A.𝑓(𝑡)=2𝑡−𝑡3f(t)=2 t −t 3 B.𝑑(𝑡)=(1.1)𝑡d(t)=(1.1) t C.𝑏(𝑡)=𝑡4−3𝑡+9b(t)=t 4 −3t+9D.ℎ(𝑡)=5𝑡+𝑡5h(t)=5 t +t 5 E.𝑐(𝑡)=𝑡2−5𝑡c(t)= t 2 −5t​

A.𝑓(𝑡)=2𝑡−𝑡3f(t)=2 t −t 3 B.𝑑(𝑡)=(1.1)𝑡d(t)=(1.1) t C.𝑏(𝑡)=𝑡4−3𝑡+9b(t)=t 4 −3t+9D.ℎ(𝑡)=5𝑡+𝑡5h(t)=5 t +t 5 E.𝑐(𝑡)=𝑡2−5𝑡c(t)= t 2 −5t​

𝑔(𝑡)={2𝑡2+2𝑡−24𝑡−3𝑖𝑓 𝑡≠3𝑏𝑖𝑓 𝑡=3g(t)={ t−32t 2 +2t−24​ b​ if t≠3if t=3​ A.14B.3C.7D.None of theseE.0

Give a vector parametric equation for the line through the point (2,4,−5)(2,4,−5) that is parallel to the line ⟨−5−5𝑡,2+𝑡,5𝑡⟩

The domain of the vector function F––(t)=t−3−−−−√i–+ln(10−2t−−−−−−√)–j+1t−3√k––𝐹_(𝑡)=𝑡−3𝑖_+𝑙𝑛(10−2𝑡)_𝑗+1𝑡−3𝑘_ is given by the interval

A PARTICLE IS MOVING ALONG A STRAIGHT LINE. ITS POSITION X AT TIME T IS GIVEN BY X=2T^3-3T^+2 FIND THE DISTANCE TRAVELLED BY THE PARTICLE IN THE FIRST 5 SECONDS