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:
one pound per dinosaur
Step-by-step explanation:
Find the cosine ratio of angle ΘΘ. Hint: Use the slash symbol ( / ) to represent the fraction bar and enter the fraction with no spaces. (4 points)
Cosine=adjacent/hypotenuse
adjacent=8
Hypotenuse =17
cosΘ=8/17
Answer:
f(x) = -5/9 x + 5 1/9
Step-by-step explanation:
f(2)=4 and f(−7)=9 means the line pass through (2,4) and (- 7,9)
f(x) = mx + b
m = (y-y') / (x-x') = (9 - 4) / (- 7 - 2) = - 5/9
for (2,4) : b = f(x) - mx = y - mx = 4 - (- 5/9) x 2 = 4 + 10/9 = 46/9 = 5 1/9
f(x) = -5/9 x + 5 1/9
check for (-7, 9) f(-7) = (-5/9) * (-7) + 5 1/9 = 35/9 + 46/9 = 81/9 = 9