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anonymous

  • 4 years ago

Please help... so confused HOLD ON! im going to draw a picture!

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  1. anonymous
    • 4 years ago
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    |dw:1328575316576:dw| solve for x

  2. precal
    • 4 years ago
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    use \[a^2+b^2=c^2\]

  3. anonymous
    • 4 years ago
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    well i think that would work but we're supposed to make it a proportion

  4. precal
    • 4 years ago
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    usually it involves the small triangle vs the large triangle

  5. anonymous
    • 4 years ago
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    ya but i cant do that... at least i dont think i can

  6. precal
    • 4 years ago
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    do you know about the angle on top? Is is split in the middle? Is it 45 degrees since the larger part is 90 degrees? I can't tell by your drawing. I know it is hard to draw

  7. anonymous
    • 4 years ago
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    well the whole triangle is 90 degrees but it doesnt tell you if its split through the middle

  8. precal
    • 4 years ago
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    |dw:1328575806467:dw| if so we can assume the side that is in between them is x...

  9. anonymous
    • 4 years ago
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    but how can you do that since x is the length of the side on the bottom?

  10. precal
    • 4 years ago
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    Is the drawing from a book or a worksheet? the problem with geometry is that you can not assume anything

  11. anonymous
    • 4 years ago
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    its in the book and ya thats why i dont like it!

  12. precal
    • 4 years ago
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    |dw:1328575902416:dw| We can assume that 90 degrees, the other one we can assume is the 45 degrees because that line that splits the triangle has a special name but I can't recall it at the moment.

  13. precal
    • 4 years ago
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    You can scan the picture and attach the file. but I think I might have it figured out

  14. anonymous
    • 4 years ago
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    well its online so give me a sec

  15. precal
    • 4 years ago
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    \[x^2+x^2=(x+3)^2\] \[2x^2=x^2+6x+9\] \[x^2-6x-9=0\] not that does not work !!!!!

  16. precal
    • 4 years ago
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    I meant that does not work!!! Sorry can't seem to type as well

  17. anonymous
    • 4 years ago
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    there is a picture of the problem!

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  18. precal
    • 4 years ago
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    ok upload it, I will be back in about 5 mins

  19. precal
    • 4 years ago
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    ok you got my attention.

  20. precal
    • 4 years ago
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    ok x+3 is the leg of the big triangle the hypotenuse is 12-x for the big triangle

  21. anonymous
    • 4 years ago
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    well the problem never tells you anything about the angles or anything and as you can tell the line doesnt go right through the right angle symbol

  22. precal
    • 4 years ago
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    ok x is the leg of the small triangle and x+3 is the hypotenuse of the small triangle there is our proportion

  23. precal
    • 4 years ago
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    |dw:1328576462186:dw|

  24. anonymous
    • 4 years ago
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    but how did you get 12-x since 12 is the whole LARGE triangle's hypotenuse?

  25. anonymous
    • 4 years ago
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    im sorry im just super confused!

  26. precal
    • 4 years ago
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    |dw:1328576638458:dw|

  27. anonymous
    • 4 years ago
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    so (x+3)(x+3)=12 x

  28. precal
    • 4 years ago
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    No, I think we are ok cross multiply \[12x=(x+3)^2\]

  29. precal
    • 4 years ago
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    yes now solve it

  30. anonymous
    • 4 years ago
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    x^2+3x+3x+9=12x x^2+6x+9=12x x^2+9=6x right? but i dont know what do to next because we havent done anything like this in a while!

  31. precal
    • 4 years ago
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    x=3 I used the quadratic formula

  32. precal
    • 4 years ago
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    \[x^2-6x+9=0\]

  33. anonymous
    • 4 years ago
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    ok but how did you get x=3?

  34. precal
    • 4 years ago
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    |dw:1328576877634:dw|

  35. precal
    • 4 years ago
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    use the quadratic formula

  36. anonymous
    • 4 years ago
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    see all of that confuses me and i dont think we've done that yet!

  37. anonymous
    • 4 years ago
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    well thanks for your help but i dont think im going to get it.... but thank you for your trying... :)

  38. precal
    • 4 years ago
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    Is this a challenge question?

  39. anonymous
    • 4 years ago
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    i dont think so.... but online they sometimes leave off something like a picture or a number so that might be what happened... thanks for you patients.

  40. precal
    • 4 years ago
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    You will understand this eventually, your teacher has to teach quadratics. One day it will make sense.

  41. anonymous
    • 4 years ago
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    haha ok thanks :)

  42. precal
    • 4 years ago
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    Oh well another reason why computers will not replace teachers. I am sure that it is an oversight of the program. Hang in there :)

  43. anonymous
    • 4 years ago
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    thank you and again thanks for your help1

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

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