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.
Learn more about Sequences on:
brainly.com/question/12246947
#SPJ4
Answer:
14.14 i hope this helps lol ;)
You'd use the equation y2-y1/x2-x1
0-(-3)/-7-5 = 3/-12 = -1/4
So -1/4 would be the slope
m= -1/4
Answer:
Step-by-step explanation:
It depends on whether the 30 cm side is the hypotenuse, or one of the legs.
If the 30 cm side is the hypotenuse, then the third side is √(30^2-10^2) = √800 = 20√2 cm
If the 30 cm side is one of the legs, then the hypotenuse is √(30^2+10^2) = √1,000 = 10√10 cm
