Answer:
![y+6=m(x+2)](https://tex.z-dn.net/?f=y%2B6%3Dm%28x%2B2%29)
where I would have to look at the table to know
.
Step-by-step explanation:
Point-slope form of a line is
![y-y_1=m(x-x_1)](https://tex.z-dn.net/?f=y-y_1%3Dm%28x-x_1%29)
where ![m \text{ is the slope and } (x_1,y_1) \text{ is a point on that line}](https://tex.z-dn.net/?f=m%20%5Ctext%7B%20is%20the%20slope%20and%20%7D%20%28x_1%2Cy_1%29%20%5Ctext%7B%20is%20a%20point%20on%20that%20line%7D)
You are given
.
So we know we are looking for an equation that looks like this:
![y-(-6)=m(x-(-2))](https://tex.z-dn.net/?f=y-%28-6%29%3Dm%28x-%28-2%29%29)
If you simplify this looks like:
![y+6=m(x+2)](https://tex.z-dn.net/?f=y%2B6%3Dm%28x%2B2%29)
Answer:
clockwise
Step-by-step explanation:
Conventionally, positive angles in the Cartesian coordinate plane are measured counterclockwise from the +x axis. Rotation in the negative direction is the opposite of that, so is clockwise.
The factorization of 12a^3b^2 +18a²b^2 – 12ab^2 is ![6 a b^{2}(a+2)(2 a-1)](https://tex.z-dn.net/?f=6%20a%20b%5E%7B2%7D%28a%2B2%29%282%20a-1%29)
<u>Solution:</u>
Given, expression is ![12 a^{3} b^{2}+18 a^{2} b^{2}-12 a b^{2}](https://tex.z-dn.net/?f=12%20a%5E%7B3%7D%20b%5E%7B2%7D%2B18%20a%5E%7B2%7D%20b%5E%7B2%7D-12%20a%20b%5E%7B2%7D)
We have to factorize the given expression completely.
Now, take the expression
![12 a^{3} b^{2}+18 a^{2} b^{2}-12 a b^{2}](https://tex.z-dn.net/?f=12%20a%5E%7B3%7D%20b%5E%7B2%7D%2B18%20a%5E%7B2%7D%20b%5E%7B2%7D-12%20a%20b%5E%7B2%7D)
Taking
as common term,
![b^{2}\left(12 a^{3}+18 a^{2}-12 a\right)](https://tex.z-dn.net/?f=b%5E%7B2%7D%5Cleft%2812%20a%5E%7B3%7D%2B18%20a%5E%7B2%7D-12%20a%5Cright%29)
Taking "a" as common term,
![b^{2}\left(a\left(12 a^{2}+18 a-12\right)\right)](https://tex.z-dn.net/?f=b%5E%7B2%7D%5Cleft%28a%5Cleft%2812%20a%5E%7B2%7D%2B18%20a-12%5Cright%29%5Cright%29)
Taking "6" as common term,
![b^{2}\left(a\left(6\left(2 a^{2}+3 a-2\right)\right)\right)](https://tex.z-dn.net/?f=b%5E%7B2%7D%5Cleft%28a%5Cleft%286%5Cleft%282%20a%5E%7B2%7D%2B3%20a-2%5Cright%29%5Cright%29%5Cright%29)
Splitting "3a" as "4a - a" we get,
![b^{2}\left(a\left(6\left(2 a^{2}+4 a-a-2\right)\right)\right)](https://tex.z-dn.net/?f=b%5E%7B2%7D%5Cleft%28a%5Cleft%286%5Cleft%282%20a%5E%7B2%7D%2B4%20a-a-2%5Cright%29%5Cright%29%5Cright%29)
![\begin{array}{l}{b^{2}(a(6(2 a(a+2)-1(a+2))))} \\\\ {b^{2}(a(6((a+2) \times(2 a-1))))} \\\\ {6 a b^{2}(a+2)(2 a-1)}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7Bb%5E%7B2%7D%28a%286%282%20a%28a%2B2%29-1%28a%2B2%29%29%29%29%7D%20%5C%5C%5C%5C%20%7Bb%5E%7B2%7D%28a%286%28%28a%2B2%29%20%5Ctimes%282%20a-1%29%29%29%29%7D%20%5C%5C%5C%5C%20%7B6%20a%20b%5E%7B2%7D%28a%2B2%29%282%20a-1%29%7D%5Cend%7Barray%7D)
Hence, the factored form of given expression is ![6 a b^{2}(a+2)(2 a-1)](https://tex.z-dn.net/?f=6%20a%20b%5E%7B2%7D%28a%2B2%29%282%20a-1%29)