Free Math Worksheets

- Convert each fraction to a terminating decimal. - Predict whether the difference should be positive or negative. - Choose one representation for the two values in each expression.

1. For a), \(\frac{1}{2} = 0.5\), so \(0.25 + 0.5 = 0.75\). 2. For b), \(\frac{3}{5} = 0.6\), so \(0.4 - 0.6 = -0.2\). 3. For c), \(0.75 - 1 = -0.25\). 4. For d), \(\frac{9}{10} = 0.9\), so \(1.1 + 0.9 = 2\).

a) \(0.75\) b) \(-0.2\) c) \(-0.25\) d) \(2\)

- Convert only values whose equivalents you know exactly. - Look for pairs that combine to a whole number. - Rearranging addends does not change their sum.

1. Lucas’s method: \(\frac{3}{4} = 0.75\) and \(\frac{1}{8} = 0.125\). Then \(0.125 + 0.75 + 0.25 + 0.125 = 1.25\). 2. Mia’s method: \(0.125 = \frac{1}{8}\) and \(0.25 = \frac{1}{4}\). Then \(\frac{1}{8} + \frac{3}{4} + \frac{1}{4} + \frac{1}{8}\). 3. Regroup: \(\left(\frac{1}{8} + \frac{1}{8}\right) + \left(\frac{3}{4} + \frac{1}{4}\right) = \frac{1}{4} + 1 = 1.25\). 4. Both methods are efficient. Mia’s grouping immediately creates a whole number from the fourths.

Both methods give \(1.25\). Mia’s method is especially efficient because \(\frac{3}{4} + \frac{1}{4} = 1\).

- Consider whether each fraction has a terminating decimal representation. - Look for fractions that add to a whole number. - A common representation is not always necessary for every term at once.

1. Converting the sixths to decimals would create repeating decimals, so keep them as fractions. 2. Group the fractions: \(\frac{5}{6} + \frac{1}{6} = 1\). 3. Group the decimals: \(0.4 + 1.1 = 1.5\). 4. Add the partial sums: \(1 + 1.5 = 2.5\).

Combine the fractions and decimals separately. The value is \(2.5\).

- Both fractions have terminating decimal representations. - Convert to one representation before combining terms. - Look for decimal pairs that are easy to add or subtract.

1. Convert the fractions to decimals: \(\frac{2}{5} = 0.4\) and \(\frac{3}{4} = 0.75\). 2. Substitute: \(0.4 + 1.35 - 0.75 + 0.1\). 3. Group efficiently: \((0.4 + 0.1) + (1.35 - 0.75) = 0.5 + 0.6 = 1.1\).

\(1.1\)

- Convert \(\frac{5}{4}\) to a decimal for the first method. - Convert both decimals to fractions for the second method. - Look for reciprocal factors.

1. Decimal method: \(\frac{5}{4} = 1.25\), so \(0.8 \cdot 1.25 + 0.25 = 1 + 0.25 = 1.25\). 2. Fraction method: \(0.8 = \frac{4}{5}\) and \(0.25 = \frac{1}{4}\). Then \(\frac{4}{5} \cdot \frac{5}{4} + \frac{1}{4} = 1 + \frac{1}{4} = \frac{5}{4}\). 3. The fraction method is especially efficient because the reciprocal factors cancel immediately.

a) \(1.25\), or \(\frac{5}{4}\) b) The fraction method is more efficient because \(\frac{4}{5} \cdot \frac{5}{4} = 1\).

- Convert each fraction to a decimal. - You can divide the numerator by the denominator. - A decimal with three places can be written as a fraction with denominator \(1000\).

1. Convert \(\frac{625}{100}\): \(625 \div 100 = 6.25\), so it is not equal to \(0.625\). 2. Convert \(\frac{5}{8}\): \(\frac{5}{8} = \frac{625}{1000} = 0.625\). 3. Convert \(\frac{3}{4}\): \(\frac{3}{4} = \frac{75}{100} = 0.75\). 4. Therefore, \(\frac{5}{8}\) is equivalent to \(0.625\).

\(\frac{5}{8}\)

- 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) \(\frac{3}{8}\)

- Interpret the fraction bar as division. - If the decimal does not terminate, identify the digit that determines rounding to the nearest hundredth. - Use \(\approx\) for rounded values in your work.

