Free math worksheets

- Multiply the whole number by the fraction's numerator, then simplify. - Look for factors that can cancel before multiplying. - In part c), pay attention to the sign of the product. - In part d), check whether the fraction can be simplified before calculating.

1. For a), \(4 \cdot \frac{3}{8}=\frac{12}{8}=\frac{3}{2}\). 2. For b), \(5 \cdot \frac{7}{10}=\frac{35}{10}=\frac{7}{2}\). 3. For c), \(-6 \cdot \frac{1}{3}=-2\). 4. For d), \(11 \cdot \frac{2}{22}=1\).

a) \(\frac{3}{2}\) b) \(\frac{7}{2}\) c) \(-2\) d) \(1\)

- Separate the familiar whole-number fact from the decimal place-value decision. - State how many tenths, hundredths, or thousandths the decimal factor represents. - Determine the sign independently of decimal placement.

1. a) \(8\cdot4=32\). Since \(0.8\) is eight tenths, the product is \(3.2\). 2. b) \(5\cdot9=45\). The product is negative and \(0.5\) is five tenths, so the result is \(-4.5\). 3. c) \(12\cdot5=60\). Since \(1.2\) has one decimal place, the product is \(6\). 4. d) \(6\cdot7=42\). Since \(0.006\) is six thousandths, the product is \(0.042\).

a) \(8\cdot4=32\); \(0.8\cdot4=3.2\) b) \(5\cdot9=45\); sign negative; \(-0.5\cdot9=-4.5\) c) \(12\cdot5=60\); \(1.2\cdot5=6\) d) \(6\cdot7=42\); \(0.006\cdot7=0.042\)

- Determine the sign of each result before calculating. - What sign does a product or quotient have when the factors or terms have different signs? - What sign does the product of two negative numbers have?

1. For a), a negative number times a positive number is negative: \((-14)\cdot3=-42\). 2. For b), a positive number divided by a negative number is negative: \(35\div(-7)=-5\). 3. For c), the product of two negative numbers is positive: \((-4)\cdot(-12)=48\). 4. For d), a positive number divided by a negative number is negative: \(100\div(-5)=-20\).

a) \(-42\) b) \(-5\) c) \(48\) d) \(-20\)

- Identify the signed change that occurs once and the number of times it occurs. - Write one compact expression for those equal repeated changes. - Interpret the sign of the total in terms of the score.

1. Four equal changes of \(-3\) points are represented by \(4\cdot(-3)\). 2. \(4\cdot(-3)=-12\). 3. The negative sign means the penalties decrease the score by \(12\) points in total.

\(4\cdot(-3)=-12\). The team's score decreases by \(12\) points.

- Rewrite a factor or dividend using nearby multiples of \(10\) or \(100\). - The required response includes the decomposition, not only the result. - Keep the negative sign attached to the factor in part d).

1. a) \(15(100+1)=1500+15=1515\). 2. b) \(18(10-1)=180-18=162\). 3. c) \((300+15)\div3=100+5=105\). 4. d) \(-22(10+1)=-220-22=-242\).

a) \(15(100+1)=1515\) b) \(18(10-1)=162\) c) \((300+15)\div3=105\) d) \(-22(10+1)=-242\)

- Look for a nearby multiple of \(10\), \(100\), or a convenient divisible part. - Show the equivalent expression that supports the mental calculation. - Determine the sign before simplifying part b).

1. a) \(2.5(10+2)=25+5=30\). 2. b) \(14(-100-1)=-1400-14=-1414\). 3. c) \((700+28)\div7=100+4=104\).

a) \(2.5(10+2)=30\) b) \(14(-100-1)=-1414\) c) \((700+28)\div7=104\)

- Distribute the subtraction sign before regrouping. - Combine the decimal terms separately from the fractional terms. - Convert the resulting fraction to a decimal only at the final step.

1. Remove the parentheses: \(4.5 - 1.25 - \frac{1}{3} + \frac{1}{12}\). 2. Combine the decimals: \(4.5 - 1.25 = 3.25\). 3. Combine the fractions: \(-\frac{1}{3} + \frac{1}{12} = -\frac{4}{12} + \frac{1}{12} = -\frac{1}{4} = -0.25\). 4. Add: \(3.25 - 0.25 = 3\).

\(3\)

- Round to convenient whole numbers for the estimate. - Convert the mixed numbers to terminating decimals for the exact calculation. - Follow the grouping symbols before adding the final term.

1. Estimate by rounding to convenient whole numbers: \(19 - (5 + 3) + 1 = 12\). 2. Convert the mixed numbers: \(4\frac{1}{2} = 4.5\) and \(1\frac{1}{4} = 1.25\). 3. Evaluate the parentheses: \(4.5 + 3.2 = 7.7\). 4. Calculate exactly: \(18.75 - 7.7 + 1.25 = 11.05 + 1.25 = 12.3\).

Estimate: about \(12\) Exact value: \(12.3\)

- Factor the whole number and denominator mentally before multiplying numerators. - Your response must show a reduced factor pair before the final multiplication. - Track the negative sign separately from cancellation.

1. a) \(15\cdot\frac{4}{25}=3\cdot\frac{4}{5}=\frac{12}{5}\). 2. b) \(24\cdot\frac{5}{16}=3\cdot\frac{5}{2}=\frac{15}{2}\). 3. c) \(-18\cdot\frac{7}{12}=-3\cdot\frac{7}{2}=-\frac{21}{2}\). 4. d) \(35\cdot\frac{3}{14}=5\cdot\frac{3}{2}=\frac{15}{2}\).

a) \(3\cdot\frac{4}{5}=\frac{12}{5}\) b) \(3\cdot\frac{5}{2}=\frac{15}{2}\) c) \(-3\cdot\frac{7}{2}=-\frac{21}{2}\) d) \(5\cdot\frac{3}{2}=\frac{15}{2}\)

- Dividing by a number is equivalent to multiplying by its reciprocal. - Rewrite mixed numbers as improper fractions before dividing. - It may help to rewrite decimals as fractions. - Keep track of the sign when dividing positive and negative rational numbers.

1. For a), multiply by the reciprocal: \(\frac{5}{6}\cdot\frac{1}{10}=\frac{1}{12}\). 2. For b), rewrite \(3 \frac{1}{3}=\frac{10}{3}\). Then \(\frac{10}{3}\cdot\frac{9}{5}=6\). 3. For c), rewrite \(-2.5=-\frac{5}{2}\). Then \(-\frac{5}{2}\cdot\frac{8}{5}=-4\). 4. For d), rewrite \(-0.6=-\frac{3}{5}\). Then \(\frac{12}{25}\cdot\left(-\frac{5}{3}\right)=-\frac{4}{5}\).

a) \(\frac{1}{12}\) b) \(6\) c) \(-4\) d) \(-\frac{4}{5}\)

- Keep track of the sign of each quotient. - Rewrite division as multiplication by a reciprocal. - Rewrite mixed numbers and decimals as fractions before dividing.

1. For a), \(-\frac{15}{16}\div5=-\frac{15}{16}\cdot\frac{1}{5}=-\frac{3}{16}\). 2. For b), \(\frac{7}{8}\div14=\frac{7}{8}\cdot\frac{1}{14}=\frac{1}{16}\). 3. For c), rewrite \(2 \frac{4}{7}=\frac{18}{7}\). Then \(\frac{18}{7}\div6=\frac{18}{7}\cdot\frac{1}{6}=\frac{3}{7}\). 4. For d), rewrite \(0.4=\frac{2}{5}\). Then \(\frac{2}{5}\div8=\frac{2}{5}\cdot\frac{1}{8}=\frac{1}{20}\).

a) \(-\frac{3}{16}\) b) \(\frac{1}{16}\) c) \(\frac{3}{7}\) d) \(\frac{1}{20}\)

- Multiply the fraction by the whole number, then simplify. - Rewrite a mixed number as an improper fraction before multiplying. - Look for opportunities to simplify before or after multiplying.

1. For a), \(-\frac{5}{12}\cdot8=-\frac{40}{12}=-\frac{10}{3}=-3 \frac{1}{3}\). 2. For b), rewrite \(2 \frac{3}{10}\) as \(\frac{23}{10}\). Then \(4\cdot\frac{23}{10}=\frac{92}{10}=\frac{46}{5}=9 \frac{1}{5}\).

a) \(-3 \frac{1}{3}\) b) \(9 \frac{1}{5}\)

- Rewrite the decimal as a fraction when useful. - Cancel common factors before multiplying. - A positive number times a negative number gives a negative product.

1. For a), \(\frac{3}{8}\cdot\frac{2}{3}=\frac{1}{4}\) after canceling common factors. 2. For b), rewrite \(1.2\) as \(\frac{6}{5}\). Then \(\frac{6}{5}\cdot\frac{5}{6}=1\). 3. For c), cancel common factors before multiplying: \(\frac{7}{10}\cdot\left(-\frac{5}{14}\right)=-\frac{1}{4}\).

a) \(\frac{1}{4}\) b) \(1\) c) \(-\frac{1}{4}\)

- Think about what happens when \(\frac{1}{2}\) is multiplied by a number less than \(1\) or greater than \(1\). - Calculate or estimate each product. - Rewrite the results with a common denominator if needed for comparison.

1. Calculate the values: \(A=\frac{1}{6}\), \(B=\frac{1}{3}\), \(C=\frac{3}{4}\), and \(D=\frac{1}{4}\). 2. Compare them, for example using denominator \(12\): \(\frac{2}{12}<\frac{3}{12}<\frac{4}{12}<\frac{9}{12}\). 3. Therefore \(A<D<B<C\), and C has the greatest value.

\(A<D<B<C\). Expression C has the greatest value.

- Rewrite decimals and percents as fractions when useful. - Rewrite a mixed number as an improper fraction before multiplying. - Cancel common factors before multiplying when possible.

1. For a), \(\frac{3}{8}\cdot\frac{4}{9}=\frac{12}{72}=\frac{1}{6}\). 2. For b), rewrite \(1 \frac{1}{4}=\frac{5}{4}\) and \(0.8=\frac{4}{5}\). Then \(\frac{5}{4}\cdot\frac{4}{5}=1\). 3. For c), rewrite \(75\%=\frac{3}{4}\). Then \(\frac{3}{4}\cdot\frac{2}{3}=\frac{1}{2}\).

a) \(\frac{1}{6}\) b) \(1\) c) \(\frac{1}{2}\)

- Cancel common factors before multiplying when possible. - Keep track of the sign of the product. - Rewrite the mixed number as an improper fraction.

1. For a), cancel common factors before multiplying: \(\frac{9}{14}\cdot\frac{7}{15}=\frac{3}{10}\). 2. For b), the product is negative. After canceling common factors, \(-\frac{3}{8}\cdot\frac{4}{9}=-\frac{1}{6}\). 3. For c), rewrite \(1 \frac{1}{5}\) as \(\frac{6}{5}\). Then \(\frac{5}{6}\cdot\frac{6}{5}=1\).

a) \(\frac{3}{10}\) b) \(-\frac{1}{6}\) c) \(1\)

- Rewrite the mixed number as an improper fraction. - Look for factors that cancel across the three fractions before multiplying. - Simplify the final product.

1. Rewrite \(3 \frac{3}{4}\) as \(\frac{15}{4}\). 2. Multiply \(\frac{15}{4}\cdot\frac{2}{5}\cdot\frac{4}{9}\), canceling common factors before multiplying. 3. The simplified product is \(\frac{2}{3}\).

\(\frac{2}{3}\)

- Interpret “of” as multiplication. - How do you multiply two fractions? - Rewrite the mixed number as an improper fraction. - Can you cancel common factors before multiplying?

1. For a), \(\frac{3}{5}\cdot\frac{10}{9}=\frac{2}{3}\). 2. For b), rewrite \(1 \frac{1}{7}=\frac{8}{7}\). Then \(\frac{7}{8}\cdot\frac{8}{7}=1\). 3. For c), \(\frac{5}{12}\cdot\frac{18}{25}=\frac{3}{10}\) after canceling common factors.

a) \(\frac{2}{3}\) b) \(1\) c) \(\frac{3}{10}\)

- Calculate the two products separately. - Simplify common factors before multiplying. - Compare the simplified results.

1. Value A is \(\frac{4}{9}\cdot\frac{27}{16}=\frac{3}{4}\) after simplifying common factors. 2. Value B is \(\frac{5}{6}\cdot\frac{9}{10}=\frac{3}{4}\). 3. Therefore the values are equal.

The values are equal; both are \(\frac{3}{4}\).

- Convert each decimal to a fraction before multiplying. - Simplify common factors before multiplying the numerators and denominators. - A fraction is in simplest form when its numerator and denominator have no common factor greater than \(1\).

1. For a), write \(0.35 = \frac{35}{100} = \frac{7}{20}\). Then \(\frac{4}{7} \cdot \frac{7}{20} = \frac{4}{20} = \frac{1}{5}\). 2. For b), write \(1.2 = \frac{12}{10} = \frac{6}{5}\). Then \(\frac{6}{5} \cdot \frac{5}{9} = \frac{6}{9} = \frac{2}{3}\). 3. For c), write \(0.16 = \frac{16}{100} = \frac{4}{25}\). Then \(\frac{3}{8} \cdot \frac{4}{25} = \frac{3}{50}\).

a) \(\frac{1}{5}\) b) \(\frac{2}{3}\) c) \(\frac{3}{50}\)

- What operation does “of” represent when working with fractions? - How do you multiply three fractions? - Think about whether repeatedly taking a fraction of a fraction should make the value larger or smaller.

1. Write each “one third” as \(\frac{1}{3}\). 2. The phrase “one third of one third of one third” means \(\frac{1}{3}\cdot\frac{1}{3}\cdot\frac{1}{3}\). 3. Multiply: \(\frac{1\cdot1\cdot1}{3\cdot3\cdot3}=\frac{1}{27}\). 4. Compare \(\frac{1}{27}\) with the claimed value \(\frac{1}{9}\). Since \(\frac{1}{27}\neq\frac{1}{9}\), Jordan's claim is false.

No. \(\frac{1}{3}\cdot\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{27}\), not \(\frac{1}{9}\).

- Calculate the value of Julia's expression first. - Then calculate one half of one eighth. - Compare the two simplified fractions.

1. Julia's expression is \(\frac{1}{4}\cdot\frac{1}{4}=\frac{1}{16}\). 2. The comparison expression is \(\frac{1}{2}\cdot\frac{1}{8}=\frac{1}{16}\). 3. Since \(\frac{1}{16}=\frac{1}{16}\), Marcus is correct.

Marcus is correct. Both expressions equal \(\frac{1}{16}\).

- Convert each mixed number first. - Look for a numerator and denominator sharing a factor before multiplying. - Keep the sign separate while simplifying magnitudes.

