Answer: D
<u>Step-by-step explanation:</u>
The first matrix contains the coefficients of the x- and y- values for both equations (top row is the top equation and the bottom row is the bottom equation. The second matrix contains what each equation is equal to.
![\begin{array}{c}2x-y\\x-6y\end{array}\qquad \rightarrow \qquad \left[\begin{array}{cc}2&-1\\1&-6\end{array}\right] \\\\\\\begin{array}{c}-6\\13\end{array}\qquad \rightarrow \qquad \left[\begin{array}{c}-6\\13\end{array}\right]](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bc%7D2x-y%5C%5Cx-6y%5Cend%7Barray%7D%5Cqquad%20%5Crightarrow%20%5Cqquad%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D2%26-1%5C%5C1%26-6%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5C%5Cbegin%7Barray%7D%7Bc%7D-6%5C%5C13%5Cend%7Barray%7D%5Cqquad%20%5Crightarrow%20%5Cqquad%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-6%5C%5C13%5Cend%7Barray%7D%5Cright%5D)
The product will result in the solution for the x- and y-values of the system.
In the problem, the first equation should be represented like this base on the variable given in the problem:
20p + 9t = 44.4
with that equation, the second equation would is given by this formula, in response with the additional number of paperback and textbook
21p + 14t = 51
to get the system equation in getting the mass of each variable, you should subtract the two equation to simplified the formula.
20p + 9t = 44.4
- 21p + 14t = 51
-------------------------
p + 5t = 6.6 // this is the system equation that could be use in getting the mass of each object
[ - 4 , 4 ) ∪ ( 4 , ∞ )
The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be.
solve x - 4 = 0 ⇒ x = 4 ( is a vertical asymptote )
There is a zero when the numerator equals zero.
x + 4 = 0 ⇒ x = - 4 ( is a zero )
domain is [-4 , 4 ) ∪ ( 4 , ∞ )
Yes that is true if you think of it as a point and the lines extending in both directions.
