10 1/3 + 4 2/6

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Answer to a math question 10 1/3 + 4 2/6

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$\frac{ 31 }{ 3 }+4 \frac{ 2 }{ 6 }$

$\frac{ 31 }{ 3 }+\frac{ 26 }{ 6 }$

$\begin{align*}&\frac{ 44 }{ 3 } \\&\begin{array} { l }14 \frac{ 2 }{ 3 },& 14.\overline{ 6 }\end{array}\end{align*}$

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Exercise 1 An ejidal association wishes to determine the distribution for the three different crops that it can plant for the next season on its available 900 hectares. Information on the total available and how many resources are required for each hectare of cultivation is shown in the following tables: Total available resource Water 15,000 m3 Fertilizer 5,000 kg Labor 125 day laborers Requirements per cultivated hectare Corn Soybeans Wheat Water 15 25 20 Fertilizer 5 8 7 Labor** 1/8 1/5 1/4 *The data in fraction means that with one day laborer it will be possible to care for 8, 5 and 4 hectares respectively. * Sales of crops 1 and 3, according to information from the Department of Agriculture, are guaranteed and exceed the capacity of the cooperative. However, soybeans must be limited to a maximum of 150 hectares. On the other hand, the profits for each hectare of crop obtained are estimated at: $7,500 for corn, $8,500 for soybeans and $8,000 for wheat. The objectives are to determine: • How many hectares of each crop must be allocated so that the profit is maximum. R= • The estimated profits for the ejidal cooperative in the next growing season. R=

0.1x8.2

392929-9

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10 1/3 + 4 2/6

Answer to a math question 10 1/3 + 4 2/6

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