The path of a rocket ship is given by the parametric equations x(t)=3t and y(t)=4t^2+1, where t is the time in seconds after launch. where is the ship after 10 seconds? a.) (3,4) b.) (3,5) c.) (30,401) d.) (30,501)

For this case we have the following equations that represent the position of the rocket:
 x (t) = 3t
 y (t) = 4t ^ 2 + 1
 Substituting t = 10 in both equations we have:
 x (10) = 3 (10) = 30
 y (10) = 4 (10) ^ 2 + 1 = 400 + 1 = 401
 Therefore, the position of the rocket is:
 (x, y) = (30, 401)
 
 the ship after 10 seconds is at:
 C.) (30,401)

Dis the solution here

21/2 or 5/2