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Marina CMI [18]
2 years ago
11

Create a recursive formula for this sequence (linear)

Mathematics
1 answer:
svetoff [14.1K]2 years ago
4 0

the answer is 3n-2 please do you get it

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Mai bank account is overdrawn by $125 which means her balance is negative 125 she gets $85 for her birthday and deposit into her
stira [4]
The answer is -125+85= -40 because she is adding -125 by 85 which equals-40
4 0
2 years ago
Please help due soon and 50pts
hjlf

Answer: A

Step-by-step explanation:

6 0
2 years ago
Can someone pls help me with this?
lara [203]

Answer:

14 is A

15 is B

Step-by-step explanation:

14 . the answer is a because first off the slope is negative so we can immediately eliminate B and D second of all the slope is 1/2 so we can eliminate D and get that the answer has to be a

15. answer is B for this one because first of all the slope is negative so we can immediately eliminate A and c and second of all the y-intercept would be 120 because 90 is what x=1 so we would have to add 30 to get what y would equal when x=0 if that makes sense

5 0
2 years ago
HELP PLEASEE!!!
lara [203]

Answer:

-10

Step-by-step explanation:

8 0
2 years ago
The sum of 30 times (1/3)^(n-1) from 1 to infinity
sergeinik [125]

Let

S_n=\displaystyle1+\frac13+\frac1{3^2}+\cdots+\frac1{3^n}

Then

\dfrac13S_n=\displaystyle\frac13+\frac1{3^2}+\frac1{3^3}+\cdots+\frac1{3^{n+1}}

and

S_n-\dfrac13S_n=\dfrac23S_n=1-\dfrac1{3^{n+1}}\implies S_n=\dfrac32-\dfrac1{2\cdot3^n}

and as n\to\infty, we end up with

\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{i=1}^{n+1}\frac1{3^{i-1}}=\lim_{n\to\infty}\left(\frac32-\frac1{2\cdot3^n}\right)=\frac32

So we have

\displaystyle\sum_{n=1}^\infty30\left(\frac13\right)^{n-1}=30\cdot\frac32=45

8 0
3 years ago
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