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:
a) 1
1
+
1
b)1
3
+
1
8
Step-by-step explanation:
Answer:
Look at the equations:
y
=
−
9
x
−
21
and
y
=
5
x
+
7
Because both equations equal (y), they both equal the same.
If an equations says "
y
=
−
9
x
−
21
" , that means we can replace (y) with
−
9
x
−
21
.
Therefore when i take the equation:
y
=
5
x
+
7
, we can replace (y) to get:
−
9
x
−
21
=
5
x
+
7
Now isolate (x):
−
9
x
−
21
=
5
x
+
7
⇔
−
21
=
14
x
+
7
⇔
−
28
=
14
x
⇔
x
=
−
2
Answer:
£24,980
Step-by-step explanation:
total pay = monthly salary + percent of annual profit + £250 bonus
monthly salary = £1,850
annual salary = 12 × £1,850 = £22,200
agency income = £356,000
agency costs = £280,000
annual profit = income - costs = £356,000 - £280,000 = £76,000
percent of annual profit = 3% × £76,000 = 0.03 × £76,000 = £2,280
£250 bonus = 2 × £250 = £500
total pay = monthly salary + percent of annual profit + £250 bonus
total pay = £22,200 + £2,280 + £500
total pay = £24,980
Answer:
The top right option.
Step-by-step explanation:
