# Question:- $$( \frac{ - 3}{4} .. \frac{2}{3} .. \frac{ - 5}{6} )rationa \: nmbers \: distunutive \\ \: of \: multipivatiom$$ ​

Math

## Question

Question:-
$$( \frac{ - 3}{4} .. \frac{2}{3} .. \frac{ - 5}{6} )rationa \: nmbers \: distunutive \\ \: of \: multipivatiom$$

• ### 1. User Answers Anonym

$$To \: understand \: this, \: consider \: the \: rational \\ numbers( \frac{ - 3}{4}.. \frac{2}{3}.. \frac{ - 5}{6} )$$

$$\frac{ - 3}{4} \times( \frac{2}{3} + ( \frac{ - 5}{6} ) = \frac{ - 3}{4} \times ( \frac{(4) + ( - 5)}{6} )$$

$$= \frac{ - 3}{4} \times ( \frac{ - 1}{6} ) = \frac{3}{24} = \frac{1}{8}$$

$$\frac{ - 3}{4} \times \frac{2}{3} = \frac{ - 3 \times 2}{4 \times 3} = \frac{ - 6}{12} = \frac{ - 1}{2}$$

$$and \: \frac{ - 3}{4} \times \frac{ - 5}{6} = \frac{5}{8}$$

$$therefore \: ( \frac{ - 3}{4} \times \frac{2}{3} ) + ( \frac{ - 3}{4} \times \frac{ - 5}{6} ) = \frac{ - 1}{2} + \frac{5}{8} = \frac{1}{8}$$

$$thus.. \frac{ - 3}{4} \times ( \frac{2}{3} + \frac{ - 5}{6} ) = ( \frac{ - 3}{4} \times \frac{2}{3} ) + ( \frac{ - 3}{4} \times \frac{ - 5}{6} )$$

$$distributivity \: of \: multiplication \: over \: \\addition \: for \: rational \: numbers$$

• ### 2. User Answers Anonym

$$= \frac{ - 3}{4} \times ( \frac{ - 1}{6} )= \frac{3}{24} = \frac{1}{8}=$$

$$\frac{ - 3}{4} \times \frac{2}{3} = \frac{ - 3 \times 2}{4 \times 3} = \frac{ - 6}{12} = \frac{ - 1}{2}$$

$$\frac{ - 3}{4} \times \frac{ - 5}{6} = \frac{5}{8}$$

$$\frac{ - 3}{4} \times \frac{2}{3} ) + ( \frac{ - 3}{4} \times \frac{ - 5}{6} ) = \frac{ - 1}{2} + \frac{5}{8} = \frac{1}{8}$$

$$\frac{ - 3}{4} \times ( \frac{2}{3} + \frac{ - 5}{6} ) = ( \frac{ - 3}{4} \times \frac{2}{3} ) + ( \frac{ - 3}{4} \times \frac{ - 5}{6} )$$

Property used here distributive