1. a) \(\frac{8}{3}\cdot\frac{9}{8}\to\frac{9}{3}=3\). 2. b) \(-\frac{9}{2}\cdot\frac{4}{9}\to-\frac{4}{2}=-2\). 3. c) \(\frac{10}{7}\cdot\left(-\frac{21}{10}\right)\to-\frac{21}{7}=-3\).

a) \(\frac{8}{3}\cdot\frac{9}{8}\to3\) b) \(-\frac{9}{2}\cdot\frac{4}{9}\to-2\) c) \(\frac{10}{7}\cdot\left(-\frac{21}{10}\right)\to-3\)

- Divide by a fraction by multiplying by its reciprocal. - Keep track of the signs in each quotient. - Rewrite mixed numbers as improper fractions first. - Cancel common factors before multiplying. - Convert an improper fraction to a mixed number at the end when appropriate.

1. For a), \(\frac{15}{16}\div\frac{5}{8}=\frac{15}{16}\cdot\frac{8}{5}=\frac{3}{2}=1 \frac{1}{2}\). 2. For b), rewrite the mixed numbers: \(-\frac{8}{3}\div\frac{10}{9}=-\frac{8}{3}\cdot\frac{9}{10}=-\frac{12}{5}=-2 \frac{2}{5}\). 3. For c), \(10\div\left(-\frac{2}{3}\right)=10\cdot\left(-\frac{3}{2}\right)=-15\). 4. For d), two negative numbers give a positive quotient: \(\frac{7}{12}\cdot\frac{9}{14}=\frac{3}{8}\).

a) \(1 \frac{1}{2}\) b) \(-2 \frac{2}{5}\) c) \(-15\) d) \(\frac{3}{8}\)

- For a missing factor, think about which inverse operation will help. - Recall how multiplying a decimal by \(10\), \(100\), or \(1000\) changes its place value. - Pay attention to the sign when a product is negative.

1. For a), use division: \(3.2\div0.4=8\). 2. For b), \(0.07\cdot8=0.56\). 3. For c), use division: \(-4.5\div1.5=-3\). 4. For d), \(0.003\cdot300=0.9\).

a) \(8\) b) \(0.56\) c) \(-3\) d) \(0.9\)

- Use place value carefully when dividing decimals. - Decide the sign of a quotient before dividing absolute values. - If the dividend is smaller in magnitude than the divisor, expect a quotient between \(-1\) and \(1\).

1. \(15.6\div6=2.6\). 2. \(8.4\div12=0.7\). 3. \(-13.5\div5=-2.7\). 4. \(0.64\div16=0.04\).

a) \(2.6\) b) \(0.7\) c) \(-2.7\) d) \(0.04\)

- Relate each decimal quotient to a familiar whole-number fact. - Scaling both dividend and divisor by the same power of \(10\) preserves a quotient. - State the sign separately when needed.

1. a) \(48\div6=8\), so \(0.48\div6=0.08\). 2. b) \(15\div3=5\), and the quotient is negative, so \(-1.5\div3=-0.5\). 3. c) Scale both numbers by \(10\): \(1.2\div0.4=12\div4=3\).

a) \(48\div6=8\); result \(0.08\) b) \(15\div3=5\); sign negative; result \(-0.5\) c) \(12\div4=3\); result \(3\)

- Convert the decimal to a fraction. - To divide by a fraction, multiply by its reciprocal. - Look for common factors before multiplying.

1. For a), simplify common factors before multiplying: \(\frac{9}{14} \cdot \frac{7}{12} = \frac{3}{8}\). 2. For b), write \(2.5 = \frac{5}{2}\). 3. Divide by multiplying by the reciprocal: \(\frac{5}{2} \cdot \frac{8}{5}\). 4. Simplify the common factors to get \(4\).

a) \(\frac{3}{8}\) b) \(4\)

- Record the benchmark dividend and divisor you choose. - Keep the estimate and exact calculation as separate lines. - Use the estimate to check sign and approximate size, not to replace exact division.

1. a) Using \(6\div0.1\) gives estimate \(60\). Exactly, \(560\div8=70\), which is reasonably close. 2. b) Using \(0.4\div2\) gives estimate \(0.2\). Exactly, \(4.41\div21=0.21\), consistent with the estimate. 3. c) Using \((-2)\div(-0.2)\) gives estimate \(10\). Exactly, \(225\div15=15\), with positive sign; the estimate has the correct sign and order of magnitude.

a) Benchmarks \(6\) and \(0.1\): estimate \(60\); exact \(70\); reasonable b) Benchmarks \(0.4\) and \(2\): estimate \(0.2\); exact \(0.21\); reasonable c) Benchmarks \(-2\) and \(-0.2\): estimate \(10\); exact \(15\); same sign and order of magnitude

- For Expression A, evaluate the difference inside the parentheses first. - For Expression B, evaluate the two products first. - Which property relates multiplying a difference to subtracting two products?

1. For Expression A, subtract inside the parentheses: \(\frac{1}{4} - \frac{1}{3} = \frac{3}{12} - \frac{4}{12} = -\frac{1}{12}\). Then \(12\left(-\frac{1}{12}\right) = -1\). 2. For Expression B, evaluate each product: \(12 \cdot \frac{1}{4} = 3\) and \(12 \cdot \frac{1}{3} = 4\). Then \(3 - 4 = -1\). 3. Both expressions equal \(-1\). 4. The two equivalent forms illustrate the distributive property.

a) Expression A: \(-1\) Expression B: \(-1\) b) The results are equal. This illustrates the distributive property.

- Determine the sign of each correct result first. - What rule do you use when dividing by a fraction? - What sign does the product of two negative numbers have?

1. For a), the product of two negative numbers must be positive. The magnitude \(0.2\cdot0.3=0.06\) is correct, so the correct result is \(0.06\). 2. For b), dividing by a fraction means multiplying by its reciprocal. The correct calculation is \(\frac{3}{4}\cdot\left(-\frac{2}{1}\right)=-\frac{6}{4}=-\frac{3}{2}\).

a) The sign is wrong. The correct result is \(0.06\). b) The divisor was not replaced by its reciprocal. The correct result is \(-\frac{3}{2}\), or \(-1.5\).

- Determine the sign before multiplying or dividing. - Simplify common factors before multiplying fractions. - To divide by a fraction, multiply by its reciprocal.

1. For a), write \(-0.9 = -\frac{9}{10}\). Then \(\frac{2}{3} \cdot \left(-\frac{9}{10}\right) = -\frac{3}{5} = -0.6\). 2. For b), write \(-0.75 = -\frac{3}{4}\). Then \(-\frac{3}{4} \div \frac{3}{4} = -1\). 3. For c), write \(1.25 = \frac{5}{4}\). Then \(\frac{5}{4} \cdot \left(-\frac{4}{5}\right) = -1\).

a) \(-0.6\), or \(-\frac{3}{5}\) b) \(-1\) c) \(-1\)

- Identify the factor shared by both terms. - Use the distributive property in reverse to factor it out. - Add the fractions before multiplying.

1. Factor out the common factor \(-15.8\): \(-15.8\left(\frac{3}{7} + \frac{4}{7}\right)\). 2. Add inside the parentheses: \(\frac{3}{7} + \frac{4}{7} = 1\). 3. Multiply: \(-15.8 \cdot 1 = -15.8\).

\(-15.8\)

- Determine the sign of a product from the number of negative factors. - For “exactly two negative,” consider every possible sign of the remaining factor. - Recall that integers are rational numbers. - Any factor of \(0\) makes the entire product \(0\).

1. For a), possible: \((-1)\cdot(-1)\cdot(-1)=-1\). 2. For b), not possible. Two negative factors have a positive product. If the third factor is positive, the product is positive; if the third factor is \(0\), the product is \(0\). Neither case gives \(-1\). 3. For c), possible: \((-1)\cdot1\cdot1=-1\). 4. For d), not possible. Any product with a factor of \(0\) equals \(0\).

a) Yes; for example, \((-1)\cdot(-1)\cdot(-1)=-1\). b) No; the third factor is nonnegative, so the product is positive or \(0\), not \(-1\). c) Yes; for example, \((-1)\cdot1\cdot1=-1\). d) No; the product would be \(0\).

- First evaluate the expression inside the parentheses as a check. - When distributing, multiply the outside factor by each term, including its sign. - Recall the sign of a product of two negative numbers.

1. Evaluate inside the parentheses: \(1.2 - 4 = -2.8\). Then \(-5(-2.8) = 14\), so Sophie is correct. 2. When distributing, the correct expression is \(-5 \cdot 1.2 - (-5 \cdot 4)\). 3. Therefore, \(-6 - (-20) = -6 + 20 = 14\). Lucas failed to account for the product of two negative factors.

1. Sophie is correct; the value is \(14\). 2. Lucas mishandled the signs. The second product contributes \(+20\), not \(-20\).

- Distribute the factor in \(A\) and compare the resulting terms with \(B\). - Pay close attention to the product of two negative numbers. - Find the value of the first two terms in \(B\), then determine its additive inverse.

1. Distributing \(-\frac{3}{4}\) in \(A\) gives \(-\frac{3}{4}\cdot2.4-\left(-\frac{3}{4}\cdot6.4\right)+1.5\), which simplifies to \(B\). 2. Evaluate \(A\): \(2.4-6.4=-4\), so \(-\frac{3}{4}(-4)+1.5=3+1.5=4.5\). 3. The part of \(B\) before the final number equals \(3\). Solve \(3+x=0\), giving \(x=-3\).

1. The distributive property shows that the expressions are equivalent. 2. \(A=4.5\) 3. Replace \(1.5\) with \(-3\).

- For Method A, consider an equivalent fraction whose denominator is \(10\). - For Method B, compare the two fraction factors after converting the decimal. - Compare the amount of multiplication and simplification required by the two methods.

1. Method A: Write \(\frac{4}{5} = 0.8\). Then \(1.25 \cdot 0.8 = 1\). 2. Method B: Write \(1.25 = \frac{5}{4}\). Then \(\frac{5}{4} \cdot \frac{4}{5} = 1\). 3. Method B makes the reciprocal factors visible, so the product simplifies immediately. Method A is also valid.

Method A: \(\frac{4}{5}=0.8\), so \(1.25\cdot0.8=1\). Method B: \(1.25=\frac{5}{4}\), so \(\frac{5}{4}\cdot\frac{4}{5}=1\). Both methods are valid. For example, Method B is especially short because the two fraction factors are reciprocals and simplify immediately. A learner may reasonably prefer Method A instead if the explanation compares the actual work in the two methods.

- In a), multiply the outside factor by each addend. - In b), identify the common factor in both terms. - Choose the equivalent form that creates simpler arithmetic.

1. For a), distribute \(42\): \(42 \cdot \frac{1}{7} + 42 \cdot \frac{1}{6} = 6 + 7 = 13\). 2. For b), factor out \(17\): \(17(3.8 + 6.2) = 17 \cdot 10 = 170\).

a) \(13\) b) \(170\)

- Multiply \(20\) by each term inside the parentheses. - Evaluate \(20 \cdot \frac{1}{4}\) and \(20 \cdot 0.2\). - Compare the resulting expression with Expression B.

1. Distribute \(20\) in Expression A: \(20 \cdot \frac{1}{4} - 20 \cdot 0.2\). 2. Evaluate the first product: \(20 \cdot \frac{1}{4} = 5\). 3. Evaluate the second product: \(20 \cdot 0.2 = 4\). 4. The expanded form is \(5 - 4\), which is exactly Expression B. Both expressions equal \(1\).

Distributing \(20\) changes Expression A to \(5 - 4\), so the expressions are equivalent and both have value \(1\).

- Identify the factor shared by both terms. - Use the distributive property in reverse. - Compare \(\frac{3}{4}\) with \(0.75\). - Recall whether the distributive property applies to sums as well as differences.

1. Both terms have the common factor \(\frac{3}{4}\). 2. Factor using the distributive property: \(\frac{3}{4} \cdot 1.2 - \frac{3}{4} \cdot 0.4 = \frac{3}{4}(1.2 - 0.4)\). 3. Since \(\frac{3}{4} = 0.75\), this is exactly the student’s method, so the claim is correct. 4. The distributive property also applies to addition: \(ab + ac = a(b + c)\). Therefore, the corresponding rule would also work with a plus sign.

The claim is correct because \(\frac{3}{4}\) can be factored: \(\frac{3}{4}(1.2 - 0.4)\). The same rule works for addition because \(ab + ac = a(b + c)\).

- Determine the sign before calculating the magnitude. - First multiply or divide the absolute values. - The product of two negative numbers is positive.

1. Determine each sign first: unlike signs give a negative result, and like signs give a positive result. 2. For a), \(145\cdot14=2030\), so the result is \(-2030\). 3. For b), \(2352\div12=196\), so the result is \(-196\). 4. For c), \(26\cdot302=7852\), and two negative factors give a positive result. 5. For d), \(4464\div18=248\), so the result is \(-248\).

a) \(-2030\) b) \(-196\) c) \(7852\) d) \(-248\)

- Identify the factor shared by both terms in each expression. - Factor it out using the distributive property in reverse. - Evaluate the sum or difference inside the parentheses first.

1. For a), factor out \(15\): \(15(48 + 52) = 15 \cdot 100 = 1500\). 2. For b), factor out \(12.5\): \(12.5(7 - 3) = 12.5 \cdot 4 = 50\). 3. For c), factor out \(-20\): \((-20)(14 + 36) = (-20) \cdot 50 = -1000\).

a) \(1500\) b) \(50\) c) \(-1000\)

- Look for factors that combine to make \(1000\). - In b), identify the common factor and add the decimal coefficients. - Rewrite \(11\) as a sum that makes multiplication easy.

1. For a), rearrange and regroup the factors: \((250 \cdot 4) \cdot 13 = 1000 \cdot 13 = 13{,}000\). 2. For b), factor out \(19\): \(19(0.8 + 0.2) = 19 \cdot 1 = 19\). 3. For c), write \(11 = 10 + 1\) and distribute: \(45(10 + 1) = 450 + 45 = 495\).

a) \(13{,}000\) b) \(19\) c) \(495\)

- Determine the sign before doing the division. - Use long division for the absolute values if needed. - What sign results when a positive number is divided by a negative number?

1. A positive number divided by a negative number gives a negative result. 2. Divide the absolute values: \(18{,}432\div16=1152\). 3. Apply the sign: \(-1152\).

\(-1152\)

- In a), distributing avoids adding fractions with unlike denominators first. - In b), identify the common factor. - In c), multiply and simplify each product before subtracting.

