Answer:
i am pretty sure 90
Step-by-step explanation:
X=9x y=-2 and the rest is filled in
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
Formula is y = a(x-h)^2 + k
Where h is 1 and k is 1
f (x) = a(x-1)^2 + 1
-3 = a(0-1)^2 + 1
-3 = a(-1)^2 + 1
-3 = a(1) + 1
-3 - 1 = a
-4 = a
a = -4
A must be equal to -4
y = -4(x-1)^2 + 1
0 = -4(x-1)^2 + 1
4(x^2 - 2x + 1) - 1 = 0
4x^2 - 8x + 4 - 1 = 0
4x^2 - 8x + 3 = 0
4x^2 - 8x = -3
Divide fpr 4 each term of the equation....x^2 - 2x = -3/4
We must factor the perfect square ax^2 + bx + c which we don't have. We must follow the rule (b/2)^2 where b is -2....(-2/2)^2 =
(-1)^2 = 1 and we add up that to both sides
x^2 - 2x + 1 = -3/4 + 1
x^2 - 2x + 1 = 1/4
(x-1)^2 = 1/4
square root both sides x-1 = (+/-) 1/2
x1 = +1/2 + 1 = 3/2
x2 = -1/2 + 1 = 1/2
x-intercepts are 1/2 and 3/2, in form (3/2,0); (1/2,0)
Answer:
B = R - M
Step-by-step explanation:
1. R = M + B Subtract M
2. R - M = B
