Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN: 9780134463216
Author: Robert F. Blitzer
Publisher: PEARSON
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**Solve the problem.**

The floor of a rectangular room is to be tiled with \( \frac{1}{3} \) foot square tiles along a \( 6 \frac{3}{8} \) foot wall. How many tiles will be needed along the wall?

- \( 2 \frac{1}{8} \) tiles
- \( 18 \frac{3}{8} \) tiles
- \( 19 \frac{1}{8} \) tiles
- 21 tiles

**Explanation:**

To solve this problem, divide the length of the wall by the size of the tile:

1. Convert \( 6 \frac{3}{8} \) feet to an improper fraction:
   \[
   6 \frac{3}{8} = \frac{51}{8}
   \]

2. Divide by the size of each tile, \( \frac{1}{3} \) foot:
   \[
   \text{Number of tiles} = \frac{\frac{51}{8}}{\frac{1}{3}} = \frac{51}{8} \times \frac{3}{1} = \frac{153}{8}
   \]

3. Convert \( \frac{153}{8} \) to a mixed number:
   \[
   \frac{153}{8} = 19 \frac{1}{8}
   \]

The answer is \( 19 \frac{1}{8} \) tiles.
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Transcribed Image Text:**Solve the problem.** The floor of a rectangular room is to be tiled with \( \frac{1}{3} \) foot square tiles along a \( 6 \frac{3}{8} \) foot wall. How many tiles will be needed along the wall? - \( 2 \frac{1}{8} \) tiles - \( 18 \frac{3}{8} \) tiles - \( 19 \frac{1}{8} \) tiles - 21 tiles **Explanation:** To solve this problem, divide the length of the wall by the size of the tile: 1. Convert \( 6 \frac{3}{8} \) feet to an improper fraction: \[ 6 \frac{3}{8} = \frac{51}{8} \] 2. Divide by the size of each tile, \( \frac{1}{3} \) foot: \[ \text{Number of tiles} = \frac{\frac{51}{8}}{\frac{1}{3}} = \frac{51}{8} \times \frac{3}{1} = \frac{153}{8} \] 3. Convert \( \frac{153}{8} \) to a mixed number: \[ \frac{153}{8} = 19 \frac{1}{8} \] The answer is \( 19 \frac{1}{8} \) tiles.
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