Answer:
Trish made 2 pans of brownies
Step-by-step explanation:
So essentially 84/12 is 7 pans if Rachel made 5 pans and we know that using subtraction we can easily solve this by subtracting 7 and 5
 
        
             
        
        
        
Answer:
A; ?; Line 3
Step-by-step explanation:
Question 1: 
-30 divided by -6 is 5. 
The negatives cancel each other out. 
And 30 divided by 6 is simply 5. 
Question 2: 

Do the operation inside the parenthesis first: 

Divide. 12 does not go into 62 evenly. 12 times 5 is 60, so the answer is 5 remainder 2: 

Or in fractions: 

(Was there a typo? From what you've given me, the correct answer is not listed.)
Question 3: 
So we have to following steps: 

The mistake is in Line 3. Danny cannot just combine 5+6(2) into 11(2). Instead, he should multiply 6(2) and then add. Thus, the correct solution is: 

 
        
             
        
        
        
Answer:
0
Step-by-step explanation:
We find the determinant of a matrix by the method below. If we have a matrix:
![\left[\begin{array}{cc}a&b\\c&d\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D)
The determinant is 
Now, using cramer's rule, we find x-value by the formula:

Where D is the determinant of the original problem and  is the determinant of the x-value matrix. How do we get those?
 is the determinant of the x-value matrix. How do we get those?
<u><em>To get original matrix and thus D, we set up the matrix as the coefficients of x and y (s) of both the equations and to get matrix of x-value and thus  , we replace the x values of the matrix with the numbers in the right hand side of the 2 equations.</em></u> We show this below:
, we replace the x values of the matrix with the numbers in the right hand side of the 2 equations.</em></u> We show this below:
<em />
<em>To get D:</em>
![\left[\begin{array}{cc}3&4\\1&-6\end{array}\right] \\D=(3)(-6)-(1)(4)=-18-4=-22](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%264%5C%5C1%26-6%5Cend%7Barray%7D%5Cright%5D%20%5C%5CD%3D%283%29%28-6%29-%281%29%284%29%3D-18-4%3D-22)
<em>To get  :</em>
:</em>
<em>![\left[\begin{array}{cc}12&4\\-18&-6\end{array}\right] \\D_x=(12)(-6)-(-18)(4)=0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D12%264%5C%5C-18%26-6%5Cend%7Barray%7D%5Cright%5D%20%5C%5CD_x%3D%2812%29%28-6%29-%28-18%29%284%29%3D0) </em>
</em>
<em />
<em>Putting into the formula, we get:</em>
<em> </em>
</em>
<em />
<em>Thus, the value of x is 0</em>
 
        
             
        
        
        
Answer:
here is the pic you wanted
Step-by-step explanation:
 
        
                    
             
        
        
        
Assume the graph is drawn with m as the x-axis, and s as the y-axis.  If it is drawn the other way, it would be just a reflection about the x=y line.
There are two unspecified but implicit conditions, namelym>=0 and s>=0, which correspond to the m- and s-axes (horiz. & vert.) when equality holds in either case.
s+m<100 is a solid line at 45 degrees intersecting the m- and s-axes at (0,100), and (100,0) respectively.
s+2m>100 is a solid line with slope -2 intersecting the axes at (0,50) and (100,0).
The feasible region (shaded) is between the two lines, but above the horizontal axis.
If the graph is drawn the other way, the feasible region is between the two lines, and to the right of the vertical axis.