Answer:
Following are the response to the given question:
Explanation:
The glamorous objective is to examine the items (as being the most valuable and "cheapest" items are chosen) while no item is selectable - in other words, the loading can be reached.
Assume that such a strategy also isn't optimum, this is that there is the set of items not including one of the selfish strategy items (say, i-th item), but instead a heavy, less valuable item j, with j > i and is optimal.
As 
, the i-th item may be substituted by the j-th item, as well as the overall load is still sustainable. Moreover, because 
and this strategy is better, our total profit has dropped. Contradiction.
Answer:
Option E is correct.
Explanation:
While the user's server forms of the specific domain Active Directory. His corporation also partnered with some firms recently. The obtained corporation seems to have a multi-domain Active Directory server for both areas, both domain operators use the following Windows Servers. This was assigned the responsibility of proposing improvements to the Active Directory system.
He needs users of all corporations will gain exposure with each other's services, however, he wants to reduce logistical activity to achieve. Thus, he would build a two-way forest relationship between both the two roots forest domains.
Answer:
graphic design and digital marketing solutions
Answer:
a. The cell tower forwards the call to the wireless provider's Mobile Telephone Switching Office
Explanation:
When a cell phone user begins to make a call, it is picked up by the cell tower located in the cell in which the cell phone is located and that belongs to the user's wireless provider. The next step in this process is that the cell tower forwards the call to the wireless provider's Mobile Telephone Switching Office.
let's say they catch-up to each other after t hour.
so
in t hour distance travelled by Renee
d = speed × time = 50t
in t hour distance travelled by Kim
d = speed × time = 60(t-1) + 0×1 = 60(t-1)
Note: here kim didn't covered any distance in first hour and in rest t-1 hour it travelled all distance
now.as distance travelled by both is same so
60(t-1) = 50t
60t -60 = 50t
adding 60 both sides
60t = 50t+60
subtracting 50t both sides
60t -50t = 50t+60 - 50t
10t = 60
dividing by 10 both sides
t = 60/10 = 6
so in 6 hour both will catch-up to each other
