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.
Answer: I would answer but the image is blocked for me- can you type the problem by any chance?
Step-by-step explanation:
Answer:
The answers are C,F
Step-by-step explanation:
Those are the coordinates where the lines intercept the x-axis.
The price was reduced by 34%, so she paid 100-34 = 66% of the original price.
16,000 x 0.66 = 10,560
She paid $10,560
I think you mean
sqrt x + 5 = sqrt (x+45)
if you plug in x = 4,
left side = sqrt 4 + 5 = 2+ 5 = 7
right side = sqrt (4+45) = sqrt 49 = 7
so the answer is x = 4 :)