Consider the matrices \( X=\left[\begin{array}{rrr}x+y-1 & 0 \\ 0 & 3\end{array}\right] \) and \( Y=\left[\begin{array}{rr}5 & 0 \\ 0 & 2 x-3 y-7\end{array}\right] \) The value(s) of \( x \) for which both matrices have no inverses is/are given by: \( \begin{array}{l}x=-2 \text { or } x=2 \\ x=-2 \\ x=-2 \text { and } x=2 \\ \text { No such value of } x \text { exists }\end{array} \).

Simplify the expression. Assume that the variable is unrestricted and use absolute value symbols when necessary. \[ \sqrt{25 y^{6}} \]

iraficar la siguiente \( y=x^{2}+4 x-5 \)

Simplify each expression. Assume that the variables represent any real numbers. Use the absolute value button only when necessary. (a) \( \sqrt{y^{16}}=\square \) (b) \( \sqrt{x^{18}}=\square \)

\( \begin{array}{l}\text { Ecuaciones Hultipicat } \\ \frac{9 x+3}{6}\end{array}=2 \)

Use the Factor Theorem to determine whether \( x-2 \) is a factor of \( P(x)=x^{3}-3 x^{2}-4 x+ \) Specifically, evaluate \( P \) at the proper value, and then determine whether \( x-2 \) is a factor. \( \begin{array}{l}P(\square)=\square \\ x-2 \text { is a factor of } P(x) \\ x-2 \text { is not a factor of } P(x)\end{array} \)