A wire was bent into the shape of a rectangle with width 5 and length 7 . If this wire
1 answer:
You might be interested in
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
Take the homogeneous part and find the roots to the characteristic equation:
This means the characteristic solution is
.
Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form
. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.
With
and 
, you're looking for a particular solution of the form 
. The functions 
satisfy
where
is the Wronskian determinant of the two characteristic solutions.
So you have
So you end up with a solution
but since
is already accounted for in the characteristic solution, the particular solution is then
so that the general solution is
This means the characteristic solution is
Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form
With
where
So you have
So you end up with a solution
but since
so that the general solution is