Free Math Worksheets

- The midpoint is the average of the two endpoints. - Add the two values first. - Divide the sum by \(2\).

1. Find the midpoint by adding the two numbers and dividing by \(2\): \(\frac{-2.4+3.6}{2}\). 2. The sum is \(-2.4+3.6=1.2\). 3. Divide by \(2\): \(1.2\div2=0.6\).

a) \(0.6\)

- 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: \(12.3\)

- Convert familiar fractions to decimals when that makes mental calculation easier. - Subtracting a negative number is equivalent to adding its opposite. - Determine the sign of each result before calculating its magnitude.

1. For a), write \(2\frac{1}{2} = 2.5\). Then \(-1.5 + 2.5 = 1\). 2. For b), write \(\frac{3}{4} = 0.75\). Subtracting a negative is the same as adding: \(0.25 + 0.75 = 1\). 3. For c), the product of two negative numbers is positive: \(-0.5 \cdot (-12) = 6\). 4. For d), write \(-3\frac{1}{2} = -3.5\). Then \(-3.5 + (-1.5) = -5\).

a) \(1\) b) \(1\) c) \(6\) d) \(-5\)

- Express all weights in the same unit. - Include the cargo and the driver. - Compare the total with the payload limit.

1. Add the existing load and the driver: \(1700+250+170=2120\,\text{lb}\). 2. Include the cement pallet: \(2120+300=2420\,\text{lb}\). 3. Since \(2420>2400\), the payload limit would be exceeded by \(20\,\text{lb}\).

Yes. The total payload would be \(2420\,\text{lb}\), which is \(20\,\text{lb}\) over the limit.

- Choose the form that makes each operation easiest. - Subtracting a negative number is equivalent to adding its opposite. - Determine the sign before multiplying or dividing. - To divide by a fraction, multiply by its reciprocal.

1. For a), write \(0.5 = \frac{1}{2}\). Then \(-\frac{3}{4} + \frac{1}{2} = -\frac{1}{4}\). 2. For b), write \(1\frac{1}{5} = 1.2\). Then \(-2.2 - (-1.2) = -2.2 + 1.2 = -1\). 3. For c), write \(1\frac{1}{2} = 1.5\). Then \(0.8 \cdot (-1.5) = -1.2\). 4. For d), divide by multiplying by the reciprocal: \(-2 \cdot \frac{5}{4} = -\frac{5}{2} = -2.5\).

a) \(-\frac{1}{4}\), or \(-0.25\) b) \(-1\) c) \(-1.2\), or \(-1\frac{1}{5}\) d) \(-2.5\), or \(-2\frac{1}{2}\)

- Choose fractions or decimals based on which form makes each calculation simpler. - Recall the sign rule for multiplying two negative numbers. - To divide by a fraction, multiply by its reciprocal.

1. For a), write \(2\frac{1}{4}=2.25\). Then \(-1.5+2.25=0.75=\frac{3}{4}\). 2. For b), write \(-0.8=-\frac{4}{5}\). The product of two negative numbers is positive: \(-\frac{3}{4}\cdot\left(-\frac{4}{5}\right)=\frac{3}{5}=0.6\). 3. For c), write \(-2.5=-\frac{5}{2}\) and \(1\frac{2}{3}=\frac{5}{3}\). Then \(-\frac{5}{2}\div\frac{5}{3}=-\frac{5}{2}\cdot\frac{3}{5}=-\frac{3}{2}=-1.5\).

a) \(0.75\), or \(\frac{3}{4}\) b) \(0.6\), or \(\frac{3}{5}\) c) \(-1.5\), or \(-\frac{3}{2}\)

- 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) \(-17\)

- Round to convenient hundreds or fifties. - Combine the negative terms first. - Compare your estimate with the choices.

1. Round the terms to \(-750\), \(100\), and \(-250\). 2. The estimated sum is \(-750+100-250=-900\). 3. The exact value is \(-897\), confirming that \(-900\) is closest.

B) \(-900\)

- Use the order of operations and evaluate the parentheses first. - Adding a negative number is the same as subtracting its absolute value. - Keep track of the sign of the intermediate result.

1. Evaluate inside the parentheses: \(52-315=-263\). 2. Add the result: \(48+(-263)=-215\).

\(-215\)

- 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: \(88-67=21\).

\(21\)

- Evaluate the parentheses first. - Subtracting a negative number becomes addition. - Compare the absolute values to determine the sign of the final sum.

1. Evaluate inside the parentheses: \(62-140=-78\). 2. Substitute the result: \(-135-(-78)=-135+78\). 3. Evaluate: \(-135+78=-57\).

\(-57\)

- Evaluate the expression inside the absolute-value bars first. - Absolute value makes \(-45\) into its distance from zero. - Then complete addition and subtraction from left to right.

1. Evaluate inside the absolute-value bars: \(30-75=-45\). 2. Take the absolute value: \(|-45|=45\). 3. Complete the remaining operations from left to right: \(45-20+15=25+15=40\).

\(40\)

- Evaluate each expression separately before comparing them. - Subtracting a negative number is equivalent to adding its opposite. - Find the difference between \(X\) and \(Y\).

1. Evaluate \(X\): \((-12)\cdot(-3)=36\), so \(X=36+(-16)=20\). 2. Evaluate \(Y\): \((-60)\div4=-15\), so \(Y=-15-(-5)=-10\). 3. Solve \(Y+z=X\): \(-10+z=20\), so \(z=30\).

\(X=20\), \(Y=-10\), and \(z=30\)

- The number halfway between two values is their mean. - Use a common denominator before adding the fractions. - Divide the sum by \(2\).

1. The midpoint is the mean: \(\frac{\frac{2}{3}+\frac{4}{5}}{2}\). 2. Use a common denominator: \(\frac{2}{3}=\frac{10}{15}\) and \(\frac{4}{5}=\frac{12}{15}\). 3. Add: \(\frac{10}{15}+\frac{12}{15}=\frac{22}{15}\). 4. Divide by \(2\): \(\frac{22}{15}\div2=\frac{22}{15}\cdot\frac{1}{2}=\frac{11}{15}\).

\(\frac{11}{15}\)

- Evaluate powers before other operations. - Keep the parentheses around a negative base when evaluating a power. - Work from the innermost grouping symbols outward.

1. For a), evaluate powers first: \(32-36-4=-8\). 2. For b), \((-2)^3=-8\), \(3^2=9\), and \(10-15=-5\). Then \(-8+9-(-5)=6\). 3. For c), \(2^3=8\), so \(8\cdot5=40\). Then \(50-40=10\), and \(100-10=90\).

a) \(-8\) b) \(6\) c) \(90\)

- Find the midpoint by adding the endpoints and dividing by \(2\). - A terminating decimal can be written as a fraction. - Move from left to right on the number line and record each integer between the endpoints.

1. Find the midpoint: \(m=\frac{-3.4+1.2}{2}=\frac{-2.2}{2}=-1.1\). 2. The value \(-1.1\) is not an integer. It is rational because \(-1.1=-\frac{11}{10}\). 3. The integers satisfying \(-3.4<z<1.2\) are \(-3, -2, -1, 0, 1\).

a) \(m=-1.1\) b) \(m\) is not an integer, but it is rational. c) \(-3, -2, -1, 0, 1\)

- The midpoint is the average of two endpoints. - Distance is always nonnegative. - Use absolute value to find a point’s distance from \(0\). - Pay attention to which side of \(0\) point \(Q\) is on.

1. For a), find the midpoint: \(\frac{-2.4+5.6}{2}=\frac{3.2}{2}=1.6\). 2. For b), subtract the lesser number from the greater number: \(\frac{5}{4}-\left(-\frac{3}{4}\right)=\frac{8}{4}=2\). 3. For c), \(P\) is \(2\) units from \(0\). Twice that distance is \(2\cdot2=4\). Since \(Q\) is positive, it represents \(4\).

a) \(1.6\) b) \(2\) c) \(4\)

- Follow grouping symbols, exponents, multiplication and division, then addition and subtraction. - A negative number squared is positive when the negative sign is part of the base. - Keep track of the sign of each multiplication and division result.

1. Evaluate the power: \((-4)^2=16\). 2. Evaluate the parentheses: \(14-18=-4\). 3. Multiply: \(5\cdot(-4)=-20\). 4. Divide: \(24\div(-6)=-4\). 5. Combine the results: \(16-(-20)+(-4)=16+20-4=32\).

\(32\)

- Evaluate the quantity inside absolute value bars before taking the absolute value. - Work from the innermost grouped expressions outward. - Determine signs before multiplying or dividing.

1. Evaluate the absolute value: \(|-12-3|=|-15|=15\). 2. Evaluate the power: \((-3)^3=-27\), so \((-27)+30=3\). 3. Multiply and divide: \(2\cdot15=30\) and \(3\div(-3)=-1\). 4. Combine: \(50-30+(-1)=19\).

\(19\)

- Break the expression into parts separated by addition and subtraction. - For powers with a negative base, check whether the exponent is even or odd. - Subtracting a negative value is the same as adding its opposite.

1. Evaluate the first power: \((-2)^5=-32\). 2. Multiply: \((-4)\cdot(-7)=28\), so the bracket is \(-32-28=-60\). 3. Evaluate the divisor: \((-2)^3=-8\), so \(120\div(-8)=-15\). 4. Evaluate the absolute value: \(|-10|=10\). 5. Combine: \(-60-(-15)+10=-35\).

\(-35\)

- Convert the fraction to a decimal. - Add a trailing zero when comparing \(2.37\) with a number having three decimal places. - The number halfway between two values is their average.

1. Convert the fraction: \(\frac{19}{8}=2.375\). 2. Compare \(2.375\) and \(2.370\). Therefore, \(\frac{19}{8}>2.37\). 3. Find the midpoint: \(\frac{2.375+2.37}{2}=\frac{4.745}{2}=2.3725\).

a) \(\frac{19}{8}\) is greater. b) \(2.3725\)

- The number exactly halfway between two values is their average. - Convert the fraction in part a) to a decimal. - In part b), use equivalent fractions with a denominator large enough to leave a numerator between the two endpoints.

1. For a), \(\frac{5}{8}=0.625\). The midpoint is \(\frac{0.625+0.63}{2}=\frac{1.255}{2}=0.6275\). 2. For b), write the endpoints as fractions: \(0.7=\frac{140}{200}\) and \(0.71=\frac{142}{200}\). 3. The fraction halfway between them is \(\frac{141}{200}\), which is already in simplest form.

a) \(0.6275\) b) \(\frac{141}{200}\)

- Convert both fractions to decimals. - Find the average of the two endpoint values. - Choose a value strictly between the lower endpoint and the midpoint.

1. Convert the fractions: \(\frac{1}{5} = 0.2\) and \(\frac{1}{4} = 0.25\). 2. Average the endpoints: \(\frac{0.2 + 0.25}{2} = \frac{0.45}{2} = 0.225\). 3. Any decimal strictly between \(0.2\) and \(0.225\) works, such as \(0.21\).

The midpoint is \(0.225\). One possible additional decimal is \(0.21\).

- Convert both fractions to decimals first. - Compare tenths, hundredths, and thousandths in order. - Write \(0.4\) as \(0.40\) if that helps you find a number between the last two values.

1. Convert the fractions: \(\frac{3}{8} = 0.375\) and \(\frac{2}{5} = 0.4\). 2. Compare the decimals: \(0.3 < 0.375 < 0.38 < 0.4\). 3. The two greatest values are \(0.38\) and \(0.4\). One decimal between them is \(0.39\).

a) \(0.3 < \frac{3}{8} < 0.38 < \frac{2}{5}\) b) Answers will vary. One possible answer is \(0.39\).

- Convert both fractions to decimals. - Find each absolute difference from \(0.15\). - The smaller distance identifies the closer value.

1. Convert the fractions: \(\frac{1}{6} = 0.1\overline{6}\) and \(\frac{1}{8} = 0.125\). 2. The distance from \(\frac{1}{6}\) to \(0.15\) is \(\frac{1}{6} - \frac{3}{20} = \frac{1}{60} = 0.01666\ldots\). 3. The distance from \(\frac{1}{8}\) to \(0.15\) is \(\frac{3}{20} - \frac{1}{8} = \frac{1}{40} = 0.025\). 4. Since \(\frac{1}{60} < \frac{1}{40}\), \(\frac{1}{6}\) is closer to \(0.15\).

\(\frac{1}{6} = 0.1\overline{6}\) and \(\frac{1}{8} = 0.125\). The value \(\frac{1}{6}\) is closer to \(0.15\).

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

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

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

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

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

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

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

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

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

- Convert \(x\) to a decimal first. - Expand the repeating decimal far enough to compare it with \(x\). - The number halfway between two values is their average.

1. Convert \(x\): \(\frac{5}{8} = 0.625\). 2. Since \(y = 0.62555\ldots\), it follows that \(x < y\). 3. Find the midpoint by averaging: \(z = \frac{0.62 + 0.625}{2}\). 4. Compute \(z = \frac{1.245}{2} = 0.6225\).

1. \(x < y\) 2. \(z = 0.6225\)

- Subtract the ending charge from the starting charge. - Use a common denominator. - Determine how many one-hour portions of that size make one full battery.

1. Find the fraction used in one hour: \(\frac{5}{6}-\frac{2}{3}=\frac{5}{6}-\frac{4}{6}=\frac{1}{6}\). 2. A full battery contains six portions of size \(\frac{1}{6}\), so it would last \(6\) hours. 3. Convert to minutes: \(6\cdot60=360\) minutes.

\(\frac{1}{6}\) of the full capacity was used. A full charge would last \(360\) minutes.

- The number halfway between two values is their average. - Convert the fractions and decimals to a common form. - On a number line, the value farther right is greater.

1. Convert \(-\frac{1}{2}\) to \(-0.5\). 2. Find the midpoint: \(x=\frac{-0.5+0.3}{2}=\frac{-0.2}{2}=-0.1\). 3. Convert the comparison value: \(-\frac{1}{8}=-0.125\). 4. Since \(-0.1>-0.125\), \(x> -\frac{1}{8}\).

\(x=-0.1\), and \(x> -\frac{1}{8}\).

- First find the fraction of the tank that was empty. - Find one third of that empty fraction. - Add the amount of water added to the original amount.

1. The empty fraction was \(1-\frac{3}{8}=\frac{5}{8}\). 2. The amount added was \(\frac{1}{3}\cdot\frac{5}{8}=\frac{5}{24}\) of the tank. 3. The new level is \(\frac{3}{8}+\frac{5}{24}=\frac{9}{24}+\frac{5}{24}=\frac{14}{24}=\frac{7}{12}\).

The tank is \(\frac{7}{12}\) full.

- Squaring a fraction means multiplying it by itself. - Divide by a fraction by multiplying by its reciprocal. - Follow the sequence of operations described in the problem.

1. Find the sum: \(1\frac{2}{3}+2\frac{5}{6}=\frac{9}{2}\). 2. Square the fraction: \(\left(\frac{3}{4}\right)^2=\frac{9}{16}\). 3. Divide: \(\frac{9}{2}\div\frac{9}{16}=\frac{9}{2}\cdot\frac{16}{9}=8\). 4. Subtract: \(8-1\frac{1}{2}=6\frac{1}{2}\).

\(6\frac{1}{2}\)

- Add the lengths of the first three pieces. - Subtract their total from the original length. - If \(\frac{2}{3}\) is removed, determine what fraction remains.

1. Add the first three lengths: \(1\frac{1}{5}+\frac{3}{4}+1\frac{3}{10}=\frac{24}{20}+\frac{15}{20}+\frac{26}{20}=3\frac{1}{4}\,\text{ft}\). 2. Subtract from the original length: \(4-3\frac{1}{4}=\frac{3}{4}\,\text{ft}\). 3. If \(\frac{2}{3}\) is cut off, \(\frac{1}{3}\) remains. The final length is \(\frac{1}{3}\cdot\frac{3}{4}=\frac{1}{4}\,\text{ft}\).

a) \(\frac{3}{4}\,\text{ft}\) b) \(\frac{1}{4}\,\text{ft}\)

- Find the total amount of fabric needed. - Multiply that length by the price per yard for Option A. - Compare the two prices and subtract to find the difference.

1. The total length needed is \(1\frac{1}{2}+2\frac{3}{4}+1\frac{1}{5}=5\frac{9}{20}=5.45\,\text{yd}\). 2. Option A costs \(5.45\cdot\$3.00=\$16.35\). 3. Option B costs \(\$16.00\), so it costs less. 4. The difference is \(\$16.35-\$16.00=\$0.35\).

Option B costs less by \(\$0.35\).

- List the possible values of \(\triangle\). - Isolate the fraction containing \(\square\). - Test each possible value of \(\triangle\) and keep only natural-number values of \(\square\).

1. Since \(\triangle<5\), \(\triangle\in\{1,2,3,4\}\). Since \(\frac{2}{\square}\) is a proper fraction, \(\square>2\). 2. Isolate the second fraction: \(\frac{2}{\square}=2-1\frac{\triangle}{5}=\frac{5-\triangle}{5}\). 3. Test the possible values of \(\triangle\). For \(\triangle=1\), \(\square=2.5\); for \(\triangle=2\), \(\square=\frac{10}{3}\); neither is a natural number. 4. For \(\triangle=3\), \(\frac{2}{\square}=\frac{2}{5}\), so \(\square=5\). For \(\triangle=4\), \(\frac{2}{\square}=\frac{1}{5}\), so \(\square=10\).

\((\triangle, \square)=(3, 5)\) and \((4, 10)\)

- A quotient is the result of division. - A difference is the result of subtraction. - Evaluate the two parts before adding them.

1. Find the quotient: \(4.8\div0.6=8\). 2. Find the difference: \(15.4-23.1=-7.7\). 3. Add the results: \(8+(-7.7)=0.3\).

\(0.3\)

- Squaring a number means multiplying it by itself. - Pay attention to which quantity is subtracted from which. - Evaluate the described parts before combining them.

1. Find the sum: \(1.2+0.8=2\). 2. Find the square: \((1.5)^2=2.25\). 3. Multiply: \(2\cdot2.25=4.5\). 4. Subtract the product from \(10\): \(10-4.5=5.5\).

\(5.5\)

- Write fractions and decimals in the same form before subtracting. - Identify the dividend and divisor carefully. - Follow the sequence described in the problem.

1. Find the product: \(0.25\cdot12=3\). 2. Find the difference: \(\frac{3}{4}-1.5=0.75-1.5=-0.75\). 3. Divide: \(3\div(-0.75)=-4\). 4. Find the square: \(4^2=16\). 5. Add: \(-4+16=12\).

\(12\)

- Write the expression you get after distributing \(12\). - Compare \(12\) with the denominators in the two fractions. - Carry out both methods, then compare the number and complexity of the steps.

1. Method 1: Convert \(0.75\) to \(\frac{3}{4}\). The least common denominator of \(3\) and \(4\) is \(12\). 2. Add inside the parentheses: \(\frac{2}{3} + \frac{3}{4} = \frac{8}{12} + \frac{9}{12} = \frac{17}{12}\). 3. Multiply: \(\frac{17}{12} \cdot 12 = 17\). 4. Method 2: Distribute \(12\): \(\frac{2}{3} \cdot 12 + 0.75 \cdot 12\). 5. Evaluate each product: \(\frac{2}{3} \cdot 12 = 8\) and \(0.75 \cdot 12 = \frac{3}{4} \cdot 12 = 9\). 6. Add: \(8 + 9 = 17\). 7. Method 2 requires less computation because \(12\) is divisible by both denominators, so each product becomes a whole number immediately.

Both methods give \(17\). Applying the distributive property first is more efficient because multiplying by \(12\) eliminates the denominators before any fraction addition is needed.

- 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: \(12.8 - 7.15 + 2.2 = 5.65 + 2.2 = 7.85\).

\(7.85\)

- Check the conversion of the fractional part to a decimal. - Recalculate the subtraction shown in the student's work. - Rewrite \(\frac{1}{4}\) as an equivalent fraction with denominator \(100\).

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

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

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

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

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

- Estimate with convenient dollar and half-dollar amounts. - Include every repeated item. - Subtract the exact total from the amount paid to find the change.

