Question 799736
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Area:  *[tex \LARGE A\ =\ xy]


Perimeter:  *[tex \LARGE P\ =\ 2x\ +\ 2y]


Since you are given that A = 21, it must be true that *[tex \LARGE y\ =\ \frac{21}{x}].  You are also given that the perimeter is *[tex \LARGE w].


Substituting in the perimeter expression:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w\ =\ 2x\ +\ 2\left(\frac{21}{x}\right)\ =\ 2x\ +\ \frac{42}{x}]


If *[tex \LARGE w\ =\ 19], then solve *[tex \LARGE 2x\ +\ \frac{42}{x}\ =\ 19] for *[tex \LARGE x].  Note:  "Breadth" is generally taken to mean "width" which is, in turn, generally taken to mean the smaller of the two dimensions of a rectangle.  The two solutions of the quadratic equation that results from the perimeter equation are the length and width of the rectangle and you should report the smaller of the two.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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