Woooow. Impressive answer, Thanks for helping, Guest #1 :D
Nope.
Proof for differentiability:
If for some a, \(\displaystyle\lim_{h\rightarrow 0}\dfrac{f(a+h)-f(a)}{h}\) exists, then f(x) is differentiable at the Cartesian point (a, f(a))
528/4 = 500/4 + 28/4 = 125 + 7 = 132
Thanks Melody :)
4X = 38, X = 9.5
8y = 57, y = 7.125
X times y = 67.6875
I know him...... But this is very urgent that I need to submit the homework tomorrow(about 9 hours later for you, and I have to sleep.)...... So I cannot wait for him to be online for a whole day.......
Your answer is not quite useful though...... But anyways thanks for attempting to help :)
\(\sin^2 x + \cos^2x = 1\\ 1 + \tan^2x = \sec^2x\\ 1+\cot^2x =\csc^2x\\ \cos(-z) = \cos z\\ \sin(-z) = -\sin z\\ \tan(-z) = -\tan z\\ \sin 2x =2\sin x \cos x\\ \cos 2x = 2\cos^2x - 1 = 1 - 2\sin^2x \tan 2x = \dfrac{2\tan x}{1-\tan^2x}\\ \sin^2x = \dfrac{1}{2}(1-\cos 2x)\\ \cos^2x = \dfrac{1}{2}(1+\cos 2x)\\ \cos(x + y) = \cos x \cos y - \sin x \sin y\\ \sin(x + y) = \sin x \cos y + \cos x \sin y\\ \tan(x+ y) = \dfrac{\tan x +\tan y}{1-\tan x \tan y}\\ \sin x - \sin y = 2\cos(\dfrac{x+y}{2})\sin)(\dfrac{x-y}{2})\\ \cos x - \cos y = -2\sin(\dfrac{x+y}{2})\sin(\dfrac{x-y}{2})\\ \cos x\cos y = \dfrac{1}{2} (\cos(x + y)+\cos(x-y))\\ \sin x \sin y = \dfrac{1}{2}(\cos(x-y)-\cos(x+y))\\ \cos x \sin y = \dfrac{1}{2}(\sin(x+y)-\sin(x-y))\)
four plus four is eight!!
or if you are using octal numbers then it is ten.
Therefore 4 + 4 = 10 LOL
me? :)
