Question

answer this questions​

Answer 1

Answer:

Question 3)

$\frac{1}{2}$ & $\frac{1}{3}$

since they do not have the same denominator thus first we will find the LCM of (2,3) and then find the factors

$= \frac{1*6}{2*6}$ & $\frac{1*4}{3*4}$

$= \frac{6}{12}$ & $\frac{4}{12}$

thus the rational number between $\frac{1}{2}$ & $\frac{1}{3}$ = $\frac{5}{12}$

III

Question 1) $\frac{3}{7}+(\frac{-4}{9})+(\frac{-11}{7}) +\frac{7}{9}$

$\frac{3}{7} -\frac{11}{7}+\frac{7}{9} -\frac{4}{9}\\\\\frac{-8}{7} + \frac{3}{9}\\\\\frac{-8}{7}+\frac{1}{3}\\\\\frac{[(-8)*3]+[1*7]}{7*3}\\\\\frac{-24+7}{21}\\\\\frac{-17}{21}$

Question 2) $\frac{-5}{6}$ & $\frac{7}{8}$

just like we did in question 3 above we will do the same here

$=\frac{-5*4}{6*4}$ & $\frac{7*3}{8*3}$

$=\frac{-20}{24}$ & $\frac{21}{24}$

thus the 5 rational numbers are :

$\frac{-22}{24}$ , $\frac{-25}{24}$ , $\frac{-30}{24}$ , $\frac{10}{24}$ , $\frac{20}{24}$

if you want you can write them in their lowest form as well

Question 3)  $(\frac{-8}{9}* \frac{1}{-5} )+(\frac{-8}{9}* \frac{-7}{11}) = \frac{-8}{9}(\frac{1}{-5} + \frac{-7}{11})$

LHS

$(\frac{-8}{9}* \frac{1}{-5} )+(\frac{-8}{9}* \frac{-7}{11})$

It is the distributive property of multiplication over addition

$= \frac{-8}{9}(\frac{1}{-5} + \frac{-7}{11})$

RHS

$= \frac{-8}{9}(\frac{1}{-5} + \frac{-7}{11})$

                                          LHS=RHS

                                       Hence proved