This requires two multiplication steps. First, find total books needed: $$147 \times 23 = 3381$$ books. Then multiply by cost per book: $$3381 \times 28 = 94668$$. The total cost is $$\94668$$. Choice A represents $$147 + 23 + 28$$. Choice B represents $$147 \times 28$$ (one classroom's cost). Choice D represents only the total number of books without multiplying by the cost per book.

Using the distributive property, $$234 \times 67 = 234 \times(70 - 3) = 234 \times 70 - 234 \times 3$$. Roberto should calculate $$234 \times 70 - 234 \times 3 = 16380 - 702 = 15678$$ to verify. Choice A adds instead of subtracts. Choice C multiplies the partial products instead of using the distributive property. Choice D uses addition throughout instead of multiplication.

This requires multiple multiplication steps. First, find boxes per day: $$156 \times 24 = 3744$$ boxes per day. Then multiply by 7 days: $$3744 \times 7 = 26208$$ boxes total. Choice B represents $$156 \times 7$$ (ignoring the 24 hours). Choice C represents only one day's production ($$156 \times 24$$). Choice D represents $$156 + 24 + 7$$, adding instead of multiplying.

To find total items, multiply packages per day by items per package by number of days: $$126 \times 37 \times 15 = 69930$$ items. First $$126 \times 37 = 4662$$, then $$4662 \times 15 = 69930$$. Choice A adds all values instead of multiplying. Choice B calculates items for one day then adds days instead of multiplying. Choice C adds packages per day to the product of items per package and days, which doesn't represent the situation correctly.

Using the distributive property, $$275 \times 48 = 275 \times(50 - 2) = 275 \times 50 - 275 \times 2$$. Since Sarah calculated $$275 \times 50 = 13750$$, she needs to subtract $$275 \times 2 = 550$$ to get $$13750 - 550 = 13200$$. Choice A adds instead of subtracts. Choice C and D don't use the distributive property correctly and would give incorrect results.

Lisa's partial products are correct: $$456 \times 9 = 4104$$ and $$456 \times 80 = 36480$$. However, when adding $$4104 + 36480$$, the correct sum is $$40584$$, not $$76584$$. Lisa made an addition error. The correct answer to $$456 \times 89$$ is $$40584$$. Choice A and B are incorrect because both partial products are calculated correctly. Choice D is wrong because $$76584$$ is not the correct final answer.

Multi-digit multiplication uses place value to correctly interpret and compute with positional digits. Partial products separate the multiplication into parts based on the multiplier's digits and their places. Aligning digits means shifting higher place partial products left to match their value, like adding zeros for tens. This connects to the area model by visualizing the product as summed rectangular areas from place value pairs. A misconception is confusing addition with multiplication in partial products, but they must be multiplied first. The algorithm works through the distributive property applied to place values. It generalizes because it systematically accounts for all interactions between digits' places.

Multi-digit multiplication uses place value to handle digits in ones, tens, and beyond systematically. Partial products come from multiplying by each digit with its place value, like 63 × 5 and 63 × 20 for 63 × 25. Aligning involves placing the tens partial product shifted left, often with a trailing zero for clarity. This ties to the area model, dividing into rectangles such as 60 × 20, 60 × 5, 3 × 20, and 3 × 5, totaling 1575. A misconception is omitting the zero in the tens partial product, which might lead to misadding and wrong totals like 126 + 315 = 441 instead. The algorithm works because it builds the product layer by layer using place values. It ensures consistent results for various multi-digit scenarios.

Multi-digit multiplication uses place value to expand numbers like 52 into 5 tens and 2 ones, which helps identify errors in computation. Partial products should be 76 × 2 = 152 for ones and 76 × 50 for tens, but the student mistakenly used 76 × 5 instead. Aligning digits requires right-alignment by place value, with the tens partial product shifted to reflect the extra factor of 10. This connects to the area model, splitting the multiplication into areas for 76 × 50 and 76 × 2, summed together. A common misconception is ignoring the tens place multiplier, treating 5 as just 5 instead of 50, leading to a smaller product. The algorithm works by breaking down numbers via place values and multiplying distributively. It generalizes well because it consistently applies these principles to yield correct results across various multi-digit problems.

Multi-digit multiplication uses place value to correctly value each digit's contribution, such as 2 in 28 representing 20. Partial products arise from multiplying by each component, like 8 ones and 2 tens, to build the total. Aligning digits involves shifting the tens product to align with its place value during addition. This relates to the area model, where place values create grid sections whose areas sum to the product. A misconception is adding without shifting for tens, treating it as a simple digit instead of a multiple of ten. The algorithm is reliable because it decomposes and reassembles based on place values. It generalizes to ensure correct products for any multi-digit combination.