1. For a), \(3 \div 4 = 0.75\). 2. For b), \(5 \div 6 = 0.8333\ldots\), which rounds to \(0.83\). 3. For c), \(12 \div 5 = 2.4\). 4. For d), \(-7 \div 9 = -0.7777\ldots\), which rounds to \(-0.78\).

a) \(0.75\) b) \(0.83\) c) \(2.4\) d) \(-0.78\)

- 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: \(\frac{7}{8}\)

- Convert each fraction by division. - Keep enough decimal places to compare values accurately before rounding. - Align the decimal points when comparing.

1. Convert \(A\): \(5 \div 7 = 0.714285\ldots\), so \(A \approx 0.714\). 2. Convert \(C\): \(11 \div 15 = 0.7333\ldots\), so \(C \approx 0.733\). 3. Compare the values: \(0.710 < 0.714285\ldots < 0.715 < 0.7333\ldots\). 4. Therefore, \(B < A < D < C\).

\(0.71 < \frac{5}{7} < 0.715 < \frac{11}{15}\)

- 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 \(3\)

- Find the prime factorization of \(126\). - Identify the prime factors that prevent a terminating decimal. - The numerator must cancel all of those factors. - Look for the smallest product that does so.

1. Factor the denominator: \(126 = 2 \cdot 3^2 \cdot 7\). 2. After simplification, the denominator may contain only factors of \(2\) and \(5\) for the decimal to terminate. 3. The numerator must cancel the factors \(3^2\) and \(7\). Their product is \(3^2 \cdot 7 = 63\). 4. The least possible value is therefore \(x = 63\). 5. Check: \(\frac{63}{126} = \frac{1}{2} = 0.5\), which terminates.

\(x = 63\)

- 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 \(5\).

- Divide the numerator by the denominator. - Compare \(0.375\) with the neighboring tenths. - Think about how measuring cups are commonly labeled in US kitchens.

1. Convert the fraction: \(3 \div 8 = 0.375\), so the amount is \(0.375\) cup. 2. The value lies between \(0.3\) and \(0.4\). Since \(0.375 - 0.3 = 0.075\), it is three-fourths of the way from \(0.3\) to \(0.4\). 3. The fraction \(\frac{3}{8}\) is exact and matches common fractional cup measurements, while \(0.375\) cup is less familiar in everyday recipe directions.

a) \(0.375\) cup b) It is three-fourths of the way from \(0.3\) to \(0.4\), closer to \(0.4\). c) \(\frac{3}{8}\) cup is exact and aligns with customary fractional cup measurements.

- Add three thirds. - Add the three terminating decimals separately. - Compare an exact fraction with a rounded decimal approximation.

1. Add the fractions: \(\frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1\). 2. Add the decimals: \(0.33 + 0.33 + 0.33 = 0.99\). 3. The fraction \(\frac{1}{3}\) equals the repeating decimal \(0.\overline{3}\). The terminating decimal \(0.33\) is an approximation, so repeated use creates a difference of \(0.01\) from the whole.

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

- 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: \(2\frac{3}{5}<\frac{27}{10}<\frac{11}{4}<2.8\)

- Convert each fraction or mixed number to a decimal. - A number must be greater than \(5.4\) and less than \(5.7\). - Keep enough digits of a repeating decimal to make a reliable comparison.

1. Convert the numbers: \(\frac{11}{2}=5.5\), \(5\frac{2}{3}=5.666\ldots\), \(\frac{53}{10}=5.3\), and \(5.65=5.65\). 2. Check the interval \(5.4<x<5.7\). The values \(5.5\), \(5.666\ldots\), and \(5.65\) satisfy both inequalities. 3. The value \(5.3\) is less than \(5.4\), so it is outside the interval.

\(\frac{11}{2}\), \(5\frac{2}{3}\), and \(5.65\)

- Convert a fraction by dividing the numerator by the denominator. - Convert a percent by dividing by \(100\). - Compare decimals from left to right by place value.

1. Convert each value: \(A = \frac{5}{8} = 0.625\), \(B = 63\% = 0.63\), \(C = 0.622\), and \(D = \frac{2}{3} = 0.6666\ldots \approx 0.667\). 2. Compare the exact values: \(0.622 < 0.625 < 0.63 < 0.6666\ldots\). 3. Therefore, \(C < A < B < D\).

a) \(A = 0.625\); \(B = 0.63\); \(C = 0.622\); \(D \approx 0.667\) b) \(C < A < B < D\)

- 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.

