Answer:
-1, 2, 6
Step-by-step explanation:
We have to solve the equation as follows: 1/(x-6) + (x/(x-2)) = (4/(x²-8x+12)).
Now, we have, ![\frac{1}{x-6} +\frac{x}{x-2} = \frac{4}{x^{2}-8x+12 }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx-6%7D%20%2B%5Cfrac%7Bx%7D%7Bx-2%7D%20%3D%20%5Cfrac%7B4%7D%7Bx%5E%7B2%7D-8x%2B12%20%7D)
⇒![\frac{(x-2)+x(x-6)}{(x-2)(x-6)} = \frac{4}{x^{2}-8x+12 }](https://tex.z-dn.net/?f=%5Cfrac%7B%28x-2%29%2Bx%28x-6%29%7D%7B%28x-2%29%28x-6%29%7D%20%3D%20%5Cfrac%7B4%7D%7Bx%5E%7B2%7D-8x%2B12%20%7D)
⇒![\frac{x-2+x^{2}-6x }{(x-2)(x-6)} =\frac{4}{(x-2)(x-6)}](https://tex.z-dn.net/?f=%5Cfrac%7Bx-2%2Bx%5E%7B2%7D-6x%20%7D%7B%28x-2%29%28x-6%29%7D%20%3D%5Cfrac%7B4%7D%7B%28x-2%29%28x-6%29%7D)
⇒![\frac{(x-2)(x-6)}{x^{2}-5x-2 }=\frac{(x-2)(x-6)}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B%28x-2%29%28x-6%29%7D%7Bx%5E%7B2%7D-5x-2%20%7D%3D%5Cfrac%7B%28x-2%29%28x-6%29%7D%7B4%7D)
⇒![(x-2)(x-6)[\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0](https://tex.z-dn.net/?f=%28x-2%29%28x-6%29%5B%5Cfrac%7B1%7D%7Bx%5E%7B2%7D%20-5x-2%7D%20-%5Cfrac%7B1%7D%7B4%7D%20%5D%3D0)
⇒
or, ![[\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1%7D%7Bx%5E%7B2%7D%20-5x-2%7D%20-%5Cfrac%7B1%7D%7B4%7D%20%5D%3D0)
If, (x-2)(x-6) =0, then x=2 or x=6
If,
, then ![x^{2} -5x-2=4](https://tex.z-dn.net/?f=x%5E%7B2%7D%20-5x-2%3D4)
and (x-6)(x+1) =0
Therefore, x=6 or -1
So the solutions for x are -1, 2 6. (Answer)
If an is a5
and
an-1 is a4
then using your recursive formula for an arithmetic sequence
an=an-1 +d
then
a5=a4+d
now, a4 =6 and common difference "d" is d=-11
hence
a5=6 -11
6-11= -5
any questions?
Answer: there are three of the holidays
Step-by-step explanation:
The prime numbers through 31 include 2,3,5,7,11,13,17,19,23,29,31,and 37 so therefore three of the holidays (Martin Luther, Easter, and Thanksgiving) fall on days that are prime numbers
Choice a.....................