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:
The sequence is:
10, 30, 50, 70, 90.....................
Step-by-step explanation:
We have,
First term (a) = 10
Common difference (d) = ?
Sum of first 5 terms (
) = 250
or, ![\frac{n}{2} [{2a+(n-1)d}] = 250](https://tex.z-dn.net/?f=%5Cfrac%7Bn%7D%7B2%7D%20%5B%7B2a%2B%28n-1%29d%7D%5D%20%3D%20250)
or, ![\frac{5}{2} [2*10 + 4d]=250](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B2%7D%20%5B2%2A10%20%2B%204d%5D%3D250)
or, ![\frac{5}{2} * 4[5+d]=250](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B2%7D%20%2A%204%5B5%2Bd%5D%3D250)
or, 10(5 + d) =250
or, 5 + d = 25
∴ d = 20
Now,
2nd term = a + d = 10 + 20 = 30
3rd term = a + 2d = 10 + 2*20 = 10 + 40 = 50
4th term = a + 3d = 10 + 3*20 = 10 + 60 = 70
5th term = a + 4d = 10 + 4*20 = 10 + 80 = 90
Answer:
Step-by-step explanation:
You have to divide -5 from each side so -5/-5x which would be x and divide -5 by 40 which is -8 so x is greater than or equal to -8
Answer:
It is 35x+20
Step-by-step explanation:
The sum of the inner angles of any triangle is always 180°, i.e. you have

In the particular case of an equilater triangle, all three angles are the same, so

and the expression becomes

which implies 
So, if you rotate the triangle with respect to its center by 60 degrees, the triangle will map into itself. In particular, if you want point A to be mapped into point B, you have to perform a counter clockwise rotation of 60 degrees with respect to the center of the triangle.
Of course, this is equivalent to a clockwise rotation of 120 degrees.
Finally, both solutions admit periodicity: a rotation of 60+k360 degrees has the same effect of a rotation of 60 degrees, and the same goes for the 120 one (actually, this is obvisly true for any rotation!)