|
Question 1121534: Hello, I'm not sure if this question goes into this section, if it's wrong I apologize. I'm posting because I need help with this question these two questions.
The first question:
g) Vertical cables are used to join the suspension cable to the roadway. How long is the vertical cable that joins the suspension cable to the roadway at a point that is 6 m away from the first tower?
The second question:
f) How far apart are the two towers? Explain how you deduced your answer.
I'm not sure if I got the right answer, the axis of symmetry is 30. Does that mean that the two towers are 30m apart?
Here is the equation:
The height of a suspension cable above the roadway of a new suspension bridge, with two equally tall towers, is described by the
equation y = 0.05(x - 30)^2 + 6. In this equation, x represents the horizontal distance, in metres, from one tower to the other tower, and y represents the height, in metres, of the cable above the roadway.
Any help would be appreciated, thank you so much.
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! I believe you mean x is the horizontal distance from the first tower
to a point on the cable (the distance between the two towers is fixed)
To answer the 1st question, find the value of y for x = 6:
y = 0.05(6 - 30)^2 + 6 -> y = 34.8 m
You are correct that the function is symmetric about x = 30, but the function
is a minimum at x = 30.
Thus the other tower is 30 m away at x = 60, so they are 60 m apart.
|
|
|
| |
