Let the width be (x-2) and the length be x: width*length = area (x-2)*x = 35 square inches x2-2x = 35 x2-2x-35 = 0 Solving the above quadratic equation works out as: x = -5 or x = 7 it must be the latter because dimensions can't be negative Therefore: length = 7 inches and width = 5 inches Check: 7*5 = 35 square inches
35
35
1 Let length and width be 7x and 5x 2 So 7x*5x = 5040 or 35x2 = 5040 3 Divide each side by 35 and then square root both sides 4 Therefore x = 12 5 Length = 7*12 = 84 inches 6 Width = 5*12 = 60 inches 7 Check: 84*60 = 5040 square inches
The dimensions are: width of 7 units and length of 35 units!
Let the width be (x-2) and the length be x: width*length = area (x-2)*x = 35 square inches x2-2x = 35 x2-2x-35 = 0 Solving the above quadratic equation works out as: x = -5 or x = 7 it must be the latter because dimensions can't be negative Therefore: length = 7 inches and width = 5 inches Check: 7*5 = 35 square inches
Area = 35*21 = 735 square inches
35 and 28
35
Perimeter = 24 inches Area = 35 square inches
It seems to be approximately 83 inches wide (+/- 6 inches) by 35 inches deep (+/- 3 inches) by 35 inches tall (+/- 4 inches)
Area = 5*7 = 35 square inches
35*4*3=420 cubic inches≈0.243055555555556 cubic feet
The width can be any number greater than zero and less than the square root of 35, and the length can be any number greater than the square root of 35, subject to the constraint that the product of the length and the width must be 35.
35
1 Let length and width be 7x and 5x 2 So 7x*5x = 5040 or 35x2 = 5040 3 Divide each side by 35 and then square root both sides 4 Therefore x = 12 5 Length = 7*12 = 84 inches 6 Width = 5*12 = 60 inches 7 Check: 84*60 = 5040 square inches
The length is 7 feet and the width is 5 feet