Question 1204076
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Let the origin be at the center of the base of the arch.  The vertex is then (0,32) and the equation is of the form<br>
{{{y=ax^2+32}}}<br>
Determine the coefficient a using the fact that the feet of the arch are at (-128,0) and (128,0).<br>
{{{0=x(128^2)+32}}}
{{{x=(-32)/(128^2)=-1/512}}}<br>
ANSWER (a): The equation of the arch is {{{y=(-1/512)x^2+32}}}<br>
To find the distance from the center where the height is 12, set y=12 in the equation and solve for x.<br>
{{{12=(-1/512)(x^2)+32}}}
{{{(1/512)(x^2)=20}}}
{{{x^2=20*512=10*1024=10(32^2)}}}
{{{x=32sqrt(10)}}}<br>
That is 101.2 meters, to the nearest tenth.<br>
ANSWER (b): approximately 101.2 meters<br>
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Note the second question answered by tutor @MathLover1 is not the question that was asked....<br>