1. Solve the inequality $2(3 - x) \leq 4 - x$. Step 1: Expand the left side: $2 \times 3 - 2 \times x = 6 - 2x$. Step 2: Write the inequality: $6 - 2x \leq 4 - x$. Step 3: Add $2x$ to both sides: $6 \leq 4 - x + 2x$ which simplifies to $6 \leq 4 + x$. Step 4: Subtract 4 from both sides: $6 - 4 \leq x$ which is $2 \leq x$. Step 5: Rewrite as $x \geq 2$. Answer: $x \geq 2$. 2. Calculate the height of the sixth student given the average height of six students is 170 cm and the heights of five students are 165 cm, 172 cm, 168 cm, 175 cm, and 163 cm. Step 1: Calculate the total height of all six students: $6 \times 170 = 1020$ cm. Step 2: Sum the heights of the five students: $165 + 172 + 168 + 175 + 163 = 843$ cm. Step 3: Subtract the sum of five students from total height to find the sixth student's height: $1020 - 843 = 177$ cm. Answer: The height of the sixth student is 177 cm. 3. Find the value of $y$ in similar triangles $KMR$ and $DAR$. Step 1: Identify corresponding sides from similarity: - $DA$ corresponds to $KM$ - $DR$ corresponds to $KR$ - $AR$ corresponds to $MR$ Step 2: Given lengths: $DA = 16$, $DR = 14$, $DK = 11$, and $KR = y$. Step 3: Since $K$ lies on $DR$, $DK = 11$ means $KR = DR - DK = 14 - 11 = 3$. Step 4: Use similarity ratio between sides $DA$ and $KM$: $$\frac{KM}{DA} = \frac{KR}{DR}$$ Step 5: Substitute known values: $$\frac{KM}{16} = \frac{y}{14}$$ Step 6: Since $KM$ corresponds to $DA$, and $KR = y$, but from step 3 $KR = 3$, so $y = 3$. Answer: $y = 3$.