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alexandr1967 [171]
3 years ago
6

Provide the first six terms of the following sequence: t1=81 tn=1/3tn-1

Mathematics
2 answers:
Andrews [41]3 years ago
8 0

We get tne n-th term with: tn=1/3tn-1

We know that the first term is t1=81, so 81=1/3t*1-1

81=1/3t-1

82=1/3t

82*3t=1

3t=1/82

t=1/82*3=1/246

The second term is: 1/

2*(1/246)*2)-1

We get the third term by replacing n with 3 and so on...

Mandarinka [93]3 years ago
7 0

Answer:

Through the window and under the door, It covers the walls and coats the floor.

Step-by-step explanation:


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having trouble with this problem, pls help :/ I know the answer, just not how to get there. (ap calc ab, integrals)
pochemuha

Answer: D) 101

Step-by-step explanation:

By linearity, we can break it up into 2 integrals. The integral and derivative of f easily cancel out

\int\limits^{10}_{-1} {(2x+0.5f'(x))} \, dx =\int\limits^{10}_{-1} {2x} \, dx +0.5\int\limits^{10}_{-1} {f'(x)} \, dx =x^2|^{^{10}}_{_{-1}}+0.5f(x)|^{^{10}}_{_{-1}}\\=(100-1)+0.5(f(10)-f(-1))=99+0.5(8-4))=101

I used the table for values of f(x) at 10 and -1. Wouldn't be surprised if this was part of a series of questions about f because I really can't see how you could use the hypothesis that f is twice differentiable on R. Same for the other table values. I'm curious about how you found the answer. Was it a different way?

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