All categories

3 years ago

How many cubes with side lengths of \dfrac14\text{ cm}

Mathematics

2 answers:

bogdanovich [222]3 years ago

5 0

Answer:

Step-by-step explanation:

Correct question

How many cubes with side lengths of ¼cm needed to fill the prism of volume 4 cubic units?

We know that,

Volume of a cube is s³

V = s³

Where 's' is length of side of a cube

Given that

The cube has a length of ¼cm, and a cube has equal length

s= ¼cm

Then, it's volume is

V = s³

V = (¼)³ = ¼ × ¼ × ¼

V = 1 / 64 cubic unit

V = 0.015625 cubic unit

Then, given that the volume of the prism to be filled is 4 cubic unit

Then,

As, we have to find the number if cubes so we will divide volume of prism by volume of one cube

Then,

n = Volume of prism / Volume of cube

n = 4 / 0.015625

n = 256

So, then required cubes to filled the prism is 256 cubes.

arsen [322]3 years ago

4 0

Answer:

135

Step-by-step explanation:

You might be interested in

1-Find the missing number:

Yuki888 [10]

1) 23

2) 3

3) 6x+5 (I am not sure what they ask for)

4) -12+x

5) -56x + 16

6) -90+80x

7) -18x+45y-27

8) 6x+24y-42

9) 2/3w+x-1

10) -1.2wy +3y

8 0

2 years ago

Find the volume of the composite solid. Round your answer to the nearest tenth.

elixir [45]

Answer:

Nasaan ang boung question

Step-by-step explanation:

Is 2,4,6,7,9,10

Yan ang nearest tenth

Pinag aralan namin Yan

I hope I help you

4 0

3 years ago

Find the integration of (1-cos2x)/(1+cos2x)

slega [8]

Given:

The expression is:

$\dfrac{1-\cos 2x}{1+\cos 2x}$

To find:

The integration of the given expression.

Solution:

We need to find the integration of $\dfrac{1-\cos 2x}{1+\cos 2x}$ .

Let us consider,

$I=\int \dfrac{1-\cos 2x}{1+\cos 2x}dx$

$I=\int \dfrac{2\sin^2x}{2\cos^2x}dx$ $[\because 1+\cos 2x=2\cos^2x,1-\cos 2x=2\sin^2x]$

$I=\int \dfrac{\sin^2x}{\cos^2x}dx$

$I=\int \tan^2xdx$ $\left[\because \tan \theta =\dfrac{\sin \theta}{\cos \theta}\right]$

It can be written as:

$I=\int (\sec^2x-1)dx$ $[\because 1+\tan^2 \theta =\sec^2 \theta]$

$I=\int \sec^2xdx-\int 1dx$

$I=\tan x-x+C$

Therefore, the integration of $\dfrac{1-\cos 2x}{1+\cos 2x}$ is $I=\tan x-x+C$ .

8 0

2 years ago

Explain how you can write 35% as the sum of two benchmark percents or as a multiple of a percent

OlgaM077 [116]

I'll just take the points thanks XD

3 0

3 years ago

Consider the functions f(x) = 12^x and g(x)= -2(12)^x. Which transformations must be applied to function F to produce the graph

maksim [4K]

Answer:

Step-by-step explanation:

g(x) = -2f(x)

Vertical stretch by virtue of the factor 2

Reflection over the x-axis by virtue of the factor -1

5 0

2 years ago