<span>The sum of the two sides of the triangle must be greater than the third side, so
11 < third side < 21</span>
If a polynomial "contains", in a multiplicative sense, a factor
, then the polynomial has a zero at
.
So, you polynomial must contain at least the following:
![(x-(-2)),\quad (x-0),\quad (x-3),\quad (x-5)](https://tex.z-dn.net/?f=%28x-%28-2%29%29%2C%5Cquad%20%28x-0%29%2C%5Cquad%20%28x-3%29%2C%5Cquad%20%28x-5%29)
If you multiply them all, you get
![x(x+2)(x-3)(x-5)=x^4 - 6 x^3 - x^2 + 30 x](https://tex.z-dn.net/?f=x%28x%2B2%29%28x-3%29%28x-5%29%3Dx%5E4%20-%206%20x%5E3%20-%20x%5E2%20%2B%2030%20x)
Now, if you want the polynomial to be zero only and exactly at the four points you've given, you can choose every polynomial that is a multiple (numerically speaking) of this one. For example, you can multiply it by 2, 3, or -14.
If you want the polynomial to be zero at least at the four points you've given, you can multiply the given polynomial by every other function.
It’s J
In the equation y=mx+b, B is the y-intercept and since the equation is y=-1/3x+2, the line crosses at two in the y-axis and since the fraction is negative it goes down.
Answer:
35.991
Step-by-step explanation:
4. Is congruent and 5. Is prove hope that helps