I believe by the first equation they are referring two a representation of the total amount of money made up by the two types of coins. I found this to be 0.05x + 0.10y = 3.50
In my work I solved for the amount of each coin- I assume this is what the final step of your problem entails. If you need any help understanding/ I misinterpreted something please feel free to comment and ask :)
#5c coins or x= 36
#10c coins or y= 17
Answer:
- A. Robin because there are 36 outcomes and only one is (6, 6).
Step-by-step explanation:
<u>You would get the probability of double sixes as:</u>
Correct choice is A
First, pull out a factor of
![x](https://tex.z-dn.net/?f=x)
.
![x^6+3x^5+x^4-5x^3-6x^2-2x=x(x^5+3x^4+x^3-5x^2-6x-2)](https://tex.z-dn.net/?f=x%5E6%2B3x%5E5%2Bx%5E4-5x%5E3-6x%5E2-2x%3Dx%28x%5E5%2B3x%5E4%2Bx%5E3-5x%5E2-6x-2%29)
Notice that when
![x=-1](https://tex.z-dn.net/?f=x%3D-1)
(which you can arrive at via the rational root theorem), you have
![(-1)^5+3(-1)^4+(-1)^3-5(-1)^2-6(-1)-2=-1+3-1-5+6-2=0](https://tex.z-dn.net/?f=%28-1%29%5E5%2B3%28-1%29%5E4%2B%28-1%29%5E3-5%28-1%29%5E2-6%28-1%29-2%3D-1%2B3-1-5%2B6-2%3D0)
which means you can pull out a factor of
![x+1](https://tex.z-dn.net/?f=x%2B1)
. Upon dividing you get
![\dfrac{x^5+3x^4+x^3-5x^2-6x-2}{x+1}=x^4+2x^3-x^2-4x-2](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E5%2B3x%5E4%2Bx%5E3-5x%5E2-6x-2%7D%7Bx%2B1%7D%3Dx%5E4%2B2x%5E3-x%5E2-4x-2)
The rational root theorem will come in handy again, suggesting that
![x=-1](https://tex.z-dn.net/?f=x%3D-1)
appears a second time as a root, which means
![\dfrac{x^4+2x^3-x^2-4x-2}{x+1}=x^3+x^2-2x-2](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E4%2B2x%5E3-x%5E2-4x-2%7D%7Bx%2B1%7D%3Dx%5E3%2Bx%5E2-2x-2)
Now this is more readily factored without having to resort to the rational root theorem. You have
![x^3+x^2-2x-2=x^2(x+1)-2(x+1)=(x^2-2)(x+1)](https://tex.z-dn.net/?f=x%5E3%2Bx%5E2-2x-2%3Dx%5E2%28x%2B1%29-2%28x%2B1%29%3D%28x%5E2-2%29%28x%2B1%29)
so in fact,
![x=-1](https://tex.z-dn.net/?f=x%3D-1)
shows up as a root for a third time.
So, you have
![x^6+3x^5+x^4-5x^3-6x^2-2x=x(x+1)^3(x^2-2)=0](https://tex.z-dn.net/?f=x%5E6%2B3x%5E5%2Bx%5E4-5x%5E3-6x%5E2-2x%3Dx%28x%2B1%29%5E3%28x%5E2-2%29%3D0)
Two roots are obvious,
![x=0](https://tex.z-dn.net/?f=x%3D0)
and
![x=-1](https://tex.z-dn.net/?f=x%3D-1)
(with multiplicity 3). The remaining two are given by