$=(a^2-10a)-(b^2+6b) +16$
$=[(a^2-2(5)a+25)-25]-[(b^2+2(3)b+9)-9]+16$
$=(a-5)^2-25-(b+3)^2+9+16$
$=(a-5)^2-(b+3)^2$
If two tangent segments to a circle share a
common endpoint outside a circle, then the two segments are congruent. This
is according to the intersection of two tangent theorem. The theorem states
that given a circle, if X is any point
within outside the circle and if Y and Z are points such that XY and XZ are
tangents to the circle, then XY is equal to XZ.
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Answer:
y=1/3x+4
Step-by-step explanation:
It should be correct as the gradient = 1/3
and the y-intercept = C
= +4
Answer:
12x + 15y=34
x=34/12 = 17/6
y=34/15
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-6x + 5y=3
x=3/-6 = 1/-2
y= 3/5
Step-by-step explanation:
