Drew creates a table of ordered pairs representing the width and area of a dog pen. which situation could describe drew’s plans for the dog pen? a) drew has 36 feet of fencing and wants to represent the possible areas of the dog pen for possible widths. b) drew has a maximum of 81 feet of fencing to use around the pen, and the length must be twice the width. c) drew is building a rectangular dog pen with a length that is up to 4 feet more than the width and has a perimeter of at least 77 feet. d) drew wants to build a dog pen that is between 7 feet and 11 feet in width and uses, at most, 18 feet of fencing.

A) Drew has 36 feet of fencing and wants to represent the possible areas of the dog pen for possible widths.  

Explanation:

The first entry in the table is 7 ft for the width.  Since the area is 77 sq ft, this means the length must be 77/7 = 11.  This means the perimeter would be 7*2+11*2 = 14+22 = 36.

The second entry in the table is 8 ft for the width.  The length would then be 80/8 = 10.  This means the perimeter would be 8*2+10*2 = 16+20 = 36.

The third entry in the table is 9 ft for the width.  The length would then be 81/9 = 9.  This means the perimeter would be 9*2+9*2 = 18+18 = 36.

The fourth entry in the table is 10 ft for the width.  The length would then be 80/10 = 8.  This means the perimeter would be 10*2+8*2 = 20+16 = 36.

The last entry in the table is 11 ft for the width.  The length would then be 77/11 = 7.  This means the perimeter would be 11*2+7*2 = 22+14 = 36.

This means Drew must have 36 feet of fence and wants to find the possible areas.

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