Answer:
The distance between the ship at N 25°E and the lighthouse would be 7.26 miles.
Step-by-step explanation:
The question is incomplete. The complete question should be
The bearing of a lighthouse from a ship is N 37° E. The ship sails 2.5 miles further towards the south. The new bearing is N 25°E. What is the distance between the lighthouse and the ship at the new location?
Given the initial bearing of a lighthouse from the ship is N 37° E. So, 
is 37°. We can see from the diagram that 
would be 
143°.
Also, the new bearing is N 25°E. So, 
would be 25°.
Now we can find 
. As the sum of the internal angle of a triangle is 180°.
Also, it was given that ship sails 2.5 miles from N 37° E to N 25°E. We can see from the diagram that this distance would be our BC.
And let us assume the distance between the lighthouse and the ship at N 25°E is 
We can apply the sine rule now.
So, the distance between the ship at N 25°E and the lighthouse is 7.26 miles.
Let's call x1: The rate per hour of the number one mechanic. X2: The rate per hour of mechanic number two. The first thing you should do is identify the system of equations that best describes the problem. In this case it is a system of 2 equations with two unknowns which when solved gives a total of x1 = 85 $ / h and x2 = 50 $ / h. Attached solution.
Answer:
10: -11n^5
12: 6k^2 -6k+7
Step-by-step explanation:
Ok so what your gonna do is add or subtract the ones with the same exponent.
so -15+4= -11 since they are both n^5 you can add them so your answer is:
-11n^2
now you have 8k^2-k-5k+7-2k. so you are gonna rearrange them so the same exponents are together.(keep the sign in front in front, if no sign it is positive)
8k^2-2k^2-5k-k+7
add like exponents
6k^2-6k+7
Complementary angles = 90
x+y =90
y=x+24
substitute this in to the equation
x+x+24=90
2x+24=90
subtract 24 from each side
2x = 66
divide by 2
x=33
y = 33+24
y=57
Answer: 33, 57
3/2 since you move 3 spaces horizontally (x) and 2 spaces up (y)
