Asked by deborah
-5jk
_______
35j^2k^2
reduce
5/35 = 1/7
j/j^2 = 1/j
k/k^2 = 1/k
answer is -1/(7jk)
_______
35j^2k^2
reduce
5/35 = 1/7
j/j^2 = 1/j
k/k^2 = 1/k
answer is -1/(7jk)
Answers
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Anonymous
75a^2b^2 over 25ab
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