Solve negative one fifth times the quantity 4 minus 14 times the quantity one fourth squared end quantity end quantity.
1 answer:
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Complete question:
Rory records the percentage of battery life remaining on his phone throughout a day. The graph represents the percentage of battery life remaining after a certain number of hours.
Rory’s battery was full at 6:00 a.m, or t = 0. Which times could Rory's phone have been plugged into the charger? Select three options.
12:00 p.m.
3:00 p.m.
5:00 p.m.
8:00 p.m.
11:00 p.m.
The graph is attached
Answer:
3:00 p.m
5:00 p.m
11:00 p.m
Step-by-step explanation:
From the graph, there are rise times, fall times and there are times the movement is steady (no rise or fall).
The rise time represents the time the phone was plugged. The fall time represents when the charger has been unplugged so the battery level starts depreciating. When it is constant, it means the battery level is at 100 but the charger is still connected to the phone.
We are told at initial condition, time was 6 a.m (t = 0). To get the exact time, we are to add the initial condition, i.e 6
From the graph, the times the phone could have been plugged are 8:00 to 12:00 hrs and 16:00 to 20:00 hrs.
Converting from 24 hr to 12 hr time, we have:
6 + 8hrs = 14:00 = 2:00 p.m
6 + 9hrs = 15:00 = 3:00 p.m
6 + 10 hrs = 16:00 = 4:00 p.m
6 + 11 hrs = 17:00 = 5:00 p.m
6 + 12 hrs = 18:00 = 6:00 p.m
6 + 16 hrs = 22:00 = 10:00 p.m
6 + 17 hrs = 23:00 = 11:00 p.m
6 + 18 hrs = 00:00 = 12:00 am
6 + 19 = 1 : 00 = 1:00 a.m
6 + 20 = 2:00 = 2:00 a.m
From the options given in the question, we have:
3:00 p.m; 5:00 p.m; 11:00 p.m
Therefore, times could Rory's phone have been plugged into the charger are:
3:00 p.m
5:00 p.m
11:00 p.m
