Algebra and Trigonometry: Structure and Method, Book 2

2000th Edition

ISBN: 9780395977255

Author: MCDOUGAL LITTEL

Publisher: McDougal Littell

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Question

Chapter 5.9, Problem 22P

To determine

To find the two positive numbers that differ by 12 and have harmonic mean 5.

Expert Solution & Answer

Answer to Problem 22P

The number be 15 and 3

Explanation of Solution

Let be the numbers be (a) and (a−12) and harmonic mean 5 Given that the average of 1a and 1a−12 is 15

Substitute the values

(1a+1a−12)2=151a+1a−12=25a−12+aa(a−12)=255(2a−12)=2a(a−12)10a−60=2a2−24a2a2−34a+60=0a2−17a+30=0(a−15)(a−2)=0a=2,15

If a=2 then

b=a−12=2−12b=−10

But both numbers be positive So, a≠2

If a=15

b=15−12=3b=3

Hence, the number be 15 and 3.

Chapter 5.9, Problem 21P

Chapter 5.9, Problem 23P

Chapter 5 Solutions

Algebra and Trigonometry: Structure and Method, Book 2

Ch. 5.1 - Prob. 1OE Ch. 5.1 - Prob. 2OE Ch. 5.1 - Prob. 3OE Ch. 5.1 - Prob. 4OE Ch. 5.1 - Prob. 5OE Ch. 5.1 - Prob. 6OE Ch. 5.1 - Prob. 7OE Ch. 5.1 - Prob. 8OE Ch. 5.1 - Prob. 9OE Ch. 5.1 - Prob. 10OE

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