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photoshop1234 [79]

4 years ago

10

What is a linear function that relates to this table?

1 answer:

natima [27]4 years ago

5 0

The linear function should be y=2 i apologize if i am wrong

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What is -3 - 15. can you please solve this for me.

shepuryov [24]

Answer:

-18

Step-by-step explanation:

-3-15 = -3 + -15

= -18

8 0

3 years ago

1 > 5 (b - 14) + 16 help me please

nydimaria [60]

Answer:

b < 11

Step-by-step explanation:

Given

1 > 5(b - 14) + 16 ← distribute and simplify right side

1 > 5b - 70 + 16

1 > 5b - 54 ( add 54 to both sides )

55 > 5b ( divide both sides by 5 )

11 > b, thus

b < 11

8 0

4 years ago

See in the picture for question

vichka [17]

The ratio will be 1:4

The actual suns temperature is <span>27,000,000 Fahrenheit
To plug it the ratio
27,000,000 : 108,000,000
</span>

6 0

3 years ago

Midpoint(-5,-20) and endpoint (-1,-11) Find other endpoint

Valentin [98]

Answer:

Step-by-step explanation: Is this grade 9th work?

8 0

4 years ago

Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

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

so that $\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}$ .

Let $\{c_n\}$ be the sequence of differences of $\{b_n\}$ ,

$c_n = b_{n+1} - b_n$

and we see that this is a constant sequence, $\{c_n\} = \{2, 2, 2, 2, \ldots\}$ . In other words, $\{b_n\}$ is an arithmetic sequence with common difference between terms of 2. That is,

$2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2$

and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

$a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)$

$a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))$

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

6 0

2 years ago