Answer:

Step-by-step explanation:
We are given that


y(0)=0
y'(0)=1
By comparing with

We get


q(x)=0
p(x),q(x) and g(x) are continuous for all real values of x except 3.
Interval on which p(x),q(x) and g(x) are continuous
and (3,
By unique existence theorem
Largest interval which contains 0=
Hence, the larges interval on which includes x=0 for which given initial value problem has unique solution=
4y + 3 ≤ y + 6
4y + 3 - 3 ≤ y + 6 - 3
4y ≤ y + 3
4y - y ≤ y - y + 3
3y ≤ 3
3y/3 ≤ 3/3
y ≤ 1
So any value of y less than or equal to 1 (so 1 is included in the solution set) satisfies the inequality. C is the correct answer.
Answer: rational number
Step-by-step explanation:
Mixed numbers are all rational numbers because they can be expressed as a fraction. To be a rational number, a number must be able to be written
No it is not quadratic. because when you graph it it is not a parabola