- Read the starting value and signed jump from the diagram. - Add tenths to tenths.

1. Start at \(1.2\) and add the jump \(0.3\). 2. \(1.2+0.3=1.5\).

\(1.5\)

- Keep the same unit: tenths. - Add the number of tenths first. - Then write that number of tenths as a decimal.

1. \(24\) tenths plus \(3\) tenths is \(27\) tenths. 2. \(27\) tenths is \(2.7\).

\(27\) tenths, or \(2.7\).

- Rewrite both amounts as tenths. - Add the tenths. - Write the result as a decimal number of liters.

1. \(1.6\) is \(16\) tenths and \(0.3\) is \(3\) tenths. 2. \(16+3=19\) tenths. 3. \(19\) tenths is \(1.9\).

\(16\) tenths \(+3\) tenths \(=19\) tenths, so there are \(1.9\,\text{L}\) in the pitcher.

- Line up digits by place value, with decimal points directly under each other. - Add trailing zeros when they help show matching place values. - Keep the decimal point aligned in the answer.

1. For a), write \(15.8\) as \(15.80\). Then \(9.42+15.80=25.22\). 2. For b), write \(14.6\) as \(14.60\). Then \(14.60-6.75=7.85\). 3. For c), write \(3.59\) as \(3.590\). Then \(0.734+3.590=4.324\). 4. For d), write \(30\) as \(30.000\). Then \(30.000-12.841=17.159\).

a) \(25.22\) b) \(7.85\) c) \(4.324\) d) \(17.159\)

- Round to nearby whole numbers for a quick estimate. - Align the decimal points when finding the exact value. - Add trailing zeros when they help show matching place values.

1. For a), an estimate is \(43+5+13=61\). 2. The exact sum is \(42.70+5.09+13.00=60.79\). 3. For b), an estimate is \(105-37=68\). 4. The exact difference is \(105.200-37.485=67.715\).

a) Estimate: \(61\); exact: \(60.79\) b) Estimate: \(68\); exact: \(67.715\)

- Find the total weight of each package first. - Align the decimal points when adding. - Subtract the smaller total from the larger total to find the difference.

1. Package A weighs \(3.25+1.80=5.05\) pounds. 2. Package B weighs \(2.65+2.45=5.10\) pounds. 3. Since \(5.10>5.05\), Package B is heavier. 4. The difference is \(5.10-5.05=0.05\) pound.

Package B is heavier. It weighs \(5.10\) pounds, which is \(0.05\) pound more than Package A.

- Add the two leg times for each team. - In a timed race, the smaller total time wins. - Subtract the two total times to find the winning margin.

1. Blue Team's total time is \(32.45+31.98=64.43\,\text{s}\). 2. Red Team's total time is \(32.12+32.26=64.38\,\text{s}\). 3. The shorter total time wins, and \(64.38<64.43\), so Red Team wins. 4. The winning margin is \(64.43-64.38=0.05\,\text{s}\).

Red Team wins with a total time of \(64.38\,\text{s}\). The winning margin is \(0.05\,\text{s}\).

- Look for a nearby whole number that is easier to add or subtract. - After using the nearby number, decide whether you added or subtracted too much. - Correct that small difference at the end.

1. For a), subtract \(20\) instead of \(19.99\), then add back \(0.01\): \(45.67-20+0.01=25.68\). 2. For b), add \(4\) instead of \(3.998\), then subtract the extra \(0.002\): \(12.456+4-0.002=16.454\).

a) \(25.68\), using \(45.67-20+0.01\) b) \(16.454\), using \(12.456+4-0.002\)

- Find the cost of the three notebooks first. - Add all the item costs to find the total purchase price. - Subtract the total cost from the amount paid. - Align decimal points carefully.

1. The three notebooks cost \(3\times\$0.60=\$1.80\). 2. The total cost is \(\$1.35+\$1.80+\$9.49=\$12.64\). 3. The change is \(\$20.00-\$12.64=\$7.36\).

\(\$7.36\)

- Align the decimal points before adding. - Add trailing zeros if they help line up place values. - To round to the nearest hundredth, check the thousandths digit.

1. Add the distances: \(7.45+5.7+9.333=22.483\) miles. 2. To round to the nearest hundredth, look at the thousandths digit, which is \(3\). The rounded total is \(22.48\) miles.

a) \(22.483\) miles b) \(22.48\) miles

- How much does each price increase when rounded up to the next whole dollar? - Add the rounded prices first. - Since the rounded prices are too high, how should the final total be adjusted?

