emma spent 42 hours on math and language arts homework last month. She spent about 29% of this total time on math estimate about
1 answer:
You might be interested in
Answer:
- L(t) = 727.775 -51.875cos(2π(t +11)/365)
- 705.93 minutes
Step-by-step explanation:
a) The midline of the function is the average of the peak values:
(675.85 +779.60)/2 = 727.725 . . . minutes
The amplitude of the function is half the difference of the peak values:
(779.60 -675.85)/2 = 51.875 . . . minutes
Since the minimum of the function is closest to the origin, we choose to use the negative cosine function as the parent function.
Where t is the number of days from 1 January, we want to shift the graph 11 units to the left, so we will use (t+11) in our function definition.
Since the period is 365 days, we will use (2π/365) as the scale factor for the argument of the cosine function.
Our formula is ...
L(t) = 727.775 -51.875cos(2π(t +11)/365)
__
b) L(55) ≈ 705.93 minutes
By letting
we get derivatives
a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to
Examine the lowest degree term , which gives rise to the indicial equation,
with roots at r = 0 and r = 4/5.
b) The recurrence for the coefficients is
so that with r = 4/5, the coefficients are governed by
c) Starting with , we find
so that the first three terms of the solution are
Remark
If you don't start exactly the right way, you can get into all kinds of trouble. This is just one of those cases. I think the best way to start is to divide both terms by x^(1/2)
Step One
Divide both terms in the numerator by x^(1/2)
y= 6x^(1/2) + 3x^(5/2 - 1/2)
y =6x^(1/2) + 3x^(4/2)
y = 6x^(1/2) + 3x^2 Now differentiate that. It should be much easier.
Step Two
Differentiate the y in the last step.
y' = 6(1/2) x^(- 1/2) + 3*2 x^(2 - 1)
y' = 3x^(-1/2) + 6x I wonder if there's anything else you can do to this. If there is, I don't see it.
I suppose this is possible.
y' = 3/x^(1/2) + 6x
y' =
Frankly I like the first answer better, but you have a choice of both.
If you don't start exactly the right way, you can get into all kinds of trouble. This is just one of those cases. I think the best way to start is to divide both terms by x^(1/2)
Step One
Divide both terms in the numerator by x^(1/2)
y= 6x^(1/2) + 3x^(5/2 - 1/2)
y =6x^(1/2) + 3x^(4/2)
y = 6x^(1/2) + 3x^2 Now differentiate that. It should be much easier.
Step Two
Differentiate the y in the last step.
y' = 6(1/2) x^(- 1/2) + 3*2 x^(2 - 1)
y' = 3x^(-1/2) + 6x I wonder if there's anything else you can do to this. If there is, I don't see it.
I suppose this is possible.
y' = 3/x^(1/2) + 6x
y' =
Frankly I like the first answer better, but you have a choice of both.