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Ivan
3 years ago
13

Find the perimeter of a field that has length 2/x + 1 and width 5/x^2 -1.

Mathematics
1 answer:
Semmy [17]3 years ago
3 0

Answer:

D)4x + 6/(x + 1)(x - 1)

Step-by-step explanation:

A field is basically a rectangle, so to find the perimeter of our field we are using the formula for the perimeter of a rectangle

p=2(l+w)

where

p is the perimeter

l is the length

w is the width

We know from our problem that the field has length 2/x + 1 and width 5/x^2 -1, so l=\frac{2}{x+1} and w=\frac{5}{x^2-1}.

Replacing values:

p=2(l+w)

p=2(\frac{2}{x+1} +\frac{5}{x^2-1})

Notice that the denominator of the second fraction is a difference of squares, so we can factor it using the formula a^2-b^2=(a+b)(a-b) where a is the first term and b is the second term. We can infer that a=x^2 and b=1^2. So, x^2-1=(x+1)(x-1). Replacing that:

p=2(\frac{2}{x+1} +\frac{5}{x^2-1})

p=2(\frac{2}{x+1} +\frac{5}{(x+1)(x-1})

We can see that the common denominator of our fractions is (x+1)(x-1). Now we can simplify our fraction using the common denominator:

p=2(\frac{2(x-1)+5}{(x+1)(x-1)} )

p=2(\frac{2x-2+5}{(x+1)(x-1)} )

p=2(\frac{2x+3}{(x+1)(x-1)} )

p=\frac{4x+6}{(x+1)(x-1)}

We can conclude that the perimeter of the field is D)4x + 6/(x + 1)(x - 1).

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