When writing a repeating decimal as a fraction does the number of repeating digits matter explain

No. Even if the number of different digits in the cycle is 1 or 1 million, the method of finding the fraction is the same.

Here's one sample

x = 0.333333333333333333 to infinity.

10x = 3.3333333333333333

x = 0.3333333333333333 Subtract

9x = 3

x = 1/3 which you likely knew already.

A more complicated one would be

x = 0.123456712345671234567 You have to indicate this goes to infinity.

1000000x = 1234567.1234567 to infinity

. . . . . . . . x = . . . . . . .0.1234567

999999x = 1234567 divide by 999999

x = 1234567/999999 This might reduce but for the purposes of this question it does not matter. Notice the method is the same.

If you mean that does the cycle contain an infinite number of digits, then yes it does matter. There have to be an infinite number of them.

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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