Answer :
The required range of for the width is (20, 28).
Given that,
The perimeter of a rectangular fence is to be at least 120 feet and no more than 168 feet. If the length of the fence is to be twice the width, what is the range of values for the width of the fence is to be determined.
What is a rectangle?
The rectangle is a four-sided polygon whose opposites sides are equal and has an angle of 90° between its sides.
Here,
Let the width of the rectangular fence be w,
according to the question,
length = 2w
The perimeter of the rectangle should be at least 120 feet and not more than 168 feet,
Inequality for the perimeter,
120 ≤ perimeter of the rectangle ≤ 168
120 ≤ 2 * length + 2 * width ≤ 168
120 ≤ 2 * 2w + 2 * w ≤ 168
120 ≤ 6w ≤ 168
dividing by 6
20 ≤ w ≤ 28
Thus, the required range of for the width is (20, 28).
Learn more about rectangles here:
https://brainly.com/question/16021628
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a rectangle has a perimeter of length + length + width + width.
L = length of the fence.
W = width of the fence.
so the perimeter will be L+L+W+W or 2L+2W or 2(L+W).
now, we know that 120 ⩽ 2(L+W).
we also know that 168 ⩾ 2(L+W)
and we also know that whatever the length is, is twice the width, or L = 2W.
[tex]\bf 120\le 2(L+W)\implies 120\le 2(2W+W)\implies 120\le 2(3W)
\\\\\\
120\le 6W\implies \cfrac{120}{6}\le W\implies \boxed{20\le W}\\\\
-------------------------------\\\\
168\ge 2(L+W)\implies 168\ge 2(2W+W)\implies 168\ge 2(3W)
\\\\\\
168\ge 6W\implies \cfrac{168}{6}\ge W\implies \boxed{28\ge W}\\\\
-------------------------------\\\\
20\le W \le 28[/tex]
L = length of the fence.
W = width of the fence.
so the perimeter will be L+L+W+W or 2L+2W or 2(L+W).
now, we know that 120 ⩽ 2(L+W).
we also know that 168 ⩾ 2(L+W)
and we also know that whatever the length is, is twice the width, or L = 2W.
[tex]\bf 120\le 2(L+W)\implies 120\le 2(2W+W)\implies 120\le 2(3W)
\\\\\\
120\le 6W\implies \cfrac{120}{6}\le W\implies \boxed{20\le W}\\\\
-------------------------------\\\\
168\ge 2(L+W)\implies 168\ge 2(2W+W)\implies 168\ge 2(3W)
\\\\\\
168\ge 6W\implies \cfrac{168}{6}\ge W\implies \boxed{28\ge W}\\\\
-------------------------------\\\\
20\le W \le 28[/tex]
