Triangle A B C is reflected and then dilated to form smaller triangle A double-prime B double-prime C double-prime Which transfo
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The dimensions such as length is 18 and width is 16 or length is 32 and width is 9 will produce an enclosed area of 288ft.
<h3>What is the dimension of length and width?</h3>We know that,
P = l + 2w
50 = l + 2w
50 - 2w = l
A = l x w
288 = (50 - 2w)w
288 = 50w -2w²
2w² - 50w + 288
By dividing throughout by 2
w² - 25w + 144
This is in quadratic equation form.
By applying quadratic formula,
w = (-(-25) ± ) / 2
w = (-(-25) ± 7)/2
w1 = (-(-25) + 7)/2
w1 = (25 + 7) / 2
w1 = 16
w2 = (-(-25) - 7)/2
w2 = (25 - 7) / 2
w2 = 9
By substituting the values in eq1,
A = l x w
if w = 16
288 = l x 16
l = 288/16
l = 18
if w = 9
288 = l x 9
l = 288/9
l = 32
The dimensions will be length is 18 and width is 16 or length is 32 and width is 9.
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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)