Answer:
D
Step-by-step explanation:
sohcahtoa
Sin x = 32/58
sin x = .5517
x = 33°
Answer:
Distance= 6.6 miles
Bearing= N 62.854°W
Step-by-step explanation:
Let's determine angle b first
Angle b=20° (alternate angles)
Using cosine rule
Let the distance between the liner and the port be x
X² =8.8²+2.4²-2(8.8)(2.4)cos20
X²= 77.44 + 5.76-(39.69)
X²= 43.51
X= √43.51
X= 6.596
X= 6.6 miles
Let's determine the angles within the triangle using sine rule
2.4/sin b = 6.6/sin20
(2.4*sin20)/6.6= sin b
0.1244 = sin b
7.146= b°
Angle c= 180-20-7.146
Angle c= 152.854°
For the bearing
110+7.146= 117.146
180-117.146= 62.854°
Bearing= N 62.854°W
1/3(9x) =3x
1/3(18) =6
3x+6-5
Then you do
-(x)=-x
-(3)=-3
5x-x-3 = 4x-3
4x-3=3x+6-5
4x-3=3x+1
x=4
Take the augmented matrix,
![\left[\begin{array}{ccc|c}2&1&-3&-20\\1&2&1&-3\\1&-1&5&19\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D2%261%26-3%26-20%5C%5C1%262%261%26-3%5C%5C1%26-1%265%2619%5Cend%7Barray%7D%5Cright%5D)
Swap the row 1 and row 2:
![\left[\begin{array}{ccc|c}1&2&1&-3\\2&1&-3&-20\\1&-1&5&19\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%262%261%26-3%5C%5C2%261%26-3%26-20%5C%5C1%26-1%265%2619%5Cend%7Barray%7D%5Cright%5D)
Add -2(row 1) to row 2, and -1(row 1) to row 3:
![\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&-3&4&22\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%262%261%26-3%5C%5C0%26-3%26-5%26-14%5C%5C0%26-3%264%2622%5Cend%7Barray%7D%5Cright%5D)
Add -1(row 2) to row 3:
![\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&0&9&36\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%262%261%26-3%5C%5C0%26-3%26-5%26-14%5C%5C0%260%269%2636%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 3 by 1/9:
![\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&0&1&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%262%261%26-3%5C%5C0%26-3%26-5%26-14%5C%5C0%260%261%264%5Cend%7Barray%7D%5Cright%5D)
Add 5(row 3) to row 2:
![\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&0&6\\0&0&1&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%262%261%26-3%5C%5C0%26-3%260%266%5C%5C0%260%261%264%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 2 by -1/3:
![\left[\begin{array}{ccc|c}1&2&1&-3\\0&1&0&-2\\0&0&1&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%262%261%26-3%5C%5C0%261%260%26-2%5C%5C0%260%261%264%5Cend%7Barray%7D%5Cright%5D)
Add -2(row 2) and -1(row 3) to row 1:
![\left[\begin{array}{ccc|c}1&0&0&-3\\0&1&0&-2\\0&0&1&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%26-3%5C%5C0%261%260%26-2%5C%5C0%260%261%264%5Cend%7Barray%7D%5Cright%5D)
So we have
.
To make the exponent positive, make a fraction. 1/3 * 1/3. This is 0.1 repeating. 3^1 is 3. So the answer is no.