the function is d= f(t)
when t1= 1 , d1= 5
when t2= 4, d2= 20
[tex]\text{rate od change = }\frac{d_2-d_1}{t_2-t_1}[/tex][tex]\text{rate of change = }\frac{20-5}{4-1}=\text{ }\frac{15}{3}=\text{ 5}[/tex][tex]\begin{gathered} ifd_{2\text{ }}=20,d_{3\text{ }}=35,t_{2\text{ }}=4,t_{3\text{ }}=7 \\ \text{rate of change = }\frac{d_3-d_2}{t_3-t_2}\text{ = }\frac{35-20}{7-4}=\text{ }\frac{15}{3}=\text{ 5} \end{gathered}[/tex]Thus if d is a function of t
and the rate of change is constant
then d = 5t is the function