1. The first items cost \(\$1.49+4\cdot\$0.55+\$6.75=\$10.44\). 2. Estimate \(\$10.44\) as about \(\$10.50\), and three more notebooks as \(3\cdot\$1.50=\$4.50\). The estimate is \(\$15.00\), so the purchase will be very close to the available amount. 3. The three additional notebooks cost \(3\cdot\$1.49=\$4.47\). 4. The exact total is \(\$10.44+\$4.47=\$14.91\). 5. The change is \(\$15.00-\$14.91=\$0.09\).

a) The estimate is \(\$15.00\), so the money appears just sufficient, but an exact calculation is needed. b) The total is \(\$14.91\), and the change is \(\$0.09\).

- 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: \(16\)

- Use nearby decimals for a quick estimate. - For an exact result, avoid replacing repeating decimals with rounded values. - Convert the terminating decimals to fractions and use common denominators.

1. Estimate using nearby convenient values: \(4.7 - 1.7 + 0.2 \approx 3.2\). 2. Fractions are useful because \(\frac{2}{3}\) and \(\frac{1}{6}\) have repeating decimal representations. 3. Convert the decimals: \(1.5 = \frac{3}{2}\) and \(0.2 = \frac{1}{5}\). 4. Evaluate the parentheses: \(\frac{3}{2} + \frac{1}{6} = \frac{10}{6} = \frac{5}{3}\). 5. Calculate: \(\frac{14}{3} - \frac{5}{3} + \frac{1}{5} = 3 + \frac{1}{5} = 3.2\).

Estimate: about \(3.2\) Exact value: \(3\frac{1}{5}\), or \(3.2\)

- Apply multiplication and division before addition and subtraction. - Multiply a fraction by a whole number using the whole number as a factor. - Divide by a fraction by multiplying by its reciprocal. - Convert mixed numbers to improper fractions when useful.

1. For a), multiply first: \(\frac{3}{8}\cdot2=\frac{3}{4}\). Then \(1 \frac{1}{4}+\frac{3}{4}=2\). 2. For b), divide first: \(10\div\frac{5}{2}=10\cdot\frac{2}{5}=4\). Then \(4-\frac{1}{2}=3 \frac{1}{2}\).

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

- Evaluate expressions inside parentheses first. - Apply multiplication and division before addition and subtraction. - Simplify factors before multiplying fractions when possible. - Rewrite whole numbers or mixed numbers as fractions when useful.

1. For a), \(\frac{3}{4}\cdot\frac{2}{3}=\frac{1}{2}\). 2. Also, \(1 \frac{1}{2}\div3=\frac{3}{2}\div3=\frac{1}{2}\). 3. Therefore, \(\frac{1}{2}+\frac{1}{2}=1\). 4. For b), \(2-\frac{1}{3}=\frac{5}{3}\), and \(\frac{5}{3}\cdot\frac{6}{5}=2\).

a) \(1\) b) \(2\)

- Work inside parentheses first. - Use a common denominator for addition or subtraction. - Divide by a fraction by multiplying by its reciprocal. - Simplify factors before multiplying when possible.

1. For a), \(\frac{7}{10}\div\frac{14}{5}=\frac{7}{10}\cdot\frac{5}{14}=\frac{1}{4}\). 2. Also, \(\frac{3}{4}\cdot\frac{1}{3}=\frac{1}{4}\), so the total is \(\frac{1}{2}\). 3. For b), multiply inside the parentheses: \(\frac{2}{3}\cdot\frac{1}{4}=\frac{1}{6}\). 4. Then \(\frac{1}{6}+\frac{1}{6}=\frac{1}{3}\), and \(3 \frac{1}{2}-\frac{1}{3}=\frac{7}{2}-\frac{1}{3}=\frac{19}{6}=3 \frac{1}{6}\).

a) \(\frac{1}{2}\) b) \(3 \frac{1}{6}\)

- Evaluate the expression inside the parentheses first. - Write the decimal and mixed number as fractions before multiplying. - Look for common factors in numerators and denominators. - Count the negative factors to determine the sign of the product.

1. Convert the mixed number and add inside the parentheses: \(\frac{2}{3} + 1\frac{1}{6} = \frac{4}{6} + \frac{7}{6} = \frac{11}{6}\). 2. Write \(-1.5 = -\frac{3}{2}\). 3. Multiply: \(\left(-\frac{3}{2}\right) \cdot \frac{11}{6} \cdot \left(-\frac{4}{11}\right)\). 4. Two negative factors give a positive product. Cancel the common factor \(11\), then simplify: \(\frac{3 \cdot 4}{2 \cdot 6} = \frac{12}{12} = 1\). 5. Fractions keep the calculation exact and make common factors visible before multiplication.

The value is \(1\). Writing all the numbers as fractions keeps the calculation exact and makes it easy to cancel common factors.

- 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) \(x=\frac{3}{5}\)

- Combine the two multiplicative comparisons. - Convert mixed numbers to improper fractions. - Divide the largest amount by the combined scale factor.

1. Combine the two scale factors in the expression \(42\div(1\frac{3}{4}\cdot1\frac{1}{3})\). 2. Convert the mixed numbers: \(1\frac{3}{4}=\frac{7}{4}\) and \(1\frac{1}{3}=\frac{4}{3}\). 3. Their product is \(\frac{7}{4}\cdot\frac{4}{3}=\frac{7}{3}\). 4. Divide: \(42\div\frac{7}{3}=42\cdot\frac{3}{7}=18\).

The bucket holds \(18\) gallons. One expression is \(42\div(1\frac{3}{4}\cdot1\frac{1}{3})\).

- 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 \(-2\frac{1}{2}=-\frac{5}{2}\): \(x=\frac{2}{5}\cdot\left(-\frac{5}{2}\right)=-1\).

\(x=-1\)

- Evaluate the parentheses before multiplying. - Determine the sign of the product before multiplying the fractions. - Simplify common factors when possible.

1. Evaluate the parentheses: \(2 \frac{2}{3}-\frac{1}{6}=\frac{8}{3}-\frac{1}{6}=\frac{5}{2}\). 2. Multiply: \(-1 \frac{1}{5}\cdot\frac{5}{2}=-\frac{6}{5}\cdot\frac{5}{2}=-3\). 3. Add: \(-3+\frac{3}{4}=-\frac{9}{4}=-2 \frac{1}{4}\).

\(-2 \frac{1}{4}\), or \(-\frac{9}{4}\)

- 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: \(-\frac{1}{2}-\frac{3}{2}=-2\).

\(-2\)

- Evaluate parentheses first. - Convert mixed numbers to improper fractions before calculating. - Use sign rules when multiplying negative rational numbers. - Use a common denominator for fraction addition or subtraction.

1. For a), \(2 \frac{1}{2}+\frac{3}{10}=\frac{5}{2}+\frac{3}{10}=\frac{14}{5}\). 2. Then \(4 \frac{2}{5}-\frac{14}{5}=\frac{22}{5}-\frac{14}{5}=\frac{8}{5}=1 \frac{3}{5}\). 3. For b), \(\frac{2}{3}-2=-\frac{4}{3}\). 4. Then \(-1 \frac{1}{4}\cdot\left(-\frac{4}{3}\right)=-\frac{5}{4}\cdot\left(-\frac{4}{3}\right)=\frac{5}{3}=1 \frac{2}{3}\).

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

- What common denominator can you use for \(6\) and \(4\)? - How do you divide by a fraction? - Determine the sign of the final result before calculating.

1. Evaluate the parentheses: \(-\frac{5}{6}+1 \frac{1}{4}=-\frac{10}{12}+\frac{15}{12}=\frac{5}{12}\). 2. Divide by multiplying by the reciprocal: \(\frac{5}{12}\div\left(-\frac{5}{12}\right)=\frac{5}{12}\cdot\left(-\frac{12}{5}\right)\). 3. Simplify: \(-1\).

\(-1\)

- 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) \(3\)

- 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: \(\frac{1}{3} \div \frac{6}{5} = \frac{1}{3} \cdot \frac{5}{6} = \frac{5}{18}\).

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

- How do you divide by a fraction? - Would converting the decimal to a fraction help? - Which operation should you perform first in b)? - Can you simplify common factors before multiplying?

1. For a), multiply by the reciprocal: \(\frac{14}{15}\cdot\frac{10}{21}\). 2. Cancel common factors and multiply: \(\frac{14}{15}\cdot\frac{10}{21}=\frac{4}{9}\). 3. For b), write \(1.2=\frac{6}{5}\). Then \(\frac{6}{5}\cdot\frac{5}{6}=1\). 4. Add: \(1+\frac{1}{2}=\frac{3}{2}=1 \frac{1}{2}\).

a) \(\frac{4}{9}\) b) \(\frac{3}{2}\), or \(1 \frac{1}{2}\)

- Follow the order of operations: parentheses and powers before multiplication or division. - What happens to the sign when a negative number is squared? - How can you write \(0.75\) as a fraction?

1. For a), add inside the parentheses: \(\frac{2}{3}+\frac{1}{4}=\frac{11}{12}\). 2. Multiply: \(\frac{11}{12}\cdot\frac{12}{11}=1\). 3. For b), evaluate the power first: \(\left(-\frac{3}{8}\right)^2=\frac{9}{64}\). 4. Write \(0.75=\frac{3}{4}\), then divide: \(\frac{3}{4}\div\frac{9}{64}=\frac{3}{4}\cdot\frac{64}{9}=\frac{16}{3}\).

a) \(1\) b) \(\frac{16}{3}\)

- Use the order of operations inside the brackets first. - Which multiplication and division should be completed before the addition? - Pay close attention to sign rules when multiplying and dividing negative numbers.

1. Divide inside the brackets: \(4.5\div(-0.5)=-9\). 2. Multiply inside the brackets: \(7\cdot(-2)=-14\). 3. Add inside the brackets: \(-9+(-14)=-23\). 4. Finish the calculation: \(-18+(-23)=-41\).

\(-41\)

- Evaluate the two expressions separately. - It may help to write fractions and decimals in the same form. - How do you divide by a fraction?

1. Expression A: \(\frac{3}{4}=0.75\), so \(0.75-1.25=-0.5\). Then \((-0.5)\cdot(-4)=2\). 2. Expression B: \(2.5\div\frac{1}{2}=5\), so \(5-6=-1\). 3. Since \(2>-1\), Expression A has the greater value.

Expression A is greater because \(2>-1\).

- Use the order of operations before adding or subtracting. - Determine the sign of each product or quotient before calculating. - For powers, think about how many negative factors are being multiplied. - Converting the simple fraction to a decimal may help in e).

1. For a), the negative product has magnitude \(0.48\), which is slightly less than \(0.5\), so predict a positive result. Then \((-0.4)\cdot1.2=-0.48\), and \(-0.48+0.5=0.02\). 2. For b), the quotient is positive and less than \(0.8\), so predict a negative result. Then \((-2.5)\div(-5)=0.5\), and \(0.5-0.8=-0.3\). 3. For c), \(0.2\cdot3\) is greater than \(0.12\), so predict a negative result. Then \(0.2\cdot3=0.6\), and \(0.12-0.6=-0.48\). 4. For d), \((-3)^3\) is negative with magnitude \(27\), which is less than \(30\), so predict a positive result. Then \((-3)^3=-27\), and \(-27+30=3\). 5. For e), \(0.3\div0.5=0.6\), which is greater than \(\frac{1}{2}\), so predict a positive result. Then \(-\frac{1}{2}+0.6=-0.5+0.6=0.1\).

a) Positive; \(0.02\) b) Negative; \(-0.3\) c) Negative; \(-0.48\) d) Positive; \(3\) e) Positive; \(0.1\)

- Sometimes the sign alone is enough to compare two results. - Think about which result lies farther to the right on a number line. - What happens to absolute value when you divide by a positive number between \(0\) and \(1\)? - Sign rules can often settle a comparison without exact arithmetic.

1. For a), \((-0.5)^2\) is positive and \((-0.5)^3\) is negative, so the left side is greater. 2. For b), adding \(2.1\) moves \(-5.5\) to the right on the number line, while subtracting \(2.1\) moves it to the left. Therefore, the left side is greater. 3. For c), the product of three negative factors is negative, so it is less than \(0\). 4. For d), dividing by \(0.5\) doubles the absolute value, while multiplying by \(0.5\) halves it. Both results are negative, so the division result is smaller.

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

- Complete powers, multiplication, and division before addition. - When does an odd power of a negative number stay negative? - In d), does the exponent apply to the negative sign? - Converting the simple fraction to a decimal may help in b).

1. For a), evaluate the power first: \((-0.2)^3=-0.008\). Then \(-0.008+0.01=0.002\). 2. For b), divide first: \(0.8\div2=0.4\). Since \(-\frac{4}{5}=-0.8\), the result is \(-0.8+0.4=-0.4\). 3. For c), divide first: \(1.5\div(-0.3)=-5\). Then \(-5+4=-1\). 4. For d), the exponent applies to \(0.7\), not to the leading negative sign: \(-0.7^2=-(0.7^2)=-0.49\). Then \(-0.49+0.5=0.01\).

a) \(0.002\) b) \(-0.4\) c) \(-1\) d) \(0.01\)

- Evaluate each expression separately first. - What is the difference between \(-x^2\) and \((-x)^2\)? - For negative numbers, the value farther left on a number line is smaller.

1. Evaluate each expression: \(A=-0.16\), \(B=0.16\), \(C=-0.8\), and \(D=-0.2\). 2. Compare the negative values: \(-0.8<-0.2<-0.16\). 3. The positive value \(0.16\) is greatest. 4. Therefore, the order is C, D, A, B.

C, D, A, B

- Convert between fractions and decimals when useful. - Write an equation for each missing divisor or factor. - Apply the sign rules before calculating.

1. For a), \(-\frac{2}{5}\cdot0.4=-0.4\cdot0.4=-0.16\). 2. For b), solve \(-0.16\div x=-1\). Thus \(x=0.16=\frac{4}{25}\). 3. For c), \(-0.16+0.2=0.04=\frac{1}{25}\). Its reciprocal is \(25\), so \(0.04\cdot25=1\).

a) \(-0.16\), or \(-\frac{4}{25}\) b) \(0.16\), or \(\frac{4}{25}\) c) \(25\)

- Determine the sign of each result first. - Think about where the results would lie on a number line. - Comparing absolute values can sometimes determine the correct symbol without exact arithmetic.

1. For a), multiplying \(-1.2\) by \(5\) increases its absolute value, while dividing by \(5\) decreases its absolute value. Both results are negative, so the multiplication result is smaller. Thus, use \(<\). 2. For b), \(-3\cdot(-4)\) is positive, while \(-3+(-4)\) is negative. Therefore, use \(>\). 3. For c), \(-\frac{1}{2}+\frac{1}{3}=-\frac{1}{6}\), and \(-\frac{1}{2}\cdot\frac{1}{3}=-\frac{1}{6}\). Therefore, use \(=\).

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

- Complete powers, multiplication, and division before addition or subtraction. - Determine the sign of each product or quotient before calculating. - Converting the familiar fraction to a decimal may help in c).

1. For a), divide first: \((-3.6)\div2=-1.8\). Then \(-12.4+(-1.8)=-14.2\). 2. For b), multiply first: \((-8)\cdot(-1.5)=12\). Then \(12-13=-1\). 3. For c), multiply first: \(1.25\cdot0.4=0.5\). Since \(\frac{3}{4}=0.75\), the result is \(0.75-0.5=0.25\). 4. For d), evaluate the power first: \((-2)^4=16\). Then \(16\div(-8)=-2\).

a) \(-14.2\) b) \(-1\) c) \(0.25\) d) \(-2\)

- To divide by a fraction, multiply by its reciprocal. - Evaluate the expression inside the parentheses before applying the negative sign outside. - Follow the order of operations: evaluate the exponent before multiplying.

1. For a), write \(-0.4 = -\frac{2}{5}\). Then \(-\frac{2}{5} \cdot \frac{5}{8} = -\frac{1}{4}\). 2. For b), division by \(-\frac{1}{4}\) is multiplication by \(-4\): \(-2.5 \cdot (-4) = 10\). 3. For c), evaluate inside the parentheses: \(\frac{1}{3} - \frac{1}{2} = -\frac{1}{6}\). The negative sign outside changes the result to \(\frac{1}{6}\). 4. For d), evaluate the exponent first: \((-0.1)^2 = 0.01\). Then \(0.01 \cdot (-10) = -0.1\).

a) \(-\frac{1}{4}\), or \(-0.25\) b) \(10\) c) \(\frac{1}{6}\) d) \(-0.1\)

- How does an even or odd exponent affect the sign of a negative base? - Apply parentheses before multiplication or division. - Can you simplify common factors before multiplying the fractions?

1. For a), \((-0.2)^2\) is positive because the exponent is even, while \((-0.2)^3\) is negative because the exponent is odd. Therefore, \((-0.2)^2\) is greater. 2. For b), \(-4.8+2.4=-2.4\), and \(-2.4\div(-0.6)=4\). 3. For c), multiply first: \(-\frac{3}{5}\cdot\left(-\frac{10}{9}\right)=\frac{2}{3}\). Then \(\frac{2}{3}+(-1)=-\frac{1}{3}\).

a) \((-0.2)^2\) is greater because it is positive while \((-0.2)^3\) is negative. b) \(4\) c) \(-\frac{1}{3}\)

- Follow the order of operations: evaluate parentheses first, then multiplication or division before subtraction. - Convert the terminating and repeating decimals to fractions. - Look for values that may cancel in the final subtraction.

1. For a), write \(0.25 = \frac{1}{4}\). Then \(\frac{1}{4} + \frac{1}{8} = \frac{3}{8}\), so \(\frac{4}{7} \cdot \frac{3}{8} = \frac{3}{14}\). 2. For b), perform the division before the subtraction. Write \(0.\overline{5} = \frac{5}{9}\). Then \(\frac{2}{3} \div \frac{5}{9} = \frac{2}{3} \cdot \frac{9}{5} = \frac{6}{5} = 1.2\). 3. Subtract: \(1.2 - 1.2 = 0\).

a) \(\frac{3}{14}\) b) \(0\)

- 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: \(-0.6 - (-0.8) = -0.6 + 0.8 = 0.2\).

\(0.2\)

- Evaluate the two expressions separately. - What happens to the sign when a negative number is squared? - Could writing the fraction in \(B\) as a decimal make the multiplication easier? - Compare the two final values.

1. Evaluate \(A\): \((-0.3)^2=0.09\), so \(0.09\div0.1=0.9\). 2. Evaluate \(B\): \(-2.5\cdot\frac{2}{5}=-1\), so \(-1+1.9=0.9\). 3. Both expressions have the same value, \(0.9\).

Yes. Both expressions equal \(0.9\).

- Evaluate parentheses before other operations. - Would converting the decimal to a fraction help with the multiplication? - Simplify common factors before multiplying when possible.

1. For a), evaluate the parentheses: \(\frac{2}{3}+0.5=\frac{2}{3}+\frac{1}{2}=\frac{7}{6}\). Then \(\frac{3}{2}-\frac{7}{6}=\frac{1}{3}\). 2. For b), \(-2.4=-\frac{12}{5}\), so \(-\frac{12}{5}\cdot\frac{5}{6}=-2\). 3. For c), write the values as tenths: \(\frac{7}{10}-\frac{3}{10}-\frac{5}{10}=-\frac{1}{10}=-0.1\).

a) \(\frac{1}{3}\) b) \(-2\) c) \(-0.1\), or \(-\frac{1}{10}\)

- Work from the innermost parentheses outward. - How do you divide by a fraction? - Decide separately for each part whether fractions or decimals make the arithmetic easier.

1. For a), \(-1.2+\frac{2}{5}=-1.2+0.4=-0.8\). Then \(-0.8\cdot1.5=-1.2\). 2. For b), \(-0.75=-\frac{3}{4}\), so \(\frac{3}{8}\div\left(-\frac{3}{4}\right)=\frac{3}{8}\cdot\left(-\frac{4}{3}\right)=-\frac{1}{2}\). Then \(-\frac{1}{2}-1.25=-1.75=-\frac{7}{4}\). 3. For c), \(0.5+\frac{1}{6}=\frac{1}{2}+\frac{1}{6}=\frac{2}{3}\). Then \(\frac{1}{3}-\frac{2}{3}=-\frac{1}{3}\), so the leading negative gives \(\frac{1}{3}\).

a) \(-1.2\), or \(-\frac{6}{5}\) b) \(-1.75\), or \(-\frac{7}{4}\) c) \(\frac{1}{3}\)

- Powers come before multiplication, division, addition, and subtraction. - What is the difference between \(-10^2\) and \((-10)^2\)? - What happens when you subtract a negative number?

