The cup, on the 9th hole of a golf course, is located, dead center, in the middle of a circular green, which is 35 feet in radius. the ball follows a straight-line path, and exits the green at the right-most edge. assume the ball travels 11 ft/sec. introduce coordinates, so that the cup is the origin of an xy-coordinate system. provide numerical answers below, with two decimal places of accuracy. ball starts at (-40,-50). suppose that l is a line, tangent to the boundary of the golf green, and parallel to the path of the ball. let q be the point where the line is tangent to the circle. notice that there are two possible positions for q. find the possible x-coordinates of q.

± 27.33 ft

Step-by-step explanation:

For the given problem, we can estimate the initial and final coordinates of the line of the ball path as (-40,-50) and (0,0). Therefore, the slope is:

(-50-0)/(-40-0) = 50/40 = 1.25

Similarly, we can estimate the slope of a perpendicular line to the line of the ball path as: -1*(1/1.25) = -0.8.

Therefore, using (0,0) and the slope -0.8, the equation of the perpendicular line is: -0.8 = (y-0)/(x-0);

-0.8 = y/x

y = -0.8x

Furthermore, we are given the circle radius as 35 ft and we can use the distance formula to find the two points 35 ft far from the origin:

35^2 = x^2 + y^2

y = -0.8x

35^2 = x^2 + (-0.8x)^2

1225 = (x^2 + 0.64x^2)

1225 = 1.64x^2

x^2 = 1225/1.64 = 746.95

x = sqrt(746.95) = ± 27.33 ft

i think the answer is seven, for if i'm being useless °ω°

step-by-step explanation: