Answer:
Step-by-step explanation:
A binary string with 2n+1 number of zeros, then you can get a binary string with 2n(+1)+1 = 2n+3 number of zeros either by adding 2 zeros or 2 1's at any of the available 2n+2 positions. Way of making each of these two choices are (2n+2)22. So, basically if b2n+12n+1 is the number of binary string with 2n+1 zeros then your
b2n+32n+3 = 2 (2n+2)22 b2n+12n+1
your second case is basically the fact that if you have string of length n ending with zero than you can the string of length n+1 ending with zero by:
1. Either placing a 1 in available n places (because you can't place it at the end)
2. or by placing a zero in available n+1 places.
0 ϵ P
x ϵ P → 1x ϵ P , x1 ϵ P
x' ϵ P,x'' ϵ P → xx'x''ϵ P
Answer:
im thinking n is 6.5 not sure wait not 6.5 its 3.5
The two answers are the one selected and the choice |x-5| ≤ 1. The steps to solve the other equation are:
|x-5| ≤ 1
x-5 ≤ 1, x-5 ≥ 0
-(x-5) ≤ 6, x-5 is less than 0
x ≤ 6, x ≥ 5
x ≥ 4, x is less than 5
x ∈ [5,6]
x ∈ [4,5]
4 ≤ x ≤ 6
(^which was the original answer^)
Answer:
here is the answer to your question
Answer: D) 18
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Explanation:
|-11| simplifies to 11
|-7| simplifies to 7
Whatever is inside the absolute value, you just remove the negative and that's the result of that absolute value expression. It represents the distance on a number line. So for instance -11 is 11 units from 0, which is why |-11| = 11.
Overall,
|-11| + |-7| = 11 + 7 = 18