1. For a), evaluate the power first: \(-10^2=-100\). 2. Then \(-100\div5=-20\) and \(4\cdot(-6)=-24\). So \(-20-(-24)+3=7\). 3. For b), \((-3)^3=-27\) and \(8\cdot(-2)=-16\). 4. Then \(-27+(-16)-(-12)=-31\).

a) \(7\) b) \(-31\)

- Evaluate both expressions separately and compare them. - What is the base of each power? - What is the sign of the product of two negative numbers?

1. For \(A\), the exponent applies only to \(6\): \(-6^2=-(6^2)=-36\). Then \(-36\div4+5=-9+5=-4\). 2. For \(B\), the parentheses make \(-6\) the base: \((-6)^2=(-6)\cdot(-6)=36\). Then \(36\div4+5=9+5=14\). 3. The parentheses determine whether the negative sign is part of the base being squared.

a) \(A=-4\); \(B=14\) b) In \(-6^2\), only \(6\) is squared. In \((-6)^2\), the entire negative number is squared, giving a positive power value.

- Convert the mixed number to a decimal for the first method. - Determine the sign of the value inside the parentheses. - The product of two negative numbers is positive. - Compare the number of conversions and simplifications in each method.

1. Decimal method: Write \(1\frac{1}{5} = 1.2\). Then \(0.8 - 1.2 = -0.4\), and \(-0.4 \cdot (-2.5) = 1\). 2. Fraction method: Write \(0.8 = \frac{4}{5}\), \(1\frac{1}{5} = \frac{6}{5}\), and \(-2.5 = -\frac{5}{2}\). Then \(\frac{4}{5} - \frac{6}{5} = -\frac{2}{5}\), and \(-\frac{2}{5} \cdot \left(-\frac{5}{2}\right) = 1\). 3. Both methods are efficient because every number has a simple terminating decimal and fraction form.

Both methods give \(1\). A valid explanation may favor either method if it refers to the simple conversions or simplifications used.

- Determine whether the fractional parts have terminating decimal representations. - Add fractions with unlike denominators using a common denominator. - To divide by a fraction, multiply by its reciprocal.

1. Fractions are more useful because \(\frac{1}{3}\) and \(\frac{1}{6}\) have repeating decimal representations. 2. Write \(1.75 = \frac{7}{4}\), \(2\frac{1}{3} = \frac{7}{3}\), and \(-1\frac{1}{6} = -\frac{7}{6}\). 3. Add inside the parentheses: \(\frac{7}{4} + \frac{7}{3} = \frac{21}{12} + \frac{28}{12} = \frac{49}{12}\). 4. Divide by multiplying by the reciprocal: \(\frac{49}{12} \cdot \left(-\frac{6}{7}\right) = -\frac{7}{2} = -3.5\).

\(-\frac{7}{2}\), or \(-3.5\). Fractions avoid rounded repeating decimals in the calculation.

- Recall the decimal form of \(\frac{5}{8}\). - Alternatively, convert each decimal to a fraction. - Determine the sign of the value inside the parentheses and the sign of the quotient.

1. A decimal method is efficient because \(\frac{5}{8} = 0.625\). Then \(-0.625 + 0.125 = -0.5\), and \(-0.5 \div (-0.2) = 2.5\). 2. A fraction method also works: \(0.125 = \frac{1}{8}\) and \(-0.2 = -\frac{1}{5}\). Then \(-\frac{5}{8} + \frac{1}{8} = -\frac{1}{2}\), and \(-\frac{1}{2} \div \left(-\frac{1}{5}\right) = \frac{5}{2}\). 3. Both methods give the same result.

\(2.5\), or \(\frac{5}{2}\)

- Check whether a fraction has a terminating decimal representation. - Look for factors that simplify before multiplication. - Use decimals when all values have simple terminating decimal forms.

1. For a), fractions are efficient because \(\frac{1}{3}\) is a repeating decimal. Write \(0.5 = \frac{1}{2}\). Then \(\frac{1}{3} + \frac{1}{2} = \frac{5}{6}\). 2. For b), fractions are efficient because \(0.25 = \frac{1}{4}\). Then \(\frac{1}{4} \cdot \frac{4}{7} = \frac{1}{7}\). 3. For c), decimals are efficient because every value has a terminating decimal form. Write \(\frac{1}{5} = 0.2\). Then \(0.4 \cdot 1.2 - 0.2 = 0.48 - 0.2 = 0.28\).

a) Fractions: \(\frac{5}{6}\) b) Fractions: \(\frac{1}{7}\) c) Decimals: \(0.28\)

- A difference has a minuend and a subtrahend. - Evaluate the quotient and product separately. - Pay attention when subtracting a negative value.

1. The quotient is \(2.4\div(-0.6)=-4\). 2. The product is \(-\frac{1}{2}\cdot5=-2.5\). 3. The difference is \(-4-(-2.5)=-1.5\).

The expression is \((2.4\div(-0.6))-\left(-\frac{1}{2}\cdot5\right)\), and its value is \(-1.5\).

- Identify the two factors. - Use parentheses so the sum and difference are evaluated first. - Recall the sign rule for multiplying two negative numbers.

1. The first factor is \(-3.5+1.25=-2.25\). 2. The second factor is \(0.8-1.2=-0.4\). 3. The product is \((-2.25)\cdot(-0.4)=0.9\).

The expression is \((-3.5+1\frac{1}{4})(0.8-1.2)\), and its value is \(0.9\).

- The dividend comes before the division symbol, and the divisor comes after it. - Convert fractions and decimals to a common form. - Check the sign of the quotient.

1. The dividend is \(-\frac{3}{10}-0.2=-0.3-0.2=-0.5\). 2. The divisor is \(\frac{1}{8}+(-0.375)=0.125-0.375=-0.25\). 3. The quotient is \((-0.5)\div(-0.25)=2\).

The expression is \((-\frac{3}{10}-0.2)\div(\frac{1}{8}+(-0.375))\), and its value is \(2\).

- Apply the order of operations inside each set of parentheses. - How do you divide by a fraction? - Simplify the final fraction or write it as a mixed number.

1. Evaluate the first parentheses: \(-4.5=-\frac{9}{2}\), so \(-\frac{9}{2}\div\frac{9}{10}=-\frac{9}{2}\cdot\frac{10}{9}=-5\). 2. Evaluate the second parentheses: \(1.2=\frac{6}{5}\), so \(\frac{2}{3}-\frac{6}{5}=\frac{10}{15}-\frac{18}{15}=-\frac{8}{15}\). 3. Multiply: \(-5\cdot\left(-\frac{8}{15}\right)=\frac{8}{3}=2 \frac{2}{3}\).

\(\frac{8}{3}\), or \(2 \frac{2}{3}\)

- Identify the product and the difference named in the description. - Pay attention to what is subtracted from what. - Write the complete expression with parentheses before evaluating.

1. The expression is \((\frac{1}{4}-0.75)-(-1.2\cdot\frac{5}{6})\). 2. The first difference is \(\frac{1}{4}-\frac{3}{4}=-0.5\). 3. The product is \(-1.2\cdot\frac{5}{6}=-1\). 4. The final value is \(-0.5-(-1)=0.5\).

The expression is \((\frac{1}{4}-0.75)-(-1.2\cdot\frac{5}{6})\), and its value is \(0.5\).

- Find the sum and difference first. - Pay attention when subtracting a negative number. - Multiply the two final results to identify their relationship.

1. The first expression is \((a-b)\div(a+b)\). Substitution gives \((0.4-(-0.6))\div(0.4+(-0.6))=1\div(-0.2)=-5\). 2. The second expression is \((a+b)\div(a-b)\). Substitution gives \((0.4+(-0.6))\div(0.4-(-0.6))=-0.2\div1=-0.2\). 3. Since \((-5)\cdot(-0.2)=1\), the results are reciprocals.

1. \((a-b)\div(a+b)=-5\) 2. \((a+b)\div(a-b)=-0.2\) The results are reciprocals.

- Which operations have priority over addition? - What does an exponent mean? - Check the sign of the value inside the parentheses.

1. The first line contains two errors: \(1.5+2.5\) was added before the multiplication, and \(3^2\) was incorrectly treated as \(3\cdot2\). 2. Evaluate the parentheses: \(4.4-6.4=-2\). 3. Evaluate the power: \(3^2=9\). 4. Multiply before adding or subtracting: \(1.5+2.5\cdot(-2)-9=1.5-5-9\). 5. Finish: \(1.5-5-9=-12.5\).

The errors are in the first shown step: addition was done before multiplication, and \(3^2\) was treated as \(3\cdot2\). The correct value is \(-12.5\).

- Evaluate \(A\) and \(B\) separately first. - What sign results when a negative number is squared? - Converting \(\frac{1}{4}\) to a decimal may make \(B\) easier to evaluate.

1. Evaluate \(A\): \((-0.6)^2=0.36\) and \(0.64\div(-2)=-0.32\). Thus, \(A=0.36-0.32=0.04\). 2. Evaluate \(B\): \(1.2-0.7=0.5\), so \(0.5\cdot0.5=0.25\). Then \(B=\frac{1}{4}-0.25=0\). 3. Since \(0.04>0\), \(A>B\).

\(A>B\), because \(A=0.04\) and \(B=0\).

- Check each line against the order of operations. - Focus on the change from line 2 to line 3. Which operation should happen first? - Does division or subtraction have priority?

1. Lines 1 and 2 are correct. The first error occurs in line 3, where subtraction was performed before the division. 2. Starting from line 2, divide first: \(1.2\div0.3=4\). 3. Then subtract: \(\frac{3}{4}-4=0.75-4=-3.25\).

The first error is in line 3 because subtraction was done before division. The correct value is \(-3.25\).

- A quotient indicates division, and a product indicates multiplication. - Evaluate the two parts separately. - Then add the results.

1. The expression is \((12.6\div3)+(\frac{1}{2}\cdot1.4)\). 2. The quotient is \(12.6\div3=4.2\). 3. The product is \(\frac{1}{2}\cdot1.4=0.7\). 4. Add: \(4.2+0.7=4.9\).

The expression is \((12.6\div3)+(\frac{1}{2}\cdot1.4)\), and its value is \(4.9\).

- Identify what is subtracted from what. - Evaluate the sum and product separately. - Pay attention when subtracting a negative number.

1. The expression is \((-1.2+2\frac{1}{4})-(-\frac{3}{5}\cdot0.5)\). 2. The sum is \(-1.2+2.25=1.05\). 3. The product is \(-\frac{3}{5}\cdot0.5=-0.3\). 4. Subtract: \(1.05-(-0.3)=1.35\).

The expression is \((-1.2+2\frac{1}{4})-(-\frac{3}{5}\cdot0.5)\), and its value is \(1.35\).

- Use parentheses to separate the difference and product. - Convert the mixed number and decimal to compatible forms. - Evaluate the numerator and denominator before dividing.

1. The expression is \((-5.25-(-2\frac{1}{4}))\div(\frac{2}{3}\cdot1.5)\). 2. The difference is \(-5.25-(-2.25)=-3\). 3. The product is \(\frac{2}{3}\cdot\frac{3}{2}=1\). 4. Divide: \(-3\div1=-3\).

The expression is \((-5.25-(-2\frac{1}{4}))\div(\frac{2}{3}\cdot1.5)\), and its value is \(-3\).

- Follow the order of operations. - Would fractions or decimals be easier for each part? - Pay close attention to the signs when subtracting. - Break the expression into smaller parts.

1. Evaluate the first brackets: \(-1.5+\frac{2}{5}=-1.5+0.4=-1.1\). 2. Multiply: \(-1.1\cdot10=-11\). 3. Evaluate the second parentheses: \(-\frac{3}{4}\div0.25=-0.75\div0.25=-3\). 4. Subtract: \(-11-(-3)=-8\).

\(-8\)

- Start with the innermost parentheses. - What sign results when two negative numbers are multiplied? - How do you divide by a fraction? - Decide whether a fraction or decimal form is more useful for the final result.

1. Evaluate the inner parentheses: \(\frac{1}{5}-0.6=0.2-0.6=-0.4\). 2. Multiply inside the brackets: \((-2.5)\cdot(-0.4)=1\). 3. Divide: \(1\div\left(-\frac{1}{2}\right)=-2\). 4. Subtract: \(\frac{5}{8}-(-2)=\frac{21}{8}=2 \frac{5}{8}=2.625\).

\(2 \frac{5}{8}\), or \(2.625\)

- Work from the innermost grouping symbols outward. - Can you simplify the division by the decimal? - What sign results from multiplying two negative numbers? - Complete the division inside the braces before subtracting.

1. Evaluate the brackets: \(4-\frac{1}{2}=3.5\). 2. Divide: \(3.5\div(-0.7)=-5\). 3. Subtract inside the braces: \(-5-2=-7\). 4. Multiply: \(-7\cdot\left(-\frac{2}{7}\right)=2\).

\(2\)

- Reverse the steps from last to first. - Use the inverse operation for each instruction. - Track decimal place value carefully.

1. Work backward from \(25\). Undo the subtraction: \(25+6.8=31.8\). 2. Undo multiplication by \(4\): \(31.8\div4=7.95\). 3. Undo addition of \(4.2\): \(7.95-4.2=3.75\). 4. Undo multiplication by \(0.5\): \(3.75\div0.5=7.5\).

\(7.5\)

- First find the fraction of the park occupied by the lake. - Apply that fraction to the total number of acres. - Convert the resulting acreage to square feet.

1. The forest and lawn fraction is \(\frac{2}{5}+\frac{3}{10}=\frac{7}{10}\). 2. The lake fraction is \(1-\frac{7}{10}=\frac{3}{10}\). 3. The lake covers \(\frac{3}{10}\cdot15=4.5\) acres. 4. Convert to square feet: \(4.5\cdot43{,}560=196{,}020\,\text{ft}^2\).

The lake covers \(196{,}020\,\text{ft}^2\).

- Decide whether money is spent or received in each story. - Notice whether the calculation applies to one person or three people. - Use the order of operations.

1. Story A matches Expression 2: \(20-3\cdot1.50-2.40=20-4.50-2.40=13.10\). 2. Story B matches Expression 3: \(20-3\cdot1.50+2.40=17.90\). 3. Story C matches Expression 1: \(3\cdot(20-1.50-2.40)=3\cdot16.10=48.30\).

A: Expression 2, \(\$13.10\) B: Expression 3, \(\$17.90\) C: Expression 1, \(\$48.30\)

- Determine whether each discount is applied once or once per child. - Evaluate both expressions. - Compare the totals.

1. Expression 1 subtracts \(\$3.00\) once from the total, so it represents Offer 1. 2. Expression 2 reduces the price for each of the four children, so it represents Offer 2. 3. Expression 1 equals \(14.00+11.00-3.00=22.00\). 4. Expression 2 equals \(4\cdot3.00+11.00=23.00\). 5. Offer 1 costs \(\$1.00\) less.

Expression 1 represents Offer 1, and Expression 2 represents Offer 2. Offer 1 costs less: \(\$22.00\) instead of \(\$23.00\).

- Multiplication and division come before addition and subtraction. - Converting between fractions and decimals may make a step easier.

1. For a), divide before adding: \(0.5\div\frac{1}{2}=1\). Then \(\frac{3}{4}+1=\frac{7}{4}=1.75\). 2. For b), multiply before subtracting: \(1.2\cdot\frac{5}{6}=\frac{6}{5}\cdot\frac{5}{6}=1\). Then \(1-0.4=0.6=\frac{3}{5}\).

a) \(1.75\), or \(1 \frac{3}{4}\) b) \(0.6\), or \(\frac{3}{5}\)

- Evaluate parentheses first. - How do you divide by a fraction? - Use either fractions or decimals consistently when that makes a calculation easier.

1. For a), add inside the parentheses: \(\frac{5}{9}+\frac{1}{3}=\frac{8}{9}\). Then \(\frac{8}{9}\div\frac{4}{3}=\frac{8}{9}\cdot\frac{3}{4}=\frac{2}{3}\). 2. For b), \(\frac{1}{2}-0.2=0.5-0.2=0.3\). Then \(0.6\div0.3=2\).

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

- Evaluate each set of parentheses separately. - Decide whether fractions or decimals make each step simpler. - In b), treat the two parenthesized expressions as separate blocks before subtracting.

1. For a), \(1.4-\frac{2}{5}=1.4-0.4=1\). Then \(\frac{1}{2}\cdot1+\frac{3}{10}=0.5+0.3=0.8=\frac{4}{5}\). 2. For b), \(\frac{2}{3}\div\frac{4}{9}=\frac{2}{3}\cdot\frac{9}{4}=\frac{3}{2}\). Also, \(0.5\cdot\frac{1}{2}=\frac{1}{4}\). Thus, \(\frac{3}{2}-\frac{1}{4}=\frac{5}{4}=1 \frac{1}{4}\).

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

- 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: \(\frac{3}{4}\cdot\frac{1}{3}=\frac{1}{4}\).

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

- Evaluate both sets of parentheses before dividing. - How do you divide by a fraction? - Can you simplify common factors before multiplying?

1. First parentheses: \(\frac{7}{8}-\frac{1}{4}=\frac{7}{8}-\frac{2}{8}=\frac{5}{8}\). 2. Second parentheses: \(\frac{1}{2}+\frac{1}{3}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6}\). 3. Divide by multiplying by the reciprocal: \(\frac{5}{8}\div\frac{5}{6}=\frac{5}{8}\cdot\frac{6}{5}=\frac{3}{4}\).

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

- Follow the order: parentheses, multiplication, then addition. - What common denominator can you use for the final addition? - Simplify intermediate fractions when possible.

1. Evaluate the parentheses: \(\frac{4}{5}-\frac{1}{10}=\frac{8}{10}-\frac{1}{10}=\frac{7}{10}\). 2. Multiply: \(\frac{7}{10}\cdot\frac{2}{3}=\frac{7}{15}\). 3. Add: \(\frac{7}{15}+\frac{1}{6}=\frac{14}{30}+\frac{5}{30}=\frac{19}{30}\).

\(\frac{19}{30}\)

- Find how many equal bags can be made from the total amount. - Add the desired profit to the original cost to find the needed revenue. - Divide that revenue equally across all the bags.

1. Find the number of bags: \(15\div0.25=60\). 2. The required total revenue is \(127.50+67.50=195.00\) dollars. 3. Divide the revenue by the number of bags: \(195.00\div60=3.25\). 4. The price should be \(\$3.25\) per bag.

\(\$3.25\) per bag

- Complete multiplication or division before addition or subtraction. - Determine the sign of each product or quotient before calculating. - For addition with unlike signs, compare absolute values.

1. For a), multiply first: \((-14)\cdot6=-84\). Then \(-84+90=6\). 2. For b), divide first: \((-72)\div(-4)=18\). Then \(18-20=-2\). 3. For c), multiply first: \(38\cdot2=76\). Then \(-105+76=-29\). 4. For d), divide first: \((-83)\div(-1)=83\). Then \(17-83=-66\).

a) \(6\) b) \(-2\) c) \(-29\) d) \(-66\)

- Evaluate parentheses first. - Inside parentheses, multiplication comes before addition. - Write each parenthesized result before completing the outside operation.

1. For a), \(-15+(-25)=-40\), and \((-40)\cdot(-4)=160\). 2. For b), \(-12-8=-20\), and \(200\div(-20)=-10\). 3. For c), multiply first inside the parentheses: \(4\cdot(-5)=-20\). Then \(-20+7=-13\), so \(-13-(-13)=0\).

a) \(160\) b) \(-10\) c) \(0\)

- 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: \(-\frac{1}{2}\div\left(-\frac{1}{4}\right)=-\frac{1}{2}\cdot(-4)=2\).

\(2\)

- Apply parentheses before division and subtraction. - Converting the fraction to a decimal may simplify the first step. - What sign results when dividing two negative numbers?

1. Evaluate the parentheses: \(-2.25+0.5=-1.75\). 2. Divide: \(-1.75\div(-0.5)=3.5\). 3. Subtract: \(3.5-4.5=-1\).

\(-1\)

- Evaluate the expression inside the brackets first. - Choose fractions or decimals based on which form makes each step easier. - What happens when a number and its opposite are added?

