Acircle is centered at the origin and contains the point (4,0). would (2 square root 2,-square root 6) also be on the circle

circle do not contains point (2√2 , -6)

Step-by-step explanation:

Equation of circle:

The centre-radius form of the circle equation is in the format

with the centre being at the point (h, k) and the radius being "r"

Plug in values given the question to find the radius of circle

(4-0)² + (0 - 0)² = r²

r² = 16

r = √16

r  = 4

To find coordinates(2√2 , -6) is in the circle

Simply plug in the x and y from your point (2√2 , -6)

If the the answer is greater than r², then the point lies outside of the circle else they are in the circle

((2√2)² – 0)² + ((-6)² – 0)² ≤ 16

 8 + 36 ≤ 16

 44 ≤ 16 False

 Hence, (2√2 , -6) don't lie in or on the circle.

$25

step-by-step explanation:

25 - 4.50 = 20.5

20.5 / 2 = 10.25

10.25 + 10 = 20.25

20.25 - 12.50 = 7.75

therefore the answer is $25 as shown as the first number.