Answer:
- is the answers for the question
Step-by-step explanation:
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Answer:
When Jamal Crawford comes to the free throw line, in the 2012-2013 NBA season, he would make 871 free throws out of 1000. Therefore, his percentage of making a free throw would by %87.1
-5(-2b-1)
Collect like terms and simplify
(Multiply the number outside the bracket with the ones in the parentheses/bracket)
(-x-)= +
(+x+)= +
(-x+)= -
(+x-)= -
Basically same signs multiplied = +
Different signs multiplied = -
-5(-2b)= 10b
-5(-1)=5
10b+5 is your answer
I assume you're asking to solve for the n-th term in the sequence,
.
From the given recursive rule,
![a_n = a_{n-1} + 2 \implies a_{n-1} = a_{n-2} + 2](https://tex.z-dn.net/?f=a_n%20%3D%20a_%7Bn-1%7D%20%2B%202%20%5Cimplies%20a_%7Bn-1%7D%20%3D%20a_%7Bn-2%7D%20%2B%202)
and by substitution,
![\implies a_n = a_{n-2} + 2\times2](https://tex.z-dn.net/?f=%5Cimplies%20a_n%20%3D%20a_%7Bn-2%7D%20%2B%202%5Ctimes2)
Similarly,
![a_n = a_{n-1} + 2 \implies a_{n-2} = a_{n-3} + 2](https://tex.z-dn.net/?f=a_n%20%3D%20a_%7Bn-1%7D%20%2B%202%20%5Cimplies%20a_%7Bn-2%7D%20%3D%20a_%7Bn-3%7D%20%2B%202)
![\implies a_n = a_{n-3} + 3\times2](https://tex.z-dn.net/?f=%5Cimplies%20a_n%20%3D%20a_%7Bn-3%7D%20%2B%203%5Ctimes2)
The pattern continues, so that we can write the n-th term in terms of the 1st one:
![a_n = a_1 + (n-1)\times2 \implies a_n = 10 + 2(n-1) = \boxed{2n+8}](https://tex.z-dn.net/?f=a_n%20%3D%20a_1%20%2B%20%28n-1%29%5Ctimes2%20%5Cimplies%20a_n%20%3D%2010%20%2B%202%28n-1%29%20%3D%20%5Cboxed%7B2n%2B8%7D)
So the first few terms of the sequence are
{10, 12, 14, 16, 18, 20, …}
Answer:
x=6
Step-by-step explanation: