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3 years ago

If each of the cubes in a Rubik's Cube represented 1 cubic inch, what would be the volume of the Rubik's Cube?

Mathematics

1 answer:

kicyunya [14]3 years ago

5 0

Answer:

The volume of the Rubik's Cube is 27 in^3

Step-by-step explanation:

Each face of a Rubik's Cube has 3x3 cubes, then each face has 9 small cubes.

We can think of a Rubik's Cube as 3 pieces, such that each piece has 9 small cubes on it, and the 3 pieces are: Front side, middle part, back part.

Then the total number of small cubes on a Rubik's cube is 9*3 = 27 small cubes.

Now we know that each one of these small cubes has a volume of 1 in^3

Then the volume of 27 of these cubes is 27 times the volume of a single smaller cube, then the volume of the Rubik's Cube is:

V = 27*( 1 in^3) = 27 in^3

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The area of the regular pentagon is 6.9 cm2. What is the perimeter? 2 cm 5 cm 10 cm 20 cm

disa [49]

Given

Area of the regular pentagon is 6.9 cm².

Find out the perimeter of a regular pentagon

To proof

Formula

Area of regular pentagon is

$= \frac{1}{4}\sqrt{5(5+2\sqrt{5})}\ a^{2}$

As given in the question

area of regular pentagon = 6.9 cm²

now equating the area value with the area formula.

$6.9 =\frac{1}{4}\sqrt{5(5+2\sqrt{5})} a^{2}$

Now put

√5 = 2.24 ( approx)

put in the above equation

$\frac{6.9\times 4}{\sqrt{5(5+2\times2.24)}}=a^{2}\\\frac{27.6}{\sqrt{47.4}}=a^{2}\\\frac{27.6}{6.88}=a^{2}$

thus

a² = 4.01

a = √ 4.01

a = 2.0 cm ( approx)

As perimeter represented the sum of all sides.

i.e regular pentagon have five sides of equal length.

Thus

perimeter of the regular pentagon = 5 × side length

= 5 ×2.00

therefore the perimeter of the regular pentagon = 10cm

option c is correct

Hence proved

3 0

3 years ago

Read 2 more answers

The distance between the points (1,2) and (x,-1) is 5 units. Find the possible values of x

Mariulka [41]

Answer:

-3,5

Step-by-step explanation:

$d = \sqrt{(x _{2} - x _1) {}^{2} + (y _{2} - y _{1} {} ) {}^{2} }$

$5 = \sqrt{(x - 1) {}^{2} + ( - 1 - 2) {}^{2} }$

$5 = \sqrt{(x - 1) {}^{2} + ( - 3) {}^{2} }$

$25 = (x - 1) {}^{2} + 9$

$16 = (x - 1) {}^{2}$

$4 = x - 1$

$x = 5$

$- 4 = x - 1$

$x = - 3$

7 0

2 years ago

I NEED HELP WITH #12-15 PLEASE

sveta [45]

12-15=3 ............................................................

5 0

3 years ago

An arch for a bridge over a highway is in the form of a semiellipse. The top of the arch is 35 feet above ground (the major axis

jeka94

For given problem:

Put midpoint of ellipse, (0,0) at epicenter of bridge at ground level.

Specified length of vertical major axis = 70=2a

a=35

a^2=1225

Equation of ellipse:

x^2/b^2+y^2=1

plug in coordinates of given point on ellipse(25, 10)

25^2/b^2 + 10^2/a^2 = 1

625/b^2 + 100/1225=1

625/b^2 = 1 - 100/ 1225 = .918

b^2 = 625/.918 ≈ 681

b ≈ 26.09

length of minor axis = 2b = 2(26.09) ≈ 52.16 ft

Span of bridge ≈ 52.16 ft

8 0

3 years ago

I don't understand the math homework I need to do tonight please help

SpyIntel [72]

3 to the -2 power is basically 3x-3, which is -9, and -5 to the -3 power is (-5)(-5)(-5) which is equal to -125. hope that helped you out, bud!

4 0

3 years ago