1. Round \(\$12.95\) up to \(\$13.00\), an increase of \(\$0.05\), and round \(\$8.98\) up to \(\$9.00\), an increase of \(\$0.02\). 2. Add the rounded amounts: \(\$13.00+\$9.00=\$22.00\). 3. The total rounding increase is \(\$0.05+\$0.02=\$0.07\). 4. Subtract the extra amount: \(\$22.00-\$0.07=\$21.93\).

Round to \(\$13.00\) and \(\$9.00\), add to get \(\$22.00\), then subtract the \(\$0.07\) total adjustment. The exact total is \(\$21.93\).

- Which whole number is very close to \(2.998\)? - Decide whether using that whole number subtracts too much or too little. - Correct the small difference at the end.

1. Replace \(2.998\) with \(3\). 2. \(34.152-3=31.152\). 3. Subtracting \(3\) removed \(0.002\) too much because \(3-2.998=0.002\). 4. Add it back: \(31.152+0.002=31.154\).

\(31.154\), using \(34.152-3+0.002\).

- Identify the full price and the amount already saved. - Find the difference between the two amounts. - Think about how much is needed to reach the next whole dollar first.

1. Subtract the amount already saved from the price: \(\$45.00 - \$27.50 = \$17.50\).

Lukas needs to save \(\$17.50\) more.

- Find the combined value of all three items. - Line up the decimal points when adding. - Add the cents and dollars carefully.

1. Add the three prices: \(\$34.90 + \$12.50 + \$45.00 = \$92.40\).

The Meyer family pays \(\$92.40\) altogether.

- Express each target weight as a sum or difference of the listed weights. - Apply the same addition or subtraction relationship to the prices. - Keep decimal places aligned when operating on the prices.

1. The price of \(15\,\text{oz}\) is \(\$5.60+\$2.80=\$8.40\). 2. The price of \(11\,\text{oz}\) is \(\$5.60+\$0.56=\$6.16\). 3. The price of \(9\,\text{oz}\) is \(\$5.60-\$0.56=\$5.04\).

a) \(\$8.40\) b) \(\$6.16\) c) \(\$5.04\)

- You can convert each amount to cents before adding. - One hundred cents equals \(\$1.00\). - Align decimal points and add by place value. - In part d), write both amounts in the same unit first.

1. a) Add the dollars and cents: \(\$3.45+\$6.55=\$10.00\). 2. b) \(\$12.80+\$4.35=\$17.15\). 3. c) \(\$0.99+\$7.05=\$8.04\), and \(\$8.04+\$1.10=\$9.14\). 4. d) \(85\) cents is \(\$0.85\). Then \(\$0.85+\$2.40=\$3.25\).

a) \(\$10.00\) b) \(\$17.15\) c) \(\$9.14\) d) \(\$3.25\)

- Use subtraction as the inverse of addition. - Count up to a nearby whole dollar when that helps. - In part c), add the known amounts first.

1. a) \(\$20.00-\$14.20=\$5.80\). 2. b) \(\$10.50-\$6.75=\$3.75\). 3. c) \(\$0.55+\$1.25=\$1.80\), then \(\$5.00-\$1.80=\$3.20\).

a) \(\$5.80\) b) \(\$3.75\) c) \(\$3.20\)

- Align the decimal points before adding. - Record each carry when a column totals more than \(9\). - You can check by converting every amount to cents.

1. Align the decimal points and add by place value. 2. Hundredths: \(0+9+5+5=19\). Write \(9\) and carry \(1\). 3. Tenths: \(5+9+0+8+1=23\). Write \(3\) and carry \(2\). 4. Ones: \(8+9+4+7+2=30\). Write \(0\) and carry \(3\). 5. Tens: \(1+0+2+0+3=6\). 6. The total is \(\$60.39\).

\(\$60.39\)

- Align decimal points when subtracting. - You can convert each amount to cents to use whole-number subtraction. - Remember that \(\$1.00\) is \(100\) cents. - Preserve two decimal places for dollar amounts.

1. a) \(\$15.20-\$6.45=\$8.75\). 2. b) \(\$40.05-\$12.30=\$27.75\). 3. c) \(\$8.10-\$0.95=\$7.15\). 4. d) \(\$200.00-\$135.55=\$64.45\).

a) \(\$8.75\) b) \(\$27.75\) c) \(\$7.15\) d) \(\$64.45\)

- Think about subtracting from a whole-dollar amount such as \(\$100.00\). - Pay close attention to zeros in the cents places. - Counting up to the next whole dollar can simplify a calculation. - Check by adding the difference to the amount subtracted.

1. a) \(\$100.00-\$45.67=\$54.33\). 2. b) \(\$23.08-\$14.50=\$8.58\). 3. c) \(\$200.00-\$123.45=\$76.55\). 4. d) \(\$11.02-\$9.95=\$1.07\).

a) \(\$54.33\) b) \(\$8.58\) c) \(\$76.55\) d) \(\$1.07\)

- Identify the two money units used in the expression. - How many cents equal one dollar? - Convert all amounts to the same unit before calculating. - Line up decimal points when adding and subtracting.

1. Convert cents to dollars: \(215\text{ cents} = \$2.15\). 2. Add the first two amounts: \(\$8.45 + \$2.15 = \$10.60\). 3. Subtract the third amount: \(\$10.60 - \$3.60 = \$7.00\).

\(\$7.00\)

- You can rewrite each amount in cents before subtracting. - You can also subtract the dollars and cents by place value. - Remember that \(\$1.00\) equals \(100\) cents. - For a subtraction such as \(\$100.00 - \$63.45\), try subtracting in parts.

1. For a), subtract the hundredths, tenths, ones, and tens: \(\$74.25 - \$28.15 = \$46.10\). 2. For b), regroup from \(\$100.00\) and subtract: \(\$100.00 - \$63.45 = \$36.55\). 3. For c), regroup \(\$12.05\) as \(11\) dollars and \(105\) cents. Then \(105 - 80 = 25\), so the difference is \(\$11.25\). 4. For d), regroup and subtract: \(\$5.10 - \$2.55 = \$2.55\).

a) \(\$46.10\) b) \(\$36.55\) c) \(\$11.25\) d) \(\$2.55\)

- Line up decimal places carefully when adding or subtracting. - A negative jump means subtracting, so it moves left on the number line. - If the starting value is missing, use the inverse operation to work backward.

1. For a), add the jump to the starting value: \(4.5+2.7=7.2\), so \(x=7.2\). 2. For b), the starting value is unknown. Undo the jump of \(-3.5\): \(6.7+3.5=10.2\), so \(x=10.2\). 3. For c), subtract the jump size from the starting value: \(12.3-4.6=7.7\), so \(x=7.7\).

a) \(x=7.2\) b) \(x=10.2\) c) \(x=7.7\)

- Work step by step and find the middle number first. - Line up the decimal places carefully.

1. First jump: \(12.4+3.8=16.2\). 2. Second jump: \(16.2-2.5=13.7\).

The missing numbers are \(16.2\) and \(13.7\).

- Equivalent trailing zeros can help line up decimal places. - Add digits that represent the same place value. - Keep the decimal point aligned in the result.

1. Writing \(6.4\) as \(6.40\) does not change its value and aligns hundredths with hundredths. 2. Add hundredths: \(0 + 7 = 7\). 3. Add tenths: \(4 + 3 = 7\). 4. Add ones: \(6 + 2 = 8\). 5. The sum is \(8.77\).

\(8.77\). Writing \(6.4\) as \(6.40\) makes the place-value columns explicit.

- Adding a trailing zero does not change the value of \(10.0\). - Make sure each decimal place has a matching column before regrouping. - Check the difference by adding it to \(3.68\).

1. Rewrite \(10.0\) as the equivalent decimal \(10.00\) so hundredths align with hundredths. 2. Subtract \(10.00 - 3.68\), regrouping across the zeros. 3. The difference is \(6.32\).

Write \(10.0\) as \(10.00\) and align the decimal points. The difference is \(6.32\).

- Equivalent decimals can have trailing zeros. - Line up equal place values before adding. - Check whether your sum is a little more than \(5\).

1. \(4.7=4.70\), so the tenths and hundredths places can be lined up. 2. Add by place value: \(4.70+0.35=5.05\). 3. The trailing zero does not change the value of \(4.7\); it makes the hundredths place visible.

\(5.05\). Writing \(4.70\) makes the hundredths place easy to line up with \(0.35\).

