#latex
1 messages · Page 2 of 1
Parallax Error
$$\begin{equation}z_1=r_1e^{i\theta},\,z_2=r_2e^{i\theta}, z_1z_2=r_1e^{i\theta}\times r_2e^{i\theta}=r_1r_2e^{i(\theta_1+\theta_2)}\:and\:\frac{z_1}{z_2}=\frac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}}=\frac{r_1}{r_2}e^{i(\theta_1-\theta_2)}\end{equation}$$
```Compilation error:```! LaTeX Error: Bad math environment delimiter.
See the LaTeX manual or LaTeX Companion for explanation.
Type H <return> for immediate help.
...
l.57 $$\begin{equation}
z_1=r_1e^{i\theta},\,z_2=r_2e^{i\theta}, z_1z_2=r_1e^...
Your command was ignored.
Type I <command> <return> to replace it with another command,```
huh
$$|z_1z_2|=r_1r_2=|z_1||z_2|:\textnormal{and}:arg(z_1z_2)=\theta_1+\theta_2=\argz_1+\argz_2$$
Parallax Error
$$|z_1z_2|=r_1r_2=|z_1||z_2|\:\textnormal{and}\:arg(z_1z_2)=\theta_1+\theta_2=\argz_1+\argz_2$$
```Compilation error:```! Undefined control sequence.
l.57 ...{and}\:arg(z_1z_2)=\theta_1+\theta_2=\argz
_1+\argz_2$$
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.```
$$|z_1z_2|=r_1r_2=|z_1||z_2|:\textnormal{and}:\arg(z_1z_2)=\theta_1+\theta_2=\arg z_1+\arg z_2$$
Parallax Error
e=mc2
$$e-mc2
$$|z_1z_2|=r_1r_2=|z_1||z_2|:\textnormal{and}:\arg(z_1z_2)=\theta_1+\theta_2=\arg z_1+\arg z_2$$
Sornio
$$e=mc2$$
Sornio
$$e=1010000002$$
Sornio
$$e=10*1000000*2$$
$$E=mc^2$$
Parallax Error
$y=mx+c$
Legend
I can't DM you! Do you have DMs disabled?
$y = mx + c$
Sheikh NotPixel Ibn sululu
$10^5$
pcb

$
$
pcb
Sheikh Aboola Al Thani
$$
Sheikh Aboola Al Thani
Sheikh Aboola Al Thani
$$|z_1z_2|=r_1r_2=|z_1||z_2|:\textnormal{and}:arg(z_1z_2)=\theta_1+\theta_2=\argz_1+\argz_2$$
```Compilation error:```! Undefined control sequence.
l.57 ...l{and}:arg(z_1z_2)=\theta_1+\theta_2=\argz
_1+\argz_2$$
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.```
$$\begin{equation}z_1=r_1e^{i\theta},,z_2=r_2e^{i\theta}, z_1z_2=r_1e^{i\theta}\times r_2e^{i\theta}=r_1r_2e^{i(\theta_1+\theta_2)}:and:\frac{z_1}{z_2}=\frac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}}=\frac{r_1}{r_2}e^{i(\theta_1-\theta_2)}\end{equation}$$
Sheikh Aboola Al Thani
$$\begin{equation}z_1=r_1e^{i\theta},,z_2=r_2e^{i\theta}, z_1z_2=r_1e^{i\theta}\times r_2e^{i\theta}=r_1r_2e^{i(\theta_1+\theta_2)}:and:\frac{z_1}{z_2}=\frac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}}=\frac{r_1}{r_2}e^{i(\theta_1-\theta_2)}\end{equation}$$
```Compilation error:```! LaTeX Error: Bad math environment delimiter.
See the LaTeX manual or LaTeX Companion for explanation.
Type H <return> for immediate help.
...
l.57 $$\begin{equation}
z_1=r_1e^{i\theta},,z_2=r_2e^{i\theta}, z_1z_2=r_1e^{...
Your command was ignored.
Type I <command> <return> to replace it with another command,```
wait do u want me to display the question too?