1. Multiply the first part: \(\frac{3}{8}\cdot(-1.6)=-0.6\). 2. Inside the brackets, \(\frac{1}{5}\div(-0.4)=0.2\div(-0.4)=-0.5\). 3. Then \(-0.5+0.5=0\). 4. Finish: \(-0.6-0=-0.6=-\frac{3}{5}\).

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

- Pay attention to signs when multiplying and dividing. - Converting a simple fraction to a decimal may make a step easier. - Follow the order of operations.

1. For a), \(\frac{3}{4}-1.25=0.75-1.25=-0.5\). Then \((-0.5)\cdot(-4)=2\), and \(2+0.5=2.5\). 2. For b), \(\frac{1}{5}+0.1=0.2+0.1=0.3\). Then \(-0.6\div0.3=-2\), and \(-2-(-2)=0\).

a) \(2.5\) b) \(0\)

- What sign results when a negative number is squared? - Work from the innermost grouping symbols outward. - Can you write \(\frac{5}{8}\) as a decimal?

1. For a), \(0.4-\frac{4}{5}=0.4-0.8=-0.4\). Then \((-1.5)\cdot(-0.4)=0.6\), and \(0.6\div(-0.3)=-2\). 2. For b), \((-0.5)^2=0.25\). Then \(1.5+0.25=1.75\), and \(\frac{5}{8}-1.75=0.625-1.75=-1.125=-\frac{9}{8}\).

a) \(-2\) b) \(-1.125\), or \(-\frac{9}{8}\)

- For an estimate, round to numbers that are easy to use mentally. - Follow the order of operations for the exact calculation. - Convert a mixed number to a decimal if that makes the arithmetic easier.

1. For a), estimate \(4.2\approx4\) and \(3 \frac{4}{5}=3.8\approx4\), so \((4+4)\cdot0.5\approx4\). 2. Exactly, \(4.2+3.8=8\), and \(8\cdot0.5=4\). 3. For b), estimate \(10.5\approx11\) and \(2.5\approx3\), so \(11-3\cdot3\approx2\). 4. Exactly, multiply before subtracting: \(10.5-2.5\cdot3=10.5-7.5=3\).

a) Estimate: about \(4\); exact value: \(4\) b) Estimate: about \(2\); exact value: \(3\)

- 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: \(-1\div0.5=-2\).

\(-2\)

- Evaluate the parentheses first. - Determine the sign of the value inside the parentheses before multiplying. - What sign results when multiplying two negative numbers?

1. Evaluate the parentheses: \(-4.2+1.85=-2.35\). 2. Multiply: \((-2.35)\cdot(-2.5)=5.875\).

\(5.875\)

- Multiplication and division come before addition. - It may help to write the fraction and decimal in the same form. - Find the two partial results before adding them.

1. Divide: \(\frac{3}{4}=0.75\), so \(0.75\div(-0.5)=-1.5\). 2. Multiply: \(1.2\cdot5=6\). 3. Add: \(-1.5+6=4.5\).

\(4.5\)

- Find the distance between the two endpoints. - Divide the total distance into three equal lengths. - Starting at the left endpoint, add one section length at a time.

1. Find the total length: \(2.4-(-1.2)=3.6\). 2. Divide by \(3\) to find each part’s length: \(3.6\div3=1.2\). 3. Starting at \(-1.2\), the first division point is \(-1.2+1.2=0\). 4. The second division point is \(0+1.2=1.2\).

\(0\) and \(1.2\)

- Convert fractions and decimals to the same form when useful. - Pay attention to signs in division and subtraction. - Apply multiplication and division before addition or subtraction. - Among negative numbers, the value closer to \(0\) is greater.

1. For a), \(1.2\div(-3)=-0.4\), so \(-0.5-0.4=-0.9\). 2. For b), \(0.4\cdot(-2)=-0.8\), and \(-0.8+0.6=-0.2\). 3. For c), \(-\frac{1}{4}-(-0.15)=-0.25+0.15=-0.1\). 4. Since \(-0.1>-0.2>-0.9\), expression c) has the greatest value.

Expression c) has the greatest value, \(-0.1\).

- Use the order of operations for each expression. - In \(-3^2\), does the exponent apply to the negative sign? - Evaluate each expression separately before ordering the results.

1. Expression A: \(0.6\div(-0.2)=-3\), so \(A=0.5-(-3)=3.5\). 2. Expression B: \(\frac{1}{4}-0.75=0.25-0.75=-0.5\), so \(B=2.5\cdot(-0.5)=-1.25\). 3. Expression C: the exponent applies before the leading negative, so \(-3^2=-9\). Then \(C=\frac{-9+5}{4}=-1\). 4. Therefore, \(-1.25<-1<3.5\), so the order is B, C, A.

B, C, A; \(-1.25<-1<3.5\)

- Identify the operation represented by each mathematical term. - In “subtract A from B,” B comes first. - Use parentheses to preserve the described structure.

1. For a), the expression is \((82.7+17.3)-(15.5\cdot4)\). The sum is \(100\), the product is \(62\), and the value is \(38\). 2. For b), the expression is \((100-12.5)\div(5\cdot0.5)\). The difference is \(87.5\), the product is \(2.5\), and the value is \(35\).

a) \((82.7+17.3)-(15.5\cdot4)=38\) b) \((100-12.5)\div(5\cdot0.5)=35\)

- Multiplication and division come before addition and subtraction. - How do you multiply a whole number by a fraction? - How do you divide by a fraction? - Fractions need common denominators before addition or subtraction. - Evaluate parentheses first.

1. For a), multiply first: \(\frac{1}{4}\cdot3=\frac{3}{4}\). Then \(\frac{3}{8}+\frac{3}{4}=\frac{9}{8}=1 \frac{1}{8}\). 2. For b), \(\frac{5}{6}\cdot\frac{9}{10}=\frac{45}{60}=\frac{3}{4}\). 3. For c), \(12\div\frac{4}{5}=12\cdot\frac{5}{4}=15\). Then \(15-7=8\). 4. For d), \(\frac{2}{3}+\frac{1}{6}=\frac{5}{6}\). Then \(\frac{5}{6}\cdot12=10\).

a) \(\frac{9}{8}\), or \(1 \frac{1}{8}\) b) \(\frac{3}{4}\) c) \(8\) d) \(10\)

- Identify each opposite and absolute value first. - Pay attention to the order in a subtraction statement. - Convert fractions to decimals if helpful.

1. For a), the opposite of \(-\frac{3}{4}\) is \(0.75\), and \(|-1.25|=1.25\). Their sum is \(2\). 2. For b), the opposite of \(\frac{1}{5}\) is \(-0.2\). Then \(-0.2-0.6=-0.8\). 3. For c), \(-\frac{5}{8}+0.125=-0.625+0.125=-0.5\). Then \(-0.5\cdot4=-2\).

a) \(2\) b) \(-0.8\) c) \(-2\)

- Find each opposite and absolute value before combining quantities. - In part c), evaluate the sum first. - Check the signs in each result.

1. For a), \(|-3.6|=3.6\), and the opposite of \(0.4\) is \(-0.4\). Thus \(3.6\div(-0.4)=-9\). 2. For b), \(|-\frac{9}{10}|=0.9\), and the opposite of \(-0.4\) is \(0.4\). Thus \(0.9-0.4=0.5\). 3. For c), the opposite of \(1.5\) is \(-1.5\), and \(-\frac{1}{3}+\frac{4}{3}=1\). Thus \(-1.5\cdot1=-1.5\).

a) \(-9\) b) \(0.5\) c) \(-1.5\)

- Represent floors below ground with negative numbers and floors above ground with positive numbers. - Find the height of each endpoint relative to ground level. - The vertical distance is the difference between the two signed heights.

1. Parking Level \(-4\) is at \(-4\cdot11.5=-46\) feet. 2. Level \(6\) is at \(6\cdot11.5=69\) feet. 3. The vertical distance is \(69-(-46)=115\) feet. 4. Equivalently, the elevator moves across \(6-(-4)=10\) floor intervals, and \(10\cdot11.5=115\) feet.

\(115\) feet

- Review the sign rules for multiplication and division. - Adding two negative numbers gives a negative sum. - Subtracting a negative number is equivalent to adding.

1. The product of two negative numbers is positive: \((-5)\cdot(-2)=10\). 2. The sum of two negative numbers is negative: \(-5+(-2)=-7\). 3. Subtracting a negative number gives \(y-x=-2-(-5)=-2+5=3\), which is positive. 4. The quotient of two negative numbers is positive: \(\frac{-5}{-2}=\frac{5}{2}=2.5\).

a) Positive; \(10\) b) Negative; \(-7\) c) Positive; \(3\) d) Positive; \(2.5\)

- List factor pairs of the required product. - In part b), make sure all three integers are different. - Determine how many negative factors are needed for a negative product.

1. For a), list integer factor pairs of \(-10\): \(1\) and \(-10\), \(-1\) and \(10\), \(2\) and \(-5\), and \(-2\) and \(5\). Only \(2+(-5)=-3\), so the integers are \(2\) and \(-5\). 2. For b), choose three different integer factors with a negative product. For example, \(1\cdot2\cdot(-6)=-12\).

a) \(2\) and \(-5\) b) For example, \(1\), \(2\), and \(-6\)

- Convert all quantities to fractions or all to decimals when helpful. - Pay attention to signs in addition and subtraction. - Apply multiplication before addition. - How can you multiply a fraction and a decimal efficiently?

1. For a), \(-3.5+5-0.75=1.5-0.75=0.75=\frac{3}{4}\). 2. For b), \(1.2=\frac{6}{5}\), so \(\frac{6}{5}\cdot\left(-\frac{5}{6}\right)=-1\). Then \(-1+2.5=1.5=\frac{3}{2}\).

a) \(0.75\), or \(\frac{3}{4}\) b) \(1.5\), or \(\frac{3}{2}\)

- 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) \(-3\)

- Break each expression into smaller parts. - Simplify common factors before multiplying fractions. - When multiplication and division are mixed, work from left to right after parentheses. - Track the sign at each step.

1. For a), \(\frac{2}{7}-2=-\frac{12}{7}\). Then \(-\frac{7}{10}\cdot\left(-\frac{12}{7}\right)=\frac{6}{5}=1.2\). Finally, \(1.2-1.4=-0.2=-\frac{1}{5}\). 2. For b), \(1 \frac{1}{3}+\frac{1}{6}=\frac{4}{3}+\frac{1}{6}=\frac{3}{2}\). Then \(\frac{3}{2}\div\left(-\frac{3}{4}\right)=-2\), and \(-2\cdot0.5=-1\).

a) \(-0.2\), or \(-\frac{1}{5}\) b) \(-1\)

- Identify the sum and difference. - Use parentheses so both are evaluated before multiplication. - Recall the sign rule for multiplying two negative numbers.

1. The sum is \(-14+6=-8\). 2. The difference is \(7-12=-5\). 3. Multiply: \((-8)\cdot(-5)=40\).

\(40\)

- Evaluate each description separately. - “Subtract A from B” means \(B-A\). - Convert to decimals if that makes comparison easier.

1. Value A is \(-\frac{3}{4}\cdot\frac{8}{5}=-\frac{6}{5}=-1.2\). 2. Value B is \(-\frac{7}{10}\div1-\frac{1}{2}=-0.7-0.5=-1.2\). 3. The values are equal.

The values are equal; both are \(-1.2\).

- Apply the order of operations to the student’s expression. - Use parentheses to make the difference and product occur first. - Evaluate those two quantities before dividing.

1. The student’s expression lacks parentheses, so division and multiplication would occur before subtraction. 2. The correct expression is \((2.5-7.5)\div((-0.25)\cdot8)\). 3. The difference is \(-5\), and the product is \(-2\). 4. The quotient is \(-5\div(-2)=2.5\).

Parentheses are needed around the difference and product. The correct value is \(2.5\).

- Multiplication and division come before addition and subtraction. - How do you divide a fraction by a whole number? - Simplify before multiplying when possible. - For b), evaluate the parentheses first.

1. For a), \(\frac{5}{6}\cdot\frac{3}{10}=\frac{1}{4}\) and \(\frac{1}{4}\div(-2)=-\frac{1}{8}\). Then \(\frac{1}{4}-\left(-\frac{1}{8}\right)=\frac{3}{8}\). 2. For b), \(-\frac{2}{3}+\frac{5}{6}=\frac{1}{6}\). Then \(\frac{1}{6}\cdot12=2\), and \(2-7.5=-5.5\).

a) \(\frac{3}{8}\) b) \(-5.5\)

- Identify the least and greatest numbers. - Find their distance by subtracting the lesser value from the greater value. - Divide by \(0.2\) to convert a numerical distance to centimeters on the drawing.

1. Since \(1\,\text{cm}\) represents \(0.2\), divide each absolute value by \(0.2\). Thus, \(-1.4\) is \(7\,\text{cm}\) left of \(0\), \(0.6\) is \(3\,\text{cm}\) right, \(-0.8\) is \(4\,\text{cm}\) left, and \(1.2\) is \(6\,\text{cm}\) right. 2. The least number is \(-1.4\), and the greatest is \(1.2\). 3. Their numerical distance is \(1.2-(-1.4)=2.6\). 4. On the drawing, the length is \(2.6\div0.2=13\,\text{cm}\).

a) \(-1.4\): \(7\,\text{cm}\) left; \(0.6\): \(3\,\text{cm}\) right; \(-0.8\): \(4\,\text{cm}\) left; \(1.2\): \(6\,\text{cm}\) right. b) The numerical distance is \(2.6\), which is \(13\,\text{cm}\) on the drawing.

- Multiply the numerical distance from \(0\) by the scale in centimeters per unit. - The midpoint is the average of the two endpoints. - Use a common denominator before adding the fractions.

1. For a), the distance from \(-1.25\) to \(0\) is \(1.25\) units. The drawing distance is \(1.25\cdot4\,\text{cm}=5\,\text{cm}\), to the left of \(0\). 2. For b), find the midpoint: \(\frac{-\frac{1}{2}+\frac{3}{4}}{2}=\frac{\frac{1}{4}}{2}=\frac{1}{8}\). 3. As a decimal, \(\frac{1}{8}=0.125\).

a) \(5\,\text{cm}\) to the left of \(0\) b) \(\frac{1}{8}=0.125\)

- Find the total numerical distance from \(A\) to \(B\). - Divide that distance by the number of equal intervals. - Use the interval value to move from one point to another.

1. The total numerical distance is \(3-(-2)=5\). 2. One interval represents \(5\div10=0.5\). 3. Four intervals to the right of \(A\) gives \(-2+4\cdot0.5=0\). 4. The distance from \(-0.5\) to \(1.5\) is \(1.5-(-0.5)=2\). Since each interval represents \(0.5\), the number of intervals is \(2\div0.5=4\).

a) \(0.5\) b) \(0\) c) \(4\) intervals

- Apply parentheses before multiplication or division. - How do you divide by a fraction? - Pay close attention to the sign in the final subtraction.

1. Evaluate the parentheses: \(\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\). 2. Multiply: \(\frac{5}{6}\cdot\frac{6}{5}=1\). 3. Divide: \(\frac{3}{4}\div\left(-\frac{1}{2}\right)=\frac{3}{4}\cdot(-2)=-\frac{3}{2}\). 4. Subtract: \(1-\left(-\frac{3}{2}\right)=\frac{5}{2}=2.5\).

\(2.5\), or \(\frac{5}{2}\)

- Convert the mixed number to an improper fraction. - Evaluate the expressions inside the brackets first. - Multiplication and division come before addition inside the brackets. - Simplify common factors before multiplying when possible.

1. Inside the brackets, \(\frac{2}{3}\cdot\left(-\frac{3}{8}\right)=-\frac{1}{4}\). 2. Also, \(\frac{1}{2}\div\frac{4}{5}=\frac{1}{2}\cdot\frac{5}{4}=\frac{5}{8}\). 3. Add inside the brackets: \(-\frac{1}{4}+\frac{5}{8}=\frac{3}{8}\). 4. Subtract from \(1 \frac{1}{4}=\frac{5}{4}\): \(\frac{5}{4}-\frac{3}{8}=\frac{7}{8}=0.875\).

\(\frac{7}{8}\), or \(0.875\)

- The fractions of all ingredients must add to one whole. - Subtract the total juice fraction from \(1\). - Once you know what fraction equals \(3\,\text{cups}\), scale up to the whole amount.

1. Add the juice fractions: \(\frac{1}{3} + \frac{1}{4} + \frac{1}{6} = \frac{4}{12} + \frac{3}{12} + \frac{2}{12} = \frac{9}{12} = \frac{3}{4}\). 2. Subtract from the whole: \(1 - \frac{3}{4} = \frac{1}{4}\). Sparkling water is one-fourth of the punch. 3. Let \(x\) be the total number of cups. Write \(\frac{1}{4}x = 3\). 4. Multiply by \(4\): \(x = 12\). 5. Find the orange juice amount: \(\frac{1}{3} \cdot 12 = 4\,\text{cups}\).

a) Sparkling water makes up \(\frac{1}{4}\) of the punch. b) Sarah makes \(12\,\text{cups}\) of punch. c) The punch contains \(4\,\text{cups}\) of orange juice.

- Consider what “rounded to the nearest whole percent” says about precision. - First calculate the reported portion of all \(500\) visitors. - A head count must be a whole number. - Check \(31\) as a percent of the roller-coaster fans and round it.

1. For part a, the calculated percent of all visitors is \(0.42 \cdot 0.15 = 0.063 = 6.3\%\). 2. For part b, \(500 \cdot 0.063 = 31.5\). This is not an exact count because people are counted in whole numbers and the \(15\%\) value was rounded. 3. There were \(500 \cdot 0.42 = 210\) roller-coaster fans. 4. If \(31\) of them preferred wooden coasters, the actual subgroup percent was \(\frac{31}{210} \cdot 100\% \approx 14.76\%\). 5. Since \(14.76\%\) rounds to \(15\%\), the employee’s statement is consistent with the survey report.

a) The calculated overall percent is \(6.3\%\). b) The calculation gives \(31.5\) people, but this is not an exact count because the subgroup percent was rounded and people are counted in whole numbers. c) Yes. \(31\) out of \(210\) is approximately \(14.76\%\), which rounds to \(15\%\).

- Look for combinations that make easy whole numbers. - Decide whether regrouping or evaluating parentheses first is more efficient. - Simplify fractions before multiplying when possible.

1. For a), \(18-50=-32\), so \(42+(-32)=10\). 2. For b), \(2.3+1.7=4\). Then regroup as \(6.7+3.3-4=10-4=6\). 3. For c), \(\frac{17}{8}-\frac{5}{8}=\frac{12}{8}=\frac{3}{2}\). Then \(\frac{4}{9}\cdot\frac{3}{2}=\frac{2}{3}\).

a) \(10\) b) \(6\) c) \(\frac{2}{3}\)

- In a), identify the common factor. - In b), pair factors that make \(10\). - In c), cancel common factors before multiplying the fractions.

1. For a), factor out \(\frac{3}{5}\): \(\frac{3}{5}(17 + 8) = \frac{3}{5} \cdot 25 = 15\). 2. For b), rearrange and regroup: \((-2.5)(-4) \cdot 7 = 10 \cdot 7 = 70\). 3. For c), simplify the product: \(\frac{5}{6} \cdot \frac{18}{5} = 3\). Then \(12 - 3 = 9\).

a) \(15\) b) \(70\) c) \(9\)

- 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) \(-60\)

- Translate sum, difference, product, and quotient carefully. - Use parentheses to preserve the described order. - Evaluate inside parentheses first.

1. For a), \((12.5-3.5)\cdot(0.4+0.6)=9\cdot1=9\). 2. For b), \((7.2+4.8)\div(2\cdot1.5)=12\div3=4\).

a) \((12.5-3.5)\cdot(0.4+0.6)=9\) b) \((7.2+4.8)\div(2\cdot1.5)=4\)

- Multiplication and division come before subtraction. - What sign results when a positive number is divided by a negative number? - Pay attention to both the subtraction sign and the negative product.

