Question

PAGE NO.
DATE:
120
Find the value
of [4^-1+3^-1 +6^-2]^-1​

Answer 1

Answer:

$\frac{12}{7}$

Step-by-step explanation:

${( {4}^{ - 1} + {3}^{ - 1} + {6}^{ - 2} })^{ - 1} \\ \\ = \frac{1}{ ( {4}^{ - 1} + {3}^{ - 1} + {6}^{ - 2} )} \\ \\ = \frac{1}{ \frac{1}{4} + \frac{1}{3} + \frac{1}{ {6}^{2} } } \\ \\ = \frac{1}{ \frac{1}{4} + \frac{1}{3} + \frac{1}{36} } \\ \\ = \frac{1}{ \frac{8 + 12 + 1}{36} } \\ \\ = \frac{1}{ \frac{21}{36} } \\ \\ = \frac{36}{21} \\ \\ = \frac{12}{7}$

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