the square.
Best to take this in two parts.
1, find the factors of 36:
1 and 36
2 and 18
3 and 12
4 and 9
6 and 6
2, Now, using those, add the factors and see which add up to 15... in this case, that would be 12 and 3!
Answer:
37°
Step-by-step explanation:
∠ECF= ∠ACG
∠ACG= 90° - ∠GAC= 90° - 53°= 37°