1. For a), distribute \(24\): \(24 \cdot \frac{3}{8} + 24 \cdot \frac{1}{6} = 9 + 4 = 13\). 2. For b), factor out \(\frac{5}{12}\): \(\frac{5}{12}\left(\frac{4}{7} + \frac{3}{7}\right) = \frac{5}{12} \cdot 1 = \frac{5}{12}\). 3. For c), factor out \(15\): \(15\left(\frac{7}{5} - \frac{2}{3}\right)\). Distributing directly gives \(21 - 10 = 11\).

a) \(13\) b) \(\frac{5}{12}\) c) \(11\)

- A negative product requires factors with opposite signs. - Start from familiar factor pairs of \(30\). - Scale one factor up and the other down to create decimal or fractional factors.

1. For a), one example is \(3\cdot(-10)=-30\). 2. For b), one example is \((-1.5)\cdot20=-30\). 3. For c), one example is \(\frac{3}{4}\cdot(-40)=-30\).

a) For example, \(3\cdot(-10)\) b) For example, \((-1.5)\cdot20\) c) For example, \(\frac{3}{4}\cdot(-40)\)

- A factor that produces \(1\) is the reciprocal of the other factor. - Use division to find a missing factor. - Recall the zero-product property.

1. For a), \(x=1\div(-2)=-\frac{1}{2}\). 2. For b), \(y=1\div0.1=10\). 3. A fraction and its reciprocal have product \(1\), such as \(\frac{2}{5}\cdot\frac{5}{2}=1\). 4. Yes. A product is \(0\) when at least one factor is \(0\), such as \(0\cdot5=0\).

a) \(x=-\frac{1}{2}\) b) \(y=10\) c) For example, \(\frac{2}{5}\) and \(\frac{5}{2}\) d) Yes; for example, \(0\cdot5=0\).

- Use place value to position the decimal point in each product. - Decide the sign of each product before multiplying. - Multiply the absolute values first, then apply the sign.

1. \(4.8\cdot1.5=7.2\). 2. \((-2.2)\cdot3.4=-7.48\). 3. \(0.05\cdot(-0.14)=-0.007\). 4. \((-1.25)\cdot(-0.8)=1\).

a) \(7.2\) b) \(-7.48\) c) \(-0.007\) d) \(1\)

- Notice how the same digit product can appear with different place values. - Count the decimal places in the factors. - Determine the sign of each product separately from its magnitude.

1. The base digit product is \(6\cdot7=42\). 2. For a), \(0.6\cdot0.7=0.42\). 3. For b), \(0.06\cdot0.7=0.042\). 4. For c), the product is positive: \((-0.6)\cdot(-0.07)=0.042\). 5. For d), the product is negative: \((-0.06)\cdot0.07=-0.0042\).

a) \(0.42\) b) \(0.042\) c) \(0.042\) d) \(-0.0042\)

- For a), consider a positive fraction between \(0\) and \(1\). - For b), recall the sign of a product of two negative numbers. - For c), identify the number that cannot be used as a divisor.

1. For a), use \(\frac{1}{2}\). Its reciprocal is \(2\), and \(2 > \frac{1}{2}\). 2. For b), use \(-4\). Its reciprocal is \(-\frac{1}{4}\), and \((-4)\left(-\frac{1}{4}\right) = 1\), which is positive. 3. For c), use \(0\). Zero has no reciprocal because division by \(0\) is undefined.

a) The reciprocal of \(\frac{1}{2}\) is \(2\), which is greater than \(\frac{1}{2}\). b) \((-4)\left(-\frac{1}{4}\right) = 1\), which is positive. c) \(0\) has no reciprocal.

- For a), test a product with exactly one zero factor. - For b), test a negative number. - For c), consider a positive number between \(0\) and \(1\).

1. For a), \(0 \cdot 5 = 0\), but only one factor is \(0\). 2. For b), \((-3)(-1) = 3\), and \(3 > -3\). 3. For c), \((0.5)^2 = 0.25\), and \(0.25 < 0.5\).

a) \(0 \cdot 5 = 0\), although \(5 \ne 0\). b) \((-3)(-1) = 3 > -3\). c) \((0.5)^2 = 0.25 < 0.5\).

- Determine the sign before calculating the magnitude. - What sign results from multiplying numbers with like or unlike signs? - Simplify common factors before multiplying fractions when possible. - Converting between fractions and decimals can make a product easier.

1. For a), unlike signs give a negative product: \((-12)\cdot11=-132\). 2. For b), two negative factors give a positive product: \((-0.5)\cdot(-0.08)=0.04\). 3. For c), \(\frac{3}{7}\cdot\left(-\frac{14}{9}\right)=-\frac{42}{63}=-\frac{2}{3}\). 4. For d), \(-2.5=-\frac{5}{2}\), so \(-\frac{5}{2}\cdot\frac{2}{5}=-1\).

a) \(-132\) b) \(0.04\) c) \(-\frac{2}{3}\) d) \(-1\)

- First determine the sign of each missing number. - Use the inverse operation to find a missing factor or dividend. - For the last part, think about factor pairs of \(18\).

1. For a), divide the product by the known factor: \(-56\div7=-8\). 2. For b), multiply the quotient by the divisor: \(4\cdot(-9)=-36\). 3. For c), divide the product by the known factor: \(10\div(-0.5)=-20\). 4. For d), choose two integer factors of \(18\) with opposite signs. One example is \(2\cdot(-9)=-18\).

a) \(-8\) b) \(-36\) c) \(-20\) d) For example, \(2\) and \(-9\)

- For a), evaluate the sum inside the parentheses first. - For b), multiply \(8\) by each addend. - Equivalent methods must produce the same value.

1. Add first: \(8(12.5 + 5) = 8(17.5) = 140\). 2. Distribute first: \(8 \cdot 12.5 + 8 \cdot 5 = 100 + 40 = 140\). 3. Both methods give the same value, as expected for equivalent expressions.

a) \(140\) b) \(140\) The results are equal.

- In a), distribute the subtraction sign to both terms inside the parentheses. - Look for additive inverses. - In b), pair two factors whose product is \(1\).

1. For a), remove the parentheses: \(-15.8 - 4.2 + 15.8\). Group additive inverses: \((-15.8 + 15.8) - 4.2 = -4.2\). 2. For b), regroup the factors: \((25 \cdot 0.04)(-17) = 1(-17) = -17\).

a) \(-4.2\) b) \(-17\)

- Rewrite \(101\) using \(100\) and \(1\). - In b), identify the common factor. - Choose the equivalent form that creates the simplest arithmetic.

1. For a), write \(101 = 100 + 1\): \(12(100 + 1) = 1200 + 12 = 1212\). 2. For b), factor out \(\frac{5}{9}\): \(\frac{5}{9}(14 - 5) = \frac{5}{9} \cdot 9 = 5\).

a) \(1212\) b) \(5\)

- For the direct method, evaluate multiplication before addition or subtraction. - For the factoring method, identify the factor shared by both terms. - Compare the two methods after completing each part.

1. For a), direct calculation gives \(154 + 126 = 280\). Factoring gives \(14(11 + 9) = 14 \cdot 20 = 280\). 2. For b), direct calculation gives \(52 - 39 = 13\). Factoring gives \(6.5(8 - 6) = 6.5 \cdot 2 = 13\). 3. For c), direct calculation gives \(-6 + (-9) = -15\). Factoring gives \((-1.5)(4 + 6) = -1.5 \cdot 10 = -15\).

a) \(280\) b) \(13\) c) \(-15\)

- Determine the sign before calculating each product or quotient. - How do you multiply fractions? - How do you divide by a fraction? - A whole number can be written as a fraction when helpful.

1. For a), product: \(-2.5\cdot0.4=-1\). Quotient: \(-2.5\div0.4=-6.25\). 2. For b), product: \(\frac{3}{10}\cdot\left(-\frac{6}{5}\right)=-\frac{9}{25}\). Quotient: \(\frac{3}{10}\div\left(-\frac{6}{5}\right)=\frac{3}{10}\cdot\left(-\frac{5}{6}\right)=-\frac{1}{4}\). 3. For c), product: \((-12)\cdot\left(-\frac{4}{3}\right)=16\). Quotient: \((-12)\div\left(-\frac{4}{3}\right)=(-12)\cdot\left(-\frac{3}{4}\right)=9\).

a) Product: \(-1\); quotient: \(-6.25\) b) Product: \(-\frac{9}{25}\); quotient: \(-\frac{1}{4}\) c) Product: \(16\); quotient: \(9\)

- Determine the sign before calculating. - Simplify common factors before multiplying when possible. - Dividing by a whole number is equivalent to multiplying by its reciprocal.

1. For a), a negative times a positive is negative: \((-15)\cdot4=-60\). 2. For b), dividing two negative numbers gives a positive result: \((-20)\div(-5)=4\). 3. For c), \(\frac{4}{7}\cdot14=8\). 4. For d), \(\frac{3}{10}\div3=\frac{3}{10}\cdot\frac{1}{3}=\frac{1}{10}\). 5. For e), two negative factors give a positive product: \(\left(-\frac{2}{5}\right)\cdot\left(-\frac{1}{2}\right)=\frac{1}{5}\).

a) \(-60\) b) \(4\) c) \(8\) d) \(\frac{1}{10}\), or \(0.1\) e) \(\frac{1}{5}\), or \(0.2\)

- Use an inverse operation to find each missing number. - A number multiplied by its reciprocal equals \(1\). - Look for factors you can cancel before multiplying fractions.

1. For a), multiply by the reciprocal of \(-\frac{5}{6}\): \(1\cdot\left(-\frac{6}{5}\right)=-\frac{6}{5}\). 2. For b), multiply the quotient by the divisor: \(\frac{8}{9}\cdot\left(-\frac{3}{4}\right)=-\frac{24}{36}=-\frac{2}{3}\).

a) \(-\frac{6}{5}\) b) \(-\frac{2}{3}\)

- Use the order of operations. - Evaluate the parentheses first, and then subtract that result from \(88\).

1. Evaluate inside the parentheses: \(112-45=67\). 2. Subtract the result: \(88-67=21\).

\(21\)

- Identify the inverse operations in each expression. - Determine which operations undo each other. - Compare the starting values after simplifying the structure.

1. In expression \(A\), multiplying by \(77\) and then dividing by \(77\) undo each other, so \(A = 333\). 2. In expression \(B\), adding \(77\) and then subtracting \(77\) undo each other, so \(B = 330\). 3. Since \(333 > 330\), expression \(A\) has the greater value.

Expression \(A\) is greater because \(A = 333\) and \(B = 330\).

- Regroup factors to make \(10\), \(100\), or \(1000\). - In part d, identify the common factor. - Use the distributive property in reverse when a factor appears in both products.

1. For a), regroup: \((25 \cdot 4) \cdot 17 = 100 \cdot 17 = 1700\). 2. For b), regroup: \((125 \cdot 8) \cdot 11 = 1000 \cdot 11 = 11{,}000\). 3. For c), regroup: \((5 \cdot 20) \cdot 42 = 100 \cdot 42 = 4200\). 4. For d), factor out the common factor: \(63(8 + 2) = 63 \cdot 10 = 630\).

a) \(1700\) b) \(11{,}000\) c) \(4200\) d) \(630\)

- Translate each phrase “the product of” into multiplication. - Identify the factor that appears in both products. - Factor it out before adding the remaining factors.

1. The expression is \(47 \cdot 19 + 47 \cdot 81\). 2. Factor out the common factor \(47\): \(47(19 + 81)\). 3. Evaluate: \(47 \cdot 100 = 4700\).

\(47 \cdot 19 + 47 \cdot 81 = 47(19 + 81) = 4700\)

- Check whether the order of terms changes. - Check whether the grouping changes. - Check whether a common factor is distributed or factored out.

1. In a), the order of \(57\) and \(72\) changes, so the commutative property of addition is used. 2. In b), the factors are reordered and regrouped, so the commutative and associative properties of multiplication are used. 3. In c), the common factor \(9\) is factored out, so the distributive property is used in reverse.

a) Commutative property of addition b) Commutative and associative properties of multiplication c) Distributive property

- Look for pairs whose product is a convenient benchmark number. - Keep signs attached to factors when rearranging. - Show the grouping that makes each product short.

1. \((25\cdot4)(-13)=100(-13)=-1300\). 2. \((-125)(-8)\cdot7=1000\cdot7=7000\). 3. \(\left(\frac{4}{9}\cdot\frac{9}{4}\right)(-17)=1(-17)=-17\). 4. \((0.2\cdot5)(-47)=1(-47)=-47\).

1) \((25\cdot4)(-13)=-1300\) 2) \((-125)(-8)\cdot7=7000\) 3) \(\left(\frac{4}{9}\cdot\frac{9}{4}\right)(-17)=-17\) 4) \((0.2\cdot5)(-47)=-47\)

- Look for factors that create simple partial products. - One property allows reordering; another allows regrouping. - Determine the sign of the final product before calculating its magnitude.

1. Marie used the commutative and associative properties to reorder and regroup factors so that the first product was \(10\). 2. For b), group \(8\) and \(125\): \(8 \cdot 125 = 1000\). 3. Group the remaining factors: \((-1.5)(-2) = 3\). 4. Multiply: \(1000 \cdot 3 = 3000\).

a) Marie used the commutative and associative properties of multiplication. b) \(3000\)

- Follow the grouping symbols in each expression. - Determine the sign of each result before calculating. - Compare all three final values.

1. \((-96 \cdot 16) \div 8 = -1536 \div 8 = -192\). 2. \((-96 \div 8) \cdot 16 = -12 \cdot 16 = -192\). 3. \(-96(16 \div 8) = -96 \cdot 2 = -192\). 4. All three expressions have the same value.

All three expressions equal \(-192\).

- Determine the sign of a positive number divided by a negative number. - For the second method, divide by the factors one at a time in their given order. - Compare the two final results.

1. Multiply first: \((-6) \cdot 4 = -24\), then \(144 \div (-24) = -6\). 2. Divide successively: \(144 \div (-6) = -24\), then \(-24 \div 4 = -6\). 3. Both methods give the same result.

Both methods give \(-6\).

- Compute each product before comparing. - Track the sign and magnitude separately. - A product of numbers with the same sign is positive. - Use the signed multiplication rule explicitly in part c).

1. a) \(9\cdot3=27\) and \(4\cdot2=8\), so \(27>8\). 2. b) \(6\cdot4=24\) and \((-3)\cdot1=-3\), so \(24>-3\). 3. c) \((-5)(-4)=20\) and \(2\cdot3=6\), so \(20>6\). 4. In part c), the product of two negative numbers is positive.

a) \(27>8\) b) \(24>-3\) c) \(20>6\) The key rule in part c) is that a negative times a negative is positive.

