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Mathematics 8 Online
OpenStudy (anonymous):

The sequence an = one half(2)^n − 1 is graphed below Find the average rate of change between n = 2 and n = 4.

OpenStudy (anonymous):

@jim_thompson5910 sorry :( you're the only one who seems to be online

jimthompson5910 (jim_thompson5910):

what is a2?

OpenStudy (anonymous):

I'm not sure what you mean

jimthompson5910 (jim_thompson5910):

what happens when you plug n = 2 into the given sequence

OpenStudy (anonymous):

I dont know :(

jimthompson5910 (jim_thompson5910):

alternatively, since this is already graphed for you, you can find the slope through (2,1) and (4,4)

jimthompson5910 (jim_thompson5910):

since those points represent when n = 2, n = 4

OpenStudy (anonymous):

oh okay, so i find the slope?

jimthompson5910 (jim_thompson5910):

yes through those two points

OpenStudy (anonymous):

3/2

OpenStudy (anonymous):

that's my answer

jimthompson5910 (jim_thompson5910):

yes it is

OpenStudy (anonymous):

thank you! one more?

jimthompson5910 (jim_thompson5910):

sure

OpenStudy (anonymous):

Given the arithmetic sequence an = 2 − 5^(n + 1), what is the domain for n? I know its the same concept as the other one, idk why i can't get it

jimthompson5910 (jim_thompson5910):

that's not arithmetic, did you mean an = 2 − 5(n + 1)

OpenStudy (anonymous):

Oh yeah sorry

jimthompson5910 (jim_thompson5910):

same idea as a geometric sequence

jimthompson5910 (jim_thompson5910):

the domain is the set of positive whole numbers

jimthompson5910 (jim_thompson5910):

although this arithmetic sequence is still odd...hmm

OpenStudy (anonymous):

I got confused cause one of the answers says All integers where n ≥ 2

jimthompson5910 (jim_thompson5910):

what else are you given

OpenStudy (anonymous):

All real numbers All integers where n ≥ 1 All integers where n > 1 All integers where n ≥ 2

jimthompson5910 (jim_thompson5910):

So most likely All integers where n ≥ 1 because again, it's the counting numbers or natural numbers

OpenStudy (anonymous):

Okay, thank you very much :)

jimthompson5910 (jim_thompson5910):

you're welcome

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