Question 1171946
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Let *[tex \Large x] represent the measure of the short piece.  Then the long piece is *[tex \Large 2x] and the middle-sized piece is *[tex \Large x\ +\ 4].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 2x\ +\ x\ +\ 4\ =\ 28]


Solve for *[tex \Large x], and then calculate *[tex \Large 2x] and *[tex \Large x\ +\ 4].

																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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