We can use the first tio ordered pairs (3,2) and (6,4) to find the rate of change:
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ (x_1,y_1)=(3,2) \\ (x_2,y_2)=(6,4) \\ \Rightarrow m=\frac{4-2}{6-3}=\frac{2}{3} \\ m=\frac{2}{3} \end{gathered}[/tex]
then, using the first ordered pair, we can find the relation function:
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ \Rightarrow y-2=\frac{2}{3}(x-3)=\frac{2}{3}x-2 \\ \Rightarrow y=\frac{2}{3}x-2+2=\frac{2}{3}x \\ y=\frac{2}{3}x \end{gathered}[/tex]
thus, the equation that describes the relationship is y = 2/3 x
Next, to find the distance of the zebra after 48 minutes, we have to make x = 48 and solve for y:
[tex]\begin{gathered} x=48 \\ \Rightarrow y=\frac{2}{3}\cdot(48)=\frac{96}{3}=32 \\ y=32 \end{gathered}[/tex]
therefore, the zebra would travel 32 miles in 48 minutes
