- In a fraction, identify which number is divided by which. - Carry out that division to obtain the decimal form.

1. \(\frac{3}{4}\) means \(3\div4\). 2. \(3\div4=0.75\).

a) \(3\div4\) b) \(0.75\)

- Treat the fraction bar as division. - Determine the quotient's magnitude and sign separately if helpful. - Keep the signed division expression visible in the response.

1. Interpret the fraction as \((-9)\div4\). 2. \(9\div4=2.25\), so \((-9)\div4=-2.25\).

\((-9)\div4=-2.25\)

- Does \(20\) divide evenly into \(100\)? - How can you keep a fraction's value unchanged when changing its denominator? - Which decimal place represents hundredths?

1. Multiply numerator and denominator by \(5\): \(\frac{7}{20}=\frac{35}{100}\). 2. Thirty-five hundredths is \(0.35\).

\(\frac{7}{20}=\frac{35}{100}=0.35\). The equivalent fraction means thirty-five hundredths, written with \(3\) in the tenths place and \(5\) in the hundredths place.

- When the dividend is smaller than the divisor, add a decimal point and bring down a zero. - Keep track of what is left over after each division step. - What is alike about the two remainder steps?

1. Extend \(1\) to \(1.0\). In the tenths step, \(10\div3\) gives digit \(3\) and remainder \(1\) tenth. 2. Bring down another zero. In the hundredths step, \(10\div3\) again gives digit \(3\) and remainder \(1\) hundredth. 3. The same remainder digit repeats at each place, so every further decimal digit is \(3\): \(\frac{1}{3}=0.\overline{3}\).

Digits: \(3,3\). Remainders: \(1,1\). The remainder repeats, so \(\frac{1}{3}=0.\overline{3}\).

- Look for the shortest group of digits that repeats. - The bar should cover the entire repeating block, not just its last digit. - Count digits in a single block.

1. The smallest block that repeats is \(36\). 2. Therefore the notation is \(0.\overline{36}\), and the block has \(2\) digits.

a) \(0.\overline{36}\) b) \(2\) digits

- Does the decimal in the question stop or continue forever? - Write more digits of the exact fractional value. - A short decimal approximation is not always an exact equality.

1. The fraction has the exact expansion \(\frac{1}{3}=0.333\dots=0.\overline{3}\). 2. The number \(0.333\) stops after three digits, so it is slightly smaller than \(\frac{1}{3}\).

No. \(0.333<0.\overline{3}=\frac{1}{3}\).

- Read the decimal using its place value. - Start with tenths, then simplify. - Divide numerator and denominator by the same common factor.

1. The digit \(6\) is in the tenths place, so \(0.6=\frac{6}{10}\). 2. Simplify: \(\frac{6}{10}=\frac{3}{5}\).

\(0.6=\frac{6}{10}=\frac{3}{5}\); the digit \(6\) means six tenths.

- Interpret the fraction bar as division. - If the decimal does not terminate, identify the digit that determines rounding to the nearest hundredth. - Distinguish exact decimal values from rounded approximations.

1. \(3\div4=0.75\). 2. \(5\div6=0.8333\ldots\), so \(\frac{5}{6}\approx0.83\) to the nearest hundredth. 3. \(12\div5=2.4\). 4. \(-7\div9=-0.7777\ldots\), so \(-\frac{7}{9}\approx-0.78\) to the nearest hundredth.

a) \(\frac{3}{4}=0.75\) b) \(\frac{5}{6}\approx0.83\) c) \(\frac{12}{5}=2.4\) d) \(-\frac{7}{9}\approx-0.78\)

- Divide the numerator by the denominator to convert a fraction to a decimal. - Convert the mixed number’s fractional part. - Compare the decimals by aligning place values.

1. Convert each number: \(\frac{11}{4}=2.75\), \(2.8=2.8\), \(2\frac{3}{5}=2.6\), and \(\frac{27}{10}=2.7\). 2. Compare the decimal forms: \(2.6<2.7<2.75<2.8\). 3. Therefore, \(2\frac{3}{5}<\frac{27}{10}<\frac{11}{4}<2.8\).

Decimal forms: \(2.75\), \(2.8\), \(2.6\), and \(2.7\) Order: \(2\frac{3}{5}<\frac{27}{10}<\frac{11}{4}<2.8\)

- Continue the division through at least the thousandths place. - Use the digit immediately to the right of the hundredths place to decide whether to round up. - Distinguish the exact repeating decimal from its rounded hundredths value.

1. \(19\div12=1.5833\ldots\). The thousandths digit is \(3\), so \(\frac{19}{12}\approx1.58\) to the nearest hundredth. 2. \(7\div11=0.636363\ldots\). The thousandths digit is \(6\), so \(\frac{7}{11}\approx0.64\) to the nearest hundredth.

a) \(\frac{19}{12}\approx1.58\); thousandths digit: \(3\) b) \(\frac{7}{11}\approx0.64\); thousandths digit: \(6\)

- Rewrite every number in the same form before comparing. - Divide the numerator by the denominator to convert each fraction. - Write zeros at the right of decimals when helpful for comparing place values.

1. Convert the fractions: \(\frac{2}{5} = 0.4\), \(\frac{1}{2} = 0.5\), and \(\frac{3}{8} = 0.375\). 2. Compare the decimals: \(0.375 < 0.4 < 0.405 < 0.45 < 0.5\). 3. Replace each converted decimal with its original fraction.

\(\frac{2}{5}=0.4\), \(\frac{1}{2}=0.5\), and \(\frac{3}{8}=0.375\) Least to greatest: \(\frac{3}{8} < \frac{2}{5} < 0.405 < 0.45 < \frac{1}{2}\)

- Carry out numerator divided by denominator before choosing the bounding hundredths. - Compare the decimal expansion digit by digit with two adjacent hundredths. - Include both the conversion and inequality in your response.

1. \(5\div6=0.8333\ldots=0.8\overline{3}\). 2. Compare with consecutive hundredths: \(0.83<0.8333\ldots<0.84\). 3. Therefore, \(0.83<\frac{5}{6}<0.84\).

\(\frac{5}{6}=0.8333\ldots=0.8\overline{3}\). Thus, \(0.83<\frac{5}{6}<0.84\).

- Convert all values to the same form. - Write each percent as a decimal by dividing by \(100\). - Convert fractions using division or an equivalent fraction with denominator \(100\).

1. Convert the fraction and percent forms: \(\frac{3}{4} = 0.75\), \(70\% = 0.70\), and \(\frac{13}{20} = \frac{65}{100} = 0.65\). 2. Compare the decimals: \(0.65 < 0.70 < 0.72 < 0.75\). 3. Replace the converted values with their original forms.

\(\frac{3}{4}=0.75\), \(70\%=0.70\), and \(\frac{13}{20}=0.65\) Least to greatest: \(\frac{13}{20} < 70\% < 0.72 < \frac{3}{4}\)

- Put every value into decimal form before making a selection. - Interpret the fraction bars as division and the percent sign as “per hundred.” - Check the place value of every conversion, especially zeros.

1. The first value is already \(0.375\). 2. Dividing \(3\) by \(8\) gives \(0.375\). 3. Converting the percent gives \(37.5\%=0.375\). 4. The remaining fraction gives \(\frac{19}{50}=0.38\), which is close but not equal to \(0.375\).

\(0.375=0.375\), \(\frac{3}{8}=0.375\), \(37.5\%=0.375\), but \(\frac{19}{50}=0.38\). The different number is \(\frac{19}{50}\).

- Try rewriting each denominator as \(10\), \(100\), or \(1000\). - You can also divide the numerator by the denominator.

1. For a), rewrite the fraction with a denominator of \(100\): \(\frac{9}{20} = \frac{45}{100} = 0.45\). 2. For b), divide the numerator by the denominator: \(3 \div 8 = 0.375\). Equivalently, \(\frac{3}{8} = \frac{375}{1000}\). 3. For c), \(\frac{1}{4} = \frac{25}{100} = 0.25\). Therefore, \(2\frac{1}{4} = 2.25\).

a) \(0.45\) b) \(0.375\) c) \(2.25\)

- Convert the fraction in each pair before comparing. - Keep enough digits of a repeating decimal to distinguish it from a terminating decimal. - Include the conversion in each sub-answer.

1. \(\frac{2}{5}=0.4\), so the values are equal. 2. \(\frac{1}{3}=0.3333\ldots\), so \(0.33<\frac{1}{3}\). 3. \(\frac{17}{20}=0.85\), so the values are equal.

a) \(\frac{2}{5}=0.4\), so \(=\) b) \(\frac{1}{3}=0.3333\ldots\), so \(0.33<\frac{1}{3}\) c) \(\frac{17}{20}=0.85\), so \(=\)

- Consider whether each denominator can be related conveniently to a power of \(10\). - When a decimal does not terminate, look for a repeating remainder pattern. - Check each decimal by converting it back to the original fraction if needed.

1. \(\frac{9}{20}=\frac{45}{100}=0.45\). 2. \(5\div6=0.8333\ldots=0.8\overline{3}\). 3. \(4\div11=0.3636\ldots=0.\overline{36}\). 4. \(\frac{17}{50}=\frac{34}{100}=0.34\).

a) \(0.45\) b) \(0.8\overline{3}\) c) \(0.\overline{36}\) d) \(0.34\)

- Write a terminating decimal as a fraction with denominator \(10\), \(100\), or \(1000\). - Simplify every fraction completely. - A denominator of \(8\) can be changed to \(1000\).

