Can u take a better pic and maybe i can help u
I'll do problem 13 to get you started.
The expression is the same as
Then we can do a bit of algebra like so to change that n into n-1
This is so we can get the expression in a(r)^(n-1) form
- a = 8/7 is the first term of the geometric sequence
- r = 2/7 is the common ratio
Note that -1 < 2/7 < 1, which satisfies the condition that -1 < r < 1. This means the infinite sum converges to some single finite value (rather than diverge to positive or negative infinity).
We'll plug those a and r values into the infinite geometric sum formula below
S = a/(1-r)
S = (8/7)/(1 - 2/7)
S = (8/7)/(5/7)
S = (8/7)*(7/5)
S = 8/5
S = 1.6
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Answer in fraction form = 8/5
Answer in decimal form = 1.6
Answer:
5n-10 < 6n-8
n - is the variable that's stands for the unspecified number
step 1. ten is less than 5 times a number
step 2. LESS THAN SIGN :)
step 3. six times the number decreased by 8
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oh and if you're looking to simplify that problem I believe it would be worked out like this....5n-10<6n-8....subtract 5 from both sides....-10
Answer:
7
Step-by-step explanation:
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