- Which digits are covered by the bar? - What happens once you reach the end of that group of digits? - Number the decimal places as you write them.

1. The digit \(2\) appears once before the repeating block. 2. The block \(35\) then repeats, giving \(0.2353535\ldots\). 3. The first six decimal digits are \(235353\).

First six digits: \(235353\). Repeating block: \(35\). Nonrepeating first digit: \(2\).

- Compare the repeating digits in the two decimals. - Can a simple multiplication turn the known decimal into the target decimal? - Apply the same multiplication to the known fraction.

1. Each digit in \(0.\overline{4}\) is four times the corresponding digit in \(0.\overline{1}\). 2. Multiply the known equality by \(4\): \(0.\overline{4}=4\cdot\frac{1}{9}=\frac{4}{9}\).

Multiply both sides of \(0.\overline{1}=\frac{1}{9}\) by \(4\): \(0.\overline{4}=\frac{4}{9}\).

- Write the first few digits of \(100x\), keeping the repeating part visible. - Which digits stand to the right of the decimal point in each expression? - Compare the decimal parts before you subtract the whole-number parts.

1. Multiplying by \(100\) moves the two-digit repeating block one full block to the left: \(100x=24.242424\ldots\). 2. The parts after the decimal point are now identical in \(100x\) and \(x\), so they cancel when subtracted. 3. Therefore \(100x-x=24\).

Both decimals end in \(.242424\ldots\), so those digits cancel and \(100x-x=24\).

- Name the repeating decimal with a variable. - Shift the decimal point by one place so the repeating tails align. - Subtract the aligned equations.

1. Let \(x=0.\overline{7}\). 2. Then \(10x=7.\overline{7}\). 3. Subtract: \(10x-x=7\), so \(9x=7\). 4. Therefore, \(x=\frac{7}{9}\).

\(0.\overline{7}=\frac{7}{9}\)

- Count the digits in the repeating block. - Use a power of ten that shifts one complete block. - Check whether the resulting numerator and denominator share a factor.

1. Let \(x=0.\overline{34}\). 2. Since the period has \(2\) digits, \(100x=34.\overline{34}\). 3. Subtracting gives \(100x-x=34\), so \(99x=34\). 4. Thus, \(x=\frac{34}{99}\), which is already in lowest terms.

\(0.\overline{34}=\frac{34}{99}\)

- Treat every position in the repeating block as significant, including the zero. - Shift by the full length of the period. - Align the repeating tails before subtracting.

1. Let \(x=0.\overline{052}\). 2. The period has \(3\) digits, so \(1000x=52.\overline{052}\). 3. Subtracting gives \(999x=52\). 4. Therefore, \(x=\frac{52}{999}\), which is in lowest terms.

\(0.\overline{052}=\frac{52}{999}\)

- The repeating tail must line up after shifting. - Subtract the original decimal from the shifted one. - Reduce the resulting improper fraction.

1. Let \(x=2.\overline{6}\). 2. Then \(10x=26.\overline{6}\). 3. Subtract: \(10x-x=24\), so \(9x=24\). 4. Therefore, \(x=\frac{24}{9}=\frac{8}{3}\).

\(2.\overline{6}=\frac{8}{3}\)

- Temporarily work with the positive magnitude. - Shift by the full repeating block. - Apply the sign after reducing the fraction.

1. First convert the positive decimal. Let \(y=0.\overline{18}\). 2. Then \(100y=18.\overline{18}\), so \(99y=18\). 3. Thus, \(y=\frac{18}{99}=\frac{2}{11}\). 4. Restore the negative sign: \(-0.\overline{18}=-\frac{2}{11}\).

\(-0.\overline{18}=-\frac{2}{11}\)

- Count all digits in the repeated block. - Choose the shift that returns the same block to the same positions. - Verify that subtraction would cancel every repeating digit.

1. The period has \(3\) digits, so the correct equation is C: \(1000x=305.\overline{305}\). 2. Subtracting the original equation gives \(999x=305\). 3. Therefore, \(x=\frac{305}{999}\), which is in lowest terms.

Equation C; \(0.\overline{305}=\frac{305}{999}\)

- Convert the decimal before comparing numerators. - Reduce the fraction to a denominator of \(11\). - Match the reduced form to the given expression.