- Write the decimals so equal place values line up. - Subtract the distance already walked from the total distance. - Add your result back to \(3.7\) to check.

1. Write \(3.7\) as \(3.70\) and subtract by place value. 2. \(8.25-3.70=4.55\). 3. Check: \(3.70+4.55=8.25\).

\(4.55\,\text{mi}\); check: \(3.70+4.55=8.25\).

- The original amount was split into the amount used and the amount left. - Choose the operation that puts those two parts back together. - Check that subtracting either part from your total gives the other part.

1. The original mass equals the amount used plus the amount left. 2. Add by place value: \(2.75+2.65=5.40\). 3. The bag originally held \(5.4\,\text{kg}\).

\(5.4\,\text{kg}\)

- Align tenths with tenths and hundredths with hundredths. - You may write \(1.8\) as \(1.80\). - Add all three pieces and check that the total is greater than each part.

1. Write \(1.8\) as \(1.80\). 2. Add \(3.25+1.80+0.45=5.50\). 3. The total length is \(5.5\,\text{m}\).

\(5.5\,\text{m}\)

- Line up the decimal points. - Add from the hundredths place to the left. - Regroup \(10\) hundredths as \(1\) tenth when needed.

1. Add hundredths and regroup: \(8+7=15\) hundredths, so write \(5\) hundredths and regroup \(1\) tenth. 2. Add tenths: \(2+6+1=9\) tenths. 3. Add ones: \(1+0=1\). The total is \(1.95\) inches.

\(1.95\) inches

- Round each decimal to a nearby whole number for the estimate. - In part b), evaluate the parentheses before subtracting. - Compare each exact result with its estimate to check whether it is reasonable.

1. For a), an estimate is \(57+14+8=79\). The exact sum is \(56.70+14.00+8.29=78.99\). 2. For b), an estimate is \(95-(24+32)=39\). For the exact value, first calculate \(24.38+31.70=56.08\), then \(95.40-56.08=39.32\).

a) Estimate: \(79\); exact: \(78.99\) b) Estimate: \(39\); exact: \(39.32\)

- Look carefully at the hundredths place in both numbers. - Write \(15.8\) as \(15.80\) and line up the decimal points. - Ask what must happen when subtracting \(5\) hundredths from \(0\) hundredths.

1. Sarah did not account correctly for the hundredths place. The first number has \(0\) hundredths, so the subtraction requires regrouping. 2. Writing \(15.8\) as \(15.80\) makes the place values visible. Then \(15.80-2.45=13.35\). 3. The trailing zero shows explicitly that the hundredths calculation begins with \(0-5\), which requires regrouping from the tenths place.

Sarah ignored the regrouping needed in the hundredths place. The correct result is \(13.35\). Writing \(15.80\) makes the \(0\) hundredths visible and helps keep the place values aligned.

- Add the morning and afternoon distances first. - Subtract the distance already traveled from the total distance. - Compare all three segment lengths using place value.

1. The hiker has completed \(4.6+4.25=8.85\) miles. 2. The remaining distance is \(11.5-8.85=2.65\) miles. 3. Compare \(4.6\), \(4.25\), and \(2.65\). The morning distance, \(4.6\) miles, is the longest.

The hiker has \(2.65\) miles left. The morning segment, \(4.6\) miles, is the longest.

- Check each proposed strategy step by step. - If you subtract more than needed, should the compensation be added or subtracted? - Think about the direction of each adjustment on a number line.

1. Statement 1 is incorrect. Subtracting \(4\) removes \(0.1\) too much, so that amount must be added back: \(17.4-4+0.1=13.5\). 2. Statement 2 is correct. Adding \(3\) adds \(0.02\) too much, so subtracting \(0.02\) compensates: \(5.65+3-0.02=8.63\).

1) Incorrect. The correct strategy is \(17.4-4+0.1=13.5\). 2) Correct. \(5.65+3-0.02=8.63\).

- Round the prices to convenient nearby amounts for the estimate. - Remember that two bottles of soda are purchased. - Subtract the coupon from the purchase total. - Add the parking cost after applying the coupon.

1. A reasonable estimate is \(\$4.60+\$2.30+\$3.10+\$2.70-\$1.50+\$3.50=\$14.70\), so \(\$15.00\) should be enough. 2. The two sodas cost \(2\times\$1.35=\$2.70\). 3. The purchases total \(\$4.58+\$2.25+\$3.12+\$2.70=\$12.65\). 4. After the coupon, the cost is \(\$12.65-\$1.50=\$11.15\). 5. Including parking, the exact total is \(\$11.15+\$3.50=\$14.65\), so \(\$15.00\) is enough.

