Answer:
-2, 3, -0.5 + 0.866i, -0.5 - 0.866i.
Step-by-step explanation:
As the last term is -6 , +/- 2 , +/- 3 are possible zeroes.
Try 2:-
(2)^4 - 6(2)^2 - 7(2) - 6 = -28 so 2 is not a zero.
3:-
(3)^4 - 6(3)^2 - 7(3) - 6 = 0 so 3 is a zero.
(-2)^4 - 6(-2)^2 - 7(-2) - 6 = 0 so -2 is also a zero.
Divide the function by (x +2)(x - 3), that is x^2 - x - 6
gives x^2 + x + 1
x^2 + x + 1
So we have x^2 + x + 1 = 0
x = [-1 +/- √(1^1 - 4*1*1)] / 2
= -1 + √(-3) / 2 , - 1 - √(-3) / 2.
= -0.5 + 0.866i, -0.5 - 0.866i
Answer:
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hope it helps.
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If the negative square root is found to be one of your solutions, then that is indicative of a pair of imaginary roots (the imaginary i). According to the conjugate rule, if you have one solution that is imaginary, you will have another but with the opposite sign. For example, if a solution to a quadratic is found to be 2 - i, then its conjugate, 2 + i is also a solution. They will ALWAYS go in pairs. Same thing with radical solutions. If one solution is found to be 
then 
will also be a solution.
Answer:
6t=9? let me know if that's correct x
Step-by-step explanation:
