What is the center of the circle given by the equation (x-3)2 + (y-9)2 = 16?
A. (-9,-3)
B. (-3,-9)
C. (3,9)
D. (9,3)

The center is (3,9) and the radius is 4

Step-by-step explanation:

We can write the equation for a circle as

(x-h)^2 + (y-k)^2 = r^2

Where (h,k) is the center and r is the radius

(x-3)^2 + (y-9)^2 = 4^2

The center is (3,9) and the radius is 4

C

Step-by-step explanation:

(x-a)^2 + (y-b)^2 = r^2

(a,b) = Center

r = radius

SO...

That means the center of the circle " (x-3)^2 + (y-9)^2 = 16" is...

(3,9) (C)

3: 2

step-by-step explanation:

shorter piece is 2: 3, so just reverse it, and you get 3: 2

£56

step-by-step explanation:

u divide 24 by 3 and u get 8 so if u do 8 time 7 or 8+8+8+8+8+8+8 u get 56 and 56 divided by 7 is 8.