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:
Using the table, the slope is 1/2
Answer:
3rd answer down
Step-by-step explanation:
ok so you take the median of all the #'s which turns out to be 126.5 and that is where your bar should point to in the box, then you should start with 105, and your last bit of the line should end at 150, the start of your box should start at the median of 105 and 117, and the end of your box should stop at the median of 145 and 150.
Answer:
A
Step-by-step explanation:
p^-8q^-20 * p^8q^-10
=q^-30
=1/q^30
Answer:

Step-by-step explanation:
Given

Required
Shorten
We have:

Rationalize

Expand



Take positive square roots
Take LCM

Collect like terms