1. Let \(x=0.\overline{54}\). 2. Then \(100x=54.\overline{54}\), so \(99x=54\). 3. Thus, \(x=\frac{54}{99}=\frac{6}{11}\). 4. Therefore, \(n=6\).

\(n=6\)

- Apply the same variable method used for any one-digit period. - Subtract the two aligned equations. - Interpret the solved value as an exact equality.

1. Let \(x=0.\overline{9}\). 2. Then \(10x=9.\overline{9}\). 3. Subtracting gives \(9x=9\). 4. Thus, \(x=1\), so \(0.\overline{9}=1\) exactly.

\(0.\overline{9}=1\)

- Let a letter represent the repeating decimal. - What happens if you multiply it by two powers of ten that shift the repeating part to matching positions? - Subtract the shifted versions to remove the repeating tail.

1. Let \(x=0.09999\ldots\). Then \(10x=0.9999\ldots\) and \(100x=9.9999\ldots\). 2. Subtract: \(100x-10x=9\), so \(90x=9\) and \(x=\frac{1}{10}\). 3. Thus \(0.0\overline{9}=\frac{1}{10}=0.1\) exactly.

Let \(x=0.0\overline{9}\). Then \(10x=0.\overline{9}\) and \(100x=9.\overline{9}\). Subtracting gives \(90x=9\), so \(x=\frac{1}{10}=0.1\) exactly.

- Convert the repeating representation before comparing. - Reduce the resulting fraction completely. - Different representations can name the same number.

1. Let \(x=0.\overline{81}\). 2. Then \(100x=81.\overline{81}\), so \(99x=81\). 3. Thus, \(x=\frac{81}{99}=\frac{9}{11}\). 4. The two expressions represent the same rational number.

The claim is false. \(0.\overline{81}=\frac{9}{11}\).

- Convert both repeating decimals separately. - Notice whether the resulting denominators match. - Perform the exact subtraction after conversion.

1. Convert \(0.\overline{7}=\frac{7}{9}\). 2. Convert \(0.\overline{2}=\frac{2}{9}\). 3. Subtract: \(\frac{7}{9}-\frac{2}{9}=\frac{5}{9}\).

\(\frac{5}{9}\)

- Form the unreduced block-over-nines fraction. - Look for a common factor shared with \(99\). - Explain the denominator change as part of reducing an equivalent fraction.

1. A two-digit period gives \(0.\overline{48}=\frac{48}{99}\). 2. The numerator and denominator share a factor of \(3\). 3. Dividing both by \(3\) gives \(\frac{48}{99}=\frac{16}{33}\). 4. No further common factor remains.

\(0.\overline{48}=\frac{16}{33}\). The denominator becomes \(33\) because both \(48\) and \(99\) are divided by \(3\).

- Look at which digit is covered by the bar. - Write both shifted decimals, including their whole-number parts, before subtracting. - After you find \(x\), check whether numerator and denominator share a factor.

1. Multiply: \(10x=2.777\ldots\) and \(100x=27.777\ldots\). 2. Subtract: \(100x-10x=25\), so \(90x=25\). 3. Therefore \(x=\frac{25}{90}=\frac{5}{18}\).

\(10x=2.777\ldots\) and \(100x=27.777\ldots\), so \(100x-10x=25\). Thus \(90x=25\) and \(0.2\overline{7}=\frac{25}{90}=\frac{5}{18}\).

- After the first quotient digit, note what remains when you subtract the largest possible multiple of \(11\). - What does it mean if the same remainder occurs a second time? - For a two-digit repeating block, what familiar denominator could help with a fraction check?

1. In long division, \(70\div11\) gives digit \(6\) with remainder \(4\), since \(70=6\cdot11+4\). 2. Next, \(40\div11\) gives digit \(3\) with remainder \(7\), since \(40=3\cdot11+7\). The remainder \(7\) starts the same steps again, so \(\frac{7}{11}=0.636363\ldots=0.\overline{63}\). 3. Check: \(0.\overline{63}=\frac{63}{99}=\frac{7}{11}\).

\(\frac{7}{11}=0.\overline{63}\). The division remainders repeat: \(4, 7, 4, 7,\ldots\). Check: \(0.\overline{63}=\frac{63}{99}=\frac{7}{11}\).

- Identify whether each decimal eventually repeats a fixed block. - For the rational decimal, look for a place-value shift that aligns the repeating tails. - A rule-generated decimal is rational only if it terminates or eventually becomes periodic.