- Compare the number of columns covered by the shaded rectangle with the total number of columns. - Compare the number of rows covered by the shaded rectangle with the total number of rows. - After multiplying the two dimension fractions, compare the result with the fraction of individual cells that are shaded.

1. The grid has \(4\) columns, and the shaded rectangle spans \(3\) of them, so the shaded width is \(\frac{3}{4}\). 2. The grid has \(3\) rows, and the shaded rectangle spans \(2\) of them, so the shaded height is \(\frac{2}{3}\). 3. Multiply: \(\frac{3}{4}\cdot\frac{2}{3}=\frac{6}{12}=\frac{1}{2}\). 4. There are \(12\) equal cells and \(6\) are shaded, so \(\frac{6}{12}=\frac{1}{2}\), which checks the product.

a) \(\frac{3}{4}\) b) \(\frac{2}{3}\) c) \(\frac{1}{2}\) d) \(6\) of \(12\) cells are shaded, so \(\frac{6}{12}=\frac{1}{2}\).

- Identify the quantity given per minute and the amount of time that passes. - Choose a relationship whose units reduce to meters. - Interpret a negative signed vertical change after calculating.

1. The quantity relationship is signed rate times elapsed time: \((-2.5)\cdot1.6\). 2. \((-2.5)\cdot1.6=-4\). 3. The signed vertical change is \(-4\,\text{m}\), meaning \(4\,\text{m}\) downward.

\((-2.5\,\text{m/min})\cdot(1.6\,\text{min})=-4\,\text{m}\). The negative sign means a \(4\,\text{m}\) downward change.

- Identify the original signed quantity and the fractional scale applied to it. - Write a single expression that gives the scaled signed change. - Interpret the sign only after finding the scaled value.

1. Trail B's change is the fraction \(\frac{5}{8}\) of \(-12.8\,\text{m}\), so model it as \(\frac{5}{8}\cdot(-12.8)\). 2. \(12.8\div8=1.6\) and \(1.6\cdot5=8\), so the result is \(-8\,\text{m}\). 3. The negative sign represents a net descent.

\(\frac{5}{8}\cdot(-12.8\,\text{m})=-8\,\text{m}\), representing a net descent of \(8\,\text{m}\).

- Let \(x\) be the signed change from one transaction and relate six copies of \(x\) to the total. - Solve that relationship for one transaction's change. - Interpret a negative transaction change in the account context.

1. Split the total signed change equally among six transactions: \((-39.00)\div6\). 2. \((-39.00)\div6=-6.50\). 3. Each transaction decreases the balance by \(\$6.50\).

\((-39.00)\div6=-6.50\). Each transaction decreases the balance by \(\$6.50\).

- Let \(n\) be the number of equal steps and write a relationship between \(n\), the change per step, and the total change. - Solve for the count \(n\). - Check that the sign of a count makes sense in context.

1. The number of equal steps is modeled by \((-12)\div(-1.5)\). 2. \((-12)\div(-1.5)=8\). 3. The result is positive because it is a count of \(8\) steps; the quotient of the two negative signed changes is also positive.

\((-12)\div(-1.5)=8\). The answer is positive because it counts \(8\) lowering steps, and the quotient of two negative values is positive.

- The requested quantity is “per second,” so relate the total change to the elapsed seconds. - Choose an expression whose units become \(\text{m/s}\). - Interpret the sign of the resulting rate in terms of altitude.

1. A per-second quantity is the total signed change distributed over \(2.4\) seconds: \((-7.2)\div2.4\). 2. \((-7.2)\div2.4=-3\). 3. The rate is \(-3\,\text{m/s}\), meaning altitude decreases by \(3\,\text{m}\) per second on average.

\((-7.2\,\text{m})\div(2.4\,\text{s})=-3\,\text{m/s}\). The negative sign indicates an average downward change.

- Let \(x\) represent the full signed change. - Write an equation saying that three fifths of \(x\) equals the observed change. - Solve that equation while keeping the signed meaning of \(x\).

1. If \(x\) is the full signed change, then \(\frac{3}{5}x=-9\). 2. Solving gives \(x=(-9)\div\frac{3}{5}=(-9)\cdot\frac{5}{3}=-15\). 3. The full signed change is \(-15\,\text{m}\), a \(15\,\text{m}\) descent.

\(\frac{3}{5}x=-9\), so \(x=(-9)\div\frac{3}{5}=-15\,\text{m}\).

- Read the common signed jump size and count how many equal jumps are shown. - Write one compact expression for the total of those equal changes. - Use the displayed start to check each successive landing.

1. Each jump is \(-0.75\), and there are \(3\) equal jumps, so the total change is \(3\cdot(-0.75)=-2.25\). 2. Starting at \(1.5\), the successive landings are \(0.75\), \(0\), and \(-0.75\).

a) \(3\cdot(-0.75)\) b) \(-2.25\) c) \(0.75\), \(0\), \(-0.75\)

- Evaluate the grouped sum first. - The fractions \(\frac{1}{8}\) and \(\frac{1}{5}\) have terminating decimal forms. - Align decimal places carefully when adding and subtracting.

1. Convert the mixed numbers to decimals: \(3\frac{1}{8} = 3.125\) and \(2\frac{1}{5} = 2.2\). 2. Evaluate inside the parentheses: \(3.125 + 4.025 = 7.15\). 3. Substitute and evaluate: \(12.8 - 7.15 + 2.2 = 5.65 + 2.2 = 7.85\).

\(7.85\)

- In a), look for a factor shared by both products. - In b), which two factors multiply to make \(100\)? - In c), rewrite \(99\) using \(100\) and \(1\).

1. For a), factor out the common factor \(16\): \(16(12 + 8) = 16 \cdot 20 = 320\). 2. For b), use the commutative and associative properties to group \(25\) and \(4\): \((25 \cdot 4) \cdot 37 = 100 \cdot 37 = 3700\). 3. For c), rewrite \(99\) as \(100 - 1\) and distribute: \(45(100 - 1) = 45 \cdot 100 - 45 \cdot 1 = 4500 - 45 = 4455\).

a) \(320\) b) \(3700\) c) \(4455\)

- Convert the fractional parts to terminating decimals. - A subtraction sign before parentheses changes both signs inside. - Look for decimal pairs that combine to whole numbers.

1. Estimate: \((12 + 5) - (2 - 1) = 16\). 2. Convert the mixed numbers: \(12\frac{3}{8} = 12.375\) and \(1\frac{9}{20} = 1.45\). 3. Remove the parentheses and regroup: \(12.375 + 4.55 - 2.375 + 1.45\). 4. Evaluate: \((12.375 - 2.375) + (4.55 + 1.45) = 10 + 6 = 16\). 5. The exact result agrees with the estimate.

Estimate: \(16\) Exact value: \(16\)

- Choose a representation that exposes cancellation or a reciprocal relationship. - Keep the chosen rewrite visible in the response. - Determine the sign independently of the magnitude calculation.

1. a) \(\frac{27}{8}\cdot\frac{4}{9}=\frac{3}{2}\). 2. b) Rewrite \(0.8=\frac{4}{5}\): \(\frac{4}{5}\div\frac{4}{15}=\frac{4}{5}\cdot\frac{15}{4}=3\). 3. c) Rewrite \(2.25=\frac{9}{4}\): \(\frac{9}{4}\cdot\left(-\frac{2}{3}\right)=-\frac{3}{2}\).

a) \(\left(\frac{3}{2}\right)^3=\frac{27}{8}\); result \(\frac{3}{2}\) b) \(0.8=\frac{4}{5}\); result \(3\) c) \(2.25=\frac{9}{4}\); result \(-\frac{3}{2}\)

- State the benchmark factors used in each estimate. - Keep the estimate separate from the exact computation. - Finish by comparing the size and sign of the exact value with the estimate.

1. a) Using \(4\) and \(2\) gives estimate \(8\). Exactly, \(\frac{7}{2}\cdot\frac{12}{5}=\frac{42}{5}=8\frac{2}{5}\), consistent with the estimate. 2. b) Using \(-5\) and \(1.5\) gives estimate \(-7.5\). Exactly, \(-\frac{14}{3}\cdot\frac{3}{2}=-7\), consistent with the estimate. 3. c) Using \(0.6\) and \(3\) gives estimate \(1.8\), about \(2\). Exactly, \(\frac{6}{10}\cdot\frac{10}{3}=2\), consistent with the estimate.

a) Benchmarks \(4\) and \(2\): estimate \(8\); exact \(8\frac{2}{5}\); consistent b) Benchmarks \(-5\) and \(1.5\): estimate \(-7.5\); exact \(-7\); consistent c) Benchmarks \(0.6\) and \(3\): estimate \(1.8\), about \(2\); exact \(2\); consistent

- Convert mixed numbers before looking for factors that cancel. - Show the reduced factors before carrying out the final multiplication. - Determine the product sign from the signs of the factors.

1. a) \(\frac{5}{8}\cdot\frac{6}{5}\cdot4\) cancels to \(\frac{1}{2}\cdot6=3\). 2. b) \(\frac{9}{4}\cdot\frac{4}{3}\cdot\frac{1}{2}\) cancels to \(3\cdot\frac{1}{2}=\frac{3}{2}\).

a) \(\frac{5}{8}\cdot\frac{6}{5}\cdot4\to\frac{1}{2}\cdot6=3\) b) \(\frac{9}{4}\cdot\frac{4}{3}\cdot\frac{1}{2}\to3\cdot\frac{1}{2}=\frac{3}{2}\)

- Record the rounded factors used in each estimate. - Make the comparison from the estimates before presenting exact products. - Then rewrite the mixed numbers as improper fractions for the exact check.

1. Estimate \(A\approx4.5\cdot1.3\approx5.9\) and \(B\approx3.3\cdot1.5\approx5.0\), so predict \(A>B\). 2. Exactly, \(A=\frac{9}{2}\cdot\frac{4}{3}=6\). 3. Exactly, \(B=\frac{10}{3}\cdot\frac{3}{2}=5\). 4. The exact comparison confirms the prediction.

Estimate: \(A\approx4.5\cdot1.3\approx5.9\), \(B\approx3.3\cdot1.5\approx5.0\), so predict \(A>B\). Exact: \(A=6\), \(B=5\). The prediction is confirmed.

- With several factors, you can look for common factors across all numerators and denominators before multiplying. - Count the negative factors first to determine the sign of the product. - Rewrite mixed numbers as improper fractions.

1. For a), rewrite the mixed numbers: \(\frac{6}{5}\cdot\frac{5}{3}\cdot\frac{5}{6}\). Cancel common factors to get \(\frac{5}{3}=1 \frac{2}{3}\). 2. For b), rewrite the mixed numbers: \(-\frac{12}{5}\cdot\frac{15}{8}\cdot\left(-\frac{1}{3}\right)\). Two negative factors give a positive product. Cancel common factors to get \(\frac{3}{2}=1 \frac{1}{2}\).

a) \(1 \frac{2}{3}\) b) \(1 \frac{1}{2}\)

- Calculate each expression separately. - Look for reciprocal factors that simplify to \(1\). - After finding the values, order the letters that represent them.

1. \(A=\frac{7}{4}\cdot\frac{8}{7}=2\). 2. \(B=\frac{8}{3}\cdot\frac{3}{8}=1\). 3. \(C=-\frac{6}{5}\cdot\left(-\frac{5}{2}\right)=3\). 4. Since \(1<2<3\), the order is \(B<A<C\).

\(B<A<C\), with values \(1<2<3\).

- Rewrite mixed numbers as improper fractions before looking for cancellation. - In products, a factor in a numerator may cancel with a factor in any denominator. - For the comparison, put both simplified products in comparable exact forms.

1. a) \(\frac{8}{5}\cdot\left(-\frac{5}{2}\right)\to-\frac{8}{2}=-4\). 2. b) \(-\frac{15}{4}\cdot\left(-\frac{8}{15}\right)\to\frac{8}{4}=2\). 3. c) \(\frac{7}{9}\cdot\frac{9}{7}\cdot(-3)=-3\). 4. d) The products simplify to \(\frac{3}{2}\) and \(\frac{5}{4}\), so the first is greater.

a) \(\frac{8}{5}\cdot\left(-\frac{5}{2}\right)\to-4\) b) \(-\frac{15}{4}\cdot\left(-\frac{8}{15}\right)\to2\) c) \(\frac{7}{9}\cdot\frac{9}{7}\cdot(-3)=-3\) d) \(\frac{3}{2}>\frac{5}{4}\), so \(2\frac{1}{4}\cdot\frac{2}{3}\) is greater.

- Rewrite each division as a fraction before evaluating. - Dividing by a fraction is the same as multiplying by its reciprocal. - For repeated division, work from left to right unless parentheses say otherwise. - Cancel common factors before multiplying. - Keep track of the negative sign in part b).

1. For a), \(7\div10=\frac{7}{10}\), so the complex fraction is \(\frac{\frac{14}{25}}{\frac{7}{10}}\). Then \(\frac{14}{25}\cdot\frac{10}{7}=\frac{4}{5}\). 2. For b), \(-18\div35=-\frac{18}{35}\), so the complex fraction is \(\frac{-\frac{18}{35}}{\frac{9}{14}}\). Then \(-\frac{18}{35}\cdot\frac{14}{9}=-\frac{4}{5}\). 3. For c), division is evaluated from left to right. The nested complex fraction is \(\frac{\frac{\frac{21}{16}}{\frac{7}{8}}}{3}\). First, \(\frac{21}{16}\div\frac{7}{8}=\frac{3}{2}\). Then \(\frac{3}{2}\div3=\frac{1}{2}\).

a) \(\frac{\frac{14}{25}}{\frac{7}{10}}=\frac{4}{5}\) b) \(\frac{-\frac{18}{35}}{\frac{9}{14}}=-\frac{4}{5}\) c) \(\frac{\frac{\frac{21}{16}}{\frac{7}{8}}}{3}=\frac{1}{2}\)

- Convert mixed numbers to improper fractions. - Identify whether \(x\) is a factor, dividend, or divisor. - Evaluate powers and complex fractions before solving.

1. For a), \(x=-\frac{10}{9}\div\frac{8}{3}=-\frac{10}{9}\cdot\frac{3}{8}=-\frac{5}{12}\). 2. For b), \(x=\frac{9}{2}\div\frac{3}{4}=6\). 3. For c), \(\left(-\frac{1}{2}\right)^3=-\frac{1}{8}\). Thus \(x=(-16)\cdot\left(-\frac{1}{8}\right)=2\). 4. For d), \(\frac{2}{5}\div\frac{4}{15}=\frac{3}{2}\). Then \(\frac{3}{2}\div x=\frac{5}{2}\), so \(x=\frac{3}{2}\div\frac{5}{2}=\frac{3}{5}\).

a) \(x=-\frac{5}{12}\) b) \(x=6\) c) \(x=2\) d) \(x=\frac{3}{5}\)

- Simplify the side that does not contain \(x\) first. - Treat the complex fraction as division. - Use the inverse operation to isolate \(x\).

