Question

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

Answer 1

i've answered number 2, but I'm not sure about that answer.

Answer 2

@Michele_Laino
@nincompoop
@campbell_st
@Loser66

Answer 3

@Michele_Laino will you help me?

Answer 4

ok!

Answer 5

exercise #1

Answer 6

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

Answer 7

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?

Answer 8

oops...
\[\Large QR = \sqrt {{{7.5}^2} + {{8.4}^2} - 2 \times 7.5 \times 8.4 \times \cos 43} = ...?\]
what is QR?

Answer 9

sqrt −231cs64+126.81

Answer 10

find QR and then use sin formula to find <R

Answer 11

I don't know how to do that..

Answer 12

I got this:
QR= 5.89

Answer 13

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

Answer 14

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

Answer 15

ok

Answer 16

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

Answer 17

0.86841896

Answer 18

ok! so what is R?

Answer 19

0.86841896 <----this?

Answer 20

no, that is sin(R)

Answer 21

I'm not sure then..

Answer 22

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

Answer 23

I got this:
R= 60.27 degrees

Answer 24

Oh, now I see. So it is D?

Answer 25

yes!

Answer 26

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

Answer 27

yes!

Answer 28

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

Answer 29

exercise #2

Answer 30

yes

Answer 31

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

Answer 32

what is C?

Answer 33

107

Answer 34

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

Answer 35

Is it A then?

Answer 36

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

Answer 37

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

Answer 38

A.

Answer 39

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

Answer 40

13.1189

Answer 41

I got
16.24

Answer 42

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

Answer 43

It's C

Answer 44

yes! I think so!

Answer 45

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

Answer 46

This is really helpful

Answer 47

now let's go to exercise #3

Answer 48

okay

Answer 49

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

Answer 50

30.4

Answer 51

yes! I got Area=30.42

Answer 52

yay!!

Answer 53

Next, let's go on exercise #4

Answer 54

okay

Answer 55

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

Answer 56

yes

Answer 57

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

Answer 58

1 -4 -3
2 -7 5

Answer 59

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.

Answer 60

THat's okay, we all make mistakes :)

Answer 61

I meant 1-4= -3

Answer 62

What?

Answer 63

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

Answer 64

ooh so what do I do next?

Answer 65

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

Answer 66

-3 0 1
2 7 1

Answer 67

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

Answer 68

mine wasn't right?

Answer 69

no, I'm sorry

Answer 70

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

Answer 71

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

Answer 72

-3
-4
-1
2
7
-1

Answer 73

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

Answer 74

0-7= -7 right?

Answer 75

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

Answer 76

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

Answer 77

Oh, I see what I was doing wrong now.

Answer 78

so, what is the right option?

Answer 79

A

Answer 80

that's right!

Answer 81

thank you so much for your help!!

Answer 82

:)