How do you simplify #(9sqrt25)/sqrt50#?

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Answer 1 · 2018-04-27T06:33:18

#9/sqrt2#

#(9 sqrt25)/sqrt50#

#= (9xx5)/sqrt 50#

#sqrt50# can be simplified to #sqrt(25xx2) = 5sqrt 2#

#45 /( 5 sqrt 2)#

= #9 /sqrt2#

Answer 2 · 2018-04-27T06:33:33

#[9sqrt2]/2#

#9sqrt25# = #9 xx 5# = 45

#sqrt50=sqrt25xx2=5sqrt2#

so #[9sqrt25]/sqrt50=45/[5sqrt2]#

dividing by top and bottom by 5 leaves

#9/sqrt2#

If we multiply top and bottom by #sqrt2# this will rationalise it (remove the surd from the denominator)

#9/sqrt2 xx sqrt2/sqrt2# =#[9sqrt2]/2#

Answer 3 · 2018-04-27T06:36:17

#(9sqrt25)/sqrt50 = (9sqrt2)/2#

#sqrt(ab) = sqrta *sqrtb#

So, # (9sqrt25)/sqrt50 =(9sqrt25)/(sqrt25*sqrt2 )=9/sqrt2#

#9/sqrt2 xx sqrt2/sqrt2 = (9sqrt2)/2#