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

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

Mathematics
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i've answered number 2, but I'm not sure about that answer.
@Michele_Laino will you help me?

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Other answers:

ok!
exercise #1
here we have to apply the Theore of carnot, and we can write: \[\Large Q{R^2} = P{Q^2} + P{R^2} - 2PQ \times PR\cos P\]
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?
oops... \[\Large QR = \sqrt {{{7.5}^2} + {{8.4}^2} - 2 \times 7.5 \times 8.4 \times \cos 43} = ...?\] what is QR?
sqrt −231cs64+126.81
find QR and then use sin formula to find
I don't know how to do that..
I got this: QR= 5.89
oh..how did I get that answer....??
next we have to apply the law of sines, so we can write: \[\Large \frac{{QR}}{{\sin P}} = \frac{{PQ}}{{\sin R}}\]
ok
substituting our data, we get: \[\Large \sin R = \frac{{PQ}}{{QR}}\sin P = \frac{{7.5}}{{5.89}} \times \sin 43 = ...?\]
0.86841896
ok! so what is R?
0.86841896 <----this?
no, that is sin(R)
I'm not sure then..
you have to use the sin^-1 function on your calculator
I got this: R= 60.27 degrees
Oh, now I see. So it is D?
yes!
Thank you!!! can you help me on the others?
yes!
for the second one, I'm not sure if that one is the correct answer.
exercise #2
yes
we have: \[\Large C = 180 - \left( {A + B} \right) = 180 - \left( {41 + 32} \right) = ...?\]
what is C?
107
\[\frac{\text{AB}}{\sin ((180-32-41) {}^{\circ})}=\frac{9}{\sin (41 {}^{\circ})} \]\[\text{AB}\to 13.1189 \]
Is it A then?
ok! now we have to apply the law of sines again, so we can write: \[\Large \frac{{AB}}{{\sin C}} = \frac{{AC}}{{\sin B}}\]
@robtobey did this part of the problem already, right?
A.
substituting your data, we get: \[\Large AB = \frac{{AC\sin C}}{{\sin B}} = \frac{{9 \times \sin 107}}{{\sin 32}} = ...?\]
13.1189
I got 16.24
I was using @robtobey 's answer. I just did the calculations, I got 16.24 also
It's C
yes! I think so!
I get that one too...thank you!!
This is really helpful
now let's go to exercise #3
okay
here we have to apply the subsequent formula: \[\Large \begin{gathered} Area = \frac{1}{2} \times PQ \times RQ \times \sin Q = \hfill \\ \hfill \\ = \frac{1}{2} \times 18 \times 8 \times \sin 25 =...? \hfill \\ \end{gathered} \]
30.4
yes! I got Area=30.42
yay!!
Next, let's go on exercise #4
okay
if: \[\Large A + B = C\] then: \[\Large B = C - A\] am I right?
yes
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
1 -4 -3 2 -7 5
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.
THat's okay, we all make mistakes :)
I meant 1-4= -3
What?
hint: \[\Large \begin{gathered} B = C - A = \left( {\begin{array}{*{20}{c}} 1&{ - 2}&8 \\ 5&0&2 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 4&2&7 \\ 3&7&{ - 3} \end{array}} \right) = \hfill \\ \hfill \\ = \left( {\begin{array}{*{20}{c}} {1 - 4}&{...}&{...} \\ {...}&{...}&{...} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 3}&{...}&{...} \\ {...}&{...}&{...} \end{array}} \right) \hfill \\ \end{gathered} \]
ooh so what do I do next?
you have to subtract every component of the second matrix, from the corresponding component of the first matrix
-3 0 1 2 7 1
for example: \[\Large \begin{gathered} B = C - A = \left( {\begin{array}{*{20}{c}} 1&{ - 2}&8 \\ 5&0&2 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 4&2&7 \\ 3&7&{ - 3} \end{array}} \right) = \hfill \\ \hfill \\ = \left( {\begin{array}{*{20}{c}} {\left( {1 - 4} \right)}&{\left( { - 2 - 2} \right)}&{\left( {8 - 7} \right)} \\ {\left( {5 - 3} \right)}&{\left( {0 - 7} \right)}&{\left( {2 - \left( { - 3} \right)} \right)} \end{array}} \right) \hfill \\ \end{gathered} \]
mine wasn't right?
no, I'm sorry
Wait, I'm onfused. How do I do this?
it is simple: 1-4= -3 -2-2=-4 and so on...
-3 -4 -1 2 7 -1
oh, no the 1's aren't negatives.
0-7= -7 right?
2-(-3)= 2+ 3= 5
8-7=1 and, finally: 5-3=2 \[\Large \begin{gathered} B = C - A = \left( {\begin{array}{*{20}{c}} 1&{ - 2}&8 \\ 5&0&2 \end{array}} \right) - \left( {\begin{array}{*{20}{c}} 4&2&7 \\ 3&7&{ - 3} \end{array}} \right) = \hfill \\ \hfill \\ = \left( {\begin{array}{*{20}{c}} {\left( {1 - 4} \right)}&{\left( { - 2 - 2} \right)}&{\left( {8 - 7} \right)} \\ {\left( {5 - 3} \right)}&{\left( {0 - 7} \right)}&{\left( {2 - \left( { - 3} \right)} \right)} \end{array}} \right) = \hfill \\ \hfill \\ = \left( {\begin{array}{*{20}{c}} { - 3}&{ - 4}&1 \\ 2&{ - 7}&5 \end{array}} \right) \hfill \\ \end{gathered} \]
Oh, I see what I was doing wrong now.
so, what is the right option?
A
that's right!
thank you so much for your help!!
:)

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