1. For a), \(0.625 = \frac{625}{1000}\). Divide the numerator and denominator by \(125\): \(\frac{625}{1000} = \frac{5}{8}\). 2. For b), \(\frac{2}{5} = \frac{4}{10} = 0.4\). 3. For c), \(2.4 = \frac{24}{10} = \frac{12}{5}\). 4. For d), \(-\frac{3}{8} = -\frac{375}{1000} = -0.375\).

a) \(\frac{5}{8}\) b) \(0.4\) c) \(\frac{12}{5}\) d) \(-0.375\)

- Find a whole-number multiplier that changes the denominator to a power of \(10\). - Multiply numerator and denominator by the same value. - The equivalent fraction must appear before the decimal in your response.

1. a) \(\frac{3}{5}=\frac{6}{10}=0.6\). 2. b) \(\frac{7}{20}=\frac{35}{100}=0.35\). 3. c) \(\frac{9}{40}=\frac{225}{1000}=0.225\). 4. d) \(\frac{11}{250}=\frac{44}{1000}=0.044\).

a) \(\frac{3}{5}=\frac{6}{10}=0.6\) b) \(\frac{7}{20}=\frac{35}{100}=0.35\) c) \(\frac{9}{40}=\frac{225}{1000}=0.225\) d) \(\frac{11}{250}=\frac{44}{1000}=0.044\)

- Write the fraction conversions before comparing the four values. - Add trailing zeros when aligning decimal places. - Give the final order using the original forms.

1. \(\frac{3}{8}=0.375\) and \(\frac{2}{5}=0.4\). 2. Compare \(0.375<0.380<0.400<0.420\). 3. Restore the original forms in the ordered list.

Conversions: \(\frac{3}{8}=0.375\), \(\frac{2}{5}=0.4\). Order: \(\frac{3}{8}<0.38<\frac{2}{5}<0.42\).

- Write each decimal as a fraction with denominator \(10\), \(100\), or \(1000\). - Simplify or scale the fraction to match the given denominator or numerator. - Compare corresponding parts of equivalent fractions.

1. For a), \(0.375 = \frac{375}{1000} = \frac{3}{8}\), so \(\square = 3\). 2. For b), \(0.64 = \frac{64}{100} = \frac{16}{25}\), so \(\square = 16\). 3. For c), \(0.07 = \frac{7}{100}\). Comparing \(\frac{7}{\square}\) with \(\frac{7}{100}\) gives \(\square = 100\).

a) \(\square = 3\) b) \(\square = 16\) c) \(\square = 100\)

- Keep a remainder list beside each long division. - A zero remainder ends the decimal; a repeated nonzero remainder restarts a quotient-digit cycle. - Include the remainder evidence, not only the decimal classification.

1. The nonzero remainders are \(14,12,8\), followed by \(0\), giving \(0.1875\); it terminates. 2. The remainders go \(2\to2\), so remainder \(2\) repeats and \(\frac{5}{6}=0.8\overline{3}\). 3. The remainders go \(7\to4\to7\), so remainder \(7\) repeats and \(\frac{4}{11}=0.\overline{36}\).

a) Remainders \(14,12,8,0\); \(\frac{3}{16}=0.1875\), terminating b) Remainders \(2,2,\ldots\); \(\frac{5}{6}=0.8\overline{3}\), repeating c) Remainders \(7,4,7,\ldots\); \(\frac{4}{11}=0.\overline{36}\), repeating

- Convert the fraction in each part before deciding the symbol. - Scaling to denominator \(10\) or \(100\) may make a conversion immediate. - Keep each conversion visible in the answer.

1. a) \(\frac{7}{10}=0.7<0.75\). 2. b) \(\frac{1}{4}=0.25<0.4\). 3. c) \(\frac{12}{50}=0.24\), so the values are equal. 4. d) \(\frac{6}{5}=1.2<1.3\).

a) \(\frac{7}{10}=0.7<0.75\) b) \(0.4>0.25=\frac{1}{4}\) c) \(\frac{12}{50}=0.24\), so \(=\) d) \(1.3>1.2=\frac{6}{5}\)

- Treat each fraction bar as division or rewrite it with a power-of-ten denominator. - Do not stop after recognizing one likely match; check all three options. - Compare each decimal exactly with \(0.625\).

1. \(\frac{625}{100}=6.25\). 2. \(\frac{5}{8}=0.625\). 3. \(\frac{3}{4}=0.75\). 4. Therefore, \(\frac{5}{8}\) matches \(0.625\).

\(\frac{625}{100}=6.25\) \(\frac{5}{8}=0.625\) \(\frac{3}{4}=0.75\) Matching fraction: \(\frac{5}{8}\)

- For some fractions, rewrite the denominator as \(10\), \(100\), or \(1000\). - A percent is a number divided by \(100\). - To simplify a fraction, divide the numerator and denominator by a common factor.

1. Multiply \(\frac{7}{25}\) by \(\frac{4}{4}\): \(\frac{7}{25} = \frac{28}{100} = 0.28\). 2. Write \(0.08\) as \(\frac{8}{100}\), then simplify: \(\frac{8}{100} = \frac{2}{25}\). 3. Convert the mixed number: \(1\frac{3}{4} = 1 + 0.75 = 1.75\). 4. Divide by \(100\): \(12.5\% = 0.125\). 5. Write \(0.375\) as \(\frac{375}{1000}\), then simplify: \(\frac{375}{1000} = \frac{3}{8}\).

(1) \(0.28\) (2) \(\frac{2}{25}\) (3) \(1.75\) (4) \(0.125\) (5) \(\frac{3}{8}\)

- Set up numerator divided by denominator and continue by bringing down zeros. - Keep a separate record of each remainder. - Use what happens to the remainder to justify termination.

1. In \(7.0000\div16\), the quotient digits are \(4,3,7,5\). 2. The successive nonzero remainders are \(6\), \(12\), and \(8\), followed by remainder \(0\). 3. Therefore, \(\frac{7}{16}=0.4375\), and it terminates because the remainder reaches \(0\).

\(\frac{7}{16}=0.4375\). The nonzero remainders are \(6\), \(12\), and \(8\); the next remainder is \(0\), so the decimal terminates.

- Continue the division by bringing down zeros and record each remainder. - Compare each new remainder with earlier remainders. - The quotient digits associated with a repeated remainder will repeat in the same cycle.

1. The division produces remainder \(2\) after the first decimal digit, then remainder \(8\). 2. The next step returns remainder \(8\), so the quotient digit \(6\) repeats. 3. Thus, \(\frac{5}{12}=0.41\overline{6}\).

a) The remainder sequence reaches \(2\), then \(8\), and then \(8\) repeats. b) \(\frac{5}{12}=0.41\overline{6}\)

- Record the remainder after each quotient digit. - Stop the cycle only when a remainder already seen returns. - Match the repeating quotient digits to the completed remainder cycle.

1. The first decimal digit is \(2\), leaving remainder \(4\). 2. The next digit is \(1\), leaving remainder \(7\), which returns to the starting remainder state. 3. Therefore, the repeating block is \(21\) and \(\frac{7}{33}=0.\overline{21}\).

The remainders cycle \(4\to7\to4\). The repeating block is \(21\), so \(\frac{7}{33}=0.\overline{21}\).

- Relate the new numerator to the known fraction's numerator. - What operation changes one third to two thirds? - Keep the repeating bar in the exact result.

1. \(\frac{2}{3}=2\cdot\frac{1}{3}\). 2. Doubling each repeating third gives \(0.\overline{6}\), so \(\frac{2}{3}=0.\overline{6}\).

\(\frac{2}{3}=0.\overline{6}\); the repeating digit \(3\) doubles to \(6\).

- In long division, what happens when the partial dividend is smaller than the divisor? - Record the remainder each time you bring down a zero. - Compare remainders to decide whether the digits will repeat.

1. The first decimal step gives \(70\div99\): \(70<99\), so the tenths digit is \(0\). 2. Bring down another zero: \(700=7\cdot99+7\). The hundredths digit is \(7\), with remainder \(7\). 3. The same steps repeat, producing the next digits \(0\) and \(7\). The first four digits are \(0707\), so \(\frac{7}{99}=0.\overline{07}\). 4. The first digit is zero because \(70\) is too small to contain a whole \(99\).

\(70<99\), so the tenths digit is \(0\). Then \(700=7\cdot99+7\), so the hundredths digit is \(7\), with remainder \(7\). These steps repeat for the next digits \(0,7\). Thus the first four digits are \(0707\) and \(\frac{7}{99}=0.\overline{07}\).

- A quotient digit tells how many times the divisor fits into the partial dividend. - Compare the remainder after each subtraction line. - What would happen if the same remainder appeared at the next step?

1. After the first decimal digit, \(10-6=4\), leaving remainder \(4\). 2. The next step gives \(40-36=4\), so the missing digit is \(36\div6=6\). 3. The same nonzero remainder repeats, and the digit \(6\) recurs. Thus \(\frac{1}{6}=0.1\overline{6}\).

a) \(6\) b) Repeating: \(\frac{1}{6}=0.1\overline{6}\). The nonzero remainder \(4\) repeats.

- Rewrite a mixed number as an improper fraction before dividing. - Look for a repeating remainder in the division. - A fully simplified fraction terminates only when its denominator has no prime factors other than \(2\) and \(5\).

1. Divide to convert \(\frac{7}{8}\): \(7 \div 8 = 0.875\), which terminates. 2. Rewrite the mixed number as \(1\frac{2}{3} = \frac{5}{3}\). Then \(5 \div 3 = 1.666\ldots = 1.\overline{6}\). 3. Divide to convert \(-\frac{13}{11}\): \(-13 \div 11 = -1.181818\ldots = -1.\overline{18}\). 4. Only \(\frac{7}{8}\) has a terminating decimal form.

