Answer:
Here we just want to find the Taylor series for f(x) = ln(x), centered at the value of a (which we do not know).
Remember that the general Taylor expansion is:
![f(x) = f(a) + f'(a)*(x - a) + \frac{1}{2!}*f''(a)(x -a)^2 + ...](https://tex.z-dn.net/?f=f%28x%29%20%3D%20f%28a%29%20%2B%20f%27%28a%29%2A%28x%20-%20a%29%20%2B%20%5Cfrac%7B1%7D%7B2%21%7D%2Af%27%27%28a%29%28x%20-a%29%5E2%20%2B%20...)
for our function we have:
f'(x) = 1/x
f''(x) = -1/x^2
f'''(x) = (1/2)*(1/x^3)
this is enough, now just let's write the series:
![f(x) = ln(a) + \frac{1}{a} *(x - a) - \frac{1}{2!} *\frac{1}{a^2} *(x - a)^2 + \frac{1}{3!} *\frac{1}{2*a^3} *(x - a)^3 + ....](https://tex.z-dn.net/?f=f%28x%29%20%3D%20ln%28a%29%20%2B%20%20%5Cfrac%7B1%7D%7Ba%7D%20%2A%28x%20-%20a%29%20-%20%5Cfrac%7B1%7D%7B2%21%7D%20%2A%5Cfrac%7B1%7D%7Ba%5E2%7D%20%2A%28x%20-%20a%29%5E2%20%2B%20%5Cfrac%7B1%7D%7B3%21%7D%20%2A%5Cfrac%7B1%7D%7B2%2Aa%5E3%7D%20%2A%28x%20-%20a%29%5E3%20%2B%20....)
This is the Taylor series to 3rd degree, you just need to change the value of a for the required value.
Answer:
3.3
Step-by-step explanation:
d = (10 - 4) + (8 - 3)
d = 6 + 5
d = square root of 11
d = 3.31662479 or 3.3
I'll be expecting that brainliest :)
U're goal is to get k alone.
So divide both sides by 9/10.
3/2 divided by 9/10 is 5/3
That would be -2 1/2 , 3 1/2
208 is the answer good luck