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:
The slope is -1/2 or -0.5
Hope this helps!
Answer:
don't know how to do sorry for that but i can tell me how to.do
Step-by-step explanation:
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Answer:
3x-6
Step-by-step explanation:
Answer:
120
Step-by-step explanation:
