Question

b overall, the state university scored higher than the private university. d the state university had more variation in scores than the private college. 10. the line represented by the equation 3x + 6y = 4 is perpendicular to the line represented by which of the following equations? a. -4x - 8y = -5 b. 3x + 2y = 4 c. 2x - y = -1 d. -3x + 6y = -4 11. on the time–distance graph, what is the meaning of a line that has a negative slope? a. the body is moving backward. b. the body is slowing down. c. the body is stationary. d. nothing. velocities, like distances, can’t be negative. 12. what is the equation for a line perpendicular to 6x - 7y = 8 and passing through the point (8, 1)? a. $y = -\frac{7}{6}x + \frac{31}{3}$ b. $y = \frac{6}{7}x - \frac{8}{7}$ c. -6x + 7y = -8 d. $y = \frac{7}{6}x - \frac{31}{3}$

Explanation:

Question 10

Step1: Find slope of \( 3x + 6y = 4 \)

Rewrite in slope - intercept form \( y=mx + b \) (where \( m \) is the slope).
\( 6y=-3x + 4\)
\( y=-\frac{3}{6}x+\frac{4}{6}=-\frac{1}{2}x+\frac{2}{3}\). So the slope \( m_1 =-\frac{1}{2}\).
For two lines to be perpendicular, the product of their slopes \( m_1\times m_2=- 1\). So \( m_2=\frac{-1}{m_1}=\frac{-1}{-\frac{1}{2}} = 2\).

Step2: Find slopes of each option

• Option A: \( - 4x-8y=-5\)

Rewrite as \( 8y=-4x + 5\)
\( y=-\frac{4}{8}x+\frac{5}{8}=-\frac{1}{2}x+\frac{5}{8}\). Slope \( m=- \frac{1}{2}\).

• Option B: \( 3x + 2y = 4\)

Rewrite as \( 2y=-3x + 4\)
\( y=-\frac{3}{2}x + 2\). Slope \( m =-\frac{3}{2}\).

• Option C: \( 2x-y=-1\)

Rewrite as \( y = 2x+1\). Slope \( m = 2\).

• Option D: \( - 3x+6y=-4\)

Rewrite as \( 6y=3x - 4\)
\( y=\frac{3}{6}x-\frac{4}{6}=\frac{1}{2}x-\frac{2}{3}\). Slope \( m=\frac{1}{2}\).

In a time - distance graph, the slope of the line is given by \( \text{slope}=\frac{\text{change in distance}}{\text{change in time}}\), which represents velocity. A negative slope means that as time increases, distance decreases. This implies that the body is moving backward (towards the starting point or in the opposite direction of the positive distance axis). Option B is incorrect because a negative slope is not related to slowing down (slowing down would be related to the magnitude of the slope changing). Option C is incorrect because a stationary body has a slope of 0 (horizontal line). Option D is incorrect because velocities can be negative (indicating direction).

Step1: Find slope of \(6x - 7y = 8\)

Rewrite in slope - intercept form: \(7y=6x - 8\)
\(y=\frac{6}{7}x-\frac{8}{7}\). The slope of this line \(m_1=\frac{6}{7}\).
For a line perpendicular to it, the slope \(m_2\) satisfies \(m_1\times m_2=-1\). So \(m_2=-\frac{7}{6}\).

Step2: Use point - slope form \(y - y_1=m(x - x_1)\) (point \((8,1)\))

The point - slope form is \(y - 1=-\frac{7}{6}(x - 8)\)
\(y-1=-\frac{7}{6}x+\frac{56}{6}\)
\(y=-\frac{7}{6}x+\frac{28}{3}+1\)
\(y=-\frac{7}{6}x+\frac{28 + 3}{3}=-\frac{7}{6}x+\frac{31}{3}\)

Answer:

C. \(2x - y=-1\)

Question 11