How do you evaluate #\sqrt { \frac { 27} { 9} } + 5#?

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Answer 1 · 2017-12-14T14:07:41

#sqrt(3) + 5#

= 6.73205

The expression can be rewritten as:

#(3^3/3^2)^(1/2)# + 5

and thus, applying the exponent outside the bracket:

#(3^(3/2)/3) + 5#

= #((3sqrt(3))/(3)) + 5#

= #sqrt(3) + 5#

= 6.73205

Answer 2 · 2017-12-14T14:08:54

#6.7321#

#sqrt(27/9) + 5#

#sqrt27/3 + 5#

#sqrt(9 xx 3)/3 + 5#

#(sqrt9 xx sqrt3)/3 + 5#

#(3 xx sqrt3)/3 + 5#

#(3 sqrt3)/3 + 5#

#(cancel3 sqrt3)/cancel3 + 5#

#sqrt3 + 5#

#1.7321 + 5#

#6.7321#

Answer 3 · 2017-12-14T14:15:29

#=sqrt3+5#

#=6.732#

Simplify under the root first:

#sqrt(27/9) +5#

#=sqrt((cancel9xx3)/cancel9)+5#

#=sqrt3+5#

#sqrt3# is an irrational number, so you have to round off if you calculate the answer as a decimal.

#= 1.732+5#

#=6.732#