Question

A bridge on a river is modeled by the equation h = -0.2d2 + 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?

Answer 1

The answer is 2.2108
Solution:
first equation:
h = -0.2d² + 2.25d-d + 6h = 21.77
2nd equation:-d + 6h = 21.77d = 6h - 21.77
Then substitute:h = -0.2(6h - 21.77)² + 2.25(6h - 21.77)h = 3.9968
Answer:d = 2.2108

Answer 2

Answer:

Point A is approximately 4 unit above the ground.

Step-by-step explanation:

A bridge on a river is modeled by the equation h = -0.2d² + 2.25d -------(1)

Where h is the height and d is the horizontal distance.

For cleaning and maintenance purpose a worker wants to tie a taut a 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

d - 6h = -21.77

d = 6h - 21.77 -------(2)

Now we have to find the value of h which is the distance from the ground level.

Now we substitute d from equation 2 to equation 1

h = -0.2(6h - 21.77)²+ 2.25(6h - 21.77)

h = -0.2(36h²+ 473.9329 - 261.24h) + 13.5h - 48.9825

h = -(7.2h² + 94.79 - 52.25h) + 13.5h - 48.98

h = -7.2h² - 94.79 + 52.25h + 13.5h - 48.98

h = -7.2h² + 65.75h - 143.77

0 = -7.2h² + 65.75h - h - 143.77

0 = -7.2h² + 64.75h - 143.77

7.2h²- 64.75h + 143.77 = 0

Now we divide the equation by 7.2

h² - 9h + 20 = 0

h² - 5h - 4h + 20 = 0

h(h - 5) - 4(h - 5) = 0

(h - 4)(h - 5) = 0

h = 4 or 5

Therefore, point A is approximately 4 unit above the ground.