1. Simplify the right side: \(\left(\frac{1}{2}\right)^2=\frac{1}{4}\), and \(\frac{1}{4}\div\frac{5}{8}=\frac{1}{4}\cdot\frac{8}{5}=\frac{2}{5}\). 2. Multiply both sides by \(-2\frac{1}{2}=-\frac{5}{2}\): \(x=\frac{2}{5}\cdot\left(-\frac{5}{2}\right)=-1\).

\(x=-1\)

- Treat the two sides of the subtraction as separate blocks. - Divide by a fraction by multiplying by its reciprocal. - Keep track of signs when subtracting rational numbers.

1. First parentheses: \(\frac{5}{8}-\frac{3}{2}=-\frac{7}{8}\). 2. Divide: \(-\frac{7}{8}\div\frac{7}{4}=-\frac{7}{8}\cdot\frac{4}{7}=-\frac{1}{2}\). 3. Second parentheses: \(2\div\frac{4}{3}=2\cdot\frac{3}{4}=\frac{3}{2}\). 4. Subtract: \(-\frac{1}{2}-\frac{3}{2}=-2\).

\(-2\)

- Look for familiar square numbers. - Can you rewrite two factors as numbers equally far from the same multiple of ten? - Break a large dividend into multiples of the divisor that are easy to divide mentally. - Decide the sign before using a known square.

1. For a), recognize that \(14^2=196\), so \(196\div14=14\). 2. For b), use \((20-2)(20+2)=20^2-2^2=400-4=396\). 3. For c), decompose \(1001=700+280+21\). Then \(1001\div7=100+40+3=143\). 4. For d), the product is negative and \(13^2=169\), so \(-13\cdot13=-169\).

a) \(14\) b) \(396\) c) \(143\) d) \(-169\)

- Look for factors that can make \(100\) or a divisor that can be split into simpler factors. - Decompose a factor or dividend rather than using a standard algorithm. - Show the equivalent expression that supports your mental method.

1. a) \((25\cdot4)\cdot7=700\). 2. b) \((126\div9)\div2=7\). 3. c) \(-15(10+2)=-180\). 4. d) \(-(70\div7+35\div7)=-15\).

a) \((25\cdot4)\cdot7=700\) b) \((126\div9)\div2=7\) c) \(-15(10+2)=-180\) d) \(-(70\div7+35\div7)=-15\)

- Use estimation or multiplication as an inverse operation to check a quotient. - Pay close attention to decimal place value. - Decide the sign before checking a quotient involving a negative number.

1. For a), \(4.5\div9=0.5\), so the equation is correct. 2. For b), \(1.44\div12=0.12\), so \(1.2\) is incorrect. 3. For c), \(-3.6\div4=-0.9\), so the equation is correct. 4. For d), \(0.07\div2=0.035\), so \(0.35\) is incorrect.

a) Correct b) Incorrect; \(0.12\) c) Correct d) Incorrect; \(0.035\)

- Choose an exact representation that creates common factors across the product. - Keep the conversion visible in the response. - Simplify before multiplying when possible.

1. a) \(0.25=\frac{1}{4}\), so \(\frac{1}{4}\cdot\frac{4}{7}=\frac{1}{7}\). 2. b) \(0.8=\frac{4}{5}\), so \(\frac{4}{5}\cdot\frac{3}{4}=\frac{3}{5}\). 3. c) \(1.25=\frac{5}{4}\) and \(0.4=\frac{2}{5}\), so the product is \(\frac{1}{2}\). 4. d) \(0.9=\frac{9}{10}\), so \(\frac{2}{3}\cdot\frac{9}{10}=\frac{3}{5}\).

a) \(0.25=\frac{1}{4}\); result \(\frac{1}{7}\) b) \(0.8=\frac{4}{5}\); result \(\frac{3}{5}\) c) \(1.25=\frac{5}{4}\), \(0.4=\frac{2}{5}\); result \(\frac{1}{2}\) d) \(0.9=\frac{9}{10}\); result \(\frac{3}{5}\)

- Decide whether estimation is enough before calculating exactly. - What happens when a positive number less than \(1\) is multiplied by another positive number less than \(1\)? - For negative numbers, which value lies farther to the right on a number line?

1. For a), \(2.5\div5=0.5\) and \(2.5\cdot0.2=0.5\), so the values are equal. 2. For b), \(-3.6\div6=-0.6\) and \(-3.6\div4=-0.9\). Since \(-0.6>-0.9\), use \(>\). 3. For c), \(0.8\cdot0.8=0.64\), and \(0.64<0.8\), so use \(<\).

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

- Convert the decimals and percent to fractions. - Use a common denominator for the subtraction. - To divide by a fraction, multiply by its reciprocal. - Simplify common factors before multiplying.

1. For a), write \(0.5 = \frac{1}{2}\). Then \(\frac{5}{6} - \frac{1}{2} = \frac{5}{6} - \frac{3}{6} = \frac{1}{3}\). 2. For b), write \(1.4 = \frac{7}{5}\). Then \(\frac{7}{5} \cdot \frac{5}{7} = 1\). 3. For c), write \(25\% = \frac{1}{4}\) and \(3\frac{1}{5} = \frac{16}{5}\). Then \(\frac{1}{4} \cdot \frac{16}{5} = \frac{4}{5}\). 4. For d), write \(0.3 = \frac{3}{10}\). Then \(\frac{9}{10} \div \frac{3}{10} = \frac{9}{10} \cdot \frac{10}{3} = 3\).

a) \(\frac{1}{3}\) b) \(1\) c) \(\frac{4}{5}\), or \(0.8\) d) \(3\)

- Look for exact rewrites that expose cancellation or familiar decimal facts. - Show the chosen rewrite before carrying out the operation. - Keep sign reasoning separate from magnitude simplification.

1. a) Write \(0.5=\frac{1}{2}\) and regroup to simplify: \(\left(\frac{12}{25}\cdot\frac{5}{3}\right)\cdot\frac{1}{2}=\frac{2}{5}\). 2. b) Write \(3.6=\frac{18}{5}\), then \(\frac{18}{5}\cdot\frac{10}{9}=4\). 3. c) Cancel common factors in \(-\frac{3}{4}\cdot\frac{16}{21}\) to get \(-\frac{4}{7}\). 4. d) Write \(80\%=0.8\), then \(0.8\div0.4=2\).

a) \(0.5=\frac{1}{2}\); \(\left(\frac{12}{25}\cdot\frac{5}{3}\right)\cdot\frac{1}{2}=\frac{2}{5}\) b) \(3.6=\frac{18}{5}\); result \(4\) c) Cancel in \(-\frac{3}{4}\cdot\frac{16}{21}\); result \(-\frac{4}{7}\) d) \(80\%=0.8\); result \(2\)

- Compare the amount of work created by converting each value to the other representation. - A conversion that creates a repeating decimal may be less convenient than keeping an exact fraction. - For division, consider whether an equivalent quotient can make the divisor easier to use.

1. For a), fractions are efficient because \(\frac{5}{6}\) is a repeating decimal. Write \(0.4 = \frac{2}{5}\). Then \(\frac{2}{5} \cdot \frac{5}{6} = \frac{1}{3}\). 2. For b), fractions are efficient because \(\frac{1}{3}\) is a repeating decimal. Write \(0.9 = \frac{9}{10}\). Then \(\frac{1}{3} \cdot \frac{9}{10} = \frac{3}{10}\). 3. For c), decimals are efficient because both numbers are terminating decimals. Multiply both numbers by \(10\): \(1.25 \div 0.5 = 12.5 \div 5 = 2.5\).

a) \(\frac{1}{3}\). Fractions are especially efficient because converting \(\frac{5}{6}\) to a decimal would produce a repeating decimal; \(0.4=\frac{2}{5}\) then simplifies cleanly with \(\frac{5}{6}\). b) \(\frac{3}{10}=0.3\). Fractions are especially efficient because \(\frac{1}{3}\) is repeating as a decimal, while \(0.9=\frac{9}{10}\) multiplies cleanly as a fraction. c) \(2.5=\frac{5}{2}\). Decimals are efficient because scaling both terminating decimals by \(10\) gives \(12.5\div5\). Fractions are also efficient here: \(\frac{5}{4}\div\frac{1}{2}=\frac{5}{2}\). Other exact choices are acceptable when the explanation accurately compares the work created by the chosen representations.

- Evaluate the expression inside the parentheses first. - A fraction form may be more useful when a fraction has a repeating decimal representation. - Break the calculation into separate multiplication and division steps.

1. Evaluate the expression inside the parentheses first. Write \(0.6 = \frac{3}{5}\). 2. Multiply: \(\frac{3}{5} \cdot \frac{5}{9} = \frac{1}{3}\). 3. Write \(1.2 = \frac{6}{5}\). 4. Divide by multiplying by the reciprocal: \(\frac{1}{3} \div \frac{6}{5} = \frac{1}{3} \cdot \frac{5}{6} = \frac{5}{18}\).

\(\frac{5}{18}\)

- Focus on how the value of a factor changes when one numerator or denominator changes. - Express that change as a multiplier applied to \(P\). - Combine the two factor changes in part c) before applying them to \(P\).

1. a) Doubling a numerator doubles that factor, so the new product is \(2P=\frac{3}{5}\). 2. b) Doubling a denominator halves that factor, so the new product is \(\frac{1}{2}P=\frac{3}{20}\). 3. c) Doubling the numerator multiplies the factor by \(2\), and halving its denominator multiplies it by another \(2\). The new product is \(4P=\frac{6}{5}\).

a) Factor change \(\times2\); new product \(2P=\frac{3}{5}\) b) Factor change \(\times\frac{1}{2}\); new product \(\frac{1}{2}P=\frac{3}{20}\) c) Factor change \(\times4\); new product \(4P=\frac{6}{5}\)

- Write the power as repeated multiplication. - How many times is the base used as a factor? - Track the decimal places when multiplying decimals. - What sign should an odd power of a negative number have?

1. Expand the power: \((-0.5)^3=(-0.5)\cdot(-0.5)\cdot(-0.5)\). 2. Multiply: \((-0.5)\cdot(-0.5)=0.25\), and \(0.25\cdot(-0.5)=-0.125\). Therefore, B is correct. 3. A results from multiplying the base by the exponent instead of using repeated multiplication. 4. C has the wrong sign; an odd power of a negative number is negative. D has a decimal place-value error.

B is correct. A: The exponent was treated as a factor. C: The sign is incorrect. D: The decimal place value is incorrect.

- Numbers whose product is \(1\) are reciprocals. - Determine the sign of the sum of two negative numbers. - Use the sign rule for division.

1. Write \(-2\frac{1}{2}=-\frac{5}{2}\). The other factor must be its reciprocal, \(-\frac{2}{5}\). 2. Their sum is \(-2.5+(-0.4)=-2.9\), which is negative. 3. Both numbers are negative, so their quotient is positive.

a) \(-\frac{2}{5}\), or \(-0.4\) b) \(-2.9\); negative c) Positive, because both numbers are negative.

- If both factors are written as sums, identify every partial product required by the distributive property. - Convert the mixed number and decimal to fractions. - Simplify common factors before multiplying.

1. The factors were multiplied part by part, which omits the cross products required when multiplying two sums. The work also loses the negative sign on the fractional part of \(-2\frac{2}{3}\). Converting both factors to fractions is more efficient here. 2. Write \(-2\frac{2}{3} = -\frac{8}{3}\) and \(1.5 = \frac{3}{2}\). 3. Multiply and simplify: \(-\frac{8}{3} \cdot \frac{3}{2} = -4\).

The calculation omits the cross products required by the distributive property and loses the negative sign on the fractional part of the mixed number. The correct result is \(-2\frac{2}{3} \cdot 1.5 = -\frac{8}{3} \cdot \frac{3}{2} = -4\).

- Look for an exact fraction form that shares factors with the other rational number. - A repeating decimal should remain exact rather than being truncated. - Show the conversion you use before the operation.

1. a) \(-0.12=-\frac{3}{25}\), giving \(-\frac{1}{10}\). 2. b) \(0.375=\frac{3}{8}\), giving \(\frac{3}{8}\cdot\frac{4}{3}=\frac{1}{2}\). 3. c) \(-1.2=-\frac{6}{5}\), giving \(-\frac{2}{3}\). 4. d) \(0.\overline{6}=\frac{2}{3}\), giving \(4\cdot\frac{3}{2}=6\).

a) \(-0.12=-\frac{3}{25}\); result \(-\frac{1}{10}\) b) \(0.375=\frac{3}{8}\); result \(\frac{1}{2}\) c) \(-1.2=-\frac{6}{5}\); result \(-\frac{2}{3}\) d) \(0.\overline{6}=\frac{2}{3}\); result \(6\)

- Apply multiplication and division before subtraction. - Convert the mixed number to a decimal or improper fraction. - Determine the sign of each product or quotient. - Subtracting a negative number is equivalent to adding its opposite.

1. Convert the mixed number: \(1\frac{1}{5} = 1.2\). 2. Multiply: \(1.2 \cdot (-0.5) = -0.6\). 3. Divide: \(2.4 \div (-3) = -0.8\). 4. Subtract: \(-0.6 - (-0.8) = -0.6 + 0.8 = 0.2\).

\(0.2\)

- Convert a repeating decimal to a fraction if useful. - Consider what kind of number results from multiplying integers. - Determine the sign from the number of negative factors.

1. For a), possible. For example, \(0.\overline{3}\cdot1.8\cdot1=0.6\). 2. For b), not possible. A product of integers is an integer, but \(0.6\) is not an integer. 3. For c), not possible. Exactly one negative factor would make the product negative.

a) Yes; for example, \(0.\overline{3}\cdot1.8\cdot1=0.6\). b) No; a product of integers must be an integer. c) No; the product would be negative.

- Find the factor shared by both terms in each expression. - Factor it out using the distributive property in reverse. - Combine the numbers inside the parentheses before multiplying.

1. For a), factor out \(\frac{4}{7}\): \(\frac{4}{7}(-13.2 - 0.8) = \frac{4}{7}(-14) = -8\). 2. For b), factor out \(\frac{2}{3}\): \(\frac{2}{3}(-15.5 + 6.5) = \frac{2}{3}(-9) = -6\).

a) \(-8\) b) \(-6\)

- Multiply the outside factor by every term inside the parentheses. - Keep each operation sign from inside the parentheses. - Each product simplifies to a whole number.

