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Sliva [168]
3 years ago
12

Can you find the height, in feet, of the stacked hay bales?

Mathematics
2 answers:
grandymaker [24]3 years ago
8 0
Yes you can. you just got to do the work.
Dmitriy789 [7]3 years ago
3 0
The height of the stacked hay bales is approxim 6feet tall
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A regular hexagon is dilated by a scale factor of 7/5 to create a new hexagon. how does the perimeter of the new hexagon compare
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This is the concept of algebra, let the original length of the hexagon be x, after the dilation the length became x×7/5=7/5x
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6*7/5x=8.4x
therefore the perimeter of the new compared to the old will be:
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A rectangular box with a volume of 684 ftcubed is to be constructed with a square base and top. The cost per square foot for the
SCORPION-xisa [38]

Answer:

The dimensions of the rectangular box is 36.23 ft×36.23 ft×4.345 ft.  

Minimum cost=2046.16 cents      

Step-by-step explanation:

Given that a rectangular box with a volume of 684 ft³.

The base and the top of the rectangular box is square in shape

Let the length and width of the rectangular box be x.

[since the base is square in shape,  length=width]

and the height of the rectangular box be h.

The volume of rectangular box is = Length ×width × height

                                                      =(x²h) ft³

x^2h=684\Rightarrow h=\frac{684}{x^2}  (1)

The area of the base and top of rectangular box is = x² ft²

The surface area of the sides= 2(length+width) height

                                               =2(x+x)h

                                               =4xh ft²

The total cost to construct the rectangular box is

=[(x²×20)+(x²×15)+(4xh×1.5)] cents  

=(20x²+15x²+6xh) cents

=(25x²+6xh) cents

Total cost= C(x).

C(x) is in cents.

∴C(x)=25x²+6xh

Putting h=\frac{684}{x^2}

C(x)=25x^2+6x\times\frac{684}{x^2} \Rightarrow C(x)=25x^2+\frac{4104}{x}

Differentiating with respect to x

C'(x)=50x-\frac{4104}{x^2}

To find minimum cost, we set C'(x)=0

\therefore50x-\frac{4104}{x^2}=0\\\Rightarrow50x=\frac{4104}{x^2}\\\Rightarrow x^3=\frac{4104}{50}\Rightarrow x\approx 4.345 ft.

Putting the value x in equation (1) we get

h=\frac{684}{(4.345)^2}

 ≈36.23 ft.

The dimensions of the rectangular box is 36.23 ft×36.23 ft×4.345 ft.  

Minimum cost C(x)=[25(4.345)²+10(4.345)(36.23)] cents

                               =2046.16 cents      

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