Question The perimeter of a rectangular playing field is 244 feet. If its length is 2 feet longer than twice its width, what are the dimensions of the field?

Answer: L=82 and W=40 Step-by-step explanation: The answer is W=40 and L=82 If you’re wondering how to solve it, then let “P” represent perimeter “L” represent length and “W” represent width. P=2L+2W L=2W+2 244=2(2W+2)+2W 244=4W+4+2W 244=6W+4 6W=244 244/6W= W=40 Now plug the 40 in for w in: L=2W+2 L=2(40)+2 L=80+2 L=82 final answer: L=82 and W=40 Reply

The dimensions of the field are; L=82 and W=40 What is the perimeter? It’s the sum of the length of the sides used to make the given figure. Given that perimeter of a rectangle playing field is 2the 44 feet. If its length is 2 feet longer than twice its width, then; let “P” represent perimeter “L” represent length and “W” represent width. Therefore, the perimeter of a rectangular playing field, P=2L+2W Also, L = 2W+2 Thus, P=2L+2W 244=2(2W+2)+2W 244=4W+4+2W 244=6W+4 6W=244 244/6W W=40 Now plug the 40 in for w; L=2W+2 L=2(40)+2 L=80+2 L=82 Hence, the dimensions of the field are; L=82 and W=40 Learn more about perimeter here: https://brainly.com/question/10466285 #SPJ2 Reply

Answer:Step-by-step explanation:dimensionsof thefieldare; L=82 and W=40## What is the perimeter?

lengthof thesidesused to make the givenfigure.perimeterof a rectangle playing field is 2the 44 feet. If its length is 2 feetlongerthantwiceits width, then;perimeter“L” representlengthand “W” representwidth.rectangular playing field,plugthe 40 in for w;dimensionsof thefieldare; L=82 and W=40perimeterhere: