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anonymous

  • one year ago

A right triangle has base (x - 7) units and height (2x - 10) units. Part A: What is the square of the length of the hypotenuse of the triangle? Show your work. (4 points) Part B: What is the area of the triangle? (3 points) Part C: Using the solution obtained in Part B, explain the closure property of multiplication of polynomials. (3 points)

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  1. anonymous
    • one year ago
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    @ganeshie8 @Nnesha @hartnn

  2. anonymous
    • one year ago
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    Please help Anyone

  3. anonymous
    • one year ago
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    @mathmate @welshfella Can you guys help? I have a general idea of what I need to do for part one, but I get really lost afterwards.

  4. welshfella
    • one year ago
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    square of hypotenuse , by Pythagoras theorem = sum of squares on the other 2 sides = (x - 7^2 + (2x - 10)^2

  5. anonymous
    • one year ago
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    I got that much, but I need to simplify it I believe

  6. anonymous
    • one year ago
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    @welshfella

  7. kobeni-chan
    • one year ago
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    What do you have that you need to simplify?

  8. anonymous
    • one year ago
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    I need the square length of the hypotenuse. so wouldn't I need to simplify (x-7)^2 + (2x+10)= C^2?

  9. anonymous
    • one year ago
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    Like I know that much but I don't know what to do afterwards. do I square everything?

  10. kobeni-chan
    • one year ago
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    You need to square (x-7) and (2x-10) right now. If you have a scientific calculator you can do this easily, but if you have a reg. calculator then do (x-7)(x-7) and use the distributive property. Then do the same with (2x-10)(2x-10)

  11. anonymous
    • one year ago
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    |dw:1433177251700:dw|

  12. anonymous
    • one year ago
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    |dw:1433177275411:dw|

  13. anonymous
    • one year ago
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    |dw:1433177299845:dw|

  14. anonymous
    • one year ago
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    |dw:1433177325031:dw| ?

  15. anonymous
    • one year ago
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    Did I do that part correct?

  16. kobeni-chan
    • one year ago
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    h/o

  17. anonymous
    • one year ago
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    Oki cx

  18. kobeni-chan
    • one year ago
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    Yes that's right :) sorry if I'm slow to respond I'm doing my exam

  19. anonymous
    • one year ago
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    Okay, so now I do (2x-10)(2x-10)?

  20. kobeni-chan
    • one year ago
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    Yep

  21. anonymous
    • one year ago
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    |dw:1433177671889:dw| Sorry for the bad writing ;-;~

  22. welshfella
    • one year ago
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    now had these 2 expansions together x&3 - 14x + 49 + 4x^2 - 40x + 100

  23. welshfella
    • one year ago
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    **correction first term is x^2

  24. anonymous
    • one year ago
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    add them?

  25. welshfella
    • one year ago
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    yes

  26. anonymous
    • one year ago
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    |dw:1433178085721:dw|

  27. welshfella
    • one year ago
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    add like terms eg x^2 + 4x^2 = 5 x^2

  28. welshfella
    • one year ago
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    yep

  29. anonymous
    • one year ago
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    is that the answer? for part A I mean

  30. welshfella
    • one year ago
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    yes

  31. welshfella
    • one year ago
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    Part B the area = 1/2 * base * height = 1/2 * (x - 7) (2x - 10)

  32. anonymous
    • one year ago
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    how do I do that o.O

  33. anonymous
    • one year ago
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    @kobeni-chan ? ;-;~

  34. welshfella
    • one year ago
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    to make it easier 1/2 * (2x - 10) = (x - 5) now work out (x - 7)(x - 5) and you gave the required area

  35. welshfella
    • one year ago
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    * have

  36. anonymous
    • one year ago
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    |dw:1433179051398:dw|

  37. anonymous
    • one year ago
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    @amistre64 @Here_to_Help15

  38. anonymous
    • one year ago
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    @mathstudent55 @paki @Luigi0210

  39. anonymous
    • one year ago
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    @welshfella can you please help me

  40. kobeni-chan
    • one year ago
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    I'm so sorry I got very busy and forgot about your question :( what part are you on?

  41. anonymous
    • one year ago
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    Part B I believe >.>

  42. anonymous
    • one year ago
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    @Aureyliant >.>

  43. anonymous
    • one year ago
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    Ok so Pt. B is asking the area.. what is your understanding of this Q?

  44. anonymous
    • one year ago
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    That for the area, you need to do a=1/2*b*h since we already have b and h, we just plug it in. Right?

  45. anonymous
    • one year ago
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    Yup!

  46. anonymous
    • one year ago
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    I got that much, but when it comes to trying to find said area, I get completely lost

  47. anonymous
    • one year ago
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    \[1/2*(2x-10)(x-7)\] is your equation

  48. anonymous
    • one year ago
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    I know that much, but I don't know how to apply the 1/2 to the (x-7)

  49. welshfella
    • one year ago
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    you've worked it out correctly alice

  50. anonymous
    • one year ago
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    so is that trinomial the Area?

  51. anonymous
    • one year ago
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    You can either take it's inverse out, but since it's an expression & not an equation, it's like dividing (x-7) by 2.

  52. welshfella
    • one year ago
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    apply the half to (2x -10) first makes it easier 1/2 * (2x - 10 ) = x - 5

  53. anonymous
    • one year ago
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    ^ Agreed

  54. welshfella
    • one year ago
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    so it comes down to (x -5)(x - 7) which you worked out correctly in an earlier post alice

  55. anonymous
    • one year ago
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    I did (x-5)(x-7). Is there no other step afterwards?

  56. anonymous
    • one year ago
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    Nope. You can check your work by multiplying again, and you should be getting the same answer you started with.

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