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2 years ago

PLEASE HELP ITS DUE SOON! WILL GIVE BRAINLIST!!

Mathematics

2 answers:

3241004551 [841]2 years ago

7 0

It is D, it just makes the most sense with the answers

Nostrana [21]2 years ago

4 0

Should be D. I’m not sure tho this question is kinda hard for me too

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Brainliest and Five Stars for correct answers

wel

$\bf \lim\limits_{x\to 9}~\cfrac{x^2-81}{x-9}\implies \cfrac{\stackrel{\textit{difference of squares}}{x^2-9^2}}{x-9}\implies \cfrac{\underline{(x-9)}(x+9)}{\underline{(x-9)}}
\\\\\\
\lim\limits_{x\to 9}~x+9\implies (9)+9\implies 18$

3 0

2 years ago

What is the domain and range of f(x)=2.5x when x goes up to 800

Mandarinka [93]

Answer:

Domain: all real numbers. Range: (-∞, 2000]

Step-by-step explanation:

f(x)=2.5x is a simple "ramp" function, a linear function and a polynomial. As such, its domain contains "all real numbers." That value 800 defines the largest value that this f(x) can have: f(800) = 2.5(800) = 2000.

Thus, the range is "all real numbers from -∞ through and including 2000."

3 0

3 years ago

Which equation is related to x+23=9?.

cestrela7 [59]

If x +23 = 9, then, x = 9-23.

So x = -14

3 0

2 years ago

Read 2 more answers

Help? I can't seem to understand arithmegic sequence.

katrin2010 [14]

A sequence $\{a_n\}$ is arithmetic if the difference between consecutive terms is some fixed number, regardless of which pair of consecutive terms you pick out of the sequence.

For example, the following sequences are arithmetic:

1, 2, 3, 4, 5, 6, ... (difference = 1)

-25, -20, -15, -10, -5, ... (difference = 5)

2. Carla's sequence is not arithmetic, because the differences between consecutive terms are all different:

13 - 11 = 2

17 - 13 = 4

25 - 17 = 8

She can adjust the sequence by changing the last two numbers to 15 and 17, since this makes the difference fixed:

13 - 11 = 2

15 - 13 = 2

17 - 15 = 2

and so on.

3. The sequence

45, 48, 51, 54, ...

is arithmetic with difference 3 between terms. Recursively, we can write the $n$ th term, $a_n$ , in terms of the previous, $(n-1)$ th term, $a_{n-1}$ :

$a_n=a_{n-1}+3$

By this definition, we can just as easily write the $(n-1)$ th term in terms of the $(n-2)$ th term:

$a_{n-1}=a_{n-2}+3$

Then, substituting this into the previous equation, we have

$a_n=(a_{n-2}+3)+3=a_{n-2}+2\cdot3$

We can continue this process to write $a_n$ in terms of $a_1$ :

$a_{n-2}=a_{n-3}+3\implies a_n=a_{n-3}+3\cdot3$

$a_{n-3}=a_{n-4}+3\implies a_n=a_{n-4}+4\cdot3$

and so on. (You might notice that the subscript of the term on the right side, and the number of 3s being added, together sum to $n$ .) The pattern continues down to

$a_n=a_1+(n-1)\cdot3$