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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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Answer:

36x - 108 kilometers²

Step-by-step explanation:

Area = length x width

36 ( x - 3 ) =

36x - 108

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Figure A is a scale image of Figure B.<br><br> What is the value of x?<br> PS. it's not 17
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Step-by-step explanation:

24÷(45÷25)= 21

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3 years ago
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Three friends each have some ribbon. Carl has 54 inches of ribbon,Tino has 13.5 feet of ribbon, and Baxter has 3.5 yards of ribb
deff fn [24]

Answer:

  342 inches, or 28.5 feet, or 9.5 yards

Step-by-step explanation:

It is perhaps easiest to start with all the dimensions in inches.

  13.5 ft = (13.5 ft)(12 in/ft) = 162 in

  3.5 yd = (3.5 yd)(36 in/yd) = 126 in

Then the total of the lengths of ribbon is ...

  54 in + 162 in + 126 in = 342 in

Converting to feet, we get ...

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And converting to yards, we get ...

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The total length of ribbon is 342 inches, or 28.5 feet, or 9.5 yards.

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3 years ago
Find the length and width of a rectangle that has the given area and a minimum perimeter. Area: 162 square feet
Gnom [1K]

Answer:

The width and length of rectangle is 12.728 m

Step-by-step explanation:

Let the length of the rectangle = L

let the width of the rectangle = W

The subjective function is given by;

F(p) = 2(L + W)

F = 2L + 2W

Area of the rectangle is given by;

A = LW

LW = 162 ft²

L = 162 / W

Substitute in the value of L into subjective function;

f = 2l + 2w\\\\f = 2(\frac{162}{w} )+2w\\\\f = \frac{324}{w} + 2w\\\\\frac{df}{dw} = \frac{-324}{w^2} +2\\\\

Take the second derivative of the function, to check if it will given a minimum perimeter

\frac{d^2f}{dw^2}= \frac{648}{w^3} \\\\Thus, \frac{d^2f}{dw^2}>0, \ since,\frac{648}{w^3} >0 \ (minimum \ function \ verified)

Determine the critical points of the first derivative;

df/dw = 0

\frac{-324}{w^2} +2 = 0\\\\-324 + 2w^2=0\\\\2w^2 = 324\\\\w^2 = \frac{324}{2} \\\\w^2 = 162\\\\w= \sqrt{162}\\\\w = 12.728 \ m

L = 162 / 12.728

L = 12.728 m

Therefore, the width and length of rectangle is 12.728 m

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