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More partial products Recall that we can use a drawing of a rectangle to help us with calculating products. The rectangle is divided into regions and we determine “partial products” which are then added to find the total.

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More partial products Recall that we can use a drawing of a rectangle to help us with calculating products. The rectangle is divided into regions and we determine “partial products” which are then added to find the total. 3 7 Blue 3 × 5 = 15 Yellow 3 × 2 = 6 Total = 21 Example 1:

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More partial products Example 2: The total number of squares can be found from 54 × 23. One of the ways to calculate 54 × 23 is to divide the rectangle into 4 regions Orange: 50 x 20 = 1000 Yellow: 4 x 20 = 80 White: 50 x 3 = 150 Blue: 4 x 3 = 12 Total: 1242 So 54 × 23 = 1242

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More partial products pinkyellowgreenorange 2 x 42 x 0.70.1 x 40.1 x 0.7 81.40.40.07 So 2.1 x 4.7 = 8+1.4+0.4+0.07 = 9.87 Example 3:

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More partial products Now we are going to explore this technique of partial products with fractions. Draw a rectangle and label the sides with 2 and 4 ½ Can you make two regions in the rectangle and label the sides?

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More partial products Now we are going to explore this technique of partial products with fractions. Draw a rectangle and label the sides with 2 and 4 ½ Can you make two regions in the rectangle and label the sides?

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More partial products Now we are going to explore this technique of partial products with fractions. Draw a rectangle and label the sides with 2 and 4 ½ Can you make two regions in the rectangle and label the sides? So we have Yellow: 2 × 4 = 8 Pink: 2 × ½ = 1 So 2 × 4 ½ = 9 Were you expecting 9?

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More partial products What if we needed to find 2 1 / 3 x 4 ½ Can you extend the rectangle underneath?

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More partial products What if we needed to find 2 1 / 3 x 4 ½ Can you extend the rectangle underneath?

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yellowpinkgreenblue 2 x 42 x 1 / 2 1 / 3 x 4 1 / 3 x 1 / 2 So now we have 4 partial products

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yellowpinkgreenblue 2 x 42 x 1 / 2 1 / 3 x 4 1 / 3 x 1 / 2 81 4 / 3 = 1 1 / 3 1/61/6 So now we have 4 partial products So 2 1 / 3 x 4 1 / 2 Can be found by adding 8 + 1 + 1 1 / 3 + 1 / 6 = 10 ½

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We can also consider lower and upper bounds to check our answers.

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2 1 / 3 x 4 ½

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Upper bound 3x5 = 15

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Lower bound 2 x 4= 8 Upper bound = 15

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Lower bound = 8 Upper bound = 15 So we know that our answer (to 2 1 / 3 x 4 ½) lies between 8 and 15.

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AREA AND CIRCUMFERENCE OF A CIRCLE. diameter radius circumference The perimeter of a circle is called the circumference (C). The diameter (d) of a circle.

AREA AND CIRCUMFERENCE OF A CIRCLE. diameter radius circumference The perimeter of a circle is called the circumference (C). The diameter (d) of a circle.

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