SOLUTION: How would you solve:
(a) 16^1/4 (b) 25^3/2 (c) 8^-2/3
^ stands for power of
Thanks.
Algebra.Com
Question 230724: How would you solve:
(a) 16^1/4 (b) 25^3/2 (c) 8^-2/3
^ stands for power of
Thanks.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
How would you solve:
(a) 16^1/4 = 2
------------------------
(b) 25^3/2 = 5^3 = 125
-----------------------
(c) 8^-2/3 = 2^-2 = (1/2)^2 = 1/4
=========================================
Cheers,
Stan H.
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