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anonymous

  • 5 years ago

how do you get this length...

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  1. amistre64
    • 5 years ago
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    i feel that your questions might be lacking some vital components :/

  2. anonymous
    • 5 years ago
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    |dw:1327247673332:dw|

  3. amistre64
    • 5 years ago
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    thats a scaled version of the base

  4. anonymous
    • 5 years ago
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    That's a secret length.

  5. amistre64
    • 5 years ago
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    if you knew some angles itd be doable as well

  6. anonymous
    • 5 years ago
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    I think we can use Euler straight edge for that.

  7. AravindG
    • 5 years ago
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    wt kind of triangle is it???

  8. amistre64
    • 5 years ago
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    did Fool just suggest a ruler? lol

  9. anonymous
    • 5 years ago
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    i don't need a numerical answer. length in question is c, height of c from base b is y

  10. anonymous
    • 5 years ago
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    not a right trianle

  11. amistre64
    • 5 years ago
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    |dw:1327247884124:dw|

  12. anonymous
    • 5 years ago
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    just need to solve in terms of y. no angle given

  13. anonymous
    • 5 years ago
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    ^^^^yes

  14. amistre64
    • 5 years ago
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    \[\frac{c}{h}=\frac{b}{h-y}\] \[c=\frac{bh}{h-y}\]is my only idea without knowing any fancy triangle thrms

  15. amistre64
    • 5 years ago
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    where h = height of the altitude

  16. anonymous
    • 5 years ago
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    hmmm i hate answering "none of the above"

  17. amistre64
    • 5 years ago
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    we have options?

  18. anonymous
    • 5 years ago
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    yes

  19. amistre64
    • 5 years ago
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    way to make life harder for us by not posting them, kudos :)

  20. anonymous
    • 5 years ago
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    Yes amstre that's a ruler :D

  21. anonymous
    • 5 years ago
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    a) (h-y)(b/h) b)(h-y)(h/b) c)(b/h)y d)(h/b)y

  22. anonymous
    • 5 years ago
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    sorry, i wasn't trying to get a straight answer in the first place, i want to solve :P

  23. anonymous
    • 5 years ago
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    You need to use basic proportionality theorem.

  24. amistre64
    • 5 years ago
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    i think my solution is adequate then, if we can rearrange it into one of those formats

  25. amistre64
    • 5 years ago
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    \[\frac{bh}{h-y}\] \[b\frac{h}{h-y}\] \[b\frac{1}{1-y/h}\] maybe the wolf can give us options?

  26. amistre64
    • 5 years ago
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    it was worth a shot, check my proportion; is it ok?

  27. anonymous
    • 5 years ago
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    about to just plug in numbers and see

  28. amistre64
    • 5 years ago
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    |dw:1327248716441:dw|

  29. amistre64
    • 5 years ago
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    this is the same type of set up but with something i can see better

  30. anonymous
    • 5 years ago
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    your proportions aren't right but i also don't know if this pic is drawn to scale. h is longer than b, c is shorter than b, of coarse. and y is shorter than b and c

  31. anonymous
    • 5 years ago
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    from your expressions, c ends up being longer than b

  32. amistre64
    • 5 years ago
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    3/4 = 2/c c = 8/3 = 2 2/3? sqrt(2^2 + 8/3^2)= 10/3 5/3 = d/2 10/3 = d thats a match in my book

  33. amistre64
    • 5 years ago
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    c is to b as h-y is to h \[c=\frac{b}{h}(h-y)\]might be better :)

  34. amistre64
    • 5 years ago
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    i think that matches your "a" option

  35. anonymous
    • 5 years ago
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    yes, it's exact. i don't get why though. so did you derive "a" from your earilier expression c=(bh)/(h-y)?

  36. amistre64
    • 5 years ago
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    no, i had to dust out the cobwebs and make sure I was putting things in their proper places.

  37. amistre64
    • 5 years ago
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    similar triangle differ by some scalar multiple; they keep the same ratios from one tri to the next

  38. amistre64
    • 5 years ago
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    so if we equate the proper ratios we can solve for them

  39. amistre64
    • 5 years ago
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    bases with repect to heights will math in silimlar tris. b/h = c/(h-y) are the respective parts to match up

  40. amistre64
    • 5 years ago
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    then we solve for c

  41. anonymous
    • 5 years ago
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    i think i need to study "basic proportiionality theorem". don't recall learning about it

  42. anonymous
    • 5 years ago
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    thx for your help

  43. amistre64
    • 5 years ago
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    your welcome, and good luck :)

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is replying to Can someone tell me what button the professor is hitting...

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