if an archway is20 feet high and 50 feet wide, and a 14 foot high and 10 foot wife truck drive under it without going into other lane,,, please show me how to work this out! correct answer will get 10 points thanks!
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(-25, 0), (25, 0), (0, 20) are points on the parabola. Plug the coordinates into the equation y = ax² + bx + c to get the system of equations:
625a - 25b + c = 0
625a + 25b + c = 0
c = 20
Solve the system to get a = -0.032, b = 0, c = 20. (It's too long to solve here)
The equation for the archway is y = -0.032x² + 20.
Let y = 14 and solve for x:
14 = -0.032x² + 20
0.032x² = 6
x² = 187.5
x = 13.69306394
13.69306394 - 10 = 3.69306394
So, as long as the wife keeps her left wheels less than 3.69306394 feet to the right of the dividing line, she will clear the arch.
http://www.flickr.com/photos/dwread/6386…
(-25, 0), (25, 0), (0, 20) are points on the parabola. Plug the coordinates into the equation y = ax² + bx + c to get the system of equations:
625a - 25b + c = 0
625a + 25b + c = 0
c = 20
Solve the system to get a = -0.032, b = 0, c = 20. (It's too long to solve here)
The equation for the archway is y = -0.032x² + 20.
Let y = 14 and solve for x:
14 = -0.032x² + 20
0.032x² = 6
x² = 187.5
x = 13.69306394
13.69306394 - 10 = 3.69306394
So, as long as the wife keeps her left wheels less than 3.69306394 feet to the right of the dividing line, she will clear the arch.
http://www.flickr.com/photos/dwread/6386…