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:
°
Step-by-step explanation:
For the upper left angle, we know that they are supplementary, so we can find the first angle of the triangle by
For the bottom right angle, we know that they are supplementary, so we can find that angle by
As a triangle has 180 degrees, we can subtract the values of the other two angles to find the final upper right angle.
Lastly, The upper right angle and x are supplementary so we can find x by subtracting the angle from 180
Answer:
5.2 Minutes, or 5 minutes and 12 seconds
Step-by-step explanation:
You can solve this problem by working backward. Since our number gets smaller by a 1/10 or 0.1, in order to work backward we must multiply 3.60 by 10/9, since one tenth has been taken away already. When we do this, we get 4.0. When we add 1.2 minutes to that, we get a final answer of 5.2 minutes, or 5 minutes and 12 seconds.
Any decimal number that is repeating can be written in the form <span>
</span> with b not equal to zero, so they are rational numbers.
The short answer is
yes, 0.6 repeating is a rational number.
Hope this helped :)
Answer: b. segment TX = 16 m.
