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Murljashka [212]

2 years ago

7

Can someone help me plzzz and thank you

1 answer:

Fynjy0 [20]2 years ago

8 0

Answer:

2nd option....

Yes; SSS

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What is 1/10 of 547,120 ?<br> O 5471.20<br> 5,471,200<br> o 547.120<br> 54,712

vodka [1.7K]

Answer:

54712

Step-by-step explanation:

Starting with 547120, move the decimal point one place to the left, obtaining 54712. (Matches last answer choice)

8 0

3 years ago

Read 2 more answers

Evaluate g(3) if g(x)=2x+2

klemol [59]

Plug in 3 for x so
2(3)+2
6+2=
8
The evaluation is 8

4 0

3 years ago

Select the expression that is equivalent to the polynomial given below.<br> (4x+3)(3x-8)

posledela

Answer:

Step-by-step explanation:

5 0

2 years ago

Can anyone do this with steps ..

Harrizon [31]

$\\ \sf\longmapsto \displaystyle{\lim_{x\to 0}}\dfrac{sinx-sin3x}{sins3x-sinx}$

$\boxed{\sf \displaystyle\lim_{x\to 0}\dfrac{f(x)}{g(x)}=\dfrac{\lim_{x\to 0}f(x)}{\lim_{x\to 0}g(x)}}$

Now

$\\ \sf\longmapsto \displaystyle{\dfrac{\lim_{x\to 0}sinx-\lim_{x\to 0}sin3x}{\lim_{x\to 0}sin3x-\lim_{x\to 0}sinx}}$

$\boxed{\sf \displaystyle{\lim_{x\to 0}}f(x)=f(a)}$

$\\ \sf\longmapsto \dfrac{sin0-sin3(0)}{sin3(0)-sin0}$

$\\ \sf\longmapsto \dfrac{0-0}{0-0}$

$\\ \sf\longmapsto \dfrac{0}{0}$

$\\ \sf\longmapsto 0$

7 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

3 years ago