Answer:
x^2 + y^2 + 16x + 6y + 9 = 0
Step-by-step explanation:
Using the formula for equation of a circle
(x - a)^2 + (y + b)^2 = r^2
(a, b) - the center
r - radius of the circle
Inserting the values given in the question
(-8,3) and r = 8
a - -8
b - 3
r - 8
[ x -(-8)]^2 + (y+3)^2 = 8^2
(x + 8)^2 + (y + 3)^2 = 8^2
Solving the brackets
( x + 8)(x + 8) + (y +3)(y+3) = 64
x^2 + 16x + 64 + y^2 + 6y + 9 = 64
Rearranging algebrally,.
x^2 + y^2 + 16x + 6y + 9+64 - 64 = 0
Bringing in 64, thereby changing the + sign to -
Therefore, the equation of the circle =
x^2 + y^2 + 16x + 6y + 9 = 0
Setup 2 problems
2x - 7 < 15 and 2x - 7 > -15
2x < 22 2x > -8
x < 11 x > -4
Or you can write it -4 < x < 11
X²+bx=c
x=[-b+√(b²+4c)]/2
2x(x+5)=4
2x²+10x=4
x²+5x=2
x=[-5+√(5²+4*4)]/2
if b=-1/2 and c=-9/2
x= {1/2+√[(1/2)²+4*(-9/2)]}/2
Well your equation would be 44-n=p (p=final price)
Figure it out stop being lazy and asking for help
