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Blizzard [7]
3 years ago
10

The length of an edge of a cube is 9 centimeters. What is the volume of the cube?

Mathematics
2 answers:
Anni [7]3 years ago
4 0

Answer:

729cm^{3}

Step-by-step explanation:

Seeing as it is a cube, all ide are equal, therefore you can do:

S^{3}    or    L x w x h

either way it works out to be:

9 x 9 x 9

which equals 729

you can't forget about the units, which is cm^{3} since its a 3d shape

S^{3}

= 9 x 9 x 9

= 729cm^{3}

KATRIN_1 [288]3 years ago
3 0

Answer:

729

Step-by-step explanation:

Volume of cube=a^3

Volume of cube=9^3=729

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There are 29235 marbles and 371 jacks. If you want to know the percentage of jacks, do you divide the first number by the second
garik1379 [7]

Answer:

Correct answer: Two decimal places to the right

Step-by-step explanation:

We will set the classic proportion:

29,235   -    100%

     371    -        x%

29,235 : 371 = 100 : x  ⇒ 29,235 · x = 371 · 100

x = (371/ 29,235) · 100 = 0.0127 · 100 = 1.27%

x = 1.27%

You divide first number by the second and then multiply with 100,

which is the same as moving the decimal point two places to the right.

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3 years ago
The sum of 86,68, and 38 is 192. What do you also know about the sum
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3 years ago
D<br> Evaluate<br> arcsin<br> (6)]<br> at x = 4.<br> dx
sineoko [7]

Answer:

\frac{1}{2\sqrt{5} }

Step-by-step explanation:

Let, \text{sin}^{-1}(\frac{x}{6}) = y

sin(y) = \frac{x}{6}

\frac{d}{dx}\text{sin(y)}=\frac{d}{dx}(\frac{x}{6})

\frac{d}{dx}\text{sin(y)}=\frac{1}{6}

\frac{d}{dx}\text{sin(y)}=\text{cos}(y)\frac{dy}{dx} ---------(1)

\frac{1}{6}=\text{cos}(y)\frac{dy}{dx}

\frac{dy}{dx}=\frac{1}{6\text{cos(y)}}

cos(y) = \sqrt{1-\text{sin}^{2}(y) }

          = \sqrt{1-(\frac{x}{6})^2}

          = \sqrt{1-(\frac{x^2}{36})}

Therefore, from equation (1),

\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}

Or \frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}

At x = 4,

\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}

\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}

                   =\frac{1}{6\sqrt{\frac{36-16}{36}}}

                   =\frac{1}{6\sqrt{\frac{20}{36} }}

                   =\frac{1}{\sqrt{20}}

                   =\frac{1}{2\sqrt{5}}

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