1. For a), \(19 \div 12 = 1.5833\ldots\). The thousandths digit is \(3\), so the value rounds to \(1.58\). 2. For b), \(7 \div 11 = 0.636363\ldots\). The thousandths digit is \(6\), so the value rounds to \(0.64\).

a) \(1.58\); thousandths digit: \(3\) b) \(0.64\); thousandths digit: \(6\)

- Divide a percent by \(100\) to write it as a decimal. - Convert fractions by division. - Check the thousandths digit when rounding to the nearest hundredth.

1. For a), \(112.4\% = 1.124\), which rounds to \(1.12\). 2. For b), \(5 \div 13 = 0.384615\ldots\), which rounds to \(0.38\). 3. For c), \(25 \div 6 = 4.1666\ldots\), which rounds to \(4.17\).

a) \(1.12\) b) \(0.38\) c) \(4.17\)

- 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.

\(\frac{3}{8} < \frac{2}{5} < 0.405 < 0.45 < \frac{1}{2}\)

- Interpret a fraction bar as division. - Determine whether the decimal form of \(\frac{4}{3}\) terminates or repeats. - Check which numbers are exactly equal before ordering them.

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 \(\frac{4}{3} > 1.333\).

- Divide the numerator by the denominator. - Identify the hundredths immediately below and above the decimal value. - Write trailing zeros when comparing.

1. Convert the fraction: \(5 \div 6 = 0.8333\ldots = 0.8\overline{3}\). 2. The decimal is greater than \(0.83 = 0.8300\ldots\). 3. It is less than the next hundredth, \(0.84 = 0.8400\ldots\). 4. Therefore, \(0.83 < \frac{5}{6} < 0.84\).

\(\frac{5}{6}\) lies between \(0.83\) and \(0.84\).

- 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. - Write each decimal to enough places to see the first difference. - Compare the digits from left to right.

1. Divide to convert the fraction: \(5 \div 11 = 0.\overline{45} = 0.454545\ldots\). 2. Write the terminating decimals with additional zeros: \(0.45 = 0.450000\ldots\) and \(0.454 = 0.454000\ldots\). 3. Comparing digits from left to right gives \(0.450000\ldots < 0.454000\ldots < 0.454545\ldots\).

\(0.45 < 0.454 < \frac{5}{11}\)

- Convert each fraction to a decimal. - Find the absolute difference between each decimal and \(0.41\). - Keep enough decimal places to compare the distances accurately.

1. Convert the fractions to decimals: \(\frac{3}{8}=0.375\), \(\frac{2}{5}=0.4\), and \(\frac{4}{9}=0.444\ldots\). 2. Find the distances from \(0.41\): \(|0.41-0.375|=0.035\) \(|0.41-0.4|=0.01\) \(\left|0.41-0.444\ldots\right|=\frac{31}{900}\approx0.0344\). 3. Since \(0.01\) is the smallest distance, \(\frac{2}{5}\) is closest to \(0.41\).

\(\frac{2}{5}\)

- 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. For a), \(0.\overline{5}=0.555\ldots\). It rounds to \(0.6\) to the nearest tenth and \(0.56\) to the nearest hundredth. 2. For b), \(1.274\) rounds to \(1.3\) to the nearest tenth and \(1.27\) to the nearest hundredth. 3. For c), \(-0.\overline{18}=-0.1818\ldots\). It rounds to \(-0.2\) to the nearest tenth and \(-0.18\) to the nearest hundredth. 4. For d), \(3.009\) rounds to \(3.0\) to the nearest tenth and \(3.01\) to the nearest hundredth.

a) Tenth: \(0.6\); hundredth: \(0.56\) b) Tenth: \(1.3\); hundredth: \(1.27\) c) Tenth: \(-0.2\); hundredth: \(-0.18\) d) Tenth: \(3.0\); hundredth: \(3.01\)

- 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, both numbers round to \(0.3\). 3. To the nearest hundredth, both numbers round to \(0.27\). 4. To the nearest thousandth, \(A\) rounds to \(0.273\), while \(B\) is written as \(0.270\). 5. Therefore, the rounded values first differ at the thousandths place.

They first differ when rounded to the nearest thousandth: \(0.273\) and \(0.270\).

- Divide each numerator by its denominator. - A remainder of zero produces a terminating decimal. - If a remainder repeats, identify where the repeating block begins.

