Ideal timeline of the dance routine = 4 minutes = 4 × 60 seconds = 240 seconds
Variation allowed in the dance routine timeline = +- 5 seconds
Let the timeline of the dance routine be T
⇒ 240 seconds - 5 seconds < T < 240 seconds + 5 seconds
⇒ 235 seconds < T < 245 seconds
⇒
minutes < T <
minutes
⇒ 3.92 minutes < T < 4.08 minutes
So the least possible time of the dance routine can be 3.92 minutes (or 235 seconds) and the greatest possible time of the dance routine can be 4.08 minutes (or 245 seconds)
Answer:
![(x-1)(2x^2+x+2)](https://tex.z-dn.net/?f=%28x-1%29%282x%5E2%2Bx%2B2%29)
Step-by-step explanation:
Factorize:
![f(x)=2x^3-x^2+x-2](https://tex.z-dn.net/?f=f%28x%29%3D2x%5E3-x%5E2%2Bx-2)
<u>Factor Theorem</u>
If f(a) = 0 for a polynomial then (x - a) is a factor of the polynomial f(x).
Substitute x = 1 into the function:
![\implies f(1)=2(1)^3-1^2+1-2=0](https://tex.z-dn.net/?f=%5Cimplies%20f%281%29%3D2%281%29%5E3-1%5E2%2B1-2%3D0)
Therefore, (x - 1) is a factor.
As the polynomial is cubic:
![\implies f(x)=(x-1)(ax^2+bx+c)](https://tex.z-dn.net/?f=%5Cimplies%20%20f%28x%29%3D%28x-1%29%28ax%5E2%2Bbx%2Bc%29)
Expanding the brackets:
![\implies f(x)=ax^3+bx^2+cx-ax^2-bx-c](https://tex.z-dn.net/?f=%5Cimplies%20%20f%28x%29%3Dax%5E3%2Bbx%5E2%2Bcx-ax%5E2-bx-c)
![\implies f(x)=ax^3+(b-a)x^2+(c-b)x-c](https://tex.z-dn.net/?f=%5Cimplies%20%20f%28x%29%3Dax%5E3%2B%28b-a%29x%5E2%2B%28c-b%29x-c)
Comparing coefficients with the original polynomial:
![\implies ax^3=2x^3 \implies a=2](https://tex.z-dn.net/?f=%5Cimplies%20ax%5E3%3D2x%5E3%20%5Cimplies%20a%3D2)
![\implies (b-a)x^2=-x^2 \implies b-2=-1 \implies b=1](https://tex.z-dn.net/?f=%5Cimplies%20%28b-a%29x%5E2%3D-x%5E2%20%5Cimplies%20b-2%3D-1%20%5Cimplies%20b%3D1)
![\implies -c=-2 \implies c=2](https://tex.z-dn.net/?f=%5Cimplies%20-c%3D-2%20%5Cimplies%20c%3D2)
Therefore:
![\implies f(x)=(x-1)(2x^2+x+2)](https://tex.z-dn.net/?f=%5Cimplies%20%20f%28x%29%3D%28x-1%29%282x%5E2%2Bx%2B2%29)
Cannot be factored any further.
I agree with the answer above
Given:
let h be the high quality bean
let c be the cheaper bean
h + c = 160
6h + 3.25c = 160*4.97
6h + 3.25c = 795.20
h = 160 - c
6(160 - c) + 3.25c = 795.20
960 - 6c + 3.25c = 795.20
-2.75c = 795.20 - 960
-2.75c = -164.80
c = -164.80 / -2.75
c = 59.92 or 60 lbs
h = 160 - c
h = 160 - 60
h = 100 lbs
Sarah should blend 60 lbs of cheap coffee bean and 100 lbs of high quality coffee bean.