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
Answer:
Juanita dissolves 46 g of MgBr2 (molar mass: 184.11 g/mol) in 0.5 kg of distilled water. What is the molality of the solution?
Solution to the system of equation (-2,5)
Answer:
1. 4 and 4sqrt2
2. 6sqrt2 and 6sqrt2
3. 5 and 5sqrt2
4. 5sqrt2 and 5sqrt2
5. 6 and 6
6. 8 and 8sqrt2
7. sqrt3 and sqrt6
8. sqrt2 and sqrt4
Step-by-step explanation:
A. Is the correct answer hope this helped support me by thanking me and give brainliest
