- Use the transformed whole-number division displayed in the image. - Recall the multiplication fact related to \(36\div6\).

1. Multiplying both numbers by \(10\) gives the equivalent division \(36\div6\), as shown. 2. \(36\div6=6\).

\(6\)

- What power of \(10\) will make the divisor a whole number? - Whatever multiplication you use on the divisor, use on the dividend too. - Think about whether the number of equal groups changes.

1. Make both numbers \(10\) times as large: \(1.2\to12\) and \(0.3\to3\). 2. This gives \(12\div3\). 3. Making both the amount and the group size \(10\) times as large does not change how many equal groups fit, so the quotient stays the same.

\(3\). Thus \(1.2\div0.3=12\div3\). It works because making both numbers \(10\) times as large keeps the number of equal groups the same.

- Look for a change that makes the decimal divisor a whole number. - Whatever change is made to the divisor must also be made to the dividend. - Your explanation should name the change to both numbers.

1. Make both \(2.4\) and \(0.6\) \(10\) times as large. 2. They become \(24\) and \(6\), so the equivalent division is \(24\div6\). 3. Therefore, choice A is correct. Both numbers were multiplied by \(10\), so the number of equal groups stays the same.

A. \(24\div6\). Both the dividend and divisor were multiplied by \(10\).

- Think of a division expression as a fraction. - Compare how the dividend and divisor are scaled in each choice. - The quotient stays the same when both are scaled by the same nonzero factor.

1. In expression 1, both dividend and divisor are multiplied by \(10\), so the quotient stays the same. 2. In expression 2, both are multiplied by \(100\), so the quotient stays the same. 3. In expression 3, both are divided by \(10\), so the quotient stays the same. 4. In expression 4, the dividend is multiplied by \(100\) but the divisor by only \(10\), so the quotient changes. 5. Expressions 1, 2, and 3 are equivalent to the original division.

Expressions \(1\), \(2\), and \(3\) have the same quotient as \(2.4\div0.08\).

- Create an equivalent division with a whole-number divisor. - Scale the dividend and divisor by the same power of \(10\). - Use familiar whole-number division facts after scaling.

1. a) \(15\div0.3=150\div3=50\). 2. b) \(0.8\div0.04=80\div4=20\). 3. c) \(0.45\div0.15=45\div15=3\).

a) \(50\) b) \(20\) c) \(3\)

- Create an equivalent quotient with a whole-number divisor. - Scale the dividend and divisor by the same power of \(10\). - Check that the size of each quotient is reasonable.

1. \(7.2\div0.8=72\div8=9\). 2. \(1.26\div0.3=12.6\div3=4.2\). 3. \(0.048\div0.006=48\div6=8\). 4. \(3.45\div0.15=345\div15=23\).

a) \(9\) b) \(4.2\) c) \(8\) d) \(23\)

- Ask how many times each piece length fits into the total length. - Create an equivalent division with a whole-number divisor if helpful. - Check that the number of pieces times the piece length gives the full strip length.

1. For a), \(3.5\div0.25=14\), so \(14\) pieces can be cut. 2. For b), \(3.5\div0.14=25\), so \(25\) pieces can be cut.

a) \(14\) pieces b) \(25\) pieces

- What power of ten would make the decimal divisor a whole number? - Whatever change you make to the divisor must also be made to the dividend. - Complete the transformed dividend before dividing.

1. Multiply both the dividend and divisor by \(10\). This changes \(4.8\div0.6\) to \(48\div6\) without changing the quotient. 2. The missing digit is \(8\). 3. \(48\div6=8\), so the quotient is \(8\).

a) The missing digit is \(8\). Multiplying both the dividend and divisor by \(10\) changes \(4.8\div0.6\) to \(48\div6\), so the divisor becomes \(6\) without changing the quotient. b) The quotient is \(8\).

- Use the spacing between neighboring ticks to identify one group size. - Count how many equal groups reach \(2.4\). - Division can ask how many groups of one size fit in a total.

1. Each interval has length \(0.3\). 2. There are \(8\) equal intervals from \(0\) to \(2.4\). 3. The matching equation is \(2.4\div0.3=8\).

\(8\) intervals; \(2.4\div0.3=8\)

- Scale the dividend and divisor by the same power of \(10\). - Interpret the whole-number remainder in tenths of a kilogram. - Check by multiplying the number of full bags by the bag size and adding the leftover birdseed.

1. Scale both quantities by \(10\): \(5.0\div0.6=50\div6\). 2. \(50\div6=8\) remainder \(2\). 3. The remainder of \(2\) tenths is \(0.2\,\text{kg}\), so \(8\) full bags can be filled with \(0.2\,\text{kg}\) left. 4. Check: \(8\times0.6+0.2=5.0\).

