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KonstantinChe [14]

2 years ago

12

which function has a constant additive rate of change of –1/4? A coordinate plane with a straight line with a negative slope. Th

e line passes through (negative 2, 2) and (2, 1). A coordinate plane with a curved line passing through (negative 1, 2), (0, negative 1), the minimum (2, negative 2), and (4, negative 1). A two column table with five rows. The first column, x, has the entries, 20, 21, 22, 23. The second column, y, has the entries negative 1, negative 1.5, negative 2, negative 2.5. A two column table with five rows. The first column, x, has the entries, negative 12, negative 11, negative 10, negative 9. The second column, y, has the entries, 7, 11, 14, 17.

2 answers:

e-lub [12.9K]2 years ago

5 0

Answer:

it's A

Step-by-step explanation:

i just did it

solong [7]2 years ago

3 0

Answer:

It is not the last one...

It's the first one...

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Determine whether the series is convergent or divergent. 1 2 3 4 1 8 3 16 1 32 3 64 convergent divergent Correct:

Vanyuwa [196]

Answer:

This series diverges.

Step-by-step explanation:

In order for the series to converge, i.e. $\lim_{n \to \infty} a_n =A$ it must hold that for any small $\epsilon$ >0, there must exist $n_0\in \mathbb{N}$ so that starting from that term of the series all of the following terms satisfy that $|a_n-A|n_0$ .

It is obvious that this cannot hold in our case because we have three sub-series of this observed series. One of them is a constant series with $a_n=1$ , the other is constant with $a_n=3$ , and the third one has terms that are approaching infinity.

Really, we can write this series like this:

$a_n=\begin{cases} 1 \ , \ n=4k+1, k\in \mathbb{N}_0\\ 2^{k}\ , \ n=2k, k\in \mathbb{N}_0\\3\ , \ n=4k+3, k\in \mathbb{N}_0\end{cases}$

If we denote the first series as $b_n=1$ , we will have that $\lim_{k \to \infty} b_k=1$ .

The second series is denoted as $c_k=2^k$ and we have that $\lim_{k \to \infty} c_k=+\infty$ .

The third sub-series $d_k=3$ is a constant series and it holds that $\lim_{k \to \infty} d_k=3$ .

Since those limits of sub-series are different, we can never find such $n_0\\$ so that every next term of the entire series is close to one number.

To make an example, if we observe the first sub-series if follows that A must be equal to 1. But if we chose $\epsilon =1$ , all those terms associated with the third sub-series will be out of this interval $(A-1, A+1)=(0, 2)$ .

Therefore, the observed series diverges.

5 0

2 years ago

Find each measure for the given set of data: 11, 13, 17, 20, 22, 25, 27, 31, 31, 33, Mean= Median= Range= interquartile range= I

Tresset [83]

Answer:

Mean=23

Median =23.5

Range=22

Interquartile range = 14

Step-by-step explanation:

Given set of data

11,13,17,20,22,25,27,31,31,33

Mean $=\frac{11+13+17+20+22+25+27+31+31+33}{10}$

$=\frac{230}{10}$

=23

The $(\frac{n}{2})$ th term= the 5th term

=22

The $(\frac{n}{2}+1)^{th}$ term = The $6^{th$ term

= 25

The median = $\frac{22+25}{2}$

=23.5

Range = Highest term - lowest term

=33- 11

=22

Here lower half {11,13,17,20,22}

The middle number of lower half is first quartile.

Q₁ = 17

And lower half is{25,27,31,31,33}

The middle number of upper half is third quartile.

Q₃=31

Interquartile Range =Q₃-Q₁

=31-17

=14

8 0

2 years ago

An astronaut on the moon throws a baseball upward. the astronaut is 6 ft, 6 in. tall, and the initial velocity of the ball is

marta [7]

Answer:

Step-by-step explanation:

The position function is

$s(t)=-2.7t^2+50t+6.5$ and if we are looking for the time(s) that the ball is 10 feet above the surface of the moon, we sub in a 10 for s(t) and solve for t:

$10=-2.7t^2+50t+6.5$ and

$0=-2.7t^2+50t-3.5$ and factor that however you are currently factoring quadratics in class to get

t = .07 sec and t = 18.45 sec

There are 2 times that the ball passes 10 feet above the surface of the moon, once going up (.07 sec) and then again coming down (18.45 sec).

For part B, we are looking for the time that the ball lands on the surface of the moon. Set the height equal to 0 because the height of something ON the ground is 0:

$0=-2.7t^2+50t+6.5$ and factor that to get

t = -.129 sec and t = 18.65 sec

Since time can NEVER be negative, we know that it takes 18.65 seconds after launch for the ball to land on the surface of the moon.

4 0

1 year ago

Please help me on this question

Serggg [28]

-4 = 8m + 18n
-18n = 8m + 4
/-18 /-18 /-18
n = 8m/-18 + 4/-18

I'm not sure so yeah

7 0

2 years ago

Read 2 more answers

Three trigonometric functions for a given angle are shown below. cosecant theta = thirteen-twelfths; secant theta = Negative thi

Kay [80]

Answer:

(–5, 12) is the correct answer.

Step-by-step explanation:

We are given the following values:

$cosec\theta = \dfrac{13}{12}\\sec\theta=-\dfrac{13}{5}\\cot\theta =-\dfrac{5}{12}$

Now, we know the following identities:

$sin \theta = \dfrac{1}{cosec\theta}\\cos \theta = \dfrac{1}{sec\theta}\\tan \theta = \dfrac{1}{cot\theta}$

Now, the values are:

$sin\theta = \dfrac{12}{13}\\cos\theta=-\dfrac{5}{13}\\tan\theta =-\dfrac{12}{5}$

Sine value is positive and cos, tan values are negative.

It can be clearly observed that $\theta$ is in 2nd quadrant.

2nd quadrant means, the value of $x$ will be negative and $y$ will be positive.

Let us have a look at the value of $tan\theta$ :

$tan\theta = \dfrac{Perpendicular}{Base}\\OR\\tan\theta = \dfrac{y-coordinate}{x-coordinate} = -\dfrac{12}{5}\\\therefore y = 12,\\x = -5$

Please refer to the attached image for clear understanding and detailed explanation.

Hence, the correct answer is coordinate (x,y) is (–5, 12)

8 0

2 years ago