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uselessatmath Group Title

Short Leg= 2x-6 Long leg= 2x Longest= 30 Find the missing sides to get 30

  • one year ago
  • one year ago

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  1. tkhunny Group Title
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    Perhaps there was a figure or a description of a figure?

    • one year ago
  2. uselessatmath Group Title
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    I can't screenshot the figure. The codes in the page prevent it.

    • one year ago
  3. tkhunny Group Title
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    And you are entirely unable to describe it?

    • one year ago
  4. uselessatmath Group Title
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    its Pythagoreon Theorem

    • one year ago
  5. uselessatmath Group Title
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    a squared plus b squared equals c squared

    • one year ago
  6. tkhunny Group Title
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    Okay, then you should apply that. Can you identify a, b, and c?

    • one year ago
  7. uselessatmath Group Title
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    I did.

    • one year ago
  8. uselessatmath Group Title
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    Reread the question.

    • one year ago
  9. tkhunny Group Title
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    You did what? Fill n the blanks... For the Pythagorean Theorem, I need to identify a, b, and c. a = ?? b = ?? c = 30

    • one year ago
  10. uselessatmath Group Title
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    I did.

    • one year ago
  11. uselessatmath Group Title
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    A equals short leg

    • one year ago
  12. uselessatmath Group Title
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    B equals the other leg

    • one year ago
  13. uselessatmath Group Title
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    C equals longest leg

    • one year ago
  14. tkhunny Group Title
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    Okay. What did the Pythagorean Theorem look like when you substituted those values?

    • one year ago
  15. uselessatmath Group Title
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    When I squared them all?

    • one year ago
  16. tkhunny Group Title
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    Did you do that? Yes. What did you get?

    • one year ago
  17. uselessatmath Group Title
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    I have not done it yet because that and the whole problem is where my confusion lies.

    • one year ago
  18. tkhunny Group Title
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    I think we have established this: a = 2x-6 b = 2x c = 30 We have also established the Pythagorean Theorem \(a^{2} + b^{2} = c^{2}\) Now our task is to substitute the given values into the Pythagorean Theorem. You do that and let's see if you wander off.

    • one year ago
  19. uselessatmath Group Title
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    I wander off in the beginning.

    • one year ago
  20. tkhunny Group Title
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    Stop saying that and simply start writing. Change your screen name. Do SOMETHING so that you can win one battle, here. You seem to be in the wrong class. How can we move you toward a more positive outcome? I can't get you to write ANYTHING. That's not good.

    • one year ago
  21. uselessatmath Group Title
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    I have Aspbergers. Either bear with my incompetence that I was born with or help me find someone who can help.

    • one year ago
  22. tkhunny Group Title
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    You didn't answer my question. Can you substitute the known values into the known formula, or not? Do you know what substitution means? Have you ever seen the process?

    • one year ago
  23. uselessatmath Group Title
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    Thank you for your help. Good day.

    • one year ago
  24. MarcLeclair Group Title
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    What tkhunny is asking is which is the hypothenuse and the adjacent side.

    • one year ago
  25. uselessatmath Group Title
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    What tk hunny was not understanding is I wrote them already and as neither side is in normal form I am confused.

    • one year ago
  26. MarcLeclair Group Title
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    Would you be able to use the draw option under the chat log to drwa the picture for us?

    • one year ago
  27. uselessatmath Group Title
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    maybe im on a laptop

    • one year ago
  28. MarcLeclair Group Title
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    Just do your best, if you have problem explaining it is much easier to see a picture. If we can't understand your question, we will not be able to help.

    • one year ago
  29. uselessatmath Group Title
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    |dw:1354487589391:dw|

    • one year ago
  30. MarcLeclair Group Title
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    okay tell me if im right:|dw:1354487622965:dw|

    • one year ago
  31. uselessatmath Group Title
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    swap the legs

    • one year ago
  32. MarcLeclair Group Title
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    the one that are perpendicular?

    • one year ago
  33. uselessatmath Group Title
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    2x should be on the bottom with 2x minus 6 on the left the 30 is in the right spot

    • one year ago
  34. uselessatmath Group Title
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    |dw:1354487934867:dw|

    • one year ago
  35. MarcLeclair Group Title
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    okay that helps, so you want to solve for x. Because you have 1 variable ( which in this case would be x, and I'm assuming you know what variable means) In this case you have a formula relating all angle. We will call it as follow ( because a square + b square = c sqare is too vague and can confuse you): Opposite + adjacent = hypothenuse. Now you're hypothenuse is a mathematical term for what you called " longest leg" which is mathematically wrong to say. Your opposite and adjacent is the two other "Legs" and is relative (therefore meaning it all depends on how you see it) to what you want. I relate the BOTTOM RIGHT angle always. THerefore my opposite side is the 2x-6 and the adjacent is 2x Now come the equation which is as follow: adjacent(squared) + opposite(squared) = hypothenuse (squared) Plug in ( therefore replace the word corresponding to the right side) with the right equation, and solve for x Can you solve for x?

