Answer:
41
Step-by-step explanation:
x=12
so y=3⋅12 +5
=36+5
=41
Thet contradict each other, that's why both of them are incorrect.
<span>Suppose that a polynomial has four roots: s, t, u, and v. If the polynomial were evaluated at any of these values, it would have to be zero. Therefore, the polynomial can be written in this form.
p(x)(x - s)(x - t)(x - u)(x - v), where p(x) is some non-zero polynomial
This polynomial has a degree of at least 4. It therefore cannot be cubic.
Now prove Kelsey correct. We have already proved that there can be no more than three roots. To prove that a cubic polynomial with three roots is possible, all we have to do is offer a single example of that. This one will do.
(x - 1)(x - 2)(x - 3)
This is a cubic polynomial with three roots, and four or more roots are not possible for a cubic polynomial. Kelsey is correct.
Incidentally, if this is a roller coaster we are discussing, then a cubic polynomial is not such a good idea, either for a vertical curve or a horizontal curve. I hope this helps</span><span>
</span>
Answer:
y-determinant = 2
Step-by-step explanation:
Given the following system of equation:
Let's represent it using a matrix:
![\left[\begin{array}{ccc}1&2\\1&-3\end{array}\right] = \left[\begin{array}{ccc}5\\7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%5C%5C1%26-3%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%5C%5C7%5Cend%7Barray%7D%5Cright%5D)
The y‐numerator determinant is formed by taking the constant terms from the system and placing them in the y‐coefficient positions and retaining the x‐coefficients. Then:
![\left[\begin{array}{ccc}1&5\\1&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%265%5C%5C1%267%5Cend%7Barray%7D%5Cright%5D%20)
y-determinant = (1)(7) - (5)(1) = 2.
Therefore, the y-determinant = 2
Answer:
3000.897
Step-by-step explanation:
Answer:
its C
Step-by-step explanation:
I just saw that