Enlarge triangle ABC by scale factor of -2 and centre (4,2)
1 answer:
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Answer:
The percentage increase in Area of rectangle is 200%
Step-by-step explanation:
Given as :
The ratio of the width to the length of a rectangle is 2:3
Let The length of rectangle = 2 x
Let The width of rectangle = 3 x
∵ Area of rectangle = length × width
So, = 2 x × 3 x
Or, = 6 x²
<u>Again</u>
The increased length of rectangle = 2 x + 2 x = 4 x
The increase width of rectangle = 3 x + 50% of 3 x
I.e The increase width of rectangle = 3 x + 1.5 x = 4.5 x
∵ Increased Area of rectangle = increased length × increased width
Or, = 4 x × 4.5 x = 18 x²
So, The percentage increase in area = × 100
Or, The percentage increase in area = × 100
Or , The percentage increase in area = × 100
∴ The percentage increase in area = 200
Hence, The percentage increase in Area of rectangle is 200% Answer
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)