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MariettaO [177]
3 years ago
6

An excavating company charges $300 an hour for the use of a backhoe and $500 an hour for the use of a bulldozer. (part of an hou

r counts as a full hour. ) The company employs one operator for 40 hours per week to operate the machinery. If the company wants to bring in at least $18,400 each week from equipment rental, how many hours per week can it schedule the operator to use a backhoe?
Mathematics
1 answer:
Fed [463]3 years ago
4 0

We have that

Cost use of a backhoe----------------------$300 an hour

<span> Estimate for equipment rental each week---------------------$18,400</span>

Total hours operator per week--------------------------------------40

<span> 40*300</span>=$12000---------------------it is the cost per week to use a backhoe

 $12000<$18400  is ok 

<span> The maximum of hours that I can rent </span>a backhoe <span> are 40 hours, since it is the limit of the operator and it do not overcome the total budget </span>

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What is Perpendicular to 5x-3y=30 and passing through (-2,7)
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k:y=m_1x+b_1;\ l:y=m_2x+b_2\\\\k\ \perp\ l\iff m_1\cdot m_2=-1\\\\k\ ||\ l\iff m_1=m_2

We have:

k:5x-3y=30\ \ \ \ |-5x\\\\-3y=-5x+30\ \ \ \ |:(-3)\\\\y=\dfrac{5}{3}x-10\to m_1=\dfrac{5}{3}\\\\l:y=m_2x+b\\\\k\ \perp l\Rightarrow \dfrac{5}{3}m_2=-1\ \ \ \ |\cdot\dfrac{3}{5}\\\\m_2=-\dfrac{3}{5}

l:y=-\dfrac{3}{5}x+b

The line l is passing through (-2, 7). Substitute the coordinates of the point to the equation of a line l:

7=-\dfrac{3}{5}\cdot(-2)+b\\\\\dfrac{6}{5}+b=7\ \ \ \ |-\dfrac{6}{5}\\\\b=\dfrac{35}{5}-\dfrac{6}{5}\\\\b=\dfrac{29}{5}

l:y=-\dfrac{3}{5}x+\dfrac{29}{5}\ \ \ \ |\cdot5\\\\5y=-3x+29\ \ \ \ |+3x\\\\3x+5y=29

<h2>Answer: 3x + 5y = 29</h2>
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