1. The decimal for \(x\) repeats: \(x = 1.\overline{23}\). Therefore, \(x\) is rational. 2. The decimal for \(y\) is nonterminating and nonrepeating because the number of consecutive \(2\)s keeps increasing. Therefore, \(y\) is irrational. 3. Let \(x = 1.\overline{23}\). Then \(100x = 123.\overline{23}\). Subtracting gives \(99x = 122\), so \(x = \frac{122}{99}\). The numerator and denominator have no common factor, so the fraction is in lowest terms.

1. \(x\) is rational, and \(y\) is irrational. 2. The decimal for \(x\) repeats, while the decimal for \(y\) is nonterminating and nonrepeating. 3. \(x = \frac{122}{99}\)

- Separate the nonrepeating part from the repeating part. - Create two shifted equations whose repeating tails match. - Reduce after solving for the variable.

1. Let \(x=0.1\overline{6}\). 2. Shift past the nonrepeating digit: \(10x=1.\overline{6}\). 3. Shift one more place: \(100x=16.\overline{6}\). 4. Subtract: \(100x-10x=15\), so \(90x=15\). 5. Thus, \(x=\frac{15}{90}=\frac{1}{6}\).

\(0.1\overline{6}=\frac{1}{6}\)

- Count the nonrepeating digits before the bar. - Build two equations that differ by one complete period. - Simplify the final fraction carefully.

1. Let \(x=0.42\overline{3}\). 2. Shift past the two nonrepeating digits: \(100x=42.\overline{3}\). 3. Shift one more digit: \(1000x=423.\overline{3}\). 4. Subtract: \(1000x-100x=381\), so \(900x=381\). 5. Therefore, \(x=\frac{381}{900}=\frac{127}{300}\).

\(0.42\overline{3}=\frac{127}{300}\)

- Account separately for the prefix and the two-digit period. - Align the repeating blocks before subtracting. - Expect an improper fraction because the decimal is greater than \(1\).

1. Let \(x=1.2\overline{45}\). 2. Shift past the one nonrepeating digit: \(10x=12.\overline{45}\). 3. Shift two additional places: \(1000x=1245.\overline{45}\). 4. Subtract: \(1000x-10x=1233\), so \(990x=1233\). 5. Thus, \(x=\frac{1233}{990}=\frac{137}{110}\).

\(1.2\overline{45}=\frac{137}{110}\)

- Compare the length of the repeating block with the number of places shifted. - The infinite tails must match exactly before subtraction. - Reduce the fraction after correcting the setup.

1. Multiplying by \(10\) shifts only one digit, so the repeating tails do not align. 2. The period has \(2\) digits, so use \(100x=27.\overline{27}\). 3. Subtracting \(x=0.\overline{27}\) gives \(99x=27\). 4. Therefore, \(x=\frac{27}{99}=\frac{3}{11}\).

Riley shifted by too few places. The correct conversion is \(0.\overline{27}=\frac{3}{11}\).

- Write both shifted decimals explicitly. - Check that the repeating parts line up. - Subtract the finite parts after the tails cancel.

1. \(10x=4.\overline{2}\) and \(100x=42.\overline{2}\). 2. Subtracting gives \(100x-10x=42.\overline{2}-4.\overline{2}=38\). 3. Thus, \(90x=38\). 4. Therefore, \(x=\frac{38}{90}=\frac{19}{45}\).

The box contains \(38\), and \(0.4\overline{2}=\frac{19}{45}\).

- Relate the number of repeating digits to a power of \(10\). - Notice what coefficient remains after subtracting the original variable. - Check for a common factor only after forming the fraction.

1. Let \(x=0.\overline{124}\). 2. A \(3\)-digit shift gives \(1000x=124.\overline{124}\). 3. Subtracting gives \((1000-1)x=124\), so \(999x=124\). 4. Therefore, \(x=\frac{124}{999}\). 5. Since \(124\) and \(999\) share no common factor greater than \(1\), the fraction is already in lowest terms.

The denominator \(999\) comes from \(1000-1\). Thus, \(0.\overline{124}=\frac{124}{999}\).

- Align the repeating nines using two shifts. - Reduce the resulting fraction fully. - Check whether the reduced denominator produces a terminating decimal.

