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Answer:
Step-by-step explanation:
[tex]r_1(t) = < t^2 , 8t-15,t^2> \\ \\ r_2(t) = < 11t-30, \ t^2 , \ 13t-40> \\ \\ at \ r_1 = r_2 \\ \\ t^2 = 11t - 30 , \ \ \ \ \ \ \ \ \ \ \ \ 8t - 15 = t^2 , \ \ \ \ \ \ \ \ \ t^2 = 13t -40 \\ \\ t^2 - 11t + 30\ \ \ \ \ \ \ \ \ \ \ \ t^2 -8t + 15 , \ \ \ \ \ \ \ \ \ \ \ \ t^2 -13t + 40 \\ \\ t^2 -8t-3t + 30=0 \ \ \ \ \ \ t^2 - 2t -4t + 15=0 \ \ \ \ \ \ t^2 - 1t - 13t + 40 =0[/tex]
[tex](t-6)(t-5) \ \ \ \ \ \ \ \ \ \ \ (t-5)(t-3) \ \ \ \ \ \ \ \ \ \ \ (t-5)(t-8)[/tex]
[tex]\mathbf{The \ common \ value \ of \ t \ = \ 5}[/tex]