SOLUTION: two transmission towers 40 ft high is 200 feet apart. if the lowest point of the cale is 10 ft above the ground, what is the vertical distance from the roadway to the cable 50 ft f

Question 829405: two transmission towers 40 ft high is 200 feet apart. if the lowest point of the cale is 10 ft above the ground, what is the vertical distance from the roadway to the cable 50 ft from the center
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
two transmission towers 40 ft high is 200 feet apart. if the lowest point of the cable is 10 ft above the ground, what is the vertical distance from the roadway to the cable 50 ft from the center
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Assuming this is a parabola,
Using ax^2 + bx + c = y form, find a, b, c
:
assign 3 ea x,y coordinates from the given info
0,40; 100,10; 200,40
:
Using the 1st coordinate, we know that c=40, find a & b using 2nd and 3rd
x=100; y=10
100^2a + 100b + 40 = 10
10000a + 100b = 10 - 40
10000a + 100b = -30
:
x=200; y=40
200^2a + 200b + 40 = 40
40000a + 200b = 40 - 40
40000a + 200b = 0
Use elimination here, multiply the 1st equation by 2, subtract from above
40000a + 200b = 0
20000a + 200b = -60
--------------------subtraction eliminate b, find a
20000a = 60
a = 60/20000
a = .003
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Use 10000a + 100b = -30, to find b, replace a with .003
10000(.003) + 100b = -30
30 + 100b = -30
100b = -30 - 30
100b = -60
b = -60/100
b = -.6
Our equation: y = .003x^2 - .6x + 40
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looks like this, green line ref 40 ft above ground

"what is the vertical distance from the roadway to the cable 50 ft from the center?"
Those points would be x=50 and x=150
Find y when x=50
y = .003(50^2) -.6(50) + 40
y = 7.5 - 30 + 40
y = 17.5 is the height 50 ft from the center (purple ref line)
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