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Juliette [100K]
2 years ago
7

Jack and Jane are married and both work. However, due to their responsibilities at home, they have decided that they do not want

to work over 65 hours per week combined. Jane is paid $12.50 per hour at her job, and Jack is paid $10 per hour at his. Neither of them are paid extra for overtime, but they are allowed to determine the number of hours per week that they wish to work. If they need to make a minimum of $750 per week before taxes, what is the maximum amount of hours that Jack can work per week according to these limits?

Mathematics
1 answer:
Zanzabum2 years ago
5 0
To solve a system of inequalities we graph both of them.
The inequality representing their combined pay would be
12.50x+10.00y \geq 750.  This is because Jane makes 12.50/hr, Jack makes 10.00/hr, and they want to make at least, so greater than or equal to, $750 combined.
The inequality representing their combined hours working would be
x+y \leq 65, since they do not want their combined hours to be over 65.  In both of these inequalities, <em>x</em> represents the number of hours Jane works and <em>y</em> represent the number of hours Jack works.
To graph these, we solve both of them for <em>y</em>:
x+y \leq 65&#10;\\ x+y-x \leq 65-x&#10;\\ y \leq 65-x
12.50x+10.00y \geq750&#10;\\12.50x+10.00y-12.50x \geq 750-12.50x&#10;\\10.00y \geq 750-12.50x&#10;\\ \frac{10.00y}{10.00} \geq \frac{750}{10.00} - \frac{12.50x}{10.00}&#10;\\y \geq 75-1.25x
The attached screenshot shows what the graph looks like.  Going to the point where they intersect, we see that the shaded region that satisfies both inequalities begins when Jane works 40 hours and Jack works 25.

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nata0808 [166]

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Step-by-step explanation:

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A box of pencils is 5 1/4 inches wide. Seven pencils, laid side by side, take up 2 5/8 inches of the width. How many inches of t
Andre45 [30]

Answer:   Width of box is not taken up by pencils =  2\dfrac58\text{ inches}

Width of each pencil =\dfrac38\text{ inches}

Step-by-step explanation:

Given: Width of pencil box = 5\dfrac14\text{ inches}=\dfrac{21}{4}\text{ inches}

Width of seven pencils = 2\dfrac58\text{ inches}=\dfrac{16+5}{8}\text{ inches}=\dfrac{21}{8}\text{ inches}

Width of box is not taken up by pencils =  Width of pencil box  - Width of seven pencils

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Width of box is not taken up by pencils =  2\dfrac58\text{ inches}

Width of each pencil = (Width of seven pencils ) ÷ 7

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Width of each pencil =\dfrac38\text{ inches}

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