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:
okkkkkkkkkkkkkkkkkkkkkkkkkkkk
So this in the exponential form of ab^x, where a is the y intercept, b is the growth factor(base) and x is your domain. From this, we can answer each question. The initial amount is 800, growth rate is 32% and factor is 1.32.
Happy learning!
Answer:
Step-by-step explanation:
<u>Given:</u>
- Initial mass is m = 80 g.
- Half life = 10 days
- Total time = 60 days
<u>Number of half-life periods:</u>
<u>Equation for remaining sample:</u>
- s = m*(1/2)^r
- s = 80*(1/2)^6 = 80/64 = 1.25 g
Answer:
k = 10
Explanation:
The initial polynomial is:
If f(-2) = 0, then when we replace x by -2, the result will be 0. It means that we can write the following equation:
Therefore, we can solve for k as follows:
So, the value of k is 10