Display the expression
k
And “problem C” at the top
Problem C
\
$log_3(\tfrac{\sin(2\pi)}{\tan(\pi/3)} + \sum_{n=0}^{10}(\frac{1+\sqrt{1+\sqrt{1}}}{1+(1^{-1} + 1^{1-1} \cdot \cos{(\pi/4)}} - \tfrac{2^{4}-2^{3}}{(1+\sqrt{1})^2)^n)$
damn
nasi
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😭
ill do the working
$\log_3{(\tfrac{0}{\sqrt{3}} + \sum_{n=0}^{\infty}(\frac{1+\sqrt{2}}{1+\sqrt{2}} - \frac{8}{4})^{n})$
nasi
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I need a whole working
It’s okay
$\log_3{(\frac{0}{\sqrt{3}} + \sum_{n=0}^{\infty}(\frac{1+\sqrt{2}}{1+\sqrt{2}} - \frac{8}{4})^{n})}$
skySLAYER
@frosty venture there were 2 issues
1st fraction u had used \tfrac it should be \frac
U forgot to close the 1st curly brackets after the log
Also I reccomend using \left and \right
It's much better
Ull get something like this
$\log_3\left(\tfrac{0}{\sqrt{3}} + \sum_{n=0}^{\infty}(\frac{1+\sqrt{2}}{1+\sqrt{2}} - \frac{8}{4})^{n}\right)$
skySLAYER
@frosty venture
ohhh thanks 🙏
Gave +1 rep to skyslayer_1#0
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$a$
pcb
$sum{i=0}^{\infty}(9*10^-i)$
pcb
pcb
$\sum_{i=0}^{\infty}(9*10^-i)$
pcb
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,calc(1+2)
Result:
3
Please delete @lofty raven
@restive sundial has been banned for nsfw discord links
thanks for letting me know
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Yw
$(\frac{5+6}{2}\ ,\ \frac{-1-7}{2})$
Itachi Uchiha
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$\frac{\sin x}{n} = six$
Stampybest333
$six=6$
Stampybest333
$\therefore \frac{\sin x}{n} = 6$
Stampybest333
I hope this changes your life
barty
epicdude99
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epicdude99
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[ \frac{8}{x} = 2x+1 ]
So,
[\frac{8}{x} = \frac{2x+1}{1} ]
[ 8=2x^2+x ]
[ 0=2x^2+x-8 ]
epicdude99
[ 8/x = 2x+1 ]
damn,
[ 8/x = \frac{2x+1}{1} ]
[ 8=2x^2+x ]
[ 0=2x^2+x-8 ]
Ames AKA Max
\begin{tikzpicture}
\draw
(0, 0) coordinate (A) node[left] {$A$} % a node A at (0,0) with a label
-- (4,0) coordinate (C) node[right] {$C$} % a node C at (4,0) with a label
node[midway, below]{$?$} % a label betwenn A and C
-- (4,2) coordinate (B) node[above right] {$B$} % a node B at (4,2) with a label
node[midway, right]{$4$} % a label between C and B
-- cycle % close the path
node[midway, above, sloped]{10}% a label between B and A
\end{tikzpicture}
epicdude99
\begin{tikzpicture}
\draw
(0, 0) coordinate (A) node[left] {$A$} % a node A at (0,0) with a label
-- (4,0) coordinate (C) node[right] {$C$} % a node C at (4,0) with a label
node[midway, below]{$?$} % a label betwenn A and C
-- (4,2) coordinate (B) node[above right] {$B$} % a node B at (4,2) with a label
node[midway, right]{$4$} % a label between C and B
-- cycle % close the path
node[midway, above, sloped]{10}% a label between B and A
\end{tikzpicture}
```Compilation error:```! Package tikz Error: Giving up on this path. Did you forget a semicolon?.
See the tikz package documentation for explanation.
Type H <return> for immediate help.
...
l.59 \end{tikzpicture}
This error message was generated by an \errmessage
command, so I can't give any explicit help.
Pretend that you're Hercule Poirot: Examine all clues,```
[ \frac{8}{x} = 2x+1 ]
So,
[\frac{8}{x} = \frac{2x+1}{1} ]
[ 8=2x^2+x ]
[ 0=2x^2+x-8 ]
epicdude99
[ \frac{6}{2x} = \frac{56+2x^2}{5} ]
[ \frac{6}{2x} = \frac{56+2x^2}{5} ]
how do you work
$$ sinx= 2 $$
Shadooo
Shadooo
$$ \frac{\sinfuljoyacts}{sin}= x$$
Shadooo
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yooo this is such a cool ass feature
Ikr
But it's frustrating tho 😖
If u like coding and stuff then this shi is definitely for ya
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,tex |z| >= |z - 10|
zx7ly
,tex |x + yi| >= |x + yi - 10|
zx7ly
,tex x^2 + y^2 >= x^2 + y^2 + 100
zx7ly
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,tex 0 >= 100
zx7ly
,tex ( k^2 - 45/360 * pi * (K/2)^2 ) / (k^2)
chmclez
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,tex
Please give me something to compile, for example latex ,tex The solutions to \(x^2 = 1\) are \(x = \pm 1\).See ,help and ,help tex for detailed usage and further examples!
$$ 10/10 $$
helpun
,tex help
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,calc 7/10
Result:
0.7
ahh
@ionic gull also this
Im not sure what I expected from a channel named 'latex,' but it definitely wasn't maths

Yeah same when for the first time I saw it 😭
$$ hi $$
theannoyingpoliticsguy
Zingsuu
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I was expecting gloves and thigh highs
$$ YOYOYOYOYOYOYOYOY $$
hi
,W HIHI
$$ Fart $$
Quin
peak efficiency
Zingsuu
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$a^x$
CoolGuy
$a = -w^2 x$
CoolGuy
Why does it always react with delete symbol when I message? Is there something wrong that I am doing?
helpun
$baddies$
constipated
Sir Gator III
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$$$\begin{align}
y &= ax^n\cdot bx^p \
f = ax^n &\qquad g'= bx^p \qquad\qquad \overbrace{\int f\cdot g' = f\cdot g' -\int f'\cdot g}^{\text{Integration by parts}}\
f' = anx^{n-1} &\qquad g= \int bx^p dx = \frac{b}{p+1}x^{p+1}\
\int (ax^n\cdot bx^p)dx &= ax^n \cdot bx^p - \int anx^{n-1}\cdot \frac{b}{p+1}x^{p+1}\
\int (ax^n\cdot bx^p)dx &= bax^{p+n} - \frac{anb}{(p+1)(n+p+1)}x^{n+p+1} + c
\end{align}$$$
Sir Gator III
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wait what tikz works?
