For these problems you have to use PEMDAS so for problem 15 you would do the parentheses first, (21-3) you get 18 and then exponents, 3^2, you get 9, no multiplication but there is division so take your answers, 18 and 9 and divide them, you get 2, so the answer to 15 is 2.
One way would be to find the distance from the point to the center of the circle and compare it to the radius
for

the center is (h,k) and the radius is r
and the distance formula is
distance between

and

is

r=radius
D=distance form (8,4) to center
if r>D, then (8,4) is inside the circle
if r=D, then (8,4) is on the circle
if r<D, then (8,4) is outside the circle
so



the radius is

center is (-2,3)
find distance between (8,4) and (-2,3)






≈4.2

≈10.04
do r<D
(8,4) is outside the circle
Answer:
This is because to square a number just means to multiply it by itself. For example, (-2) squared is (-2)(-2) = 4. Note that this is positive because when you multiply two negative numbers you get a positive result.
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