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:
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.
B: z=80___________________________________
Add each term and divide it by 2.
5+7 = 12
12/2 = 6
-2 + 6 = 4
4/2 = 2, then add the i back on to get 2i.
The midpoint is 6 + 2i
3 hours I think?
There’s not enough context for this.
It can rlly be anything
Answer:
No it is not a solution
Step-by-step explanation:
To solve this, you should replace your (x,y) with (8,3)- 8 being your x value and 3 being your y value.
So this inequality, your y value is supposed to be bigger than the value of whatever 3x-4 equals.
So, your inequality should be 3>3(8)-4
3*8=24-4=20
So is 3>(greater than)20? No. Which then means that (8,3) is not a solution because it would make the inequality not true.
Hope this helps!