The farmer will need:
[tex]\boxed{191\frac{11}{12}yd}[/tex]
In order to enclose the area shown in the figure below.
Explanation:
The diagram below shows the representation of this problem. Let:
[tex]x: The \ length \ of \ the \ rectangular \ pastures \\ \\ y: The \ width \ of \ the \ rectangular \ pastures[/tex]
We know that:
[tex]x=5\frac{5}{8}yd \\ \\ y=30\frac{2}{9}yd[/tex]
So the fencing the farmer needs can be calculated as the perimeter of the two adjacent rectangular pastures:
[tex]P=2(x+y)+y \\ \\ P=2(50\frac{5}{8}+30\frac{2}{9})+30\frac{2}{9} \\ \\ P=100\frac{10}{8}+60\frac{4}{9}+30\frac{2}{9} \\ \\ P=100\frac{10}{8}+90\frac{6}{9} \\ \\ P=100\frac{5}{4}+90\frac{2}{3} \\ \\ P=190(\frac{15+8}{12}) \\ \\ P=190(\frac{23}{12}) \\ \\ \\ Expressing \ as \ a \ mixed \ fraction: \\ \\ P=190+1+\frac{11}{12} \\ \\ P=191+\frac{11}{12} \\ \\ \boxed{P=191\frac{11}{12}yd}[/tex]
Learn more:
Enclosing areas: https://brainly.com/question/1904034
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