\(\frac{7}{8} = 0.875\), terminating \(1\frac{2}{3} = 1.\overline{6}\) \(-\frac{13}{11} = -1.\overline{18}\) Terminating decimal: \(\frac{7}{8}\)

- Put all four values into comparable decimal form. - Keep enough digits to distinguish close values before using the requested rounded forms. - Check the final order against the unrounded values.

1. Convert \(A\): \(5\div7=0.714285\ldots\), so \(A\approx0.714\) to the nearest thousandth. 2. Convert \(C\): \(11\div15=0.7333\ldots\), so \(C\approx0.733\) to the nearest thousandth. 3. Compare the unrounded values: \(0.710<0.714285\ldots<0.715<0.7333\ldots\). 4. Therefore, \(B<A<D<C\).

\(A\approx0.714\) \(C\approx0.733\) Least to greatest: \(0.71<\frac{5}{7}<0.715<\frac{11}{15}\)

- Simplify a fraction before examining its denominator. - A terminating decimal comes from a denominator whose only prime factors are \(2\) and \(5\). - Check whether each fraction is already in simplest form.

1. A fraction in simplest form has a terminating decimal exactly when its denominator has no prime factors other than \(2\) and \(5\). 2. For a), \(25 = 5^2\), so \(\frac{7}{25}\) has a terminating decimal. 3. For b), \(\frac{9}{12} = \frac{3}{4}\), and \(4 = 2^2\), so the decimal terminates. 4. For c), \(40 = 2^3 \cdot 5\), so \(\frac{13}{40}\) has a terminating decimal. 5. For d), \(30 = 2 \cdot 3 \cdot 5\). The factor \(3\) remains, so the decimal repeats rather than terminates.

a) Terminating, because \(25 = 5^2\) b) Terminating, because the simplified denominator is \(4 = 2^2\) c) Terminating, because \(40 = 2^3 \cdot 5\) d) Repeating, because the simplified denominator contains a factor of \(3\)

- Factor the denominator completely. - Identify which denominator primes are incompatible with termination after simplification. - The least numerator must contain enough factors to cancel all of those primes.

1. Factor \(126=2\cdot3^2\cdot7\). 2. A terminating decimal requires the simplified denominator to contain only factors \(2\) and \(5\). 3. The numerator must therefore cancel \(3^2\cdot7=63\). 4. The least positive choice is \(x=63\), and \(\frac{63}{126}=\frac{1}{2}=0.5\).

\(126=2\cdot3^2\cdot7\). The factors \(3^2\cdot7\) must cancel, so the least positive numerator is \(x=63\). Then \(\frac{63}{126}=\frac{1}{2}=0.5\), which terminates.

- List primes other than \(2\) and \(5\). - Use those primes to form two-digit numbers. - Factor each proposed denominator to justify it.

1. Each chosen denominator must have at least one prime factor other than \(2\) or \(5\). 2. One choice is \(12\), because \(12 = 2^2 \cdot 3\). The factor \(3\) makes the decimal repeat. 3. A second choice is \(21\), because \(21 = 3 \cdot 7\). Both prime factors produce a repeating decimal. 4. A third choice is \(11\), which is itself a prime other than \(2\) or \(5\). 5. Many other answers are possible, provided the denominator has a prime factor other than \(2\) or \(5\).

Answers will vary. One possible set is \(11\), \(12\), and \(21\), because each number has a prime factor other than \(2\) or \(5\).

- Treat the fraction bar as a division sign. - Find the two neighboring tenth values and subtract to compare distances. - What do the numerator and denominator tell you about one cup?

1. Divide \(3\) by \(8\): \(\frac{3}{8}=0.375\). 2. The value lies between \(0.3\) and \(0.4\). It is \(0.075\) above \(0.3\) and \(0.025\) below \(0.4\). 3. The fraction shows three of eight equal parts of one cup. The decimal expresses the same exact quantity using tenths, hundredths, and thousandths.

a) \(0.375\) cup b) Between \(0.3\) and \(0.4\); distances \(0.075\) and \(0.025\), respectively. c) \(\frac{3}{8}\) directly names three of eight equal parts, while \(0.375\) uses decimal place value.

- Compare repeated use of the exact fraction with repeated use of the shortened decimal. - Decide whether \(0.33\) is the complete decimal expansion of \(\frac{1}{3}\). - Exactness matters when the three shares must recombine to one whole.

1. \(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\). 2. \(0.33+0.33+0.33=0.99\). 3. Since \(\frac{1}{3}=0.\overline{3}\), the terminating decimal \(0.33\) is only an approximation.

a) \(1\) b) \(0.99\) c) \(\frac{1}{3}\) is exact, while \(0.33\) is a rounded approximation of \(0.\overline{3}\).

- Write each nondecimal number in decimal form before comparing it with the endpoints. - A value must satisfy both strict inequalities. - Keep enough repeating digits to distinguish the value from \(5.7\).

1. \(\frac{11}{2}=5.5\), so it lies in the interval. 2. \(5\frac{2}{3}=5.666\ldots\), so it lies in the interval. 3. \(\frac{53}{10}=5.3\), so it does not lie in the interval. 4. \(5.65\) already lies between \(5.4\) and \(5.7\).

\(\frac{11}{2}=5.5\) — in the interval \(5\frac{2}{3}=5.666\ldots\) — in the interval \(\frac{53}{10}=5.3\) — not in the interval \(5.65\) — in the interval

- Express all four quantities in the same numerical form before comparing them. - Keep enough digits from a nonterminating decimal to round and order correctly. - Compare corresponding decimal places from left to right.

1. Convert each value: \(A=\frac{5}{8}=0.625\), \(B=63\%=0.63\), \(C=0.622\), and \(D=\frac{2}{3}=0.6666\ldots\approx0.667\). 2. Compare the exact values: \(0.622<0.625<0.63<0.6666\ldots\). 3. Therefore, \(C<A<B<D\).

a) \(A=0.625\); \(B=0.63\); \(C=0.622\); \(D\approx0.667\) b) \(C<A<B<D\)

- A percent means a number out of one hundred. - Divide the numerator by the denominator for each fraction. - Look at the thousandths digit when rounding to hundredths.

1. \(112.4\%=1.124\), so to the nearest hundredth it is \(\approx1.12\). 2. \(5\div13=0.\overline{384615}\), so \(\frac{5}{13}\approx0.38\). 3. \(25\div6=4.1\overline{6}\), so \(\frac{25}{6}\approx4.17\).

a) Exact decimal \(1.124\); rounded \(1.12\) b) Exact: \(0.\overline{384615}\); rounded: \(0.38\) c) Exact: \(4.1\overline{6}\); rounded: \(4.17\)

- Put every value into a form that makes close decimal comparisons possible. - Pay attention to the difference between a terminating decimal and an infinite repeating decimal. - Identify any exact equalities before ordering the distinct values.

1. Convert the mixed number: \(1\frac{3}{10}=1.3\). 2. Convert the improper fraction: \(\frac{4}{3}=1.3333\ldots=1.\overline{3}\). 3. Compare the decimal forms: \(1.300<1.330<1.3330<1.3333\ldots\). 4. Therefore, \(1.3=1\frac{3}{10}<1.33<1.333<\frac{4}{3}\). 5. The decimal \(1.333\) stops after three decimal places, while \(\frac{4}{3}\) has infinitely many repeating \(3\)s.

a) \(1.3=1\frac{3}{10}<1.33<1.333<\frac{4}{3}\) b) \(\frac{4}{3}=1.\overline{3}\) is a repeating decimal, while \(1.333\) is a terminating decimal, so \(\frac{4}{3}>1.333\).

- Because the numerator is \(3\), what must be true about \(n\) if the fraction is in simplest form? - List numbers between \(10\) and \(30\) whose only prime factors are \(2\) and \(5\). - Check powers and products of \(2\) and \(5\) systematically.

1. Since \(\frac{3}{n}\) is in simplest form, \(n\) is not divisible by \(3\). 2. A terminating decimal requires the denominator to have only prime factors of \(2\) and \(5\). 3. The numbers between \(10\) and \(30\) with only those prime factors are \(16 = 2^4\), \(20 = 2^2 \cdot 5\), and \(25 = 5^2\). 4. None of these values is divisible by \(3\), so all three keep the fraction in simplest form. 5. Therefore, \(n = 16\), \(20\), or \(25\).

\(n = 16\), \(20\), or \(25\)

- Divide the numerator by the denominator to see the repeating block. - Add trailing zeros to terminating decimals to compare them fairly. - Compare digits only as far as the first difference.

1. Divide: \(5\div11=0.454545\ldots=0.\overline{45}\). 2. Write the terminating values as \(0.450000\ldots\) and \(0.454000\ldots\). 3. Compare their first differing digits: \(0.45<0.454<0.454545\ldots\).

\(\frac{5}{11}=0.\overline{45}\) \(0.45<0.454<\frac{5}{11}\)

- Simplify each fraction first. - A terminating decimal has only factors of \(2\) and \(5\) in its simplified denominator. - The larger exponent of \(2\) and \(5\) determines the number of decimal places.

1. For a), \(\frac{42}{105} = \frac{2}{5}\). The denominator is \(5^1\), so the decimal terminates with \(1\) decimal place. 2. For b), \(\frac{33}{120} = \frac{11}{40}\), and \(40 = 2^3 \cdot 5\). The decimal terminates with \(\max(3, 1) = 3\) decimal places. 3. For c), \(\frac{9}{72} = \frac{1}{8}\), and \(8 = 2^3\). The decimal terminates with \(3\) decimal places.

a) Terminating; \(1\) decimal place b) Terminating; \(3\) decimal places c) Terminating; \(3\) decimal places

- Simplify each fraction before examining its denominator. - A prime factor other than \(2\) or \(5\) causes repetition. - Distinguish total decimal places from nonrepeating digits before a repeating block.

