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zalisa [80]
3 years ago
8

The width of a rectangle is 6 in. less than its length. The perimeter is 68 in. What is the width of the rectangle?

Mathematics
2 answers:
stepan [7]3 years ago
7 0
Length= y in.
Width= (y-6) in.
Perimeter = 2(L*W)
Perimeter=2L+2W
68in.=2(y)+2(y-6)
68=2y+2y-12
68=4y-12
68+12=4y
4y=80
y=20
Length= 20 inches, width =14 inches
kolezko [41]3 years ago
7 0
To answer let the length of the rectangle be represented by L. With the condition from the problem, the width is L - 6. The perimeter of the rectangle is twice the sum of the length and width. With the given values, the equation becomes,
                                  P = 2L + 2W
                                 68 in = 2L + 2(L - 6)
The value of L from the equation is 20 inches. Therefore, the width is equal to 14 inches. 
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Find the smallest possible value of$$\frac{(y-x)^2}{(y-z)(z-x)} \frac{(z-y)^2}{(z-x)(x-y)} \frac{(x-z)^2}{(x-y)(y-z)},$$where $x
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Step-by-step explanation:

For this case we have the following expression:

\frac{(y-x)^2}{(y-z)(z-x)} \frac{(z-y)^2}{(z-x)(x-y)} \frac{(x-z)^2}{(x-y)(y-z)}

And we can rewrite this expression like this:

\frac{(y-x)^2}{(x-y)^2} \frac{(x-z)^2}{(z-x)^2}\frac{(z-y)^2}{(y-z)^2}

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On this case we can see that (a-b)^2 = (b-a)^2

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So then we can simplify all the expression and we got this:

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