For the number 4n to end with digit zero for any natural number n, it should be divisible by 5. This means that the prime factorisation of 4n should contain the prime number 5
Given

,

is in Quadrant IV,

, and

is in Quadrant III, find

We can use the angle subtraction formula of sine to answer this question.

We already know that

.
We can use the Pythagorean identity

to find

.

Since

is in Quadrant IV, and sine is represented as y value on the unit circle, we must assume the negative value

.
As similar process is then done with

.

And since

is in Quadrant III, and cosine in represented as x value on the unit cercle, we must assume the negative value

.
Now we can fill in our angle subtraction formula!
Step-by-step explanation:
We know that if the equation of the circle is (x-h)^2 + (y-k)^2 = r^2, it has a center of (h,k) and a radius of r.
Matching this up with the given equation, we get the center is (-5, 4) and the radius is 9.
Answer:
B) at an x-intercept
Step-by-step explanation:
Hope this helps :))