A square has a perimeter of 40 inches. What is it’s area?

What is the circumference of circle P?<br> Express your answer in terms of .<br> AB = 14 in.

MA_775_DIABLO [31]

Answer:

$C=14\pi\ in$

Step-by-step explanation:

we know that the circumference is equal to

$C=\pi D$

we have

$D=14\ in$

substitute

$C=\pi (14)$

$C=14\pi\ in$

5 0

2 years ago

Read 2 more answers

A bicyclist slow with a force of 3.5 × 10² N. if the bicyclist and bicycle have a total mass of 1.0 × 10² kg, what is the accele

Aleonysh [2.5K]

Answer:

3.5 m/s²

Step-by-step explanation:

We are given;

The force of a bicycle as 3.5 × 10² N
Mass of the bicycle as 1.0 × 10² kg

We are required to determine the acceleration of the bicycle;

From Newton's second law of motion;

F = ma

Rearranging this;

a = F÷ m

Therefore;

Acceleration = 3.5 × 10² N ÷ 1.0 × 10² kg

= 3.5 m/s²

Thus, the acceleration of the bicycle is 3.5 m/s²

7 0

3 years ago

Suppose you have been assigned to measure the height of the local water tower.

Usimov [2.4K]

The water tower is 62.8 meters tall :))

5 0

3 years ago

You roll a number cube. How likely is it that you will roll a number less than 1? less than 7? Explain.

geniusboy [140]

Answer:

There is a zero percent that you will roll a number less than one but a 100% chance of rolling a number less than 7.

Step-by-step explanation:

6 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