Answer:
![\left[\begin{array}{ccc}3&-5  &|12\\4&-2  &|15\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-5%20%20%26%7C12%5C%5C4%26-2%20%20%26%7C15%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
When making a matrix of two equations with the variables x and y, the result will be a matrix with three columns:
- a column for the values of x in each equation
- a column for the values of y in each equation
- a column for the independent values of each equation
since our system of equations is:

we can see that the value for x in the first equation is 3 and in the second equation is 4, thus the first column will have the numbers 3 and 4:
![\left[\begin{array}{ccc}3&&\\4&&\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26%26%5C%5C4%26%26%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Now for the values of y we hvae -5 in the first equation and -2 in the second equation, we update the matrix with another column with the values of -5 and -2:
![\left[\begin{array}{ccc}3&-5&\\4&-2&\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-5%26%5C%5C4%26-2%26%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Finally, the last column is the independent values of each equation (or the results) in the first equation that number is 12 and in the second equation is 15, thus the matrix is:
![\left[\begin{array}{ccc}3&-5&12\\4&-2&15\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-5%2612%5C%5C4%26-2%2615%5C%5C%5Cend%7Barray%7D%5Cright%5D)
usually there is a line separating the columns for the values of x and y, and the independent values: 
![\left[\begin{array}{ccc}3&-5  &|12\\4&-2  &|15\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-5%20%20%26%7C12%5C%5C4%26-2%20%20%26%7C15%5C%5C%5Cend%7Barray%7D%5Cright%5D)
this is the matrix of the system of equations
 
        
             
        
        
        
Answer:
25.61 ?
Step-by-step explanation:
use Pythagoreans theorem
a^2+b^2=c^2
20^2+16^2=656
sqrt 656=25.61
 
        
             
        
        
        
Answer:
(a) 17°
(b) 78°
Step-by-step explanation:
(a) sin A= 0.2896 
 to calculate for "A" check for sin inverse of 0.2896
A= sin^-1(0.2896) = 16.83
to the nearest degree A= 17°
 
(b) tan A = 4.7046
A= tan^-1(4.7046)
A = 77.99
to the nearest degree A = 78°
 
        
             
        
        
        
To factor quadratic equations of the form ax^2+bx+c=y, you must find two values, j and k, which satisfy two conditions.
jk=ac and j+k=b
The you replace the single linear term bx with jx and kx. Finally then you factor the first pair of terms and the second pair of terms. In this problem...
2k^2-5k-18=0
2k^2+4k-9k-18=0
2k(k+2)-9(k+2)=0
(2k-9)(k+2)=0
so k=-2 and 9/2
k=(-2, 4.5)