1. For a), \(7 \div 9 = 0.777\ldots = 0.\overline{7}\). The repeating block begins immediately. 2. For b), \(13 \div 40 = 0.325\). The division ends, so the decimal terminates. 3. For c), \(5 \div 6 = 0.8333\ldots = 0.8\overline{3}\). 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

- Expand repeating decimals by several digits. - Convert fractions by dividing the numerator by the denominator. - Add trailing zeros when comparing terminating decimals.

1. For a), \(\frac{9}{11} = 0.818181\ldots\), so \(0.81 < \frac{9}{11}\). 2. For b), \(0.\overline{12} = 0.121212\ldots\), which is greater than \(0.121000\ldots\). 3. For c), \(2.\overline{6} = 2.6666\ldots\), which is greater than \(2.6600\ldots\). 4. For d), \(\frac{3}{8} = 0.375\), so the values are equal.

a) \(0.81 < \frac{9}{11}\) b) \(0.\overline{12} > 0.121\) c) \(2.66 < 2.\overline{6}\) d) \(\frac{3}{8} = 0.375\)

- Convert \(\frac{2}{3}\) by division. - Recall what a bar over a decimal digit means. - Compare enough decimal places to find the first difference.

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 to decimals before comparing. - Be careful with the order of negative numbers. - Recall the value of \(0.\overline{9}\). - Expand both repeating decimals in d) until you see the first different digit.

1. For a), \(\frac{5}{11} = 0.4545\ldots\), so \(0.45 < \frac{5}{11}\). 2. For b), \(-\frac{5}{8} = -0.625\). Since \(-0.620 > -0.625\), \(-0.62 > -\frac{5}{8}\). 3. For c), \(0.\overline{9} = 1\). 4. For d), \(1.2\overline{3} = 1.2333\ldots\) and \(1.\overline{23} = 1.2323\ldots\). At the third decimal place, \(3 > 2\), so \(1.2\overline{3} > 1.\overline{23}\).

a) \(<\) b) \(>\) c) \(=\) d) \(>\)

- Divide the numerator by the denominator to check each claim. - Distinguish terminating decimals from repeating decimals. - For a one-digit repeating block that begins immediately, consider a denominator of \(9\).

1. For a), \(4 \div 9 = 0.444\ldots\), so the repeating bar is missing. The correct statement is \(\frac{4}{9} = 0.\overline{4}\). 2. For b), \(0.\overline{7}\) is repeating, not terminating. The correct fraction is \(\frac{7}{9}\). 3. For c), \(1 \div 11 = 0.090909\ldots\), 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. For a), \(\frac{2}{3} = 0.666\ldots\). The thousandths digit is \(6\), so the value rounds to \(0.67\). 2. For b), the bar is only over the digit \(5\), so \(0.4\overline{5} = 0.4555\ldots\). 3. For c), \(\frac{1}{3} = 0.\overline{3}\), not \(0.3\). Therefore, \(\frac{2}{3} = 0.\overline{6}\). 4. For d), \(8 \div 99 = 0.080808\ldots = 0.\overline{08}\).

a) The value was rounded down incorrectly: \(\frac{2}{3} \approx 0.67\). 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.

- Try rewriting each fraction with a denominator made of nines. - You can also divide each numerator by its denominator. - Compare the 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 each repeating block with a multiple of \(9\). - Multiply the numerator by \(9\) to predict the two-digit block. - Check whether denominator \(9\) produces a one-digit or two-digit repeating block.

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 \(7 \cdot 9 = 63\), \(\frac{7}{11} = 0.\overline{63}\). Since \(9 \cdot 9 = 81\), \(\frac{9}{11} = 0.\overline{81}\). 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.

- Convert each fraction by dividing the numerator by the denominator. - Expand repeating decimals by several digits. - Compare the digits from left to right.

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, \(-1.2555\ldots < -1.25 < -1.20\).

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

- Factor \(60\). - Identify the prime factor that prevents termination. - Determine what must be true of \(k\) for that factor to cancel. - List all matching values in the given range.

1. Factor the denominator: \(60 = 2^2 \cdot 3 \cdot 5\). 2. A fraction in simplest form has a terminating decimal only when its denominator has no prime factors other than \(2\) and \(5\). 3. Therefore, the factor \(3\) must cancel, so \(k\) must be divisible by \(3\). 4. The possible values are \(3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57\).

