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Brut [27]
2 years ago
7

The formula for the volume of a cube is V = s³​ , where s​ is the length of one side of the cube. if s= 1/3​​ of a unit, what is

the volume, in cubic units, of the cube?
Mathematics
2 answers:
grandymaker [24]2 years ago
8 0

Answer:

Solution given:

volume

V=s³

if s=1/3 of units

new volume will be

V new=(1/3)³=1/27 or 0.037unit³ is your answer

AleksandrR [38]2 years ago
4 0

FORMULA:

  • Volume of cube = s³

ANSWER:

If s is 1/3, then

Volume = (1/3)³

  • 1/27 units³

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sweet-ann [11.9K]
The answer to this is negative
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The probability that an event will occur is 15%. Which of these best describes the likelihood of the event occurring? (5 points)
olga_2 [115]
Unlikely because certain would be 100% which is wrong
6 0
2 years ago
A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot
ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box  is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = (x\times x)

                               =x^2

The volume of the box  is = area of the base × height

                                           =x^2h

Therefore,

x^2h=272

\Rightarrow h=\frac{272}{x^2}

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

=20x^2 cents

The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

=10x^2 cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

=4xh\times 1.5 cents

=6xh cents

Total cost = (20x^2+10x^2+6xh)

                =30x^2+6xh

Let

C=30x^2+6xh

Putting the value of h

C=30x^2+6x\times \frac{272}{x^2}

\Rightarrow C=30x^2+\frac{1632}{x}

Differentiating with respect to x

C'=60x-\frac{1632}{x^2}

Again differentiating with respect to x

C''=60+\frac{3264}{x^3}

Now set C'=0

60x-\frac{1632}{x^2}=0

\Rightarrow 60x=\frac{1632}{x^2}

\Rightarrow x^3=\frac{1632}{60}

\Rightarrow x\approx 3

Now C''|_{x=3}=60+\frac{3264}{3^3}>0

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is =\frac{272}{3^2}

                                          =30.22 ft.

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

7 0
3 years ago
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daser333 [38]

Answer:

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Step-by-step explanation:

(-2)(-3)^2-2(2-5). Original

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4 0
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Oxana [17]
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Therefore the answer is a.
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Next you divide it by 6, 18 divided by 6 =3
Therefore 1/6=3
I hope that this helped

6 0
3 years ago
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