Find three consecutive odd integers such that the square of the second decreased by four times the first is seven more than six
1 answer:
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Answer:
(8) d. 4/3 (9) a. (X - 4)^2 + (y-4)^2= 25
Step-by-step explanation:
8. CENTER OF CIRCLE: O (2,6)
radius of circle : r = √4 = 2
if m = 4/3 line: y = 4/3 x
perpendicular distance (d) from center of circle to line y = 4/3 x is
d = |((4/3 x 2) + (-1) x 6)| / √((4/3)² + (-1)²) = |(-10/3)| / √(25/9) = (10/3) / (5/3) = 2
d = r means line touch with circle
9. center of circle is the intersection of the diagonals of rectangle (4,4) and its radius is 5 (illustrated)
the equation of circle: (X - 4)^2 + (y-4)^2= 25
