The mode is the number that occurs the most so if you look at the numbers, thee only one that occurs twice is 5 so 5 is the mode.
2 times as much.
(In the cereal aisle at Costco they always have 2 boxes a pack)
The mode is 0 since it is the most frequent.
Answer:
domain:2,-7,11,5
range:-8,6,-8,10
to explain more the domain would be your independent variable (y) and the range would be your dependent variable (x)
hope I helped
Answer:
Step-by-step explanation:
1.
To write the form of the partial fraction decomposition of the rational expression:
We have:
![\mathbf{\dfrac{8x-4}{x(x^2+1)^2}= \dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7B8x-4%7D%7Bx%28x%5E2%2B1%29%5E2%7D%3D%20%5Cdfrac%7BA%7D%7Bx%7D%2B%5Cdfrac%7BBx%2BC%7D%7Bx%5E2%2B1%7D%2B%5Cdfrac%7BDx%2BE%7D%7B%28x%5E2%2B1%29%5E2%7D%7D)
2.
Using partial fraction decomposition to find the definite integral of:
![\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}dx](https://tex.z-dn.net/?f=%5Cdfrac%7B2x%5E3-16x%5E2-39x%2B20%7D%7Bx%5E2-8x-20%7Ddx)
By using the long division method; we have:
![x^2-8x-20 | \dfrac{2x}{2x^3-16x^2-39x+20 }](https://tex.z-dn.net/?f=x%5E2-8x-20%20%7C%20%5Cdfrac%7B2x%7D%7B2x%5E3-16x%5E2-39x%2B20%20%7D)
![- 2x^3 -16x^2-40x](https://tex.z-dn.net/?f=-%202x%5E3%20-16x%5E2-40x)
<u> </u>
![x+ 20](https://tex.z-dn.net/?f=x%2B%2020)
So;
![\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x+\dfrac{x+20}{x^2-8x-20}](https://tex.z-dn.net/?f=%5Cdfrac%7B2x%5E3-16x%5E2-39x%2B20%7D%7Bx%5E2-8x-20%7D%3D%202x%2B%5Cdfrac%7Bx%2B20%7D%7Bx%5E2-8x-20%7D)
By using partial fraction decomposition:
![\dfrac{x+20}{(x-10)(x+2)}= \dfrac{A}{x-10}+\dfrac{B}{x+2}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%2B20%7D%7B%28x-10%29%28x%2B2%29%7D%3D%20%5Cdfrac%7BA%7D%7Bx-10%7D%2B%5Cdfrac%7BB%7D%7Bx%2B2%7D)
![= \dfrac{A(x+2)+B(x-10)}{(x-10)(x+2)}](https://tex.z-dn.net/?f=%3D%20%5Cdfrac%7BA%28x%2B2%29%2BB%28x-10%29%7D%7B%28x-10%29%28x%2B2%29%7D)
x + 20 = A(x + 2) + B(x - 10)
x + 20 = (A + B)x + (2A - 10B)
Now; we have to relate like terms on both sides; we have:
A + B = 1 ; 2A - 10 B = 20
By solvong the expressions above; we have:
![B = \dfrac{3}{2}](https://tex.z-dn.net/?f=B%20%3D%20%20%5Cdfrac%7B3%7D%7B2%7D)
Now;
![\dfrac{x+20}{(x-10)(x+2)} = \dfrac{5}{2(x-10)} + \dfrac{3}{2(x+2)}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%2B20%7D%7B%28x-10%29%28x%2B2%29%7D%20%3D%20%5Cdfrac%7B5%7D%7B2%28x-10%29%7D%20%2B%20%5Cdfrac%7B3%7D%7B2%28x%2B2%29%7D)
Thus;
![\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)}](https://tex.z-dn.net/?f=%5Cdfrac%7B2x%5E3-16x%5E2-39x%2B20%7D%7Bx%5E2-8x-20%7D%3D%202x%20%2B%20%5Cdfrac%7B5%7D%7B2%28x-10%29%7D%2B%20%5Cdfrac%7B3%7D%7B2%28x%2B2%29%7D)
Now; the integral is:
![\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = \int \begin {bmatrix} 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)} \end {bmatrix} \ dx](https://tex.z-dn.net/?f=%5Cint%20%5Cdfrac%7B2x%5E3-16x%5E2-39x%2B20%7D%7Bx%5E2-8x-20%7D%20%5C%20dx%20%3D%20%20%5Cint%20%5Cbegin%20%7Bbmatrix%7D%202x%20%2B%20%5Cdfrac%7B5%7D%7B2%28x-10%29%7D%2B%20%5Cdfrac%7B3%7D%7B2%28x%2B2%29%7D%20%5Cend%20%7Bbmatrix%7D%20%5C%20dx)
![\mathbf{\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = x^2 + \dfrac{5}{2}In | x-10|\dfrac{3}{2} In |x+2|+C}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cint%20%5Cdfrac%7B2x%5E3-16x%5E2-39x%2B20%7D%7Bx%5E2-8x-20%7D%20%5C%20dx%20%3D%20%20x%5E2%20%2B%20%5Cdfrac%7B5%7D%7B2%7DIn%20%7C%20x-10%7C%5Cdfrac%7B3%7D%7B2%7D%20In%20%7Cx%2B2%7C%2BC%7D)
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