Question

So obviously the first function would be:

40t when 0 < t < and = to 2
But what is the second one?

Answer 1

Answer: 30(t - 2) + 80

Explanation:

`a) f(t) =\left \{ {{40t}\qquad \qquad 0\leq t\leq 2 \atop {30(t-2)+80}\quad 2< t\leq 6} \right.`

`b)\ \text{Plot the following coordinates}\\\begin {array}\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}`

c) (i) 40(1.5) = 60

(ii) 30(2.5 - 2) + 80 = 95

(iii) 30(3.5 - 2) + 80 = 125

d) 140 ÷ 4 = 35