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stellarik [79]
3 years ago
13

Look at photo (WITH THE PHOTO)

Mathematics
2 answers:
frez [133]3 years ago
8 0
The answer to that would be 3/8
padilas [110]3 years ago
5 0

Answer:

I got 11/12

Step-by-step explanation:

Because you would take 2/3 × 1/4

They don't have the same denominators so you have to find a multiple. 12 is a multiple of both 3 and 4. In order to get a denominator of 12 for 2/3 you would have to multiply 3 by 4 and 2 by 4 leaving you with 8/12. Then you would have to multiply 1 by 3 and 4 by 3 leaving you with 3/12. Then you add them giving you 11/12. Hope this helps

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A closed box with a square base is to have a volume of 171 comma 500 cm cubed. The material for the top and bottom of the box co
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C(x)=\dfrac{20x^3+1715000}{x}\\$Minimum cost, C(35)=\$29,400

The dimensions that will lead to minimum cost of the box are a base length of 35 cm and a height of 140 cm.

Step-by-step explanation:

Volume of the Square-Based box=171,500 cubic cm

Let the length of a side of the base=x cm

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x^2h=171,500\\h=\dfrac{171500}{x^2}

The material for the top and bottom of the box costs ​$10.00 per square​ centimeter.

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Therefore, Cost  of the Top and Bottom =\$10X2x^2=20x^2

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Surface Area of the Sides=4xh

Cost of the sides=$2.50 X 4xh =10xh

\text{Substitute h}$=\dfrac{171500}{x^2} $into 10xh\\Cost of the sides=10x(\dfrac{171500}{x^2})=\dfrac{1715000}{x}

Therefore, total Cost of the box

= 20x^2+\dfrac{1715000}{x}\\C(x)=\dfrac{20x^3+1715000}{x}

To find the minimum total cost, we solve for the critical points of C(x). This is obtained by equating its derivative to zero and solving for x.

C'(x)=\dfrac{40x^3-1715000}{x^2}\\\dfrac{40x^3-1715000}{x^2}=0\\40x^3-1715000=0\\40x^3=1715000\\x^3=1715000\div 40\\x^3=42875\\x=\sqrt[3]{42875}=35

Recall that:

h=\dfrac{171500}{x^2}\\Therefore:\\h=\dfrac{171500}{35^2}=140cm

The dimensions that will lead to minimum costs are base length of 35cm and height of 140cm.

Therefore, the minimum total cost, at x=35cm

C(35)=\dfrac{20(35)^3+1715000}{35}=\$29,400

8 0
3 years ago
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