1. For a), \(\frac{13}{65} = \frac{1}{5}\). The simplified denominator has only the prime factor \(5\), so the decimal terminates with \(1\) decimal place. 2. For b), \(\frac{14}{350} = \frac{1}{25} = \frac{1}{5^2}\). The decimal terminates with \(2\) decimal places. 3. For c), \(\frac{7}{30}\) is in simplest form, and \(30 = 2 \cdot 3 \cdot 5\). The factor \(3\) makes the decimal repeat. In fact, \(\frac{7}{30} = 0.2\overline{3}\), with one nonrepeating digit before the repeating part.

a) Terminating; \(1\) decimal place b) Terminating; \(2\) decimal places c) Repeating; \(0.2\overline{3}\) has one nonrepeating decimal digit

- Simplify each fraction first. - Factor the simplified denominator into powers of \(2\) and \(5\). - The larger exponent tells how many decimal places are needed.

1. Simplify the first fraction: \(\frac{27}{1200} = \frac{9}{400}\). 2. Factor the denominator: \(400 = 2^4 \cdot 5^2\). 3. The greater exponent is \(4\), so \(\frac{27}{1200}\) has a terminating decimal with \(4\) decimal places. 4. The fraction \(\frac{19}{400}\) is already in simplest form and has the same denominator, so it also has \(4\) decimal places. 5. Indeed, \(\frac{9}{400} = 0.0225\) and \(\frac{19}{400} = 0.0475\).

a) \(4\) decimal places b) Both fractions have \(4\) decimal places because, after simplifying, both denominators are \(400 = 2^4 \cdot 5^2\).

- Convert only the fractional part first. - Identify the length of the repeating block. - Decide whether the target position is odd or even.

1. Convert the fractional part: \(5 \div 11 = 0.454545\ldots = 0.\overline{45}\). 2. Therefore, \(2\frac{5}{11} = 2.\overline{45}\). 3. The repeating block has length \(2\): \(4\) occurs in odd-numbered decimal places and \(5\) occurs in even-numbered decimal places. 4. Since \(50\) is even, the \(50\)th digit is \(5\).

\(2\frac{5}{11} = 2.\overline{45}\). The \(50\)th digit after the decimal point is \(5\).

- The negative sign does not change the repeating digit pattern. - Separate any nonrepeating digits from the repeating block. - Count only the digits in the repeating block.

1. Convert \(\frac{5}{12}\): \(5 \div 12 = 0.41666\ldots = 0.41\overline{6}\). Its repeating block is \(6\), so the period length is \(1\). 2. Convert \(-\frac{7}{11}\): \(-7 \div 11 = -0.636363\ldots = -0.\overline{63}\). Its repeating block is \(63\), so the period length is \(2\). 3. Therefore, \(-\frac{7}{11}\) has the longer period.

\(\frac{5}{12} = 0.41\overline{6}\), period length \(1\) \(-\frac{7}{11} = -0.\overline{63}\), period length \(2\) \(-\frac{7}{11}\) has the longer period.

- Write several digits of each repeating decimal. - Look at the digit immediately to the right of the requested place. - Preserve trailing zeros that show the requested rounding place.

1. \(0.\overline{5}=0.555\ldots\). Thus \(0.\overline{5}\approx0.6\) to the nearest tenth and \(0.\overline{5}\approx0.56\) to the nearest hundredth. 2. \(1.274\approx1.3\) to the nearest tenth and \(1.274\approx1.27\) to the nearest hundredth. 3. \(-0.\overline{18}=-0.1818\ldots\). Thus \(-0.\overline{18}\approx-0.2\) to the nearest tenth and \(-0.\overline{18}\approx-0.18\) to the nearest hundredth. 4. \(3.009\approx3.0\) to the nearest tenth and \(3.009\approx3.01\) to the nearest hundredth.

a) Tenth: \(0.\overline{5}\approx0.6\); hundredth: \(0.\overline{5}\approx0.56\) b) Tenth: \(1.274\approx1.3\); hundredth: \(1.274\approx1.27\) c) Tenth: \(-0.\overline{18}\approx-0.2\); hundredth: \(-0.\overline{18}\approx-0.18\) d) Tenth: \(3.009\approx3.0\); hundredth: \(3.009\approx3.01\)

- Write both numbers with at least four decimal places. - Round each number to each requested place in order. - Add trailing zeros to the terminating decimal when needed to show the same precision.

1. Write the decimals as \(A=0.272727\ldots\) and \(B=0.270000\ldots\). 2. To the nearest tenth, \(A\approx0.3\) and \(B\approx0.3\). 3. To the nearest hundredth, \(A\approx0.27\), while \(B=0.27\) exactly. 4. To the nearest thousandth, \(A\approx0.273\), while \(B=0.270\) exactly. 5. Therefore, the rounded values first differ at the thousandths place.

They first differ when rounded to the nearest thousandth: \(A\approx0.273\) and \(B=0.270\).

- Start with exact conversions of the two fractions. - Closeness depends on positive distances, not the size of the values alone. - Put each distance into a comparable fraction or decimal form.

1. Convert: \(\frac{1}{6}=0.1\overline{6}\) and \(\frac{1}{8}=0.125\). 2. The distances from \(0.15=\frac{3}{20}\) are \(\frac{1}{6}-\frac{3}{20}=\frac{1}{60}\) and \(\frac{3}{20}-\frac{1}{8}=\frac{1}{40}\). 3. Since \(\frac{1}{60}<\frac{1}{40}\), \(\frac{1}{6}\) is closer.

\(\frac{1}{6}=0.1\overline{6}\), \(\frac{1}{8}=0.125\). Distances: \(\frac{1}{60}\) and \(\frac{1}{40}\), respectively. \(\frac{1}{6}\) is closer to \(0.15\).

- Write \(0.625\) as a fraction over \(1000\) and simplify. - Increase both the simplified numerator and denominator by \(1\). - Convert the new fraction by division before comparing.

1. Convert the decimal: \(0.625 = \frac{625}{1000} = \frac{5}{8}\). Thus, \(a = 5\) and \(b = 8\). 2. Form the new fraction: \(\frac{a+1}{b+1} = \frac{6}{9} = \frac{2}{3}\). 3. Convert the new fraction: \(\frac{2}{3} = 0.\overline{6}\). 4. Since \(0.\overline{6} > 0.625\), the new value is greater.

Original fraction: \(\frac{5}{8}\) New fraction: \(\frac{6}{9} = \frac{2}{3}\) The new value is greater because \(0.\overline{6} > 0.625\).

- Track the decimal and the remainder pattern for each fraction. - Distinguish a division that ends from one whose remainder cycle repeats. - For a repeating result, identify whether any digits occur before the cycle begins.

1. \(7\div9=0.777\ldots=0.\overline{7}\). The repeating block begins immediately. 2. \(13\div40=0.325\). The division ends, so the decimal terminates. 3. \(5\div6=0.8333\ldots=0.8\overline{3}\). The digit \(8\) comes before the repeating block.

a) \(0.\overline{7}\), repeating with no nonrepeating digits b) \(0.325\), terminating c) \(0.8\overline{3}\), repeating with a nonrepeating beginning

- Convert any fraction before making the comparison. - Expand a repeating decimal until the first decisive digit appears. - Add trailing zeros to a terminating decimal when helpful.

1. \(0.8100\ldots\) and \(\frac{9}{11}=0.8181\ldots\), so \(<\). 2. \(0.\overline{12}=0.121212\ldots\) and \(0.121=0.121000\ldots\), so \(>\). 3. \(2.6600\ldots\) and \(2.\overline{6}=2.6666\ldots\), so \(<\). 4. \(\frac{3}{8}=0.375\), so \(=\).

a) \(0.8100\ldots<0.8181\ldots=\frac{9}{11}\) b) \(0.121212\ldots>0.121000\ldots\) c) \(2.6600\ldots<2.6666\ldots\) d) \(\frac{3}{8}=0.375\)

- Convert \(\frac{7}{9}\) to a repeating decimal. - Compare enough places to distinguish the values. - Extend the decimals to more places to find a number between the last two.

1. Convert the fraction: \(\frac{7}{9} = 0.7777\ldots = 0.\overline{7}\). 2. Compare the decimals: \(0.7000\ldots < 0.7700\ldots < 0.7777\ldots < 0.7780\ldots\). 3. The two greatest values are \(\frac{7}{9}\) and \(0.778\). One decimal between them is \(0.7779\).

a) \(0.7 < 0.77 < \frac{7}{9} < 0.778\) b) Answers will vary. One possible answer is \(0.7779\).

- Rewrite repeating decimals as familiar fractions when possible. - When two products share the same positive factor, compare the other factors. - Expand \(\frac{1}{3}\) far enough to compare it with \(0.333\).

1. \(0.\overline{3}=\frac{1}{3}\) and \(0.\overline{6}=\frac{2}{3}\). Their sum is \(1\), so the expressions are equal. 2. \(\frac{1}{2}=0.5<0.55\). Multiplying both values by the same positive factor \(0.4\) preserves the inequality, so the left product is less. 3. \(\frac{1}{3}=0.3333\ldots>0.3330\ldots\).

a) \(=\), because \(\frac{1}{3}+\frac{2}{3}=1\) b) \(<\), because \(0.5<0.55\) and the common factor is positive c) \(>\), because \(0.3333\ldots>0.333\)

- Put all four values into comparable decimal form. - Pay attention to the difference between a terminating decimal and one that repeats forever. - Check for exact equality before ordering the distinct values.

1. Convert the fraction: \(\frac{2}{3}=0.6666\ldots=0.\overline{6}\). 2. Write the terminating decimals with additional zeros: \(0.6=0.6000\ldots\) and \(0.666=0.6660\ldots\). 3. The two repeating decimals are equal and are greater than \(0.666\), which is greater than \(0.6\).

