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2 years ago

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Mathematics

1 answer:

pantera1 [17]2 years ago

5 0

Answer: answer is c pls mark me brain

Step-by-step explanation:

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What is the measure of angle R? * 7 points R (10n + 19)" (12n - 4) S

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Answer:

(10n + 19) (12n - 4)º ... The midsegment of a trapezoid is parallel to each base and its measure is one half the ...

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Step-by-step explanation:

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2 years ago

Three times a number, x, increased by four is equal to five times the number, x. Which equation can be used to solve for x?

Dimas [21]

3x+4=5x
4=5x-3x
4=2x
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2 years ago

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There are 184 goldfish at the pet store the gold fish in 4 tanks each with the same number of fish . How many fish are in 3 tank

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Answer:

Step-by-step explanation:

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3 years ago

.Sammie paid $249.99 for a game system and then bought two equally-priced games. If she spent a total of $339.97, what is the pr

bulgar [2K]

Answer: $44.99

Step-by-step explanation:

Let P = equal prices games

The equation would be

249.99+2p=339.97

Subtract 249.99 from each side

2p=339.97

Divide by two and you got the answer.

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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.

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3 years ago