Answer:
<h2>√2,2√4</h2>
Step-by-step explanation:
<h2>√2,√8</h2><h2>√2,√4×2</h2><h2>√2,√2×2×2</h2><h2>√2,√2×2 2</h2><h2>√2,2√4 answer</h2>
72 because I helped you show do you Andreas show me too
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:
A. 25%
25% is equivalent to 1/4 and when you multiply 75 by 4 you get 300. Which means 75 is equivalent to 1/4 and 1/4 is equivalent to 25%. So the answer is 25%. I'm so sorry if this explanation is bad