1. Distribute \(12\) to each term: \(12 \cdot \frac{3}{4} - 12 \cdot 0.5 + 12 \cdot \frac{1}{6}\). 2. Evaluate the products: \(12 \cdot \frac{3}{4} = 9\), \(12 \cdot 0.5 = 6\), and \(12 \cdot \frac{1}{6} = 2\). 3. Combine the results: \(9 - 6 + 2 = 5\).

\(5\)

- Search for pairs whose product is \(1\). - Keep track of which property permits reordering and which permits regrouping. - Show the grouped product in the answer.

1. Regroup as \(\left(\frac{5}{9}\cdot1.8\right)\left((-0.4)(-2.5)\right)\). 2. Each group equals \(1\), so the product is \(1\). 3. The commutative property changes order and the associative property changes grouping.

\(\left(\frac{5}{9}\cdot1.8\right)\left((-0.4)(-2.5)\right)=1\cdot1=1\). Properties: commutative and associative properties of multiplication.

- Look for reciprocal factors or factors that make a power of \(10\). - You may change order and grouping but not the factors themselves. - Include the reordered/regrouped form in each answer.

1. a) \(\left(\frac{2}{9}\cdot\frac{9}{2}\right)\left(\frac{5}{11}\cdot22\right)=1\cdot10=10\). 2. b) \((12.5\cdot8)\cdot0.7=100\cdot0.7=70\).

a) \(\left(\frac{2}{9}\cdot\frac{9}{2}\right)\left(\frac{5}{11}\cdot22\right)=10\) b) \((12.5\cdot8)\cdot0.7=70\)

- A fraction bar acts like division and groups the numerator and denominator. - Which operation is performed last in the entire expression? - Evaluate the numerator and denominator separately. - Use sign rules when dividing rational numbers.

1. The overall expression is a quotient. Its numerator is the sum of \(-3.6\) and \(1.2\), and its denominator is the quotient of \(-3.6\) and \(1.2\). 2. Numerator: \(-3.6+1.2=-2.4\). 3. Denominator: \(-3.6\div1.2=-3\). 4. Final division: \(-2.4\div(-3)=0.8\).

a) The expression is the quotient of the sum \(-3.6+1.2\) and the quotient \(-3.6\div1.2\). b) \(T=0.8\)

- What happens when \(-1\) is raised to an odd power? - Does an even or odd exponent change the sign? - Does changing the order of factors change a product?

1. Since \(3\) is odd, \((-1)^3=-1\). Therefore, Expression A is \(-1\cdot4.2=-4.2\). 2. Since \(7\) is odd, \((-1)^7=-1\). Therefore, Expression B is \(4.2\cdot(-1)=-4.2\). 3. Both expressions have the same value.

Yes. Both expressions equal \(-4.2\) because both odd powers of \(-1\) equal \(-1\), and multiplication can be written in either factor order.

- In a), rearrange the factors to pair the two decimals. - Convert the decimals to fractions to check whether their product is \(1\). - In b), compare dividing by twice a number with dividing by the number and then doubling.

1. For a), rearrange and regroup the factors: \((0.8 \cdot 1.25)\frac{3}{7}\). Since \(0.8 = \frac{4}{5}\) and \(1.25 = \frac{5}{4}\), their product is \(1\). The equation is true. 2. For b), dividing by a product is equivalent to dividing successively by both factors: \(a \div (bc) = a \div b \div c\). The right side divides by \(\frac{1}{3}\) and then multiplies by \(2\), so the equation is false.

a) True, because \(0.8 \cdot 1.25 = 1\). b) False, because dividing by \(\frac{1}{3} \cdot 2\) is not the same as dividing by \(\frac{1}{3}\) and then multiplying by \(2\).

- For multiplication only, you can regroup factors to make the arithmetic easier. - Determine the sign as you work through each expression. - When multiplication and division are mixed, work from left to right.

1. For a), regroup the factors: \((-4)\cdot(-8)=32\), then \(32\cdot125=4000\). 2. For b), work from left to right: \((-3600)\div(-15)=240\), then \(240\div(-4)=-60\). 3. For c), \(12\cdot(-15)=-180\), then \(-180\div(-9)=20\), and \(20\cdot(-2)=-40\).

a) \(4000\) b) \(-60\) c) \(-40\)

- Evaluate the parentheses first. - Use a common denominator for the subtraction. - Simplify a fraction before the next step when possible.

1. Subtract inside the parentheses: \(\frac{5}{6}-\frac{1}{2}=\frac{5}{6}-\frac{3}{6}=\frac{2}{6}=\frac{1}{3}\). 2. Multiply: \(\frac{3}{4}\cdot\frac{1}{3}=\frac{1}{4}\).

\(\frac{1}{4}\)

- In a), both terms are divided by the same number. - In b), all three terms share the same factor. - Factor first, then evaluate the expression inside the parentheses.

1. For a), view division by \(9\) as multiplication by \(\frac{1}{9}\) and factor: \((720 - 180) \div 9 = 540 \div 9 = 60\). 2. For b), factor out \(1.2\): \(1.2(15 + 5 - 10) = 1.2 \cdot 10 = 12\).

a) \(60\) b) \(12\)

- Would fractions or decimals be easier for this expression? - Use one form consistently before calculating. - Determine the sign of the product before multiplying.

1. Write the quantities in a common form: \(1 \frac{1}{5}=1.2\) and \(\frac{7}{10}=0.7\). 2. Evaluate the parentheses: \(0.2-0.7=-0.5\). 3. Multiply: \(1.2\cdot(-0.5)=-0.6=-\frac{3}{5}\).

\(-0.6\), or \(-\frac{3}{5}\)

- Evaluate each set of parentheses separately. - How do you divide by a fraction? - What sign results when a negative number is divided by a negative number?

1. First parentheses: \(\frac{5}{6}-\frac{4}{3}=\frac{5}{6}-\frac{8}{6}=-\frac{1}{2}\). 2. Second parentheses: \(0.75-1=-0.25=-\frac{1}{4}\). 3. Divide: \(-\frac{1}{2}\div\left(-\frac{1}{4}\right)=-\frac{1}{2}\cdot(-4)=2\).

\(2\)

- What order of operations applies when there are no parentheses? - How do parentheses change which quantity is the divisor? - Evaluate each expression step by step.

1. For \(A\), evaluate the parentheses first: \(1.2+0.9=2.1\). Then \(8.4\div2.1=4\). 2. For \(B\), division comes before addition: \(8.4\div1.2=7\), then \(7+0.9=7.9\). 3. The parentheses in \(A\) make the entire sum the divisor, while in \(B\) only \(1.2\) is the divisor.

\(A=4\); \(B=7.9\). The parentheses change the structure and order of operations.

- Converting the fraction to a decimal may simplify the first subtraction. - Pay attention to the sign when dividing. - Evaluate both sets of parentheses first.

1. First parentheses: \(1.2-\frac{3}{4}=1.2-0.75=0.45\). 2. Second parentheses: \(0.2-0.5=-0.3\). 3. Divide: \(0.45\div(-0.3)=-1.5=-\frac{3}{2}\).

\(-1.5\), or \(-\frac{3}{2}\)

- Treat the numerator and denominator as separate expressions first. - Convert between fractions and decimals when that makes a step easier. - A fraction bar means divide the completed numerator by the completed denominator.

1. Numerator: \(\frac{1}{5}-0.6=0.2-0.6=-0.4\), so \(2.5\cdot(-0.4)=-1\). 2. Denominator: \(0.25\cdot8=2\), then \(2-1.5=0.5\). 3. Divide numerator by denominator: \(-1\div0.5=-2\).

\(-2\)

- Follow the grouping and work from left to right. - Determine the sign at each division step. - What sign results when dividing two negative numbers?

1. First divide: \(-2160\div18=-120\). 2. Then divide again: \(-120\div(-3)=40\).

\(40\)

- Count negative factors to determine the sign of a product. - You may regroup factors in a product to make the arithmetic easier. - What sign results when a negative number in parentheses is squared?

1. For a), four negative factors give a positive product: \(1\cdot2\cdot3\cdot4=24\). 2. For b), \(-\frac{1}{2}\cdot8=-4\), then \((-4)\cdot(-0.25)=1\). 3. For c), \((-0.1)^2=0.01\), then \(0.01\cdot(-1000)=-10\). 4. For d), \((-5)\cdot\left(-\frac{2}{5}\right)=2\), then \(2\cdot(-3)=-6\).

a) \(24\) b) \(1\) c) \(-10\) d) \(-6\)

- Determine the sign before calculating the magnitude. - Simplify fractions with whole-number factors when possible. - Expand a power such as \((-1)^4\) as repeated multiplication if needed. - Converting a simple decimal to a fraction may help.

1. For a), \((-12)\cdot\frac{3}{4}=-9\), then \((-9)\cdot(-2)=18\). 2. For b), \(-\frac{2}{5}\cdot(-10)=4\), then \(4\cdot(-0.5)=-2\). 3. For c), \((-1)^4=1\), so \(1\cdot7\cdot(-3)=-21\). 4. For d), \(\frac{4}{9}\cdot(-18)=-8\), then \((-8)\cdot\left(-\frac{1}{2}\right)=4\).

a) \(18\) b) \(-2\) c) \(-21\) d) \(4\)

- Evaluate each expression separately. - What sign results when a negative number is squared? - Among negative numbers, which one lies farther to the right on a number line?

1. For \(X\), \((-4)^2=16\), so \(X=16\cdot(-0.5)=-8\). 2. For \(Y\), \((-2)\cdot(-2)=4\), then \(4\cdot(-2.5)=-10\). 3. Since \(-8>-10\), \(X\) is greater.

\(X\) is greater because \(-8>-10\).

- Work from the bottom row upward. - Each missing block depends only on the two blocks directly below it. - Pay attention to the sign of each product.

1. For the left middle block, multiply the two blocks below it: \((-1.2)\cdot(-5)=6\). 2. For the right middle block, \((-5)\cdot0.25=-1.25\). 3. For the top block, multiply the two middle values: \(6\cdot(-1.25)=-7.5\).

Middle row: \(6\) and \(-1.25\) Top block: \(-7.5\)

- Order the numbers before identifying the least and greatest. - Look for a decimal whose magnitude is the reciprocal of an integer in the set. - A positive product comes from factors with the same sign.

1. The least number is \(-8\), and the greatest is \(4\). Their product is \(-32\). 2. Since \(0.125=\frac{1}{8}\), \((-8)\cdot0.125=-1\). 3. The possible positive products from equal signs are \((-8)\cdot(-0.5)=4\) and \(0.125\cdot4=0.5\). The greater product is \(4\).

a) \(-32\) b) \(-8\) and \(0.125\) c) \(-8\) and \(-0.5\); the product is \(4\).

- Check whether each side uses addition or multiplication. - Determine the sign of a product before comparing values. - Evaluate both sides before choosing the comparison symbol.

1. For a), \(0.2\cdot0.3=0.06\) and \(0.2+0.3=0.5\), so \(0.06<0.5\). 2. For b), \(1.2\cdot0.5=0.6\) and \(1.2\cdot1.5=1.8\), so \(0.6<1.8\). 3. For c), \((-0.4)\cdot0.5=-0.2\) and \((-0.4)\cdot(-0.5)=0.2\), so \(-0.2<0.2\). 4. For d), \(0.25\cdot4=1\) and \(0.5\cdot2=1\), so the values are equal.

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

- The sign depends on whether the number of negative factors is even or odd. - In c), represent the number of negative factors with a variable. - Use the total of \(15\) factors to determine how many are negative.

1. A product with an odd number of negative factors is negative; a product with an even number of negative factors is positive. 2. For a), \(7\) is odd, so the product is negative. 3. For b), all \(15\) factors are negative. Since \(15\) is odd, the product is negative. 4. For c), let \(n\) be the number of negative factors. Then there are \(2n\) positive factors, so \(n + 2n = 15\). Thus, \(n = 5\). Since \(5\) is odd, the product is negative.

a) Negative, because \(7\) is odd. b) Negative, because \(15\) is odd. c) Negative, because there are \(5\) negative factors and \(10\) positive factors.

- Count the negative factors and decide whether that count is even or odd. - In b), subtract the number of positive factors from \(12\). - In c), determine how changing one sign affects the number of negative factors.

1. For a), yes. With \(12\) negative factors, the number of negative factors is even, so the product is positive. 2. For b), the other \(12 - 3 = 9\) factors are negative. Since \(9\) is odd, the product is negative. 3. For c), changing the sign of one factor changes the number of negative factors by \(1\). Its parity switches from even to odd or from odd to even, so the sign of the product always reverses. Jonah is correct.

a) Yes; the product would be positive. b) Negative, because there are \(9\) negative factors. c) Jonah is correct. Reversing one factor’s sign changes the parity of the number of negative factors, so the product’s sign reverses.

- Count the negative factors in each product. - Use the sign rule for multiplying a positive number by a negative number. - A product of negative factors depends on whether their count is even or odd.

1. Product \(A\) has \(10\) negative factors. Since \(10\) is even, \(A\) is positive. Product \(B\) has \(11\) negative factors, so \(B\) is negative. 2. A positive number times a negative number is negative, so \(A \cdot B\) is negative. 3. Equivalently, \(A \cdot B\) contains \(10 + 11 = 21\) negative factors, and \(21\) is odd. 4. A product of \(n\) negative numbers is negative exactly when \(n\) is odd.

a) \(A\) is positive. b) \(A \cdot B\) is negative. c) The product is negative when \(n\) is odd.

- Multiply the outside factor by every term inside the parentheses. - Keep each addition or subtraction sign. - Simplify products by canceling common factors when possible.

1. For a), distribute: \(\frac{4}{5} \cdot 10 - \frac{4}{5} \cdot 2.5 = 8 - 2 = 6\). 2. For b), distribute: \(18 \cdot \frac{1}{2} + 18 \cdot \frac{1}{3} - 18 \cdot \frac{1}{6} = 9 + 6 - 3 = 12\). 3. For c), distribute: \(\frac{7}{11} \cdot 22 - \frac{7}{11} \cdot 5.5 = 14 - 3.5 = 10.5\).

a) \(6\) b) \(12\) c) \(10.5\)

- Identify the factor shared by every term in each expression. - Factor first, then evaluate the sum or difference inside the parentheses. - Keep negative signs attached to their factors.

1. For a), factor out \(27\): \(27(13 + 87) = 27 \cdot 100 = 2700\). 2. For b), factor out \(0.4\): \(0.4(35 - 15) = 0.4 \cdot 20 = 8\). 3. For c), factor out \(\frac{3}{8}\): \(\frac{3}{8}(17 - 9) = \frac{3}{8} \cdot 8 = 3\). 4. For d), factor out \(0.25\): \(0.25(-12 - 8) = 0.25 \cdot (-20) = -5\).

a) \(2700\) b) \(8\) c) \(3\) d) \(-5\)

- Evaluate parentheses first. - How do you divide by a fraction? - Multiplication and division come before subtraction. - What sign results when multiplying two negative numbers?

