All categories

3 years ago

Which of these 3-dimensional shapes is composed of only square faces?

Mathematics

2 answers:

elena-14-01-66 [18.8K]3 years ago

7 0

Answer:

d.Cube

Step-by-step explanation:

The correct answer is cube, a cube is made up of 6 squares

Luden [163]3 years ago

6 0

The answer is the cube

You might be interested in

How do i factor 7x^2+3x-4

Bas_tet [7]

Answer:

(x+1) (7x - 4)

Step-by-step explanation:

7x² + 3x - 4

Use the diamond...

7 times -4 equals -28 and 7 plus -4 = 3

Use factor by grouping b/c a does not equal 1

7x² + 7x - 4x - 4

7x(x + 1) -4(x + 1)

(x+1) (7x - 4)

6 0

3 years ago

Problem Page

leva [86]

1/8 + 2/3= 19/24

I/you have walked 19/24 miles total, since 19/24 can't be simplified.

7 0

3 years ago

Read 2 more answers

Enter the pair of fractions as a pair of fractions with a common denominator.

leva [86]

18 would be the common denominator

7 0

3 years ago

Read 2 more answers

Solve the system:

exis [7]

The answer is C) x=11/19, y=7/19

8 0

3 years ago

True or false. Tan^2 x = 1 - cos2x/ 1 + cos 2x

koban [17]

ANSWER

True

EXPLANATION

The given trigonometric equation is

$\tan^{2} (x) = \frac{1 - \cos(2x) }{1 + \cos(2x) }$

Recall the double angle identity:

$\cos(2x) = \cos^{2} x - \sin^{2}x$

We apply this identity to obtain:

$\tan^{2} (x) = \frac{1 - (\cos^{2} x - \sin^{2}x) }{1 + (\cos^{2} x - \sin^{2}x) }$

We maintain the LHS and simplify the RHS to see whether they are equal.

Expand the parenthesis

$\tan^{2} (x) = \frac{1 - \cos^{2} x + \sin^{2}x }{1 + \cos^{2} x - \sin^{2}x}$

$\implies\tan^{2} (x) = \frac{1 - \cos^{2} x + \sin^{2}x }{1 - \sin^{2}x + \cos^{2} x }$

Recall that:

$1 - \sin^{2}x = \cos^{2}x$

$1 - \cos^{2}x = \sin^{2}x$

We apply these identities to get:

$\implies\tan^{2} (x) = \frac{\sin^{2}x + \sin^{2}x }{\cos^{2} x + \cos^{2} x }$

$\implies\tan^{2} (x) = \frac{2\sin^{2}x }{ 2\cos^{2} x }$

$\implies\tan^{2} (x) = \frac{\sin^{2}x }{ \cos^{2} x }$

$\implies \tan^{2} (x) =( \frac{\sin x }{ \cos x })^{2}$

Also

$\frac{\sin x }{ \cos x } = \tan(x)$

$\implies \tan^{2} (x) =( \tan x )^{2}$

$\implies \tan^{2} (x) =\tan^{2} (x)$

Therefore the correct answer is True

5 0

3 years ago