Answer: 30(t - 2) + 80
Step-by-step explanation:
[tex]a) f(t) =\left \{ {{40t}\qquad \qquad 0\leq t\leq 2 \atop {30(t-2)+80}\quad 2< t\leq 6} \right.[/tex]
[tex]b)\ \text{Plot the following coordinates}\\\begin {array}{c|l||c}\underline{\quad t\quad }&\underline{\quad 40t\qquad }&\underline{Coordinate}\\ 0&40(0)=0&(0,0)\\1&40(1)=40&(1,40)\\2&40(2)=80&(2, 80)\\ \\\underline{\quad t\quad }&\underline{\quad 30(t-2)+80\quad}&\underline{Coordinate}\\3&30(3-2)+80=110&(3,110)\\4&30(4-2)+80=140&(3,140)\\5&30(5-2)+80=170&(3,170)\\6&30(6-2)+80=200&(3,200)\\\end{array}[/tex]
c) (i) 40(1.5) = 60
(ii) 30(2.5 - 2) + 80 = 95
(iii) 30(3.5 - 2) + 80 = 125
d) 140 ÷ 4 = 35
