Yes.
Length = 20 m and width = 9 m
Perimeter of a rectangle is 2 times (length + width) P = 2 (L+W) In this case the length is 2m more than twice the width. That is written L = 2 + 2W P = 2 (L + W) 58 = 2 (2+ 2W + W) 58 = 2 (2 + 3W) 58/2 = (2(2+3W))/2 29 = 2 + 3W 29 - 2 = 2 + 3W - 2 27 = 3W 9 = W L = 2 + 2W L = 2 +2(9) L = 2 + 18 L = 20 For a rectangle with a perimeter of 58 m and a length that is 2 m more than twice the width, the length is 20 m and the width is 9 m.
it is 2
1cm:15m If 1 cm = 20 m then 6 cm = 120 m real length
The perimeter is L(length) x 2 + W(width) x 2 So 2L + 2W = 88 And L = W+4 So we can now replace the L in the equation and we get 2(W+4) + 2W = 88 Work out the brackets and we get 2W + 8 + 2 W = 88 Subtract 8 from both sides, and we get 2W + 2W = 80 or, 4W = 80 so we can divide both sides by 4 and we get W = 20 So now we can add 4 to the width and we get the length as 24. Now, to check the calculation, we add them all up Length 1 = 24 Length 2 = 24 Width 1 ==20 Width 2 ==20 Perimeter - 88
You can cut 10 pieces of 2-meter length from a 20-meter long material. This is calculated by dividing the total length (20 meters) by the length of each piece (2 meters). So, 20 ÷ 2 = 10. This means you can get ten 2-meter segments from a 20-meter length.
length ≥ 10 m and length ≤ 20 m
Its impossible as on comparing 1 m and 20 m we can get 1 m <20 m .so,20 meter cannot be cut out of 1 meter .But,yes 1 meter can be cut off 20 meter.
-18
For unit of length and mass and conversion . The conversion relations between dm and m are given .The relation is as follows we can say that 1 m=10 dm. thus,20 dm=2 m
Length = 20 m and width = 9 m
7/1.4 = 5 pieces
The length available for cutting into pieces is 4.50 - 0.42 = 4.08 m If this length is now cut into 6 pieces then each piece measures 4.08 ÷ 6 = 0.68 m or 68 cm
about 20 cm in length
Perimeter of a rectangle is 2 times (length + width) P = 2 (L+W) In this case the length is 2m more than twice the width. That is written L = 2 + 2W P = 2 (L + W) 58 = 2 (2+ 2W + W) 58 = 2 (2 + 3W) 58/2 = (2(2+3W))/2 29 = 2 + 3W 29 - 2 = 2 + 3W - 2 27 = 3W 9 = W L = 2 + 2W L = 2 +2(9) L = 2 + 18 L = 20 For a rectangle with a perimeter of 58 m and a length that is 2 m more than twice the width, the length is 20 m and the width is 9 m.
20 m = 65.62 feet 12 m = 39.37 feet (rounded)
It is m^2 square units.