All categories

4 years ago

How many planes of symmetry does a cube have?

Mathematics

2 answers:

Goryan [66]4 years ago

6 0

Answer:

Step-by-step explanation:

Anestetic [448]4 years ago

3 0

Answer:

Nine Planes

Step-by-step explanation:

Three planes are parallel to the six faces of the cube. The other six are diagonal.

You might be interested in

If A = 3x power 2+2y + 2 and B=6x power 2 - 8y + 1 then A+B and A-B

ollegr [7]

Answer:

see explanation

Step-by-step explanation:

Given A = 3x² + 2y + 2 and B = 6x² - 8y + 1 , then

A + B

= 3x² + 2y + 2 + 6x² - 8y + 1 ← collect like terms

= 9x² - 6y + 3

-------------------------------

A - B

= 3x² + 2y + 2 - (6x² - 8y + 1) ← distribute parenthesis by - 1

= 3x² + 2y + 2 - 6x² + 8y - 1 ← collect like terms

= - 3x² + 10y + 1

5 0

3 years ago

The solution of [5x(4+6)-9]

Anna35 [415]

Answer:

Step-by-step explanation:

BIDMAS

[5 x (4 + 6) - 9]

Brackets first

4 + 6 = 10

[5 x (10) - 9]

Multiply next

5 x 10 = 50

Subtract last

50 - 9 = 41

3 0

3 years ago

Please HELP!!:) 2. Of the 200 species of birds found in the refuge, 96 have been

dusya [7]

Answer:

48%

Step-by-step explanation:

96 out of the 200 species of birds have been seen nesting. To make it easier, let's make it out of a hundred. Just divide both numbers by 2, this will result in the ratio being 48 out of 100. Now, we have our percentage, 48%.

6 0

3 years ago

Read 2 more answers

Determine if the given mapping phi is a homomorphism on the given groups. If so, identify its kernel and whether or not the mapp

shtirl [24]

Answer:

(a) No. (b)Yes. (c)Yes. (d)Yes.

Step-by-step explanation:

(a) If $\phi: G \longrightarrow G$ is an homomorphism, then it must hold

that $b^{-1}a^{-1}=(ab)^{-1}=\phi(ab)=\phi(a)\phi(b)=a^{-1}b^{-1}$ ,

but the last statement is true if and only if G is abelian.

(b) Since G is abelian, it holds that

$\phi(a)\phi(b)=a^nb^n=(ab)^{n}=\phi(ab)$

which tells us that $\phi$ is a homorphism. The kernel of $\phi$

is the set of elements g in G such that $g^{n}=1$ . However,

$\phi$ is not necessarily 1-1 or onto, if $G=\mathbb{Z}_6$ and

n=3, we have

$kern(\phi)=\{0,2,4\} \quad \text{and} \quad\\\\Im(\phi)=\{0,3\}$

(c) If $z_1,z_2 \in \mathbb{C}^{\times}$ remeber that

$|z_1 \cdot z_2|=|z_1|\cdot|z_2|$ , which tells us that $\phi$ is a

homomorphism. In this case

$kern(\phi)=\{\quad z\in\mathbb{C} \quad | \quad |z|=1 \}$ , if we write a

complex number as $z=x+iy$ , then $|z|=x^2+y^2$ , which tells

us that $kern(\phi)$ is the unit circle. Moreover, since

$kern(\phi) \neq \{1\}$ the mapping is not 1-1, also if we take a negative

real number, it is not in the image of $\phi$ , which tells us that

$\phi$ is not surjective.

(d) Remember that $e^{ix}=\cos(x)+i\sin(x)$ , using this, it holds that

$\phi(x+y)=e^{i(x+y)}=e^{ix}e^{iy}=\phi(x)\phi(x)$

which tells us that $\phi$ is a homomorphism. By computing we see

that $kern(\phi)=\{2 \pi n| \quad n \in \mathbb{Z} \}$ and

$Im(\phi)$ is the unit circle, hence $\phi$ is neither injective nor

surjective.

7 0

3 years ago

What is the end behavior of the graph?

ANTONII [103]

Answer:

As $x$ approaches infinity $f(x)$ approaches negative infinity.

Step-by-step explanation:

As you can see in the graph as $x$ grows larger in the negative direction, the $y$ increases in the negative direction too.

Thus:

$\lim_{x \to -\infty} f(x) = -\infty$

5 0

2 years ago