Algebra
posted by mv .
h=7/5dd^2/100 where h is the height of the arch above the road in meters.
Not actually looking to have all completed, but give advice on how to achieve the results
a. where does the arch and road come together?
b.How far is it from where the arch begins on the left and where it ends on the right end of the bridge?
c.If you are onefourth of the way along the bridge how high above you is the arch?
d.How far along the bridge do you need to walk from the left end to get to apoint where the arch is 20m above the road?
Is this the correct approach to start?
h=7/5dd^2/100
clear fractios by multiplying all by 500
0=700d5d^2
5d^2700d=0
5d(d140)=0

ok so far
so if 5d(d140) = 0
d = 0 or d = 140
(visualize the bridge starting at d=0 and going to d=140
so it is 140 m wide at the base
c) if you are 1/4 way along the road, d = 35
h = (7/5)(35))  35^2/100 = 36.75
d) to have h = 20
(7/5)d  d^2/100 = 20
multiply by 100
140d  d^2 = 2000
d^2  140d + 2000 = 0
d = 123.85 or d = 16.148
Notice you would be 20 m below the bridge at both ends of the bridge.
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