Answer:
(a) We have, \[(-8)\,+\_\_\_\_=0\] \[(-8)+8=0\] [\[\because \]Additive inverse of ?8 is 8] (b) We have, \[~13+\_\_\_=0\] \[13+(-13)=0\] [\[\because \]Additive inverse of ?13 is 13] (c) We have, \[12+(-12)=\_\_\_12+(-12)=0\] [\[\because \]12 and ?12 are additive inverse of each other] (d) We have, \[(-4)+\_\_\_\_=-12\] Here, sum of two integers is not zero, so here we cannot write additive inverse of ?4. \[-4+\_\_\_\_=-12\Rightarrow -4+\_\_\_=-(4+8)\]\[\Rightarrow \] \[-4+\_\_=(-4)+(-8)\] So, on comparing both sides, we get ?8 to fill the blank space. Thus, \[(-4)+(-8)=-12\] (e) We have, \[\_\_15=-10\Rightarrow \_\_-(10+5)=-10\] \[\Rightarrow \] \[\_\_-10-5=-10\] To get RHS = ?10, we need to add additive inverse of ?5, i.e. 5 in RHS. So, \[+5-10-5=-10\] Hence, + 5 ? 15 = ?10
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