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A real estate purveyor purchases a 60{,}00060,00060, comma, 000 square foot \left(\text{ft}^2\right)(ft 2 )(, start text, f, t, end text, squared, )warehouse and decides to turn it into a storage facility. The warehouse's width is exactly \dfrac 2 3 3 2 ​ start fraction, 2, divided by, 3, end fraction of its length. What is the warehouse's width? Round your answer to the nearest foot.

Respuesta :

Answer:

200 feet

Step-by-step explanation:

Area of the warehouse [tex]=60,000$ ft^2[/tex]

Let the length of the warehouse=l

The warehouse's width is exactly  [tex]\dfrac23[/tex] of its length

Therefore: Width of the warehouse[tex]=\dfrac23l[/tex]

Area =Length X Width

Therefore:

[tex]\dfrac23l*l=60000\\$Cross multiply\\2l^2=60000*3\\2l^2=180000\\$Divide both sides by 2\\2l^2 \div 2=180000 \div 2\\l^2=90000\\l^2=300^2\\$Length, l=300 feet\\Recall: Width =\dfrac23l\\$Therefore, Width of the warehouse=\dfrac23*300=200$ feet[/tex]

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