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Algebra

by lucas211 » Wed Apr 06, 2016 3:13 am
Hello everybody

Would appreciate a little help on the following question.

(0,08)^-4
(0,04)^-3 ?

a) -16/25
b) 16/25
c) 25/16
d) 16
e) 25

Thanks in advance
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by regor60 » Wed Apr 06, 2016 5:55 am
Can be less confusing if you call .04 "X" and therefore, .08 = 2X

Then, (2X)^-4/x^-3 = (1/2^4) x X^-4/X^-3 = (1/16) x X^-1 = 1/16X

Substituting .04 back in for X yields 1/(16 x 4/100) = 1/(16/25) =25/16
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