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3 years ago

11

-2.5y + 8y - 10.5y 5.5 0 -21 -5

1 answer:

zubka84 [21]3 years ago

6 0

The answer of the equstion is -5

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157×23 show work pls

Sergeeva-Olga [200]

Step-by-step explanation:

157×23
3611

hope it helps

<h3>stay safe healthy and happy.</h3>

8 0

3 years ago

Read 2 more answers

En la fábrica de Don Toño hoy fabricaron 320 pantalones, si los guardan en cajas donde caben 50 ¿cuántas cajas llenaron?

Scrat [10]

Answer:

6.4 Cajas

Step-by-step explanation:

320/50 = 6.4 Cajas

8 0

3 years ago

Help with math pleasr

Nat2105 [25]

To solve this you need to take the answer from the previous iteration.

Iteration 1: The answer is given
a1 = 300

Iteration 2: Solve using the equation. a(n-1) is the answer from the previous iteration. n = the number of the current iteration. So, a(n-1) = 300 and n = 2

a2 = 300/2 = 150

Iteration 3: a(n-1) = 150 and n = 3
a3 = 150/3 = 50

Iteration 4: a(n-1) = 50 and n = 4
a4 = 50/4 = 12.5

Answer: 12.5

3 0

3 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

(6, 3), (7-4) find the slope<br>

Hunter-Best [27]

Answer:

-7

Step-by-step explanation:

Y-Y $\frac{-4-3}{7-6}$

4 0

3 years ago