If arithmetic, then there is a common difference
if geometric, then there is a common ratio
to find the common ratio or difference
if you subtract successive terms from each other, they should have the same result if it is aritmetic
if you divide successive terms by each other, they should have the same result if it is geometric
2,3 and 5
3-2=1
5-3=2
2≠1
not aritmetic
3/2=1.5
5/3=1.66666
1.5≠1.6666
not geometric
she is wrong
Answer:
Consider a circle centered at C(-2,-4). If the point P(1,-1) lies on the circle, then which of the following points also lies on the circle?
<h3>A. (-2, -√18)</h3>
B. (-2+√18, -4)
C. (√18, -4)
D. (-2, 4+√18)
Step-by-step explanation:
- If the distance is greater than the radius, the point lies outside. If it's equal to the radius, the point lies on the circle. And if it's less than the radius, you guessed it right, the point will lie inside the circle.
hope it's help you
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: the answer is down bellow
Step-by-step explanation:
x3-10oo Dimensions ft each S.
Answer:
It is b
Step-by-step explanation:
