-2.
It is in y=mx+b form. And 'b' is the y-inter. -2 is in the 'b' place, therefore, -2 is the y-inter
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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:g(f(x))= x^2 +6x+5
Step-by-step explanation:
g(f(x))= x^2 -4
g(f(x))= (x+3)^2 -4
g(f(x))= (x+3)(x+3) -4
g(f(x))= x^2 +6x+9-4
g(f(x))= x^2 +6x+5
I'm pretty sure the answer is 259