Answer:
The area of the triangle is 18 square units.
Step-by-step explanation:
First, we determine the lengths of segments AB, BC and AC by Pythagorean Theorem:
AB
![AB = \sqrt{(5-2)^{2}+[6-(-1)]^{2}}](https://tex.z-dn.net/?f=AB%20%3D%20%5Csqrt%7B%285-2%29%5E%7B2%7D%2B%5B6-%28-1%29%5D%5E%7B2%7D%7D)

BC


AC
![AC = \sqrt{(-1-2)^{2}+[4-(-1)]^{2}}](https://tex.z-dn.net/?f=AC%20%3D%20%5Csqrt%7B%28-1-2%29%5E%7B2%7D%2B%5B4-%28-1%29%5D%5E%7B2%7D%7D)

Now we determine the area of the triangle by Heron's formula:
(1)
(2)
Where:
- Area of the triangle.
- Semiparameter.
If we know that
,
and
, then the area of the triangle is:


The area of the triangle is 18 square units.
(2k² + 5k - 6)(3k - 1)
(6k³ - 2k² + 15k² - 5k - 18k + 6)
6k² + 13k² - 23k + 6
Use the FOIL method to simplify the problem.
(FOIL stands for first outer, inner, last)
Answer:
it doesn't
Step-by-step explanation:
hehe, try again
I think it wold be b
please tell me if i wrong