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
The answer is 21.52 so I hope that helped
Answer: Senior: $8, Child: 14
Step-by-step explanation:
3s + c = 38
3s + 2c = 52
c = 14
3s + 14 = 38
3s = 24
s = 8
use s=r0
but first convert 360 degree to radian
you will get 6.284 radian
then you substitute and get the answer!
12= r (6.284)
r= 1.91 cm
Answer:
A=52
B=48
C=45
D=58
Step-by-step explanation:
To find the variables just do 60 minus the number they give. For example, to find A, you do 60-8.
Hope this helps :)
