Take the linear coefients easy
11,-12, 7
5,12,-7
so
![\left[\begin{array}{ccc}11&-12&7\\5&12&-7\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%26-12%267%5C%5C5%2612%26-7%5Cend%7Barray%7D%5Cright%5D%20)
or you could do also
![\left[\begin{array}{ccc}11&-12\\5&12\end{array}\right] \left[\begin{array}{ccc}7\\-7\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%26-12%5C%5C5%2612%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%5C%5C-7%5Cend%7Barray%7D%5Cright%5D%20)
All the sides added up! That's how to find the perimeter.
I like to start by dividing the given number by 2, unless a larger factor is evident.
1386 = 2(693). It's pretty obvious that we can divide 693 by 3: 693=3(231).
Then we have 1386 = (2)(3)(231). Try the same procedure. Does 231 have any integer, prime factors? See how far you can take this factoring.
99.3 cause form 100 percent you minus 0.7 which makes it 99.3 percent chance of failure
