## 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

2. anonymous

@Michele_Laino @nincompoop @campbell_st @Loser66

3. anonymous

@Michele_Laino will you help me?

4. Michele_Laino

ok!

5. Michele_Laino

exercise #1

6. Michele_Laino

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

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

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

sqrt −231cs64+126.81

10. anonymous

find QR and then use sin formula to find <R

11. anonymous

I don't know how to do that..

12. Michele_Laino

I got this: QR= 5.89

13. anonymous

oh..how did I get that answer....??

14. Michele_Laino

next we have to apply the law of sines, so we can write: $\Large \frac{{QR}}{{\sin P}} = \frac{{PQ}}{{\sin R}}$

15. anonymous

ok

16. Michele_Laino

substituting our data, we get: $\Large \sin R = \frac{{PQ}}{{QR}}\sin P = \frac{{7.5}}{{5.89}} \times \sin 43 = ...?$

17. anonymous

0.86841896

18. Michele_Laino

ok! so what is R?

19. anonymous

0.86841896 <----this?

20. Michele_Laino

no, that is sin(R)

21. anonymous

I'm not sure then..

22. Michele_Laino

you have to use the sin^-1 function on your calculator

23. Michele_Laino

I got this: R= 60.27 degrees

24. anonymous

Oh, now I see. So it is D?

25. Michele_Laino

yes!

26. anonymous

Thank you!!! can you help me on the others?

27. Michele_Laino

yes!

28. anonymous

for the second one, I'm not sure if that one is the correct answer.

29. Michele_Laino

exercise #2

30. anonymous

yes

31. Michele_Laino

we have: $\Large C = 180 - \left( {A + B} \right) = 180 - \left( {41 + 32} \right) = ...?$

32. Michele_Laino

what is C?

33. anonymous

107

34. anonymous

$\frac{\text{AB}}{\sin ((180-32-41) {}^{\circ})}=\frac{9}{\sin (41 {}^{\circ})}$$\text{AB}\to 13.1189$

35. anonymous

Is it A then?

36. Michele_Laino

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

@robtobey did this part of the problem already, right?

38. anonymous

A.

39. Michele_Laino

substituting your data, we get: $\Large AB = \frac{{AC\sin C}}{{\sin B}} = \frac{{9 \times \sin 107}}{{\sin 32}} = ...?$

40. anonymous

13.1189

41. Michele_Laino

I got 16.24

42. anonymous

I was using @robtobey 's answer. I just did the calculations, I got 16.24 also

43. anonymous

It's C

44. Michele_Laino

yes! I think so!

45. anonymous

I get that one too...thank you!!

46. anonymous

47. Michele_Laino

now let's go to exercise #3

48. anonymous

okay

49. Michele_Laino

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

30.4

51. Michele_Laino

yes! I got Area=30.42

52. anonymous

yay!!

53. Michele_Laino

Next, let's go on exercise #4

54. anonymous

okay

55. Michele_Laino

if: $\Large A + B = C$ then: $\Large B = C - A$ am I right?

56. anonymous

yes

57. Michele_Laino

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

1 -4 -3 2 -7 5

59. anonymous

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

THat's okay, we all make mistakes :)

61. Michele_Laino

I meant 1-4= -3

62. anonymous

What?

63. Michele_Laino

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

ooh so what do I do next?

65. Michele_Laino

you have to subtract every component of the second matrix, from the corresponding component of the first matrix

66. anonymous

-3 0 1 2 7 1

67. Michele_Laino

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

mine wasn't right?

69. Michele_Laino

no, I'm sorry

70. anonymous

Wait, I'm onfused. How do I do this?

71. Michele_Laino

it is simple: 1-4= -3 -2-2=-4 and so on...

72. anonymous

-3 -4 -1 2 7 -1

73. anonymous

oh, no the 1's aren't negatives.

74. Michele_Laino

0-7= -7 right?

75. Michele_Laino

2-(-3)= 2+ 3= 5

76. Michele_Laino

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

Oh, I see what I was doing wrong now.

78. Michele_Laino

so, what is the right option?

79. anonymous

A

80. Michele_Laino

that's right!

81. anonymous

thank you so much for your help!!

82. Michele_Laino

:)