\begin{tikzpicture}[scale=1]
\begin{axis}
\addplot[name path=A,color=blue, very thick, smooth, domain=0:11] {3*x^0.5};
\addplot[name path=B,color={rgb:red,107;green,142;blue,35}, very thick, smooth, domain=0:11] {10-x};
\path[name path=C] (\pgfkeysvalueof{/pgfplots/xmin},0) -- (\pgfkeysvalueof{/pgfplots/xmax},0);
\path[name path=D] (\pgfkeysvalueof{/pgfplots/xmin},0) -- (\pgfkeysvalueof{/pgfplots/xmax},0);
\addplot [blue!30] fill between [
of=A and C,soft clip={domain=0:4},
];
\addplot [blue!30] fill between [
of=B and D,soft clip={domain=4:10},
];
\end{axis}
\end{tikzpicture}
Sir Gator III
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dang ig axis drawings don't work
\begin{tikzpicture}
\draw
(0,0) -- (1,0);
\draw[x={(2cm,0.5cm)},color=red] (0,0) -- (1,0);
\end{tikzpicture}
Sir Gator III
$fuck_chemistry$
Tamas
$fuck{chemistry}$
shadoo
$\text{why f chem? I like chem....}$
Adam
$no$
Sheikh Maxina Bin Him
$$$\begin{align}
y &= ax^n\cdot bx^p
f = ax^n &\qquad g'= bx^p \qquad\qquad \overbrace{\int f\cdot g' = f\cdot g' -\int f'\cdot g}^{\text{Integration by parts}}
f' = anx^{n-1} &\qquad g= \int bx^p dx = \frac{b}{p+1}x^{p+1}
\int (ax^n\cdot bx^p)dx &= ax^n \cdot bx^p - \int anx^{n-1}\cdot \frac{b}{p+1}x^{p+1}
\int (ax^n\cdot bx^p)dx &= bax^{p+n} - \frac{anb}{(p+1)(n+p+1)}x^{n+p+1} + c
\end{align}$$$
.help
$$\begin{align}
y &= ax^n\cdot bx^p
f = ax^n &\qquad g'= bx^p \qquad\qquad \overbrace{\int f\cdot g' = f\cdot g' -\int f'\cdot g}^{\text{Integration by parts}}
f' = anx^{n-1} &\qquad g= \int bx^p dx = \frac{b}{p+1}x^{p+1}
\int (ax^n\cdot bx^p)dx &= ax^n \cdot bx^p - \int anx^{n-1}\cdot \frac{b}{p+1}x^{p+1}
\int (ax^n\cdot bx^p)dx &= bax^{p+n} - \frac{anb}{(p+1)(n+p+1)}x^{n+p+1} + c
\end{align}$$
shadoo
$$\begin{align}
y &= ax^n\cdot bx^p \
f = ax^n &\qquad g'= bx^p \qquad\qquad \overbrace{\int f\cdot g' = f\cdot g' -\int f'\cdot g}^{\text{Integration by parts}}\
f' = anx^{n-1} &\qquad g= \int bx^p dx = \frac{b}{p+1}x^{p+1}\
\int (ax^n\cdot bx^p)dx &= ax^n \cdot bx^p - \int anx^{n-1}\cdot \frac{b}{p+1}x^{p+1}\
\int (ax^n\cdot bx^p)dx &= bax^{p+n} - \frac{anb}{(p+1)(n+p+1)}x^{n+p+1} + c
\end{align}$$
```Compilation error:```! Package amsmath Error: Erroneous nesting of equation structures;
(amsmath) trying to recover with `aligned'.
See the amsmath package documentation for explanation.
Type H <return> for immediate help.
...
l.55 \end{align}
$$
Try typing <return> to proceed.
If that doesn't work, type X <return> to quit.```
didn't I send that 💀
,help
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,list
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Render LaTeX code and configure rendering options.
tex: Render LaTeX code.
ctan: Searches the ctan
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$(P + \frac{a}{V^2} (V-b) = nRT$
mush
$E_{Cell} = E^o_{cell} - \frac{RT}{nF} ln (\frac{[Zn^{2+}]{[Cu^{2+}^2]}$
mush
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$\frac{RT}{nF} ln \frac{[Zn^{2+}]} {[Cu^{2+}]^2}$
mush
how does this thing even work?
Real
so bored out of your mind youre checking all the channels?
yes so sbluld i
$\delta$\
Little mishal
Little mishal
Oni
$alpha+beta=\frac{-b}{a}$
shadoo
$\alpha+\beta=\frac{-b}{a}$
Saif
alr ty
Gave +1 rep to saifurrahman!
$\Omega$
Little mishal
$\text{R}=\frac{V}{I}$
Little mishal
Fire
you
you're so low ur not even on the alpha scale
ok buddy
$C_2H_4$
mush
Can anyone tell me how to learn like this things
@torn frigate, can you please tell me
@solid oriole help
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If you still have questions, talk to our friendly support team here.
Render LaTeX code and configure rendering options.
tex: Render LaTeX code.
ctan: Searches the ctan
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texconfig: View or modify your personal LaTeX rendering options.
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,help tex
,, <equations>
,tex <code>
,align <align block>
,texsp <code>
,texw <code>
,tikz <code>
Compiles and displays LaTeX document code. For a quick introduction to using LaTeX, see one of the resources linked below.
The output is extensively configurable, see ,help texconfig for more information about the possible configuration options.
LaTeX macros and packages may also be used in this command via inclusion into the preamble, see ,help preamble for more information.
