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2 years ago

13

Please help! I will give brainliest to the best answer!

1 answer:

Ahat [919]2 years ago

7 0

Answer:

A

Step-by-step explanation:

y < 1/2x + 1

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The high temperature one day was -1 °F. The low temperature was -5 °F. What was the difference between the high and low temperat

insens350 [35]

Answer:

The difference is 4 degrees

Step-by-step explanation:

If its difficult for you to subtract negatives just switch them to a positive 1 and 5 now subtract 1 from 5=4 and because the measure being used is degrees the final answer would be 4 degrees.

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2 years ago

What is the measure (in radians) of central angle & in the circle below?

MA_775_DIABLO [31]

Answer:

7pi/6

Step-by-step explanation:

Using the formula for arc length in radians, we get:

14pi = 12(theta)
theta = 7pi/6 [divide both sides by 12]

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1 year ago

What are the partial products of 34 and 57

Brut [27]

<span>2257987 is the answer</span>

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3 years ago

To adopt a dog from an animal shelter, you must pay $80 for vaccinations, $65 to spay or neuter the dog, and $50 for a wellness

Leya [2.2K]

The equation with (x) in dogs and (y) in dollars is $y=195x$

First, you must set up your equation.

$y=80x+65x+50x$

Then, you must add like terms to simplify the equation

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4 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