In the following sequence of numbers each number has one more "1" than the preceding number: 1, 11, 111, 1111, 11111, ... . What is the tens digit of the sum of the first 30 numbers of the sequence?

2

Step-by-step explanation:

we have the sequence 1, 11, 111, 1111 and so on.

We can write this sequence as:

A0 = 1

An = A(n -1) + 10^n.

such that:

A1 = A0 + 10^1 = 1 + 10 = 11.

A2 = 11 + 10^2 = 11 + 100 = 111.

and we want to find the tens digit of the sum of the first 30 terms.

ok, in the 30 terms, in the units digit we have a 1, so we have:

30*1 = 30

we leave the zero and the 3 goes to the tens place.

For the tens place. the first term does not aport nothing, and the other 29 terms aport a 1, so we have:

tens = 3 + 29*1 = 32

then we leave a 2 here and pass a 3 to the hundreds place.

But we already answered the question, the tens digit in the sum of the first 30 numbers of the sequence is 2.

  y= -3x is the required function.

step-by-step explanation:

we have been given a table:

x             y

0             0

1             -3

3             -9

-2             6

we can see that the values of y is -3 times the value of x

hence, the function would be: y= -3x.

put the values of x will give the same y as in the table that means function is satisfying the given table.