1. Let \(x=1.24\overline{9}\). 2. \(100x=124.\overline{9}\) and \(1000x=1249.\overline{9}\). 3. Subtracting gives \(900x=1125\). 4. Thus, \(x=\frac{1125}{900}=\frac{5}{4}\). 5. Therefore, the equivalent terminating decimal is \(1.25\).

\(1.24\overline{9}=\frac{5}{4}=1.25\)

- Replace the repeating decimal with an exact fraction first. - Isolate the unknown after the conversion. - Use a common denominator for the remaining subtraction.

1. Convert \(0.\overline{4}=\frac{4}{9}\). 2. Subtract to isolate \(x\): \(x=\frac{5}{6}-\frac{4}{9}\). 3. Using denominator \(18\), \(x=\frac{15}{18}-\frac{8}{18}=\frac{7}{18}\).

\(x=\frac{7}{18}\)

- Express both repeating decimals as fractions. - Find their sum before applying the meaning of average. - Keep the calculation exact.

1. Convert \(0.\overline{3}=\frac{1}{3}\) and \(0.\overline{6}=\frac{2}{3}\). 2. Their sum is \(\frac{1}{3}+\frac{2}{3}=1\). 3. Divide by \(2\): the average is \(\frac{1}{2}\).

\(\frac{1}{2}\)

- Convert the repeating side length before using the shape information. - Use the number of equal sides in the frame. - State both requested exact values with units.

1. Convert \(0.\overline{27}=\frac{27}{99}=\frac{3}{11}\). 2. An equilateral triangle has \(3\) equal sides. 3. The perimeter is \(3\cdot\frac{3}{11}=\frac{9}{11}\,\text{m}\).

Side length: \(\frac{3}{11}\,\text{m}\) Perimeter: \(\frac{9}{11}\,\text{m}\)

- Convert the repeating time to an exact fraction first. - Use the standard relationship between hours and minutes. - Check whether the converted number of minutes is reasonable for more than one hour.

1. Let \(x=1.\overline{6}\). Then \(10x-x=15\), so \(9x=15\). 2. Thus, \(x=\frac{15}{9}=\frac{5}{3}\) hours. 3. Multiply by \(60\) minutes per hour: \(\frac{5}{3}\cdot60=100\) minutes.

a) \(\frac{5}{3}\) hours b) \(100\) minutes

- Convert the repeating decimal before comparing all three forms. - Look for two representations of the same value. - Compare the remaining decimal by place value.

1. Convert \(0.\overline{45}=\frac{45}{99}=\frac{5}{11}\). 2. The first two numbers are equal. 3. \(\frac{5}{11}=0.4545\ldots<0.46\). 4. Therefore, the two equal values come before \(0.46\).

\(0.\overline{45}=\frac{5}{11}<0.46\)

- Use two shifts because the decimal has a prefix and a period. - Align the repeating tails exactly. - Compare only after obtaining the reduced fraction.

1. Let \(x=0.1\overline{8}\). 2. Then \(10x=1.\overline{8}\) and \(100x=18.\overline{8}\). 3. Subtracting gives \(90x=17\). 4. Therefore, \(x=\frac{17}{90}\), so the values are equal.

\(0.1\overline{8}=\frac{17}{90}\)

- Count the prefix digits and the repeating digits separately. - Build the smaller shift first, then move one full period farther. - Reduce the improper fraction at the end.

1. Let \(x=2.03\overline{12}\). 2. Shift past the \(2\) nonrepeating digits: \(100x=203.\overline{12}\). 3. Shift one full period farther: \(10000x=20312.\overline{12}\). 4. Subtracting gives \(9900x=20109\). 5. Thus, \(x=\frac{20109}{9900}=\frac{6703}{3300}\).

\(2.03\overline{12}=\frac{6703}{3300}\)

- Shift past both zeros before aligning a complete period. - Compare the two powers of \(10\) used in the subtraction. - Reduce only after forming the exact fraction.

1. Let \(x=0.00\overline{72}\). 2. Then \(100x=0.\overline{72}\) and \(10000x=72.\overline{72}\). 3. Subtracting gives \(9900x=72\). 4. Thus, \(x=\frac{72}{9900}=\frac{2}{275}\). 5. The two prefix places contribute two factors of \(10\) to the unreduced denominator.

\(0.00\overline{72}=\frac{2}{275}\)

- Write the shifted decimals and compare the positions of the repeating digits. - The smaller multiplier must move past the entire nonrepeating prefix. - Subtract only equations with identical infinite tails.

