Answer:
correct answer is b
Step-by-step explanation:
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: 11 year
P(1) = 37,100
P(4) = 58,400
The linear equation (for x ≥ 1)
P(x) = 37,100 + a(x-1)
For x = 4
58,400 = 37,100 + a(4-1)
58,400 - 37,100 = 3a
21300 = 3a
a = 7100
So, the linear equation:
P(x) = 37100 + 7100*(x-1)
P(x) = 37100 + 7100x - 7,100
P(x) = 7100x + 30000
To find when the profit should reach 108100, we can substitute P(x) by 108100.
108100 = 7100x + 30000
108100 - 30000 = 7100x
78100 = 7100x
x = 78100/7100
x = 11
Answer: 11 year
500. But 1000 cannot go in 2, the result would be 0.002.
Answer:
it's 400
Step-by-step explanation:
just do it like this I guess 5 × 8 × 10
