Answer:
A) 
B) 
C) for n = 2
= 1
for n = 3
= 3
for n = 4
= 8
for n = 5
= 19
Step-by-step explanation:
A) A recurrence relation for the number of bit strings of length n that contain a pair of consecutive Os can be represented below
if a string (n ) ends with 00 for n-2 positions there are a pair of consecutive Os therefore there will be :
strings
therefore for n ≥ 2
The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os
b ) The initial conditions
The initial conditions are : 
C) The number of bit strings of length seven containing two consecutive 0s
here we apply the re occurrence relation and the initial conditions
for n = 2
= 1
for n = 3
= 3
for n = 4
= 8
for n = 5
= 19
A= 1/2(7 + 11)(5)
A = 1/2(18)(5)
A = 45 square units
Answer:
a = 12
Step-by-step explanation:
add 2 to each side
5a/3 = a/4 + 17
distribute 12 to remove fractions
20a = 3a + 204
17a = 204
a = 12
You subtract and find x=10
Answer:
1) 15-4x
2) 3x+2
Step-by-step explanation:
1) product of x and 4 is 4x then subtract that from 15
2) use x as your age in this situation so your age times 3 is 3x and since he's 2 years older than that u add 2 more