Answer:
The word "Of" in an equation is interpreted to be multiplication. Also, you must move the decimal two places to the left to change a percent to a decimal.
1. 20 * .12 = 2.4
2. .3 * 70 = 21
3. 300 * .25 = 75 Blue marbles
4. 40 * .35 = $14
Answer:
C
Step-by-step explanation:
hope it helps
![\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
She added the check fee instead of subtracting it