2) -7x + 5y=10 4x - 5y=-10
2 answers:
To solve this, the problem is solvable by elimination. So you see how the 5y and -5y cancel out? You basically add those two and you add the -7x and 4x and 10 and -10 leaving you with -3x=0. Dividing by each side gives you x=0. Then, plug 0 in for x in -7x+5y=10. Giving you
-7(0)+5y=10.
0+5y=10
5y=10
Divide by 5 on each side giving you y=2. So your answer is (0,2).
-7(0)+5y=10.
0+5y=10
5y=10
Divide by 5 on each side giving you y=2. So your answer is (0,2).
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Brainliest please if you found my answer helpful! :)
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.