How do you evaluate #Y=2\cdot (0.25)^{-1}#?

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Answer 1 · 2018-01-30T13:20:07

See a solution process below:

Use these rules of exponents to rewrite the expression on the right:

#x^color(red)(a) = 1/x^color(red)(-a)# and #a^color(red)(1) = a#

#Y = 2 * (0.25)^color(red)(-1)#

#Y = 2 * 1/(0.25)^color(red)(- -1)#

#Y = 2 * 1/(0.25)^color(red)(1)#

#Y = 2 * 1/(0.25)#

#Y = 2/(0.25)#

#Y = 8#