\( \left[\begin{array}{rrr|r}2 & 1 & -3 & 2 \\ 2 & -3 & 1 & 10 \\ -2 & 1 & 1 & -6\end{array}\right] \) Use the following steps to reduce the augmented matrix to generalized row echelon form: Step 1: \( R_{3}+R_{1} \longrightarrow R_{3} \) \( R_{2}-R_{1} \longrightarrow R_{2} \) Step \( 2: \) \( \left(-\frac{1}{4}\right) R_{2} \longrightarrow R_{2} \) \( \left(-\frac{1}{2}\right) R_{3} \longrightarrow R_{3} \) Step \( 3: \) \( R_{3}+R_{2} \longrightarrow R_{3} \) Deduce that the solutions are:

Which of the following equations has no solution? \( \begin{array}{c}5 x+20-7 x=7(3+x)-3 x \\ x+20-5 x=7(3+x)-3 x \\ 5 x+21-x=7(3+x)-3 x \\ 5 x+20-x=7(3+x)-3 x\end{array} \)

Which of the following equations has no solution \( \begin{array}{l}x+20-5 x=7(3+x)-3 x \\ \begin{array}{l}5 x+20-7 x=7(3+x)-3 x \\ 5 x+21-x=7(3+x)-3 x \\ 5 x+20-x=7(3+x)-3 x\end{array} \\ \begin{array}{l}5\end{array}\end{array} \$ \begin{array}{l}\text { - }\end{array} \)

5) \( 4(2-x)=x+3 \)

a) \( \log _{2} x=1-x \);

Suppose the linear regression line \( y=3.009 x \) - 77.131 predicts a pizza parlor's profits based on the number of pizzas sold. If \( x \) represents the number of pizzas sold, and \( y \) represents the pizza partor's profits in dollars, about how much can the plizza parlor expect in profits if it sells 275 pizzas? A. \( \$ 750 \) B. \( \$ 900 \) C. \( \$ 825 \) D. \( \$ 675 \)