This problem tackles the place values of numbers. From the rightmost end of the number to the leftmost side, these place values are ones, tens, hundreds, thousands, ten thousands, hundred thousands, millions, ten millions, one hundred millions, and so on and so forth. My idea for the solution of this problem is to add up all like multiples. In this problem, there are 5 multiples expressed in ones, thousands, hundred thousands, tens and hundreds. Hence, you will add up 5 like terms. The solution is as follows
30(1) + 82(1,000) + 4(100,000) + 60(10) + 100(100)
The total answer is 492,630. Therefore, the number's identity is 492,630.
Answer: -30
Step-by-step explanation:
Answer:
No solution
Step-by-step explanation:
24a-22= -4(1-6a)
24a-22= -4+24a
-24a -24a
0-22= -4
-22= -4
4 consecutive integers...z, z + 1, z + 2, z + 3
z + (z + 1) + (z + 2) + (z + 3) = 166...combine like terms
4z + 6 = 166
4z = 166 - 6
4z = 160
z = 160/4
z = 40
z + 1 = 40 + 1 = 41
z + 2 = 40 + 2 = 42
z + 3 = 40 + 3 = 43
so ur numbers are : 40,41,42,43 with the largest being 43
