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.
The answer is 31,500.
Let's clarify some things :
1% of 45,000 is 450
Maybe you're wondering why , it's actually very simple .
Just move the decimal to the left once if you want to find 10% , move the decimal twice if you want to find 1%.
We now that 1% equals 450. Then 2% equals to....
Well just add another 1% .
450 + 450 or 450 times 2
Which equals 900.
Now we want to find what's the population after 15 years.
900 times 15= 13,500
But that's not the answer ....
That's what is decreasing.
45,000 - 13,500= 31,500
That's why the answer is 31,500.
Answer:
(x+6)(y+3)
Step-by-step explanation:
- xy + 3x + 6y + 18
- x(y + 3) + 6(y+3)
- <u>(</u><u>x</u><u>+</u><u>6</u><u>)</u><u>(</u><u>y</u><u>+</u><u>3</u><u>)</u>
Answer:
Draw a 3-D of a terrain indicating different fault behaviors. Use the terrain below. Be sure that the entire terrain should cover at least three type of fault. Labcl the fault arcas and its part