Respuesta :
the picture in the attached figure
Let
x--------> the length of the rectangular pastures
y--------> the width of the rectangular pastures
we know that
[tex] x=50\frac{5}{8}yd\\ \\ y=30\frac{2}{9} yd [/tex]
Perimeter of the two adjacent rectangular pastures is equal to
[tex] P=2*(x+y)+y [/tex]
[tex] P=2*(50\frac{5}{8} +30\frac{2}{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} yd [/tex]
[tex] P=190\frac{23}{12}=191\frac{11}{12} yd [/tex]
therefore
the answer is
[tex] 191\frac{11}{12} yd [/tex]
The farmer needs 191 11/12 yards of fencing.
Given that a farmer needs to enclose two adjacent rectangular pastures, to determine how much fencing does the farmer need, knowing that the length of both fields combined is 50 5/8 yards, and the width of each of them (also including the division between them) is 30 2/9 yards, the following calculation must be made:
Both fields must be considered as the same rectangle, with an internal division the size of its width.
- (2 x 50 5/8) + (3 x 30 2/9) = X
- (2 x 50,625) + (3 x 30,222) = X
- 101.25 + 90.66 = X
- 191.916 = X
- 191 11/12 = X
Therefore, the farmer needs 191 11/12 yards of fencing.
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