Answer:

Step-by-step explanation:
A second order linear , homogeneous ordinary differential equation has form
.
Given: 
Let
be it's solution.
We get,

Since
, 
{ we know that for equation
, roots are of form
}
We get,

For two complex roots
, the general solution is of form 
i.e 
Applying conditions y(0)=1 on
, 
So, equation becomes 
On differentiating with respect to t, we get

Applying condition: y'(0)=0, we get 
Therefore,

Answer:
(1,2)
Step-by-step explanation:
thats the point of intersection so they have the same outcome
I'm a little confused by the question but I hope this helps :)
Answer:
the 2nd on (B)
Step-by-step explanation:
The x does not repeat
Answer: y= 0.074x + 50.48
Step-by-step explanation: