Answer:
its (5/2)x
Step-by-step explanation:
(hint: y=mx+b, "m" is the slope)
Hope this helps
Answer:
D.
Step-by-step explanation:
adding 8 cancels it out on side with x, the left side
Taylor series is ![f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }](https://tex.z-dn.net/?f=f%28x%29%20%3D%20ln2%20%2B%20%5Csum_%7Bn%3D1%29%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn-1%7D%20%5Cfrac%7B%28n-1%29%21%7D%7Bn%21%289%29%5E%7Bn%7D%28x9%29%5E%7B2%7D%20%20%7D)
To find the Taylor series for f(x) = ln(x) centering at 9, we need to observe the pattern for the first four derivatives of f(x). From there, we can create a general equation for f(n). Starting with f(x), we have
f(x) = ln(x)
![f^{1}(x)= \frac{1}{x} \\f^{2}(x)= -\frac{1}{x^{2} }\\f^{3}(x)= -\frac{2}{x^{3} }\\f^{4}(x)= \frac{-6}{x^{4} }](https://tex.z-dn.net/?f=f%5E%7B1%7D%28x%29%3D%20%5Cfrac%7B1%7D%7Bx%7D%20%5C%5Cf%5E%7B2%7D%28x%29%3D%20-%5Cfrac%7B1%7D%7Bx%5E%7B2%7D%20%7D%5C%5Cf%5E%7B3%7D%28x%29%3D%20-%5Cfrac%7B2%7D%7Bx%5E%7B3%7D%20%7D%5C%5Cf%5E%7B4%7D%28x%29%3D%20%5Cfrac%7B-6%7D%7Bx%5E%7B4%7D%20%7D)
.
.
.
Since we need to have it centered at 9, we must take the value of f(9), and so on.
f(9) = ln(9)
![f^{1}(9)= \frac{1}{9} \\f^{2}(9)= -\frac{1}{9^{2} }\\f^{3}(x)= -\frac{1(2)}{9^{3} }\\f^{4}(x)= \frac{-1(2)(3)}{9^{4} }](https://tex.z-dn.net/?f=f%5E%7B1%7D%289%29%3D%20%5Cfrac%7B1%7D%7B9%7D%20%5C%5Cf%5E%7B2%7D%289%29%3D%20-%5Cfrac%7B1%7D%7B9%5E%7B2%7D%20%7D%5C%5Cf%5E%7B3%7D%28x%29%3D%20-%5Cfrac%7B1%282%29%7D%7B9%5E%7B3%7D%20%7D%5C%5Cf%5E%7B4%7D%28x%29%3D%20%5Cfrac%7B-1%282%29%283%29%7D%7B9%5E%7B4%7D%20%7D)
.
.
.
Following the pattern, we can see that for
,
This applies for n ≥ 1, Expressing f(x) in summation, we have
![\sum_{n=0}^{\infinite} \frac{f^{n}(9) }{n!} (x-9)^{2}](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfinite%7D%20%5Cfrac%7Bf%5E%7Bn%7D%289%29%20%7D%7Bn%21%7D%20%28x-9%29%5E%7B2%7D)
Combining ln2 with the rest of series, we have
![f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }](https://tex.z-dn.net/?f=f%28x%29%20%3D%20ln2%20%2B%20%5Csum_%7Bn%3D1%29%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn-1%7D%20%5Cfrac%7B%28n-1%29%21%7D%7Bn%21%289%29%5E%7Bn%7D%28x9%29%5E%7B2%7D%20%20%7D)
Taylor series is ![f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2} }](https://tex.z-dn.net/?f=f%28x%29%20%3D%20ln2%20%2B%20%5Csum_%7Bn%3D1%29%5E%7B%5Cinfty%7D%28-1%29%5E%7Bn-1%7D%20%5Cfrac%7B%28n-1%29%21%7D%7Bn%21%289%29%5E%7Bn%7D%28x9%29%5E%7B2%7D%20%20%7D)
Find out more information about taylor series here
brainly.com/question/13057266
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Answer:
First choice
Step-by-step explanation:
<u>Given</u>:
If you are dealt 4 cards from a shuffled deck of 52 cards.
We need to determine the probability of getting two queens and two kings.
<u>Probability of getting two queens and two kings:</u>
The number of ways of getting two queens is ![4C_2](https://tex.z-dn.net/?f=4C_2)
The number of ways of getting two kings is ![4C_2](https://tex.z-dn.net/?f=4C_2)
Total number of cases is ![52C_4](https://tex.z-dn.net/?f=52C_4)
The probability of getting two queens and two kings is given by
![\text {probability}=\frac{\text {No.of fanourable cases}}{\text {Total no.of cases}}](https://tex.z-dn.net/?f=%5Ctext%20%7Bprobability%7D%3D%5Cfrac%7B%5Ctext%20%7BNo.of%20fanourable%20cases%7D%7D%7B%5Ctext%20%7BTotal%20no.of%20cases%7D%7D)
Substituting the values, we get;
![probability=\frac{4C_2 \cdot 4C_2}{52C_4}](https://tex.z-dn.net/?f=probability%3D%5Cfrac%7B4C_2%20%5Ccdot%204C_2%7D%7B52C_4%7D)
Simplifying, we get;
![probability=\frac{6 (6)}{270725}](https://tex.z-dn.net/?f=probability%3D%5Cfrac%7B6%20%286%29%7D%7B270725%7D)
![probability=\frac{36}{270725}](https://tex.z-dn.net/?f=probability%3D%5Cfrac%7B36%7D%7B270725%7D)
![probability=0.000133](https://tex.z-dn.net/?f=probability%3D0.000133)
Thus, the probability of getting two queens and two kings is 0.000133