65 sequences.
Lets solve the problem,
The last term is 0.
To form the first 18 terms, we must combine the following two sequences:
0-1 and 0-1-1
Any combination of these two sequences will yield a valid case in which no two 0's and no three 1's are adjacent
So we will combine identical 2-term sequences with identical 3-term sequences to yield a total of 18 terms, we get:
2x + 3y = 18
Case 1: x=9 and y=0
Number of ways to arrange 9 identical 2-term sequences = 1
Case 2: x=6 and y=2
Number of ways to arrange 6 identical 2-term sequences and 2 identical 3-term sequences =8!6!2!=28=8!6!2!=28
Case 3: x=3 and y=4
Number of ways to arrange 3 identical 2-term sequences and 4 identical 3-term sequences =7!3!4!=35=7!3!4!=35
Case 4: x=0 and y=6
Number of ways to arrange 6 identical 3-term sequences = 1
Total ways = Case 1 + Case 2 + Case 3 + Case 4 = 1 + 28 + 35 + 1 = 65
Hence the number of sequences are 65.
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First is yes
Second is no
Third is no
Fourth is no
Answer:
y = 3x +4
Step-by-step explanation:
The equation for the parallel line will have the same x- and y-coefficients, but a different constant. You can put the given point values into the equation to see what the constant needs to be:
y = 3x + b
10 = 3·2 + b . . . . . . substitute x=2, y=10
4 = b . . . . . . . . . . . . subtract 6
The equation of the line is ...
y = 3x +4
Answer:
15 games
Step-by-step explanation:
12X3=36 5X3=15
<span>-8x-3(1+2x)
= </span><span>-8x -3 - 6x
= -14x - 3</span>
