4.5{4_x}+36=202-2.5{3+28}
you first multiply 4.5 and the numbers in the bracket.
4.5*4=18
4.5*x=4.5x
you will obtain 18-4.5x=202_2.5{3+28}
2.5*3=7.5 2.5*28=70
18_4.5x=202_7.5_70
18_4.5x=124.5_18
-4.5x=106.5
x=23.67
![\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B5x%5E4-7x%5E3-12x%5E2%2B6x%2B21%7D%7B%28x-3%29%28x%5E2-2%29%5E2%7D%3D%5Cdfrac%7Ba_1%7D%7Bx-3%7D%2B%5Cdfrac%7Ba_2x%2Ba_3%7D%7Bx%5E2-2%7D%2B%5Cdfrac%7Ba_4x%2Ba_5%7D%7B%28x%5E2-2%29%5E2%7D)
![\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)](https://tex.z-dn.net/?f=%5Cimplies%205x%5E4-7x%5E3-12x%5E2%2B6x%2B21%3Da_1%28x%5E2-2%29%5E2%2B%28a_2x%2Ba_3%29%28x-3%29%28x%5E2-2%29%2B%28a_4x%2Ba_5%29%28x-3%29)
When
![x=3](https://tex.z-dn.net/?f=x%3D3)
, you're left with
![147=49a_1\implies a_1=\dfrac{147}{49}=3](https://tex.z-dn.net/?f=147%3D49a_1%5Cimplies%20a_1%3D%5Cdfrac%7B147%7D%7B49%7D%3D3)
When
![x=\sqrt2](https://tex.z-dn.net/?f=x%3D%5Csqrt2)
or
![x=-\sqrt2](https://tex.z-dn.net/?f=x%3D-%5Csqrt2)
, you're left with
![\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D17-8%5Csqrt2%3D%28%5Csqrt2a_4%2Ba_5%29%28%5Csqrt2-3%29%26%5Ctext%7Bfor%20%7Dx%3D%5Csqrt2%5C%5C17%2B8%5Csqrt2%3D%28-%5Csqrt2a_4%2Ba_5%29%28-%5Csqrt2-3%29%5Cend%7Bcases%7D%5Cimplies%5Cbegin%7Bcases%7D-5%2B%5Csqrt2%3D%5Csqrt2a_4%2Ba_5%5C%5C-5-%5Csqrt2%3D-%5Csqrt2a_4%2Ba_5%5Cend%7Bcases%7D)
Adding the two equations together gives
![-10=2a_5](https://tex.z-dn.net/?f=-10%3D2a_5)
, or
![a_5=-5](https://tex.z-dn.net/?f=a_5%3D-5)
. Subtracting them gives
![2\sqrt2=2\sqrt2a_4](https://tex.z-dn.net/?f=2%5Csqrt2%3D2%5Csqrt2a_4)
,
![a_4=1](https://tex.z-dn.net/?f=a_4%3D1)
.
Now, you have
![5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)](https://tex.z-dn.net/?f=5x%5E4-7x%5E3-12x%5E2%2B6x%2B21%3D3%28x%5E2-2%29%5E2%2B%28a_2x%2Ba_3%29%28x-3%29%28x%5E2-2%29%2B%28x-5%29%28x-3%29)
![5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)](https://tex.z-dn.net/?f=5x%5E4-7x%5E3-12x%5E2%2B6x%2B21%3D3x%5E4-11x%5E2-8x%2B27%2B%28a_2x%2Ba_3%29%28x-3%29%28x%5E2-2%29)
![2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)](https://tex.z-dn.net/?f=2x%5E4-7x%5E3-x%5E2%2B14x-6%3D%28a_2x%2Ba_3%29%28x-3%29%28x%5E2-2%29)
By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that
![a_2x^4=2x^4](https://tex.z-dn.net/?f=a_2x%5E4%3D2x%5E4)
and
![a_3(-3)(-2)=6a_3=-6](https://tex.z-dn.net/?f=a_3%28-3%29%28-2%29%3D6a_3%3D-6)
. These alone tell you that you must have
![a_2=2](https://tex.z-dn.net/?f=a_2%3D2)
and
![a_3=-1](https://tex.z-dn.net/?f=a_3%3D-1)
.
So the partial fraction decomposition is
Answer:
how many numbers have absolute value 3? zero is your answer
how many numbers have absolute value a, if a>0? your answer is also zero
Step-by-step explanation:
Answer:
The first choice
Step-by-step explanation:
Multiplyil by 6 to get 4y - 1 =21