Algebra
posted by Erin .
Determine the approximate maximum distance
y=0.01x2 +0.7x+6.1

100 y = x^2 70 x  610
x^2  70 x = 100 y + 610
x^2  70 x + 1225 = 100y + 1835
(x35)^2 = 100 y + 1835
vertex at x = 35
then y =18.35
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