If a guild or user has latex recognition enabled (see ,config latex and ,help autotex), messages containing LaTeX will automatically be compiled and this command is generally not required.
tex: Code is compiled in the default document environment.
, or mtex: Code is rendered in math mode, in a gather* environment.
align: Code is rendered in math mode, aligned in an align* environment.
texsp: Same as tex, but ||spoiler|| the output image.
texw: Don't pad the output (with transparent pixels) after compilation.
tikz: Code is rendered in a tikzpicture environment.
autotex: Toggle whether your LaTeX is automatically rendered.
preamble: View or modify your LaTeX preamble.
texconfig: View or modify your personal LaTeX rendering options.
Our own LaTeX cheat-sheet
LaTeX Mathematical symbols for undergrads (and everyone else)
Find a LaTeX symbol by drawing it on Detexify
Friendly introduction to mathematical LaTeX, with links
TeX Stackexchange, where every question has been asked before
The LaTeX Support Discord server, origin of the LaTeX Support Network!
,tex This is a fraction: \(\frac{1}{2}\)
,, \int^\infty_0 f(x)~dx
,align a + 1 &= 2\\ a &= 1
,tikz \draw(0,0) circle (1);
Visit our support server here to speak to our friendly support team!
@fringe scarab You may go through these resources
$$ \mathbb{Z} $$
Saif
Wow
$$ \mathbb{ZNOTES}$$
mushancholic
$$\sqrt[3]{27} = 3$$
Melancholic shadoo
@lusty raft, nice to meet you I am from bd. I need help with learning latex
@urban flint, do you know where to learn LaTex
@torn frigate, any resources to learning latex and use it to solve a problem from start to end in softpaper
As-salamu alaykum Ayman, Tomar jonno taza guideline:
-
Install a Distribution
- TeX Live (cross-platform)
- MiKTeX (Windows)
- MacTeX (Mac)
-
Choose an Editor:
- Overleaf (online)
- TeXShop (Mac)
- TeXworks (Windows)
-
Learn the Basics
- Start with a simple document:
\documentclass{article} \begin{document} Hello, World! \end{document} - Learn to compile to PDF
- Start with a simple document:
-
Key Concepts
- Document Structure: Sections, formatting, lists
- Math Typesetting: Use
\( ... \)for inline,\[ ... \]for display. - References: Use
\label{}and\ref{}for citations
-
Resources
- YouTube: Find tutorial videos
-
Practice
- Create simple documents, then add figures, tables, and more
-
Experiment
- Explore packages (e.g.,
amsmath,graphicx) for advanced features
- Explore packages (e.g.,
kurfu
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Watch yt videos
Walikum Salam, thank you @lusty raft, ami shesta korbo. Ar tumi ki kon classe. Ar tumi to onaek bhalo extracurriculars paiso. Tumi ke middle easte. Ar tomaki ami privaletly message de
Gave +1 rep to spyderbillu!
Thank you @ripe niche
Gave +1 rep to dingus_0_0!
How the f you learned then
magic
how do u know about my extracurriculars? no im from Dhaka, sure u can dm me
@lusty raft you there
$$ 1) Sin^2x + Cos^2x = 1$$
$$ Sin^2x - 1 = Cos^2x$$
$$ Cos^2x - 1 = Sin^2x$$
$$ 2) 1 - Sec^2x = Tan^2x$$
$$ 1 + Tan^2x = Sec^2x$$
$$ Sec^2x - Tan^2x = 1$$
$$ 3) Cosec^2x - Cot^2x = 1$$
$$ 1 + Cot^2x = Cosec^2x$$
$$ 1 - Cosec^2x = Cot^2x$$
$$ 4) Sin(A \pm B) = SinACosB \pm CosASinB$$
$$ 5) Sin(2A) = 2SinACosA$$
$$ 6) Cos(A \pm B) = CosACosB \pm SinASinB$$
$$ 7) Cos(2A) = (CosA)^2 - (SinA)^2$$
$$ 8) Tan(A+B) = \frac{TanA \pm TanB}{ 1 \mp TanATanB}$$
$$ 9) Tan(2A) = \frac{2TanA}{1-Tan^2A}$$
$$ 10) Cot(A+B) = \frac{CotACotB-1}{CotB+CotA}$$
$$ 11) Cot(2A) = \frac{Cot^2A+1}{2CotA}$$
$$ 12) Sin(A+B)Sin(A-B) = Sin^2A - Sin^2B (or) Cos^2B - Cos^2A$$
$$ 13) Cos(A+B)Cos(A-B) = Cos^2A - Sin^2B (or) Cos^2B - Sin^2A $$
Melancholic shadoo
$Cl_2 + 2Br^- -> 2Cl^- + Br_2$
mush
$Cl_2 + 2Br^- \to 2Cl^- + Br_2$
BETRAAAYAAAAAL g..
$$ y = 2x+1 -x^2 $$
$$ y = x^2-4x+1 $$
$$ y = 2x^2-8x + 1 $$
$$ x^2 - 4x+1 = 2x+1 -x^2 $$
$$ 2x^2-6x=0 $$
$$ 2x(x-3)=0 $$
$$ x = 0\ or\ 3 $$
$$ 2x^2-8x+1 = x^2-4x+1 $$
$$ x^2-4x =0 $$
$$ x(x-4)=0 $$
$$x=0\ or\ x= 4 $$
BETRAAAYAAAAAL g..
done.
$$ Sin(2) $$
$$ Sin(2) $$
done.
