i
saale pcm waale
it wasmonth ago 
saale pcm waale
3 Likes
Jee bhi nikala hai lmao
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waiting for gradHappy right now(not real)
so ja 1 baj rhe hai
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Pichle October se sleep cycle non existent ho gayi thi, not gonna sleep till 2 lol
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mai toh chala chain churane
i
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All I have to say is 
!
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Good job! Now, how about this to everyone:
x = 2
Find the value of a if 2z = 4.16.
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first 100 digits of square root of 2 but you cant look them up
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QUESTION: \lim_{x\rightarrow\infty}\frac{e^{8x}-1}{x}-((\lim_{x\rightarrow\infty}\frac{x^2-25}{x+5})+\prod^3_{k=3}k)
\displaystyle{
\lim_{x\rightarrow\infty}\frac{e^{8x}-1}{x}
\\
\\\lim_{x\rightarrow\infty}\frac{\frac{d}{dx}[e^{8x}-1]}{\frac{d}{dx}[x]} \qquad\text{Apply L'Hopital's Rule}
\\\lim_{x\rightarrow\infty}\frac{8e^{8x}}{1}
\\8\lim_{x\rightarrow\infty}e^{8x}=8\cdot\infty=\infty
}
\displaystyle{
\lim_{x\rightarrow\infty}\frac{x^2-25}{x+5}
\\\lim_{x\rightarrow\infty}\frac{(x+5)(x-5)}{x+5}
\\
\lim_{x\rightarrow\infty}x-5=\infty
}
\displaystyle{
\prod^3_{k=3}k=1\cdot2\cdot3=6
}
\displaystyle{
\infty-(\infty+6)=\text{Undefined}
}
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I don’t really get this question.
x=2\\ z=\frac{4.16}{2}=2.08\\ a=?
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also known as the L hospital rule (we all did this in calc to annoy our teacher xD… then we dubbed plus minus plinus and our teacher started questioning our sanity)
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Exactly. But I will give you a hint:
Alphabet letters correspond to what x should be divided by (e.g. b should be divided by 2 because it is the 2nd letter of the alphabet).
Basically, divide 2.08 by __ to find a.
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Bruh, this ain’t just math then. This is some cryptology shit idk about.
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z is the 26th letter of the alphabet, so you divide 2.08 by 26 to find a = 0.08.
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damn i didnt know The Hospital had a rule
also welcome back fad :]
1 Like
QUESTION: \int^{\tan x}_{\frac{1}{e}}\frac{t}{1+t^2} \ dt+\int^{\cot x}_{\frac{1}{e}}\frac{1}{t(1+t^2)} \ dt=?
\text{Computing Antiderivative} \\
\int\frac{t}{1+t^2}dt
\\ \text{Substitution}
\\ u=1+t^2 \rightarrow du=2t \ dt
\\ \frac{1}{2}\int\frac{1}{u} \ du
\\ \frac{1}{2}(\ln(u))
\\ \text{Undoing Substitution}
\\ \frac{\ln(1+t^2)}{2}
\\ \text{Applying Limits}
\\ [\frac{\ln(1+t^2)}{2}]^{\tan x}_{\frac{1}{e}}=\frac{1}{2}(\ln (1+\tan^2(x))-\ln(1+\frac{1}{e^2}))
\\ \frac{1}{2}(\ln (\sec^2(x))-\ln(1+e^{-2}))
\text{Computing Antiderivative} \\
\int\frac{1}{t(1+t^2)} \ dt
\\ \text{Partial Fraction Decomposition}
\\ \frac{1}{t(1+t^2)}=\frac{A}{t}+\frac{B}{1+t^2}
\\ 1=A(1+t^2)+Bt
\\ t=0 \rightarrow A=1
\\ \therefore B=-t
\\ \frac{1}{t(1+t^2)}=\frac{1}{t}-\frac{t}{1+t^2}
\\ \int\frac{1}{t}-\frac{t}{1+t^2} \ dt=\int\frac{1}{t} \ dt - \int\frac{t}{1+t^2} \ dt
\int\frac{1}{t} \ dt=\ln(t)
\int\frac{t}{1+t^2} \ dt
\\ \text{Substitution}
\\ u=1+t^2 \rightarrow du=2t \ dt
\\ \frac{1}{2}\int\frac{1}{u} \ du
\\ \frac{1}{2}(\ln(u))
\\ \text{Undoing Substitution}
\\ \frac{\ln(1+t^2)}{2}
\ln(t)- \frac{\ln(1+t^2)}{2}
\\ \text{Applying Limits}
\\ [\ln(t)- \frac{\ln(1+t^2)}{2}]^{\cot x}_\frac{1}{e}
\\ \frac{1}{2}(\ln(1+e^{-2})-\ln(\csc^2x))+\ln|\cot x|+1
\text{Answer}
\\ \frac{1}{2}(\ln (\sec^2(x))-\ln(1+e^{-2}))+\frac{1}{2}(\ln(1+e^{-2})-\ln(\csc^2x))+\ln|\cot x|+1
\\ \underline{\frac{1}{2}(\ln(\sec^2x)-\ln(\csc^2x))+\ln|\cot x|+1}
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I didn’t think someone will do it, but yeh
Send trade request, ign is matter
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I solved the equation just cause. You can keep the reward.
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Daisy had 5 apples. She gave 2 to David. Calculate the size of the moon and multiply it by your receding hairline divided by how much you’ve seen your dad (0) and subtact and add to the Size of the Sun before World War II by adding with subtraction.
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