X=1/2-3y/4
and
x=10/3+2y/3
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.
Using the law os cosines formula b^2 = a^2 + c^2 - 2*a*c*cos(B)
a = 17, b = 8, c = 16
8^2 = 17^2 + 16^2 - 2*17*16* cos(B)
64 = 289 + 256 - 544 * cos(B)
544*cos(B) = 289 + 256 - 64
544 * cos(B) = 481
cos (B) = 481/544
B = arccos(481/544)
B = 27.8 degrees
Answer:
1960
Step-by-step explanation:
1960
times 56 times 7 time 5 = 1960
