Answer: um 86%
Step-by-step explanation:
Answer:
650=65% of 1000
Step-by-step explanation:
650-600=65 or 65%
Answer:
y = (x -5)² + 3.
Step-by-step explanation:
Given : parabola with a vertex at (5,3).
To find : Which equation has a graph that is a parabola.
Solution : We have given vertex at (5,3).
Vertex form of parabola : y = (x -h)² + k .
Where, (h ,k ) vertex .
Plug h = 5 , k= 3 in vertex form of parabola.
Equation :y = (x -5)² + 3.
Therefore, y = (x -5)² + 3.
It would be
(X^4+2X^2) -119(X^2-2X) then you would simplify
X^4+2X^2-119X^2-2X
So it would be
X^4+117X^2-2X=0
So basically X is undefined
Looks like your function is

Rewrite it as

Recall that for
, we have

If we replace
with
, we get

By the ratio test, the series converges if

Solving for
gives the interval of convergence,

We can confirm that the interval is open by checking for convergence at the endpoints; we'd find that the resulting series diverge.