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
192.
2(8+4)÷2(8+8) = 1(4+8)×(8+8)
1×12×16=192.
Answer:
2
Step-by-step explanation:
Opposite angle of a quadrilateral add up to 180 degrees.
This means Angle A plus Angle C equal 180.
We can solve for X using that, then solve for Angle B.
2x-7 + x +4 = 180
Simplify:
3x -3 = 180
Add 3 to each side:
3x = 183
Divide both sides by 3:
x = 183 /3
x = 61
Now we know x, replace x with 61 in the equation for Angle B:
Angle B = 2x+3 = 2(61) +3 = 122 +3 = 125 degrees.
