It’s already in the simplest form.
Answer:
The solution of the inequation
is
.
Step-by-step explanation:
First of all, let simplify and factorize the resulting polynomial:



Roots are found by Quadratic Formula:
![r_{1,2} = \frac{\left[-\left(-\frac{11}{6}\right)\pm \sqrt{\left(-\frac{11}{6} \right)^{2}-4\cdot (1)\cdot \left(-\frac{10}{6} \right)} \right]}{2\cdot (1)}](https://tex.z-dn.net/?f=r_%7B1%2C2%7D%20%3D%20%5Cfrac%7B%5Cleft%5B-%5Cleft%28-%5Cfrac%7B11%7D%7B6%7D%5Cright%29%5Cpm%20%5Csqrt%7B%5Cleft%28-%5Cfrac%7B11%7D%7B6%7D%20%5Cright%29%5E%7B2%7D-4%5Ccdot%20%281%29%5Ccdot%20%5Cleft%28-%5Cfrac%7B10%7D%7B6%7D%20%5Cright%29%7D%20%5Cright%5D%7D%7B2%5Ccdot%20%281%29%7D)
and 
Then, the factorized form of the inequation is:

By Real Algebra, there are two condition that fulfill the inequation:
a) 


b) 


The solution of the inequation
is
.
Answer:
The answer is C
Step-by-step explanation:
Because... if you start in point A and do what C say it will bring you to the same point as the second triangle
Hopes this helps❤️
Answer:
Check below, please
Step-by-step explanation:
Step-by-step explanation:
1.For which values of x is f '(x) zero? (Enter your answers as a comma-separated list.)
When the derivative of a function is equal to zero, then it occurs when we have either a local minimum or a local maximum point. So for our x-coordinates we can say

2. For which values of x is f '(x) positive?
Whenever we have

then function is increasing. Since if we could start tracing tangent lines over that graph, those tangent lines would point up.

3. For which values of x is f '(x) negative?
On the other hand, every time the function is decreasing its derivative would be negative. The opposite case of the previous explanation. So

4.What do these values mean?

5.(b) For which values of x is f ''(x) zero?
In its inflection points, i.e. when the concavity of the curve changes. Since the function was not provided. There's no way to be precise, but roughly
at x=-4 and x=4