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anonymous

  • one year ago

I've got 4 algebra 2 questions that I'm really stuck on. can someone please help me??

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  1. anonymous
    • one year ago
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    i've answered number 2, but I'm not sure about that answer.

  2. anonymous
    • one year ago
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    @Michele_Laino @nincompoop @campbell_st @Loser66

  3. anonymous
    • one year ago
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    @Michele_Laino will you help me?

  4. Michele_Laino
    • one year ago
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    ok!

  5. Michele_Laino
    • one year ago
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    exercise #1

  6. Michele_Laino
    • one year ago
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    here we have to apply the Theore of carnot, and we can write: \[\Large Q{R^2} = P{Q^2} + P{R^2} - 2PQ \times PR\cos P\]

  7. Michele_Laino
    • one year ago
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    substituting your data, we get: \[\Large QR = \sqrt {{{7.5}^2} + {{8.4}^2} - 27.5 \times 8.4 \times \cos 43} = ...?\] what is QR?

  8. Michele_Laino
    • one year ago
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    oops... \[\Large QR = \sqrt {{{7.5}^2} + {{8.4}^2} - 2 \times 7.5 \times 8.4 \times \cos 43} = ...?\] what is QR?

  9. anonymous
    • one year ago
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    sqrt −231cs64+126.81

  10. anonymous
    • one year ago
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    find QR and then use sin formula to find <R

  11. anonymous
    • one year ago
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    I don't know how to do that..

  12. Michele_Laino
    • one year ago
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    I got this: QR= 5.89

  13. anonymous
    • one year ago
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    oh..how did I get that answer....??

  14. Michele_Laino
    • one year ago
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    next we have to apply the law of sines, so we can write: \[\Large \frac{{QR}}{{\sin P}} = \frac{{PQ}}{{\sin R}}\]

  15. anonymous
    • one year ago
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    ok

  16. Michele_Laino
    • one year ago
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    substituting our data, we get: \[\Large \sin R = \frac{{PQ}}{{QR}}\sin P = \frac{{7.5}}{{5.89}} \times \sin 43 = ...?\]

  17. anonymous
    • one year ago
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    0.86841896

  18. Michele_Laino
    • one year ago
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    ok! so what is R?

  19. anonymous
    • one year ago
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    0.86841896 <----this?

  20. Michele_Laino
    • one year ago
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    no, that is sin(R)

  21. anonymous
    • one year ago
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    I'm not sure then..

  22. Michele_Laino
    • one year ago
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    you have to use the sin^-1 function on your calculator

  23. Michele_Laino
    • one year ago
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    I got this: R= 60.27 degrees

  24. anonymous
    • one year ago
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    Oh, now I see. So it is D?

  25. Michele_Laino
    • one year ago
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    yes!

  26. anonymous
    • one year ago
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    Thank you!!! can you help me on the others?

  27. Michele_Laino
    • one year ago
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    yes!

  28. anonymous
    • one year ago
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    for the second one, I'm not sure if that one is the correct answer.

  29. Michele_Laino
    • one year ago
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    exercise #2

  30. anonymous
    • one year ago
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    yes

  31. Michele_Laino
    • one year ago
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    we have: \[\Large C = 180 - \left( {A + B} \right) = 180 - \left( {41 + 32} \right) = ...?\]

  32. Michele_Laino
    • one year ago
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    what is C?

  33. anonymous
    • one year ago
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    107

  34. anonymous
    • one year ago
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    \[\frac{\text{AB}}{\sin ((180-32-41) {}^{\circ})}=\frac{9}{\sin (41 {}^{\circ})} \]\[\text{AB}\to 13.1189 \]

  35. anonymous
    • one year ago
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    Is it A then?

  36. Michele_Laino
    • one year ago
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    ok! now we have to apply the law of sines again, so we can write: \[\Large \frac{{AB}}{{\sin C}} = \frac{{AC}}{{\sin B}}\]

  37. anonymous
    • one year ago
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    @robtobey did this part of the problem already, right?

  38. anonymous
    • one year ago
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    A.

  39. Michele_Laino
    • one year ago
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    substituting your data, we get: \[\Large AB = \frac{{AC\sin C}}{{\sin B}} = \frac{{9 \times \sin 107}}{{\sin 32}} = ...?\]

  40. anonymous
    • one year ago
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    13.1189

  41. Michele_Laino
    • one year ago
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    I got 16.24

  42. anonymous
    • one year ago
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    I was using @robtobey 's answer. I just did the calculations, I got 16.24 also

  43. anonymous
    • one year ago
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    It's C

  44. Michele_Laino
    • one year ago
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    yes! I think so!

  45. anonymous
    • one year ago
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    I get that one too...thank you!!

  46. anonymous
    • one year ago
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    This is really helpful

  47. Michele_Laino
    • one year ago
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    now let's go to exercise #3

  48. anonymous
    • one year ago
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    okay

  49. Michele_Laino
    • one year ago
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    here we have to apply the subsequent formula: \[\Large \begin{gathered} Area = \frac{1}{2} \times PQ \times RQ \times \sin Q = \hfill \\ \hfill \\ = \frac{1}{2} \times 18 \times 8 \times \sin 25 =...? \hfill \\ \end{gathered} \]

  50. anonymous
    • one year ago
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    30.4

  51. Michele_Laino
    • one year ago
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    yes! I got Area=30.42

  52. anonymous
    • one year ago
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    yay!!