\(k \in \{3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57\}\)

- Expand a repeating decimal by several digits. - Write terminating decimals with the same number of places when helpful. - Convert the fraction to a decimal. - For negative values, consider which number lies farther right on the number line.

1. For a), \(0.\overline{7} = 0.777\ldots\), so \(0.\overline{7} > 0.77\). 2. For b), write \(-1.23 = -1.230\). Since \(-1.230 < -1.229\), \(-1.23 < -1.229\). 3. For c), \(\frac{1}{20} = \frac{5}{100} = 0.05\), so the values are equal. 4. For d), \(-0.001\) is closer to zero than \(-0.01\), so \(-0.001 > -0.01\).

a) \(0.\overline{7} > 0.77\) b) \(-1.23 < -1.229\) c) \(0.05 = \frac{1}{20}\) d) \(-0.001 > -0.01\)

- Convert fractions or mixed numbers to decimals first. - Expand repeating decimals by several digits. - For negative values, the number with greater absolute value is less.

1. For a), \(-\frac{5}{4} = -1.25\), and \(-1.25 < -1.2\). 2. For b), \(0.\overline{15} = 0.151515\ldots\), which is greater than \(0.151000\ldots\). 3. For c), \(-3.45 = -3.450\), which is greater than \(-3.455\). 4. For d), \(2\frac{2}{25} = 2 + \frac{8}{100} = 2.08\), so the values are equal.

a) \(-\frac{5}{4} < -1.2\) b) \(0.\overline{15} > 0.151\) c) \(-3.45 > -3.455\) d) \(2.08 = 2\frac{2}{25}\)

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

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

- A denominator of \(9\) or \(99\) often reveals the repeating block directly. - Dividing by \(10\) shifts every decimal digit one place to the right. - Compare the number of nines in the denominator with the length of the repeating block.

1. For a), \(4 \div 9 = 0.444\ldots = 0.\overline{4}\). 2. For b), \(4 \div 99 = 0.040404\ldots = 0.\overline{04}\). 3. For c), \(\frac{4}{90} = \frac{4}{9} \div 10\). Dividing by \(10\) shifts the digits one place to the right, so the result is \(0.0\overline{4}\). 4. For d), \(41 \div 99 = 0.414141\ldots = 0.\overline{41}\).

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

- Track the remainders in \(1 \div 11\). - Multiply the decimal from part a) by \(3\). - Think about what dividing a decimal by \(10\) does to its digits.

1. Long division gives \(1 \div 11 = 0.0909\ldots = 0.\overline{09}\). 2. Multiply the result by \(3\): \(3 \cdot 0.\overline{09} = 0.\overline{27}\). Thus, \(\frac{3}{11} = 0.\overline{27}\). 3. Since \(\frac{1}{110} = \frac{1}{11} \div 10\), every digit shifts one place to the right. Therefore, \(\frac{1}{110} = 0.0\overline{09}\).

a) \(0.\overline{09}\) b) \(0.\overline{27}\) c) \(0.0\overline{09}\), because dividing by \(10\) shifts the digits one place to the right.

- 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.

\(\frac{13}{20} < 70\% < 0.72 < \frac{3}{4}\)

- Convert all four numbers to decimals. - To convert a percent to a decimal, divide by \(100\). - Notice that \(0.375 = \frac{375}{1000}\), not \(\frac{375}{100}\).

1. Convert each value to a decimal. The first value is already \(0.375\). 2. Divide to convert the fraction: \(\frac{3}{8} = 3 \div 8 = 0.375\). 3. Divide the percent by \(100\): \(37.5\% = 0.375\). 4. Convert the last fraction: \(\frac{375}{100} = 3.75\). 5. The first three values are equal. The value \(3.75\) is ten times \(0.375\), so \(\frac{375}{100}\) does not belong.

\(\frac{375}{100}\) does not belong because it equals \(3.75\), while each of the other three values equals \(0.375\).

- 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.

- Simplify the fraction before analyzing its denominator. - Identify the prime factors that remain in the simplified denominator. - A terminating decimal has only factors of \(2\) and \(5\) in its denominator.

1. Student B is correct because the denominator must be examined after the fraction is simplified. 2. Simplify: \(\frac{18}{75} = \frac{6}{25}\). 3. The simplified denominator is \(25 = 5^2\), which has no prime factors other than \(5\). 4. Therefore, the decimal terminates. In fact, \(\frac{6}{25} = 0.24\).

