Solve the problem.
-A Ferris wheel has a diameter of 280 feet and the bottom of the Ferris wheel is 12 feet above the ground. Find the equation of the wheel if the origin is placed on the ground directly below the center of the wheel, as illustrated.

A) $x^{2}+(y-140)^{2}=19,600$
B) $\mathrm{x}^{2}+\mathrm{y}^{2}=19,600$
C) $x^{2}+(y-140)^{2}=78,400$
D) $x^{2}+(y-152)^{2}=19,600$

Question

Solve the problem.
-A Ferris wheel has a diameter of 280 feet and the bottom of the Ferris wheel is 12 feet above the ground. Find the equation of the wheel if the origin is placed on the ground directly below the center of the wheel, as illustrated.

A) $x^{2}+(y-140)^{2}=19,600$ 
B) $\mathrm{x}^{2}+\mathrm{y}^{2}=19,600$ 
C) $x^{2}+(y-140)^{2}=78,400$ 
D) $x^{2}+(y-152)^{2}=19,600$

Quizplus · Accepted Answer

The Answer of Solve the problem.
-A Ferris wheel has a diameter of 280 feet and the bottom of the Ferris wheel is 12 feet above the ground. Find the equation of the wheel if the origin is placed on the ground directly below the center of the wheel, as illustrated.

A) $x^{2}+(y-140)^{2}=19,600$ 
B) $\mathrm{x}^{2}+\mathrm{y}^{2}=19,600$ 
C) $x^{2}+(y-140)^{2}=78,400$ 
D) $x^{2}+(y-152)^{2}=19,600$

Solve the Problem x2+(y−140)2=19,600x^{2}+(y-140)^{2}=19,600x2+(y−140)2=19,600 B) x2+y2=19,600\mathrm{x}^{2}+\mathrm{y}^{2}=19,600x2+y2=19,600 C) x2+(y−140)2=78,400x^{2}+(y-140)^{2}=78,400x2+(y−140)2=78,400 D) x2+(y−152)2=19,600x^{2}+(y-152)^{2}=19,600x2+(y−152)2=19,600

Solve the Problem $x^{2}+(y-140)^{2}=19,600$ B) $\mathrm{x}^{2}+\mathrm{y}^{2}=19,600$ C) $x^{2}+(y-140)^{2}=78,400$ D) $x^{2}+(y-152)^{2}=19,600$