1. Divide: \(18\div(-3)=-6\). 2. Multiply: \(4\cdot(-2.5)=-10\). 3. Subtract: \(-6-(-10)=4\).

\(4\)

- How do you multiply fractions? - How do you divide by a fraction? - Use a common denominator before adding the two fraction results.

1. Multiply: \(\left(-\frac{5}{6}\right)\cdot\frac{3}{10}=-\frac{1}{4}\). 2. Divide: \(\frac{2}{5}\div\left(-\frac{4}{5}\right)=\frac{2}{5}\cdot\left(-\frac{5}{4}\right)=-\frac{1}{2}\). 3. Add: \(-\frac{1}{4}+\left(-\frac{1}{2}\right)=-\frac{3}{4}\).

\(-\frac{3}{4}\), or \(-0.75\)

- 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 \(-0.6\)

- Treat the numerator and denominator as separate expressions first. - Apply powers before multiplication or subtraction. - A fraction bar groups the entire numerator and denominator. - Simplify common factors before multiplying fractions when possible.

1. Numerator: \(\left(\frac{1}{3}\right)^2=\frac{1}{9}\), so \(\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\). 2. Denominator: \(1.25=\frac{5}{4}\), so \(\frac{5}{4}\cdot\frac{8}{3}=\frac{10}{3}\). Then \(\frac{10}{3}-4=-\frac{2}{3}\). 3. Divide: \(\frac{7}{18}\div\left(-\frac{2}{3}\right)=\frac{7}{18}\cdot\left(-\frac{3}{2}\right)=-\frac{7}{12}\).

\(-\frac{7}{12}\)

- Use common denominators for the subtractions inside the parentheses. - What sign results when a positive number is divided by a negative number? - Decide whether fractions or decimals are easier for the final addition.

1. First parentheses: \(\frac{5}{6}-\frac{3}{4}=\frac{10}{12}-\frac{9}{12}=\frac{1}{12}\). 2. Second parentheses: \(\frac{1}{2}-\frac{2}{3}=\frac{3}{6}-\frac{4}{6}=-\frac{1}{6}\). 3. Divide: \(\frac{1}{12}\div\left(-\frac{1}{6}\right)=-\frac{1}{2}\). 4. Add: \(-0.5+1.5=1\).

\(1\)

- Follow the order of operations. - Convert between fractions and decimals when that makes a calculation easier. - For division, choose the representation that makes the quotient easiest to compute.

1. For a), multiply first: \(0.3\cdot\frac{1}{3}=\frac{1}{10}\). Then \(\frac{2}{5}+\frac{1}{10}=\frac{1}{2}=0.5\). 2. For b), \(1.2-\frac{3}{4}=1.2-0.75=0.45\). Then \(0.45\div0.5=0.9=\frac{9}{10}\). 3. For c), \(0.75^2=\left(\frac{3}{4}\right)^2=\frac{9}{16}\). Then \(\frac{9}{16}-\frac{1}{8}=\frac{7}{16}=0.4375\).

a) \(0.5\), or \(\frac{1}{2}\) b) \(0.9\), or \(\frac{9}{10}\) c) \(0.4375\), or \(\frac{7}{16}\)

- What is the difference between \(-x^2\) and \((-x)^2\)? - Work from inside parentheses outward when powers are involved. - Track signs carefully in multiplication.

1. For a), \(-2^4=-16\) because the exponent applies before the leading negative. Also, \((-3)^2=9\), so \(9\cdot\frac{1}{9}=1\). Thus, \(-16+1=-15\). 2. For b), \(\frac{1}{2}-\frac{2}{3}=-\frac{1}{6}\). Squaring gives \(\frac{1}{36}\), so \(\frac{5}{6}-\frac{1}{36}=\frac{29}{36}\). 3. For c), \(10-10^2=10-100=-90\). Then \(-0.1\cdot(-90)=9\).

a) \(-15\) b) \(\frac{29}{36}\) c) \(9\)

- A fraction bar groups the entire numerator and denominator. - Work from the innermost grouping symbols outward. - Choose fractions or decimals based on which form makes each step easier.

1. For a), the numerator is \(\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\), and the denominator is \(\frac{1}{4}+\frac{1}{6}=\frac{5}{12}\). Then \(\frac{1}{6}\div\frac{5}{12}=\frac{2}{5}=0.4\). 2. For b), \(1-\frac{4}{5}=0.2\). Then \(0.5\cdot0.2+0.1=0.2\), so \(2-0.2=1.8=\frac{9}{5}\). 3. For c), \(\left(-\frac{2}{3}\right)^3=-\frac{8}{27}\). Then \(-\frac{8}{27}\div\frac{4}{9}=-\frac{2}{3}\), and \(-\frac{2}{3}+1=\frac{1}{3}\).

a) \(0.4\), or \(\frac{2}{5}\) b) \(1.8\), or \(\frac{9}{5}\) c) \(\frac{1}{3}\)

- Apply multiplication and division before addition and subtraction when there are no grouping symbols. - To divide by a fraction, multiply by its reciprocal. - Simplify fractions when it makes the arithmetic easier. - For fraction addition or subtraction, use a common denominator.

1. For a), multiply first: \(\frac{5}{6}\cdot\frac{9}{10}=\frac{3}{4}\). Then \(\frac{3}{8}-\frac{3}{4}=\frac{3}{8}-\frac{6}{8}=-\frac{3}{8}\). 2. For b), evaluate the parentheses first: \(\frac{2}{5}+\frac{1}{2}=\frac{9}{10}\). Then divide by multiplying by the reciprocal: \(\frac{9}{10}\div\left(-\frac{3}{20}\right)=\frac{9}{10}\cdot\left(-\frac{20}{3}\right)=-6\). 3. For c), divide first: \(\frac{4}{7}\div\frac{8}{21}=\frac{4}{7}\cdot\frac{21}{8}=\frac{3}{2}\). Then \(2-\frac{3}{2}=\frac{1}{2}\).

a) \(-\frac{3}{8}\) b) \(-6\) c) \(\frac{1}{2}\)

- Converting the decimal to a fraction can make the arithmetic easier. - Work inside the parentheses before performing the division and subtraction. - To divide by a fraction, multiply by its reciprocal.

1. Write the decimal as a fraction: \(0.5=\frac{1}{2}\). 2. Evaluate the parentheses: \(\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\). 3. Divide next: \(\frac{5}{6}\div\left(-\frac{1}{2}\right)=\frac{5}{6}\cdot(-2)=-\frac{5}{3}\). 4. Subtract: \(\frac{5}{6}-\left(-\frac{5}{3}\right)=\frac{5}{6}+\frac{10}{6}=\frac{15}{6}=\frac{5}{2}=2.5\).

\(2.5\), or \(\frac{5}{2}\)

- Apply multiplication and division before addition and subtraction. - Review the sign rules for multiplying and dividing positive and negative numbers. - To divide by a fraction, multiply by its reciprocal. - Keep place values aligned when working with decimals.

1. For a), multiply first: \(12\cdot(-5)=-60\). Then \(-60+72=12\). 2. For b), divide by multiplying by the reciprocal: \(\frac{4}{5}\div\left(-\frac{8}{15}\right)=\frac{4}{5}\cdot\left(-\frac{15}{8}\right)=-\frac{3}{2}\). Then \(-\frac{3}{2}+1=-\frac{1}{2}=-0.5\). 3. For c), multiply first: \(0.4\cdot0.3=0.12\). Then \(0.12-0.2=-0.08\).

a) \(12\) b) \(-0.5\), or \(-\frac{1}{2}\) c) \(-0.08\)

- Evaluate each expression separately before comparing them. - Think of each result as a point on a number line. - The distance of a number from \(0\) is its absolute value.

1. Evaluate \(A\): \(-3\cdot4+10=-12+10=-2\). Its distance from \(0\) is \(2\). 2. Evaluate \(B\): \(15\div(-3)+4=-5+4=-1\). Its distance from \(0\) is \(1\). 3. Evaluate \(C\): \(0.5\cdot(-6)+2.8=-3+2.8=-0.2\). Its distance from \(0\) is \(0.2\). 4. Since \(0.2\) is the smallest distance, expression \(C\) has the value closest to \(0\).

Expression \(C\), with value \(-0.2\), is closest to \(0\).

- Use grouping symbols first, then exponents, then multiplication and division, then addition and subtraction. - Pay attention to whether a negative sign is included in the base of a power. - When subtracting a negative number, rewrite the subtraction carefully before simplifying.

1. Evaluate the parentheses first: \(2-2\cdot3=2-6=-4\). Then \(5\cdot(-4)=-20\). 2. Evaluate the power first: \((-2)^2=4\). Then \(4\cdot3=12\), so \(4+12=16\). 3. Evaluate the power: \((-3)^3=-27\). Then \(-27+20=-7\). 4. Evaluate the parentheses: \(2-7=-5\). Then \(15\div(-5)=-3\). 5. Evaluate the parentheses: \(5-8=-3\). Then \(-10-(-3)=-10+3=-7\).

1. \(-20\) 2. \(16\) 3. \(-7\) 4. \(-3\) 5. \(-7\)

- In a), apply multiplication before subtraction. - For fractions with unlike denominators, use a common denominator before adding. - Review the sign rule for multiplying two negative numbers. - To divide by a fraction, multiply by its reciprocal.

1. For a), multiply first: \(\frac{2}{3}\cdot\frac{9}{8}=\frac{3}{4}\). Then \(\frac{3}{4}-\frac{3}{4}=0\). 2. For b), evaluate the parentheses: \(\frac{1}{2}+\frac{3}{5}=\frac{11}{10}\). Then \(\frac{11}{10}\div\frac{11}{10}=1\). 3. For c), multiply in the numerator: \((-4)\cdot(-9)=36\). Then \(36\div6=6\).

a) \(0\) b) \(1\) c) \(6\)

- Identify the operation named by each signal word. - In part b), determine what is subtracted from what. - Find the absolute value before adding in part c.

1. For a), \(-8\cdot(14+(-20))=-8\cdot(-6)=48\). 2. For b), \((-6\cdot7)-(-45)=-42+45=3\). 3. For c), \(2\cdot(-15)+|-50|=-30+50=20\).

a) \(48\) b) \(3\) c) \(20\)

- Distinguish an opposite from an absolute value. - Apply sign rules when multiplying and squaring. - “Subtract A from B” means \(B-A\).

1. For a), \((-15)\cdot(-3-7)=(-15)\cdot(-10)=150\). 2. For b), \((-\frac{1}{2})^2\cdot|-16|=\frac{1}{4}\cdot16=4\). 3. For c), \((-5)\cdot(-9)-(-12+18)=45-6=39\).

a) \(150\) b) \(4\) c) \(39\)

- Translate product and quotient into operations. - Apply the sign rules for multiplication and division. - Add the two intermediate results.

1. The product is \(-4\cdot12=-48\). 2. The quotient is \(100\div(-5)=-20\). 3. Add: \(-48+(-20)=-68\).

\(-68\)

- Identify what is subtracted from what. - Use common denominators when adding or subtracting fractions. - Subtracting a negative number is equivalent to adding its opposite.

1. The sum is \(-\frac{1}{2}+\frac{1}{6}=-\frac{1}{3}\). 2. The product is \(\frac{2}{3}\cdot(-\frac{3}{4})=-\frac{1}{2}\). 3. Subtract: \(-\frac{1}{3}-(-\frac{1}{2})=\frac{1}{6}\).

The expression is \((-\frac{1}{2}+\frac{1}{6})-(\frac{2}{3}\cdot(-\frac{3}{4}))\), and its value is \(\frac{1}{6}\).

- “Decrease by” indicates subtraction. - Find the opposite before adding in part b). - In “the difference of A and B,” calculate \(A-B\).

1. For a), \(-18.5-(12.4-15.9)=-18.5-(-3.5)=-15\). 2. For b), \((-14.6+21.3)+(-7.2)=6.7-7.2=-0.5\). 3. For c), \((-10-(-25))-3\cdot4.5=15-13.5=1.5\).

a) \(-15\) b) \(-0.5\) c) \(1.5\)

- Work backward through the operations. - Which operation undoes division by \(4\)? - What operation undoes subtracting \(24\)? - Check your answer by running it through the machine from the beginning.

1. Work backward from \(10\). Undo division by \(4\): \(10 \cdot 4 = 40\). 2. Undo subtracting \(24\): \(40 + 24 = 64\). 3. Undo multiplication by \(8\): \(64 \div 8 = 8\). 4. Check: \((8 \cdot 8 - 24) \div 4 = 40 \div 4 = 10\).

\(8\)

- Reverse the chain of operations one step at a time. - What must you do first to undo division by \(12\)? - At each step, use the inverse operation.

1. Work backward from \(11\). Undo division by \(12\): \(11 \cdot 12 = 132\). 2. Undo subtracting \(28\): \(132 + 28 = 160\). 3. Undo multiplication by \(4\): \(160 \div 4 = 40\). 4. Undo adding \(17\): \(40 - 17 = 23\). 5. Check: \(((23 + 17) \cdot 4 - 28) \div 12 = 132 \div 12 = 11\).

\(23\)

- First identify the least five-digit whole number. - Work backward and use the inverse of each operation. - Keep the place values aligned when calculating.

1. The least five-digit whole number is \(10{,}000\). 2. Work backward and undo adding \(325\): \(10{,}000 - 325 = 9675\). 3. Undo subtracting \(1100\): \(9675 + 1100 = 10{,}775\). 4. Undo adding \(2450\): \(10{,}775 - 2450 = 8325\). 5. Check: \(8325 + 2450 - 1100 + 325 = 10{,}000\).

\(8325\)

- Round each addend to the nearest thousand for the estimate. - Evaluate the expressions separately. - The negative value farther left is less.

1. Rounding to the nearest thousand gives about \(-3000+(-2000)=-5000\) for Expression 1 and \(-8000+4000=-4000\) for Expression 2, so Expression 1 should be less. 2. Exactly, Expression 1 is \(-3250-1780=-5030\). 3. Expression 2 is \(-8410+3520=-4890\). 4. Since \(-5030<-4890\), Expression 1 is less.

Expression 1 is less. Its exact value is \(-5030\), and Expression 2 equals \(-4890\).

- Round each number to the nearest thousand. - Add the two negative numbers first. - Check that the exact value is close to the estimate.

1. The estimate is \(-5000+(-3000)+7000=-1000\). 2. The exact sum of the first two addends is \(-4890+(-3120)=-8010\). 3. Then \(-8010+6950=-1060\).

Estimate: \(-1000\) Exact value: \(-1060\)

- Find the opposite by changing the sign. - Evaluate absolute value as distance from zero. - Follow the stated operation order.

1. The opposite of \(-350\) is \(350\), and \(|-120|=120\). Thus, \(|-120|+350=120+350=470\). 2. The sum is \(-200+800=600\). Then \(600-450=150\).

a) \(|-120|+350=470\) b) \((-200+800)-450=150\)

- Evaluate the difference, sum, absolute value, and opposite separately. - Subtracting a negative is adding a positive. - Follow the order stated in each description.

1. In part a), \(1500-850=650\), and the opposite of \(150\) is \(-150\). Thus, \(650-(-150)=800\). 2. In part b), \(-125+(-200)=-325\), and \(|-75|=75\). Thus, \(-325+75=-250\).

a) \((1500-850)-(-150)=800\) b) \([-125+(-200)]+|-75|=-250\)

- Identify which quantity is being subtracted from which. - Evaluate the two grouped expressions separately. - Then subtract the second result from the first.

1. The difference is \(300-(-50)=350\). 2. The sum is \(-48+122=74\). 3. Subtracting the sum from the difference gives \(350-74=276\).

\([300-(-50)]-[-48+122]=276\)

- Translate “absolute value” and “opposite” carefully. - Determine which operation happens first. - Evaluate only after matching each instruction.

1. Instruction (1) matches B and gives \(18+12=30\). 2. Instruction (2) matches D and gives \(18+(-12)=6\). 3. Instruction (3) matches A and gives \(-18+(-12)=-30\). 4. Instruction (4) matches C and gives \(-[-30]=30\).

(1) B; \(30\) (2) D; \(6\) (3) A; \(-30\) (4) C; \(30\)

- Translate each instruction into symbols first. - Absolute value is nonnegative. - A number and its opposite sum to zero.

1. The opposite of \(24\) is \(-24\), so \(-16+(-24)=-40\). 2. \(|-35|+(-15)=35-15=20\). 3. \(|-12|+|18|=12+18=30\). 4. \(-[-10+10]=-[0]=0\).

a) \(-16+(-24)=-40\) b) \(|-35|+(-15)=20\) c) \(|-12|+|18|=30\) d) \(-[-10+10]=0\)

- Calculate both sides of each claim separately. - Distinguish the sum of absolute values from the absolute value of a sum. - A number and its opposite add to zero.

1. \(|-7|+|-3|=7+3=10\), and \(-[-7+(-3)]=-(-10)=10\), so a) is true. 2. The opposite of \(-5\) is \(5\), and \(5+(-5)=0\), so b) is true. 3. \(|-15+5|=|-10|=10\), but \(|-15|+|5|=15+5=20\), so c) is false.

a) True b) True c) False

- Round to nearby hundreds. - Track the signs carefully. - Match each estimate before checking exactly.

1. Part a) is about \(-300-400-100=-800\), and exactly equals \(-805\), so it matches 3. 2. Part b) is about \(-900+300+200=-400\), and exactly equals \(-395\), so it matches 4. 3. Part c) is about \(500-250-250=0\), and exactly equals \(-1\), so it matches 1. 4. Part d) is about \(-1200-400-100=-1700\), and exactly equals \(-1700\), so it matches 2.

a) 3 b) 4 c) 1 d) 2

- Use the order of operations. - Evaluate the expression inside the brackets first. - Replace the bracketed expression with its value before continuing.

1. Evaluate inside the brackets: \(52+(-97)=-45\). 2. Substitute the result: \(-128-(-45)\). 3. Rewrite the subtraction and evaluate: \(-128+45=-83\).

\(-83\)

- Break the expression into two smaller calculations. - Evaluate each bracketed expression separately. - Keep track of both intermediate results before the final subtraction. - Subtracting a negative number is the same as adding a positive number.

1. Evaluate the first bracketed expression: \(-34+86=52\). 2. Evaluate the second bracketed expression: \(-42-58=-100\). 3. Subtract the two results: \(52-(-100)=52+100=152\).

\(152\)

- A sum is the result of addition. - The opposite of a number is the same distance from zero on the other side. - Pay close attention to the signs and parentheses in part b).

1. The sum is \((-34)+(-16)=-50\). 2. The opposite of \(50\) is \(-50\). 3. Subtract the sum from that number: \(-50-(-50)=0\).

a) \(-50\) b) \(0\)

- Find the sum and the difference separately. - Subtracting a negative number is the same as adding a positive number. - Read carefully to determine which value is subtracted from which.

1. The sum is \(12+(-18)=-6\). 2. The difference is \(12-(-18)=30\). 3. Subtract the difference from the sum: \(-6-30=-36\).

\(-36\)

- Translate each instruction into one operation in a single expression. - Subtracting a negative number becomes addition. - Perform the operations in order after simplifying the signs.

1. The expression is \(-50-(-30)+(-45)-(+15)\). 2. Simplify the signs: \(-50+30-45-15\). 3. Evaluate: \(-50+30=-20\), \(-20-45=-65\), and \(-65-15=-80\).

Expression: \(-50-(-30)+(-45)-(+15)\) Simplified: \(-50+30-45-15\) Final value: \(-80\)

- Evaluate each expression carefully using the order of operations. - Record the sign of each intermediate value. - Compare all four final values.

1. A has value \(-22+10-8=-20\). 2. B has value \(-22+(8-10)=-24\). 3. C has value \(10-(22+8)=-20\). 4. D has value \(-(22-10)-8=-20\). 5. Expression B is the only expression with a different value.

Expression B does not belong. Its value is \(-24\), while A, C, and D each have value \(-20\).

- Look for numbers that combine to make convenient multiples of \(10\) or \(100\). - When rearranging, keep each number together with the sign before it. - Numbers with the same ones digit may be easy to subtract.

1. Estimate: \(50+60-30+40-80=40\). 2. Rearrange and group: \((48-28)+(63+37)-80\). 3. Evaluate: \(20+100-80=40\).

Estimate: \(40\) Exact value: \(40\)

- Group the terms with negative signs. - Look for pairs that combine to make convenient multiples of \(100\). - Then combine the positive total and the negative total.

