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:
c ln (x^3/y^3)
Step-by-step explanation:
3 ln x - 3 lny
We know that a ln b = ln b^a
ln x^3 - ln y^3
We also know ln a - ln b = ln (a/b)
ln (x^3/y^3)
Answer: The graph is shifted vertically, but not horizontally
Step-by-step explanation:
Since the added constant is outside the x term, the constant will only shift the graph vertically.
I put the solution on the paper
<span> 7x+2y=5;13x+14y=-1 </span>Solution :<span><span> {x,y} = {1,-1}</span>
</span>System of Linear Equations entered :<span><span> [1] 7x + 2y = 5
</span><span> [2] 13x + 14y = -1
</span></span>Graphic Representation of the Equations :<span> 2y + 7x = 5 14y + 13x = -1
</span>Solve by Substitution :
// Solve equation [2] for the variable y
<span> [2] 14y = -13x - 1
[2] y = -13x/14 - 1/14</span>
// Plug this in for variable y in equation [1]
<span><span> [1] 7x + 2•(-13x/14-1/14) = 5
</span><span> [1] 36x/7 = 36/7
</span><span> [1] 36x = 36
</span></span>
// Solve equation [1] for the variable x
<span><span> [1] 36x = 36</span>
<span> [1] x = 1</span> </span>
// By now we know this much :
<span><span> x = 1</span>
<span> y = -13x/14-1/14</span></span>
<span>// Use the x value to solve for y
</span>
<span> y = -(13/14)(1)-1/14 = -1 </span>Solution :<span><span> {x,y} = {1,-1}</span>
<span>
Processing ends successfully</span></span>
