Answer:
ok I can. what is the question?
Answer:
I think it's B.
Step-by-step explanation:
Sorry if I'm wrong :/
<h3>
Answer: n = -11</h3>
=========================================================
Explanation:
Since x-2 is a factor of f(x), this means f(2) = 0.
More generally, if x-k is a factor of p(x), then p(k) = 0. This is a special case of the remainder theorem.
So if we plugged x = 2 into f(x), we'd get
f(x) = x^3+x^2+nx+10
f(2) = 2^3+2^2+n(2)+10
f(2) = 8+4+2n+10
f(2) = 2n+22
Set this equal to 0 and solve for n
2n+22 = 0
2n = -22
n = -22/2
n = -11 is the answer
Therefore, x-2 is a factor of f(x) = x^3+x^2-11x+10
Plug x = 2 into that updated f(x) function to find....
f(x) = x^3+x^2-11x+10
f(2) = 2^3+2^2-11(2)+10
f(2) = 8+4-22+10
f(2) = 0
Which confirms our answer.
5/3 is greater improper fraction comes out to 1 2/3
Answer:
a) 
b) 
Step-by-step explanation:
Given integrals are:
---(1)
---- (2)
Standard form
<h3>Part A</h3>
compare (1) with standard form
![[f'(x)]^{2} = 64x^{8}\\f'(x)=\pm 8x^{4}\\f(x)= \pm8\frac{x^{5}}{5}+C](https://tex.z-dn.net/?f=%5Bf%27%28x%29%5D%5E%7B2%7D%20%3D%2064x%5E%7B8%7D%5C%5Cf%27%28x%29%3D%5Cpm%208x%5E%7B4%7D%5C%5Cf%28x%29%3D%20%5Cpm8%5Cfrac%7Bx%5E%7B5%7D%7D%7B5%7D%2BC)
<h3>Part B</h3>
Compare (2) with standard form
![[f'(x)]^{2}=64cos^{2}(2x)\\f'(x)= \pm 8cos(2x)\\f(x)=\pm 8\frac{sin(2x)}{2}+C\\f(x)= \pm 4sin(2x)+C](https://tex.z-dn.net/?f=%5Bf%27%28x%29%5D%5E%7B2%7D%3D64cos%5E%7B2%7D%282x%29%5C%5Cf%27%28x%29%3D%20%5Cpm%208cos%282x%29%5C%5Cf%28x%29%3D%5Cpm%208%5Cfrac%7Bsin%282x%29%7D%7B2%7D%2BC%5C%5Cf%28x%29%3D%20%5Cpm%204sin%282x%29%2BC)
