[size=150]Find [b]all[/b] the solutions to each equation.[/size][br][br][math]x(x-1)=0[/math]
[math](5-x)(5+x)=0[/math]
[math](2x+1)(x+8)=0[/math]
[math](3x-3)(3x-3)=0[/math]
[math](7-x)(x+4)=0[/math]
[math]x^2+18x+81=0[/math]
[math]x^2+0.2x+0.01=0[/math]
[size=150]Here is how Elena solves the quadratic equation [math]x^2-3x-18=0[/math].[/size][br][br][math]\displaystyle \begin{align} x^2 -3x -18 &=0\\ (x-3)(x+6)&=0\\ x-3=0 \quad \text { or } &\quad x+6=0\\ x=3\quad \text{ or } &\quad x= \text- 6\\ \end{align}\\[/math][br][br]Is her work correct? If you think there is an error, explain the error and correct it.[br][br]Otherwise, check her solutions by substituting them into the original equation and showing that the equation remains true.
[table][tr][td][math]\begin{align} p^2-5p&=0\\p(p-5)&=0\\p-5&=0\\p&=5\end{align}[/math][br][/td][td][size=150]She thinks that her solution is correct because substituting 5 for [math]p[/math] in [br]the original expression [math]p^2-5p[/math] gives [math]5^2-5(5)[/math], which is [math]25-25[/math] or 0.[/size][/td][/tr][/table][br]Explain the mistake that Jada made and show the correct solutions.
[size=150]Choose a statement to correctly describe the zero product property. [br][br]If [math]a[/math] and [math]b[/math] are numbers, and [math]a\cdot b=0[/math], then:[/size]
[size=150] Which expression is equivalent to [math]x^2-7x+12[/math]?[/size][br]
[size=150]Select [b]all[/b] the functions whose output values will eventually overtake the output values of function [math]f[/math] defined by [math]f(x)=25x^2[/math].[/size]
[math]p(x)=\begin{cases} x-1, \quad x\leq \text- 2 \\ 2x-1,\quad x>\text{-}2\\ \end{cases}[/math][br][br][size=150]Find the value of [math]p[/math] at each given input.[br][br][math]p(\text{-}20)[/math] [br][/size]
[math]p(\text{-}2)[/math]