W=15 no
W= 21 no
W= 1 Yes
If im right the question is just asking which variable is less than 15.
so the only variable that is less than 15 is 1.
I'll assume the ODE is actually

Look for a series solution centered at
, with



with
and
.
Substituting the series into the ODE gives





- If
for integers
, then




and so on, with

- If
, we have
for all
because
causes every odd-indexed coefficient to vanish.
So we have

Recall that

The solution we found can then be written as


Answer:
32°
Step-by-step explanation:
16 × 2 = 32
Keywords:
Bisected; means to divide into two parts.
Mr. Trevor's laptop as he has 400 minutes compared to Mrs Lanes 360 minuted.(6x60=360). So, Mr. Trevor had a total time of 6 hours and 40 minutes.