Let

Differentiating twice gives


When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.
Substitute these into the given differential equation:


Then the coefficients in the power series solution are governed by the recurrence relation,

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.
• If n is even, then n = 2k for some integer k ≥ 0. Then




It should be easy enough to see that

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then




so that

So, the overall series solution is


The number that equals 280 tenths + 19 thousandths is 28.019
<h3>How to determine the number?</h3>
The statement is given as:
280 tenths + 19 thousandths
Rewrite the terms of the expression as fractions
So, we have
280/10 + 19/1000
Evaluate the quotients
28.0 + 0.019
Add both numbers
28.019
Hence, the number that equals 280 tenths + 19 thousandths is 28.019
Read more about expressions at:
brainly.com/question/723406
#SPJ1
You have not given us any of the steps that Ricardo took to simplify the
expression, and you also haven't given us the list of choices that includes
the description of his mistake, so you're batting O for two so far.
Other than those minor details, the question is intriguing, and it certainly
draws me in.
If Ricardo made a mistake in simplifying that expression, I'm going to say that
it was most likely in the process of removing the parentheses in the middle.
Now you understand that this is all guess-work, because of all the stuff that you
left out when you copied the question, but I think he probably forgot that the 3x
operates on everything inside the parentheses.
He probably wrote that 3x (x-3) is
either 3x² - 3
or x - 9x .
In reality, when properly simplified,
3x (x - 3) = 3x² - 9x .
Answer:
The answer is D, 80$ for three 1 hour lessons. I hope this helps :D
Well combine:
2x + 3x + x + 60 = 180
combine like terms:
6x + 60 =180
then move the 60:
6x = -120
now divide both sides by 6:
x= -20
i hope this helped! :)