An engineer designed a suspension bridge with three hanging cables, with the end of each cable attached to a tower. He defined
cable AB by the function f(x) = 0.1(x+11)^2+5 The bridge has a clearance of 24ft above the water's surface.
Knowing that the width of each tower is 2 ft, what is the height of each tower?
(Assume that the tower at point B is centered on the y-axis and that the cables are attached to the ends of the tower.)
Knowing that the width of each tower is 2 ft, what is the height of each tower?
(Assume that the tower at point B is centered on the y-axis and that the cables are attached to the ends of the tower.)
1 answer:
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Answer:
0.05 seconds; 2.575 m off the ground
Step-by-step explanation:
This is a system of equations that you need to solve for the variables t and h.
h = -5t² + 32t + 2
h = 31.5t + 1
Since h will be the same when the bullet and skeet collide, then h is the same in both equations and can be set equal to each other and then you can solve for t.
-5t² + 32t + 2 = 31.5t + 1
-5t² + 32t + 2 - 31.5t - 1= 31.5t + 1 - 31.5t -1 (move everything to the left side)
-5t² + 0.5t + 1 = 0 (combine like terms)
Now we can use the quadratic formula to solve for the time.
t = (substitute)
t = (multiply)
t = (add)
t = (square root)
t = -0.5/-10 or -9.5/-10 (combine like terms in each situation)
t = 0.05 or .95 (divide)
Since it will only hit it once, the first time it hits is the amount of time, therefore it takes .05 seconds to hit the skeet.
Next, substitute t into one of the equations to find the height at which it occurs.
h = 31.5(0.05) + 1 (substitute)
h = 1.575 + 1 (multiply)
h = 2.575 (add)
Therefore the skeet will be 2.575 meters off the ground.
Here is how to find the intercepts, if the graph is hovering above the x axis (k<0) there will be no real solutions for the x intercept
