Question

trinity does this work to find the decimal equivalent of \\(\frac{34}{99}\\). what does her work tell you about the decimal equivalent?
\\(\

$$\begin{array}{r}0.34 \\\\ 99 \\overline{)34.000} \\\\ -297 \\\\ \\hline 430 \\\\ -396 \\\\ \\hline 340 \\\\ \\end{array}$$

\\)
options:

• it is 0.34. the decimal stops when she divides into the same number twice.
• it is \\(0.\overline{34}\\). the decimal repeats when she divides into the same number twice.
• it is \\(0.3\overline{4}\\). the decimal repeats when she divides into the same number twice.
• she needs to keep going to find more digits.

Explanation:

To determine the decimal equivalent of \(\frac{34}{99}\) (note: likely a typo, should be \(\frac{34}{99}\) not \(\frac{14}{99}\) from the long division), we analyze the long - division steps. In the long - division of \(34\div99\), after the first subtraction, we get a remainder, and then after the second subtraction, the remainder is \(340\), which is the same as the original dividend (when we consider the decimal expansion, we are essentially bringing down zeros). This means that the digits \(34\) will start to repeat. So the decimal is a repeating decimal \(0.\overline{34}\), and it repeats when we divide into the same number (the remainder \(340\) is related to the original \(34\) when considering the decimal places) twice.

Answer:

It is \(0.\overline{34}\). The decimal repeats when she divides into the same number twice. (Assuming the option with this description is the correct one; since the options are presented with this kind of description, this is the answer based on the long - division of a fraction with denominator \(99\) which is a factor of \(99 = 9\times11\), and fractions with denominators that are factors of \(9, 99, 999,\cdots\) have repeating decimals with the number of repeating digits equal to the number of \(9\)s in the denominator. Here, denominator \(99\) has two \(9\)s, so the repeating block has two digits, \(34\).)