\(8\) full bags, with \(0.2\,\text{kg}\) left; check: \(8\times0.6+0.2=5.0\).

- Read the amount using the ones and tenths columns. - Express \(0.5\) as a number of tenths. - Ask how many groups of that many tenths fit in the total.

1. The model shows \(1\) one and \(5\) tenths. Trade the one for \(10\) tenths to get \(15\) tenths altogether. 2. Each group of \(0.5\) is \(5\) tenths. 3. \(15\div5=3\), so \(1.5\div0.5=3\).

There are \(3\) groups. The model represents \(15\) tenths, and \(0.5\) is \(5\) tenths, so \(15\div5=3\). Thus \(1.5\div0.5=3\).

- Each division asks how many \(0.3\)-meter lengths fit in a larger length. - Scale the dividend and divisor by the same power of \(10\) if that makes the division easier. - Multiply the number across by the number of rows to find the total.

1. Across the width: \(1.2\div0.3=4\) stones. 2. Along the length: \(6\div0.3=20\) rows. 3. The total is \(4\times20=80\) stones.

a) \(4\) stones b) \(20\) rows c) \(80\) stones

- Divide the stack height by \(25\) to find one sheet's thickness. - Use the stated conversion for part a). - For part b), divide the target height by one sheet's thickness in centimeters.

1. \(13.75\div25=0.55\,\text{cm}\) per sheet. 2. \(0.55\,\text{cm}=5.5\,\text{mm}\). 3. \(44\div0.55=80\), so \(80\) sheets are needed.

a) \(5.5\,\text{mm}\) b) \(80\) sheets

- Start with the divisions by \(10\) and \(1\). - Look for a pattern in the quotients. - Track how the divisor changes from one line to the next. - Compare the place value of the digits in consecutive quotients.

1. The quotients are \(0.045\), \(0.45\), \(4.5\), \(45\), and \(450\). 2. Each time the divisor becomes one tenth as large, the quotient becomes ten times as large.

a) \(0.045\) b) \(0.45\) c) \(4.5\) d) \(45\) e) \(450\) The quotient becomes \(10\) times as large each time the divisor becomes one tenth as large.

- Compare each expression with the original pair of numbers. - Scale the dividend and divisor by the same factor. - Use the simplest equivalent division to find the final quotient. - Check that every completed expression gives the same value.

1. From \(1.2\) to \(12\), multiply by \(10\), so also multiply \(0.3\) by \(10\): the first blank is \(3\). 2. From \(0.3\) to \(30\), multiply by \(100\), so also multiply \(1.2\) by \(100\): the second blank is \(120\). 3. From \(1.2\) to \(0.12\), divide by \(10\), so also divide \(0.3\) by \(10\): the third blank is \(0.03\). 4. Use the easiest equivalent expression: \(12\div3=4\).

\(1.2\div0.3=12\div3=120\div30=0.12\div0.03\). The common quotient is \(4\).

- Think of the division as a fraction and recall how equivalent fractions are formed. - Compare the factor used to scale the dividend with the factor used to scale the divisor. - Make the divisor a whole number while keeping the quotient unchanged.

1. To keep a quotient unchanged, the dividend and divisor must be multiplied by the same nonzero number. 2. Making \(0.5\) into \(5\) multiplies the divisor by \(10\), so the dividend must also be multiplied by \(10\): \(0.45\) becomes \(4.5\), not \(45\). 3. The equivalent division is \(4.5 \div 5 = 0.9\). 4. Jordan's quotient \(9\) is ten times too large because the dividend was scaled more than the divisor.

The method is incorrect because the dividend and divisor were scaled by different factors. Correctly, \(0.45 \div 0.5 = 4.5 \div 5 = 0.9\).

- Compare how the decimal place changes from one expression to another. - Compare a) with b) to isolate the change in the divisor. - Compare a) with d) to see what happens when both numbers are scaled equally.

1. For a), \(2.4\div0.6=4\). 2. For b), \(2.4\div0.06=40\). 3. For c), \(0.24\div0.6=0.4\). 4. For d), \(0.24\div0.06=4\). 5. Making only the divisor ten times smaller makes the quotient ten times larger. Making only the dividend ten times smaller makes the quotient ten times smaller. Scaling both by the same factor leaves the quotient unchanged.

a) \(4\) b) \(40\) c) \(0.4\) d) \(4\) Scaling the dividend and divisor by the same nonzero factor leaves the quotient unchanged.

- Rewrite each division with a whole-number divisor when useful. - Compare how the dividend and divisor are scaled across the expressions. - Calculate each quotient and compare the results.

