![Nanoscale Phase Segregation in Supramolecular π-Templating for Hybrid Perovskite Photovoltaics from NMR Crystallography | Journal of the American Chemical Society Nanoscale Phase Segregation in Supramolecular π-Templating for Hybrid Perovskite Photovoltaics from NMR Crystallography | Journal of the American Chemical Society](https://pubs.acs.org/cms/10.1021/jacs.0c11563/asset/images/medium/ja0c11563_0008.gif)
Nanoscale Phase Segregation in Supramolecular π-Templating for Hybrid Perovskite Photovoltaics from NMR Crystallography | Journal of the American Chemical Society
![Let f : (0, pi)→ R be a twice differentiable function such that limit t→x f(x)sint - f(t)sinxt - x = sin^2x for all xepsilon (0, pi) .If f (pi6) = - Let f : (0, pi)→ R be a twice differentiable function such that limit t→x f(x)sint - f(t)sinxt - x = sin^2x for all xepsilon (0, pi) .If f (pi6) = -](https://dwes9vv9u0550.cloudfront.net/images/1106744/a5c369a6-9df2-476b-92fe-d0c82c9b64c0.jpg)
Let f : (0, pi)→ R be a twice differentiable function such that limit t→x f(x)sint - f(t)sinxt - x = sin^2x for all xepsilon (0, pi) .If f (pi6) = -
![If f (x) and g (x) are differentiable functions for 0< x< 1 such that f (0) = 2, g (0) = 0, f (1) = 6, g (1) = 2 , then in the interval (0,1) If f (x) and g (x) are differentiable functions for 0< x< 1 such that f (0) = 2, g (0) = 0, f (1) = 6, g (1) = 2 , then in the interval (0,1)](https://dwes9vv9u0550.cloudfront.net/images/683762/dd45f7d6-aad5-4910-b34c-f5a8587f4417.jpg)
If f (x) and g (x) are differentiable functions for 0< x< 1 such that f (0) = 2, g (0) = 0, f (1) = 6, g (1) = 2 , then in the interval (0,1)
![Let f:R -> ( 0,(2pi)/2] defined as f(x) = cot^-1 (x^2-4x + alpha) Then the smallest integral value of alpha such that, f(x) is into function is Let f:R -> ( 0,(2pi)/2] defined as f(x) = cot^-1 (x^2-4x + alpha) Then the smallest integral value of alpha such that, f(x) is into function is](https://d10lpgp6xz60nq.cloudfront.net/ss/web/1668560.jpg)
Let f:R -> ( 0,(2pi)/2] defined as f(x) = cot^-1 (x^2-4x + alpha) Then the smallest integral value of alpha such that, f(x) is into function is
The expression 3 [ sin ^4 { 3pi2 - alpha } + sin ^4 (3pi + alpha ) ] - 2 [ sin ^6 ( pi2 + alpha ) . + sin ^6 (5pi - alpha )] is equal to :
![Match the function with its graph. State the period of the function. [The graphs are labeled (a), (b), (c), (d), (e), and (f).] \qquad y = -2\sec(\pi x/2) \, (a) src='6682280-556417884486163178157.png' alt='' Match the function with its graph. State the period of the function. [The graphs are labeled (a), (b), (c), (d), (e), and (f).] \qquad y = -2\sec(\pi x/2) \, (a) src='6682280-556417884486163178157.png' alt=''](https://homework.study.com/cimages/multimages/16/6682280-556417884486163178157.png)
Match the function with its graph. State the period of the function. [The graphs are labeled (a), (b), (c), (d), (e), and (f).] \qquad y = -2\sec(\pi x/2) \, (a) src='6682280-556417884486163178157.png' alt=''
![Unitary transformation for Poincaré beams on different parts of Poincaré sphere | Scientific Reports Unitary transformation for Poincaré beams on different parts of Poincaré sphere | Scientific Reports](https://media.springernature.com/lw685/springer-static/image/art%3A10.1038%2Fs41598-020-71189-2/MediaObjects/41598_2020_71189_Fig5_HTML.png)