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4.  Find
(a)  (-7)-8-(-25)
(b)  (-13)+32-8-1
(c)  (-7)+(-8)+(-90)
(d)  50-(-40)-(-2)

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We already know that the sum of the additive inverse is Zero (a+ (-a) = 0

(a)  (-7)-8-(-25)
      \\= -7 -8 +25\\ =-(7+8)+(15)+10\\ =-15+(-15)+10\\ =0+10 =10


(b)  (-13)+32-8-1 = (-13)+ 32 -(8+1)
                                                  \\= (-13)+ 32 -(9)\\ =-(13+9)+32\\ =(-22)+22+10\\ =0+10=10


(c)  (-7)+(-8)+(-90)
       \\= (-7)-(8)-(90)\\ =-(7+8+90)\\ =-(105)


(d)  50-(-40)-(-2)
       = 50 +40 +2
       =92

Posted by

manish

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