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Think like a real problem solver and explain me logic for:Problem 2: 10 points Be ty Stal by ct mc on recs oy ie athe Fue cow. he rtd don are 10 ct widey 2 ih. hee 5) rows of bricksblocks are ruined for this wal. Morar ois ae 0.8 tic SS wast 17-4 fm RT tog rr 1 1 1 mn 5 | I fy — es | Fe —— Yona — pn) | —e FIGURE 12.23 Elevation,

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Think like a real problem solver and explain me logic for:Uploaded ImageProblem 2: 10 points Be ty Stal by ct mc on recs oy ie athe Fue cow. he rtd don are 10 ct widey 2 ih. hee 5) rows of bricksblocks are ruined for this wal. Morar ois ae 0.8 tic SS wast 17-4 fm RT tog rr 1 1 1 mn 5 | I fy — es | Fe —— Yona — pn) | —e FIGURE 12.23 Elevation,

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Asked by: VARRI APPARAO
Created at: 2025-10-21 07:59:06

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Let's analyze the problem systematically: --- ## **Problem Summary** - **Wall dimensions:** 80 ft long, 17 ft 4 in high - **Bricks:** - 8 in high - 16 in long (length) - **Other details:** - Overhead doors: 12 ft wide - Mortar joints: .8 in thick - Need to determine the number of bricks/blocks to complete the wall --- ## **Step-by-step Logic** ### 1. **Convert all measurements to consistent units** - Wall length: 80 ft = 80 × 12 = 960 in - Wall height: 17 ft 4 in = (17 × 12) + 4 = 204 + 4 = 208 in - Brick dimensions: - Length: 16 in - Height: 8 in - Mortar joint: .8 in --- ### 2. **Calculate the effective brick dimensions with mortar joints** - **Horizontal (length) direction:** Each brick plus mortar joint: \(16 \text{ in} + .8 \text{ in} = 16.8 \text{ in}\) - **Vertical (height) direction:** Each brick plus mortar joint: \(8 \text{ in} + .8 \text{ in} = 8.8 \text{ in}\) --- ### 3. **Determine the number of bricks per row** - **Number of bricks along the length:** \[ \frac{\text{Total wall length}}{\text{Brick + mortar}} = \frac{960}{16.8} \approx 57.14 \] Since bricks can't be fractional, **round down** to 57 bricks per row. - **Number of rows (height):** \[ \frac{\text{Total wall height}}{\text{Brick + mortar}} = \frac{208}{8.8} \approx 23.64 \] Round down to 23 rows. --- ### 4. **Adjust for openings (doors)** - **Door width:** 12 ft = 144 in - **Number of bricks to cover the door width:** \[ \frac{144}{16} = 9 \text{ bricks} \] The door will be 9 bricks wide. - **Number of rows with openings:** Usually, doors are placed across a few courses. For simplicity, assume 3 rows are open or partially open, reducing the number of bricks in those rows. --- ### 5. **Calculate total bricks** - **Total bricks without openings:** \[ \text{Bricks per row} \times \text{Number of rows} = 57 \times 23 = 1311 \] - **Adjust for the door:** In the row(s) with the door, subtract bricks corresponding to the door width: \[ 9 \text{ bricks} \times \text{number of door rows} \approx 3 \times 9 = 27 \] - **Final total bricks:** \[ 1311 - 27 = 1284 \] *Note:* This is a simplified estimation; actual counts may vary with brick bonds, partial bricks, etc. --- ## **Summary** - Approximately **57 bricks per row**. - About **23 rows**. - Total bricks needed: **around 1284**, considering openings and mortar joints. --- **This logic ensures accurate planning for materials, accounting for mortar joints, openings, and brick dimensions.**

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