Student B 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.

\(-0.8 < -\frac{3}{4} = -0.75 < \frac{2}{3} < 0.7 < 75\%\)

- 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: \(0.25 < 0.35\).

\(-0.45 < -0.4 = -\frac{2}{5} < 0 < \frac{1}{4} < 0.35\)

- Examine the whole expression before choosing a representation. - Avoid converting a fraction to a repeating decimal when fraction addition is simpler. - Look for values that combine to a whole number or a simple decimal.

1. For a), use \(\frac{1}{4} = 0.25\): \((0.25 + 0.5) - 1 = 0.75 - 1 = -0.25\). 2. For b), use \(1\frac{1}{2} = 1.5\): \(2.75 - (1.5 + 0.25) = 2.75 - 1.75 = 1\). 3. For c), add the fractions first: \(\frac{1}{3} + \frac{1}{6} = \frac{1}{2} = 0.5\). Then \(0.5 + 0.2 = 0.7\).

a) \(-0.25\) b) \(1\) c) \(0.7\)

- Consider the decimal representation of \(\frac{1}{3}\). - Convert the terminating decimal to a fraction. - Rewrite the fractions with a common denominator.

1. Fractions are more efficient because \(\frac{1}{3}\) has a repeating decimal representation. 2. Convert \(0.25 = \frac{1}{4}\) and \(1\frac{1}{3} = \frac{4}{3}\). 3. Use denominator \(12\): \(\frac{16}{12} + \frac{3}{12} - \frac{7}{12} = \frac{12}{12} = 1\).

\(1\). Fractions are preferable because \(\frac{1}{3}\) is a repeating decimal.

- 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) \(\frac{4}{3}\)

- Rounding during a calculation can change the final value. - Look for fractions with the same denominator that combine to a whole. - You may reorder and regroup addends.

1. The fractions have repeating decimal representations. The student truncated both decimals, losing information. Also, \(\frac{5}{9}\) rounded to the nearest hundredth is \(0.56\), not \(0.55\); either rounded calculation would still be approximate. 2. Use the commutative and associative properties to add the fractions first: \(\left(\frac{4}{9} + \frac{5}{9}\right) + 0.3\). 3. Evaluate exactly: \(\frac{9}{9} + 0.3 = 1.3\).

a) The student used truncated repeating decimals, so the sum is not exact. b) \(\frac{4}{9} + \frac{5}{9} = 1\), so \(S = 1.3\).

- 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) \(1\)

- Apply division before subtraction. - Convert every value to a form that supports an exact calculation. - Examine the decimal representation of \(\frac{7}{9}\).

1. Convert \(0.25\) to \(\frac{1}{4}\). 2. Perform the division first: \(\frac{1}{4} \div \frac{3}{4} = \frac{1}{4} \cdot \frac{4}{3} = \frac{1}{3}\). 3. Subtract: \(\frac{7}{9} - \frac{1}{3} = \frac{7}{9} - \frac{3}{9} = \frac{4}{9}\). 4. Both \(\frac{7}{9} = 0.\overline{7}\) and \(\frac{1}{3} = 0.\overline{3}\) have repeating decimal representations. Replacing them with terminating decimals would require rounding and would not preserve the exact value.

a) \(\frac{4}{9}\) b) \(\frac{7}{9}\) and \(\frac{1}{3}\) have repeating decimal representations, so terminating decimal approximations would not produce an exact calculation.

- Recall the fraction equivalent of \(0.375\). - Complete both methods before comparing them. - Compare \(0.375\) with \(0.75\).

1. Method 1: Write \(\frac{3}{4} = 0.75\) and \(\frac{1}{2} = 0.5\). Then \(0.375 \div 0.75 = 0.5\), and \(0.5 + 0.5 = 1\). 2. Method 2: Write \(0.375 = \frac{3}{8}\). Then \(\frac{3}{8} \div \frac{3}{4} = \frac{3}{8} \cdot \frac{4}{3} = \frac{1}{2}\), and \(\frac{1}{2} + \frac{1}{2} = 1\). 3. Both methods are valid. The fraction method may feel easier because common factors simplify immediately; the decimal method may feel easier if you recognize that \(0.375\) is half of \(0.75\).

Both methods give \(1\). A valid explanation should compare the simplification in the fraction method with the recognizable decimal relationship in the decimal method.

