Answer:
They are doing OK some are getting answers and some are not like me. For some people don't even know what they're asking.
Because

therefore
f(x) = (x-3)(2x² + 10x - 1) + k, where k = constant.
Because f(3) = 4, therefore k =4.
The polynomial is
f(x) = 2x³ + 10x² - x - 6x² - 30x + 3 + 4
= 2x³ + 4x² - 31x + 7
Answer: f(x) = 2x³ + 4x² - 31x + 7
Answer:
Step-by-step explanation:
(x - 4)
What you are being asked to do is find the exact same binomial in the top as is in the bottom.
But there's a small catch. You must stipulate that x cannot equal 4. If it does, then you will get 0/0 which is undefined. You can't have that happening -- not at this level.
Any other value for x is fine.
Answer:
The value of the constant C is 0.01 .
Step-by-step explanation:
Given:
Suppose X, Y, and Z are random variables with the joint density function,

The value of constant C can be obtained as:



![C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0{e^{-0.2y}([\frac{-e^{-0.1z} }{0.1} ]\limits^\infty__0 }) \, dy }) \, dx = 1](https://tex.z-dn.net/?f=C%5Cint%5Climits%5E%5Cinfty_0%20%7Be%5E%7B-0.5x%7D%28%5Cint%5Climits%5E%5Cinfty_0%7Be%5E%7B-0.2y%7D%28%5B%5Cfrac%7B-e%5E%7B-0.1z%7D%20%7D%7B0.1%7D%20%5D%5Climits%5E%5Cinfty__0%20%7D%29%20%5C%2C%20dy%20%20%7D%29%20%5C%2C%20dx%20%3D%201)
![C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}([\frac{-e^{-0.1(\infty)} }{0.1}+\frac{e^{-0.1(0)} }{0.1} ]) } \, dy }) \, dx = 1](https://tex.z-dn.net/?f=C%5Cint%5Climits%5E%5Cinfty_0%20%7Be%5E%7B-0.5x%7D%28%5Cint%5Climits%5E%5Cinfty_0%20%7Be%5E%7B-0.2y%7D%28%5B%5Cfrac%7B-e%5E%7B-0.1%28%5Cinfty%29%7D%20%7D%7B0.1%7D%2B%5Cfrac%7Be%5E%7B-0.1%280%29%7D%20%7D%7B0.1%7D%20%5D%29%20%20%7D%20%5C%2C%20dy%20%20%7D%29%20%5C%2C%20dx%20%3D%201)
![C\int\limits^\infty_0 {e^{-0.5x}(\int\limits^\infty_0 {e^{-0.2y}[0+\frac{1}{0.1}] } \, dy }) \, dx =1](https://tex.z-dn.net/?f=C%5Cint%5Climits%5E%5Cinfty_0%20%7Be%5E%7B-0.5x%7D%28%5Cint%5Climits%5E%5Cinfty_0%20%7Be%5E%7B-0.2y%7D%5B0%2B%5Cfrac%7B1%7D%7B0.1%7D%5D%20%20%7D%20%5C%2C%20dy%20%20%7D%29%20%5C%2C%20dx%20%3D1)
![10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2y} }{0.2}]^\infty__0 }) \, dx = 1](https://tex.z-dn.net/?f=10C%5Cint%5Climits%5E%5Cinfty_0%20%7Be%5E%7B-0.5x%7D%28%5B%5Cfrac%7B-e%5E%7B-0.2y%7D%20%7D%7B0.2%7D%5D%5E%5Cinfty__0%20%20%7D%29%20%5C%2C%20dx%20%3D%201)
![10C\int\limits^\infty_0 {e^{-0.5x}([\frac{-e^{-0.2(\infty)} }{0.2}+\frac{e^{-0.2(0)} }{0.2}] } \, dx = 1](https://tex.z-dn.net/?f=10C%5Cint%5Climits%5E%5Cinfty_0%20%7Be%5E%7B-0.5x%7D%28%5B%5Cfrac%7B-e%5E%7B-0.2%28%5Cinfty%29%7D%20%7D%7B0.2%7D%2B%5Cfrac%7Be%5E%7B-0.2%280%29%7D%20%7D%7B0.2%7D%5D%20%20%20%7D%20%5C%2C%20dx%20%3D%201)
![10C\int\limits^\infty_0 {e^{-0.5x}[0+\frac{1}{0.2}] } \, dx = 1](https://tex.z-dn.net/?f=10C%5Cint%5Climits%5E%5Cinfty_0%20%7Be%5E%7B-0.5x%7D%5B0%2B%5Cfrac%7B1%7D%7B0.2%7D%5D%20%20%7D%20%5C%2C%20dx%20%3D%201)
![50C([\frac{-e^{-0.5x} }{0.5}]^\infty__0}) = 1](https://tex.z-dn.net/?f=50C%28%5B%5Cfrac%7B-e%5E%7B-0.5x%7D%20%7D%7B0.5%7D%5D%5E%5Cinfty__0%7D%29%20%3D%201)
![50C[\frac{-e^{-0.5(\infty)} }{0.5} + \frac{-0.5(0)}{0.5}] =1](https://tex.z-dn.net/?f=50C%5B%5Cfrac%7B-e%5E%7B-0.5%28%5Cinfty%29%7D%20%7D%7B0.5%7D%20%2B%20%5Cfrac%7B-0.5%280%29%7D%7B0.5%7D%5D%20%3D1)
![50C[0+\frac{1}{0.5} ] =1](https://tex.z-dn.net/?f=50C%5B0%2B%5Cfrac%7B1%7D%7B0.5%7D%20%5D%20%3D1)
⇒ 
C = 0.01
Answer:
1912
Step-by-step explanation: