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Elenathehomeschooler

  • one year ago

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.

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  1. Vocaloid
    • one year ago
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    |dw:1444344809146:dw|

  2. Vocaloid
    • one year ago
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    use the pythagorean theorem to find x

  3. 31356
    • one year ago
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    Just a reminder, the Pythagorean theorem is a^2+b^2=c^2

  4. Elenathehomeschooler
    • one year ago
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    @31356 can you help me all i have so far is |dw:1444345667230:dw| im stuck

  5. Elenathehomeschooler
    • one year ago
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    @jim_thompson5910 can you help me

  6. jim_thompson5910
    • one year ago
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    |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

  7. Elenathehomeschooler
    • one year ago
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    so put |dw:1444346934670:dw| in thr quadratic formula @jim_thompson5910

  8. jim_thompson5910
    • one year ago
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    |dw:1444347129681:dw|

  9. jim_thompson5910
    • one year ago
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    |dw:1444347161312:dw|

  10. jim_thompson5910
    • one year ago
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    |dw:1444347184674:dw|

  11. jim_thompson5910
    • one year ago
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    |dw:1444347206109:dw|

  12. Elenathehomeschooler
    • one year ago
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    okay i got x=-22.5 or x=76 did i get it right? @jim_thompson5910

  13. jim_thompson5910
    • one year ago
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    both are incorrect

  14. jim_thompson5910
    • one year ago
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    2x^2 + 2x - 840 = 0 a = 2 b = 2 c = -840 \[\Large x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\] \[\Large x=\frac{-2 \pm \sqrt{2^2-4*2*(-840)}}{2*2}\] \[\Large x=\frac{-2 \pm \sqrt{6724}}{4}\] \[\Large x=\frac{-2 \pm 82}{4}\] \[\Large x=\frac{-2 + 82}{4} \ \text{ or } \ x=\frac{-2 - 82}{4}\] \[\Large x=\frac{80}{4} \ \text{ or } \ x=\frac{-84}{4}\] \[\Large x=20 \ \text{ or } \ x=-21\]

  15. jim_thompson5910
    • one year ago
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    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

  16. Elenathehomeschooler
    • one year ago
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    oh okay i noticed what i did wrong

  17. Elenathehomeschooler
    • one year ago
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    for number 3 the ladder will reach 29ft. on the wall right @jim_thompson5910

  18. jim_thompson5910
    • one year ago
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    29 isn't the vertical distance 29 is the length of the ladder

  19. jim_thompson5910
    • one year ago
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    again, look back at Vocaloid's drawing

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