\(\frac{2}{3}=0.\overline{6}>0.666>0.6\)

- Expand each repeating decimal by several digits. - Convert the fraction to a decimal. - Remember how order reverses when comparing negative values by absolute value.

1. Convert the fraction: \(-\frac{3}{4} = -0.75\). 2. Expand the repeating decimals: \(-0.7\overline{5} = -0.75555\ldots\) and \(-0.\overline{75} = -0.757575\ldots\). 3. For negative numbers, the number with the greater absolute value is less. 4. Therefore, \(-0.757575\ldots < -0.75555\ldots < -0.75\).

\(-0.\overline{75} < -0.7\overline{5} < -0.75 = -\frac{3}{4}\)

- Convert the fractions exactly before comparing. - For unequal repeating decimals, display enough digits to reach the first difference. - For part c), use the stated one-third fact to justify the comparison rather than looking for a first differing digit. - Be careful that greater magnitude means a smaller value for negative numbers.

1. \(\frac{5}{11}=0.4545\ldots\), so \(0.4500\ldots<0.4545\ldots\). 2. \(-\frac{5}{8}=-0.625\), so \(-0.620>-0.625\). 3. Since \(\frac{1}{3}=0.\overline{3}\), multiplying both equal quantities by \(3\) gives \(1=0.\overline{9}\). 4. \(1.2\overline{3}=1.2333\ldots\) and \(1.\overline{23}=1.2323\ldots\), so the first is greater.

a) \(0.4500\ldots<0.4545\ldots=\frac{5}{11}\) b) \(-0.620>-0.625=-\frac{5}{8}\) c) \(0.\overline{9}=1\), because \(\frac{1}{3}=0.\overline{3}\) implies \(3\cdot\frac{1}{3}=3\cdot0.\overline{3}\), so \(1=0.\overline{9}\) d) \(1.2333\ldots>1.2323\ldots\)

- Check each claimed equality by comparing enough decimal digits or using a familiar fraction pattern. - Distinguish terminating decimals from repeating decimals. - For a repeating result, identify the complete repeating block rather than stopping after the shown digits.

1. \(4\div9=0.444\ldots\), so the repeating bar is missing. The correct statement is \(\frac{4}{9}=0.\overline{4}\). 2. \(0.\overline{7}\) is repeating, not terminating. The correct fraction is \(\frac{7}{9}\). 3. \(1\div11=0.090909\ldots\), so the correct decimal is \(0.\overline{09}\).

a) The repeating bar is missing: \(\frac{4}{9}=0.\overline{4}\). b) A repeating decimal was treated as a terminating decimal: \(0.\overline{7}=\frac{7}{9}\). c) The repeating pattern was omitted: \(\frac{1}{11}=0.\overline{09}\).

- Check the digit immediately to the right of the rounding place. - Look carefully at which digits are covered by the repeating bar. - Decide whether each fraction can have a terminating decimal form.

1. \(\frac{2}{3}=0.666\ldots\). The thousandths digit is \(6\), so the value rounds to \(0.67\). 2. The bar is only over the digit \(5\), so \(0.4\overline{5}=0.4555\ldots\). 3. \(\frac{1}{3}=0.\overline{3}\), not \(0.3\). Therefore, \(\frac{2}{3}=0.\overline{6}\). 4. \(8\div99=0.080808\ldots=0.\overline{08}\).

a) The value was rounded down incorrectly: \(\frac{2}{3}\approx0.67\). b) The repeating block was misread: \(0.4\overline{5}=0.4555\ldots\). c) The repeating bar was omitted: \(\frac{2}{3}=0.\overline{6}\). d) Repetition was ignored: \(\frac{8}{99}=0.\overline{08}\).

- Factor \(140\). - Identify the prime factor that must disappear from the denominator. - Determine what divisibility condition this places on \(n\). - Count the multiples in the allowed range.

1. Factor the denominator: \(140 = 2^2 \cdot 5 \cdot 7\). 2. After simplification, a terminating decimal may have only prime factors of \(2\) and \(5\) in the denominator. 3. Therefore, the factor \(7\) must cancel, so \(n\) must be a multiple of \(7\). 4. The multiples of \(7\) from \(1\) through \(139\) are \(7, 14, 21, \ldots, 133\). 5. Since \(133 \div 7 = 19\), there are \(19\) possible values of \(n\).

\(n \in \{7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, 133\}\). There are \(19\) such fractions.

- Look for a convenient common form for the three fractions before comparing them. - Pay attention to the repeating block each fraction produces. - Compare the resulting decimal digits from left to right.

1. Divide \(2\) by \(9\): \(\frac{2}{9}=0.222\ldots=0.\overline{2}\). 2. Rewrite \(\frac{7}{33}\) with a denominator of \(99\): \(\frac{7}{33}=\frac{21}{99}=0.\overline{21}\). 3. A numerator over \(99\) gives the two-digit repeating block directly: \(\frac{25}{99}=0.\overline{25}\). 4. Compare the decimal forms: \(0.2121\ldots<0.2222\ldots<0.2525\ldots\). 5. Therefore, \(\frac{7}{33}<\frac{2}{9}<\frac{25}{99}\).

\(\frac{2}{9}=0.\overline{2}\), \(\frac{7}{33}=0.\overline{21}\), and \(\frac{25}{99}=0.\overline{25}\). Least to greatest: \(\frac{7}{33}<\frac{2}{9}<\frac{25}{99}\).

- Calculate the first few divisions and compare each numerator with its repeating block. - Compare the number of nines in the denominator with the number of digits in the repeating block. - Remember to include a leading zero in a two-digit block when needed.

1. The first decimals are \(\frac{1}{9} = 0.\overline{1}\), \(\frac{2}{9} = 0.\overline{2}\), and \(\frac{3}{9} = 0.\overline{3}\). 2. The numerator becomes the one-digit repeating block. 3. Therefore, \(\frac{7}{9} = 0.\overline{7}\) and \(\frac{8}{9} = 0.\overline{8}\). 4. With denominator \(99\), the repeating block has two digits. Thus, \(\frac{1}{99} = 0.\overline{01}\) and \(\frac{12}{99} = 0.\overline{12}\). 5. Long division confirms both predictions.

a) \(\frac{1}{9} = 0.\overline{1}\), \(\frac{2}{9} = 0.\overline{2}\), and \(\frac{3}{9} = 0.\overline{3}\). The numerator becomes the repeating digit. b) \(\frac{7}{9} = 0.\overline{7}\); \(\frac{8}{9} = 0.\overline{8}\) c) \(\frac{1}{99} = 0.\overline{01}\); \(\frac{12}{99} = 0.\overline{12}\)

- Record every remainder during the division of \(1\) by \(13\). - Compare that remainder list with the numerators of the other fractions. - If a numerator appears as a remainder, the same digit cycle restarts from that point.

1. Long division gives \(\frac{1}{13} = 0.\overline{076923}\). Its successive nonzero remainders are \(10, 9, 12, 3, 4, 1\). 2. Also, \(\frac{3}{13} = 0.\overline{230769}\). Its repeating block is a cyclic shift of \(076923\). 3. This happens because the remainder \(3\) occurs in the remainder cycle for \(1 \div 13\). Starting with \(3\) continues the same cycle from that point. 4. Long division gives \(\frac{2}{13} = 0.\overline{153846}\). 5. The remainder \(2\) does not appear in the remainder cycle for \(\frac{1}{13}\), so \(\frac{2}{13}\) begins a different cycle and does not use a cyclic shift of the same repeating block.

a) \(\frac{1}{13} = 0.\overline{076923}\) and \(\frac{3}{13} = 0.\overline{230769}\). Their repeating blocks are cyclic shifts. b) \(\frac{2}{13} = 0.\overline{153846}\). It is not in the same family because the starting remainder \(2\) does not occur in the remainder cycle for \(1 \div 13\).

- Compare the first few repeating blocks and look for a numerical relationship with their numerators. - Test that relationship on the new elevenths before writing the decimals. - For denominator \(9\), compare the structure of the repeating block with the numerator.

1. The first three values are \(\frac{1}{11}=0.\overline{09}\), \(\frac{2}{11}=0.\overline{18}\), and \(\frac{3}{11}=0.\overline{27}\). 2. Each two-digit repeating block is \(9\) times the numerator. 3. Since \(7\cdot9=63\), \(\frac{7}{11}=0.\overline{63}\). Since \(9\cdot9=81\), \(\frac{9}{11}=0.\overline{81}\). 4. For denominator \(9\), \(\frac{1}{9}=0.\overline{1}\) and \(\frac{2}{9}=0.\overline{2}\). The one-digit repeating block equals the numerator.

a) \(\frac{1}{11}=0.\overline{09}\), \(\frac{2}{11}=0.\overline{18}\), and \(\frac{3}{11}=0.\overline{27}\). The repeating block is \(9\) times the numerator. b) \(\frac{7}{11}=0.\overline{63}\); \(\frac{9}{11}=0.\overline{81}\) c) Yes. For denominator \(9\), the repeating digit equals the numerator.

- Put the fractions and repeating decimal into comparable decimal form. - Keep enough digits to identify the first place where close values differ. - Check the final ordering against those decimal expansions.

1. Convert the fractions: \(\frac{7}{20}=0.35\) and \(\frac{1}{3}=0.3333\ldots=0.\overline{3}\). 2. Expand the other repeating decimal: \(0.3\overline{5}=0.3555\ldots\). 3. Compare the decimals: \(0.3333\ldots<0.34<0.35<0.3555\ldots<0.36\).