1. Estimate by rounding to the nearest hundred: \(-100+400-100+600-300=500\). 2. Group the negative terms: \(-135-65=-200\). 3. Group the positive terms: \(420+580=1000\). 4. Combine the results: \(1000-200-300=500\).

Estimate: \(\approx 500\) Exact value: \(500\)

- Look for numbers with matching ending digits. - Keep each number together with the sign before it when rearranging. - Group terms so that the intermediate values are easy to calculate.

1. Estimate by rounding to the nearest hundred: \(2800-400-1300+600-500=1200\). 2. Rearrange and group: \((2750-1250)+(638-438)-500\). 3. Evaluate: \(1500+200-500=1200\).

Estimate: \(\approx 1200\) Exact value: \(1200\)

- Look for pairs that combine easily. - Keep track of the sign before each number. - To make the total zero, the last number must be the opposite of the sum of the first three terms.

1. For a), group convenient terms: \((-45-55)+(128+72)=-100+200=100\). 2. For b), the first three terms total \(-45+128-55=28\). 3. The new last number \(x\) must satisfy \(28+x=0\), so \(x=-28\).

a) \(100\) b) \(-28\)

- Add the positive terms first. - Find the value of the expression without the final term. - To make the total zero, use the opposite of that value.

1. For a), group the positive terms: \((1250+750)-3400-600=2000-4000=-2000\). 2. For b), the first three terms total \(1250-3400+750=-1400\). 3. The new last term must be \(1400\), because \(-1400+1400=0\).

a) \(-2000\) b) \(+1400\)

- Evaluate the innermost parentheses first. - Replace each set of parentheses with its value before continuing. - Keep careful track of signs in the final subtraction.

1. In a), \(41-55=-14\) and \(82-14=68\). Then \(-14+68=54\), so \(26-54=-28\). 2. In b), \(72-95=-23\). Then \(-18+(-23)=-41\), and \(31+29=60\). Finally, \(-41-60=-101\).

a) \(-28\) b) \(-101\)

- Evaluate each parenthetical expression first. - Pay close attention to subtraction between two grouped expressions. - Write down intermediate values to reduce sign errors.

1. In a), \(74+26=100\) and \(-85+35=-50\). Then \(-32-100+(-50)=-182\). 2. In b), \(-94-126=-220\) and \(-52+318=266\). Then \(-220-266=-486\).

a) \(-182\) b) \(-486\)

- Look for numbers with matching ending digits. - Keep each number together with its sign when rearranging. - Form groups that are easy to evaluate mentally.

1. Rewrite the expression as a sum: \(781+(-495)+(-281)+195+(-100)\). 2. Rearrange and group: \((781-281)+(-495+195)-100\). 3. Evaluate: \(500-300-100=100\).

\(100\)

- Evaluate the two parenthetical expressions separately. - In the final step, a positive number is being subtracted from a negative number.

1. Evaluate the first parentheses: \(12-45=-33\). 2. Evaluate the second parentheses: \(150+5=155\). 3. Subtract the results: \(-33-155=-188\).

\(-188\)

- 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 the sum inside the absolute-value bars first. - Find the opposite of \(-18\). - Pay attention to which quantity is being subtracted.

1. The sum is \(-54+21=-33\). 2. Its absolute value is \(|-33|=33\). 3. The opposite of \(-18\) is \(18\). 4. Therefore, the expression is \(|-54+21|-18\), and its value is \(33-18=15\).

\(|-54+21|-18=15\)

- Find the difference in the order the numbers are given. - Then find and add the two opposites. - Combine the two intermediate results last.

1. Their difference in the given order is \(-125-75=-200\), so its absolute value is \(200\). 2. Their opposites are \(125\) and \(-75\), whose sum is \(125+(-75)=50\). 3. Adding the results gives \(200+50=250\).

\(250\)

- Translate each verbal description into symbols separately. - Follow the order of operations in each expression. - Compare the two final values.

1. Expression A is \(|15-40|=|-25|=25\). 2. Expression B is \(|15|-|40|=15-40=-25\). 3. Since \(25\ne-25\), the expressions do not have the same value.

No. Expression A equals \(25\), and Expression B equals \(-25\).

- 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.

- Break the expression into parenthetical values. - Decide whether the changed parentheses are added or subtracted. - Increasing a subtracted quantity decreases the total by the same amount.

1. An estimate is \(10-100-(-20)=-70\). 2. Exactly, \(38+62=100\) and \(24-44=-20\). 3. Therefore, \(12-100-(-20)=-68\). 4. Replacing \(38\) with \(48\) increases the first parenthetical value by \(10\). Because that value is subtracted, the entire expression decreases by \(10\).

a) Estimate: \(\approx -70\); exact value: \(-68\) b) The value decreases by \(10\).

- Determine the role of \(42\) inside its parentheses. - What happens to a difference when the number being subtracted becomes smaller? - Then account for the subtraction sign before the parentheses.

1. In a), \(92-42=50\) and \(16+34=50\). 2. Therefore, \(25-50-50=-75\). 3. Replacing \(42\) with \(22\) increases the first parenthetical value by \(20\), because less is subtracted from \(92\). 4. Since that parenthetical value is subtracted from the total, the final value decreases by \(20\).

a) \(-75\) b) The value decreases by \(20\).

- Pay attention to the subtraction sign before the second parentheses. - Determine how changing \(12\) changes the value inside the parentheses. - Subtracting a quantity that is \(10\) greater makes the total \(10\) less.

1. In a), \(17+33=50\) and \(12-42=-30\). 2. Therefore, \(-10-50-(-30)=-30\). 3. Replacing \(12\) with \(22\) increases the second parenthetical value by \(10\). 4. Since that parenthetical value is subtracted, the total decreases by \(10\).

a) \(-30\) b) The value decreases by \(10\).

- Group positive and negative changes. - Look for values that cancel. - Combine the remaining changes with the starting temperature.

1. The total is \(-4+7-2+3-8-5\). 2. Group \(7+3=10\) and \(-2-8=-10\), which cancel. 3. The remaining value is \(-4-5=-9\).

\(-9\,^\circ\text{F}\)

- Combine the upward changes. - Compare their total with the starting depth. - Then account for the remaining descents.

1. The total is \(-1250+450-200+800-150\). 2. The rises total \(450+800=1250\), which cancels the starting position. 3. The remaining changes are \(-200-150=-350\).

The submarine is at \(-350\,\text{m}\), or \(350\,\text{m}\) below sea level.

- Group income and expenses. - Look for amounts that make convenient hundreds or thousands. - Combine all signed changes with the starting balance.

1. The balance is \(-240+1550-600-760+450-300\). 2. Group \(-240-760=-1000\), \(1550+450=2000\), and \(-600-300=-900\). 3. Then \(2000-1000-900=100\).

The new balance is \(\$100\).

- Write each expression with variables before substituting values. - Use parentheses around the sum that is subtracted as a group. - Compare the two results and their signs.

1. In a), \((a+b)-c=(-45+20)-(-30)=-25+30=5\). 2. In b), \(c-(a+b)=-30-(-45+20)=-30-(-25)=-5\). 3. The results are opposites. In general, reversing the order of a subtraction changes the sign of the difference.

a) \((-45+20)-(-30)=5\) b) \(-30-(-45+20)=-5\) c) The results are opposites. Reversing the order of a subtraction negates the difference.

- Work backward through the operations. - Undo the last operation first. - Record each intermediate value. - Check by applying the original steps.

1. Undo the subtraction: \(160 + 30 = 190\). 2. Undo the addition: \(190 - 150 = 40\). 3. Undo the division: \(40 \cdot 5 = 200\). 4. Check: \(200 \div 5 + 150 - 30 = 160\).

The original number is \(200\).

- Treat a temperature difference as a distance on a number line. - For part b), subtract the starting temperature from the ending temperature. - To find the midpoint, add the two endpoint values and divide by \(2\).

1. The difference between Compartments 1 and 3 is \(6-(-15)=21\,^\circ\text{C}\). 2. The increase from Compartment 4 to Compartment 2 is \(-2-(-28)=26\,^\circ\text{C}\). 3. The midpoint is the average of the two temperatures: \(\frac{6+(-28)}{2}=\frac{-22}{2}=-11\,^\circ\text{C}\).

a) \(21\,^\circ\text{C}\) b) \(26\,^\circ\text{C}\) c) \(-11\,^\circ\text{C}\)

- List the first four prime numbers. - Evaluate the absolute value separately. - Subtract the prime-number sum from the absolute value.

1. The first four prime numbers are \(2,3,5,7\), and their sum is \(17\). 2. The absolute value of \(-90\) is \(90\). 3. Therefore, \(|-90|-(2+3+5+7)=90-17=73\).

\(|-90|-(2+3+5+7)=73\)

- Translate “sum” and “difference” into operations. - Use parentheses to keep the two intermediate expressions separate. - The wording “from the sum, subtract the difference” determines the order.

1. The sum is \(-412+187=-225\). 2. The difference is \(-256-(-58)=-198\). 3. Subtract the results: \(-225-(-198)=-27\).

\((-412+187)-(-256-(-58))=-27\)

- Evaluate the absolute value separately. - Find the sum before taking its opposite. - Add the two intermediate results.

1. \(|-815|=815\). 2. \(462+(-194)=268\), whose opposite is \(-268\). 3. Therefore, \(815+(-268)=547\).

\(|-815|-[462+(-194)]=547\)

- Translate “sum” into addition. - The wording “subtract \(-150\) from” determines the order. - Subtracting a negative number is the same as adding a positive number.

1. The expression is \((-65+38)-(-150)\). 2. The sum is \(-65+38=-27\). 3. Then \(-27-(-150)=-27+150=123\).

\((-65+38)-(-150)=123\)

- Separate the description into two differences. - “The difference of \(a\) and \(b\)” means \(a-b\). - Evaluate each grouped difference before adding them.

1. The expression is \((210-450)+[16-(-34)]\). 2. The first difference is \(210-450=-240\). 3. The second difference is \(16-(-34)=50\). 4. Add the results: \(-240+50=-190\).

\((210-450)+[16-(-34)]=-190\)

- Work from the innermost grouping symbols outward. - Write each intermediate value to keep track of the signs. - Subtracting a negative number is the same as adding a positive number.

1. Evaluate the inner parentheses: \(-12+25=13\) and \(8-15=-7\). 2. Evaluate the brackets: \(13-(-7)=20\). 3. Complete the calculation: \(30-20=10\).

\(10\)

- Evaluate parentheses first. - A negative sign outside parentheses takes the opposite of the parenthetical value. - Write each intermediate value before continuing.

1. In a), \(112-250=-138\), so \(84-(-138)=222\). 2. In b), \(-44+19=-25\), so \(-37-25+60=-2\). 3. In c), \(55-92=-37\) and \(18+24=42\), so \(-(-37)-42=-5\).

a) \(222\) b) \(-2\) c) \(-5\)

- Work from the innermost grouping symbols outward. - Record each intermediate result. - Pay close attention when subtracting a negative value.

1. In a), \(14-50=-36\). Then \(32-(-36)=68\), and \(15-68=-53\). 2. In b), \(-45+18=-27\). Then \(-27-(-12)=-15\), and \(-21+(-15)=-36\).

a) \(-53\) b) \(-36\)

- Evaluate the parentheses first. - Subtracting a negative number becomes addition. - Work one operation at a time.

1. Evaluate inside the parentheses: \(15-60=-45\). 2. Substitute the result: \(-35-(-45)+(-20)\). 3. Evaluate: \(-35+45-20=-10\).

\(-10\)

- Evaluate the two parenthetical expressions separately. - Subtracting a negative number becomes addition. - Add the two intermediate results with their signs.

1. Evaluate the first parentheses: \(120-250=-130\). 2. Evaluate the second parentheses: \(85-(-15)=100\). 3. Add the results: \(-130+100=-30\).

\(-30\)

- Evaluate each expression in parentheses first. - What happens when a number is multiplied by \(0\)? - Is division by every number defined? - Decide whether each equation is mathematically possible before solving.

1. a) Since \(12 - 12 = 0\), the equation becomes \(x \cdot 0 = 8\). The left side is always \(0\), so there is no solution. 2. b) Since \(5 - 4 = 1\), the equation becomes \(x \div 1 = 10\), so \(x = 10\). 3. c) Since \(10 - 2 \cdot 5 = 0\), the expression requires division by \(0\), which is undefined. There is no valid value of \(x\). 4. d) Divide by \(2\): \(x + 3 = 6\). Then subtract \(3\): \(x = 3\).

There is no valid value of \(x\) in a) or c). In a), \(x \cdot 0\) cannot equal \(8\). In c), the expression requires division by \(0\).

- 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: \(15-10=5\).

\(5\)

- Combine the two multiplication steps. - Then account for the division by \(2\). - Determine how many times the original number appears in the final result.

1. Combine the operations: multiplying by \(4\), then by \(5\), and dividing by \(2\) is equivalent to multiplying by \((4 \cdot 5) \div 2 = 10\). 2. If the original number is \(x\), then \(10x = 70\). 3. Divide by \(10\): \(x = 7\).

\(7\)

- A withdrawal decreases the balance. - Find the balance after the first transaction before applying the second. - Spending more money when the balance is already negative makes the balance more negative.

1. After the purchase, \(120-150=-30\), so the balance is \(-\$30\). 2. After the ATM withdrawal, \(-30-40=-70\), so the balance is \(-\$70\).

After the purchase: \(\$120-\$150=-\$30\) After the ATM withdrawal: \(-\$30-\$40=-\$70\)

- Picture the numbers on a number line before ordering them. - To make the least sum, consider the two least values. - The greatest distance occurs between the least and greatest values.

1. Ordering the numbers from greatest to least gives \(7>2>0>-3>-8>-12\). 2. The least sum comes from the two least numbers: \(-12+(-8)=-20\). The pair is \(-12\) and \(-8\). 3. The greatest distance is between the least and greatest numbers: \(7-(-12)=19\). The pair is \(-12\) and \(7\).

a) \(7>2>0>-3>-8>-12\) b) \(-12\) and \(-8\) c) \(-12\) and \(7\)

- “Strictly between” means that the endpoints are not included. - Place the values mentally on a number line to order them. - For the sum, test pairs containing one positive and one negative number.

1. The numbers greater than \(-6\) and less than \(4\) are \(-5,-1,3\). 2. From least to greatest, the numbers are \(-10<-5<-1<3<4<6\). 3. Since \(6+(-1)=5\), one qualifying pair is \(6\) and \(-1\).

a) \(-5,-1,3\) b) \(-10<-5<-1<3<4<6\) c) \(6\) and \(-1\)

- Round each term to the nearest ten. - Look for pairs that make \(200\) and \(-100\). - Compare the exact value with the estimate.

1. The estimate is \(130-60-40+70-110=-10\). 2. Regroup the exact expression as \((128+72)+(-57-43)-110\). 3. This gives \(200-100-110=-10\).

Estimate: approximately \(-10\) Exact value: \(-10\)

- Evaluate the parentheses first. - A positive sign before the parentheses leaves the result unchanged. - Then add from left to right.

1. Evaluate the parentheses: \(18-50=-32\). 2. Then \(-42+(-32)+60=-74+60=-14\).

\(-14\)

- Work from the innermost grouping symbols outward. - Pay attention to every subtraction sign before a grouped value. - Subtracting a negative number becomes addition.

1. Evaluate the inner parentheses: \(25-45=-20\). 2. Evaluate the brackets: \(-30-(-20)=-10\). 3. Complete the calculation: \(80-(-10)=90\).

\(90\)

- Work backward from the final number of passengers. - Use the inverse of “get off” and the inverse of “double.” - Identify what happened immediately before the final stop.

1. Work backward from \(15\) passengers. 2. Undo the \(5\) passengers getting off: \(15 + 5 = 20\). 3. Undo the doubling: \(20 \div 2 = 10\). 4. Undo the \(8\) passengers getting on: \(10 - 8 = 2\).

\(2\) passengers.

- Translate “difference” and “sum” into operations. - Write the full expression with parentheses. - Evaluate each grouped part first.

1. The difference is \(45-120=-75\). 2. The sum is \(-33+(-17)=-50\). 3. Adding the results gives \(-75+(-50)=-125\).

\(-125\)

- First find the opposite of \(45\). - The wording “subtract the sum from” determines the order. - Subtracting a negative number is the same as adding a positive number.

1. The opposite of \(45\) is \(-45\). 2. The sum is \(-215+90=-125\). 3. Subtract the sum: \(-45-(-125)=80\).

\(80\)

- Evaluate the sum and the difference separately. - The final subtraction is difference minus sum. - Subtracting a positive number from a negative number makes the result more negative.

1. The sum is \(-15+42+(-10)=17\). 2. The difference is \(-100-50=-150\). 3. Subtract the sum from the difference: \(-150-17=-167\).

\(-167\)

- Evaluate exponents and parentheses first. - Track the subtraction of a negative quantity. - For a complex fraction, simplify the numerator and denominator separately. - Dividing by a fraction means multiplying by its reciprocal.

1. \(2\cdot0.5^2-3\cdot(0.5-1.5)=2\cdot0.25-3\cdot(-1)=0.5+3=3.5\). 2. \(\frac{\frac{1}{2}+\frac{1}{4}}{\frac{1}{2}\cdot\frac{1}{4}}=\frac{\frac{3}{4}}{\frac{1}{8}}=\frac{3}{4}\cdot 8=6\).

1) \(3.5\) 2) \(6\)

- Follow the order of operations. - Evaluate powers and parentheses before multiplying. - Simplify each quotient separately. - Write intermediate results to track decimal calculations.

1. \(4\cdot(0.5^2-0.5)+\frac{2}{4}=4\cdot(0.25-0.5)+0.5=4\cdot(-0.25)+0.5=-0.5\). 2. \(\frac{12}{1.5}-\frac{0.04}{0.2}=8-0.2=7.8\).

1) \(-0.5\) 2) \(7.8\)

- Apply the transactions in the order given. - Purchases and bill payments decrease the balance; credits increase it. - A positive balance represents available money, while a negative balance represents an overdraft.

1. After the grocery purchase, the balance is \(85-120=-35\), or \(-\$35\). 2. After the credit, the balance is \(-35+50=15\), or \(\$15\). 3. After the bill payment, the balance is \(15-30=-15\), or \(-\$15\). 4. The negative sign means the account is overdrawn by \(\$15\).

After the first transaction, the balance is \(-\$35\); after the second, it is \(\$15\); and after the third, it is \(-\$15\). The negative sign means the account is overdrawn by \(\$15\).

- Picture the temperatures on a vertical number line. - For each change, decide whether to move up or down. - Apply the changes in order from the starting temperature.

1. Monday: \(5+8-10=3\), so the final temperature is \(3\,^\circ\text{C}\). 2. Tuesday: \(-3+6-7=-4\), so the final temperature is \(-4\,^\circ\text{C}\). 3. Wednesday: \(-1-4+5=0\), so the final temperature is \(0\,^\circ\text{C}\).

1. \(3\,^\circ\text{C}\) 2. \(-4\,^\circ\text{C}\) 3. \(0\,^\circ\text{C}\)

- What sign should the net have when expenses are greater than revenue? - Calculate the difference for each day first. - Then add all four signed daily results.

1. Monday: \(145-162=-17\), so the daily net is \(-\$17\). 2. Tuesday: \(210-185=25\), so the daily net is \(\$25\). 3. Wednesday: \(95-130=-35\), so the daily net is \(-\$35\). 4. Thursday: \(178-178=0\), so the daily net is \(\$0\). 5. The total net is \(-17+25-35+0=-27\), or \(-\$27\).

a) Monday: \(-\$17\); Tuesday: \(\$25\); Wednesday: \(-\$35\); Thursday: \(\$0\) b) The total net is \(-\$27\).

- A negative balance is less than a positive balance. - A purchase decreases a balance, while a deposit increases it. - A difference between balances is their distance on a number line. - Subtracting a negative number is equivalent to adding its opposite.

1. The initial difference is \(12-(-8)=20\), or \(\$20\). 2. Julia's new balance is \(12-15=-3\), or \(-\$3\). 3. Tom's new balance is \(-8+10=2\), or \(\$2\). 4. Since \(2>-3\), Tom has the greater balance. The new difference is \(2-(-3)=5\), or \(\$5\).

a) \(\$20\) b) \(-\$3\) c) \(\$2\) d) Tom has the greater balance, and the difference is \(\$5\).

