Answer:
Graph A?
Step-by-step explanation:
Just trying to help, but I don't really know. I would say A because its a HUGE jump, but B is a steady incline, but assuming that A has the little incline in the beginning, and then the jump, then back to the incline, probably A would be more profitable.
2.434 to 3 sigfigs = 2.43 .
Answer:
I. m = 2401
II. ((n+1) ∆ y)/n = 1/n[(n – y + 2)(n – y) + 1]
Step-by-step explanation:
I. Determination of m
x ∆ y = x² − 2xy + y²
2 ∆ − 5 = √m
2² − 2(2 × –5) + (–5)² = √m
4 – 2(–10) + 25 = √m
4 + 20 + 25 = √m
49 = √m
Take the square of both side
49² = m
2401 = m
m = 2401
II. Simplify ((n+1) ∆ y)/n
We'll begin by obtaining (n+1) ∆ y. This can be obtained as follow:
x ∆ y = x² − 2xy + y²
(n+1) ∆ y = (n+1)² – 2(n+1)y + y²
(n+1) ∆ y = n² + 2n + 1 – 2ny – 2y + y²
(n+1) ∆ y = n² + 2n – 2ny – 2y + y² + 1
(n+1) ∆ y = n² – 2ny + y² + 2n – 2y + 1
(n+1) ∆ y = n² – ny – ny + y² + 2n – 2y + 1
(n+1) ∆ y = n(n – y) – y(n – y) + 2(n – y) + 1
(n+1) ∆ y = (n – y + 2)(n – y) + 1
((n+1) ∆ y)/n = [(n – y + 2)(n – y) + 1] / n
((n+1) ∆ y)/n = 1/n[(n – y + 2)(n – y) + 1]
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:
2 in
Step-by-step explanation:
