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3 years ago

7

If there is 37 cars and 10 are speeding, what percentage of cars is speeding?

1 answer:

Helen [10]3 years ago

6 0

27.03% is your answer

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12a +5 identify the coefficient

ioda

Answer:

The coefficient is "12"

Step-by-step explanation:

Step 1: Encircle the variable along with its power whose coefficient we are finding.

Step 2: Leave that variable and consider all other numbers or variables written with it. That will be the coefficient.

4 0

2 years ago

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Please help, i really dont understand. will give brainliest.

hoa [83]

Answer:

I've done the math and it's 120, people say it's 120 and online 120, I really wanna help but I just

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3 years ago

Point Q is plotted on the coordinate grid. Point P is at (10, −20). Point R is vertically above point Q. It is at the same dista

bixtya [17]

Answer:

Point R is at (−20, 10), a distance of 30 units from point Q

Step-by-step explanation:

Q has coordinates (-20,-20).

P has coordinates (10,-20)

Since point R is vertically above point Q, it will have the same x-coordinate as Q.

Let R have coordinates (-20,y).

It was given that;

$|RQ|=|PQ|$

$\Rightarrow |y--20|=|10--20|$

$\Rightarrow y+20=10+20$

$\Rightarrow y=10+20-20$

$\Rightarrow y=10$ .

The coordinates of R are (-20,10).

The dstance from Q is 30 units.

3 0

3 years ago

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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 please! I need this question done super fast.

vekshin1

Answer: The answer is yes because figure B is figure A divided by 2

5 0

3 years ago

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