- 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 each pair to the same form before comparing. - Pay attention to whether a fraction has a terminating or repeating decimal form.

1. For a), \(\frac{2}{5} = \frac{4}{10} = 0.4\), so the values are equal. 2. For b), \(\frac{1}{3} = 0.3333\ldots\), while \(0.33 = 0.3300\ldots\). Therefore, \(0.33 < \frac{1}{3}\). 3. For c), \(\frac{17}{20} = \frac{85}{100} = 0.85\), so the values are equal.

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

- Try rewriting a denominator as a power of ten. - When that is not possible, divide the numerator by the denominator. - Watch for repeated remainders during division.

1. For a), \(\frac{9}{20} = \frac{45}{100} = 0.45\). 2. For b), \(5 \div 6 = 0.8333\ldots = 0.8\overline{3}\). 3. For c), \(4 \div 11 = 0.3636\ldots = 0.\overline{36}\). 4. For d), \(\frac{17}{50} = \frac{34}{100} = 0.34\).

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

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

- Find a factor that changes each denominator to \(10\), \(100\), or \(1000\). - Multiply the numerator by the same factor. - Use the number of zeros in the denominator to place the decimal point.

1. For a), multiply the numerator and denominator by \(2\): \(\frac{3}{5} = \frac{6}{10} = 0.6\). 2. For b), multiply the numerator and denominator by \(5\): \(\frac{7}{20} = \frac{35}{100} = 0.35\). 3. For c), multiply the numerator and denominator by \(25\): \(\frac{9}{40} = \frac{225}{1000} = 0.225\). 4. For d), multiply the numerator and denominator by \(4\): \(\frac{11}{250} = \frac{44}{1000} = 0.044\).

a) \(0.6\) b) \(0.35\) c) \(0.225\) d) \(0.044\)

- Convert all numbers to the same form before comparing them. - Add trailing zeros so the decimals have the same number of places.

1. Convert the fractions to decimals: \(\frac{3}{8} = 3 \div 8 = 0.375\) and \(\frac{2}{5} = \frac{4}{10} = 0.4\). 2. Compare all four decimals: \(0.375 < 0.380 < 0.400 < 0.420\). 3. Replace the converted decimals with the original fractions: \(\frac{3}{8} < 0.38 < \frac{2}{5} < 0.42\).

\(\frac{3}{8} < 0.38 < \frac{2}{5} < 0.42\)

- 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) \(\square = 100\)

- Recognize common fraction-decimal equivalents. - In part b), look for terms that become additive inverses after conversion. - Convert the decimal to a fraction before dividing in part c).

1. For a), write \(0.25=\frac{1}{4}\). Then \(\frac{1}{4}\cdot\frac{4}{7}=\frac{1}{7}=0.\overline{142857}\). 2. For b), write \(0.125=\frac{1}{8}\). Then \(\frac{3}{4}+\frac{1}{8}-\frac{1}{8}=\frac{3}{4}=0.75\). 3. For c), write \(0.6=\frac{3}{5}\). Then \(\frac{3}{5}\div\frac{2}{3}=\frac{3}{5}\cdot\frac{3}{2}=\frac{9}{10}=0.9\).

a) \(\frac{1}{7}=0.\overline{142857}\) b) \(\frac{3}{4}=0.75\) c) \(\frac{9}{10}=0.9\)

- Divide each numerator by its denominator. - A remainder of zero means the decimal terminates. - A repeated remainder means the decimal repeats.

1. For a), \(3 \div 16 = 0.1875\). The remainder becomes zero, so the decimal terminates. 2. For b), \(5 \div 6 = 0.8333\ldots = 0.8\overline{3}\). A remainder repeats, so the decimal repeats. 3. For c), \(4 \div 11 = 0.3636\ldots = 0.\overline{36}\). The remainders cycle, so the decimal repeats.

a) \(0.1875\), terminating b) \(0.8\overline{3}\), repeating c) \(0.\overline{36}\), repeating

- Convert both numbers in each pair to decimal form. - Compare decimal digits by place value. - Try scaling a denominator to \(10\) or \(100\).