The estimate shows that \(\$15.00\) should be enough. The exact total is \(\$14.65\), so the money is sufficient.

- Find the equipment cost before and after the discount. - Subtract the equipment cost from the total budget. - Find the total cost of the three towels. - Compare the remaining budget with the towel cost.

1. The jump ropes cost \(5\times\$4.25=\$21.25\). 2. Before the discount, the equipment costs \(\$21.25+\$18.90+\$12.50=\$52.65\). 3. After the discount, the cost is \(\$52.65-\$5.75=\$46.90\). 4. The remaining budget is \(\$50.00-\$46.90=\$3.10\). 5. The towels cost \(3\times\$2.80=\$8.40\). 6. Since \(\$3.10<\$8.40\), there is not enough money. The budget is short by \(\$8.40-\$3.10=\$5.30\).

No. After the equipment purchase, \(\$3.10\) remains. The towels cost \(\$8.40\), so the budget is short by \(\$5.30\).

- Convert the yard measurement to miles. - Convert the fractions to decimals. - Look for pairs that add to whole numbers.

1. Convert every length to miles. Since \(1760\,\text{yd}=1\,\text{mi}\), \(880\,\text{yd}=0.5\,\text{mi}\). Also, \(\frac{3}{4}\,\text{mi}=0.75\,\text{mi}\) and \(1\frac{1}{2}\,\text{mi}=1.5\,\text{mi}\). 2. Form convenient pairs: \(1.45+2.55=4\), \(0.75+0.25=1\), and \(1.5+0.5=2\). 3. Add the pair sums: \(4+1+2=7\).

The total length is \(7\,\text{mi}\). One efficient grouping is \((1.45+2.55)+(0.75+0.25)+(1.5+0.5)\).

- Work one place value at a time, as you would in the standard addition or subtraction algorithm. - Account for any regrouping from one place to the next. - Each letter represents one digit from \(0\) through \(9\).

1. In part a), the hundredths digits give \(2+b=9\), so \(b=7\). In the tenths place, \(a+7=15\), so \(a=8\) and \(1\) is regrouped to the ones place. Then \(1+4+3=8\), which confirms the sum. 2. In part b), regroup in the tenths place: \(10+b-5=7\), so \(b=2\). After that regrouping, the ones digit is \(1\). Regroup again from the tens place: \(11-a=8\), so \(a=3\). The remaining tens digit is \(0\), giving \(8.7\).

a) \(a=8\), \(b=7\) b) \(a=3\), \(b=2\)

- Find how far each distance is above or below \(5\), \(3\), and \(3\). - Write each small adjustment with a positive or negative sign. - Add the three adjustments before changing the total of \(11\).

1. Compare each distance with the nearby whole number: \(4.97=5-0.03\), \(3.05=3+0.05\), and \(2.98=3-0.02\). 2. The nearby whole numbers sum to \(5+3+3=11\). 3. The total adjustment is \(-0.03+0.05-0.02=0\). 4. Therefore, the exact total is \(11\) miles.

The adjustment is \(0\) because \(-0.03+0.05-0.02=0\). The exact total is \(11\) miles.

- What is the total weight, and which parts of it are already known? - How can you find the part that remains after removing the known weights? - Add all the contents before subtracting from the total.

1. Add the weights of the contents: \(4.25+1.1+0.8+2=8.15\,\text{lb}\). 2. Subtract the contents from the total: \(10.5-8.15=2.35\,\text{lb}\).

Expression: \(10.5-(4.25+1.1+0.8+2)\) The empty backpack weighs \(2.35\,\text{lb}\).

- Look for pairs that make whole-dollar totals. - Decide whether a written column layout would help you track hundredths and regrouping. - Check each total by estimating its size.

1. For Receipt A, mental math is convenient because pairs make whole-dollar totals: \(\$12.00 + \$8.00 = \$20.00\) and \(\$5.50 + \$4.50 = \$10.00\), for a total of \(\$30.00\). 2. For Receipt B, the standard algorithm is a reasonable way to organize the hundredths and regrouping: \(\$13.47 + \$22.89 + \$5.12 + \$17.04 = \$58.52\).

