a. There are 3 feet in a yard. If a rug is 5 yards long, how long is it in feet? b. There are 5280 feet in a mile. How long in feet is a 4-mile-long stretch of road? c. Will is driving 65 miles per hour. If he contin- ues driving at that speed, how far will he drive in 3 hours?

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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**Educational Resource: Understanding Multiplication Through Practical Problems**

### Problems for Section

1. **Use the meaning of multiplication to explain why each of the following problems can be solved by multiplying:**

   a. There are 3 feet in a yard. If a rug is 5 yards long, how long is it in feet?

   b. There are 5280 feet in a mile. How long in feet is a 4-mile-long stretch of road?

   c. Will is driving 65 miles per hour. If he continues driving at that speed, how far will he drive in 3 hours?

**Explanation:**

Each of these problems can be approached using multiplication because they involve repeated addition of equal groups. For instance, converting from yards to feet involves multiplying the number of yards by the number of feet per yard. Similarly, calculating distance traveled at a constant speed over multiple hours involves multiplying speed by time.
Transcribed Image Text:**Educational Resource: Understanding Multiplication Through Practical Problems** ### Problems for Section 1. **Use the meaning of multiplication to explain why each of the following problems can be solved by multiplying:** a. There are 3 feet in a yard. If a rug is 5 yards long, how long is it in feet? b. There are 5280 feet in a mile. How long in feet is a 4-mile-long stretch of road? c. Will is driving 65 miles per hour. If he continues driving at that speed, how far will he drive in 3 hours? **Explanation:** Each of these problems can be approached using multiplication because they involve repeated addition of equal groups. For instance, converting from yards to feet involves multiplying the number of yards by the number of feet per yard. Similarly, calculating distance traveled at a constant speed over multiple hours involves multiplying speed by time.
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