1. For a), \(\frac{7}{10} = 0.7\). Since \(0.7 < 0.75\), the correct symbol is \(<\). 2. For b), \(\frac{1}{4} = 0.25\). Since \(0.4 > 0.25\), the correct symbol is \(>\). 3. For c), \(\frac{12}{50} = \frac{24}{100} = 0.24\). The values are equal. 4. For d), \(\frac{6}{5} = \frac{12}{10} = 1.2\). Since \(1.3 > 1.2\), the correct symbol is \(>\).

a) \(<\) b) \(>\) c) \(=\) d) \(>\)

- 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 \(C\)

- Convert \(\frac{4}{5}\) and \(\frac{3}{4}\) to decimals. - Multiply the decimal values to find the fraction of the wall’s area. - Write the decimal product as a fraction with a power of \(10\) in the denominator, then simplify.

1. Convert the fractions: \(\frac{4}{5}=0.8\) and \(\frac{3}{4}=0.75\). 2. Multiply: \(0.8\cdot0.75=0.6\). 3. Convert back to a fraction: \(0.6=\frac{6}{10}=\frac{3}{5}\).

\(\frac{3}{5}\)

- 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 \(80\)

- Continue long division until a remainder repeats. - The digits produced before the remainder cycle restarts form the repeating block. - Count the digits under each repeating bar.

1. For \(A\), \(4 \div 11 = 0.3636\ldots = 0.\overline{36}\). 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.

- For a repeating block with two digits, think about a denominator made of two nines. - Consider what dividing a decimal by \(10\) does to its digits. - Try assigning a variable to \(0.\overline{9}\) and multiplying by \(10\). - Write \(\frac{1}{6}\) as a decimal and compare digits from left to right.

1. Statement a) 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. Statement b) 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. Statement c) is false. Let \(x = 0.\overline{9}\). Then \(10x = 9.\overline{9}\), so \(9x = 9\) and \(x = 1\). Therefore, \(0.\overline{9} = 1\). 4. Statement d) 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. \(\frac{1}{6} = 0.1\overline{6} > 0.16\).

- Find a fraction whose factor of \(3\) cancels. - The denominator must be examined after the fraction is simplified. - Factor \(120\) and identify the factor that must cancel.

1. For a), \(\frac{3}{6} = \frac{1}{2} = 0.5\). The original denominator is divisible by \(3\), but the decimal terminates. 2. For b), a fraction has a repeating decimal exactly when its denominator, after the fraction is simplified, has a prime factor other than \(2\) or \(5\). 3. For c), \(120 = 2^3 \cdot 3 \cdot 5\). The factor \(3\) must cancel for the simplified denominator to contain only \(2\) and \(5\). 4. Therefore, \(k\) must be divisible by \(3\). Every multiple of \(3\) from \(3\) through \(117\) works.

a) One counterexample is \(\frac{3}{6} = 0.5\). b) A fraction repeats exactly when its simplified denominator has a prime factor other than \(2\) or \(5\). c) \(k\) must be a multiple of \(3\).

- Arrange the digits of \(142857\) in a cycle. - Use the first decimal digit of each quotient to choose where the cycle begins. - Split the six-digit block into two halves and compare corresponding positions.

1. The repeating blocks for fractions with denominator \(7\) are cyclic shifts of \(142857\). 2. Since \(2 \div 7\) begins with \(0.2\ldots\), start the cycle with \(2\): \(\frac{2}{7} = 0.\overline{285714}\). 3. Since \(3 \div 7\) 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) \(\frac{7}{990}\)

- 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

- Factor the numerator and denominator. - A simplified denominator may contain only \(2\) and \(5\) for the decimal to terminate. - Check whether each extra prime factor in \(n\) cancels completely with \(21\).

1. Factor the fixed parts: \(21 = 3 \cdot 7\) and \(125 = 5^3\). 2. After simplification, the denominator may contain only prime factors of \(2\) and \(5\). 3. If \(n = 3\), the factor \(3\) cancels, leaving denominator \(5^3\), so the decimal terminates. 4. If \(n = 4 = 2^2\), the denominator contains only factors of \(2\) and \(5\), so the decimal terminates. 5. If \(n = 9 = 3^2\), only one factor of \(3\) cancels, so a factor of \(3\) remains. The decimal repeats. 6. If \(n = 11\), the factor \(11\) remains in the denominator. The decimal repeats. 7. If \(n = 14 = 2 \cdot 7\), the factor \(7\) cancels, leaving only factors of \(2\) and \(5\). The decimal terminates.

\(n = 3\), \(4\), or \(14\)

- 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 \(3\).