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.

a. 3/4 ÷ 1/8 = 6

step-by-step explanation:

in the question, the number line represents 3/4 divided into 6 equal units. each arrow represents a unit and there are a total of six arrows.

the size of each unit is 1/8 as shown.

showing that 1/8 divides 3/4 into 6 divisions:

3/4 ÷ 1/8

= 3/4 x 8/1

= 3/1 x 8/4

= 3 x 2

= 6

hence, the number line represents six divisions given by 3/4 ÷ 1/8.