Kimberly got 6 problems wrong on a test of 25 questions. What percent of the questions did she got correct?

Mathematics

2 answers:

Licemer1 [7]3 years ago

6 0

Answer:75%

Step-by-step explanation:

SCORPION-xisa [38]3 years ago

6 0

6/25 wrong
19/25
correct
19/25= n/100
cross multiply
25n= 19(100)
25n=1900
divide both sides by 25
n= 76%

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Pls just give coordinates i will give brainliest

MArishka [77]

Answer:

(-2,2) see attached diagram

Step-by-step explanation:

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8 0

2 years ago

Using power series, solve the LDE: (2x^2 + 1) y" + 2xy' - 4x² y = 0 --- - -- -

sattari [20]

We're looking for a solution of the form

$y=\displaystyle\sum_{n\ge0}a_nx^n$

with derivatives

$y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n$

$y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n$

Substituting these into the ODE gives

$\displaystyle\sum_{n\ge0}\left(\bigg(2(n+2)(n+1)a_{n+2}-4a_n\bigg)x^{n+2}+2(n+1)a_{n+1}x^{n+1}+(n+2)(n+1)a_{n+2}x^n\right)=0$

Shifting indices to get each term in the summand to start at the same power of $x$ and pulling the first few terms of the resulting shifted series as needed gives

$2a_2+(2a_1+6a_3)x+\displaystyle\sum_{n\ge2}\bigg((n+2)(n+1)a_{n+2}+2n^2a_n-4a_{n-2}\bigg)x^n=0$

Then the coefficients in the series solution are given according to the recurrence

$\begin{cases}a_0=y(0)\\\\a_1=y'(0)\\\\a_2=0\\\\2a_1+6a_3=0\implies a_3=-\dfrac{a_1}3\\\\a_n=\dfrac{-2(n-2)^2a_{n-2}+4a_{n-4}}{n(n-1)}&\text{for }n\ge4\end{cases}$

Given the complexity of this recursive definition, it's unlikely that you'll be able to find an exact solution to this recurrence. (You're welcome to try. I've learned this the hard way on scratch paper.) So instead of trying to do that, you can compute the first few coefficients to find an approximate solution. I got, assuming initial values of $y(0)=y'(0)=1$ , a degree-8 approximation of

$y(x)\approx1+x-\dfrac{x^3}3+\dfrac{x^4}3+\dfrac{x^5}2-\dfrac{16x^6}{45}-\dfrac{79x^7}{125}+\dfrac{101x^8}{210}$

Attached are plots of the exact (blue) and series (orange) solutions with increasing degree (3, 4, 5, and 65) and the aforementioned initial values to demonstrate that the series solution converges to the exact one (over whichever interval the series converges, that is).