All categories

2 years ago

Consider the algebraic expression:

Mathematics

2 answers:

kotykmax [81]2 years ago

4 0

Answer:

Step-by-step explanation:

Mark me as brainlist

Hope it will help you

guapka [62]2 years ago

4 0

Answer:

Step-by-step explanation:

I got it right 2021 edge

You might be interested in

Help pls ill gve brainlist <3 or however you spell it

telo118 [61]

Answer:

1) Is 3 for K

2) is 4 for X

3) is 56 for Y

8 0

2 years ago

Read 2 more answers

Find the area of the shade regions. Give your answer as a completely simplify exact value in terms of pie(no approximation). a=

mr Goodwill [35]

Answer:

125.6 in²

Step-by-step explanation:

Area shaded :

2 × Sector (72°)
2 x πr² x θ/360
2 x 3.14 x 100 x 72/360
6.28 x 100 x 1/5
20 x 6.28
125.6 in²

4 0

1 year ago

The ratio of ducks to geese in a pond is 3 : 5. There are 27 ducks. How many geese are there?

Gemiola [76]

Answer:

D:G-------> 3:5

27/3=9

9 x 5= 45

45 geese

7 0

2 years ago

What is the output value for the following function if the input value is 3.2?

mars1129 [50]

For functions, there is only many input values, but there can be only one output value, and there can be no identical x values. For this, x=3.2. If y= 2*3.2 - 1, then y=5.4.

3 0

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.

7 0

3 years ago