$$ y = 2x+1 -x^2 $$
$$ y = x^2-4x+1 $$
$$ y = 2x^2-8x + 1 $$
$$ x^2 - 4x+1 = 2x+1 -x^2 $$
$$ 2x^2-6x=0 $$
$$ 2x(x-3)=0 $$
$$ x = 0\ or\ 3 $$
$$ 2x^2-8x+1 = x^2-4x+1 $$
$$ x^2-4x =0 $$
$$ x(x-4)=0 $$
$$x=0\ or\ x= 4 $$$$ y = 2x+1 -x^2 $$
$$ y = x^2-4x+1 $$
$$ y = 2x^2-8x + 1 $$
$$ x^2 - 4x+1 = 2x+1 -x^2 $$
$$ 2x^2-6x=0 $$
$$ 2x(x-3)=0 $$
$$ x = 0\ or\ 3 $$
$$ 2x^2-8x+1 = x^2-4x+1 $$
$$ x^2-4x =0 $$
$$ x(x-4)=0 $$
$$x=0\ or\ x= 4 $$$$ y = 2x+1 -x^2 $$
$$ y = x^2-4x+1 $$
$$ y = 2x^2-8x + 1 $$
$$ x^2 - 4x+1 = 2x+1 -x^2 $$
$$ 2x^2-6x=0 $$
$$ 2x(x-3)=0 $$
$$ x = 0\ or\ 3 $$
$$ 2x^2-8x+1 = x^2-4x+1 $$
$$ x^2-4x =0 $$
$$ x(x-4)=0 $$
$$x=0\ or\ x= 4 $$$$ y = 2x+1 -x^2 $$
$$ y = x^2-4x+1 $$
$$ y = 2x^2-8x + 1 $$
$$ x^2 - 4x+1 = 2x+1 -x^2 $$
$$ 2x^2-6x=0 $$
$$ 2x(x-3)=0 $$
$$ x = 0\ or\ 3 $$
$$ 2x^2-8x+1 = x^2-4x+1 $$
$$ x^2-4x =0 $$
$$ x(x-4)=0 $$
$$x=0\ or\ x= 4 $$$$ y = 2x+1 -x^2 $$
$$ y = x^2-4x+1 $$
$$ y = 2x^2-8x + 1 $$
$$ x^2 - 4x+1 = 2x+1 -x^2 $$
$$ 2x^2-6x=0 $$
$$ 2x(x-3)=0 $$
$$ x = 0\ or\ 3 $$
$$ 2x^2...
$$i^2$$
doublerone
$$sig(1/((n^2)(b-1)^2)$$
doublerone
doublerone
$$ sin^2x + cos^2x = 1 $$
$$ cos^2x = 1-sin^2x $$
$$ sin^2x = 1-cos^2x $$
$$ tanx = \frac{sinx}{cosx} $$
$$ cotx = \frac{cosx}{sinx} $$
$$ secx = \frac{1}{cosx} (note: sec\ means\ secant\ and\ cosec\ means\ cosecant) $$
$$ cosecx = \frac{1}{sinx} $$
$$ cotx = \frac{1}{tanx} $$
$$ cot^2x = 1+cosec^2x $$
$$ tan^2x = 1+sec^2x $$
shadoo g..
shadoo g..
?
I can type black listed words using this bot?
$$[ \text{Price Elasticity of Demand (PED)} = \frac{\text{Percentage change in quantity demanded}}{\text{Percentage change in price}} ]$$
Ved Karjagi
\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
\section*{Price Elasticity of Demand (PED)}
\textbf{Price elasticity of demand (PED)} is defined as:
[
\text{PED} = \frac{\text{Percentage change in quantity demanded}}{\text{Percentage change in price}}
]
If we abbreviate change in by the Greek letter $\Delta$, this formula can be rewritten as:
[
\text{PED} = \frac{%\Delta Q}{%\Delta P}
]
Simplifying, the above formula can be rewritten as:
[
\text{PED} = \frac{\frac{\Delta Q}{Q} \times 100}{\frac{\Delta P}{P} \times 100} = \frac{\frac{\Delta Q}{Q}}{\frac{\Delta P}{P}}
]
where:
\begin{itemize}
\item $\Delta Q = Q_{\text{final value}} - Q_{\text{initial value}}$
\item $Q = Q_{\text{initial value}}$
\item $\Delta P = P_{\text{final value}} - P_{\text{initial value}}$
\item $P = P_{\text{initial value}}$
\end{itemize}
\end{document}
Ved Karjagi
Compile Error! Click the
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(You may edit your message to recompile.)
$BRUZZ$
epicdude99
$$i^2 = 2 🙀 $$
doublerone
lol we can enter emojis?
