Please solve this?
question in the attachment
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"What balanced equation represents nuclear fusion?"
We will use the fact that #sqrt(a/b)=sqrta/sqrtb#
#7*sqrt1/sqrt3-7/3*sqrt1/sqrt3+3sqrt147#
#=7/sqrt3-7/(3sqrt3)+3sqrt147#
#=(7sqrt3)/3-(7sqrt3)/9+3sqrt147#
#=(21sqrt3)/9-(7sqrt3)/9+3sqrt147#
#=(14sqrt3)/9+3sqrt147#
I've left the #3sqrt147# up till now, cos its nasty to deal with. However, seeing that we have an expression with #sqrt3#, there's a good chance that 147 will be equal to three times a square number. Some good ol' bus shelter long division (or using a calculator if you're allowed) will show you that #147=3xx49# Since 49 is a square number, we can work on simplifying this more
#=(14sqrt3)/9+3[sqrt(3xx49)]#
#=(14sqrt3)/9+3(7sqrt3)#
#=(14sqrt3)/9+21sqrt3#
#=(14sqrt3)/9+(189sqrt3)/9#
#=203/9sqrt3#
#sqrt(1/3) =sqrt3/3#
#sqrt(147)=sqrt49 xx sqrt3=7sqrt3#
#7sqrt(1/3)-2 1/3sqrt(1/3)+3sqrt(147)#
=#(7sqrt3)/3-(7sqrt3)/9+21sqrt3#
#(14sqrt3)/9+21sqrt3=(203sqrt3)/9#
=39.06736822
