Question

I. $x=\sqrt{\left[\right(36{)}^{\frac{1}{2}}\times \left(1296{)}^{\frac{1}{4}}\right]}$
II. $2y+3z=33$
III. $6y+5z=71$

• A. $x<y=z$
• B. $x\le y<z$
• C. $x<y>z$
• D. $x=y>z$
• E. $x=y=z$ or none of the above relationship is established

Answer 1

The correct option is B $x\le y<z$

I.
$x=\sqrt{\left[\right(36{)}^{\frac{1}{2}}\times \left(1296{)}^{\frac{1}{4}}\right]}=\sqrt{6\times 6}=\pm 6ByequationII\times 3-equationI6y+9z=996y+5z=71\underset{–}{---}\Rightarrow z=94z=28$
From equation II,
$2y+3\times 7=33\Rightarrow 2y=33-21=12\Rightarrow y=6x\le y<z$