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anonymous

  • 5 years ago

http://img862.imageshack.us/i/page596.jpg/ I need help # 37,38

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  1. anonymous
    • 5 years ago
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    Which one do you want to tackle first?

  2. anonymous
    • 5 years ago
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    37 first

  3. anonymous
    • 5 years ago
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    Ok, what do you know about the length of AD?

  4. anonymous
    • 5 years ago
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    the answer is d, How I can find the length?

  5. anonymous
    • 5 years ago
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    What can you tell me about the length of AD by looking at the picture?

  6. anonymous
    • 5 years ago
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    the question is what the measure of AD?

  7. anonymous
    • 5 years ago
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    Yes I know. I'm asking what you think about the problem

  8. anonymous
    • 5 years ago
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    can you see the picture?

  9. anonymous
    • 5 years ago
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    This is math. Some thinking is required ;)

  10. anonymous
    • 5 years ago
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    Yes I can see it. I'm asking what does it tell you? I know what it's telling me.

  11. anonymous
    • 5 years ago
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    yes it math

  12. anonymous
    • 5 years ago
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    given the lenghth AB, BC, CD, I don't get what you asking

  13. anonymous
    • 5 years ago
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    I wonder do you know how to get answer find length AD

  14. anonymous
    • 5 years ago
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    I know how to get the answer yes. I want to know what you know? That way I can see how to help you.

  15. anonymous
    • 5 years ago
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    Nobody cares if I can do it.

  16. anonymous
    • 5 years ago
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    I don't know

  17. anonymous
    • 5 years ago
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    if given radius is may easy for my to do

  18. anonymous
    • 5 years ago
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    I been thinking many hours but I can't solve

  19. anonymous
    • 5 years ago
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    if you can help me ,

  20. anonymous
    • 5 years ago
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    Good! You know how to solve it with the radius?

  21. anonymous
    • 5 years ago
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    yes, I can do if I know radius or given tangents or secants

  22. anonymous
    • 5 years ago
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    Ok, well for this we don't actually need the radius. This is called a tangential quadrilateral. It has the property that the two opposite sides will add to the same length.

  23. anonymous
    • 5 years ago
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    what side?

  24. anonymous
    • 5 years ago
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    AB + CD = AD + BC

  25. anonymous
    • 5 years ago
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    great, thank

  26. anonymous
    • 5 years ago
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    What do you think about 38?

  27. anonymous
    • 5 years ago
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    let me think

  28. anonymous
    • 5 years ago
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    I think H

  29. anonymous
    • 5 years ago
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    Why?

  30. anonymous
    • 5 years ago
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    w is the weight the total weight have to be less or equal

  31. anonymous
    • 5 years ago
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    r no more than 12

  32. anonymous
    • 5 years ago
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    r no more than 12 comes from what statement?

  33. anonymous
    • 5 years ago
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    If I tell you, there are 4 reams of paper in the box, what is the weight w?

  34. anonymous
    • 5 years ago
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    the given r is the box can carry no more than 12 reams of paper

  35. anonymous
    • 5 years ago
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    right.

  36. anonymous
    • 5 years ago
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    so my r is less than 12 it's 4, what is w?

  37. anonymous
    • 5 years ago
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    4.4

  38. anonymous
    • 5 years ago
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    if the box itself weighs 1.3 and each ream weighs 4.4 and there are 4 reams, what is the total weight?

  39. anonymous
    • 5 years ago
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    total; 18.9 w

  40. anonymous
    • 5 years ago
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    You sure the weight isn't 1 ?

  41. anonymous
    • 5 years ago
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    22.8w

  42. anonymous
    • 5 years ago
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    I think 18.9 was correct actually.

  43. anonymous
    • 5 years ago
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    But wait, why isn't the weight just 3.1 or 1.3 or 10

  44. anonymous
    • 5 years ago
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    Aren't all those < 1.3 + 4.4r ?

  45. anonymous
    • 5 years ago
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    w<1.3 +4.4r, r<12

  46. anonymous
    • 5 years ago
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    Right. So if I tell you that r=1, what is the weight? It's 1.1111111111111 right?

  47. anonymous
    • 5 years ago
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    because 1.111111 < 1.3 + 4.4

  48. anonymous
    • 5 years ago
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    r=1? w<5.7

  49. anonymous
    • 5 years ago
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    Sure, and 1.1111 is less than 5.7, so 1.1111 is the weight? But 0.1274 is also less than 5.7, so which one is the weight?

  50. anonymous
    • 5 years ago
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    Or is the weight both 1.1111 and 0.1274 or maybe the weight is \(\pi\). I do like \(\pi\). (Here is where you tell me I'm being dumb and the weight is 5.7)

  51. anonymous
    • 5 years ago
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    \(\pi\) is less than 5.7 so I think \(\pi\) is the weight. Cause you said anything less than 1.3 + 4.4r was a solution for the weight of a box.

  52. anonymous
    • 5 years ago
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    and I said there was 1 ream, so the box can weigh anything less than 5.7

  53. anonymous
    • 5 years ago
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    And you say "No dummy the box weighs 5.7"

  54. anonymous
    • 5 years ago
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    ;)

  55. anonymous
    • 5 years ago
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    ok, thank you so much

  56. anonymous
    • 5 years ago
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    Wait. What? Did you find the solution?

  57. anonymous
    • 5 years ago
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    yes I think w<1.3 + 4.4 r , r<12

  58. anonymous
    • 5 years ago
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    no.. that's what I'm trying to explain

  59. anonymous
    • 5 years ago
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    Once we pick an r, we know exactly how much the box weighs. The restriction is on r not on the weight.

  60. anonymous
    • 5 years ago
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    w is not any number less than 1.3 + 4.4r. If I tell you r =1, w is not any number less than 5.7. It is 5.7

  61. anonymous
    • 5 years ago
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    the key not reqite r

  62. anonymous
    • 5 years ago
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    hrm? I didn't follow that. The answer is G.

  63. anonymous
    • 5 years ago
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    If we have r, we know exactly what w is. Not that it's less than something, we know exactly. The only restriction we have is that r must be less than or equal to 12. But will be exactly 1.3 + 4.4r for any r you pick that is allowed.

  64. anonymous
    • 5 years ago
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    err w will be exactly 1.3 + 4.4r

  65. anonymous
    • 5 years ago
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    Ok , you said w = 1.3+4.4r ,r<12

  66. anonymous
    • 5 years ago
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    yes. Because once you pick how many reams in the box, the box has an exact wieght.

  67. anonymous
    • 5 years ago
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    if you pick 1 ream goes in the box, the box will wiegh 5.7. It will not weigh 1.1 or 0.213. If w < 5.7 were the definition then w could be any of those.

  68. anonymous
    • 5 years ago
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    OK, it help

  69. anonymous
    • 5 years ago
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    I understand now

  70. anonymous
    • 5 years ago
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    thank you

  71. anonymous
    • 5 years ago
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    no problem

  72. anonymous
    • 5 years ago
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    It late I go to the bed now bye

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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