So k in the first problem would =1/2
and i guess in the second problem just put k in place of the first space and in the second space put 8 for x. y would then =4. Idk if I did this right, though, I'm honestly a little confused too.
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:
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So, she has 550 dollars saved. If the junior season pass costs 315, you subtract that from 550. You should then have 235 dollars. Next, you divide 235 by 45 dollars per lesson, and you should get 5.22. You can't buy half a lesson, so she only has enough for 5 lessons.
The answer to this is 67.00
Answer: 6.1 > 1.6 (greater)
Step-by-step explanation:
6 in 6.1 is greater than the 1 in 1.6
