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3 years ago

Graph the system of inequalities. Y > x+3 y > -2x+1

Mathematics

1 answer:

Zepler [3.9K]3 years ago

6 0

9514 1404 393

Answer:

see attached

Step-by-step explanation:

The boundary lines of both inequalities are dashed lines, because neither includes the "or equal to" case. Shading is above the boundary line in both cases.

The boundary line of the first inequality has a y-intercept of +3 and a slope of 1, so goes up 1 unit for each 1 unit to the right of (0, 3).

The boundary line of the second inequality has a y-intercept of +1 and a slope of -2, so goes down 2 units for each 1 unit to the right of (0, 1).

The solution space is the top quadrant of the X where the boundary lines cross. That is where the two solution spaces overlap. It includes parts of quadrants I and II.

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Option D

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Determine if the given mapping phi is a homomorphism on the given groups. If so, identify its kernel and whether or not the mapp

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(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.

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