\(\frac{1}{3}<0.34<\frac{7}{20}<0.3\overline{5}<0.36\)

- Convert the fraction to a decimal. - Expand the repeating decimal by several digits. - Remember that more negative numbers lie farther left on the number line.

1. Convert the fraction: \(b = -\frac{5}{4} = -1.25\). 2. Expand the repeating decimal: \(c = -1.2555\ldots\). 3. Write \(d = -1.20\) to compare place values. 4. For negative numbers, the value with the greater absolute value is less. Thus, \(-1.2555\ldots < -1.25 < -1.20\).

\(-1.2\overline{5} < -1.25 = -\frac{5}{4} < -1.2\)

- Add trailing zeros to terminating decimals when aligning place values. - Expand the repeating decimal until a decisive digit appears. - Convert the fraction in part c) before comparing.

1. \(0.\overline{7}=0.777\ldots\) and \(0.77=0.770\ldots\), so \(>\). 2. \(-1.23=-1.230\), and \(-1.230<-1.229\), so \(<\). 3. \(\frac{1}{20}=0.05\), so \(=\). 4. \(-0.001=-0.001\) and \(-0.01=-0.010\), so \(-0.001>-0.010\).

a) \(0.777\ldots>0.770\ldots\) b) \(-1.230<-1.229\) c) \(0.050=0.050=\frac{1}{20}\) d) \(-0.001>-0.010\)

- Convert nondecimal rational numbers before comparing. - Use trailing zeros to align place values. - For negative numbers, compare carefully after the decimal forms are aligned.

1. \(-\frac{5}{4}=-1.25<-1.20\). 2. \(0.\overline{15}=0.151515\ldots>0.151000\ldots\). 3. \(-3.45=-3.450>-3.455\). 4. \(2\frac{2}{25}=2.08\), so the values are equal.

a) \(-\frac{5}{4}=-1.25<-1.20\) b) \(0.151515\ldots>0.151000\ldots\) c) \(-3.450>-3.455\) d) \(2\frac{2}{25}=2.08\), so \(=\)

- Convert every number to decimal form. - Align the decimal points and add trailing zeros when helpful. - Check whether any two values are equal.

1. Convert the fraction: \(-\frac{1}{3} = -0.3333\ldots = -0.\overline{3}\). 2. Write the terminating decimals with additional zeros: \(-0.3 = -0.3000\ldots\), \(-0.33 = -0.3300\ldots\), and \(-0.34 = -0.3400\ldots\). 3. For negative numbers, the value with the greater absolute value is less. 4. Therefore, \(-0.34 < -0.\overline{3} = -\frac{1}{3} < -0.33 < -0.3\).

\(-0.34 < -0.\overline{3} = -\frac{1}{3} < -0.33 < -0.3\)

- Try rewriting the decimal and the percent as fractions first. - Find a common denominator for the fractions in the expression. - Think about whether every rational number has a terminating decimal representation.

1. Rewrite the decimal and percent as fractions: \(0.4 = \frac{2}{5}\) and \(15\% = \frac{3}{20}\). 2. The expression becomes \(\frac{2}{5} + \frac{1}{3} - \frac{3}{20}\). 3. Use a common denominator of \(60\): \(\frac{24}{60} + \frac{20}{60} - \frac{9}{60} = \frac{35}{60} = \frac{7}{12}\). 4. The fraction \(\frac{1}{3}\) has the repeating decimal representation \(0.\overline{3}\), so it cannot be represented exactly by a terminating decimal. Any terminating decimal replacement would be an approximation.

The result is \(\frac{7}{12}\). Using only terminating decimals cannot give an exact calculation because \(\frac{1}{3}\) has a repeating decimal representation.

- Look for how the denominator structure relates to the length and position of the repeating block. - Compare the effect of an extra factor of \(10\) in a denominator. - Check each decimal by multiplying or converting back to the original fraction.

1. \(4\div9=0.444\ldots=0.\overline{4}\). 2. \(4\div99=0.040404\ldots=0.\overline{04}\). 3. \(\frac{4}{90}=\frac{4}{9}\div10\). Dividing by \(10\) shifts the digits one place to the right, so the result is \(0.0\overline{4}\). 4. \(41\div99=0.414141\ldots=0.\overline{41}\).

a) \(0.\overline{4}\) b) \(0.\overline{04}\) c) \(0.0\overline{4}\) d) \(0.\overline{41}\)

- Keep a separate remainder list during \(1\div11\). - A repeated remainder marks the start of the same quotient-digit cycle. - Parts b) and c) should be derived from part a), not recomputed independently.

1. In \(1\div11\), the remainders after successive quotient digits go \(10\to1\to10\). The quotient digits repeat as \(09\), so \(\frac{1}{11}=0.\overline{09}\). 2. Multiply the result by \(3\): \(\frac{3}{11}=0.\overline{27}\). 3. Since \(\frac{1}{110}=\frac{1}{11}\div10\), the decimal shifts one place right: \(0.0\overline{09}\).

a) Remainders \(10\to1\to10\); \(\frac{1}{11}=0.\overline{09}\) b) \(\frac{3}{11}=3\cdot0.\overline{09}=0.\overline{27}\) c) \(\frac{1}{110}=0.0\overline{09}\), because dividing \(\frac{1}{11}\) by \(10\) shifts the decimal expansion one place to the right.

- Look for a common factor of the numerator and denominator. - Remember to consider prime factors such as \(13\) and \(17\). - After simplifying, divide the numerator by the denominator.

1. For a), divide the numerator and denominator by \(15\): \(\frac{45}{60} = \frac{3}{4}\). Then \(3 \div 4 = 0.75\). 2. For b), divide the numerator and denominator by \(13\): \(\frac{26}{65} = \frac{2}{5}\). Then \(2 \div 5 = 0.4\). 3. For c), divide the numerator and denominator by \(12\): \(\frac{132}{48} = \frac{11}{4}\). Then \(11 \div 4 = 2.75\). 4. For d), divide the numerator and denominator by \(17\): \(\frac{51}{85} = \frac{3}{5}\). Then \(3 \div 5 = 0.6\).

a) \(\frac{3}{4} = 0.75\) b) \(\frac{2}{5} = 0.4\) c) \(\frac{11}{4} = 2.75\) d) \(\frac{3}{5} = 0.6\)

- Simplify each fraction by dividing the numerator and denominator by a common factor. - Convert each simplified fraction to a decimal. - Compare the decimals digit by digit.

1. The greatest common factor of \(18\) and \(24\) is \(6\): \(\frac{18}{24} = \frac{18 \div 6}{24 \div 6} = \frac{3}{4}\). 2. Convert the simplified fraction: \(3 \div 4 = 0.75\). 3. The greatest common factor of \(14\) and \(20\) is \(2\): \(\frac{14}{20} = \frac{14 \div 2}{20 \div 2} = \frac{7}{10}\). 4. Convert the simplified fraction: \(7 \div 10 = 0.7\). 5. Since \(0.75 > 0.7\), \(\frac{18}{24} > \frac{14}{20}\).

\(\frac{18}{24} = \frac{3}{4} = 0.75\) and \(\frac{14}{20} = \frac{7}{10} = 0.7\). Therefore, \(\frac{18}{24}\) is greater.

- First write each fraction in simplest form. - A fraction in simplest form has a terminating decimal when its denominator has no prime factors other than \(2\) and \(5\). - In long division, a repeated remainder produces a repeating decimal.

1. For \(\frac{36}{48}\), divide the numerator and denominator by \(12\): \(\frac{36}{48} = \frac{3}{4}\). The denominator \(4\) has only the prime factor \(2\), so the decimal terminates. The value is \(0.75\). 2. For \(\frac{36}{45}\), divide the numerator and denominator by \(9\): \(\frac{36}{45} = \frac{4}{5}\). The denominator \(5\) has only the prime factor \(5\), so the decimal terminates. The value is \(0.8\). 3. For \(\frac{36}{54}\), divide the numerator and denominator by \(18\): \(\frac{36}{54} = \frac{2}{3}\). The denominator has a prime factor other than \(2\) or \(5\), so the decimal repeats. The value is \(0.\overline{6}\).

\(\frac{36}{48} = \frac{3}{4} = 0.75\), terminating \(\frac{36}{45} = \frac{4}{5} = 0.8\), terminating \(\frac{36}{54} = \frac{2}{3} = 0.\overline{6}\), repeating

- Simplify the fraction before analyzing its denominator. - Identify the prime factors that remain after simplification. - Verify the classification by writing the decimal value.

1. Lucas is correct because the denominator must be examined after simplification. 2. \(\frac{18}{75}=\frac{6}{25}\). 3. Since \(25=5^2\), the decimal terminates; \(\frac{6}{25}=0.24\).

Lucas is correct. \(\frac{18}{75}=\frac{6}{25}=0.24\), so the decimal terminates because the simplified denominator is \(25=5^2\).

- Convert every value to decimal form. - Compare the negative values separately from the positive values. - Check whether any two forms represent the same number.

1. Convert the nondecimal forms: \(-\frac{3}{4} = -0.75\), \(75\% = 0.75\), and \(\frac{2}{3} = 0.6666\ldots\). 2. Order the negative values: \(-0.8 < -0.75\), and \(-\frac{3}{4} = -0.75\). 3. Order the positive values: \(0.6666\ldots < 0.7 < 0.75\). 4. Combine the two groups.

\(-\frac{3}{4}=-0.75\), \(75\%=0.75\), and \(\frac{2}{3}=0.\overline{6}\) Least to greatest: \(-0.8 < -\frac{3}{4} = -0.75 < \frac{2}{3} < 0.7 < 75\%\)

- Convert all three values to decimals to determine their order. - Write the percent as a fraction over \(100\) before subtracting. - Use a common denominator and simplify the result.

