Question

Answer me question number 5,6,7 fast please....
please answer me...urgent i will give you Brainliest.......and will get 20 points..

Answer 1

Step-by-step explanation:

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Answer 2

Question 5 :

Verify that :

$\sf(x+y) ^{ - 1} \neq \: ( {x}^{ - 1} ) + ( {y}^{ - 1} )$

When x = $\sf\dfrac{5}{9}$

and y = $\sf\dfrac{-4}{3}$

Pur the values :

$\longrightarrow \sf \: [ \dfrac{5}{9} + ( \dfrac{ - 4}{3} )] ^{ - 1} \neq \: (\dfrac{5}{9} )^{ - 1} + ( \dfrac{ - 4}{3} )^{ - 1}$

$\longrightarrow \sf (\dfrac{5}{9} - \dfrac{4}{3}) ^{ - 1} \neq( \dfrac{5}{9} ) ^{ - 1} - (\dfrac{4}{3} )^{ - 1}$

$\longrightarrow \sf( \dfrac{5 - 12}{9}) ^{ - 1} \neq( \dfrac{9}{5} - \dfrac{3}{4} )$

$\longrightarrow \sf( \dfrac{ - 7}{9}) ^{ - 1} \neq( \dfrac{36 - 15}{20} )$

$\longrightarrow \sf( \dfrac{ - 9}{7}) \neq( \dfrac{21}{20} )$

Hence,Proved!

Question 6 :

Verify that :

$\sf(x \times y) ^{ - 1} = \: ( {x}^{ - 1} ) \times ( {y}^{ - 1} )$

When x = $\sf\dfrac{-2}{3}$

and y = $\sf\dfrac{-3}{4}$

Put the values :

$\longrightarrow \sf \: [ \dfrac{ - 2}{3} \times ( \dfrac{ - 3}{4} )] ^{ - 1} = \: (\dfrac{ - 2}{3} )^{ - 1} \times ( \dfrac{ - 3}{4} )^{ - 1}$

$\longrightarrow \sf \: ( \dfrac{ 6}{12} )^{ - 1} = \: (\dfrac{ -3}{2} ) \times ( \dfrac{ -4}{3} )$

$\longrightarrow \sf \: \dfrac{ 12}{6} = \dfrac{ 12}{6}$

Hence,Proved!

Question 7 :

i) The number zero has no reciprocal

ii) 1 and -1 are their own reciprocals

iii) If a is the reciprocal of b then b is the reciprocal of a

$\sf \: iv)( 11 \times 5) ^{ - 1} = ( {11)}^{ - 1} \times ( \bold {5})^{ - 1}$

$\sf \: v) \dfrac{ - 1}{8} \times \bold{\dfrac{ - 8}{1} } = 1$

$\sf \: vi) -5 \dfrac{1}{3} = \dfrac{-16}{3}$

$\sf\therefore \dfrac{-3}{16} \times\:\dfrac{-16}{3} = 1$