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 which lines are perpendicular to the graph of \(3x + 6y = 5\), we first need to find the slope of this line. By rearranging into slope-intercept form \(y = mx + b\): 1. Start with the given equation: \[ 3x + 6y = 5 \] 2. Solve for \(y\): \[ 6y = -3x + 5 \] \[ y = -\frac{1}{2}x + \frac{5}{6} \] The slope \(m\) of this line is \(-\frac{1}{2}\). Lines that are perpendicular will have slopes that are negative reciprocals. Thus, we need to find lines with a slope of \(2\) (the negative reciprocal of \(-\frac{1}{2}\)). Now, let's analyze each line: A. \(3x + 6y = 10\): Rearranging gives the same slope as A, which is \(-\frac{1}{2}\). Not perpendicular. B. \(y = 2x - 1\): The slope here is \(2\). This is perpendicular. C. \(y - 4 = 2(x + 7)\): Distributing gives \(y = 2x + 14 + 4\), so the slope is \(2\). This is perpendicular. D. \(x - 2y = -6\): Rearranging gives \(2y = x + 6\) or \(y = \frac{1}{2}x + 3\), so the slope is \(\frac{1}{2}\). Not perpendicular. E. \(y = 3x + 5\): The slope is \(3\). Not perpendicular. In summary, the lines that are perpendicular to \(3x + 6y = 5\) are: **B. \(y = 2x - 1\)** **C. \(y - 4 = 2(x + 7)\)**