The zeroes of the equation are (1,4)
Step-by-step explanation:
Given equation is:
We have to put y=0 in the equation to find zeroes of the equation
Hence,
The zeroes of the equation are (1,4)
Keywords: Equation, zeroes of an equation
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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
Polynomial equation solverStandard form:s2 + 5s + 4 = 0Factorization:(s + 4)(s + 1) = 0Solutions based on factorization:<span><span>s + 4 = 0 ⇒ <span>s1</span> = −4</span><span>s + 1 = 0 ⇒ <span>s2</span> = −1</span></span>Extrema:Min = (−2.5, −2.25)<span><span>-4-2-4-22</span><span>x:-4y:0</span></span>
Im pretty sure that the answer is B.
