What is the value of #2x^(1/2) +x^0 +x^(1/4)# when #x=16#?

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Answer 1 · 2017-04-17T04:59:01

#11#

Sub in #x=16# to the equation.

Recall that #n^0=1# where #n# is any number.

Doing so,

#2(16)^(1/2)+(16)^0+(16)^(1/4)#

Simplify,

#2(4)+1+(2)#

#8+1+2#

#11#