Find the multiplicative inverse (reciprocal) of each of the following rational numbers:

(i) 9

(ii) -7

(iii) $$\dfrac{12}{5}$$

(iv) $$\dfrac{ -7}{9}$$

(v) $$\dfrac{-3}{-5}$$

(vi) $$\dfrac{ 2}{3} × \dfrac{9}{4}$$

(vii) $$\dfrac{-5}{8} × \dfrac{16}{15}$$

(viii) -2 × $$\dfrac{-3}{5}$$

(ix) -1

(x) $$\dfrac{0}{3}$$

(xi) 1

Asked by Aaryan | 1 year ago |  85

##### Solution :-

(i) The reciprocal of 9 is $$\dfrac{ 1}{9}$$

(ii) The reciprocal of -7 is $$\dfrac{ -1}{7}$$

(iii) The reciprocal of $$\dfrac{ 12}{5}$$ is $$\dfrac{5}{12}$$

(iv) The reciprocal of $$\dfrac{ -7}{9}$$ is $$\dfrac{ 9}{-7}$$

(v) The reciprocal of $$\dfrac{ -3}{-5}$$is $$\dfrac{ 5}{3}$$

(vi) The reciprocal of $$\dfrac{ 2}{3}$$ × $$\dfrac{ 9}{4}$$ is

Firstly solve for $$\dfrac{ 2}{3}× \dfrac{ 0}{4}= \dfrac{ 1}{1}× \dfrac{ 3}{2} = \dfrac{ 3}{2}$$

The reciprocal of $$\dfrac{ 3}{2}$$ is $$\dfrac{ 2}{3}$$

(vii) The reciprocal of $$\dfrac{ -5}{8} × \dfrac{16}{15}$$

Firstly solve for $$\dfrac{ -5}{8} × \dfrac{16}{15}=\dfrac{ -1}{1} × \dfrac{ 2}{3} = \dfrac{ -2}{3}$$

The reciprocal of $$\dfrac{- 2}{3}$$ is $$\dfrac{ 3}{-2}$$

(viii) The reciprocal of $$-2 × \dfrac{-3}{5}$$

Firstly solve for $$-2 × \dfrac{-3}{5}=\dfrac{6}{5}$$

The reciprocal of $$\dfrac{6}{5}$$is $$\dfrac{5}{6}$$

(ix) The reciprocal of -1 is -1

(x) The reciprocal of $$\dfrac{0}{3}$$ does not exist

(xi) The reciprocal of 1 is 1

Answered by Sakshi | 1 year ago

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