Question 145116
{{{(1/3)sqrt(24)-2sqrt(2/3)}}}


{{{(1/3)^2=1/9}}}, so {{{(1/3)sqrt(24)=sqrt(24/9)=sqrt((8*3)/(3*3))=sqrt(8/3)}}}, so:


{{{(1/3)sqrt(24)-2sqrt(2/3)=green(sqrt(8/3)-2sqrt(2/3))=sqrt(4*(2/3))-2sqrt(2/3)=green(sqrt(4)*sqrt(2/3)-2sqrt(2/3))=2sqrt(2/3)-2sqrt(2/3)=green(0)}}}



2.  Area of a rectangle: {{{A=LW}}}.  If length is twice the width and the area is 64, then {{{L=2W}}} and the area is {{{A=(2W)W=2W^2=64}}}


Check your work by multiplying the value you determined for W by the value you determined for L.  You have solved the problem correctly if your result is 64 and L is twice the value of W.


Divide both sides by 2 and take the square root of both sides giving you the value of W.  Multiply that by 2 to get L.