Answer:
i think base. i might be wrong.
Step-by-step explanation:
Answer:
One real root ![x_{1}=5](https://tex.z-dn.net/?f=x_%7B1%7D%3D5)
Two imaginary roots
Step-by-step explanation:
Given polynomial,
![f\left ( x \right )=x^{3}-5x^{2}+6x-30](https://tex.z-dn.net/?f=f%5Cleft%20%28%20x%20%5Cright%20%29%3Dx%5E%7B3%7D-5x%5E%7B2%7D%2B6x-30)
![f\left ( x \right )=x^{3}-5x^{2}+6x-30=0](https://tex.z-dn.net/?f=f%5Cleft%20%28%20x%20%5Cright%20%29%3Dx%5E%7B3%7D-5x%5E%7B2%7D%2B6x-30%3D0)
![f\left ( x-k \right )=0](https://tex.z-dn.net/?f=f%5Cleft%20%28%20x-k%20%5Cright%20%29%3D0)
Apply hit and trial method to find zero
![x=5](https://tex.z-dn.net/?f=x%3D5)
![f\left ( 5 \right )=5^{3}-5\times 5^{2}+6\times 5-30](https://tex.z-dn.net/?f=f%5Cleft%20%28%205%20%5Cright%20%29%3D5%5E%7B3%7D-5%5Ctimes%205%5E%7B2%7D%2B6%5Ctimes%205-30)
![=125-125+30-30](https://tex.z-dn.net/?f=%3D125-125%2B30-30)
![f\left ( 5 \right ) =0](https://tex.z-dn.net/?f=f%5Cleft%20%28%205%20%5Cright%20%29%20%20%20%3D0)
This polynomial has one factor is ![\left ( x-5 \right )](https://tex.z-dn.net/?f=%5Cleft%20%28%20x-5%20%5Cright%20%29)
We can find another factor
![f\left ( x \right )/\left ( x-5 \right )=\left ( x^{3}-5x^{2}+6x-30\right )/\left ( x-5 \right )](https://tex.z-dn.net/?f=f%5Cleft%20%28%20x%20%5Cright%20%29%2F%5Cleft%20%28%20x-5%20%5Cright%20%29%3D%5Cleft%20%28%20x%5E%7B3%7D-5x%5E%7B2%7D%2B6x-30%5Cright%20%29%2F%5Cleft%20%28%20x-5%20%5Cright%20%29)
![=x^{2}+6](https://tex.z-dn.net/?f=%3Dx%5E%7B2%7D%2B6)
![f\left ( x \right )=\left ( x-5 \right )\left ( x^{2}+6 \right )=0](https://tex.z-dn.net/?f=f%5Cleft%20%28%20x%20%5Cright%20%29%3D%5Cleft%20%28%20x-5%20%5Cright%20%29%5Cleft%20%28%20x%5E%7B2%7D%2B6%20%5Cright%20%29%3D0)
![\left ( x-5 \right )\left ( x^{2}+6 \right )=0](https://tex.z-dn.net/?f=%5Cleft%20%28%20x-5%20%5Cright%20%29%5Cleft%20%28%20x%5E%7B2%7D%2B6%20%5Cright%20%29%3D0)
![\left ( x-5 \right )=0](https://tex.z-dn.net/?f=%5Cleft%20%28%20x-5%20%5Cright%20%29%3D0)
Real root ![x_{1}=5](https://tex.z-dn.net/?f=x_%7B1%7D%3D5)
is always positive so it have no real roots
Roots are imaginary
![x_{2}=2.449489i](https://tex.z-dn.net/?f=x_%7B2%7D%3D2.449489i)
![x_{3}=-2.449489i](https://tex.z-dn.net/?f=x_%7B3%7D%3D-2.449489i)
has one real root and two imaginary roots
Answer:
5 fish
Step-by-step explanation:
10-3=7
7-2=5
5 fish
Answer:
The equation that goes through this set of points is y = -x + 5
Step-by-step explanation:
In order to find this, we need to start by finding the slope. For that we use the slope formula.
m(slope) = (y2 - y1)/(x2 - x1)
m = (6 - -2)/(-1 - 3)
m = 4/-4
m = -1
Now that we have this, we can use the slope and a point in point-slope form to get the equation.
y - y1 = m(x - x1)
y - 6 = -1(x - -1)
y - 6 = -1(x + 1)
y - 6 = -x - 1
y = -x + 5
A reflection in the x-axis means the y-coordinate will be negated (the opposite sign). This means that g(x)=-f(x)=-(x^2+5=-x^2-5.