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
I think it is the second one but I’m not for sure I’m really just doing this for points
Answer:
7.5 ab and cd
4 bd and ac
Step-by-step explanation:
Since the quotient of the two decimal numbers is 1, that tells use that they are the same number.
Since the sum of the two numbers equals 1 and both numbers are the same, then the answer must be 1 divided by 2. or .5
the fractions would be 1/2 and 1/2
Answer:
315
Step-by-step explanation:
Using a scale factor of 3 basically means just multiply the length and the width by 3.
So instead of 5 * 7, we are doing 15 * 21.
15 * 21 = 315
