\( X=\left[\begin{array}{rr}x+y-1 & 0 \\ 0 & 3\end{array}\right] \) and \( Y=\left[\begin{array}{lr}5 & 0 \\ 0 & 2 x-3 y-7\end{array}\right] \) The value(s) of \( y \) for which the matrices \( X \) has no inverse and \( Y=-5 I_{2} \) is/are given by: \( \begin{array}{l}y=1 \\ 0 \\ y=-2\end{array} \) No such value of \( y \) exists \( y=0 \)

1 Graph \( y-2=\frac{2}{3}(x-3) \)

Your answer Is Incorrect. \( 9 \times(3+4)=(\square \times 3)+(\square \times 4) \) Use the distributive property to fill in the blanks below.

Use the distributive property to fill in the blanks below. \( 6 \times(9-4)=(\llbracket \times 9)-(\square \times 4) \) Introduction to the distributive property

Consider the line \( y=\frac{5}{2} x-3 \). Find the equation of the line that is perpendicular to this line and passes through the point \( (2,-5) \). Find the equation of the line that is parallel to this line and passes through the point (2, -5). Note that the ALEKS graphing calculator may be helpful in checking your answer. Equation of perpendicular line:

b. \( -0.5 p+\frac{1}{2} p-2.75+\frac{2}{3} \)