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:
26%
Step-by-step explanation:
Given that Dan has 59 new email messages and 13 have attachment then the proportion that have attachments may be expressed as a ratio of the number with attachments to the total number of emails.
Hence proportion of the email messages have attachments as a percentage
= 13/50 * 100%
= 26%
This means that 26% of the emails received have attachments
Answer:
25 ft
Step-by-step explanation:
c = √a^2 + b^2 = √20^2 + 15^2 = 25ft
Hooke me up with a 5 star and a thanks :)
Let c represent the weight of cashews and p the weight of pecans.
Then c + 10 = total weight of the nut mixture.
An equation for the value of the mixture follows:
$1.50(10 lb) + $0.75c = (c+10)($1.00)
Solve this equation for c: 15 + .75c = c + 10. Subtract .75c from both sides:
15 = 1c - 0.75c + 10. Then 5=0.25c, and c = 5/0.25, or 20.
Need 20 lb of cashews.
Check: the pecans weigh 10 lb and are worth $1.50 per lb, so the total value of the pecans is $15. The total value of the cashews is (20 lb)($0.75/lb), or $15. Does (20 lb + 10 lb)($1/lb) = $15 + $15? Yes. So c= 20 lb is correct.
It represents the first one 3+(-1)