Which sentence describes why polygon MNOP is congruent to polygon JKLP?
1 answer:
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Answer:
y + 20 = 9(x + 2)
Step-by-step explanation:
The equation of a line in point- slope form is
y - b = m(x - a)
where m is the slope and (a, b) a point on the lie
Calculate m using the slope formula
m = (y₂ - y₁ ) / (x₂ - x₁ )
with (x₁, y₁ ) = (- 2, - 20) and (x₂, y₂ ) = (9, 79)
m = =
= 9
Using (a, b) = (- 2, - 20), then
y - (- 20) = 9(x - (- 2)), that is
y + 20 = 9(x + 2)
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)