Answer:
Step-by-step explanation:
y = sin(t^2)
y' = 2tcos(t^2)
y'' = 2cos(t^2) - 4t^2sin(t^2)
so the equation become
2cos(t^2) - 4t^2sin(t^2) + p(t)(2tcos(t^2)) + q(t)sin(t^2) = 0
when t=0, above eqution is 2. That is, there does not exist the solution. so y can not be a solution on I containing t=0.
Answer:
Step-by-step explanation:
Answer:
for the question all the way on the bottom, for the pizza everyone paid 2.59$ each
Answer:
answer is 5.3
Step-by-step explanation:
just easy
