Question

Please do help me answer this! >.< This is DUE TOMORROW AND WORTH 10 POINTS!

y = -0.007x² + 0.84 + 0.8
Where the y represents the distance above water in feet and x represents the distance in feet between the arch supports, at the point where each arch supports enters the water.

If the distance in feet between the arch supports is 60 feet,what is the height above water level of the arch support?

Answer 1

Okay, the equation is y = -.007x^2 + 0.84 + 0.8 where y is the distance above the water and x is the distance between the supports, both measurements in imperial feet.
If the distance between the supports is 60 ft, then the equation is:
$y=-.007x^{2}+.84+.8\\ y=-.007(60)^{2}+.84+.8=-23.56\ feet$

Now since you can't practically have an arch 23.56 feet under the water, I think you forgot an x after the .84 in the original equation.
Using the added x, the equation works out to be:
$y=-.007x^{2}+.84x+.8\\ y=-.007(60)^{2}+.84(60)+.8=26\ feet$
Your answer should be 26 feet above the water.

Hope I helped :)