Receipt A: \(\$30.00\) Receipt B: \(\$58.52\) Method choices may vary if they are explained and used correctly.

- Find the combined cost of the two science items. - Find the total amount Tim has. - Subtract the cost from the amount he has. - Keep track of dollars and cents.

1. Find the total cost: \(\$44.50 + \$12.00 = \$56.50\). 2. Find the total amount Tim has: \(\$35.00 + \$25.00 = \$60.00\). 3. Find the amount left: \(\$60.00 - \$56.50 = \$3.50\).

Tim will have \(\$3.50\) left.

- Decide which amounts increase the money in the jar and which amounts decrease it. - Add all money received and all money spent separately. - Line up the decimal points when adding or subtracting.

1. Add the money received: \(\$15.00 + \$10.50 = \$25.50\). 2. Add the money spent: \(\$8.40 + \$3.25 + \$12.80 + \$0.95 = \$25.40\). 3. Find the final amount: \(\$50.00 + \$25.50 - \$25.40 = \$50.10\).

Maya has \(\$50.10\) in her savings jar at the end of the week.

- Find the balance after each known transaction. - Compare the balance after the art supplies with the final balance. - The difference is the cost of the bus tickets.

1. Find the balance after the bake sale: \(\$84.50 + \$45.00 = \$129.50\). 2. Subtract the cost of the art supplies: \(\$129.50 - \$18.75 = \$110.75\). 3. The bus tickets account for the difference between this balance and the final balance: \(\$110.75 - \$92.45 = \$18.30\).

The bus tickets cost \(\$18.30\) altogether.

- Keep cash and the store-specific gift card separate. - Use the value and number of each type of bill or coin to find the cash total. - The amount before the purchase equals the amount spent plus the cash remaining.

1. Find the cash remaining: \(2 \times \$100 + 1 \times \$50 + 3 \times \$10 + 8 \times \$1 + 10 \times \$0.25 = \$200 + \$50 + \$30 + \$8 + \$2.50 = \$290.50\). 2. Find the cash he had before the purchase: \(\$290.50 + \$412.60 = \$703.10\). 3. Compare the remaining cash with the drill price. Since \(\$290.50 < \$295\), the cash is not enough. The bookstore gift card cannot be used at the hardware store.

a) \(\$290.50\) b) \(\$703.10\) c) No. He is \(\$4.50\) short because \(\$295 - \$290.50 = \$4.50\), and the gift card is valid only at the bookstore.

- Multiply to find the total value of each type of bill, then add. - Subtract the cost of the accessories from the original amount. - Compare the remaining money with the price of the bicycle.

1. Find the original amount: \(3 \times \$50 + 6 \times \$20 + 8 \times \$5 + 24 \times \$1 = \$150 + \$120 + \$40 + \$24 = \$334\). 2. Subtract the cost of the accessories: \(\$334 - \$47.35 = \$286.65\). 3. Compare the remaining amount with the bicycle price. Since \(\$286.65 > \$285\), she has enough money.

a) \(\$334\) b) \(\$286.65\) c) Yes. She has \(\$1.65\) more than the bicycle costs.

- Find the buffet cost first. - Add the drink and buffet costs. - Subtract the total spent from the amount raised.

1. The buffet costs \(\$85.40 + \$42.75 = \$128.15\). 2. The total spent is \(\$85.40 + \$128.15 = \$213.55\). 3. The amount remaining is \(\$350.00 - \$213.55 = \$136.45\).

\(\$136.45\) remains.

- Use “less than” to find the cost of the balls. - Add the two purchase amounts. - Subtract the total spent from the donation.

1. The balls cost \(\$645.80 - \$212.30 = \$433.50\). 2. The two purchases cost \(\$645.80 + \$433.50 = \$1079.30\). 3. The amount remaining is \(\$1500.00 - \$1079.30 = \$420.70\).

\(\$420.70\) remains.

- In the first calculation, work from left to right. - In the second calculation, evaluate the parentheses first. - Compare the total amount subtracted in the two methods.

1. Subtract the first amount: \(\$100.00 - \$25.50 = \$74.50\). 2. Subtract the second amount: \(\$74.50 - \$14.75 = \$59.75\). 3. For the comparison, first add the two amounts being subtracted: \(\$25.50 + \$14.75 = \$40.25\). Then \(\$100.00 - \$40.25 = \$59.75\). 4. Both methods give the same result because both subtract a total of \(\$40.25\) from \(\$100.00\).

