Use graphing to approximate the zeros of \( f(x)=x^{2}+2 x-5 \) to the nearest tenth.

Solve the absolute value inequality. \( |x-6| \geq 1 \) A. The solution set in interval notation is (Simplify your answer.) B. The solution set is \( \varnothing \).

Given \( f(x)=-3 x+4 \) and \( g(x)=x+7 \), what is the value of \( g(f(2)) ? \)

Let \( A=\left[\begin{array}{cc}4 & 5 \\ 1 & -5\end{array}\right] \) and \( B=\left[\begin{array}{cc}2 & -6 \\ 2 & 3\end{array}\right] \) Find \( -2 A+5 B \) \( -2 A+5 B= \)

lify \( \left(\frac{x^{-2} y^{-1 / 3}}{x^{-5 / 3} y^{-2 / 3}}\right)^{3} \) so there

\[ \begin{array}{l}|x+6| \leq 7\end{array} \] Solve the absolute value inequality. A. The solution set in interval notation is (Simplify your answer.) The solution set is \( \varnothing \).