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Find the dimensions of a rectangle whose length is a foot longer than twice its width and whose perimeter is 56 feet.

Sagot :

25veroprisyazhnyuk

Answer:

length=19, width=9

Step-by-step explanation:

1) So if the length is a foot longer than twice its width, then lets say the width is x and the length is 2x+1

2) Since we are trying to figure out the perimeter we must add 2 lengths+2 widths which is supposed to give us 56:

                                2(2x+1) + 2(x) = 56

3) Simplify:

                 2(2x+1) + 2(x) = 56

                    4x+2 + 2x = 56             Distribute

                      6x + 2 =56                add 2x and 4x

                         6x = 54                   subtract 2

                             x=9                        divide    

4) Put the x in:

                       length=2(9)+1                width=9

                          length=19

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