2 yd = 72 in.
1 yd = 36 in 6 yd 5 in - 2 yd 7 in = (5 + 1) yd 5 in - 2 yd 7 in = 5 yd (36 + 5) in - 2 yd 7 in = (5-2) yd (36 + 5 - 7) in = 3 yd 34 in.
2 yd 2 ft
1 mile = 1,760 yd 2 miles = 3,520 yd
52 yd 1 ft - 17 yd 2 ft = 51 yd 4 ft - 17 yd 2 ft = 34 yd 2 ft
2 yd = 6 ft
1 yd = 36 in 6 yd 5 in - 2 yd 7 in = (5 + 1) yd 5 in - 2 yd 7 in = 5 yd (36 + 5) in - 2 yd 7 in = (5-2) yd (36 + 5 - 7) in = 3 yd 34 in.
L = ( 2 yd ) ( 36 in/yd ) = 72 in <-------------------------------
2 yd 2 ft
2 yd - 1ft = 1
1 mile = 1,760 yd 2 miles = 3,520 yd
52 yd 1 ft - 17 yd 2 ft = 51 yd 4 ft - 17 yd 2 ft = 34 yd 2 ft
2 yd = 6 ft
7 yd 1 ft - 5 yd 2 ft = 22 ft - 17 ft = 5 ft = 1 yd 2 ft.
9 yd 2 ft. Incidentally, 2 yd 4 ft is a strange way of representing a distance of 3 yd 1 ft.
3 ft = 1 yd → 8 ft = 8 ÷ 3 yd = 2 2/3 yd ≈ 2.67 yd
14 feet or 4 yd 2 ft
No, you cannot construct a triangle with side lengths 2 yd, 9 yd, and 10 yd. This is because the sum of the lengths of the two shorter sides (2 yd + 9 yd = 11 yd) must be greater than the length of the longest side (10 yd) to satisfy the triangle inequality theorem. Since 11 yd is greater than 10 yd, these lengths do not form a triangle.