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2 years ago

9

Would You Rather? Snack Attack!

1 answer:

irga5000 [103]2 years ago

7 0

Answer:

I would buy 8 bags of everyone's favorite snacks because it's cheaper and everyone gets what that want.

Step-by-step explanation:

This is just my opinion...Hope it helps

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The school is having a fundraiser they are running in Carnival tickets sell for $0.50 each they are playing you're buying suppli

garri49 [273]

They would need to sell at least 500

5 0

3 years ago

If a solution has 24g of sugar in 130ml of water, what is its concentration?

Jobisdone [24]

Answer: concentration = 15.58%

Step-by-step explanation:

Given: mass of sugar = 24 g = 24 ml [ ∵ 1 g = 1 ml]

Volume of solution = Quantity of water + quantity of solute

= 130 +24 ml

= 154 ml

Concentration = $\dfrac{mass\ of\ solute}{Volume\ of\ solution}\times100$

$=\dfrac{24}{154}\times 100\\\\\approx15.58\%$

Hence, the concentration =15.58%

3 0

3 years ago

You have 6 dollars +12 pennies +to quarters + 4 dimes=2 nickels how much do you have

alina1380 [7]

You may have $26 I not a genius but this is easy

3 0

3 years ago

Which statement best describes the function below? F(x)=2x^2-3x+1

PtichkaEL [24]

<span>It is a many-to-one function.
</span>

7 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