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Chapter 14: Problem 6

\(k>0\) \(l>1\) Quantity \(\mathbf{A}\) \(\frac{1}{\frac{1}{k}+\frac{1}{l}}\) Quantity \(\underline{B}\) \(\frac{k l}{\frac{1}{k}+\frac{1}{l}}\) a. Quantity A is greater. b. Quantity B is greater. c. The two quantities are equal. d. The relationship cannot be determined from the information given.

Short Answer

Expert verified
b. Quantity B is greater.

Step by step solution

01

Simplify Quantity A

Simplify \(\frac{1}{\frac{1}{k}+\frac{1}{l}}\) using the reciprocal rule. The formula for the reciprocal of a fraction is \(\frac{1}{\frac{a}{b}} = b/a\). Hence \(\frac{1}{\frac{1}{k}+\frac{1}{l}}\) becomes \(\frac{k l}{k + l}\).
02

Simplify Quantity B

Simplify \(\frac{k l}{\frac{1}{k}+\frac{1}{l}}\) using the reciprocal rule. The formula for the reciprocal of a fraction is \(\frac{a}{\frac{b}{c}} = a \cdot \frac{c}{b}\). Hence \(\frac{k l}{\frac{1}{k}+\frac{1}{l}}\) becomes \(k l \cdot (k + l)\) which simplifies further to \(k^2 l + k l^2\).
03

Compare both quantities

By comparing quantity A and B, it can be observed that quantity A is \(\frac{k l}{k + l}\) and quantity B is \(k^2 l + k l^2\). Since both \(k\) and \(l\) are greater than 0 and 1 respectively, it can be concluded that quantity B is greater than quantity A.

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