Abridge on a river is modeled by the equation h = -0.2d^2 + 2.25d, where h is the height and d is the horizontal distance. for cleaning and maintenance purposes a worker wants to tie a taut rope on two ends of the bridge so that he can slide on the rope. the rope is at an angle defined by the equation -d + 6h = 21.77. if the rope is attached to the bridge at points a and b, such that point b is at a higher level than point a, at what distance from the ground level is point a?

We are given the equations
h = -0.2d² + 2.25d
-d + 6h = 21.77

Solving for d in the second equation:
-d + 6h = 21.77
d = 6h - 21.77
Substituting to the first equation:
h = -0.2(6h - 21.77)² + 2.25(6h - 21.77)
The value of h can then be solved with is
h = 3.9968
so
d = 2.2108

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12