20 points
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
We can see that the graph touches without crossing the x-axis (i.e. it is a double solution), and then there's another zero at
(this time it's a crossing zero, so a single solution).
This leads, up to multiple, to the polynomial
If we impose the passing through we have
So, the polynomial is
Finally, to solve , simply look at the graph, searching for the points, where the graph is below the x-axis. You can see that this happens only if
, so that's the solution to your question.