1. For a), \(\frac{2}{5}-1=-\frac{3}{5}\) and \(-2 \frac{1}{2}=-\frac{5}{2}\). Then \(-\frac{3}{5}\cdot\left(-\frac{5}{2}\right)=\frac{3}{2}=1.5\). 2. For b), \(\frac{3}{8}\div\left(-\frac{1}{4}\right)=\frac{3}{8}\cdot(-4)=-1.5\). Also, \(0.5\cdot3=1.5\). Therefore, \(-1.5-1.5=-3\).

a) \(1.5\), or \(1 \frac{1}{2}\) b) \(-3\)

- Substitute each value for \(a\). - Divide before subtracting. - Dividing by \(\frac{1}{2}\) is equivalent to multiplying by \(2\). - Track the negative sign carefully in part b).

1. For \(a=\frac{1}{8}\), \(T(a)=\frac{3}{4}-\frac{1}{8}\div\frac{1}{2}\). 2. \(\frac{1}{8}\div\frac{1}{2}=\frac{1}{8}\cdot 2=\frac{1}{4}\), so \(T(a)=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}\). 3. For \(a=-1.5\), \(T(a)=\frac{3}{4}-(-1.5)\div\frac{1}{2}\). 4. \(-1.5\div 0.5=-3\), so \(T(a)=\frac{3}{4}-(-3)=3\frac{3}{4}\).

a) \(\frac{1}{2}\) b) \(3\frac{3}{4}\)

- In a), remove the parentheses and look for pairs that make whole numbers. - In b), cancel common factors before multiplying. - In c), form simple partial products and track the total number of negative factors.

1. For a), remove the parentheses: \(15.3 - 4.8 + 2.7 - 5.2\). Group compatible terms: \((15.3 + 2.7) - (4.8 + 5.2) = 18 - 10 = 8\). 2. For b), simplify before multiplying: \(\frac{7}{11} \cdot \frac{22}{14} = \frac{1}{2} \cdot 2 = 1\). Then \(1 - 1 = 0\). 3. For c), regroup: \((-8 \cdot 1.25)((-3)(-2)) = (-10)(6) = -60\).

a) \(8\) b) \(0\) c) \(-60\)

- Use inverse operations to find each missing number. - For division, ask which divisor produces the given quotient. - Pay close attention to the signs when rearranging each equation.

1. For a), divide by the known factor: \(1.5\div(-0.5)=-3\). 2. For b), let the missing number be \(x\). From \(\frac{2}{7}\div x=-\frac{4}{21}\), compute \(x=\frac{2}{7}\div\left(-\frac{4}{21}\right)=\frac{2}{7}\cdot\left(-\frac{21}{4}\right)=-\frac{3}{2}\). 3. For c), multiply the quotient by the divisor: \(\frac{9}{4}\cdot\left(-\frac{2}{3}\right)=-\frac{3}{2}\).

a) \(-3\) b) \(-\frac{3}{2}\) c) \(-\frac{3}{2}\)

- Find the reciprocal by switching the numerator and denominator. - Use the sign rules for multiplying and dividing rational numbers. - Compare the division in b) with multiplication by the reciprocal in c).

1. For a), \(-\frac{3}{4}\cdot\frac{2}{3}=-\frac{6}{12}=-\frac{1}{2}\). 2. For b), \(-\frac{3}{4}\div\frac{2}{3}=-\frac{3}{4}\cdot\frac{3}{2}=-\frac{9}{8}\). 3. For c), the reciprocal of \(\frac{2}{3}\) is \(\frac{3}{2}\). Then \(-\frac{3}{4}\cdot\frac{3}{2}=-\frac{9}{8}\). 4. The results in b) and c) are equal, confirming that dividing by a nonzero fraction is equivalent to multiplying by its reciprocal.

a) \(-\frac{1}{2}\) b) \(-\frac{9}{8}\) c) \(-\frac{9}{8}\); it matches the result from part b).

- How do the parentheses change the order of operations? - Compare the structures of the two expressions carefully. - Predict the sign of each result before calculating.

1. For \(A\), evaluate the parentheses first: \(2.4-5.4=-3\). Then \(-3\cdot(-3)=9\). 2. For \(B\), multiply first: \(-3\cdot2.4=-7.2\). Then \(-7.2-5.4=-12.6\). 3. Since \(9>-12.6\), expression \(A\) has the greater value.

Expression \(A\) has the greater value because \(9>-12.6\).

- Rewrite division by the fraction as multiplication by its reciprocal. - Think about how a positive factor less than \(1\) affects the magnitude of a negative product. - Compare the two negative results by their positions on a number line.

1. \(x=(-10)\cdot\frac{2}{5}=-4\). 2. \(y=(-10)\div\frac{2}{5}=(-10)\cdot\frac{5}{2}=-25\). 3. Since \(-4>-25\), \(x\) lies farther to the right.

\(x=-4\) and \(y=-25\). The value \(x\) lies farther to the right because \(-4>-25\).

- Convert the mixed number and decimal to fractions. - Evaluate parentheses first. - How do you divide two fractions? - Track signs carefully.

1. First parentheses: \(\frac{4}{3}-\frac{1}{2}=\frac{8}{6}-\frac{3}{6}=\frac{5}{6}\). 2. Second parentheses: \(\frac{1}{6}-1=-\frac{5}{6}\). 3. Divide: \(\frac{5}{6}\div\left(-\frac{5}{6}\right)=-1\). 4. Multiply: \(-1\cdot\frac{3}{5}=-\frac{3}{5}\).

\(-\frac{3}{5}\), or \(-0.6\)

- Replace division by a fraction with multiplication by its reciprocal. - Evaluate each expression separately for the given values. - For part b), rewrite \(T_1\) symbolically and compare its numerator and denominator with \(T_2\).

1. For a), \(T_1=\frac{2}{3}\div\frac{4}{5}=\frac{2}{3}\cdot\frac{5}{4}=\frac{5}{6}\). 2. Also, \(T_2=\frac{2\cdot5}{3\cdot4}=\frac{5}{6}\). 3. For b), dividing by \(\frac{c}{d}\) is equivalent to multiplying by its reciprocal \(\frac{d}{c}\): \(\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\cdot\frac{d}{c}=\frac{ad}{bc}\). 4. This is exactly \(T_2\), so the expressions are equivalent whenever the stated nonzero conditions hold.

a) \(T_1=T_2=\frac{5}{6}\) b) \(\frac{a}{b}\div\frac{c}{d}=\frac{a}{b}\cdot\frac{d}{c}=\frac{ad}{bc}\), so the expressions are equivalent.

- Evaluate the difference inside the parentheses first. - The negative sign outside the parentheses means take the opposite of the value inside. - Then subtract \(65\).

1. Evaluate inside the parentheses: \(240-1240=-1000\). 2. Take the opposite of the parenthetical value: \(-(-1000)=1000\). 3. Subtract \(65\): \(1000-65=935\).

\(935\)

- Evaluate parentheses before the rest of the expression. - A subtraction sign before a grouped expression affects the entire value of that group. - Evaluate the original and modified expressions separately before comparing them.

1. In a), \(-15-5-12+8=-24\). 2. In b), evaluate the new parentheses first: \(12-(-8)=20\). 3. Then \(-15-5-20=-40\). 4. The value changed from \(-24\) to \(-40\), so it decreased by \(16\).

a) \(-24\) b) \(-40\); the value is \(16\) less than the original value.

- Identify the factor that appears in both products. - Factor that number out of the sum. - Add the remaining factors before multiplying.

1. Both products have a common factor of \(14\). 2. Factor out \(14\): \(14 \cdot (63 + 37)\). 3. Evaluate: \(14 \cdot 100 = 1400\).

\(14 \cdot 63 + 14 \cdot 37 = 14 \cdot (63 + 37) = 1400\)

- Count how many times each addend appears. - Write each repeated addend as a product. - Look for a common factor in the two products.

1. The number \(18\) appears five times, and \(12\) appears five times, so the expression is \(5 \cdot 18 + 5 \cdot 12\). 2. Factor out the common factor \(5\): \(5(18 + 12)\). 3. Evaluate: \(5 \cdot 30 = 150\).

\(5 \cdot 18 + 5 \cdot 12 = 5(18 + 12) = 150\)

- Look for factors that make \(10\), \(100\), or \(1000\). - Use place value to rewrite products that contain trailing zeros. - In the quotient, divide both numbers by the same power of \(10\).

1. For a), regroup the factors: \((4 \cdot 25) \cdot 19 = 100 \cdot 19 = 1900\). 2. For b), write the factors by place value: \(80 \cdot 600 = 8 \cdot 6 \cdot 1000 = 48{,}000\). 3. For c), divide the dividend and divisor by \(10\): \(54{,}000 \div 90 = 5400 \div 9 = 600\). 4. For d), regroup the factors: \((2 \cdot 50) \cdot 37 = 100 \cdot 37 = 3700\).

a) \(1900\) b) \(48{,}000\) c) \(600\) d) \(3700\)

- Find the scale factor caused by each change. - Multiply the scale factors when both factors change. - Test with simple factors if needed.

1. Multiplying one factor by \(4\) multiplies the product by \(4\). 2. The combined scale factor is \(10 \div 2 = 5\), so the product is multiplied by \(5\). 3. The combined scale factor is \(3 \cdot 3 = 9\), so the product is multiplied by \(9\).

a) The product is multiplied by \(4\). b) The product is multiplied by \(5\). c) The product is multiplied by \(9\).

- Compare each factor with the corresponding factor in \(12 \cdot 15\). - Identify whether each factor was multiplied or divided by a scale factor. - Combine the effects when both factors change.

1. The first factor is doubled, so the product doubles: \(180 \cdot 2 = 360\). 2. The second factor is multiplied by \(3\), so the product is multiplied by \(3\): \(180 \cdot 3 = 540\). 3. The first factor is halved and the second is doubled. The scale factors cancel, so the product remains \(180\). 4. The first factor is divided by \(3\), so the product is divided by \(3\): \(180 \div 3 = 60\).

a) \(360\) b) \(540\) c) \(180\) d) \(60\)

- Generate several examples with different positive divisors. - For a chosen divisor, determine the dividend that gives quotient \(10\). - Decide whether there is a greatest positive integer.

1. Choose any positive integer \(n\) as the divisor. 2. Use \(10n\) as the dividend. Then \(10n \div n = 10\). 3. Because there are infinitely many positive integers \(n\), there are infinitely many such division equations.

There are infinitely many. For every positive integer \(n\), \(10n \div n = 10\).

- Evaluate the innermost parentheses first. - Keep the sign of the parenthetical result when adding it inside the brackets. - Finish with the subtraction outside the brackets.

1. Evaluate the inner parentheses: \(5-15=-10\). 2. Evaluate the brackets: \(20+(-10)=10\). 3. Complete the calculation: \(15-10=5\).

\(5\)

- Look for factors that make \(10\), \(100\), or \(1000\). - Reorder and regroup factors before multiplying. - Use powers of \(10\) to simplify the final multiplication.

1. For a), regroup: \((4 \cdot 5 \cdot 5) \cdot 19 = 100 \cdot 19 = 1900\). 2. For b), regroup: \((8 \cdot 125) \cdot 15 = 1000 \cdot 15 = 15{,}000\). 3. For c), regroup: \((2 \cdot 5) \cdot 10 \cdot 13 = 100 \cdot 13 = 1300\). 4. For d), regroup: \((250 \cdot 4) \cdot 7 = 1000 \cdot 7 = 7000\).

a) \(1900\) b) \(15{,}000\) c) \(1300\) d) \(7000\)

- When a factor is outside grouping symbols, distribute it to each term inside. - When two terms share a factor, factor it out. - Choose the form that makes the arithmetic easiest.

1. For a), expand: \(8(100 + 4) = 8 \cdot 100 + 8 \cdot 4 = 800 + 32 = 832\). 2. For b), factor out \(17\): \(17 \cdot 12 + 17 \cdot 8 = 17(12 + 8) = 17 \cdot 20 = 340\). 3. For c), expand: \((50 - 2) \cdot 6 = 50 \cdot 6 - 2 \cdot 6 = 300 - 12 = 288\). 4. For d), factor out \(15\): \(44 \cdot 15 - 34 \cdot 15 = (44 - 34)15 = 10 \cdot 15 = 150\).

a) \(832\); expanded b) \(340\); factored c) \(288\); expanded d) \(150\); factored

- Check whether each term inside the grouping symbols is divisible by the divisor. - Compare one larger division with two simpler divisions. - Keep the addition or subtraction operation when splitting the quotient.

1. For a), Method 1 gives \(456 \div 8 = 57\). Method 2 gives \(400 \div 8 + 56 \div 8 = 50 + 7 = 57\). 2. For b), Method 1 gives \(558 \div 6 = 93\). Method 2 gives \(600 \div 6 - 42 \div 6 = 100 - 7 = 93\). 3. Method 2 may be easier because each term inside the grouping symbols is a convenient multiple of the divisor.

a) \(57\). Method 2 is easier because both \(400\) and \(56\) are divisible by \(8\). b) \(93\). Method 2 is easier because both \(600\) and \(42\) are divisible by \(6\).

- In part a, identify the common factor. - Rewrite \(99\) using \(100\), and rewrite \(11\) using \(10\). - Use the form that creates the easiest products.

1. For a), factor out \(13\): \(13(57 + 43) = 13 \cdot 100 = 1300\). 2. For b), write \(99\) as \(100 - 1\): \((100 - 1)14 = 1400 - 14 = 1386\). 3. For c), write \(11\) as \(10 + 1\): \(62(10 + 1) = 620 + 62 = 682\).

a) \(1300\) b) \(1386\) c) \(682\)

- Split a dividend into parts that are each divisible by the divisor. - When two quotients have the same divisor, consider combining their dividends. - Preserve the addition or subtraction operation when rewriting.

1. For a), split the dividend: \(245 \div 5 = 200 \div 5 + 45 \div 5 = 40 + 9 = 49\). 2. For b), divide each term: \(144 \div 12 - 36 \div 12 = 12 - 3 = 9\). 3. For c), combine the dividends: \((132 + 88) \div 11 = 220 \div 11 = 20\).

a) \(49\) b) \(9\) c) \(20\)

- Translate each phrase “the product of” into multiplication with grouping symbols. - Look for factors that multiply to \(1000\). - Use the commutative and associative properties to reorder and regroup.

1. The expression is \((125 \cdot 9)(8 \cdot 3)\). 2. Reorder and regroup the factors: \((125 \cdot 8)(9 \cdot 3)\). 3. Evaluate: \(1000 \cdot 27 = 27{,}000\).

