Theorems are proven with geometric proof
Swapping rows alters the sign of the determinant:
![\begin{vmatrix} x & y & z \\ -8 & 2 & -12 \\ u & v & w \end{vmatrix} = - \begin{vmatrix} x & y & z \\ u & v & w \\ -8 & 2 & -12 \end{vmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bvmatrix%7D%20x%20%26%20y%20%26%20z%20%5C%5C%20-8%20%26%202%20%26%20-12%20%5C%5C%20u%20%26%20v%20%26%20w%20%5Cend%7Bvmatrix%7D%20%3D%20-%20%5Cbegin%7Bvmatrix%7D%20x%20%26%20y%20%26%20z%20%5C%5C%20u%20%26%20v%20%26%20w%20%5C%5C%20-8%20%26%202%20%26%20-12%20%5Cend%7Bvmatrix%7D)
Multiplying a single row by a scalar scales the determinant by the same amount:
![\begin{vmatrix} x & y & z \\ u & v & w \\ -8 & 2 & -12 \end{vmatrix} = -2 \begin{vmatrix} x & y & z \\ u & v & w \\ 4 & -1 & 6 \end{vmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bvmatrix%7D%20x%20%26%20y%20%26%20z%20%5C%5C%20u%20%26%20v%20%26%20w%20%5C%5C%20-8%20%26%202%20%26%20-12%20%5Cend%7Bvmatrix%7D%20%3D%20-2%20%5Cbegin%7Bvmatrix%7D%20x%20%26%20y%20%26%20z%20%5C%5C%20u%20%26%20v%20%26%20w%20%5C%5C%204%20%26%20-1%20%26%206%20%5Cend%7Bvmatrix%7D)
Then
![\begin{vmatrix} x & y & z \\ -8 & 2 & -12 \\ u & v & w \end{vmatrix} = -(-2) \begin{vmatrix} x & y & z \\ u & v & w \\ 4 & -1 & 6 \end{vmatrix} = 2\times(-6) = \boxed{-12}](https://tex.z-dn.net/?f=%5Cbegin%7Bvmatrix%7D%20x%20%26%20y%20%26%20z%20%5C%5C%20-8%20%26%202%20%26%20-12%20%5C%5C%20u%20%26%20v%20%26%20w%20%5Cend%7Bvmatrix%7D%20%3D%20-%28-2%29%20%5Cbegin%7Bvmatrix%7D%20x%20%26%20y%20%26%20z%20%5C%5C%20u%20%26%20v%20%26%20w%20%5C%5C%204%20%26%20-1%20%26%206%20%5Cend%7Bvmatrix%7D%20%3D%202%5Ctimes%28-6%29%20%3D%20%5Cboxed%7B-12%7D)
ANSWER
The restrictions are
![a\ne -3,a\ne -\frac{1}{5}](https://tex.z-dn.net/?f=a%5Cne%20-3%2Ca%5Cne%20-%5Cfrac%7B1%7D%7B5%7D)
EXPLANATION
We were given the rational function,
![\frac{2a^2+a-15}{5a^2+16a+3}](https://tex.z-dn.net/?f=%5Cfrac%7B2a%5E2%2Ba-15%7D%7B5a%5E2%2B16a%2B3%7D)
The function is defined for all values of a, except
![5a^2+16a+3=0](https://tex.z-dn.net/?f=5a%5E2%2B16a%2B3%3D0)
This has become a quadratic trinomial, so we need to split the middle term.
We do that by multiplying the coefficient of
which is 5 by the constant term which is 3. This gives us 15.
The factors of 15 that adds up to 16 are 1 and 15.
We use these factors to split the middle term.
![5a^2+15a+a+3=0](https://tex.z-dn.net/?f=5a%5E2%2B15a%2Ba%2B3%3D0)
We now factor to get,
![5a(a+3)+1(a+3)=0](https://tex.z-dn.net/?f=5a%28a%2B3%29%2B1%28a%2B3%29%3D0)
We factor further to get,
![(a+3)(5a+1)=0](https://tex.z-dn.net/?f=%28a%2B3%29%285a%2B1%29%3D0)
This implies that,
![(a+3)=0,(5a+1)=0](https://tex.z-dn.net/?f=%28a%2B3%29%3D0%2C%285a%2B1%29%3D0)
This gives
![a=-3,a=-\frac{1}{5}](https://tex.z-dn.net/?f=a%3D-3%2Ca%3D-%5Cfrac%7B1%7D%7B5%7D)
These are the restrictions.
Answer:
7 g.
Step-by-step explanation:
7g = 7000 mg