    • one year ago
  36. uselessatmath Group Title
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    I get legs. I don't get math terms. Can you show me a visual?

    • one year ago
  37. MarcLeclair Group Title
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    |dw:1354488141269:dw|

    • one year ago
  38. uselessatmath Group Title
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    o.O

    • one year ago
  39. MarcLeclair Group Title
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    does this help?

    • one year ago
  40. uselessatmath Group Title
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    No.

    • one year ago
  41. MarcLeclair Group Title
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    yikes. Hum alright, ill try another way. When you solve for an equation, at your level, you have only one variable : x

    • one year ago
  42. uselessatmath Group Title
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    Ok stop just for a moment. You are going way too bloody complex. I don't get complex and I don't get abstract.

    • one year ago
  43. MarcLeclair Group Title
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    Ah I'm sorry my friend then, I hope you find someone who can give you a better explanation

    • one year ago
  44. uselessatmath Group Title
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    I have Aspbergers I am used to not being able to find help really.

    • one year ago
  45. tkhunny Group Title
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    Why as Aspergers a restriction on your ability to visualize, follow abstraction, or communicate symbolically? It that the only diagnosis? You are so SURE that you CAN'T get it, and you know EXACTLY what it is you can't get, that I have to wonder if we don't have other pathologies at play, here. If you don't get abstract, complex, or math symbols, how do you know you have encountered such a thing?

    • one year ago
  46. uselessatmath Group Title
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    Have you ever looked up Aspbergers? Researched it? Lived it?

    • one year ago
  47. uselessatmath Group Title
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    By your replies I can say no you have not. Aspbergers is a Social Disorder in the DMV book and can affect the ability to comprehend abstract concepts in Math, English, Reading, Writing, and also affects Social Skills.

    • one year ago
  48. tkhunny Group Title
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    Yes, Yes, and Yes. But I had no cognitive difficulty related to mathematics. That's why I'm puzzled. It's quite a different presentation than I've previously encountered. I've only seen social aspects.

    • one year ago
  49. uselessatmath Group Title
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    Not every Aspbergers person is the same.

    • one year ago
  50. tkhunny Group Title
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    Clearly. Are you aware of what level of abstration can you handle? You seemed to be okay calling a leg of a triangle '2x' and another leg '2x-6'.

    • one year ago
  51. uselessatmath Group Title
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    It is obvious you can't help me. Good day.

    • one year ago
  52. tkhunny Group Title
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    Not if you don't talk to me and you just punt.

    • one year ago
  53. uselessatmath Group Title
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    New problem got it wrong :( let me write the new one

    • one year ago
  54. uselessatmath Group Title
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    nvm same problem as this one lol

    • one year ago
  55. mathstudent55 Group Title
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    short leg = 2x - 6 long leg = 2x longest side = 30 Pythagorean theorem: short leg squared + long leg squared = longest side squared (2x - 6)^2 + (2x)^2 = 30^2 (2x - 6)(2x - 6) + 4x^2 = 900 4x^2 - 12x - 12x + 36 + 4x^2 = 900 8x^2 -24x - 864 = 0 x^2 - 3x - 108 = 0 (x - 12)(x + 9) = 0 x - 12 = 0 or x + 9 = 0 x = 12 or x = -9 Since long leg is 2x, 2(-9) = -18, and a leg can't have a negative length, discard the negative solution. x = 12 short leg: 2x - 6 = 2(12) - 6 = 18 long leg: 2x = 2(12) = 24 The lengths are 18 for the short leg, and 24 for the long leg.

    • one year ago
  56. uselessatmath Group Title
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    Holy crap thank you. Been trying to get a straight answer on this problem for 3 hours. Thank you again.

    • one year ago
  57. uselessatmath Group Title
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    Oh great word problems ugh hate those :/ will post that next

    • one year ago
  58. mathstudent55 Group Title
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    I figured you got all kinds of bs answers except a simple answer to your problem, so it was about time you got one.

    • one year ago
  59. uselessatmath Group Title
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    I appreciate it greatly. Some just don't want to hear the word simple these days. To me simple is better and easier for me to understand. It is not I refuse to try I just don't always get it. Thank you again.

    • one year ago
  60. uselessatmath Group Title
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    This question is closed sir.

    • one year ago
  61. MarcLeclair Group Title
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    To be simple is being fed the answer which I don't believe in. Even if you have Aspergers, Autism, I will never give the answer to somebody.

    • one year ago
  62. uselessatmath Group Title
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    @MarcLeClair

    • one year ago
  63. uselessatmath Group Title
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    Look sir your view is obviously different. I see simple as a way of understanding. Don't like it. Don't respond. Good day.

    • one year ago
  64. tkhunny Group Title
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    I see. You wanted someone to do the work for you, quite contrary to the forum rules. Is there ANY of that reply that you could not have produced? I couldn't get you to write a single symbol past the original post. Why? I'm pretty sure this forum is for learning, not for only answers to be copied and turned in as one's own work. Not a good day.

    • one year ago
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