Question 242764
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Let *[tex \Large w] represent the width.  Then the length, because it is a consecutive integer, must be *[tex \Large w\ +\ 1] -- because you always add 1 to find the next consecutive integer.


The perimeter of a rectangle is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P\ =\ 2l\ +\ 2w]


But since we have established a relationship between the length and the width, we can express the perimeter as a function of the width:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(w)\ =\ 2(w\ +\ 1)\ +\ 2w\ =\ 4w\ +\ 2]


Since we are given that the perimeter is equal to 70, we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4w\ +\ 2\ = 70]


Now all you need to do is solve for *[tex \Large w] to get the width, and then add 1 to get the length.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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