$$🇺🇳🇦🇫🇦🇽🇦🇱🇩🇿🇦🇸🇦🇩🇦🇴🇦🇮🇦🇶🇦🇬🇧🇩🇦🇷🇦🇲🇦🇼🇦🇺🇦🇹🇦🇿🇧🇸🇧🇭🇧🇧🇧🇾🇧🇪🇧🇿🇧🇯🇧🇲🇧🇹🇧🇴🇧🇦🇧🇼🇧🇷🇮🇴🇻🇬🇧🇳🇧🇬🇧🇫🇧🇮🇰🇭🇨🇲🇨🇦🇨🇻🇧🇶🇰🇾🇨🇫🇹🇩🇨🇱🇨🇳🇨🇽🇨🇨🇨🇴🇰🇲🇨🇬🇨🇩🇨🇰🇨🇷🇨🇮🇭🇷🇨🇺🇨🇼🇨🇿🇨🇾🇩🇰🇩🇯🇩🇲🇩🇴🇪🇨🇪🇬🇬🇶🇸🇻🇪🇷🇪🇪🇪🇹🇪🇺🇫🇰🇫🇴🇫🇯🇫🇮🇫🇷🇬🇫🇵🇫🇹🇫🇬🇦🇬🇲🇬🇪🇩🇪🇬🇭🇬🇮🇬🇷🇬🇱🇬🇩🇬🇵🇬🇺🇬🇹🇬🇬🇬🇳🇬🇼🇬🇾🇭🇹🇭🇳🇭🇰🇭🇺🇮🇸🇮🇳🇮🇩🇮🇷🇮🇶🇮🇪🇮🇲🇮🇹🇯🇲🇯🇵🇯🇪🇯🇴🇰🇿🇰🇪🇰🇮🇽🇰🇰🇼 🇰🇬🇱🇦🇱🇻🇱🇧🇱🇸🇱🇷🇱🇾🇱🇮🇱🇹🇲🇴🇲🇰🇲🇬🇲🇼🇲🇾🇲🇻🇲🇱🇲🇹🇲🇭🇲🇷🇲🇶🇲🇺🇾🇹🇲🇽🇫🇲🇲🇩🇲🇨🇲🇳🇲🇪🇲🇸🇲🇦🇲🇿🇲🇲🇳🇦🇳🇷🇳🇵🇳🇱🇳🇨🇳🇿🇳🇮🇳🇪🇳🇬🇳🇺🇳🇫🇰🇵🇲🇵🇳🇴🇴🇲🇵🇰🇵🇼🇵🇸🇵🇦🇵🇬🇵🇾🇵🇪🇵🇭🇵🇳🇵🇱🇵🇹🇵🇷🇶🇦🇷🇪🇷🇴🇷🇼🇷🇺🇼🇸🇸🇹🇸🇲🇸🇦🇸🇳🇷🇸🇸🇨🇸🇱🇸🇬🇸🇰🇸🇽🇸🇮🇬🇸🇸🇧🇸🇴🇿🇦🇰🇷$$
$$The world$$
doublerone
bruhh
omg
$$🇦🇫🇦🇽🇦🇱🇩🇿🇦🇸🇦🇩🇦🇴🇦🇮🇦🇶🇦🇬🇧🇩🇦🇷🇦🇲$$
$$🇦🇺🇦🇹🇦🇿🇧🇸🇧🇭🇧🇧🇧🇾🇧🇪🇧🇿🇧🇯🇧🇲🇧🇹🇧🇴$$
$$🇧🇦🇧🇼🇧🇷🇮🇴🇻🇬🇧🇳🇧🇬🇧🇫🇧🇮🇰🇭🇨🇲🇨🇦🇨🇻$$
$$🇧🇶🇰🇾🇨🇫🇹🇩🇨🇱🇨🇳🇨🇽🇨🇨🇨🇴🇰🇲🇨🇬🇨🇩🇨🇰$$
$$🇨🇷🇨🇮🇭🇷🇨🇺🇨🇼🇨🇿🇨🇾🇩🇰🇩🇯🇩🇲🇩🇴🇪🇨🇪🇬$$
$$🇬🇶🇸🇻🇪🇷🇪🇪🇪🇹🇪🇺🇫🇰🇫🇴🇫🇯🇫🇮🇫🇷🇬🇫🇵🇫$$
$$🇹🇫🇬🇦🇬🇲🇬🇪🇩🇪🇬🇭🇬🇮🇬🇷🇬🇱🇬🇩🇬🇵🇬🇺🇬🇹$$
$$🇬🇬🇬🇳🇬🇼🇬🇾🇭🇹🇭🇳🇭🇰🇭🇺🇮🇸🇮🇳🇮🇩🇮🇷🇮🇶$$
$$🇮🇪🇮🇲🇮🇹🇯🇲🇯🇵🇯🇪🇯🇴🇰🇿🇰🇪🇰🇮🇽🇰🇰🇼🇰🇬$$
$$🇱🇦🇱🇻🇱🇧🇱🇸🇱🇷🇱🇾🇱🇮🇱🇹🇲🇴🇲🇰🇲🇬🇲🇼🇲🇾$$
$$🇲🇻🇲🇱🇲🇹🇲🇭🇲🇷🇲🇶🇲🇺🇾🇹🇲🇽🇫🇲🇲🇩🇲🇨🇲🇳$$
$$🇲🇪🇲🇸🇲🇦🇲🇿🇲🇲🇳🇦🇳🇷🇳🇵🇳🇱🇳🇨🇳🇿🇳🇮🇳🇪$$
$$🇳🇬🇳🇺🇳🇫🇰🇵🇲🇵🇳🇴🇴🇲🇵🇰🇵🇼🇵🇸🇵🇦🇵🇬🇵🇾$$
$$🇵🇪🇵🇭🇵🇳🇵🇱🇵🇹🇵🇷🇶🇦🇷🇪🇷🇴🇷🇼🇷🇺🇼🇸🇸🇹$$
