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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??
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i've answered number 2, but I'm not sure about that answer.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino @nincompoop @campbell_st @Loser66

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino will you help me?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2here we have to apply the Theore of carnot, and we can write: \[\Large Q{R^2} = P{Q^2} + P{R^2}  2PQ \times PR\cos P\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2substituting your data, we get: \[\Large QR = \sqrt {{{7.5}^2} + {{8.4}^2}  27.5 \times 8.4 \times \cos 43} = ...?\] what is QR?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2oops... \[\Large QR = \sqrt {{{7.5}^2} + {{8.4}^2}  2 \times 7.5 \times 8.4 \times \cos 43} = ...?\] what is QR?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sqrt −231cs64+126.81

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0find QR and then use sin formula to find <R

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't know how to do that..

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I got this: QR= 5.89

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh..how did I get that answer....??

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2next we have to apply the law of sines, so we can write: \[\Large \frac{{QR}}{{\sin P}} = \frac{{PQ}}{{\sin R}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2substituting our data, we get: \[\Large \sin R = \frac{{PQ}}{{QR}}\sin P = \frac{{7.5}}{{5.89}} \times \sin 43 = ...?\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2ok! so what is R?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.00.86841896 <this?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2no, that is sin(R)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2you have to use the sin^1 function on your calculator

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I got this: R= 60.27 degrees

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, now I see. So it is D?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you!!! can you help me on the others?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for the second one, I'm not sure if that one is the correct answer.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2we have: \[\Large C = 180  \left( {A + B} \right) = 180  \left( {41 + 32} \right) = ...?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{\text{AB}}{\sin ((1803241) {}^{\circ})}=\frac{9}{\sin (41 {}^{\circ})} \]\[\text{AB}\to 13.1189 \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2ok! now we have to apply the law of sines again, so we can write: \[\Large \frac{{AB}}{{\sin C}} = \frac{{AC}}{{\sin B}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@robtobey did this part of the problem already, right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2substituting your data, we get: \[\Large AB = \frac{{AC\sin C}}{{\sin B}} = \frac{{9 \times \sin 107}}{{\sin 32}} = ...?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I was using @robtobey 's answer. I just did the calculations, I got 16.24 also

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! I think so!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I get that one too...thank you!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is really helpful

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2now let's go to exercise #3

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2here we have to apply the subsequent formula: \[\Large \begin{gathered} Area = \frac{1}{2} \times PQ \times RQ \times \sin Q = \hfill \\ \hfill \\ = \frac{1}{2} \times 18 \times 8 \times \sin 25 =...? \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! I got Area=30.42

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2Next, let's go on exercise #4

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2if: \[\Large A + B = C\] then: \[\Large B = C  A\] am I right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Very 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0THat's okay, we all make mistakes :)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2hint: \[\Large \begin{gathered} B = C  A = \left( {\begin{array}{*{20}{c}} 1&{  2}&8 \\ 5&0&2 \end{array}} \right)  \left( {\begin{array}{*{20}{c}} 4&2&7 \\ 3&7&{  3} \end{array}} \right) = \hfill \\ \hfill \\ = \left( {\begin{array}{*{20}{c}} {1  4}&{...}&{...} \\ {...}&{...}&{...} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {  3}&{...}&{...} \\ {...}&{...}&{...} \end{array}} \right) \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooh so what do I do next?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2you have to subtract every component of the second matrix, from the corresponding component of the first matrix

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2for example: \[\Large \begin{gathered} B = C  A = \left( {\begin{array}{*{20}{c}} 1&{  2}&8 \\ 5&0&2 \end{array}} \right)  \left( {\begin{array}{*{20}{c}} 4&2&7 \\ 3&7&{  3} \end{array}} \right) = \hfill \\ \hfill \\ = \left( {\begin{array}{*{20}{c}} {\left( {1  4} \right)}&{\left( {  2  2} \right)}&{\left( {8  7} \right)} \\ {\left( {5  3} \right)}&{\left( {0  7} \right)}&{\left( {2  \left( {  3} \right)} \right)} \end{array}} \right) \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait, I'm onfused. How do I do this?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2it is simple: 14= 3 22=4 and so on...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, no the 1's aren't negatives.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.287=1 and, finally: 53=2 \[\Large \begin{gathered} B = C  A = \left( {\begin{array}{*{20}{c}} 1&{  2}&8 \\ 5&0&2 \end{array}} \right)  \left( {\begin{array}{*{20}{c}} 4&2&7 \\ 3&7&{  3} \end{array}} \right) = \hfill \\ \hfill \\ = \left( {\begin{array}{*{20}{c}} {\left( {1  4} \right)}&{\left( {  2  2} \right)}&{\left( {8  7} \right)} \\ {\left( {5  3} \right)}&{\left( {0  7} \right)}&{\left( {2  \left( {  3} \right)} \right)} \end{array}} \right) = \hfill \\ \hfill \\ = \left( {\begin{array}{*{20}{c}} {  3}&{  4}&1 \\ 2&{  7}&5 \end{array}} \right) \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, I see what I was doing wrong now.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so, what is the right option?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you so much for your help!!
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