\((125 \cdot 9)(8 \cdot 3) = (125 \cdot 8)(9 \cdot 3) = 27{,}000\)

- Pay attention to which quotient is being subtracted from which. - Notice that both quotients have the same divisor. - Subtract the dividends before dividing.

1. The expression is \(1024 \div 8 - 224 \div 8\). 2. Since both quotients have the same divisor, combine the dividends: \((1024 - 224) \div 8\). 3. Evaluate: \(800 \div 8 = 100\).

\(1024 \div 8 - 224 \div 8 = (1024 - 224) \div 8 = 100\)

- Look for a factor close to a multiple of \(10\) or \(100\). - Rewrite that factor as a sum or difference. - Distribute the other factor to both terms.

1. For a), write \(102\) as \(100 + 2\): \(7(100 + 2) = 700 + 14 = 714\). 2. For b), write \(98\) as \(100 - 2\): \(9(100 - 2) = 900 - 18 = 882\). 3. For c), write \(21\) as \(20 + 1\): \(15(20 + 1) = 300 + 15 = 315\).

a) \(7(100 + 2) = 714\) b) \(9(100 - 2) = 882\) c) \(15(20 + 1) = 315\)

- Identify the factor shared by both terms. - Place the remaining factors inside grouping symbols. - Add or subtract inside the grouping symbols before multiplying.

1. For a), factor out \(8\): \((14 + 6)8 = 20 \cdot 8 = 160\). 2. For b), factor out \(12\): \((37 - 27)12 = 10 \cdot 12 = 120\). 3. For c), factor out \(5\): \(5(43 + 57) = 5 \cdot 100 = 500\).

a) \(8(14 + 6) = 160\) b) \(12(37 - 27) = 120\) c) \(5(43 + 57) = 500\)

- Compare each expanded expression with its factored form. - Identify the common factor or divisor. - In part d, determine which number added to \(8\) gives \(20\).

1. For a), factor out \(4\): \((36 + 14) \cdot 4\). Thus, \(\square = 14\) and \(\triangle = 4\). 2. For b), compare with \((25 - 15) \cdot 7\). Thus, \(\triangle = 25\). 3. For c), both quotients have divisor \(8\), so \(\square = 8\). 4. For d), the factors inside the implied sum must satisfy \(\square + 8 = 20\), so \(\square = 12\).

a) \(\square = 14\); \(\triangle = 4\) b) \(\triangle = 25\) c) \(\square = 8\) d) \(\square = 12\)

- Rewrite a factor as a convenient sum or difference. - In part c, identify the common factor. - Choose a rewrite that creates multiples of \(10\) or \(100\).

1. For a), write \(104\) as \(100 + 4\): \(7(100 + 4) = 700 + 28 = 728\). 2. For b), write \(9\) as \(10 - 1\): \(18(10 - 1) = 180 - 18 = 162\). 3. For c), factor out \(35\): \(35(16 + 4) = 35 \cdot 20 = 700\).

a) \(7(100 + 4) = 728\) b) \(18(10 - 1) = 162\) c) \(35(16 + 4) = 700\)

- Write \(14\) as both a sum and a product. - Apply the named property in each method. - Compare the number and difficulty of the steps.

1. Distributive method: \((10 + 4) \cdot 5 = 10 \cdot 5 + 4 \cdot 5 = 50 + 20 = 70\). 2. Associative method: \((7 \cdot 2) \cdot 5 = 7 \cdot (2 \cdot 5) = 7 \cdot 10 = 70\). 3. Both methods give \(70\). A reasonable comparison is that the second method is shorter because it creates the factor \(10\).

1) \((10 + 4) \cdot 5 = 70\) 2) \(7 \cdot (2 \cdot 5) = 70\) 3) The associative method is easier because it creates \(2 \cdot 5 = 10\) and uses fewer steps.

- Expand the sum in Strategy A. - Regroup the factors in Strategy B. - Match each change to the property that permits it.

1. Strategy A: \(25(10 + 2) = 250 + 50 = 300\). This uses the distributive property. 2. Strategy B: \(25(4 \cdot 3) = (25 \cdot 4) \cdot 3 = 100 \cdot 3 = 300\). This uses the associative property.

a) Strategy A: \(25(10 + 2) = 300\). Strategy B: \((25 \cdot 4) \cdot 3 = 300\). b) Strategy A uses the distributive property; Strategy B uses the associative property.

- A factor outside grouping symbols must multiply every term inside. - Identify the missing product in Sarah’s expansion. - Add the two correct partial products.

1. Sarah multiplied \(18\) by \(100\) but did not multiply \(18\) by \(2\). 2. The correct expansion is \(18(100 + 2) = 18 \cdot 100 + 18 \cdot 2\). 3. Therefore, the product is \(1800 + 36 = 1836\).

a) Sarah failed to distribute \(18\) to the second term. b) \(18(100 + 2) = 1800 + 36 = 1836\)

- Match each term inside the grouping symbols with its product outside. - Decompose \(105\) using \(100\). - Distribute the factor to both terms.

1. In a), each addend inside the grouping symbols is multiplied by \(6\), so the missing number is \(7\). 2. For b), write \(105\) as \(100 + 5\): \(8(100 + 5) = 8 \cdot 100 + 8 \cdot 5 = 800 + 40 = 840\).

a) \(7\) b) \(8(100 + 5) = 800 + 40 = 840\)

- Change the sign to find an opposite. - Rewrite decimals and mixed numbers as fractions. - Interchange the numerator and denominator to find a reciprocal. - The sign of a reciprocal stays the same as the original number.

1. For \(5 = \frac{5}{1}\), the opposite is \(-5\), and the reciprocal is \(\frac{1}{5}\). Check: \(5 \cdot \frac{1}{5} = 1\). 2. For \(-0.75 = -\frac{3}{4}\), the opposite is \(0.75\), and the reciprocal is \(-\frac{4}{3}\). Check: \(-\frac{3}{4} \cdot -\frac{4}{3} = 1\). 3. For \(\frac{4}{9}\), the opposite is \(-\frac{4}{9}\), and the reciprocal is \(\frac{9}{4}\). Check: \(\frac{4}{9} \cdot \frac{9}{4} = 1\). 4. For \(-3\frac{1}{5} = -\frac{16}{5}\), the opposite is \(3\frac{1}{5}\), and the reciprocal is \(-\frac{5}{16}\). Check: \(-\frac{16}{5} \cdot -\frac{5}{16} = 1\).

Opposites: \(-5\), \(0.75\), \(-\frac{4}{9}\), \(3\frac{1}{5}\) Reciprocals: \(\frac{1}{5}\), \(-\frac{4}{3}\), \(\frac{9}{4}\), \(-\frac{5}{16}\) Each number multiplied by its reciprocal equals \(1\).

- Test a fraction times its flipped form. - Reverse the sign to undo an opposite. - Write \(2.5\) as a fraction before taking its reciprocal. - Recall the rule for division by zero.

1. For a), if \(a \ne 0\), then \(a \cdot \frac{1}{a} = \frac{a}{a} = 1\). 2. For b), reverse the sign. The original number is \(-4.5\). 3. For c), \(2.5 = \frac{5}{2}\). Taking the reciprocal again gives \(\frac{2}{5} = 0.4\). 4. For d), the reciprocal of \(0\) would be \(\frac{1}{0}\), but division by zero is undefined.

a) \(1\) b) \(-4.5\) c) \(0.4\) or \(\frac{2}{5}\) d) \(0\) has no reciprocal because \(\frac{1}{0}\) is undefined.

- Count the total columns and the shaded columns to determine the width fraction. - Count the total rows and the shaded rows to determine the height fraction. - Convert each fraction to a decimal, multiply, and then write the product as a simplified fraction.

1. The shaded rectangle spans \(4\) of the \(5\) columns, so it covers \(\frac{4}{5}=0.8\) of the wall’s width. It spans \(3\) of the \(4\) rows, so it covers \(\frac{3}{4}=0.75\) of the wall’s height. 2. Multiply the decimal forms: \(0.8\cdot0.75=0.6\). 3. Convert the product to a fraction: \(0.6=\frac{6}{10}=\frac{3}{5}\).

Width: \(\frac{4}{5}=0.8\) Height: \(\frac{3}{4}=0.75\) Decimal product: \(0.6\) Fraction of the wall: \(\frac{3}{5}\)

- Identify which of the three quantities in the given relationship is unknown in each part. - Substitute the known quantities into the relationship before deciding how to isolate the unknown. - Use units as a check on the quantity you solve for.

1. a) The unknown is total change, so substitute into the relationship: \((-0.8)\cdot7.5=-6\). Multiplication combines rate with elapsed time. 2. b) The unknown is rate, so \(5.4=r\cdot1.8\). Solving gives \(r=5.4\div1.8=3\). Division isolates the per-second rate.

a) Unknown: total change. \((-0.8\,\text{m/s})(7.5\,\text{s})=-6\,\text{m}\). Multiplication is appropriate because rate and time are known. b) Unknown: rate. \(5.4=r(1.8)\), so \(r=5.4\div1.8=3\,\text{m/s}\). Division is appropriate because the total and time are known and the per-unit rate is unknown.

- In each part, identify whether the unknown is a total signed change or a count of equal changes. - Write the relationship among change per command, number of commands, and total change. - Interpret the kind of quantity requested before deciding what its sign should mean.

1. a) \((-1.25)\cdot6=-7.5\), so Robot A's total signed change is \(-7.5\,\text{m}\). 2. b) \((-7.5)\div(-1.5)=5\), so Robot B received \(5\) commands. 3. Part a) asks for a signed downward change; part b) asks for a count of equal changes, which is nonnegative.

a) \((-1.25)\cdot6=-7.5\,\text{m}\) b) \((-7.5)\div(-1.5)=5\) commands c) Part a) is a signed downward change, while part b) is a count of equal negative changes.

- Rewrite the fraction with a denominator of \(10\). - Make the divisor a whole number before dividing decimals. - Check a quotient by multiplying it by the divisor.

1. The fraction was converted incorrectly. Since \(\frac{2}{5} = \frac{4}{10}\), its decimal form is \(0.4\), not \(0.25\). 2. The displayed decimal division was also calculated incorrectly. In fact, \(1.5 \div 0.25 = 150 \div 25 = 6\), not \(60\). 3. For the original expression, \(1.5 \div 0.4 = 15 \div 4 = 3.75\). Equivalently, \(\frac{3}{2} \div \frac{2}{5} = \frac{3}{2} \cdot \frac{5}{2} = \frac{15}{4} = 3.75\).

The first error is that \(\frac{2}{5} = 0.4\), not \(0.25\). The second error is that \(1.5 \div 0.25 = 6\), not \(60\). The correct value is \(3.75\).

- Write the original product as a single fraction before considering the changes. - Think about the commutative property of multiplication for part b). - If one factor becomes three times as large, what change to another factor would keep the overall product unchanged?

1. The original product is \(P=\frac{2}{5}\cdot\frac{3}{7}=\frac{6}{35}\). 2. For a), the new product is \(\frac{4}{5}\cdot\frac{3}{14}=\frac{6}{35}\). The factor of \(2\) in the numerator is canceled by the factor of \(2\) in the denominator, so the product is unchanged. 3. For b), swapping the numerators gives \(\frac{3}{5}\cdot\frac{2}{7}=\frac{6}{35}\). The product is unchanged because \(2\cdot3=3\cdot2\). 4. For c), tripling the first factor must be balanced by multiplying the second factor by \(\frac{1}{3}\). The second factor becomes \(\frac{3}{7}\cdot\frac{1}{3}=\frac{1}{7}\), and \(\frac{6}{5}\cdot\frac{1}{7}=\frac{6}{35}\).

a) The product stays the same: \(\frac{6}{35}\). b) Yes. The product remains \(\frac{6}{35}\). c) Multiply the second factor by \(\frac{1}{3}\), so it becomes \(\frac{1}{7}\).

- Treat the numerator and denominator as separate expressions first. - Is there a common factor in both products in the numerator? - Simplify the denominator before doing the final division.

1. Factor the common factor in the numerator: \(12.5\cdot3.7+12.5\cdot4.3=12.5(3.7+4.3)=12.5\cdot8=100\). 2. Simplify the denominator: \(\frac{2}{3}+\frac{1}{6}=\frac{5}{6}\), so \(\frac{5}{6}\div\frac{1}{12}=\frac{5}{6}\cdot12=10\). 3. Divide: \(100\div10=10\).

\(10\)

- Find a number whose cube is \(-0.125\). - Use the sign rule for a product with two negative factors. - Track the factor contributed by doubling each of three factors.

1. For a), \((-0.5)^3=-0.125\), so \(a=b=c=-0.5\). 2. For b), no. Two negative factors and one positive factor produce a positive product. 3. For c), \((2a)(2b)(2c)=8abc\). Thus the new product is \(8(-0.125)=-1\).

a) \(a=b=c=-0.5\) b) No; the product would be positive. c) \(-1\)

- Calculate the expression both ways. - Compare the difficulty of the divisions in each method. - A single counterexample is enough to show that an “always” statement is false.

1. Subtract first: \((1100 - 55) \div 11 = 1045 \div 11 = 95\). 2. Divide each term first: \(1100 \div 11 - 55 \div 11 = 100 - 5 = 95\). 3. The statement is false. In this example, splitting the quotient is easier because both \(1100\) and \(55\) are convenient multiples of \(11\).

The value is \(95\). The statement is false because \(1100 \div 11 - 55 \div 11 = 100 - 5\) is simpler than \(1045 \div 11\).

- Evaluate the left and right sides separately. - Note whether the grouped expression is being divided or is doing the dividing. - One counterexample disproves an “always” claim.

1. In Example A, the left side is \(120 \div 10 = 12\), and the right side is \(8 + 4 = 12\). This equality is true. 2. In Example B, the left side is \(120 \div 6 = 20\), and the right side is \(30 + 60 = 90\). This equality is false. 3. Division can be distributed over a sum or difference in the dividend, but not over a sum or difference in the divisor.

Paul is incorrect. Example A is true, but Example B is false because \(20 \ne 90\). Division can distribute over a sum or difference in the dividend, not in the divisor.

- Distinguish subtraction signs from negative signs. - Evaluate nested grouping symbols from the inside out. - Break each expression into smaller parts. - Check the sign of every product.

1. \(-8-\left(\frac{-8-(-4)}{-2}\right)\cdot(-6)=-8-2\cdot(-6)=4\). 2. \((-5-2)\cdot\left[-5-(-4)\cdot(-4)\right]+(-5+4)\cdot(-3)=(-7)\cdot(-21)+3=150\).

1) \(4\) 2) \(150\)