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:
9 boxes
Step-by-step explanation:
Boxes of Cereal 3 for the price of 2, the the price of the 3 boxes of cereals must be : 2 × £2.89 = £5.78
Boxes Megan can buy for the offer = £20 ÷ £5.78 = 3.46 (to 3 s.f.)≈ 3
3 × 3 =9 packs
But they actually only cost : 3 × £5.78 = £17.34
Therefore after buying the offer Megan has £20 - £17.34 = £ 2.66 left, which is not enough to buy a single pack, so the answer is 9.
Answer:
30 40 percent
Step-by-step explanation:
it is less that 50
Answer:
18
Step-by-step explanation:
2(6) + 2(3)
12 + 6 = 18
