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3 years ago

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:

729 $cm^{3}$

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

= 729 $cm^{3}$

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

Andru [333]

The sum of these numbers is an even number and it is prime.
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3 years ago

D Evaluate arcsin (6)] at x = 4. 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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