Question

rachel says that graph r has a greater constant of variation than graph s. which statement explains whether rachel is correct?
○ rachel is correct; graph s is nonproportional.
○ rachel is correct; graph r goes farther along the x - axis.
○ rachel is not correct; both graphs go through the origin and have the same constant of variation.
○ rachel is not correct; graph s is steeper than graph r.

Explanation:

Step1: Recall constant of variation

For proportional graphs (through origin), the constant of variation \( k \) is the slope \( \frac{y}{x} \).

Step2: Analyze Graph R

Take two points on Graph R, e.g., \((0,0)\) and \((4,2)\). Slope \( k_R=\frac{2 - 0}{4 - 0}=\frac{1}{2}\).

Step3: Analyze Graph S

Take two points on Graph S, e.g., \((0,0)\) and \((2,4)\). Slope \( k_S=\frac{4 - 0}{2 - 0}=2\).

Step4: Compare slopes and evaluate Rachel's claim

Graph S has a steeper slope (greater \( k \)) than Graph R. So Rachel is wrong. Check options:

• First option: Graph S is proportional (through origin), so wrong.
• Second option: X - axis length doesn't determine slope, wrong.
• Third option: Slopes are different, wrong.
• Fourth option: Graph S is steeper (greater \( k \)), so Rachel is not correct.

Answer:

Rachel is not correct; Graph S is steeper than Graph R.