Answer:
it must also have the root : - 6i
Step-by-step explanation:
If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.
This is because in order to render a polynomial with Real coefficients, the binomial factor (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:
![(x-(a+bi))*(x-(a-bi))=\\(x-a-bi)*(x-a+bi)=\\([x-a]-bi)*([x-a]+bi)=\\(x-a)^2-(bi)^2=\\(x-a)^2-b^2(-1)=\\(x-a)^2+b^2](https://tex.z-dn.net/?f=%28x-%28a%2Bbi%29%29%2A%28x-%28a-bi%29%29%3D%5C%5C%28x-a-bi%29%2A%28x-a%2Bbi%29%3D%5C%5C%28%5Bx-a%5D-bi%29%2A%28%5Bx-a%5D%2Bbi%29%3D%5C%5C%28x-a%29%5E2-%28bi%29%5E2%3D%5C%5C%28x-a%29%5E2-b%5E2%28-1%29%3D%5C%5C%28x-a%29%5E2%2Bb%5E2)
where the imaginary unit has disappeared, making the expression real.
So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)
Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.
Answer:
notebook: $2.5
pen: $1.5
part B, he can buy 3 pens
Step-by-step explanation:
A
let the cost of notebook be n, cost of pen be p
1. 4n + 3p = 14.5
2. 5n + 6p = 21.5
by 1, you can get 3p = 14.5 - 4n, put it into 2
5n + 2(14.5 - 4n) = 21.5
5n + 29 - 8n = 21.5
3n = 29 - 21.5 = 7.5
n = 2.5
put n = 2.5 into 1
4(2.5) + 3p = 14.5
3p = 4.5
p = 1.5
B
Waseem buy 3 notebooks for each pen, let number of pen be m, then the number of notebook will be 3m
m(2.5) + 3m(1.5) < 27
2.5m + 4.5m = 7m < 27
m < 27/7
maximum integer value for m = 3
First one is 7.73 (23.19/3)
Second one is 14.04 ((19.71/7.3) x 5.2)
It's really important that you aim to understand how to do these problems yourself, rather than "pay" others to do them for you. Why not ask for "directions" or "guidance," instead of for "answers?"
This is a tricky question because it's not required that you have the full 12 beads on the string in every case. You could, for example, have 6 beads, 2 of which would be red, making the ratio of red beads 2 to 6, or 1 to 3.
What about 3 red beads in 9?
What about 1 red bead in 3?
Do these "answers" satisfy the requirement that the ratio of red beads to all beads is 1:3?
