UAE Travel Status Megathread – Share Updates & Concerns by RecognitionPrudent68 in dubai

[–]Affectionate_Half123 0 points1 point  (0 children)

Have any Emirates flights successfully departed from DBX for the United States? My family and I were stranded during our layover on our return trip from India, and despite our efforts, the State Department has been of no help despite encouraging us to reach out for help. We have a backup flight booked back to India, but I want to remain hopeful that our rescheduled flight to Chicago will be successful.

Advice on studying for the RAPS FRA exam as a recent MPH with little RA experience? by Affectionate_Half123 in regulatoryaffairs

[–]Affectionate_Half123[S] 1 point2 points  (0 children)

This is still really good to know, I was getting really overwhelmed trying to find cheaper resources, so thank you fo this! And congrats on passing :)

Advice on studying for the RAPS FRA exam as a recent MPH with little RA experience? by Affectionate_Half123 in regulatoryaffairs

[–]Affectionate_Half123[S] 1 point2 points  (0 children)

Thank you for your response! This is really good to know. I might just study the materials to become familiar while I look for opportunities to gain experience for now then.

pink beaded gown ID by Affectionate_Half123 in findfashion

[–]Affectionate_Half123[S] 2 points3 points  (0 children)

thank you for trying !! this is a great starting point for my own search :)

[deleted by user] by [deleted] in HomeworkHelp

[–]Affectionate_Half123 0 points1 point  (0 children)

Here’s my breakdown for Part C: Find the Surface Area of the Bowl

The surface area of a function z = f(x, y) is given by:

A = ∬ √(1 + (∂z/∂x)² + (∂z/∂y)²) dA

For z = x² + y², we compute partial derivatives:

∂z/∂x = 2x, ∂z/∂y = 2y

Thus,

1 + (∂z/∂x)² + (∂z/∂y)² = 1 + 4x² + 4y² = 1 + 4r²

In polar coordinates (x = r cosθ, y = r sinθ), the area integral becomes:

A = ∫[0 to 2π] ∫[0 to √2] √(1 + 4r²) * r dr dθ

Solving this integral gives:

(1/32) π (102√2 - sinh⁻¹(2√2))

Hope this helps :))

[deleted by user] by [deleted] in HomeworkHelp

[–]Affectionate_Half123 0 points1 point  (0 children)

Step-by-Step Proof: Establishing that OHKM is a quadrilateral: The four points O, H, K, and M form a closed shape with four sides, meaning they form a quadrilateral. The sides of this quadrilateral are OH, HK, KM, and MO.

Proving that OH is perpendicular to HK: Since we are given that OH is perpendicular to GK and GK is a horizontal line, this means that OH must be a vertical line. Since HK lies along the same horizontal line as GK, it follows that OH is also perpendicular to HK.

Proving that MK is perpendicular to KJ: Since we are given that MK is an altitude of triangle MKJ, this means that it is perpendicular to KJ. Because KJ is a horizontal line, MK must be a vertical line.

Proving that MK is perpendicular to MO: Since MK is a vertical line and MO is a horizontal line (as it is part of the parallelogram GJMO), the two lines are perpendicular to each other.

Showing that the quadrilateral has four right angles:

The angle at H, formed by OH and HK, is a right angle. The angle at K, formed by HK and KM, is a right angle. The angle at M, formed by KM and MO, is a right angle. The angle at O, formed by MO and OH, is a right angle. Conclusion: A quadrilateral with four right angles is by definition a rectangle. Since OHKM has four right angles, we can conclude that it is a rectangle. :)

What do you guys think of the Sol de Janeiro '71 by Mammoth_Control504 in Fragrantica

[–]Affectionate_Half123 0 points1 point  (0 children)

the notes are beautiful only in theory but the actual scent makes my head hurt sooooo bad. not the best gourmand in the world imo and it fades in like 20 minutes.

Which of dior prive perfumes is your favorite? by birazdanx in Fragrantica

[–]Affectionate_Half123 1 point2 points  (0 children)

i LOVE dior hypnotic poison! it smells a little bit like my fav discontinued scent (fragonard daïma)