An image is translated and reflected.
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.
One side of the rectangle is x=2, the other side is 2x-5
Add up all the four sides: (x+2) +(x+2)+(2x-5)+(2x-5)=54
6x-6=54
x=10
#3: suppose the first integer is x, then the second one is x+2
x(x+2)=255
x^2 +2x -255 =0
factor the quadratic equation: (x+17)(x-15)=0
x=-17, which is impossible, or x=15
so the two positive integers are 15 and 17
Answer:
1. 0=0
2. 0=17
Step-by-step explanation:
Answer:
0.22222222222
Step-by-step explanation:
To divide two by nine, you need to use your long division skills. Unfortunately, this interface doesn't lend itself well to showing you the actual long division problem, so try to follow the steps with me.
draw a division symbol (should look like half a rectangle). Put the 9 in front of the rectangle and a 2 inside it. Make the top of the rectangle long.
Note that the result of the division 2÷9 is not an exact value and is equivalent to recurring decimal 0.22... (which has 2 as the period).
