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2 years ago

5

PLEASE HELP ! I need the answers asap! :( I will mark brainlest!

1 answer:

expeople1 [14]2 years ago

6 0

Answer:

10. G) point, line, plane

11. A) 6

12. H) Postulates

Step-by-step explanation:

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Solve for k.<br> (k)4+ 3= 14<br> k=<br>

r-ruslan [8.4K]

Solving for "K" , got it! :)
I'll make it easier to solve

(k)4+3-14=?
? = 0

4(k)-11 = 0

Take off the '0' and the addition symbol so we can add from 11 on your sides

4(k) - 11
+11 +11
---- ----
15k. 22
⬆ This would be your answer for k

4(k) = 11
÷4 ÷4
--- ----
1K ?

K = 11/4 (for fraction)
K = 2.75 (as a decimal)

5 0

2 years ago

Read 2 more answers

If you double the lengths of a rectangle what can you say about the new area of it.

ivolga24 [154]

Answer:

B

Step-by-step explanation:

because u increased it by 2x

3 0

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

2 years ago

Iteru [2.4K]

Answer:

a

Step-by-step explanation:

4 0

2 years ago

I don’t understand this. Can you please help me? thank you.

il63 [147K]

Answer:

Step-by-step explanation:

Sfbwertb

4 0

1 year ago