QUESTION IMAGE
Question
- circle “linear” or “nonlinear.” if linear, state the rate of change.
rate of change:
linear nonlinear
4 0 -2 -3 x
20 12 8 6 y
rate of change:
linear nonlinear
2 1 0 -1 x
18 11 5 0 y
rate of change:
linear nonlinear
3 2 1 0 x
3 -2 -7 -12 y
Step1: Recall the definition of linear relationship
A relationship is linear if the rate of change (slope) between any two - points is constant. The formula for the rate of change (slope) between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$.
Step2: Analyze the first table
For the first table, let's take the first two non - x,y points $(4,20)$ and $(0,12)$. The rate of change $m_1=\frac{12 - 20}{0 - 4}=\frac{-8}{-4}=2$. Now take the second and third non - x,y points $(0,12)$ and $(-2,8)$. The rate of change $m_2=\frac{8 - 12}{-2-0}=\frac{-4}{-2}=2$. Take the third and fourth non - x,y points $(-2,8)$ and $(-3,6)$. The rate of change $m_3=\frac{6 - 8}{-3+2}=\frac{-2}{-1}=2$. Since the rate of change is constant, it is linear. The rate of change is 2.
Step3: Analyze the second table
For the second table, take the first two non - x,y points $(2,18)$ and $(1,11)$. The rate of change $m_1=\frac{11 - 18}{1 - 2}=\frac{-7}{-1}=7$. Take the second and third non - x,y points $(1,11)$ and $(0,5)$. The rate of change $m_2=\frac{5 - 11}{0 - 1}=\frac{-6}{-1}=6$. Since the rate of change is not constant, it is non - linear.
Step4: Analyze the third table
For the third table, take the first two non - x,y points $(3,3)$ and $(2, - 2)$. The rate of change $m_1=\frac{-2 - 3}{2 - 3}=\frac{-5}{-1}=5$. Take the second and third non - x,y points $(2,-2)$ and $(1,-7)$. The rate of change $m_2=\frac{-7 + 2}{1 - 2}=\frac{-5}{-1}=5$. Take the third and fourth non - x,y points $(1,-7)$ and $(0,-12)$. The rate of change $m_3=\frac{-12 + 7}{0 - 1}=\frac{-5}{-1}=5$. Since the rate of change is constant, it is linear. The rate of change is 5.
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First table: Linear, Rate of change: 2
Second table: Nonlinear
Third table: Linear, Rate of change: 5