1. A: \(0.45\div0.09=5\). 2. B: \(4.5\div0.9=5\). 3. C: \(45\div9=5\). 4. D: \(0.045\div0.9=0.05\). 5. Expression D is different because the other three all equal \(5\).

D has the different value: \(0.05\). A, B, and C each equal \(5\).

- Compare each dividend and divisor with \(156\) and \(5\). - Scaling both numbers by the same nonzero factor leaves the quotient unchanged. - If only one number changes by a power of \(10\), determine how the quotient changes.

1. For a), the dividend is one tenth as large, so the quotient is one tenth as large: \(3.12\). 2. For b), multiply dividend and divisor by \(10\): \(1.56\div0.5=15.6\div5=3.12\). 3. For c), the divisor is one hundredth as large as \(5\), so the quotient is \(100\) times as large: \(31.2\times100=3120\). 4. For d), the dividend is one thousandth as large as \(156\), so the quotient is one thousandth as large: \(0.0312\).

a) \(3.12\) b) \(3.12\) c) \(3120\) d) \(0.0312\)

- How much do the glitter stickers cost altogether? - How much money remains for regular stickers? - How many regular stickers can be bought with that amount? - Remember to add both kinds of stickers at the end.

1. Find the cost of the glitter stickers: \(3 \times \$0.70 = \$2.10\). 2. Find the amount spent on regular stickers: \(\$3.60 - \$2.10 = \$1.50\). 3. Find the number of regular stickers: \(\$1.50 \div \$0.30 = 5\). 4. Find the total number of stickers: \(3 + 5 = 8\).

Jonas buys \(8\) stickers altogether.

- Find how much money was handed to the cashier. - Subtract the change to find the purchase cost. - Divide by the price per kilogram, then subtract the purchased mass from the container's capacity.

1. Kojo paid \(2 \times \$50 = \$100\). 2. The honey cost \(\$100 - \$14.50 = \$85.50\). 3. The purchased mass was \(\$85.50 \div \$4.50 = 19\,\text{kg}\). 4. The starting mass was \(25\,\text{kg} - 19\,\text{kg} = 6\,\text{kg}\).

\(6\,\text{kg}\)

- First find the cost of one sticker. - How many times does one sticker's price fit into the total amount of money?

1. Find the cost of one sticker: \(\$0.80 \div 4 = \$0.20\). 2. Divide the available money by the price of one sticker: \(\$2.00 \div \$0.20 = 10\).

\(10\) stickers can be bought.

- Use the transformed divisor \(14\). - Focus on the displayed partial dividend \(126\). - The multiplier that produces that subtraction row is the missing quotient digit.

1. Equivalent scaling gives \(54.6\div14\). 2. The first quotient digit is \(3\), because \(14\times3=42\); bringing down \(6\) gives \(126\). 3. Since \(126=14\times9\), the missing tenths quotient digit is \(9\). The missing digit in \(1\square6\) is \(2\). 4. The quotient is \(3.9\).

Since \(126=14\times9\), the missing quotient digit is \(9\), and the subtraction row is \(126\), so its missing digit is \(2\). The quotient is \(3.9\).

- Use the scaled whole-number divisor shown in the image. - Focus on the second displayed partial dividend. - The one-digit multiplier that produces that partial dividend is the missing quotient digit.

1. Equivalent scaling changes \(6.72\div0.24\) to \(672\div24\). 2. The first quotient digit is \(2\), since \(24\times2=48\). Subtracting from \(67\) leaves \(19\); bringing down \(2\) gives the displayed partial dividend \(192\). 3. Since \(192=24\times8\), the missing quotient digit is \(8\). 4. The quotient is \(28\), so \(28\) full pieces can be cut.

The displayed second partial dividend is \(192\), and \(24\times8=192\), so the missing quotient digit is \(8\). The quotient is \(28\), so \(28\) full pieces can be cut.

- Work backward from the displayed next partial dividend \(16\) to determine the previous remainder. - Add that remainder to the first subtraction row \(24\) to reconstruct the partial dividend containing a box. - Use the transformed divisor \(8\) with \(16\) to determine the missing quotient digit.

1. After scaling by \(10\), the divisor is \(8\). The first quotient digit is \(3\), so the first subtraction row is \(24\). 2. For the next partial dividend to be \(16\), the remainder before bringing down the final \(6\) must be \(1\). Therefore the first partial dividend is \(25\), so the missing original digit is \(5\). 3. \(16\div8=2\), so the missing quotient digit is \(2\). 4. The completed division is \(2.56\div0.8=3.2\).

The missing dividend digit is \(5\), the missing quotient digit is \(2\), and \(2.56\div0.8=3.2\).