Find z (floor). Look at picture! Please?

Question

753 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-30T19:14:00

$0.$

$z=(sqrt3^2-2*sqrt2^2)/(sqrt3-2*sqrt2)=(3-4)/(sqrt3-2sqrt2)$

$=-1/(sqrt3-2sqrt2)xx(sqrt3+2sqrt2)/(sqrt3+2sqrt2)$

$=-(sqrt3+2sqrt2)/{sqrt3^2-(2sqrt2)^2}$

$:. z=-(sqrt3+2sqrt2)/(3-8), or,$

$z=1/5(sqrt3+2sqrt2).$

Taking, $sqrt3=1.7321, and, sqrt2=1.4142,$ we have,

$z=1/5{1.7321+2(1.4142)}=1/5(1.7321+2.8284)$

$:. z=1/5(4.5605)=0.9121.$

$:. floor z=0.$