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3 years ago

2. Which example does NOT show the

Mathematics

2 answers:

skelet666 [1.2K]3 years ago

7 0

Answer:

The answer is A....im pretty sure

Step-by-step explanation:

ikadub [295]3 years ago

3 0

Answer:

I think it would be b.ab=ba but I'm, not 100% sure

Step-by-step explanation:

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lbvjy [14]

The vaue of x is 23 cm

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PLEASE ANSWER ASAP!!!

julsineya [31]

Answer:

The answer to your question is A) x = 1

Step-by-step explanation:

We know that f(x) = 2x + 3 and g(x) = -x + 6

But f(x) = g(x)

Then, 2x + 3 = - x + 6

2x + x = 6 - 3

3x = 3

x = 3/3

x = 1

The solution is circle in the graph below.

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-2x^2-12x-10 standard form

siniylev [52]

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3 years ago

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never [62]

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2 years ago

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