I'm pretty sure he/she copied it from somewhere, I think it should be \(16c^4-16c^{12}. \)
yeah thx, i got it, its 120
Well we can simplify both of them starting with √50.
\(\sqrt{50}\)
\(=\sqrt{2\cdot \:5^2}\)
\(=\sqrt{2}\sqrt{5^2}\)
\(=5\sqrt{2}\)
\(\sqrt{18}\)
\(=\sqrt{2\cdot \:3^2}\)
\(=\sqrt{2}\sqrt{3^2}\)
\(=3\sqrt{2}\)
So adding \(3\sqrt{2}+5\sqrt{2},\) you get \(8\sqrt{2}\)
oh thats wut u mean, lemme fix it
i'm kinda sick but one of my relative answered it here:
https://www.quora.com/What-is-the-number-of-ways-of-distributing-4-identical-balls-among-4-different-boxes
first of all, welcome to webcalc 2.0!
1. solve this system of equations: x+y = 10, xy = 16, you get \(x=8, \: y=2 \). 8 - 2 = 6. keep in mind that you could do 2 - 8 and get -6 as well.
idk if this is right, here's my best shot.
this is just a system of equations:
\(\begin{pmatrix}x=5+\sqrt{69},\:&y=5-\sqrt{69}\\ x=5-\sqrt{69},\:&y=5+\sqrt{69}\end{pmatrix}\)
\(\frac{y}{3}+1=\frac{y+3}{y}\)
\(y^2+3y=3\left(y+3\right)\) (multiply both sides by 3y)
\(y^2+3y=3y+9\) (distribute right side)
\(y^2=9\) (subtract both sides by 3y)
\(y=3,\:y=-3\) (final answer)
