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2 years ago

Does this graph represent a function? Why or why not?

Mathematics

2 answers:

Ugo [173]2 years ago

8 0

Answer:

Yes

Step-by-step explanation:

Draw vertical lines parallel to the "y" axis and you will notice that they only cross the graph once. This only happens in function graphs.

Hope this helps

LiRa [457]2 years ago

3 0

Answer:

The graph represents a function

Step-by-step explanation:

As you can observe, there is only one unique output assigned to every input. If an input contained 2 or more outputs, the relationship wouldn't be a function.

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Find the greatest common factor 28 and 12<br><br> use the GCF to factor 28+12.

fgiga [73]

Answer:

(a) 4

(b) 4 * (7+3)

Step-by-step explanation:

(a) Let's first start by prime factorizing (finding the prime factors) 28 and 12:

12: 2*2*3

28: 2*2*7

We can see here they both share a common factor of 2*2, which is 4, therefore the GCF is 4.

(b) Factor a 4 out from the equation:

$4*(\frac{28}{4}+\frac{12}{4})$

This gives us:

4 * (7+3)

5 0

2 years ago

Identify the domain of the function shown in the graph.

BARSIC [14]

$x \geqslant 0$
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5 0

3 years ago

Read 2 more answers

Which statement best describes the polynomial 2x^3+x+5?

nalin [4]

The polynomial is not factorable with rational numbers

4 0

3 years ago

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8x+9-4x=25. I for real need help

Anni [7]

8x+9-4x=25
4x+9=25
4x+9-9=25-9
4x=16
4x/4=16/4
X=4

4 0

3 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

Does this graph represent a function? Why or why not?​

Does this graph represent a function? Why or why not?