Elenathehomeschooler
  • Elenathehomeschooler
Can someone help! A handy man knows from experience that his 29-foot ladder rests in its most stable position when the distance of its base from a wall is 1 foot farther than the height it reaches up the wall. 1. Draw a diagram to represent this situation. Be sure to label all unknown values in the problem. 2. Write an equation that you can use to find the unknown lengths in this situation. This equation should come directly from the labels you used in your diagram. 3. How far up a wall does this ladder reach? Show your work.
Mathematics
chestercat
  • chestercat
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Vocaloid
  • Vocaloid
|dw:1444344809146:dw|
Vocaloid
  • Vocaloid
use the pythagorean theorem to find x
31356
  • 31356
Just a reminder, the Pythagorean theorem is a^2+b^2=c^2

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Elenathehomeschooler
  • Elenathehomeschooler
@31356 can you help me all i have so far is |dw:1444345667230:dw| im stuck
Elenathehomeschooler
  • Elenathehomeschooler
@jim_thompson5910 can you help me
jim_thompson5910
  • jim_thompson5910
|dw:1444346208997:dw| \[\Large a^2 + b^2 = c^2\] \[\Large x^2 + (x+1)^2 = 29^2\] \[\Large x^2 + \color{red}{(x+1)^2} = 841\] \[\Large x^2 + \color{red}{(x+1)(x+1)} = 841\] \[\Large x^2 + \color{red}{x^2+2x+1} = 841\] \[\Large 2x^2+2x+1 = 841\] Now get everything to one side and use the quadratic formula to find x
Elenathehomeschooler
  • Elenathehomeschooler
so put |dw:1444346934670:dw| in thr quadratic formula @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
|dw:1444347129681:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1444347161312:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1444347184674:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1444347206109:dw|
Elenathehomeschooler
  • Elenathehomeschooler
okay i got x=-22.5 or x=76 did i get it right? @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
both are incorrect
jim_thompson5910
  • jim_thompson5910
2x^2 + 2x - 840 = 0 a = 2 b = 2 c = -840 \[\Large x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\] \[\Large x=\frac{-2 \pm \sqrt{2^2-4*2*(-840)}}{2*2}\] \[\Large x=\frac{-2 \pm \sqrt{6724}}{4}\] \[\Large x=\frac{-2 \pm 82}{4}\] \[\Large x=\frac{-2 + 82}{4} \ \text{ or } \ x=\frac{-2 - 82}{4}\] \[\Large x=\frac{80}{4} \ \text{ or } \ x=\frac{-84}{4}\] \[\Large x=20 \ \text{ or } \ x=-21\]
jim_thompson5910
  • jim_thompson5910
go back to Vocaloid's drawing notice how x is a height or distance we can't have negative distance, so x = -21 makes no sense. The only practical answer is x = 20
Elenathehomeschooler
  • Elenathehomeschooler
oh okay i noticed what i did wrong
Elenathehomeschooler
  • Elenathehomeschooler
for number 3 the ladder will reach 29ft. on the wall right @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
29 isn't the vertical distance 29 is the length of the ladder
jim_thompson5910
  • jim_thompson5910
again, look back at Vocaloid's drawing

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