1. Convert the values for comparison: \(x = \frac{5}{6} = 0.8333\ldots\), \(y = 83\% = 0.83\), and \(z = 0.833\). 2. Therefore, \(83\% < 0.833 < \frac{5}{6}\). 3. The difference between the greatest and least values is \(\frac{5}{6} - \frac{83}{100}\). 4. Use denominator \(300\): \(\frac{250}{300} - \frac{249}{300} = \frac{1}{300}\).

Order: \(83\% < 0.833 < \frac{5}{6}\) Difference: \(\frac{1}{300}\)

- Separate negative, zero, and positive values. - Convert each fraction to a decimal. - Check whether any decimal and fraction are equal.

1. Convert the fractions: \(-\frac{2}{5} = -0.4\) and \(\frac{1}{4} = 0.25\). 2. Order the negative values: \(-0.45 < -0.4 = -\frac{2}{5}\). 3. Place zero between the negative and positive values. 4. Order the positive values: \(0.25 < 0.35\).

\(-\frac{2}{5}=-0.4\) and \(\frac{1}{4}=0.25\) Least to greatest: \(-0.45 < -0.4 = -\frac{2}{5} < 0 < \frac{1}{4} < 0.35\)

- Determine which fractions have terminating decimal representations. - A repeating decimal cannot be written exactly as a terminating decimal. - Convert the decimal in B to a fraction and use a common denominator.

1. In A, \(\frac{1}{4} = 0.25\), and in C, \(\frac{2}{5} = 0.4\). These are terminating decimals. 2. In B, both \(\frac{2}{3}\) and \(\frac{1}{6}\) have repeating decimal representations, so using only terminating decimals would require rounding. 3. Convert \(0.5 = \frac{1}{2}\). Then \(\frac{2}{3} + \frac{1}{2} + \frac{1}{6} = \frac{4}{6} + \frac{3}{6} + \frac{1}{6} = \frac{8}{6} = \frac{4}{3}\).

a) Expression B b) \(\frac{4}{3}\)

- Distinguish an exact repeating decimal from a finite approximation. - Look for exact fractions that combine before any decimal conversion. - Show the regrouped exact expression in part b).

1. The fractions have repeating decimal expansions, so replacing them with finite decimals loses information. Also, \(\frac{5}{9}\) rounded to the nearest hundredth would be \(0.56\), not \(0.55\). 2. Regroup exactly: \(\left(\frac{4}{9}+\frac{5}{9}\right)+0.3\). 3. The fractions sum to \(1\), so \(S=1.3\).

a) Omar replaced repeating decimals with finite approximations, so the sum cannot be exact; \(0.55\) is also not the nearest-hundredth rounding of \(\frac{5}{9}\). b) \(\left(\frac{4}{9}+\frac{5}{9}\right)+0.3=1+0.3=1.3\).

- Check the decimal value of the fractional part before doing the subtraction. - Separately verify the arithmetic in the subtraction Renata actually wrote. - After correcting both issues, check that the result is reasonable compared with subtracting a number between \(1\) and \(2\) from \(4.5\).

1. Renata converted \(\frac{1}{4}\) incorrectly. Since \(\frac{1}{4} = \frac{25}{100} = 0.25\), the mixed number is \(1.25\), not \(1.14\). 2. Renata also subtracted incorrectly. Even the incorrect expression would give \(4.50 - 1.14 = 3.36\), not \(3.46\). 3. Using the correct decimal, \(4.5 - 1.25 = 3.25\).

The first error is that \(\frac{1}{4} = 0.25\), not \(0.14\). The second error is that \(4.50 - 1.14 = 3.36\), not \(3.46\). The correct result is \(3.25\).

- Recall the decimal form of \(\frac{1}{3}\). - Add \(0.8\) and \(2.3\) by aligning place values.

1. \(\frac{1}{3}=0.333\ldots=0.\overline{3}\). Therefore, \(2\frac{1}{3}=2.\overline{3}\), while \(2.3\) is only an approximation. 2. The displayed decimals were added incorrectly: \(0.8+2.3=3.1\), not \(3.3\).

a) \(\frac{1}{3}\) has the repeating decimal form \(0.\overline{3}\), so \(2.3\) is not the exact value of \(2\frac{1}{3}\). b) \(0.8+2.3=3.1\), not \(3.3\).

- Examine the decimal representation of \(\frac{1}{6}\). - Convert \(0.\overline{7}\) to a fraction.

1. For a), write \(0.2 = \frac{1}{5}\). Then \(\frac{1}{6} + \frac{1}{5} = \frac{5}{30} + \frac{6}{30} = \frac{11}{30}\). The decimal representation of \(\frac{1}{6}\) repeats, so any terminating decimal used in its place would be rounded. 2. For b), write \(0.\overline{7} = \frac{7}{9}\). Then \(\frac{7}{9} + \frac{2}{9} = 1\).

a) \(\frac{11}{30}\); \(\frac{1}{6}\) does not have a terminating decimal representation. b) \(1\)

- To convert a decimal to a percent, multiply by \(100\). - A percent such as \(0.4\%\) is less than \(1\%\), so its decimal form must be less than \(0.01\). - Use division to convert a fraction to a decimal. - Consider which prime factors a reduced denominator may have for its decimal form to terminate.

1. Row 1: \(\frac{1}{400} = 1 \div 400 = 0.0025\). As a percent, \(0.0025 \cdot 100 = 0.25\%\). 2. Row 2: \(0.4\% = \frac{0.4}{100} = 0.004\). As a fraction, \(0.004 = \frac{4}{1000} = \frac{1}{250}\). 3. Row 3: \(0.125 = 12.5\%\). As a fraction, \(0.125 = \frac{125}{1000} = \frac{1}{8}\). 4. Row 4: \(\frac{5}{6} = 5 \div 6 = 0.8\overline{3}\). As a percent, this is \(83.\overline{3}\%\). 5. In lowest terms, the denominator \(6\) contains the prime factor \(3\). Because the denominator has a prime factor other than \(2\) or \(5\), its decimal form repeats instead of terminating.

\(\frac{1}{400} = 0.0025 = 0.25\%\) \(\frac{1}{250} = 0.004 = 0.4\%\) \(\frac{1}{8} = 0.125 = 12.5\%\) \(\frac{5}{6} = 0.8\overline{3} = 83.\overline{3}\%\) The decimal for \(\frac{5}{6}\) repeats because its reduced denominator contains the prime factor \(3\).

- Factor each denominator after simplifying the fraction. - Only factors of \(2\) and \(5\) allow a terminating decimal. - You may also check by dividing the numerator by the denominator.

1. A simplified fraction has a terminating decimal when its denominator has no prime factors other than \(2\) and \(5\). 2. For \(\frac{7}{20}\), \(20 = 2^2 \cdot 5\), so the decimal terminates: \(\frac{7}{20} = \frac{35}{100} = 0.35\). 3. For \(\frac{11}{40}\), \(40 = 2^3 \cdot 5\), so the decimal terminates: \(\frac{11}{40} = \frac{275}{1000} = 0.275\). 4. The denominators of \(\frac{5}{6}\) and \(\frac{2}{3}\) contain the prime factor \(3\) after simplification, so their decimals repeat.

\(\frac{7}{20} = 0.35\) and \(\frac{11}{40} = 0.275\) terminate. The fractions \(\frac{5}{6}\) and \(\frac{2}{3}\) repeat because their simplified denominators contain a factor of \(3\).

- Determine the value of one small interval. - Points left of \(0\) have negative values. - Each interval represents one fourth, which can also be written as \(0.25\). - Simplify each fraction.

1. There are \(4\) equal intervals from \(0\) to \(1\), so each interval represents \(0.25=\frac{1}{4}\). 2. Point \(A\) is \(-1.25=-1\frac{1}{4}=-\frac{5}{4}\). 3. Point \(B\) is \(-0.5=-\frac{1}{2}\). 4. Point \(C\) is \(0.75=\frac{3}{4}\). 5. Point \(D\) is \(1.5=1\frac{1}{2}=\frac{3}{2}\).

\(A=-1.25=-1\frac{1}{4}\), \(B=-0.5=-\frac{1}{2}\), \(C=0.75=\frac{3}{4}\), \(D=1.5=1\frac{1}{2}\)

- Determine the value of one small interval. - Write each decimal as a fraction with denominator \(10\), \(100\), or \(1000\). - Simplify each fraction.

1. From \(0\) to \(0.5\), there are \(4\) equal intervals, so each interval represents \(0.5\div4=0.125=\frac{1}{8}\). 2. Point \(A\) is \(-0.75=-\frac{3}{4}\). 3. Point \(B\) is \(-0.375=-\frac{3}{8}\). 4. Point \(C\) is \(0.125=\frac{1}{8}\). 5. Point \(D\) is \(0.5=\frac{1}{2}\). 6. Point \(E\) is \(0.875=\frac{7}{8}\).

\(A=-0.75=-\frac{3}{4}\), \(B=-0.375=-\frac{3}{8}\), \(C=0.125=\frac{1}{8}\), \(D=0.5=\frac{1}{2}\), \(E=0.875=\frac{7}{8}\)

- Determine the value of one interval from \(0\) to \(1\). - Count left for negative values and right for positive values. - Write tenths as fractions and simplify.

1. The interval from \(0\) to \(1\) is divided into \(5\) equal parts, so each interval represents \(0.2=\frac{1}{5}\). 2. Point \(P\) is \(-1.2=-\frac{6}{5}\). 3. Point \(Q\) is \(-0.8=-\frac{4}{5}\). 4. Point \(R\) is \(-0.4=-\frac{2}{5}\). 5. Point \(S\) is \(0.2=\frac{1}{5}\). 6. Point \(T\) is \(0.6=\frac{3}{5}\). 7. Point \(U\) is \(1.4=\frac{7}{5}\).

