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3 years ago

Which of the following statements are true?

Mathematics

1 answer:

Aneli [31]3 years ago

4 0

Answer: It’s A

Step-by-step explanation:

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Electroboy has been charged up to +96 C which of course causes an electric field to emanate around him. An electroball charged t

inessss [21]

Answer:

static electricty

Step-by-step explanation:

it is preserved charge

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2 years ago

Steve borrowed $2000 at 8% annual interest for 6 months. How much interest did he pay? Steve paid ? in interest.

Novosadov [1.4K]

Answer:

I = (2,000) (0.08) (6)

I = (160) (6)

Steve pays $960 in interest

Step-by-step explanation:

I completed the quiz :)

4 0

2 years ago

A container has 200 liters and is losing 6 liters per minute. How many will it have after 10 minutes

IgorLugansk [536]

Answer:

140 liters will remain after 10 mins

Step-by-step explanation:

The easy way is just multiply the amount lost per min by the number of mins.

3 0

3 years ago

Read 2 more answers

Choose the equation that represents the line that passes through the point (−1, 6) and has a slope of −3.

Marat540 [252]

A) y= -3x +3 and the slope is -3

7 0

2 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