When two chords intersect "inside" a circle, what two sets of angles are formed.
1 answer:
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We know that
(ad)/(bd)=d/d time a/b=a/b since d's cancel
also
if a/b=c/d in simplest form, then a=c and b=d
we have
p/(x^2-5x+6)=(x+4)/(x-2)
therefor
p/(x^2-5x+6)=d/d times (x+4)/(x-2)
p/(x^2-5x+6)=d(x+4)/d(x-2)
therefor
p=d(x+4) and
x^2-5x+6=d(x-2)
we can solve last one
factor
(x-6)(x+1)=d(x-2)
divide both sides by (x-2)
[(x-6)(x+1)]/(x-2)=d
sub
p=d(x+4)
p=([(x-6)(x+1)]/(x-2))(x+4)
(ad)/(bd)=d/d time a/b=a/b since d's cancel
also
if a/b=c/d in simplest form, then a=c and b=d
we have
p/(x^2-5x+6)=(x+4)/(x-2)
therefor
p/(x^2-5x+6)=d/d times (x+4)/(x-2)
p/(x^2-5x+6)=d(x+4)/d(x-2)
therefor
p=d(x+4) and
x^2-5x+6=d(x-2)
we can solve last one
factor
(x-6)(x+1)=d(x-2)
divide both sides by (x-2)
[(x-6)(x+1)]/(x-2)=d
sub
p=d(x+4)
p=([(x-6)(x+1)]/(x-2))(x+4)