\(P=-1.2=-\frac{6}{5}\), \(Q=-0.8=-\frac{4}{5}\), \(R=-0.4=-\frac{2}{5}\), \(S=0.2=\frac{1}{5}\), \(T=0.6=\frac{3}{5}\), \(U=1.4=\frac{7}{5}\)

- Divide each numerator by its denominator. - Note whether the decimal terminates or repeats. - Order the three negative values before matching them to the number line. - Values farther left are smaller.

1. Convert the mixed number: \(-1\frac{1}{2} = -1.5\). 2. Convert the fractions: \(-\frac{5}{6} = -0.8333\ldots = -0.8\overline{3}\) and \(-\frac{1}{3} = -0.3333\ldots = -0.\overline{3}\). 3. On the number line, point \(A\) is at \(-1.5\), point \(B\) is at about \(-0.83\), and point \(C\) is at about \(-0.33\). 4. Therefore, \(A\) matches \(-1\frac{1}{2}\), \(B\) matches \(-\frac{5}{6}\), and \(C\) matches \(-\frac{1}{3}\).

\(-1\frac{1}{2} = -1.5\), point \(A\) \(-\frac{5}{6} = -0.8\overline{3}\), point \(B\) \(-\frac{1}{3} = -0.\overline{3}\), point \(C\)

- Keep quotient digits and remainders in separate running lists. - Compare what happens to the remainder sequences, not the denominator factorizations. - Base the classification in part c) on those recorded remainder events.

1. For \(\frac{9}{40}\), the quotient is \(0.225\) and the remainders are \(10,20,0\). 2. For \(\frac{7}{12}\), the quotient is \(0.58\overline{3}\) and the remainders are \(10,4,4,\ldots\). 3. A zero remainder ends the division, while a repeated nonzero remainder restarts the same quotient cycle.

a) \(\frac{9}{40}=0.225\); remainders \(10,20,0\) b) \(\frac{7}{12}=0.58\overline{3}\); remainders \(10,4,4,\ldots\) c) The first terminates because its remainder reaches \(0\); the second repeats because a nonzero remainder repeats.

- One counterexample is enough to disprove a universal claim. - Look for a numerator that cancels a prime factor in the denominator. - Try a small denominator such as \(3\), \(6\), or \(7\). - After simplifying, check which prime factors remain in the denominator.

1. If \(\frac{1}{n}\) repeats, then the prime factorization of \(n\) includes at least one prime other than \(2\) or \(5\). 2. However, a numerator \(k\) may share that factor with \(n\), allowing it to cancel when the fraction is simplified. 3. For example, \(\frac{1}{6} = 0.1\overline{6}\), so \(\frac{1}{6}\) repeats. 4. Choose \(k = 3\). Then \(\frac{3}{6} = \frac{1}{2} = 0.5\), which terminates. 5. This counterexample proves that Lucas's claim is false.

Lucas is not correct. For example, \(\frac{1}{6} = 0.1\overline{6}\) repeats, but \(\frac{3}{6} = \frac{1}{2} = 0.5\) terminates because the factor \(3\) cancels.

- Exactly four decimal places means the larger exponent of \(2\) and \(5\) is \(4\). - Consider separately the cases where the exponent of \(2\) is \(4\) and where the exponent of \(5\) is \(4\). - Keep only two-digit results.

1. Write \(n = 2^a \cdot 5^b\). Exactly four decimal places requires \(\max(a,b) = 4\). 2. If \(a = 4\), then \(n = 16 \cdot 5^b\). For \(b = 0\), \(n = 16\). For \(b = 1\), \(n = 80\). For \(b \ge 2\), \(n\) is at least \(400\). 3. If \(b = 4\), then \(n = 625 \cdot 2^a\), which is never two-digit. 4. Therefore, the only two-digit values are \(16\) and \(80\). 5. Check: \(\frac{1}{16} = 0.0625\) and \(\frac{1}{80} = 0.0125\), each with four decimal places.

\(16\) and \(80\)

- Track the remainder sequence while converting each fraction to a decimal. - A repeating block is determined by the cycle before a remainder repeats. - Compare the lengths of the two cycles after identifying them.

1. For \(A\), \(4\div11=0.3636\ldots=0.\overline{36}\). Its repeating block has \(2\) digits. 2. For \(B\), long division gives the digits \(3,8,4,6,1,5\) before the first remainder returns. Thus, \(\frac{5}{13}=0.\overline{384615}\). 3. The repeating block of \(B\) has \(6\) digits, so it is longer than the \(2\)-digit repeating block of \(A\).

a) \(A=0.\overline{36}\); \(B=0.\overline{384615}\) b) \(B\) has the longer repeating block. c) The repeating block of \(B\) has \(6\) digits.

- Examine how the location and length of a repeating block relate to the value represented. - For a shifted repeating decimal, compare the effect of dividing the original value by \(10\). - For \(0.\overline{9}\), look for an algebraic way to compare the repeating value with \(1\). - For the final statement, compare enough decimal digits to settle the inequality.

1. The first statement is false. A two-digit repeating block that begins immediately corresponds to a denominator of \(99\): \(0.\overline{12}=\frac{12}{99}=\frac{4}{33}\). 2. The second statement is true. Since \(\frac{5}{11}=0.\overline{45}\), dividing by \(10\) shifts every digit one place to the right: \(\frac{5}{110}=0.0\overline{45}\). 3. The third statement is false. Let \(x=0.\overline{9}\). Then \(10x=9.\overline{9}\), so \(9x=9\) and \(x=1\). Therefore, \(0.\overline{9}=1\). 4. The fourth statement is true. Since \(\frac{1}{6}=0.1666\ldots\), it is greater than \(0.1600\ldots\).

a) False. \(0.\overline{12}=\frac{12}{99}=\frac{4}{33}\). b) True. c) False. \(0.\overline{9}=1\). d) True. \(\frac{1}{6}=0.1\overline{6}>0.16\).

- Test whether a factor in the original denominator can disappear after simplification. - State the correct rule in terms of the simplified denominator. - Factor \(120\) and identify the factor that must be canceled.

1. A counterexample is \(\frac{3}{6}=\frac{1}{2}=0.5\). 2. A rational number has a terminating decimal exactly when its simplified denominator has no prime factors other than \(2\) and \(5\). 3. Since \(120=2^3\cdot3\cdot5\), the factor \(3\) must cancel, so \(k\) must be divisible by \(3\).

a) For example, \(\frac{3}{6}=0.5\). b) A fraction terminates exactly when its simplified denominator has only prime factors \(2\) and \(5\); otherwise its decimal repeats. c) \(k\) must be a multiple of \(3\).

- Treat the shown six-digit block as a repeating cycle and look for how related sevenths reuse it. - Use the beginning of each decimal only to locate the appropriate place in that cycle. - For the digit-pair observation, compare positions separated by three places.

1. The repeating blocks for fractions with denominator \(7\) are cyclic shifts of \(142857\). 2. Since \(2\div7\) begins with \(0.2\ldots\), start the cycle with \(2\): \(\frac{2}{7}=0.\overline{285714}\). 3. Since \(3\div7\) begins with \(0.4\ldots\), start the cycle with \(4\): \(\frac{3}{7}=0.\overline{428571}\). 4. Pairing opposite positions gives \(1+8=9\), \(4+5=9\), and \(2+7=9\). Every pair sums to \(9\).

a) \(\frac{2}{7}=0.\overline{285714}\) and \(\frac{3}{7}=0.\overline{428571}\) b) Each paired sum is \(9\): \(1+8=9\), \(4+5=9\), and \(2+7=9\).

- Compare the number of nines with the number of digits in each repeating block. - A zero added to the denominator multiplies it by \(10\). - Rewrite \(0.0\overline{07}\) as \(0.\overline{07} \div 10\).

1. For a), \(\frac{13}{99} = 0.\overline{13}\), with a two-digit repeating block, and \(\frac{123}{999} = 0.\overline{123}\), with a three-digit repeating block. These examples support the idea. 2. For b), \(\frac{13}{990} = \frac{13}{99} \div 10\). The digits shift one place to the right, giving \(0.0\overline{13}\). 3. For c), \(0.\overline{07} = \frac{7}{99}\). Dividing by \(10\) gives \(0.0\overline{07} = \frac{7}{990}\).

a) \(\frac{13}{99} = 0.\overline{13}\); \(\frac{123}{999} = 0.\overline{123}\) b) The repeating block shifts one place to the right: \(\frac{13}{990} = 0.0\overline{13}\). c) \(\frac{7}{990}\)

- Factor \(21\), \(125\), and each candidate \(n\). - Analyze the denominator after cancellation, not before it. - Record the surviving prime factor that makes any rejected case repeat.

1. Use \(21=3\cdot7\) and \(125=5^3\). 2. \(n=3\): the factor \(3\) cancels; denominator is \(5^3\), so it terminates. 3. \(n=4=2^2\): denominator contains only \(2\) and \(5\), so it terminates. 4. \(n=9=3^2\): only one factor \(3\) cancels, leaving a \(3\), so it repeats. 5. \(n=11\): factor \(11\) remains, so it repeats. 6. \(n=14=2\cdot7\): factor \(7\) cancels, leaving only \(2\) and \(5\), so it terminates.

\(n=3\): terminates; the \(3\) cancels and the denominator is a power of \(5\). \(n=4\): terminates; the denominator has only factors \(2\) and \(5\). \(n=9\): repeats; a factor \(3\) remains. \(n=11\): repeats; a factor \(11\) remains. \(n=14\): terminates; the \(7\) cancels and only factors \(2\) and \(5\) remain. Therefore, \(n=3,4,14\).