You are multiplying the numbers by 8 so it would most likely be patrick
Answer:
Step-by-step explanation:
2p/2 = 2x+7
1) 2x+7 -x - 5 =
width = x + 2
2) 2p/2 = 21 in
x + 5 + x + 2 = 21
2x = 14
x = 7
lenght = 7 + 5 = 12 in
width = 7 + 2 = 9 in
3 )
A = (x+5)(x+2)
4)
A = 12 x 9 = 108
Answer:
A) ![a_{n} = a_{n-1} + a_{n-2} + 2^{n-2}](https://tex.z-dn.net/?f=a_%7Bn%7D%20%3D%20a_%7Bn-1%7D%20%2B%20a_%7Bn-2%7D%20%2B%202%5E%7Bn-2%7D)
B) ![a_{0} = a_{1} = 0](https://tex.z-dn.net/?f=a_%7B0%7D%20%3D%20a_%7B1%7D%20%3D%200)
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 : ![a_{0} = a_{1} = 0](https://tex.z-dn.net/?f=a_%7B0%7D%20%3D%20a_%7B1%7D%20%3D%200)
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
Rmemeber you can do anything to an equaiton as longn as you do it to both sides
also x times 1/x=1 since x/x=1
also x times 1=x
72=r90
divide both sides by 90
72/90=r
0.8=r
r=0.8
Answer:
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Step-by-step explanation:
vbnm,.