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:
ertyghujikol
Step-by-step explanation:
3456t7ujn
Answer: From what I understand you want me to do 7*(-15)??? and the answer to that is -105.
Step-by-step explanation: All you have to do is 7*-15
Step-by-step explanation:
-7(y-1)= 21
-7y+7=21
-7y=21-7
-7y=14
divide by-7
y=-2
The general way to work this out is to solve the general expression for
the remaining quantity versus half-life, using logarithms. But that's not
necessary with these numbers.
Look at the numbers:
-- 3 mg is 1/4 of 12 mg.
-- 1/4 is the product of (1/2) x (1/2).
-- So the 3 mg is what's left of 12 mg after 2 half-lives.
The 26 minutes must be two half-lives.
-- The half-life of that substance is 26/2 = <em>13 minutes</em>.
Go Maggie !