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
PQRS is an isosceles trapezoid and m < Q=127
so m<Q = m<P =127
answer
127
Answer:
C
Step-by-step explanation:
No solutions, because the graphs do not intersect.
