Question

Can anyone just tell me what this means in interval notation?
{x element R : 6 x+sqrt(33)<=0 or 6 x>=sqrt(33)}
{x element R : 2 x>=1}
{x element R : 32 x>=17}

Answer 1

I'll do the 1st for you:
$$
6x+\sqrt{33}\le 0\\
6x \le -\sqrt{33}\\
x\le \cfrac{-\sqrt{33}}{6}
$$
OR
$$
6 x>=\sqrt{33}\\
x\ge\cfrac{\sqrt{33}}{6}
$$
Which means in interval notation:
$$
(-\infty,\cfrac{-\sqrt{33}}{6}]\cup[x\ge\cfrac{\sqrt{33}}{6},\infty)
$$

Answer 2

Correction:
$$
\left (-\infty,\cfrac{-\sqrt{33}}{6}\right ]\cup\left [\cfrac{\sqrt{33}}{6},\infty\right )
$$

Answer 3

http://www.coolmath.com/algebra/07-solving-inequalities/03-interval-notation-04.htm