#\sum_ {n=2}^infty \frac {y^n} {n}# and #\sum_ {n=2}^infty \frac {y^n} {n^2}# have the same convergence radius, #abs y < 1# then we conclude that

If Y is normally distributed s stated, then: Y ~ N (mu, sigma^2) We are given that the mean is equal to five times the standard deviation. So: mu = 5sigma ....

Apr 26, 2018 · a^ (m-n)b^ (n-l)c^ (l-m) = (x^ (m+n)y^l)^ (m-n) (x^ (n+l)y^m)^ (n-l) (x^ (l+m)y^n)^ (l-m) =x^ ( (m+n) (m-n)+ (n+l) (n-l)+ (l+m) (l-m) times y^ (l (m-n)+m (n-l)+n (l-m)) =x^ (m^2 …

The question is malformed: Further, if we we work back from the given DE: x^2 y (d^2y)/ (dx^2) = n (n+1)y We either get a trivial solution y=0 or we have: x^2 (d^2y)/ (dx^2) = n (n+1) Yielding …

Please see my attempted proof below. (I have also used some other theorems from real analysis and functional analysis in order to prove this particular theorem, so please ask me if you want …

See a solution process below: First, let's call the number of millimeters of 10% solution we are looking for: n And, let's call the number of millimeters of 25% solution we are looking for: y We …

See below. (e^(x-y)cosy+cosx+2e^(2x-y))dx + (sinx-e^(x-y)siny+3y^2e^-y)dy We know that e^y ne 0 so we can simplify the former equation to obtain (e^x cosy+e^y cosx+2e^(2x))dx + (e^y sinx …

Then the zeros of our original quartic are given by: y_n = t_n-2 Hence we find a factorisation: y^4+8y^3+12 = (y-y_1) (y-y_2) (y-y_3) (y-y_4) where y_1, y_2, y_3, y_4 are the four zeros.

Oct 6, 2023 · BTS' Golden Maknae Jungkook&nbsp;continues to prepare for the release of his solo album 'Golden.'On October 7 at midnight KST, Jungkook treated…

Jan 22, 2019 · 'Shipping' is not something new in 2019 since it has been around for years. However, before I make my point, I will try to give a specific defini…