Anonymous

The dimensions of the sides are Perimeter/4 ± Difference/2 340/4 + 30/2 = 85 + 15 = 100 yds long 340/4 - 30/2 = 85 - 15 = 70 yds wide

Anonymous

P = 2(l + w) l = w + 30 Substitute P = 2( w + 30 + w) 340 = 2(2w + 30) 170 = 2w + 30 140 = 2w w = 70 Hence l = 100

Anonymous

The perimeter of a rectangular field is 340 yds. The length is 30 yds longer than the width. 2(l + w) = 340 l + w = 170 w + 30 + w = 170 2w = 140 w = 70 Hence the dimensions: Length: 100 yds. Width: 70 yds.

Anonymous

4 * x + 60 yd = 340 yd Find x

Anonymous

Let the width be x, so length = x + 30 So x + x + x + 30 + x + 30 = 340 4x + 60 = 340 4x = 340 - 60 4x = 280 x = 70

Anonymous

Let L and w be length and width, respectively. 2L + 2w = 340 L = w + 30 Solve the system of equations. 2(w + 30) + 2w = 340 2w + 60 + 2w = 340 4w = 280 w = 70yd L = 70 + 30 = 100yd