Patio Dimensions

harpazo

Pure Mathematics
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Mar 20, 2018
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#1
A contractor orders 8 cubic yards of premixed cement, all of which is to be used to pour a patio that will be 4 inches thick. If the length of the patio is specified to be twice the width, what will be the patio dimensions?

Note: 1 cubic yard = 27 cubic feet

Seeking the first-two steps.
 

MarkFL

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#2
The volume \(V\) of the patio, in cubic feet, is:

\(\displaystyle V=w\ell\frac{1}{3}\)

We are told:

\(\displaystyle \ell=2w\)

Hence:

\(\displaystyle V=w(2w)\frac{1}{3}=\frac{2w^2}{3}\)

Ad we know:

\(\displaystyle V=8\cdot27\)

Thus:

\(\displaystyle \frac{2w^2}{3}=8\cdot27\)

Solve for \(w\).
 

harpazo

Pure Mathematics
Banned
Mar 20, 2018
5,857
361
83
NYC
#3
The volume \(V\) of the patio, in cubic feet, is:

\(\displaystyle V=w\ell\frac{1}{3}\)

We are told:

\(\displaystyle ell=2w\)

Hence:

\(\displaystyle V=w(2w)\frac{1}{3}=\frac{2w^2}{3}\)

Ad we know:

\(\displaystyle V=8\cdot27\)

Thus:

\(\displaystyle \frac{2w^2}{3}=8\cdot27\)

Solve for \(w\).
1. What is ell?

2. Where did 1/3 come from?

I understand why you multiplied 8 by 27.
 

MarkFL

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#4
1.) A typo, where I forgot to escape the string. Post corrected.

2.) 4 inches is 1/3 of a foot.
 
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harpazo

Pure Mathematics
Banned
Mar 20, 2018
5,857
361
83
NYC
#6
The volume \(V\) of the patio, in cubic feet, is:

\(\displaystyle V=w\ell\frac{1}{3}\)

We are told:

\(\displaystyle \ell=2w\)

Hence:

\(\displaystyle V=w(2w)\frac{1}{3}=\frac{2w^2}{3}\)

Ad we know:

\(\displaystyle V=8\cdot27\)

Thus:

\(\displaystyle \frac{2w^2}{3}=8\cdot27\)

Solve for \(w\).
sDraw_2018-10-15_13-27-34.png
 

MarkFL

La Villa Strangiato
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Jan 25, 2018
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St. Augustine
#7
\(\displaystyle \frac{2w^2}{3}=8\cdot27\)

\(\displaystyle w^2=2^2\cdot3^4\implies w=2\cdot3^2=18\quad\checkmark\)