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Mathematics 11 Online
OpenStudy (anonymous):

How do you find the height of the arch?? A bridge is built in the shape of a semiplliptical arch. The bridge has a span of 80 feet and a mas height of 20 feet. Choose a suitable rectangular coordinate system and find the height of the arch at a distance of 30 feet from the center. The height is about ______ feet (round to two decimal places)

OpenStudy (amistre64):

draw it out

OpenStudy (amistre64):

you want x=0, y=20 y=0, x=80/2 then what is the value of that when x=20?

OpenStudy (amistre64):

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OpenStudy (anonymous):

what do you mean x=20?

OpenStudy (anonymous):

@amistre64

OpenStudy (amistre64):

i meant x=30, just hit the wrong key on the keyboard.

OpenStudy (amistre64):

\[\frac{x^2}{p^2}+\frac{y^2}{q^2}=1\] \[\frac{y^2}{q^2}=1-\frac{x^2}{p^2}\] \[y^2=q^2-\frac{(qx)^2}{p^2}\]

OpenStudy (amistre64):

q=20, p=40, and x=30 if i se it right

OpenStudy (anonymous):

so you are trying ot find for y? since 30 is x

OpenStudy (amistre64):

yes

OpenStudy (anonymous):

so what am I suppose to do with y? wait thats the height?

OpenStudy (amistre64):

lol, yes. that will be the height at any value of x

OpenStudy (amistre64):

y^2 = some # y = sqrt(some #)

OpenStudy (amistre64):

when x=p, we have q^2 = q^2 = 0 which is what we would expect at a distance of p from the center.

OpenStudy (anonymous):

so would it be 18.97 for the height of the arch

OpenStudy (amistre64):

maybe, let me check my calculator ....

OpenStudy (amistre64):

i get something closer to 13

OpenStudy (amistre64):

\[y=\sqrt{20^2-\frac{(20*30)^2}{40^2}}\]

OpenStudy (anonymous):

okay thank you for the correcting me!

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