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A rectangle has a length of x inches and a width 2 inches less than the length.If the dimensions were doubled, what would be the area, in square inches, of the new rectangle in terms of x?2x−42 x − 48x−88 x − 82x2−4x2 x 2 − 4 x4x2−8x4 x 2 − 8 x

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Lanuel

If the dimensions of the rectangle were doubled, the area in square inches is: [tex]Area = 4x^2 - 8x[/tex]

  • Let the length of the rectangle be L.
  • Let the width of the rectangle be W.

Given the following data:

  • Length = x inches
  • Width = x - 2 inches

If the dimensions of the rectangle were doubled, the value of L and W becomes;

[tex]L = 2x[/tex]

[tex]W = 2(x - 2)[/tex]

Mathematically, the area of a rectangle is given by the formula;

[tex]Area = Length[/tex] × [tex]Width[/tex]

[tex]Area = 2x[/tex] × [tex]2(x-2)[/tex]

[tex]Area = 2x[/tex] × [tex]2x - 4[/tex]

[tex]Area = 4x^2 - 8x[/tex]  square inches.

Find more information: brainly.com/question/897975

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