$$🇸🇲🇸🇦🇸🇳🇷🇸🇸🇨🇸🇱🇸🇬🇸🇰🇸🇽🇸🇮🇬🇸🇸🇧🇸🇴$$
$$🇦🇼🇿🇦🇰🇷$$
$$ 🇺🇳 the world $$🇦🇫🇦🇽🇦🇱🇩🇿🇦🇸🇦🇩🇦🇴🇦🇮🇦🇶🇦🇬🇧🇩🇦🇷🇦🇲$$
$$🇦🇺🇦🇹🇦🇿🇧🇸🇧🇭🇧🇧🇧🇾🇧🇪🇧🇿🇧🇯🇧🇲🇧🇹🇧🇴$$
$$🇧🇦🇧🇼🇧🇷🇮🇴🇻🇬🇧🇳🇧🇬🇧🇫🇧🇮🇰🇭🇨🇲🇨🇦🇨🇻$$
$$🇧🇶🇰🇾🇨🇫🇹🇩🇨🇱🇨🇳🇨🇽🇨🇨🇨🇴🇰🇲🇨🇬🇨🇩🇨🇰$$
$$🇨🇷🇨🇮🇭🇷🇨🇺🇨🇼🇨🇿🇨🇾🇩🇰🇩🇯🇩🇲🇩🇴🇪🇨🇪🇬$$
$$🇬🇶🇸🇻🇪🇷🇪🇪🇪🇹🇪🇺🇫🇰🇫🇴🇫🇯🇫🇮🇫🇷🇬🇫🇵🇫$$
$$🇹🇫🇬🇦🇬🇲🇬🇪🇩🇪🇬🇭🇬🇮🇬🇷🇬🇱🇬🇩🇬🇵🇬🇺🇬🇹$$
$$🇬🇬🇬🇳🇬🇼🇬🇾🇭🇹🇭🇳🇭🇰🇭🇺🇮🇸🇮🇳🇮🇩🇮🇷🇮🇶$$
$$🇮🇪🇮🇲🇮🇹🇯🇲🇯🇵🇯🇪🇯🇴🇰🇿🇰🇪🇰🇮🇽🇰🇰🇼🇰🇬$$
$$🇱🇦🇱🇻🇱🇧🇱🇸🇱🇷🇱🇾🇱🇮🇱🇹🇲🇴🇲🇰🇲🇬🇲🇼🇲🇾$$
$$🇲🇻🇲🇱🇲🇹🇲🇭🇲🇷🇲🇶🇲🇺🇾🇹🇲🇽🇫🇲🇲🇩🇲🇨🇲🇳$$
$$🇲🇪🇲🇸🇲🇦🇲🇿🇲🇲🇳🇦🇳🇷🇳🇵🇳🇱🇳🇨🇳🇿🇳🇮🇳🇪$$
$$🇳🇬🇳🇺🇳🇫🇰🇵🇲🇵🇳🇴🇴🇲🇵🇰🇵🇼🇵🇸🇵🇦🇵🇬🇵🇾$$
$$🇵🇪🇵🇭🇵🇳🇵🇱🇵🇹🇵🇷🇶🇦🇷🇪🇷🇴🇷🇼🇷🇺🇼🇸🇸🇹$$
$$🇸🇲🇸🇦🇸🇳🇷🇸🇸🇨🇸🇱🇸🇬🇸🇰🇸🇽🇸🇮🇬🇸🇸🇧🇸🇴$$
$$🇦🇼🇿🇦🇰🇷$$
$$ 🇺🇳 the world 🇺🇳 $$
doublerone
$$🇦🇫🇦🇽🇦🇱🇩🇿🇦🇸🇦🇩🇦🇴🇦🇮🇦🇶🇦🇬🇧🇩🇦🇷🇦🇲$$
$$🇦🇺🇦🇹🇦🇿🇧🇸🇧🇭🇧🇧🇧🇾🇧🇪🇧🇿🇧🇯🇧🇲🇧🇹🇧🇴$$
$$🇧🇦🇧🇼🇧🇷🇮🇴🇻🇬🇧🇳🇧🇬🇧🇫🇧🇮🇰🇭🇨🇲🇨🇦🇨🇻$$
$$🇧🇶🇰🇾🇨🇫🇹🇩🇨🇱🇨🇳🇨🇽🇨🇨🇨🇴🇰🇲🇨🇬🇨🇩🇨🇰$$
$$🇨🇷🇨🇮🇭🇷🇨🇺🇨🇼🇨🇿🇨🇾🇩🇰🇩🇯🇩🇲🇩🇴🇪🇨🇪🇬$$
$$🇬🇶🇸🇻🇪🇷🇪🇪🇪🇹🇪🇺🇫🇰🇫🇴🇫🇯🇫🇮🇫🇷🇬🇫🇵🇫$$
$$🇹🇫🇬🇦🇬🇲🇬🇪🇩🇪🇬🇭🇬🇮🇬🇷🇬🇱🇬🇩🇬🇵🇬🇺🇬🇹$$
$$🇬🇬🇬🇳🇬🇼🇬🇾🇭🇹🇭🇳🇭🇰🇭🇺🇮🇸🇮🇳🇮🇩🇮🇷🇮🇶$$
$$🇮🇪🇮🇲🇮🇹🇯🇲🇯🇵🇯🇪🇯🇴🇰🇿🇰🇪🇰🇮🇽🇰🇰🇼🇰🇬$$
$$🇱🇦🇱🇻🇱🇧🇱🇸🇱🇷🇱🇾🇱🇮🇱🇹🇲🇴🇲🇰🇲🇬🇲🇼🇲🇾$$
