Step-by-step explanation:
The Taylor series expansion is:
Tₙ(x) = ∑ f⁽ⁿ⁾(a) (x − a)ⁿ / n!
f(x) = 1/x, a = 4, and n = 3.
First, find the derivatives.
f⁽⁰⁾(4) = 1/4
f⁽¹⁾(4) = -1/(4)² = -1/16
f⁽²⁾(4) = 2/(4)³ = 1/32
f⁽³⁾(4) = -6/(4)⁴ = -3/128
Therefore:
T₃(x) = 1/4 (x − 4)⁰ / 0! − 1/16 (x − 4)¹ / 1! + 1/32 (x − 4)² / 2! − 3/128 (x − 4)³ / 3!
T₃(x) = 1/4 − 1/16 (x − 4) + 1/64 (x − 4)² − 1/256 (x − 4)³
f(x) = 1/x has a vertical asymptote at x=0 and a horizontal asymptote at y=0. So we can eliminate the top left option. That leaves the other three options, where f(x) is the blue line.
Now we have to determine which green line is T₃(x). The simplest way is to notice that f(x) and T₃(x) intersect at x=4 (which makes sense, since T₃(x) is the Taylor series centered at x=4).
The bottom right graph is the only correct option.
I can simplify the formula with the data you gave me to do a simple formula like this: <span>4.25 k + 2 p = 12 I think this one explains in an easy way this. Hope it's good for you</span>
The first (and most typical) way to find distance of two points is by using the distance formula.
One alternative is the Manhattan metric, also called the taxicab metric. This option is much more complicated, and rarely used in high school math. d(x,y)=∑i|xi-yi|
<h3>Answer:</h3>
room 3 or 4
<h3>Explanation:</h3>
We assume the inequalities tell the rooms that were checked and found empty.
The solution to 3) is ...
... 2x + 3 > 11
... 2x > 8 . . . . . subtract 3
... x > 4 . . . . . . .divide by 2
The solution to 4) is ...
... -3x > -9
... x < 3 . . . . . . divide by -3
Thus, rooms greater than 4 and less than 3 were found empty. Rooms 3 and 4 were not checked, so either one could hold Nessie.
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The other inequalities have solutions that are already covered by the solutions to these.
1) x < -1 . . . . after subtacting 4
2) x > 5 . . . . after multiplying by 5