1. The decimal has two nonrepeating digits, so \(x\) and \(1000x\) do not have aligned repeating tails. 2. Use \(100x=58.\overline{3}\) and \(1000x=583.\overline{3}\). 3. Subtracting gives \(900x=525\). 4. Therefore, \(x=\frac{525}{900}=\frac{7}{12}\).

The repeating tails align for \(1000x-100x\), not \(1000x-x\). Thus, \(0.58\overline{3}=\frac{7}{12}\).

- First convert the repeating decimal to a reduced fraction. - How could you use the chance of winning one play to predict a count across many plays? - Think about whether a probability determines an exact outcome in a small set of plays.

1. Convert \(0.\overline{36}=\frac{36}{99}=\frac{4}{11}\). 2. The expected number of wins is \(55\cdot\frac{4}{11}=20\). 3. An expected count is an average predicted by the probability model, not a guarantee for any particular \(55\) plays.

Probability: \(\frac{4}{11}\) Expected wins: \(20\) No. The actual number of wins may be different.

- Rewrite the repeating scale factor exactly. - Preserve the units attached to the factor. - Look for cancellation when applying the factor to the actual length.

1. Convert \(0.\overline{24}=\frac{24}{99}=\frac{8}{33}\). 2. Multiply by \(33\,\text{ft}\): \(33\cdot\frac{8}{33}=8\). 3. The model length is \(8\,\text{in}\).

a) \(\frac{8}{33}\,\text{in}\) per foot b) \(8\,\text{in}\)

- Convert the repeating decimal to lowest terms. - Focus on what must happen to the denominator for the product to be a whole number. - Use the smallest multiplier that provides all needed factors.

1. Convert \(0.\overline{15}=\frac{15}{99}=\frac{5}{33}\). 2. The product is \(n\cdot\frac{5}{33}\). 3. Since \(5\) and \(33\) are relatively prime, the smallest positive \(n\) that cancels the denominator is \(33\). 4. The product is \(33\cdot\frac{5}{33}=5\).

\(n=33\), and the product is \(5\).

- A three-digit repeating block first gives a denominator of \(999\). - Test each option by reducing its block-over-nines fraction. - Stop once the reduced fraction matches exactly.

1. Convert choice B: \(0.\overline{378}=\frac{378}{999}\). 2. Divide numerator and denominator by \(27\): \(\frac{378}{999}=\frac{14}{37}\). 3. Therefore, choice B is the required decimal.

B. \(0.\overline{378}\)

- Notice the effect of placing one zero before the repeating block. - Convert each decimal with its correct place value. - Compare the fractions using a quotient.

1. \(A=\frac{27}{99}=\frac{3}{11}\). 2. The extra nonrepeating zero shifts the value one place right, so \(B=\frac{27}{990}=\frac{3}{110}\). 3. \(\frac{A}{B}=\frac{3/11}{3/110}=10\). 4. Therefore, \(A\) is \(10\) times as large as \(B\).

\(A=\frac{3}{11}\), \(B=\frac{3}{110}\), and \(A\) is \(10\) times as large as \(B\).

- Keep the leading zero in the first repeating block. - Convert both two-digit periods using the same initial denominator. - Look at how the numerators combine.

1. \(0.\overline{09}=\frac{9}{99}=\frac{1}{11}\). 2. \(0.\overline{90}=\frac{90}{99}=\frac{10}{11}\). 3. Their sum is \(\frac{1}{11}+\frac{10}{11}=1\). 4. The repeating blocks add to \(99\), matching the unreduced denominator \(99\).

\(0.\overline{09}+0.\overline{90}=1\)

- Treat the zero as part of the two-digit period. - Subtract the whole original decimal after shifting one complete block. - Check whether the final numerator shares a factor with \(99\).

1. Let \(x=4.\overline{05}\). 2. Then \(100x=405.\overline{05}\). 3. Subtracting the original equation gives \(99x=401\). 4. Therefore, \(x=\frac{401}{99}\), which is in lowest terms.

\(4.\overline{05}=\frac{401}{99}\)

- Identify exactly which digits lie under the bar. - Compare the claimed fraction with the decimal it would actually produce. - Use separate shifts for the nonrepeating prefix and the repeating period.

