B6 are en amVir af a 2 dlecimals use a caleulator te the eqm
Solution
9)
A(-4,-2) and B(6,4)
AB = sqrt((6+4)^2 + (4+2)^2)
= sqrt(100 + 36)
= sqrt(136)
= 2sqrt(34)
This is the diameter
And thus, the radius = sqrt(34)
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11)
x - y + 1 = 0
y= x + 1
has slope = 1
So, perpendicular line has slope = -1/1 = -1
m = -1
and point (x1,y1) = (-3,1)
y - y1 = m(x - x1)
y - 1 = -1(x + 3)
y - 1 = -x - 3
y = -x - 2
Option C
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12)
4x^2 + 4y^2 - 4x - 63 = 0
Divide by 4 all over :
x^2 + y^2 - x = 63/4
Completing the square :
x^2 - x + 1/4 + y^2 = 63/4 + 1/4
(x - 1/2)^2 + (y - 0)^2 = 16
Center = (1/2 , 0)
Radius = srt16 = 4
Option B
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13)
Option B using quadratic formula
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14)
For one soln,
discriminant, D = 0
b^2 - 4ac = 0
2k - 4(3)(6) = 0
2k - 72 = 0
2k = 72
k = 36