- Write all numbers in the same form. - The reciprocal of a nonzero fraction is found by switching its numerator and denominator. - Pay attention when adding negative numbers.

1. The sum is \(-4.5+(-1.75)=-6.25\). 2. Since \(0.8=\frac{4}{5}\), its reciprocal is \(\frac{5}{4}=1.25\). 3. Add: \(1.25+(-6.25)=-5\).

\(-5\)

- A smaller number lies farther left on the number line. - Find the opposite and reciprocal in Value A. - Evaluate the grouped sum in Value B before subtracting it.

1. For Value A, the opposite of \(2\frac{3}{5}=2.6\) is \(-2.6\), and the reciprocal of \(0.5\) is \(2\). Thus A is \(-2.6+2=-0.6\). 2. For Value B, first find the sum: \(0.6+(-3.5)=-2.9\). Then \(-1.4-(-2.9)=1.5\). 3. Since \(-0.6<1.5\), Value A lies farther left.

Value A lies farther left because \(-0.6<1.5\).

- Substitute each value using parentheses. - Work from the innermost parentheses outward. - Track the signs when multiplying negative numbers. - Simplify the final fraction.

1. \(4\cdot(-7+3)-(2-5)\cdot(-3)=4\cdot(-4)-(-3)\cdot(-3)=-16-9=-25\). 2. \(\frac{(-4)\cdot 3+8}{-4-2}=\frac{-4}{-6}=\frac{2}{3}\).

1) \(-25\) 2) \(\frac{2}{3}\)

- Use parentheses when substituting negative values. - Evaluate the parentheses in \(B\) before subtracting. - Compare the final values on a number line.

1. For a), \(A=10-(-4)+2=16\), while \(B=10-((-4)+2)=12\). Thus, \(A>B\). 2. For b), \(A=-3.5-1.5+(-5)=-10\), while \(B=-3.5-(1.5+(-5))=0\). Thus, \(A<B\).

a) \(A=16\), \(B=12\), so \(A>B\). b) \(A=-10\), \(B=0\), so \(A<B\).

- Put negative values in parentheses before applying exponents. - Evaluate the numerator and denominator separately. - Even and odd powers affect the sign differently.

1. \(2\cdot(-3)^2-3\cdot\frac{1}{2}=18-1.5=16.5\). 2. \(\frac{-0.4+0.6}{-0.4-0.6}=\frac{0.2}{-1}=-0.2\). 3. \((-2)^3+2\cdot(-2)^2-5=-8+8-5=-5\).

1) \(16.5\) 2) \(-0.2\) 3) \(-5\)

- Evaluate parentheses before exponents. - Divide fractions by multiplying by the reciprocal. - Keep negative values in parentheses when squaring. - Work one term at a time.

1. \(4\cdot(-1.5-0.5)^2=4\cdot(-2)^2=16\). 2. \(\frac{\left(-\frac{1}{3}\right)^2-1}{-\frac{1}{3}+1}=\frac{-\frac{8}{9}}{\frac{2}{3}}=-\frac{8}{9}\cdot\frac{3}{2}=-\frac{4}{3}\). 3. \(0.5\cdot 2^2\cdot(-3)-2\cdot(-3)^2=-6-18=-24\).

1) \(16\) 2) \(-\frac{4}{3}\) 3) \(-24\)

- Record the current level after each move. - Moving down is a negative change, and moving up is a positive change. - Include the starting level when identifying the highest and lowest points. - Find the distance between the highest and lowest signed values.

1. Track each position: \(5-7=-2\), \(-2+3=1\), \(1-4=-3\), and \(-3+2=-1\). The final position is Level \(-1\). 2. The visited levels are \(5,-2,1,-3,-1\). The least value is \(-3\), so the lowest level reached is Level \(-3\). 3. The highest level is \(5\), and the lowest is \(-3\). Their distance is \(5-(-3)=8\) levels.

a) Level \(-1\) b) Level \(-3\) c) \(8\) levels

- Use a variable for the total trail length. - Write an expression for the distance remaining after the morning. - After half of a remaining distance is walked, what fraction of that distance is still left? - You can also work backward from the final \(6\,\text{miles}\).

1. Let \(x\) be the total trail length in miles. 2. The morning distance is \(\frac{1}{3}x + 2\). 3. The distance remaining after the morning is \(x - (\frac{1}{3}x + 2) = \frac{2}{3}x - 2\). 4. After the hiker walks half of that remaining distance, the other half is still left. Write \(0.5(\frac{2}{3}x - 2) = 6\). 5. Simplify: \(\frac{1}{3}x - 1 = 6\). 6. Add \(1\): \(\frac{1}{3}x = 7\). Multiply by \(3\): \(x = 21\).

The trail is \(21\,\text{miles}\) long.

- If \(60\%\) is removed, what percent remains? - Write an expression for the amount left after the first removal. - You can work backward to find the amount present before the car was washed. - Each percent is applied to the amount remaining at that stage.

1. Let \(V\) be the original volume in gallons. 2. After the first removal, the volume is \(V - (\frac{1}{4}V + 20) = \frac{3}{4}V - 20\). 3. Removing \(60\%\) leaves \(40\%\), so write \(0.40(\frac{3}{4}V - 20) = 28\). 4. Divide by \(0.40\): \(\frac{3}{4}V - 20 = 70\). 5. Add \(20\): \(\frac{3}{4}V = 90\). 6. Multiply by \(\frac{4}{3}\): \(V = 120\).

The tank originally contained \(120\,\text{gal}\) of water.

- Test positive numbers, negative numbers, and zero. - Simplify each expression before deciding. - Recall the sign rules for multiplication and what happens when a number is subtracted from itself.

1. The inequality \(a+a<a\) is equivalent to \(2a<a\). Subtracting \(a\) from both sides gives \(a<0\). Therefore, the statement is true exactly when \(a\) is negative. 2. A product is negative when its two factors have opposite signs. Therefore, one of \(x\) and \(y\) is positive and the other is negative. 3. For every rational number \(k\), \(k\cdot0=0\) and \(k-k=0\). Therefore, the equation is always true.

a) Yes; \(a<0\). b) \(x\) and \(y\) have opposite signs. c) Yes; both sides equal \(0\) for every rational number \(k\).

- Think about what happens when you subtract a negative number. - Recall when a fraction or quotient equals zero. - Test the statement using a positive number, a negative number, and zero.

1. The inequality \(x-y>x\) simplifies to \(-y>0\), so \(y<0\). Thus the difference is greater than \(x\) when the subtrahend is negative. For example, \(5-(-2)=7>5\). 2. A quotient with a nonzero divisor equals \(0\) only when its dividend is \(0\). Since \(a\ne0\), \(a\div b\) cannot equal \(0\). 3. If \(x>0\), then \(-x<x\). If \(x<0\), then \(-x>x\). If \(x=0\), then \(-x=x\). Therefore, \(-x<x\) only when \(x\) is positive.

a) Yes; this happens when \(y<0\). For example, \(5-(-2)=7\). b) No; a quotient equals \(0\) only when the dividend is \(0\). c) No. The statement is true only when \(x>0\).

- Start with the known final result. - Use the inverse of each operation as you move upward. - Check by evaluating the tree from top to bottom.

1. Undo the final addition: \(68 - 18 = 50\). 2. Undo multiplication by \(5\): \(50 \div 5 = 10\). 3. Undo subtracting \(12\): \(10 + 12 = 22\). 4. Check: \((22 - 12) \cdot 5 + 18 = 68\).

\(? = 22\)

- Determine the value of one small interval. - Absolute value is distance from \(0\). - A decimal in tenths can be written over \(10\). - The midpoint is the average of two values.

1. Each small interval represents \(0.1\). Reading the marked points gives \(A=-1.2\), \(B=-0.3\), \(C=0.7\), and \(D=1.4\). 2. Their absolute values are \(1.2\), \(0.3\), \(0.7\), and \(1.4\). Therefore, \(D=1.4\) has the greatest absolute value. 3. Since \(B=-0.3\), \(B=-\frac{3}{10}\). 4. The midpoint of \(A\) and \(C\) is \(\frac{-1.2+0.7}{2}=\frac{-0.5}{2}=-0.25\).

a) \(A=-1.2\), \(B=-0.3\), \(C=0.7\), \(D=1.4\) b) \(D=1.4\) c) \(-\frac{3}{10}\) d) \(-0.25\)

- Determine the value of one small interval. - Use the point’s position relative to \(0\) to determine its sign. - Distance is the absolute value of the difference between two coordinates. - Convert quarters to decimals or decimals to fractions as needed.

1. There are \(4\) equal intervals between consecutive integers, so each interval represents \(0.25=\frac{1}{4}\). 2. Point \(A\) is at \(-1.75=-\frac{7}{4}=-1\frac{3}{4}\). 3. Point \(B\) is at \(-0.5=-\frac{1}{2}\). 4. Point \(C\) is at \(0.25=\frac{1}{4}\). 5. The distance from \(A\) to \(B\) is \(\left|-0.5-(-1.75)\right|=1.25=\frac{5}{4}\). 6. The distance from \(B\) to \(C\) is \(|0.25-(-0.5)|=0.75=\frac{3}{4}\).

a) \(A=-1.75=-\frac{7}{4}\); \(B=-0.5=-\frac{1}{2}\); \(C=0.25=\frac{1}{4}\) b) \(AB=1.25=\frac{5}{4}\); \(BC=0.75=\frac{3}{4}\)

- Start with the point for Day 1. - Relate the positive and negative numbers to increases and decreases in the graph. - Compare the value of \(T_1\) with the last plotted temperature. - Interpret the grouped increases and grouped decreases separately.

1. For \(T_1\), \(-4 + 2 + 3 - 1 - 3 + 2 + 3 = 2\). The expression starts with the Day 1 temperature and applies each daily change, so \(2\,^{\circ}\text{C}\) is the temperature on Day 7. 2. For \(T_2\), \((2 + 3 + 2 + 3) - (1 + 3) = 10 - 4 = 6\). The first group is the total of all increases, and the second group is the total of all decreases. The result \(6\,\text{K}\) is the net temperature change from Day 1 to Day 7.

\(T_1 = 2\,^{\circ}\text{C}\); this is the temperature on Day 7. \(T_2 = 6\,\text{K}\); this is the net temperature change from Day 1 to Day 7.

- Determine the value of one small interval. - Count from a nearby labeled value. - Distance is the absolute value of the difference between two coordinates.

1. The interval from \(0\) to \(0.5\) is divided into \(5\) equal parts, so each small interval represents \(0.1\). 2. Point \(A\) is two intervals right of \(-1.5\), so \(A=-1.3\). 3. Point \(B\) is two intervals right of \(-1\), so \(B=-0.8\). 4. Point \(C\) is three intervals right of \(-0.5\), so \(C=-0.2\). 5. Point \(D\) is four intervals right of \(0\), so \(D=0.4\). 6. The distance from \(B\) to \(C\) is \(|-0.2-(-0.8)|=0.6\).

a) \(A=-1.3\), \(B=-0.8\), \(C=-0.2\), \(D=0.4\) b) \(0.6\)

- Separate per-person costs from the one-time group fee. - Check whether the \(\$10.00\) fee is multiplied by the number of people. - Translate each tree into an expression before deciding.

1. Tree A first finds the discounted ticket price: \(18-3=15\). It multiplies by \(15\) people and then adds the one-time fee: \(15\cdot15+10=235\). 2. Tree B adds the \(\$10.00\) fee to each person’s ticket before multiplying by \(15\), incorrectly charging the shared fee \(15\) times. 3. Therefore, Tree A is correct, and the total is \(\$235.00\).

Tree A is correct. The total cost is \(\$235.00\).

- Find the fraction of each batch that is blue. - Find the fraction that is white. - Subtract those fractions from \(1\) to find the red fraction. - Apply the red fraction to each total mass.

1. The blue fraction is \(\frac{180}{450}=\frac{2}{5}\). 2. The white fraction is \(\frac{100}{750}=\frac{2}{15}\). 3. The red fraction is \(1-\frac{2}{5}-\frac{2}{15}=\frac{15}{15}-\frac{6}{15}-\frac{2}{15}=\frac{7}{15}\). 4. Batch A contains \(450\cdot\frac{7}{15}=210\,\text{g}\) of red paint. 5. Batch B contains \(750\cdot\frac{7}{15}=350\,\text{g}\) of red paint.

Batch A contains \(210\,\text{g}\) of red paint, and Batch B contains \(350\,\text{g}\).

- Find the mean of \(A\) and \(B\). - After finding one midpoint, use the interval between \(A\) and that midpoint. - Ask how the new midpoint compares with the preceding midpoint and with \(A\).

1. The midpoint is \(M=\frac{A+B}{2}=\frac{\frac{5}{8}+\frac{6}{8}}{2}=\frac{11}{16}\). 2. Check its position: \(\frac{5}{8}=\frac{10}{16}\) and \(\frac{6}{8}=\frac{12}{16}\), so \(\frac{10}{16}<\frac{11}{16}<\frac{12}{16}\). 3. At each later stage, use \(A\) and the newest midpoint as the two endpoints. Their midpoint is rational and lies strictly between them. Therefore, it is greater than \(A\), smaller than the previous midpoint, and different from every midpoint found earlier. This process can continue indefinitely.

1. \(M=\frac{11}{16}\) 2. Each new midpoint lies strictly between \(A\) and the preceding midpoint, so it is a new rational number inside the original interval. The process can be repeated indefinitely.

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

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

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

- Find the total length already removed. - Find \(25\%\) of the original length. - Determine the total amount that must be removed, then subtract the amount already removed.

1. The pieces already cut total \(4.5+7.75+2.2+5.4=19.85\,\text{ft}\). 2. The target remaining length is \(25\%\) of \(30\,\text{ft}\): \(0.25\cdot30=7.5\,\text{ft}\). 3. Therefore, a total of \(30-7.5=22.5\,\text{ft}\) must be removed. 4. The final piece must be \(22.5-19.85=2.65\,\text{ft}\).

The final piece must be \(2.65\,\text{ft}\) long.

- List the allowed values of \(\triangle\). - Write both terms with denominator \(12\). - Determine when the numerator is divisible by \(12\).

1. Since \(1\le\triangle\le11\), write the expression with denominator \(12\): \(3\frac{\triangle}{12}=\frac{36+\triangle}{12}\) and \(\frac{5}{6}=\frac{10}{12}\). 2. The sum is \(\frac{46+\triangle}{12}\). 3. For the result to be a natural number, \(46+\triangle\) must be divisible by \(12\). 4. The only multiple of \(12\) from \(47\) through \(57\) is \(48\), so \(46+\triangle=48\). 5. Thus \(\triangle=2\), and \(\square=4\).

\(\triangle=2\) and \(\square=4\)

- Translate each proper-fraction condition into an inequality between the symbols. - Compare the two positive fractions by cross-multiplying. - Test small values beginning with the result from part a).

1. The proper-fraction conditions require \(\triangle<\square\) and \(2<\triangle\). Therefore, the smallest possible value of \(\triangle\) is \(3\). 2. Because all values are positive, \(\frac{\triangle}{\square}>\frac{2}{\triangle}\) is equivalent to \(\triangle^2>2\square\), so \(\square<\frac{\triangle^2}{2}\). 3. Combine the conditions: \(\triangle<\square<\frac{\triangle^2}{2}\). 4. For \(\triangle=3\), \(3<\square<4.5\), so \(\square=4\). For \(\triangle=4\), any of \(5, 6, 7\) works. Thus examples include \((3, 4)\) and \((4, 5)\).

a) \(\triangle=3\) b) For example, \((3, 4)\) and \((4, 5)\)

- Find the total cost of the jump ropes first. - Subtract the rope cost from the order total to find the cost of all three medicine balls. - Divide that amount by \(3\) for the unit price. - For part c), use rounded values rather than an exact calculation.

1. The jump ropes cost \(5\cdot\$4.99=\$24.95\). 2. The three medicine balls cost \(\$71.45-\$24.95=\$46.50\). 3. For a), one medicine ball costs \(\$46.50\div3=\$15.50\). 4. For b), the correct order total is \(\$71.45-\$12.50=\$58.95\). 5. For c), the jump ropes cost about \(\$25\), leaving a little more than \(\$46\) for three medicine balls. Since \(3\cdot\$15=\$45\), each ball must cost more than \(\$15\).

a) \(\$15.50\) b) \(\$58.95\) c) About \(\$46\) remains for three balls, and \(3\cdot\$15=\$45\), so each ball must cost more than \(\$15\).

- Find the purchase total before applying the store credits. - Add the three store-credit values. - Compare the correct amount owed with \(\$16.58\). - Check whether the difference matches an item price.

1. The notepads cost \(4\cdot\$2.35=\$9.40\). 2. The highlighters cost \(2\cdot\$4.89=\$9.78\). 3. The purchase total is \(\$9.40+\$9.78+\$12.50=\$31.68\). 4. The store credits total \(\$5.50+\$3.75+\$8.20=\$17.45\). 5. For a), she owes \(\$31.68-\$17.45=\$14.23\). 6. For b), the incorrect amount is \(\$16.58-\$14.23=\$2.35\) too high. That is exactly the price of one notepad, so one notepad was likely scanned twice or a fifth notepad was entered.

a) \(\$14.23\) b) One extra \(\$2.35\) notepad was likely charged.

- Convert the percent to a decimal before dividing. - Converting \(0.75\) to a fraction can simplify the first product. - Evaluate the power and multiplication or division before adding and subtracting.

1. Write \(0.75=\frac{3}{4}\): \(\frac{2}{3}\cdot\frac{3}{4}=\frac{1}{2}=0.5\). 2. Write \(1 \frac{1}{2}=1.5\) and \(75\%=0.75\): \(1.5\div0.75=2\). 3. Evaluate the power: \(\left(\frac{1}{2}\right)^3=\frac{1}{8}=0.125\). 4. Combine: \(0.5+2-0.125=2.375\).

\(2.375\)

- Write the square of \(\frac{1}{n}\) as a single fraction. - Compare positive unit fractions by comparing their denominators. - List the perfect squares strictly between \(10\) and \(50\).

1. Write the condition as \(\frac{1}{50}<\frac{1}{n^2}<\frac{1}{10}\). 2. Since all quantities are positive, taking reciprocals reverses the inequalities: \(10<n^2<50\). 3. The natural-number squares in this interval are \(16,25,36,49\). 4. Therefore, \(n=4,5,6,7\).

\(n\in\{4,5,6,7\}\)

- Focus on the signs of the factors and addends first. - Is the value inside the parentheses in \(A\) positive or negative? - In \(B\), which addend inside the parentheses has the greater absolute value? - Dividing by \(\frac{1}{2}\) is the same as multiplying by \(2\).

1. For \(A\), the quantity \(\frac{1}{2}-\frac{3}{4}\) is negative. A negative number multiplied by a negative number is positive, so \(A\) is positive. 2. For \(B\), \(-2 \frac{2}{3}+1 \frac{1}{6}\) is negative because the negative addend has the greater absolute value. Dividing by the positive number \(\frac{1}{2}\) keeps the result negative, so \(B\) is negative. 3. Evaluate \(A\): \(-\frac{25}{8}\cdot\left(-\frac{1}{4}\right)=\frac{25}{32}\). 4. Evaluate \(B\): \(\left(-\frac{8}{3}+\frac{7}{6}\right)\div\frac{1}{2}=-\frac{3}{2}\cdot2=-3\).

a) \(B\) has a negative value. b) \(A=\frac{25}{32}\); \(B=-3\)

- Converting mixed numbers and decimals to fractions may make the arithmetic easier. - In a), determine the sign of the quantity in parentheses before multiplying. - After applying the order of operations, perform multiplication and division from left to right. - What sign results when a larger positive quantity is subtracted from a smaller one?

