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:
Options A-C-F
Step-by-step explanation:
we know that
<em>The circumference of a circle is equal to</em>
or 
where
D is the diameter and r is the radius
therefore
or 
The number pi is the ratio circumference - diameter or is the ratio circumference - 2 times radius
<em>The area of the circle is equal to</em>
therefore
The number pi is the ratio Area - radius squared
The Pythagorean theorem is a^2+b^2=c^2 where a and b are legs and c is the hypotenuse(longest side). you haven’t put a picture so I don’t know where 35 is a leg or the hypotenuse, but all you have to do is square the 2 shorter sides and add them and set it equal to the square of the longer side. You can do it!
Answer:
PQ = 3.58, and RQ = 10.4
Step-by-step explanation:
We are given the hypotenuse of the triangle, and an angle. Use sin and cos to solve.
Hypotenuse = 11,
Opposite side is PQ
Adjacent side is RQ
x = 19
Sin x = (opposite side)/(hypotenuse)
Cos x = (adjacent side)/(hypotenuse)
For PQ, this is the side opposite to the angle, so use sin,
Sin 19 = x/11
11(Sin 19) = x
3.58 = x (rounded to the nearest hundredth)
For RQ, this is the side adjacent to the angle, so use cos,
Cos 19 = x/11
11(Cos 19) = x
10.4 = x (rounded to the nearest hundredth)
