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2 years ago

14

Jim is building a rectangular deck and wants the length to be 1 ft greater than the width. what will be the dimensions of the de

ck if the perimeter is to be 66 ft?

1 answer:

Travka [436]2 years ago

3 0

Let x = width
x+1 is then the length

2x+2(x+1)=66
2x+2x+2=66
4x=64
x=16
deck will be 16x17, nice for a BBQ. :)

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3 years ago

Read 2 more answers

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$\bf \qquad \qquad \textit{double proportional variation}\\\\
\begin{array}{llll}
\textit{\underline{y} varies directly with \underline{x}}\\
\textit{and inversely with \underline{z}}
\end{array}\implies y=\cfrac{kx}{z}\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------\\\\$

$\bf \begin{cases}
p=pedalisi\\
w=weight\\
s=\textit{sitting height}
\end{cases}\quad 
\begin{array}{llll}
%pelidisi, varies directly as the cube root of a person's weight in grams and inversely as the person's sitting height in centimeters.
\textit{pelidisi varies directly}\\
\textit{as cube root of weight}\\
\textit{and inversely to }\\
\textit{sitting height}
\end{array}\implies p=\cfrac{k\sqrt[3]{w}}{s}\\\\
-------------------------------$

$\bf \textit{we know that }
\begin{cases}
w=48,820\\
s=78.7\\
p=100
\end{cases}\implies 100=\cfrac{k\sqrt[3]{48820}}{78.7}
\\\\\\
100\cdot 78.7=k\sqrt[3]{48820}\implies \cfrac{7870}{\sqrt[3]{48820}}=k
\\\\\\
thus\qquad \boxed{p=\cfrac{\frac{7870}{\sqrt[3]{48820}}\sqrt[3]{w}}{s}}
\\\\\\
\textit{now, what is \underline{p} when }
\begin{cases}
w=54,688\\
s=72.6
\end{cases}?\implies p=\cfrac{\frac{7870}{\sqrt[3]{48820}}\sqrt[3]{54688}}{72.6}$

now, if that value is less than 100, then the fellow is "undernourished", otherwise, is overfed.

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3 years ago