<u>Statement </u> <u>Reason</u>
1. AB = x + 16 1. Given
BC = 4x + 11
AC = 77
2. AB + BC = AC 2. Segment Addition Postulate
3. x + 16 + 4x + 11 = 77 3. Substitution Property
4. 5x + 27 = 77 4. Simplification <em>(added like terms)</em>
5. 5x = 50 5. Subtraction Property of Equality
6. x = 10 6. Division Property of Equality
7. AB = 10 + 16 7. Substitution Property
8. AB = 26 8. Simplification <em>(added like terms)</em>
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The words that might fit will be
always
sometimes
never
sometimes
Hope this helps.
Answer:
Step-by-step explanation:
The set {1,2,3,4,5,6} has a total of 6! permutations
a. Of those 6! permutations, 5!=120 begin with 1. So first 120 numbers would contain 1 as the unit digit.
b. The next 120, including the 124th, would begin with '2'
c. Then of the 5! numbers beginning with 2, there are 4!=24 including the 124th number, which have the second digit =1
d. Of these 4! permutations beginning with 21, there are 3!=6 including the 124th permutation which have third digit 3
e. Among these 3! permutations beginning with 213, there are 2 numbers with the fourth digit =4 (121th & 122th), 2 with fourth digit 5 (numbers 123 & 124) and 2 with fourth digit 6 (numbers 125 and 126).
Lastly, of the 2! permutations beginning with 2135, there is one with 5th digit 4 (number 123) and one with 5 digit 6 (number 124).
∴ The 124th number is 213564
Similarly reversing the above procedure we can determine the position of 321546 to be 267th on the list.
Answer:
v=-2/5 hope this is helpful
