Answer:
the least integer for n is 2
Step-by-step explanation:
We are given;
f(x) = ln(1+x)
centered at x=0
Pn(0.2)
Error < 0.01
We will use the format;
[[Max(f^(n+1) (c))]/(n + 1)!] × 0.2^(n+1) < 0.01
So;
f(x) = ln(1+x)
First derivative: f'(x) = 1/(x + 1) < 0! = 1
2nd derivative: f"(x) = -1/(x + 1)² < 1! = 1
3rd derivative: f"'(x) = 2/(x + 1)³ < 2! = 2
4th derivative: f""(x) = -6/(x + 1)⁴ < 3! = 6
This follows that;
Max|f^(n+1) (c)| < n!
Thus, error is;
(n!/(n + 1)!) × 0.2^(n + 1) < 0.01
This gives;
(1/(n + 1)) × 0.2^(n + 1) < 0.01
Let's try n = 1
(1/(1 + 1)) × 0.2^(1 + 1) = 0.02
This is greater than 0.01 and so it will not work.
Let's try n = 2
(1/(2 + 1)) × 0.2^(2 + 1) = 0.00267
This is less than 0.01.
So,the least integer for n is 2
G(x) = (x+4)^4 is the answer. It's D. last choice
G(x) = (x+4)^4 shifted F(x) = x^4 4 units to the left
To determine the amount of water in the first 8 beakers, you will subtract the 19 ml that are in the 9th beaker from the total to see how much water was actually put into the first 8 beakers.
91 ml - 19 ml = 72 ml.
The 72 ml are divided evenly between 8 beakers.
72/8 = 9 ml
There would be 9 ml of water in each of the first 8 beakers.
8/13 = 0.6153
y = cos-1(0.6153)
y = 52.0
I'm not totally sure if this is what you mean or if this is.
y = cos(8/13) = 0.9999 which to the nearest tenth is 1.0