  53. Michele_Laino
    • one year ago
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    Next, let's go on exercise #4

  54. anonymous
    • one year ago
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    okay

  55. Michele_Laino
    • one year ago
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    if: \[\Large A + B = C\] then: \[\Large B = C - A\] am I right?

  56. anonymous
    • one year ago
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    yes

  57. Michele_Laino
    • one year ago
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    so we can write this: \[\Large \begin{gathered} B = C - A = \left( {\begin{array}{*{20}{c}} 1&{ - 2}&8 \\ 5&0&2 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 4&2&7 \\ 3&7&{ - 3} \end{array}} \right) = \hfill \\ \hfill \\ = \left( {\begin{array}{*{20}{c}} {1 - 4}&{...}&{...} \\ {...}&{...}&{...} \end{array}} \right) \hfill \\ \end{gathered} \] please complete

  58. anonymous
    • one year ago
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    1 -4 -3 2 -7 5

  59. anonymous
    • one year ago
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    Very sorry for my mistake. Labeled BC and AC as both 9 each. 16.24 is correct. I believe that this is the first or second of a problem error or mine discovered by the other participants.

  60. anonymous
    • one year ago
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    THat's okay, we all make mistakes :)

  61. Michele_Laino
    • one year ago
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    I meant 1-4= -3

  62. anonymous
    • one year ago
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    What?

  63. Michele_Laino
    • one year ago
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    hint: \[\Large \begin{gathered} B = C - A = \left( {\begin{array}{*{20}{c}} 1&{ - 2}&8 \\ 5&0&2 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 4&2&7 \\ 3&7&{ - 3} \end{array}} \right) = \hfill \\ \hfill \\ = \left( {\begin{array}{*{20}{c}} {1 - 4}&{...}&{...} \\ {...}&{...}&{...} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 3}&{...}&{...} \\ {...}&{...}&{...} \end{array}} \right) \hfill \\ \end{gathered} \]

  64. anonymous
    • one year ago
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    ooh so what do I do next?

  65. Michele_Laino
    • one year ago
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    you have to subtract every component of the second matrix, from the corresponding component of the first matrix

  66. anonymous
    • one year ago
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    -3 0 1 2 7 1

  67. Michele_Laino
    • one year ago
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    for example: \[\Large \begin{gathered} B = C - A = \left( {\begin{array}{*{20}{c}} 1&{ - 2}&8 \\ 5&0&2 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 4&2&7 \\ 3&7&{ - 3} \end{array}} \right) = \hfill \\ \hfill \\ = \left( {\begin{array}{*{20}{c}} {\left( {1 - 4} \right)}&{\left( { - 2 - 2} \right)}&{\left( {8 - 7} \right)} \\ {\left( {5 - 3} \right)}&{\left( {0 - 7} \right)}&{\left( {2 - \left( { - 3} \right)} \right)} \end{array}} \right) \hfill \\ \end{gathered} \]

  68. anonymous
    • one year ago
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    mine wasn't right?

  69. Michele_Laino
    • one year ago
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    no, I'm sorry

  70. anonymous
    • one year ago
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    Wait, I'm onfused. How do I do this?

  71. Michele_Laino
    • one year ago
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    it is simple: 1-4= -3 -2-2=-4 and so on...

  72. anonymous
    • one year ago
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    -3 -4 -1 2 7 -1

  73. anonymous
    • one year ago
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    oh, no the 1's aren't negatives.

  74. Michele_Laino
    • one year ago
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    0-7= -7 right?

  75. Michele_Laino
    • one year ago
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    2-(-3)= 2+ 3= 5

  76. Michele_Laino
    • one year ago
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    8-7=1 and, finally: 5-3=2 \[\Large \begin{gathered} B = C - A = \left( {\begin{array}{*{20}{c}} 1&{ - 2}&8 \\ 5&0&2 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 4&2&7 \\ 3&7&{ - 3} \end{array}} \right) = \hfill \\ \hfill \\ = \left( {\begin{array}{*{20}{c}} {\left( {1 - 4} \right)}&{\left( { - 2 - 2} \right)}&{\left( {8 - 7} \right)} \\ {\left( {5 - 3} \right)}&{\left( {0 - 7} \right)}&{\left( {2 - \left( { - 3} \right)} \right)} \end{array}} \right) = \hfill \\ \hfill \\ = \left( {\begin{array}{*{20}{c}} { - 3}&{ - 4}&1 \\ 2&{ - 7}&5 \end{array}} \right) \hfill \\ \end{gathered} \]

  77. anonymous
    • one year ago
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    Oh, I see what I was doing wrong now.

  78. Michele_Laino
    • one year ago
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    so, what is the right option?

  79. anonymous
    • one year ago
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    A

  80. Michele_Laino
    • one year ago
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    that's right!

  81. anonymous
    • one year ago
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    thank you so much for your help!!

  82. Michele_Laino
    • one year ago
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    :)

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