giggy1115 giggy1115
  • 14-08-2018
  • Mathematics
contestada

if the perimeter is 168 and the width is 6 feet longer than thr length what are the dimensions

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jdoe0001 jdoe0001
  • 14-08-2018

[tex] \bf \textit{perimeter of a rectangle}\\\\
P=2(L+w)~~
\begin{cases}
L=length\\
w=width\\
----------\\
P=168\\
w=\stackrel{\textit{6 feet longer than L}}{L+6}
\end{cases}\implies 168=2(L+\stackrel{w}{L+6})
\\\\\\
84=2L+6\implies 78=2L\implies \cfrac{78}{2}=L\implies \boxed{39=L}
\\\\\\
w=39+6\implies \boxed{w=45} [/tex]

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