Answer:
64x³+4
-------------
2x-1
Step-by-step explanation:
I hope this helps. :)
Answer:
d. 15
Step-by-step explanation:
Putting the values in the shift 2 function
X1 + X2 ≥ 15
where x1= 13, and x2=2
13+12≥ 15
15≥ 15
At least 15 workers must be assigned to the shift 2.
The LP model questions require that the constraints are satisfied.
The constraint for the shift 2 is that the number of workers must be equal or greater than 15
This can be solved using other constraint functions e.g
Putting X4= 0 in
X1 + X4 ≥ 12
gives
X1 ≥ 12
Now Putting the value X1 ≥ 12 in shift 2 constraint
X1 + X2 ≥ 15
12+ 2≥ 15
14 ≥ 15
this does not satisfy the condition so this is wrong.
Now from
X2 + X3 ≥ 16
Putting X3= 14
X2 + 14 ≥ 16
gives
X2 ≥ 2
Putting these in the shift 2
X1 + X2 ≥ 15
13+2 ≥ 15
15 ≥ 15
Which gives the same result as above.
Answer:
We have the equation
![c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]](https://tex.z-dn.net/?f=c_1%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C0%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_2%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_3%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C4%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%2Bc_4%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D8%5C%5C4%5C%5C3%5C%5C1%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C0%5C%5C0%5Cend%7Barray%7D%5Cright%5D)
Then, the augmented matrix of the system is
![\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D0%260%260%268%5C%5C0%260%264%264%5C%5C0%263%263%263%5C%5C1%261%261%261%5Cend%7Barray%7D%5Cright%5D)
We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:
![\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%261%261%5C%5C0%263%263%263%5C%5C0%260%264%264%5C%5C0%260%260%268%5Cend%7Barray%7D%5Cright%5D)
This matrix is in echelon form. Then, now we apply backward substitution:
1.
2.
3.
4.
Then the system has unique solution that is 
and this imply that the vectors 
are linear independent.
Answer:
x = 29
Step-by-step explanation:
If the equation is :
4.2(9-x)+36=102-2.5(2x+2)
→distribute 4.2 and -2.5 in parenthesis
4.2(9 - x) + 36 = 102 -2.5(2x + 2)
37.8 - 4.2x +36 = 102 -5x -5
→ have terms with x = terms without x
Keep the terms you need the way they are, and move the terms you need on the other side of equal sign with changed sign.
4.2(9 - x) + 36 = 102 -2.5(2x + 2)
37.8 - 4.2x +36 = 102 -5x -5
- 4.2x +5 x = - 37.8 - 36 + 102 - 5
→ Combine like terms
0.8x = 23.2
→ Divide both sides by 0.8
x = 29
