Select all the lines that are perpendicular to the graph of \( 3 x+6 y=5 \). A. \( 3 x+6 y=10 \) B. \( y=2 x-1 \) C. \( y-4=2(x+7) \) D. \( x-2 y=-6 \) E \( y=3 x+5 \)

To determine the lines perpendicular to the graph of \( 3x + 6y = 5 \), we first find its slope. **Step 1. Convert \( 3x + 6y = 5 \) to slope-intercept form.** \[ 3x + 6y = 5 \quad \Rightarrow \quad 6y = 5 - 3x \quad \Rightarrow \quad y = -\frac{1}{2}x + \frac{5}{6} \] The slope of the given line is \( -\frac{1}{2} \). **Step 2. Find the slope of lines perpendicular to the given line.** For two lines to be perpendicular, the product of their slopes must equal \(-1\). If the slope of one line is \( m \), the slope of any line perpendicular to it is the negative reciprocal, i.e., \[ m_{\text{perp}} = -\frac{1}{m} \] Thus, for \( m = -\frac{1}{2} \): \[ m_{\text{perp}} = -\frac{1}{-\frac{1}{2}} = 2 \] Any line with slope \( 2 \) is perpendicular to the given line. **Step 3. Determine the slopes of the provided options.** - **Option A: \( 3x + 6y = 10 \)** Rewrite in slope-intercept form: \[ 6y = 10 - 3x \quad \Rightarrow \quad y = -\frac{1}{2}x + \frac{10}{6} \] Slope = \( -\frac{1}{2} \) (not \( 2 \)). - **Option B: \( y = 2x - 1 \)** This is already in slope-intercept form with slope \( 2 \), which is perpendicular. - **Option C: \( y - 4 = 2(x + 7) \)** Expand: \[ y - 4 = 2x + 14 \quad \Rightarrow \quad y = 2x + 18 \] Slope = \( 2 \), which is perpendicular. - **Option D: \( x - 2y = -6 \)** Solve for \( y \): \[ x - 2y = -6 \quad \Rightarrow \quad -2y = -6 - x \quad \Rightarrow \quad y = \frac{6 + x}{2} = \frac{1}{2}x + 3 \] Slope = \( \frac{1}{2} \) (not \( 2 \)). - **Option E: \( y = 3x + 5 \)** Slope = \( 3 \) (not \( 2 \)). **Conclusion** The lines with slope \( 2 \) are in **Option B** and **Option C**.