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Angelina_Jolie [31]
3 years ago
9

Help me with this please

Mathematics
2 answers:
notka56 [123]3 years ago
5 0
Greetings!

To convert a mixed number into an improper fraction, you multiply the whole number by the denominator. Then, you add that number to the numerator.

Examples:
a)
8 1/4

8*4=32
32+1=33
=33/4

b)
3 4/5
3*5=15
15+4=19
=19/5

I'll let you attempt the last one on your own. :)

Hope this helps.
-Benjamin
qwelly [4]3 years ago
5 0
I 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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Answer:

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

Volume =x^2h

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.

Surface Area of the Top and Bottom =2x^2

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

The material for the sides costs ​$2.50 per square centimeter.

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

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3 years ago
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