Find the values of each of the following:(i) \(3^{-1}\)+\(4^{-1}\)(ii

1.	Find the values of each of the following:(i) \(3^{-1}\)+\(4^{-1}\)(ii) ( \(3^{0}\)+\(4^{-1}\))x\(2^{2}\)(iii) ( \(3^{-1}\)+\(4^{-1}\)+\(5^{-1}\))\(^{0}\)(iv) \(\{{(\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}+(\frac{1}{4})^{-2}}\}\)
Answer» (i) \(3^{-1}+4^{-1}\) ⇒ \(\frac{1}{3}+\frac{1}{4}\)= \(\frac{4+2}{12}\)= \(\frac{7}{12}\)(LCM of 3 and 4 is 12) [Using \(a^{-n}\) = \(\frac{1}{a^{n}}\)] (ii) (\(3^{0}+4^{-1}\)) x \(2^{2}\) ⇒ \((1+\frac{1}{4})\times 4\) = \((\frac{4+1}{4})\times 4\) = \(\frac{5\times 4}{4}\) = 5(LCM of 1 and 4 is 4) [Using \(a^{-n}\) = \(\frac{1}{a^{n}}\); \(a^{0}\) = 1;\(a^{2}\) = \(a\times a\)] (iii) \((3^{-1}+4^{-1}+5^{-1})^{0}\) ⇒ \((3^{-1}+4^{-1}+5^{-1})^{0}\) = 1 [Using \(a^{0}\) = 1] We know that any number to power zero is always equal to 1. (iv) \(\{(\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}+(\frac{1}{4})^{-2}\}\) ⇒ (3)-(4)+\((4^{2})\) [Using \(a^{-n}\) = \(\frac{1}{a^{n}}\); \(a^{2}\) = \(a\times a\)] 3-4+16 = 15

Find the values of each of the following:(i) \(3^{-1}\)+\(4^{-1}\)(ii) ( \(3^{0}\)+\(4^{-1}\))x\(2^{2}\)(iii) ( \(3^{-1}\)+\(4^{-1}\)+\(5^{-1}\))\(^{0}\)(iv) \(\{{(\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}+(\frac{1}{4})^{-2}}\}\)

Answer»

(i) \(3^{-1}+4^{-1}\)

⇒ \(\frac{1}{3}+\frac{1}{4}\)= \(\frac{4+2}{12}\)= \(\frac{7}{12}\)(LCM of 3 and 4 is 12)

[Using \(a^{-n}\) = \(\frac{1}{a^{n}}\)]

(ii) (\(3^{0}+4^{-1}\)) x \(2^{2}\)

⇒ \((1+\frac{1}{4})\times 4\) = \((\frac{4+1}{4})\times 4\) = \(\frac{5\times 4}{4}\) = 5(LCM of 1 and 4 is 4)

[Using \(a^{-n}\) = \(\frac{1}{a^{n}}\); \(a^{0}\) = 1;\(a^{2}\) = \(a\times a\)]

(iii) \((3^{-1}+4^{-1}+5^{-1})^{0}\)

⇒ \((3^{-1}+4^{-1}+5^{-1})^{0}\) = 1

[Using \(a^{0}\) = 1]

We know that any number to power zero is always equal to 1.

(iv) \(\{(\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}+(\frac{1}{4})^{-2}\}\)

⇒ (3)-(4)+\((4^{2})\)

[Using \(a^{-n}\) = \(\frac{1}{a^{n}}\); \(a^{2}\) = \(a\times a\)]

3-4+16 = 15