Insert 6 A . M's between 15 and -13

15, 11, 7, 3, -1, -5, -9, -13

Step-by-step explanation:

Inserting Arithmetic Means (AM's)

Suppose we want to insert n equally-spaced numbers inside a given interval with endpoints [x,y].

To place n numbers, we need to open n+1 spaces, at the end of which, a new number will be placed. The given interval is then divided into n+1 rational numbers of a value (y-x)/(n+1).

We need to insert 6 AM's between 15 and -13. Each AM will be inserted at a space of (-13-15)(6+1)=-28/7=-4.

Thus the sequence is formed subtracting 4:

15

15-4=11

11-4=7

7-4=3

3-4=-1

-1-4=-5

-5-4=-9

-9-4=-13

The sequence is (numbers in bold are the AM's):

15, 11, 7, 3, -1, -5, -9, -13

37 = ending with 7

unit's digit in the cube..3

56 = ending with 6

unit's digit in the cube..2

133 = ending with

unit's digit in the cube..7

hope it : ( : )

answer: i ! the answer is   9000 seconds, also d

hope i

-lavira : )