# SOLUTION: Can someone help me with this one? I hope I write this out right... (-3*10^-3) (2*10^2)= I think I have some of it. I get to -6*1-^-5= (and then I am lost - if that is even

Algebra ->  -> SOLUTION: Can someone help me with this one? I hope I write this out right... (-3*10^-3) (2*10^2)= I think I have some of it. I get to -6*1-^-5= (and then I am lost - if that is even       Log On

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 Click here to see ALL problems on Exponents Question 748714: Can someone help me with this one? I hope I write this out right... (-3*10^-3) (2*10^2)= I think I have some of it. I get to -6*1-^-5= (and then I am lost - if that is even right) Can someone show me how to do this. My text book is useless in explanations. Thanks!Found 2 solutions by stanbon, rothauserc:Answer by stanbon(60771)   (Show Source): You can put this solution on YOUR website!(-3*10^-3) (2*10^2) -------- = -3*2*10^-3*10^2 -------- = -6*10^-1 ---- = -6/10 ---- = -0.6 ============ Cheers, Stan H. ============ Answer by rothauserc(643)   (Show Source): You can put this solution on YOUR website!we are asked to evaluate (-3*10^-3) (2*10^2)= (-3/10^3) * (2*10^2) note that 10^-3 = 1/10^3 = (-3/10) * (2) cancel 10^2 from numerator and denominator = -6/10 = -3/5