Answer:

Step-by-step explanation:
sry if im wrong
Answer:
Power of a product.
Step-by-step explanation:
Hope this helps!
=)
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: All real numbers
Step-by-step explanation: If you plug in any number, it will work. 4(a+3) = 12+4a. a=-3 0=0
Answer:
TC (A) = 40x , TC (B) = 500 + 20x
Step-by-step explanation:
Let the number of students be = x
Hall A Total Cost
Relationship Equation, where TC (A) = f (students) = f (x) 40 per person (student) = 40x
Hall B Total Cost
Relationship Equation, where TC (B) = f (students) = f (x) 500 fix fee & 20 per person (student) = 500 + 20x