1. **Problem statement:** Given the sequence $$x[n] = (6 - n)[u[n] - u[n - 6]]$$ where $$u[n]$$ is the unit step function, we want to find and sketch the sequences: 1) $$y_1[n] = x[4 - n]$$ 2) $$y_2[n] = x[2n - 3]$$ 3) $$y_3[n] = x[8 - 3n]$$ 2. **Recall the original sequence:** $$x[n] = (6 - n)[u[n] - u[n - 6]]$$ means $$x[n] = 6 - n$$ for $$0 \leq n \leq 6$$ and zero otherwise. 3. **Step 1: Find $$y_1[n] = x[4 - n]$$** - Substitute $$k = 4 - n$$ into $$x[k]$$. - The sequence is nonzero only if $$0 \leq k \leq 6$$, so: $$0 \leq 4 - n \leq 6 \Rightarrow -2 \leq n \leq 4$$ - For these $$n$$ values: $$y_1[n] = 6 - (4 - n) = 6 - 4 + n = 2 + n$$ - Outside this range, $$y_1[n] = 0$$. 4. **Step 2: Find $$y_2[n] = x[2n - 3]$$** - Substitute $$k = 2n - 3$$. - Nonzero when $$0 \leq 2n - 3 \leq 6 \Rightarrow 3/2 \leq n \leq 9/2$$ - Since $$n$$ is integer, $$n = 2, 3, 4$$. - For these $$n$$: $$y_2[n] = 6 - (2n - 3) = 6 - 2n + 3 = 9 - 2n$$ - Elsewhere, $$y_2[n] = 0$$. 5. **Step 3: Find $$y_3[n] = x[8 - 3n]$$** - Substitute $$k = 8 - 3n$$. - Nonzero when $$0 \leq 8 - 3n \leq 6 \Rightarrow 2/3 \leq n \leq 8/3$$ - Integer $$n$$ values: $$n = 1, 2$$. - For these $$n$$: $$y_3[n] = 6 - (8 - 3n) = 6 - 8 + 3n = -2 + 3n$$ - Elsewhere, $$y_3[n] = 0$$. 6. **Summary of sequences:** - $$y_1[n] = 2 + n$$ for $$-2 \leq n \leq 4$$, else 0. - $$y_2[n] = 9 - 2n$$ for $$n = 2, 3, 4$$, else 0. - $$y_3[n] = -2 + 3n$$ for $$n = 1, 2$$, else 0. 7. **Sketch notes:** - $$y_1[n]$$ is a line increasing from 0 at $$n = -2$$ to 6 at $$n = 4$$. - $$y_2[n]$$ has points: $$y_2[2] = 5$$, $$y_2[3] = 3$$, $$y_2[4] = 1$$. - $$y_3[n]$$ has points: $$y_3[1] = 1$$, $$y_3[2] = 4$$. These are discrete sequences defined only at the specified integer points.