All categories

2 years ago

What is the plan for the set of stacked cubes?

Mathematics

1 answer:

Nadusha1986 [10]2 years ago

3 0

$\large \gray { \boxed{{ \colorbox{g}{answer}}}}$

it only depends on the cubes in the pic

You might be interested in

Does 2 1/4-6/7=1 3/4

dimulka [17.4K]

Answer: No, it would be 2/5.

Step-by-step explanation: 2 1/4 - 6/7 is equal to about 1.4 when rounded to the nearest tenth. 1.4 as a fraction would be 4/10. 4/10 is simplified to 2/5 by dividing each number by 2.

7 0

3 years ago

Can you please help me solve these 6 problems? I don't know where to begin. I also don't know how to solve my work. Thank you.

allochka39001 [22]

Answer:

is it absolute vaule?

Step-by-step explanation:

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

Identify which of the following variables are discrete and which are continuous: a. Number of warts on a toad b. Survival time a

gulaghasi [49]

Answer:

1. Number of warts on a toad ⇒ discrete variable

2. Survival time after poisoning ⇒ continuous variable

3. Temperature of porridge ⇒ continuous variable

4. Number of bread crumbs in 10 meters of trail ⇒ discrete variable

5. Length of wolves’ canines ⇒ continuous variable

A discrete variable is the one which takes over range of real number, for a value in range that variables are permitted to take, there is minimum distance to nearest permissible value. In contrast, continuous variables are the one which can take infinitely many values.

4 0

3 years ago

2. Decrease £40 by 15%

Ulleksa [173]

Answer:

£34

Step-by-step explanation:

15/100*£40

=£6

£40_£6

=£34

7 0

3 years ago