Answer:
I believe you use SOH CAH TOA
Answer:
Step-by-step explanation:
y = x - 5
2x - 5y = 1
2x - 5(x - 5) = 1
2x - 5x + 25 = 1
-3x + 25 = 1
-3x = -24
x = 8
y = 8 - 5
y = 3
(8, 3)
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
Hey there! :)
2(3/5n + 3) - (2/3n - 1)
Simplify.
(2 * 3/5n) + (2 * 3) - (2/3n -1)
Simplify.
6/5n + 6 - 2/3n + 1
Add like terms.
(6/5n - 2/3n) + (6 + 1)
Make the fractions have the same denominator by multiplying fraction 1 (6/5n) by 3/3 and fraction 2 (-2/3n) by 5/5.
6/5n * 3/3 = 18/15n
-2/3n * 5/5 = -10/15n
So, our new equation looks like this : (18/15n - 10/15n) + (6 + 1)
Simplify.
8/15n + 6 —> final answer.
Hope this helped! :)