a. To explain how the partial products are shown in the figure, we need to break down the multiplication of 11 and 12 into simpler components. This involves using the distributive property of multiplication over addition.
b. To draw a similar model for 12⋅2312 \cdot 2312⋅23, we can use the area model or grid method, breaking down each number into tens and units, and then multiplying each part.
c. For the base-five model of 14five⋅22five14_{\text{five}} \cdot 22_{\text{five}}14five⋅22five, we need to convert the numbers to base ten, perform the multiplication, and then convert the result back to base five. The model can be used to visualize the multiplication process in base five.
To find the partial products for 11⋅1211 \cdot 1211⋅12, we break down the numbers as follows: 11=10+1and12=10+2 11 = 10 + 1 \quad \text{and} \quad 12 = 10 + 2 11=10+1and12=10+2 Using the distributive property: 11⋅12=(10+1)⋅(10+2)=10⋅10+10⋅2+1⋅10+1⋅2 11 \cdot 12 = (10 + 1) \cdot (10 + 2) = 10 \cdot 10 + 10 \cdot 2 + 1 \cdot 10 + 1 \cdot 2 11⋅12=(10+1)⋅(10+2)=10⋅10+10⋅2+1⋅10+1⋅2 The partial products are: 100,20,10, and 2 100, 20, 10, \text{ and } 2 100,20,10, and 2 Summing these partial products: 100+20+10+2=132 100 + 20 + 10 + 2 = 132 100+20+10+2=132
To find the partial products for 12⋅2312 \cdot 2312⋅23, we break down the numbers as follows: 12=10+2and23=20+3 12 = 10 + 2 \quad \text{and} \quad 23 = 20 + 3 12=10+2and23=20+3 Using the distributive property: 12⋅23=(10+2)⋅(20+3)=10⋅20+10⋅3+2⋅20+2⋅3 12 \cdot 23 = (10 + 2) \cdot (20 + 3) = 10 \cdot 20 + 10 \cdot 3 + 2 \cdot 20 + 2 \cdot 3 12⋅23=(10+2)⋅(20+3)=10⋅20+10⋅3+2⋅20+2⋅3 The partial products are: 200,30,40, and 6 200, 30, 40, \text{ and } 6 200,30,40, and 6 Summing these partial products: 200+30+40+6=276 200 + 30 + 40 + 6 = 276 200+30+40+6=276
First, convert the base-five numbers to base-ten: 14five=1⋅51+4⋅50=5+4=9ten 14_{\text{five}} = 1 \cdot 5^1 + 4 \cdot 5^0 = 5 + 4 = 9_{\text{ten}} 14five=1⋅51+4⋅50=5+4=9ten 22five=2⋅51+2⋅50=10+2=12ten 22_{\text{five}} = 2 \cdot 5^1 + 2 \cdot 5^0 = 10 + 2 = 12_{\text{ten}} 22five=2⋅51+2⋅50=10+2=12ten Next, multiply the base-ten equivalents: 9ten⋅12ten=108ten 9_{\text{ten}} \cdot 12_{\text{ten}} = 108_{\text{ten}} 9ten⋅12ten=108ten Finally, convert the result back to base-five: 108ten=4⋅52+1⋅51+3⋅50=413five 108_{\text{ten}} = 4 \cdot 5^2 + 1 \cdot 5^1 + 3 \cdot 5^0 = 413_{\text{five}} 108ten=4⋅52+1⋅51+3⋅50=413five
a. The partial products are 100,20,10, and 2100, 20, 10, \text{ and } 2100,20,10, and 2, which gives the product 132\boxed{132}132. The answer is A.
b. The partial products for 12⋅2312 \cdot 2312⋅23 are 200,30,40, and 6200, 30, 40, \text{ and } 6200,30,40, and 6, which gives the product 276\boxed{276}276.
c. The product of 14five⋅22five14_{\text{five}} \cdot 22_{\text{five}}14five⋅22five in base five is 413five\boxed{413_{\text{five}}}413five.
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