(0,0) you could use x=-b/2a but since your b term is nonexistant the whole term goes to 0 for the x value. Plug x=0 into the original equation which gives you another 0. ezpz
Answer:
It is relatively easy to determine whether an equation is a function by solving for y. When you are given an equation and a specific value for x, there should only be one corresponding y-value for that x-value. For example, y = x + 1 is a function because y will always be one greater than x.

Used:

Functions have the same derivatives if they differ a constant.
h(x) = f(x); g(x) = f(x) + const.
h'(x) = f'(x)
g'(x) = (f(x) + const.)' = f'(x) + (const.)' = f'(x) + 0 = f'(x)
h'(x) = g'(x)
Therefore yuor answer is A. f'(x)=g'(x)

Used:

<span>the statement is ambiguous, it must be written as (p ∧ q) ∨ r
or p ∧ (q ∨ r).</span>