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3 years ago

Which of the following statement does not describe people with self control

Mathematics

1 answer:

PolarNik [594]3 years ago

4 0

Answer:

None of them.

Step-by-step explanation:

People with self-control are not easily described.

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What is the area of this figure?

Dmitrij [34]

Answer:

the area is 30

Step-by-step explanation:

area of triangle =base×height

so ,

base=7.5

height=4

why , shouldn't we take 5(or)6 as height because we shouldn't take slanting lengths as height.

so apply base and height in the formula

7.5×4=30

please follow me

6 0

3 years ago

What is the solution to x 3 + x 2 ≤ 10x – 8? {–4, 1, 2} (–ꝏ, –4] (–ꝏ, –4) U [1, 2] (–ꝏ, –4] U [1, 2]

alisha [4.7K]

Answer:

(–ꝏ, –4] U [1, 2] or D

Step-by-step explanation:

Answer for edgenuity

3 0

3 years ago

Which of the following is a solution of x2 − 10x = −36? (6 points)

Softa [21]

Answer:

Step-by-step explanation:

Given

x² - 10x = - 36

Solve using the method of completing the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(- 5)x + 25 = - 36 + 25

(x - 5)² = - 11 ( take the square root of both sides )

x - 5 = ± $\sqrt{-11}$ ( add 5 to both sides )

x = 5 ± i $\sqrt{11}$

Thus

x = 5 - i $\sqrt11}$ or x = 5 + i $\sqrt{11}$

6 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

Help me solve this equation please

Sladkaya [172]

The answer to this one is W=-3/5 I think

4 0

3 years ago

Which of the following statement does not describe people with self control​

Which of the following statement does not describe people with self control