1. The final amount is \(\$59.75\). 2. Both calculations equal \(\$59.75\) because the same total amount is subtracted in each calculation.

- Add the two purchase amounts first. - Subtract the total cost from the amount paid. - Align the decimal points.

1. The total cost is \(\$6.75 + \$11.40 = \$18.15\). 2. The change is \(\$50.00 - \$18.15 = \$31.85\).

Ms. Meyer receives \(\$31.85\) in change.

- Add the money Mila has saved and received. - Subtract that total from the car’s price. - Align the decimal points.

1. Mila now has \(\$27.65 + \$10.50 = \$38.15\). 2. She still needs \(\$45.00 - \$38.15 = \$6.85\).

Mila still needs to save \(\$6.85\).

- Rewrite each equation using the inverse operation when helpful. - Align decimal points and calculate by place value. - Check by substituting each answer into the original equation.

1. a) The missing amount is \(\$100.00-\$37.55=\$62.45\). 2. b) The missing amount is \(\$73.20-\$28.45=\$44.75\). 3. c) The starting amount is \(\$45.50+\$19.99=\$65.49\).

a) \(\$62.45\) b) \(\$44.75\) c) \(\$65.49\)

- A positive jump increases the value; a negative jump decreases it. - When working backward, use the inverse operation for each jump. - Check your answers by following the jumps forward.

1. a) \(2.8+1.5=4.3\), then \(4.3+0.9=5.2\). 2. b) \(7.3-2.6=4.7\), then \(4.7-1.8=2.9\). 3. c) Work backward from \(5.5\): \(5.5+1.6=7.1\), then \(7.1-3.4=3.7\).

a) \(4.3\), \(5.2\) b) \(4.7\), \(2.9\) c) \(3.7\), \(7.1\)

- Use the labeled whole numbers to work out the value of one smaller interval. - A rail's length is the difference between its ending position and its starting position. - For part c), find rail \(C\)'s length before comparing it with the other rails.

1. There are \(4\) equal small intervals between consecutive whole-number marks, so each small interval represents \(0.25\,\text{cm}\). 2. Rail \(A\) starts at \(0.5\) and ends at \(3.25\). Its length is \(3.25-0.5=2.75\,\text{cm}\). 3. Rail \(B\) starts at \(1.25\) and ends at \(5\). Its length is \(5-1.25=3.75\,\text{cm}\). 4. Rail \(C\) starts at \(2.75\) and ends at \(5.25\). Its length is \(5.25-2.75=2.5\,\text{cm}\). 5. Compare the other rails with rail \(C\): \(3.75-2.5=1.25\,\text{cm}\), so rail \(B\) is exactly \(1.25\,\text{cm}\) longer than rail \(C\). 6. The total length of rails \(A\) and \(C\) is \(2.75+2.5=5.25\,\text{cm}\).

a) \(2.75\,\text{cm}\) b) \(3.75\,\text{cm}\) c) Rail \(B\), because \(3.75-2.5=1.25\,\text{cm}\) d) \(5.25\,\text{cm}\)

- Round the item prices and store credits to whole dollars for part a). - Find the exact purchase total and exact credit total separately. - Compare the correct amount owed with \(\$3.27\). - See whether that difference matches a plausible error in the tablecloth price.

1. For a), estimate the purchase as \(3\times\$1+\$7+2\times\$2=\$14\). Estimate the store credits as \(\$5+\$2+\$4=\$11\). The credits should not cover the full purchase. 2. For b), the exact purchase total is \(3\times\$1.49+\$6.95+2\times\$2.25=\$15.92\). 3. The store credits total \(\$5.00+\$2.20+\$3.50=\$10.70\). 4. Lucas still owes \(\$15.92-\$10.70=\$5.22\). 5. For c), the difference between the correct amount and the register amount is \(\$5.22-\$3.27=\$1.95\). 6. Entering the tablecloth as \(\$5.00\) instead of \(\$6.95\) lowers the purchase total by exactly \(\$1.95\), which explains the displayed \(\$3.27\).

a) No. The purchase is about \(\$14\), while the credits total about \(\$11\). b) Lucas still owes \(\$5.22\). c) The tablecloth was likely entered as \(\$5.00\) instead of \(\$6.95\).