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3 years ago

Which of the sets of ordered pairs represents a function? (5 points)

Mathematics

1 answer:

Evgen [1.6K]3 years ago

7 0

Answer:

The answer is choices Only A

Step-by-step explanation:

hope this helps

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Linda was selling tickets for the school play. She sold 10 more adult tickets than children's tickets. Write an expression for t

eimsori [14]

Answer:

10t

Step-by-step explanation:

10 people

t = ticket

3 0

3 years ago

Read 2 more answers

A man mixed 2 teaspoons of sugar into his large coffee but then added one more teaspoon of sugar because it was not sweet enough

PIT_PIT [208]

Answer:

the ratio of teaspoons of sugar to one large coffee is 3: 1

Step-by-step explanation:

The computation of the ratio of sugar teaspoons to one large coffee is as follows:

Sugar teaspoons = 2 + 1 = 3

And, the one large coffee = 1

So, based on the above information

The ratio between them is

= 3:1

= 3 to 1

hence, the ratio of teaspoons of sugar to one large coffee is 3: 1

3 0

3 years ago

Help me with this fast pleaseee

sveta [45]

I’m confused on what it is asking

4 0

3 years ago

Kelli’s bank raised their ATM transaction fee from $0.25 per use to $0.40 per use. Kelli has made 13 ATM transactions since the

RUDIKE [14]

Answer:

the answer to this would be $5.20

Step-by-step explanation:

0.40 times 13 = 5.20

4 0

3 years ago

HELP PICS INCLUDED! WILL GIVE BRAINLIEST! SHOW WORK AND EXPLAIN

Naddik [55]

I assume you know about the dot product, and that for two vectors $\mathbf a$ and $\mathbf b$ , the angle between them $\theta$ satisfies

$\mathbf a\cdot\mathbf b=\|\mathbf a\|\|\mathbf b\|\cos\theta\iff\cos\theta=\dfrac{\mathbf a\cdot\mathbf b}{\|\mathbf a\|\|\mathbf b\|}$

Then the vectors are parallel if the angle between them is 0 or 180 degrees (0 or pi radians), which would make $\cos\theta=1$ or $\cos\theta=-1$ , respectively.

Part A)

$\vec v_1=\langle\sqrt3,1\rangle\implies\|\vec v_1\|=\sqrt{(\sqrt3)^2+1^2}=\sqrt4=2$

$\vec v_2=\langle-\sqrt3,-1\rangle=-\vec v_1\implies\|\vec v_2\|=\|\vec v_1\|=2$

$\vec v_1\cdot\vec v_2=(\sqrt3)(-\sqrt3)+(1)(-1)=-4$

Then the angle between $\vec v_1,\vec v_2$ is such that

$\cos\theta=\dfrac{-4}{(2)(2)}=-1\implies\theta=\pi\,\mathrm{rad}$

so these vectors are parallel ("antiparallel", more specifically, which means they are parallel but point in opposite directions).

Part B) involves the same computations:

$\vec u_1=\langle2,3\rangle\implies\|\vec u_1\|=\sqrt{2^2+3^2}=\sqrt{13}$

$\vec u_2$ has the same components but differing by sign and order, as $\vec u_1$ ; its magnitude remains the same, though:

$\vec u_2=\langle-3,-2\rangle\implies\|\vec u_2\|=\sqrt{(-3)^2+(-2)^2}=\sqrt{13}$

$\vec u_1\cdot\vec u_2=(2)(-3)+(3)(-2)=-12$

$\implies\cos\theta=\dfrac{-12}{(\sqrt{13})(\sqrt{13})}=-\dfrac{12}{13}\implies\theta=\cos^{-1}\left(-\dfrac{12}{13}\right)$

which is neither 0 nor pi, which means these vectors are not parallel.

4 0

4 years ago