1. The expression \(\frac{12}{99}\) represents \(0.\overline{12}\), where both digits repeat. 2. In \(0.1\overline{2}\), only the digit \(2\) repeats. 3. Let \(x=0.1\overline{2}\). Then \(10x=1.\overline{2}\) and \(100x=12.\overline{2}\). 4. Subtracting gives \(90x=11\), so \(x=\frac{11}{90}\).

Pim repeated the prefix digit incorrectly. The correct value is \(0.1\overline{2}=\frac{11}{90}\).

- Count the two prefix digits before the repeating digit. - Use two equations whose repeating tails are identical. - Compare the exact fraction with the choices.

1. Let \(x=0.07\overline{4}\). 2. Then \(100x=7.\overline{4}\) and \(1000x=74.\overline{4}\). 3. Subtracting gives \(900x=67\). 4. Thus, \(x=\frac{67}{900}\), which is choice B.

B. \(\frac{67}{900}\)

- Convert each repeating decimal independently. - Reduce before finding a common denominator. - Keep the final subtraction exact.

1. Convert \(1.\overline{3}=\frac{4}{3}\). 2. Convert \(0.\overline{81}=\frac{81}{99}=\frac{9}{11}\). 3. Subtract using denominator \(33\): \(\frac{4}{3}-\frac{9}{11}=\frac{44}{33}-\frac{27}{33}=\frac{17}{33}\).

\(\frac{17}{33}\)

- Use the repeating structure of \(x\) to derive its fraction without approximating. - Compare \(x\) and \(y\) digit by digit until their first difference. - A terminating decimal can be a rational candidate if it lies strictly between the two numbers.

1. Since \(x = 0.121212\ldots\), \(x = \frac{12}{99} = \frac{4}{33}\). 2. The decimal for \(y\) contains blocks of \(1\)s and \(2\)s whose lengths keep increasing. Therefore, no fixed block repeats forever. The decimal is nonterminating and nonrepeating, so \(y\) is irrational. 3. Compare \(x = 0.121212\ldots\) with \(y = 0.121122\ldots\). At the fourth decimal place, \(x\) has \(2\) and \(y\) has \(1\), so \(y < x\). 4. One rational number between them is \(z = 0.1212\). Indeed, \(0.121122\ldots < 0.121200\ldots < 0.121212\ldots\).

a) \(x = \frac{4}{33}\) b) \(y\) is irrational because its increasing block lengths make its decimal expansion nonterminating and nonrepeating. c) One example is \(z = 0.1212\), since \(y < 0.1212 < x\).

- Keep the unknown digit symbolic while shifting the decimal. - Express both sides with the same denominator after conversion. - Compare the resulting numerators.

1. Let \(x=0.2\overline{d}\). 2. Then \(10x=2.\overline{d}\) and \(100x=20+d+0.\overline{d}\). 3. Subtracting gives \(90x=18+d\). 4. Since \(x=\frac{13}{45}=\frac{26}{90}\), \(18+d=26\). 5. Therefore, \(d=8\).

\(d=8\)

- Represent the two-digit block as one whole number. - Shift exactly one full period. - Keep the block symbolic until after the subtraction.

1. Let \(x=0.\overline{ab}\). 2. Then \(100x=N.\overline{ab}\). 3. Subtracting gives \(99x=N\), so \(x=\frac{N}{99}\). 4. For \(62\), \(N=62\), so \(0.\overline{62}=\frac{62}{99}\).

The rule is \(0.\overline{ab}=\frac{10a+b}{99}\). Therefore, \(0.\overline{62}=\frac{62}{99}\).

- In the first method, choose shifts that place identical repeating tails under each other. - In the second method, separate the whole, terminating, and repeating portions. - Compare the reduced results rather than their unreduced denominators.

1. a) \(1000x=3124.\overline{24}\) and \(10x=31.\overline{24}\), so \(990x=3093\). Thus \(x=\frac{3093}{990}=\frac{1031}{330}\). 2. b) Let \(y=0.0\overline{24}\). Then \(1000y-10y=24\), so \(y=\frac{24}{990}=\frac{4}{165}\). 3. \(3+\frac{1}{10}+\frac{4}{165}=\frac{990+33+8}{330}=\frac{1031}{330}\), matching part a).

a) \(\frac{1031}{330}\) b) The decomposition also gives \(\frac{1031}{330}\).