\( x^{2}-5 \geqslant 3 \) 2) \( 4 \sqrt{x}<100 \) 3) \( 4 x^{3}+11<43 \) 4) \( \frac{1}{x}+9 \geqslant-1 \)

To solve \( x^{2}-5 \geqslant 3 \), first, simplify it to \( x^{2} \geqslant 8 \). This means \( x \) can be either \( \sqrt{8} \) or \( -\sqrt{8} \) (or in simpler terms \( x \geqslant 2\sqrt{2} \) or \( x \leqslant -2\sqrt{2} \)). It’s important to denote these intervals on a number line for visibility! For \( 4 \sqrt{x}<100 \), divide both sides by 4 to get \( \sqrt{x} < 25 \). Squaring both sides yields \( x < 625 \). Always remember to check the original inequality when squaring, as it can introduce extraneous solutions! In the case of \( 4 x^{3}+11<43 \), simplify to find \( 4 x^{3} < 32 \), then divide by 4 to get \( x^{3} < 8 \). Finally, take the cube root, yielding \( x < 2 \). Keep in mind that any negative values are also acceptable! The inequality \( \frac{1}{x}+9 \geqslant-1 \) boils down to \( \frac{1}{x} \geqslant-10 \). This implies \( x \) can either be positive or cater to the scenario where \( x < 0 \). Be cautious with the signs when interpreting! Enjoy solving these inequalities! They can often lead to fun discussions about functions and their graphs. Don’t forget to check for extraneous solutions and plot your intervals correctly!