All categories

3 years ago

T&L - Math 7 - Volume 7.4 -19-20

Mathematics

1 answer:

Mashcka [7]3 years ago

5 0

Answer: You would need 512 cubic centimetres

Step-by-step explanation: The first approach to this question would be to understand the properties of the shape given in the question.

If a cube has an edge with length 8 cm, then all edges measure 8 cm as well. That is one property of a cube. Hence, the length, width and height all measure 8 cm each.

The volume of a cube is given as follows;

V = L x W x H (and the answer is expressed as V³)

Since the length , width and height all measure 8 cm, the volume can simply be expressed as

V = L x L x L

V = L³

V = 8³

V = 512 cm³

Therefore to completely fill a cube with edge length of 8 cm you would need 512 cubic centimetres.

You might be interested in

Brainiest to whoever right

sasho [114]

Is there more context to the question?

6 0

3 years ago

Read 2 more answers

Rewrite log2^8=3 in exponential form

kolezko [41]

2^3=8 is log2^8=3 in exponential form

7 0

3 years ago

What is the distance between the following two points?<br> (-5, -1) and (-8, -4)

liberstina [14]

The answer is -3 , -3

5 0

3 years ago

Help with practice problem

padilas [110]

Answer:

attached above with detail explanation.

5 0

2 years ago

Find all solutions of each equation on the interval 0 ≤ x < 2π.

Korvikt [17]

Answer:

$x = 0$ or $x = \pi$ .

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

$\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}$ .

$\displaystyle \sec{x} = \frac{1}{\cos{x}}$ .

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, $\sin^{2}{x} = 1 - \cos^{2}{x}$ . Therefore, for the square of tangents,

$\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}$ .

This equation will thus become:

$\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2$ .

To simplify the calculations, replace all $\cos^{2}{x}$ with another variable. For example, let $u = \cos^{2}{x}$ . Keep in mind that $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$ .