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3 years ago

Which is not an improper fraction equal to eight

Mathematics

2 answers:

Oliga [24]3 years ago

7 0

The fraction that is improper and it equals 8 is 64/8

Nady [450]3 years ago

6 0

It would be 64/8 hope that helped

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PLZ HELP WILL MARK BRAINLIEST!

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Step-by-step explanation:

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focus = (-3, 2)

vertex = (-3, 3)

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2 years ago

What is 8640 minutes in days

Zolol [24]

Answer:

it is 6 days

Step-by-step explanation:

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Please answer thxxxx

Llana [10]

B. 13

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Simple interest of $725 at 3.25% for 6 months

Rzqust [24]

The simple interest on the principal amount of $725 at a rate of 3.25% for 6 months is $736.78

<h3>What is simple interest? </h3>

Simple interest is calculated on the loan principal amount or the initial deposit into a savings account. It can be determined by using the formula:

$\mathbf{S.I = P(1 + rt)}$

where:

P = principal = $725
R = rate = 3.25%
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$\mathbf{S.I = 725(1 + 0.0325(0.5))}$

S.I = $736.78

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1 year 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