1. For a), \(2 \frac{1}{3}=\frac{7}{3}\) and \(4.5-6=-1.5=-\frac{3}{2}\). 2. Then \(\frac{7}{3}\cdot\left(-\frac{3}{2}\right)=-\frac{7}{2}\), and \(-\frac{7}{2}\div\frac{7}{4}=-\frac{7}{2}\cdot\frac{4}{7}=-2\). 3. For b), \(\frac{5}{12}\div\frac{25}{18}=\frac{5}{12}\cdot\frac{18}{25}=\frac{3}{10}\). 4. Also, \(0.4\cdot1 \frac{1}{2}=\frac{2}{5}\cdot\frac{3}{2}=\frac{3}{5}\). Therefore, \(\frac{3}{10}-\frac{3}{5}=-\frac{3}{10}\).

a) \(-2\) b) \(-\frac{3}{10}\), or \(-0.3\)

- Find the weight of each crate in order. - Subtract the crate total from the elevator capacity. - The final answer must be a whole number of packages that does not exceed the limit.

1. Crate B weighs \(\frac{3}{4}\cdot420=315\,\text{lb}\). 2. Crate C weighs \(315+90=405\,\text{lb}\). 3. The crates weigh \(420+315+405=1140\,\text{lb}\) altogether. 4. The remaining capacity is \(1800-1140=660\,\text{lb}\). 5. Since \(660\div25=26.4\), at most \(26\) whole packages can be added.

a) \(1140\,\text{lb}\) b) \(26\) packages

- Convert the percent to a decimal. - Use the completed diagonal to find the common product. - Then use rows, columns, or diagonals with only one missing value.

1. Use the main diagonal to find the common product: \(0.25\cdot0.5\cdot1=0.125\). 2. Convert \(10\%=0.1\). 3. Top middle: \(0.25\cdot x\cdot0.1=0.125\), so \(x=5\). 4. Bottom left: \(0.1\cdot0.5\cdot x=0.125\), so \(x=2.5\). 5. Middle left: \(0.25\cdot x\cdot2.5=0.125\), so \(x=0.2\). 6. Middle right: \(0.2\cdot0.5\cdot x=0.125\), so \(x=1.25\). 7. Bottom middle: \(5\cdot0.5\cdot x=0.125\), so \(x=0.05\).

<table> <tr><td>\(0.25\)</td><td>\(5\)</td><td>\(10\%\)</td></tr> <tr><td>\(0.2\)</td><td>\(0.5\)</td><td>\(1.25\)</td></tr> <tr><td>\(2.5\)</td><td>\(0.05\)</td><td>\(1\)</td></tr> </table> The common product is \(0.125\), or \(\frac{1}{8}\).

- “Filled completely” means only whole containers count. - Find the amount used by the full containers, then subtract from the total. - For part c), first find how much juice the \(20\) large jugs use.

1. For a), \(26\cdot0.75=19.5\), so \(26\) large jugs can be filled and \(20-19.5=0.5\) gallon remains. 2. With small jugs, \(60\cdot0.33=19.8\), so \(0.2\) gallon remains. Since \(0.2<0.5\), the manager's claim is true. 3. For c), \(20\) large jugs hold \(20\cdot0.75=15\) gallons, leaving \(5\) gallons. 4. Fifteen small jugs use \(15\cdot0.33=4.95\) gallons, leaving \(0.05\) gallon. A sixteenth small jug cannot be filled completely.

a) \(26\) large jugs; \(0.5\) gallon remains b) The claim is true. Small jugs leave \(0.2\) gallon, while large jugs leave \(0.5\) gallon. c) \(15\) small jugs

- The parentheses can represent a combined total. - Dividing by \(2\) can represent sharing equally between two people. - Use a context in which all quantities and units make sense.

1. One possible story is: Two friends share the cost of a \(\$12.50\) pizza and three drinks that cost \(\$4.50\) each. 2. Evaluate the multiplication: \(3\cdot4.50=13.50\). 3. Add inside the parentheses: \(12.50+13.50=26.00\). 4. Divide equally: \(26.00\div2=13.00\).

a) Answers will vary. One example is two friends equally sharing a \(\$12.50\) pizza and three \(\$4.50\) drinks. b) \(13\), or \(\$13.00\) in the example

- Find the total production cost first. - Add the target profit to find the total revenue needed. - Determine the remaining revenue and remaining pounds after the morning sales. - Compare the required afternoon price with the morning price.

1. Total production cost: \(120\cdot1.80=216\) dollars. 2. Required total revenue: \(216+240=456\) dollars. 3. Morning revenue: \(70\cdot4.50=315\) dollars. 4. Remaining revenue needed: \(456-315=141\) dollars. 5. Remaining amount: \(120-70=50\) lb. 6. Afternoon price: \(141\div50=2.82\) dollars per pound. 7. Price difference: \(4.50-2.82=1.68\) dollars per pound.

The afternoon price must be \(\$1.68\) per pound lower.

- Track only the signs first for part a). - Subtracting a negative number is equivalent to adding its opposite. - How do you divide by a fraction?

1. For a), \(\frac{1}{4}\cdot(-3)\) is negative, so subtracting it makes the bracket value larger. In fact, its magnitude is \(\frac{3}{4}>\frac{5}{12}\), so the bracket value is positive. A positive number divided by a negative number is negative. 2. For b), \(\frac{1}{4}\cdot(-3)=-\frac{3}{4}\). 3. Then \(-\frac{5}{12}-\left(-\frac{3}{4}\right)=-\frac{5}{12}+\frac{9}{12}=\frac{1}{3}\). 4. Divide: \(\frac{1}{3}\div\left(-\frac{1}{6}\right)=\frac{1}{3}\cdot(-6)=-2\).

a) The final result is negative. b) \(-2\)

- Convert mixed numbers and decimals to fractions if that helps you calculate exactly. - Work from inside the parentheses outward. - Simplify common factors before multiplying when possible.

1. Divide inside the parentheses: \(6 \frac{2}{3}=\frac{20}{3}\) and \(2.5=\frac{5}{2}\), so \(\frac{20}{3}\div\frac{5}{2}=\frac{8}{3}\). 2. Subtract: \(\frac{8}{3}-0.8=\frac{8}{3}-\frac{4}{5}=\frac{28}{15}\). 3. Multiply: \(\frac{28}{15}\cdot1.5=\frac{28}{15}\cdot\frac{3}{2}=\frac{14}{5}\). 4. Add \(2 \frac{1}{4}=\frac{9}{4}\): \(\frac{14}{5}+\frac{9}{4}=\frac{101}{20}=5 \frac{1}{20}=5.05\).

\(5.05\), or \(5 \frac{1}{20}\)

- Find the distance from the known endpoint to the midpoint. - The unknown endpoint is the same distance from the midpoint on the other side. - Check by averaging the two endpoints.

1. The distance from \(-3.5\) to the midpoint \(0.5\) is \(0.5-(-3.5)=4\). 2. Move the same distance to the other side of the midpoint: \(0.5+4=4.5\). 3. Therefore, \(b=4.5\). This also satisfies \(\frac{-3.5+4.5}{2}=0.5\).

\(b=4.5\)

- For estimation, replace complicated values with nearby easy numbers. - In c), determine the sign by counting the negative factors. - Convert between fractions and decimals when that makes the exact calculation easier.

1. For a), \((-1.5)^3\approx-3.4\) and \(0.375\approx0.4\), so the result is about \(-3\). Exactly, \((-1.5)^3+0.375=-3.375+0.375=-3\). 2. For b), the parentheses are about \(-1.5\), and dividing by \(-0.5\) gives about \(3\). Exactly, \((2.25-3.75)\div(-0.5)=3\). 3. For c), the parentheses are about \(-3.8\), so \(0.1\cdot(-3.8)\cdot(-3)\approx1.1\). Exactly, \(0.1\cdot\left(-\frac{23}{6}\right)\cdot(-3)=\frac{23}{20}=1.15\).

a) Estimate: about \(-3\); exact value: \(-3\) b) Estimate: about \(3\); exact value: \(3\) c) Estimate: about \(1.1\); exact value: \(1.15\)

- Evaluate the expression inside the parentheses first. - For part 2, find how much the first number changes. - Track how that change is affected by the division that follows.

1. Write \(\frac{3}{4}=0.75\). Then \(1.25-0.75=0.5\), \(0.5\div0.5=1\), and \(1+0.1=1.1\). 2. Replacing \(1.25\) with \(1.5\) increases the value inside the parentheses by \(0.25\). Dividing that increase by \(0.5\) doubles it, so the final value increases by \(0.5\). The new value is \(1.6\).

1. \(1.1\) 2. The value increases by \(0.5\), from \(1.1\) to \(1.6\).

- To subtract a number, add its opposite. - The opposite of a number has the same distance from zero and the opposite sign. - For parts b) and c), write an equation in the form \(\text{starting value}-x=\text{target value}\).

1. In a), (I) is \((-48)+122=74\), and (II) is \((-210)+(-90)=-300\). 2. In b), solve \(74-x=10\) to get \(x=64\). Solve \(-300-x=10\) to get \(x=-310\). 3. In c), the opposite of \(74\) is \(-74\). Solving \(74-x=-74\) gives \(x=148\). The opposite of \(-300\) is \(300\). Solving \(-300-x=300\) gives \(x=-600\).

a) (I) \(74\); (II) \(-300\) b) (I) \(64\); (II) \(-310\) c) (I) \(148\); (II) \(-600\)

- Evaluate both expressions before answering parts b) and c). - Find the opposite of the value of Expression (I). - Represent the missing number in each subtraction with a variable.

1. In a), (I) is \((-15)+(-85)=-100\), and (II) is \(40+(-110)=-70\). 2. In b), solve \(-100-x=-70\). This gives \(x=-30\). 3. In c), the opposite of \(-100\) is \(100\). Solve \(-70-y=100\) to get \(y=-170\).

a) (I) \(-100\); (II) \(-70\) b) \(-30\) c) \(-170\)

- The sum contains one positive and one negative two-digit number. - To maximize the result, maximize the positive number and minimize the absolute value of the negative number. - To minimize the result, reverse that strategy.

1. The two terms must be one positive two-digit number and one negative two-digit number. 2. To maximize the sum, make the positive number as large as possible while keeping the negative number’s absolute value as small as possible: \((+98)+(-45)=53\). 3. To minimize the sum, reverse the signs: \((-98)+(+45)=-53\).

a) \((+98)+(-45)=53\) b) \((-98)+(+45)=-53\)

- Divide the four digits into two pairs systematically. - Form two-digit numbers from each pair and compare their absolute values. - The sum is closest to zero when those absolute values are as close as possible.

1. The expression is the sum of one positive and one negative two-digit number, so its distance from zero is the difference between their absolute values. 2. Check the possible digit pairings. The closest pair is \(31\) and \(24\), whose difference is \(7\). 3. Therefore, \((+31)+(-24)=7\) or \((-31)+(+24)=-7\). Both results are \(7\) units from zero.

\((+31)+(-24)=7\) or \((-31)+(+24)=-7\)

- Evaluate the innermost parentheses first. - A negative sign outside parentheses changes the sign of the value inside. - Check the sign of every intermediate result.

1. In a), \(34+156-20=170\) and \(55+45=100\). Then \(145-170-100=-125\). 2. In b), \(-760+140=-620\), so \(-(-620)=620\). Also, \(24-110=-86\), and \(-86+480=394\). Therefore, \(620-394=226\).

a) \(-125\) b) \(226\)

- A subtraction sign before parentheses subtracts the value of the entire group. - Try grouping two or three consecutive terms. - The new result must be less than the original result.

1. In a), \(12-8+15-5=14\). 2. For b), group the last three signed terms: \(12-((+8)-(-15)+(-5))\). 3. The grouped expression has value \(8+15-5=18\). 4. Therefore, \(12-18=-6\).

a) \(14\) b) \(12-((+8)-(-15)+(-5))=-6\)

- The new parentheses must enclose at least one operation to affect the value. - Parentheses following a subtraction sign are especially likely to change the result. - Try several placements and evaluate each modified expression.

1. In a), \(-40+10-20-5=-55\). 2. One possible new expression is \(-40-((-10)-(+20))+(-5)\). Its value is \(-40-(-30)-5=-15\). 3. Another possible new expression is \(-40-(-10)-((+20)+(-5))\). Its value is \(-40+10-15=-45\). 4. Both values differ from \(-55\).

a) \(-55\) b) Possible answers: \(-40-((-10)-(+20))+(-5)=-15\) \(-40-(-10)-((+20)+(-5))=-45\)

- Evaluate both parenthetical expressions first. - Determine how far the original value is from \(20\). - Increasing a value in an added group increases the total, while decreasing a value in a subtracted group increases the total.

1. In a), \(12-45=-33\) and \(-18+5=-13\), so \(36-33-(-13)=16\). 2. To increase the total from \(16\) to \(20\), increase the first parenthetical value by \(4\). Replace \(12\) with \(16\). 3. The second parentheses are subtracted. To increase the total by \(4\), decrease that parenthetical value by \(4\). Replace \(-18\) with \(-22\).

a) \(16\) b) Replace \(12\) with \(16\). c) Replace \(-18\) with \(-22\).

- Taking the opposite twice returns to the original number. - List the valid digit arrangements systematically, remembering that zero cannot be the hundreds digit. - Add the four valid numbers before performing the subtraction.

1. The opposite of the opposite of \(-350\) is \(-350\). 2. The valid three-digit numbers are \(104\), \(140\), \(401\), and \(410\). A three-digit number cannot begin with \(0\). 3. Their sum is \(104+140+401+410=1055\). 4. The required expression is \(-350-(104+140+401+410)=-1405\).

\(-350-(104+140+401+410)=-1405\)

- Identify the smallest four-digit number with four identical digits. - Determine the greatest prime less than \(100\). - Pay attention to which quantity is subtracted from which.

1. The smallest positive four-digit number with identical digits is \(1111\), and the greatest two-digit prime is \(97\). 2. Their difference is \(1111-97=1014\). 3. The opposite of \(-235\) is \(235\). 4. Therefore, \((1111-97)-235=779\).

\((1111-97)-235=779\)

- The opposite of a number has the same absolute value and the opposite sign. - Use grouping symbols to separate the sum and the difference. - Evaluate the grouped expressions before the final subtraction.

1. The sum is \(-315+105=-210\), so its opposite is \(210\). 2. The difference is \(45-(-82)=127\). 3. The expression is \(-[(-315)+105]-[45-(-82)]\). 4. Evaluate: \(210-127=83\).

\(-[(-315)+105]-[45-(-82)]=83\)

- A negative result can be made by subtracting a greater value from a smaller value. - Consider whether multiplication can create the value that must be subtracted. - Parentheses can make an entire sum the quantity being subtracted.

1. In a), \(10-4\cdot7=10-28=-18\). 2. In b), \(3\cdot5-20=15-20=-5\). 3. In c), \(10-(25+5)=10-30=-20\).

a) \(10-4\cdot7=-18\) b) \(3\cdot5-20=-5\) c) \(10-(25+5)=-20\)

- Start with the final amount and reverse each step. - If half remained after a purchase, how many were there before it? - Add back any fixed number that was removed.

1. Work backward from the final \(10\) pretzels. 2. Before the third customer bought half, there were \(10 \cdot 2 = 20\) pretzels. 3. Before the second customer bought \(15\), there were \(20 + 15 = 35\) pretzels. 4. Before the first customer bought half, there were \(35 \cdot 2 = 70\) pretzels. 5. Checking forward leaves \(10\) pretzels, so the result is correct.

\(70\) pretzels.

- 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) \(150\)

- Convert mixed numbers and decimals to fractions. - Use common denominators for addition and subtraction. - Pay close attention to the negative sign in front of \(x\). - Work inside the parentheses first.

1. For part a), \(T=-\frac{2}{3}-\left(-\frac{1}{6}-\frac{1}{2}\right)=-\frac{2}{3}-\left(-\frac{2}{3}\right)=0\). 2. For part b), convert the values to fractions: \(x=-\frac{9}{4}\), \(y=\frac{11}{8}\), and \(z=-\frac{1}{2}\). 3. Then \(T=-\left(-\frac{9}{4}\right)-\left(\frac{11}{8}-\left(-\frac{1}{2}\right)\right)=\frac{18}{8}-\frac{15}{8}=\frac{3}{8}\).

a) \(0\) b) \(\frac{3}{8}\)

- Find the fraction of the route remaining after the first day. - Express the second-day distance as a fraction of the total route. - Determine what fraction remains for the third day. - Find each day’s distance before comparing them.

1. Let \(d\) be the total route length in miles. The first-day distance is \(\frac{2}{5}d\). 2. After the first day, \(d - \frac{2}{5}d = \frac{3}{5}d\) remains. 3. The second-day distance is \(\frac{1}{3} \cdot \frac{3}{5}d = \frac{1}{5}d\). 4. The third-day distance is \(\frac{3}{5}d - \frac{1}{5}d = \frac{2}{5}d\). 5. Write \(\frac{2}{5}d = 12\). Multiply by \(\frac{5}{2}\): \(d = 30\). 6. The daily distances are \(12\,\text{miles}\), \(6\,\text{miles}\), and \(12\,\text{miles}\). Therefore, the first and third days have equal distances.

The route is \(30\,\text{miles}\) long. The group hikes \(12\,\text{miles}\) on both the first and third days.

- Begin with the amount left at the end of the third day. - If \(x\) is the morning amount, the evening amount is \(\frac{x}{2} - 0.25\). - Reverse “use half plus one-fourth quart” one day at a time. - A three-row table may help organize the backward steps.

1. Work backward from the end of the third day. If \(x_3\) is the morning amount, then \(\frac{x_3}{2} - 0.25 = 0.5\). Thus \(\frac{x_3}{2} = 0.75\), so \(x_3 = 1.5\,\text{qt}\). 2. Before the second day, solve \(\frac{x_2}{2} - 0.25 = 1.5\). Then \(\frac{x_2}{2} = 1.75\), so \(x_2 = 3.5\,\text{qt}\). 3. Before the first day, solve \(\frac{x_1}{2} - 0.25 = 3.5\). Then \(\frac{x_1}{2} = 3.75\), so \(x_1 = 7.5\,\text{qt}\).

The hiker started with \(7.5\,\text{qt}\) of water.

- Work backward from the final \(\$300\). - After one-fourth of an amount is spent, three-fourths remains before the fixed fee is considered. - Find the amount present before the second month, then find the original amount. - Keep the percentage-based expenses and fixed expenses separate.

1. Let \(R\) be the amount remaining after the first month. After the second-month expenses, \(R - (\frac{1}{4}R + 30) = 300\). 2. Simplify: \(\frac{3}{4}R - 30 = 300\), so \(\frac{3}{4}R = 330\) and \(R = 440\). 3. Let \(x\) be the original fund. After the first month, \(x - (\frac{1}{3}x + 50) = 440\). 4. Simplify: \(\frac{2}{3}x - 50 = 440\), so \(\frac{2}{3}x = 490\). 5. Multiply by \(\frac{3}{2}\): \(x = 735\).

The fund originally contained \(\$735\).

- Follow the tree from the raw score toward the final result. - To work backward, apply inverse operations in reverse order. - Use parentheses so the subtraction occurs before division.

1. The calculation is \((4x-200)\div5\). For \(x=100\): \((400-200)\div5=40\). For \(x=250\): \((1000-200)\div5=160\). For \(x=75\): \((300-200)\div5=20\). 2. Work backward from \(120\): \(120\cdot5=600\), \(600+200=800\), and \(800\div4=200\). The raw score was \(200\). 3. The expression is \((4x-200)\div5\).

a) 1) \(40\) 2) \(160\) 3) \(20\) b) \(200\) raw-score points c) \((4x-200)\div5\)

- The base is lower than the peak by the full height of the mountain. - Subtract \(2600\) from the peak’s elevation. - Depth is reported as a positive distance below sea level.

1. The peak is at \(-50\,\text{m}\). 2. The base is \(2600\,\text{m}\) lower, so its elevation is \(-50-2600=-2650\,\text{m}\). 3. Therefore, the base is \(2650\,\text{m}\) below sea level.

The base is \(2650\,\text{m}\) below sea level.

- Evaluate the branch with only known numbers first. - Find the value the other branch must have to make the final sum. - Then use an inverse operation to find \(x\).

1. Evaluate the known branch: \(15 \cdot 8 = 120\). 2. The other branch must equal \(150 - 120 = 30\). 3. Solve \(x - 25 = 30\): \(x = 30 + 25 = 55\). 4. Check: \(15 \cdot 8 + (55 - 25) = 150\).

\(x = 55\)