anonymous
  • anonymous
I've got 4 algebra 2 questions that I'm really stuck on. can someone please help me??
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
i've answered number 2, but I'm not sure about that answer.
anonymous
  • anonymous
@Michele_Laino @nincompoop @campbell_st @Loser66
anonymous
  • anonymous
@Michele_Laino will you help me?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.