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
The correct question is
<span>Given cos theta=4/9 and csc theta < 0 find sin theta and tan theta
</span>
we know that
csc theta=1/sin theta
if csc theta < 0
then
sin theta < 0
we have that
<span>cos theta=4/9
we know that
sin</span>² theta+cos² theta=1
so
sin² theta=1-cos² theta-----> 1-(4/9)²----> 1-(16/81)----> 65/81
sin theta=-√(65/81)---->-√65/9
the answer Part a) is
sin theta=-√65/9
Part b) find tan theta
tan theta=sin theta/cos theta
tan theta=(-√65/9)/(4/9)-----> tan theta=-√65/4
the answer part b) is
tan theta=-√65/4
(3/4) / (1/11) =
3/4 * 11/1 =
33/4 <== improper
Answer:
2
Step-by-step explanation: