Answer:
B
M=2
Step-by-step explanation:
rise over run
you rise 2 and go over 1
2 over 1 is just 2.
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.
They are not the same thing.
Answer:
<u>Option </u><u>D</u> (y = 5/6x -12).
Step-by-step explanation:
Hey there!
The equation of the line which passes through the point (12,-2) is (y+2) = m2(x-12)………(i) [Using one point formula].
According to the question, the first line passes through point (12,6) and (0,-4).
So,
Therefore, the slope of the line is 5/6.
Now as per the condition of parallel lines, m1 =m2 = 5/6.
So, keeping the value of m2 in equation (i), we get;
(y+2) = 5/6(x-12)
or, y = 5/6x - 12.
Therefore, the required equation is y = 5/6 X - 12.
<u>Hope</u><u> </u><u>it </u><u>helps</u><u>!</u>
