anonymous
  • anonymous
4200 ft. apart with height of 546 ft above roadway...cable is 15ft above roadway...I have vertex at (0,15) and (2100, 546) , (-2100, 546) but can't get equation to work
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
gonna have to chase the question am i...... stay put! lol
amistre64
  • amistre64
those are some awfully big number, this could get messy :)
anonymous
  • anonymous
yes I've worked it out several times

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amistre64
  • amistre64
your values are good... (2100^2)a +(2100)b + c = 546 (-2100^2)a -(2100)b + c = 546 0a +0b + c = 15 we see the c = 15 here right?
amistre64
  • amistre64
(2100^2)a +(2100)b + c-c = 546 -15 (2100^2)a +(2100)b = 531 (-2100^2)a -(2100)b + c-c = 546-15 (2100^2)a -(2100)b = 531 --------------------------- 531 -2100b a = ----------- (2100)^2 substitute this into the next equation right? (2100^2)a +(2100)b = 531 ------------------------------
amistre64
  • amistre64
(2100^2)(531 -2100b) -------------------- -(2100)b = 531 (2100)^2 531 -2100b - 2100b = 531 it says b = 0 :)
amistre64
  • amistre64
which is what you would expect it to be since we are on the y axis already and just raised by 15 right?
anonymous
  • anonymous
so we are using y-y1=c(x-x1)2 yes, y-intercept is (0, 15)
amistre64
  • amistre64
531 a = ---------- (2100)^2 \[y = \frac{531}{2100^2} x^2 + 15\]
amistre64
  • amistre64
see if this fits :)
amistre64
  • amistre64
when x = 2100 we get: 531 + 15 = 546 right?
amistre64
  • amistre64
come on..you know im right....tell me i did good :)
anonymous
  • anonymous
I was squaring 2100 and got lost in math... YES you're right...AWESOME!!!! THANKS ALOT!
anonymous
  • anonymous
I was comparing my work to yours...all different attempts , lol
amistre64
  • amistre64
lol...... thanx :) you had it tight to begin with, dont let the numbers scare you ;)
anonymous
  • anonymous
Thanks!
anonymous
  • anonymous
you have time for another

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