$$🇲🇻🇲🇱🇲🇹🇲🇭🇲🇷🇲🇶🇲🇺🇾🇹🇲🇽🇫🇲🇲🇩🇲🇨🇲🇳$$
$$🇲🇪🇲🇸🇲🇦🇲🇿🇲🇲🇳🇦🇳🇷🇳🇵🇳🇱🇳🇨🇳🇿🇳🇮🇳🇪$$
$$🇳🇬🇳🇺🇳🇫🇰🇵🇲🇵🇳🇴🇴🇲🇵🇰🇵🇼🇵🇸🇵🇦🇵🇬🇵🇾$$
$$🇵🇪🇵🇭🇵🇳🇵🇱🇵🇹🇵🇷🇶🇦🇷🇪🇷🇴🇷🇼🇷🇺🇼🇸🇸🇹$$
$$🇸🇲🇸🇦🇸🇳🇷🇸🇸🇨🇸🇱🇸🇬🇸🇰🇸🇽🇸🇮🇬🇸🇸🇧🇸🇴$$
$$🇦🇼🇿🇦🇰🇷$$
$$ 🇺🇳 the world $$🇦🇫🇦🇽🇦🇱🇩🇿🇦🇸🇦🇩🇦🇴🇦🇮🇦🇶🇦🇬🇧🇩🇦🇷🇦🇲$$
$$🇦🇺🇦🇹🇦🇿🇧🇸🇧🇭🇧🇧🇧🇾🇧🇪🇧🇿🇧🇯🇧🇲🇧🇹🇧🇴$$
$$🇧🇦🇧🇼🇧🇷🇮🇴🇻🇬🇧🇳🇧🇬🇧🇫🇧🇮🇰🇭🇨🇲🇨🇦🇨🇻$$
$$🇧🇶🇰🇾🇨🇫🇹🇩🇨🇱🇨🇳🇨🇽🇨🇨🇨🇴🇰🇲🇨🇬🇨🇩🇨🇰$$
$$🇨🇷🇨🇮🇭🇷🇨🇺🇨🇼🇨🇿🇨🇾🇩🇰🇩🇯🇩🇲🇩🇴🇪🇨🇪🇬$$
$$🇬🇶🇸🇻🇪🇷🇪🇪🇪🇹🇪🇺🇫🇰🇫🇴🇫🇯🇫🇮🇫🇷🇬🇫🇵🇫$$
$$🇹🇫🇬🇦🇬🇲🇬🇪🇩🇪🇬🇭🇬🇮🇬🇷🇬🇱🇬🇩🇬🇵🇬🇺🇬🇹$$
$$🇬🇬🇬🇳🇬🇼🇬🇾🇭🇹🇭🇳🇭🇰🇭🇺🇮🇸🇮🇳🇮🇩🇮🇷🇮🇶$$
$$🇮🇪🇮🇲🇮🇹🇯🇲🇯🇵🇯🇪🇯🇴🇰🇿🇰🇪🇰🇮🇽🇰🇰🇼🇰🇬$$
$$🇱🇦🇱🇻🇱🇧🇱🇸🇱🇷🇱🇾🇱🇮🇱🇹🇲🇴🇲🇰🇲🇬🇲🇼🇲🇾$$
$$🇲🇻🇲🇱🇲🇹🇲🇭🇲🇷🇲🇶🇲🇺🇾🇹🇲🇽🇫🇲🇲🇩🇲🇨🇲🇳$$
$$🇲🇪🇲🇸🇲🇦🇲🇿🇲🇲🇳🇦🇳🇷🇳🇵🇳🇱🇳🇨🇳🇿🇳🇮🇳🇪$$
$$🇳🇬🇳🇺🇳🇫🇰🇵🇲🇵🇳🇴🇴🇲🇵🇰🇵🇼🇵🇸🇵🇦🇵🇬🇵🇾$$
$$🇵🇪🇵🇭🇵🇳🇵🇱🇵🇹🇵🇷🇶🇦🇷🇪🇷🇴🇷🇼🇷🇺🇼🇸🇸🇹$$
$$🇸🇲🇸🇦🇸🇳🇷🇸🇸🇨🇸🇱🇸🇬🇸🇰🇸🇽🇸🇮🇬🇸🇸🇧🇸🇴$$
$$🇦🇼🇿🇦🇰🇷$$
$$ 🇺🇳 the world 🇺🇳 $$
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$$🇦🇫🇦🇽🇦🇱🇩🇿🇦🇸🇦🇩🇦🇴🇦🇮🇦🇶🇦🇬🇧🇩🇦🇷🇦🇲$$
$$🇦🇺🇦🇹🇦🇿🇧🇸🇧🇭🇧🇧🇧🇾🇧🇪🇧🇿🇧🇯🇧🇲🇧🇹🇧🇴$$
$$🇧🇦🇧🇼🇧🇷🇮🇴🇻🇬🇧🇳🇧🇬🇧🇫🇧🇮🇰🇭🇨🇲🇨🇦🇨🇻$$
$$🇧🇶🇰🇾🇨🇫🇹🇩🇨🇱🇨🇳🇨🇽🇨🇨🇨🇴🇰🇲🇨🇬🇨🇩🇨🇰$$
$$🇨🇷🇨🇮🇭🇷🇨🇺🇨🇼🇨🇿🇨🇾🇩🇰🇩🇯🇩🇲🇩🇴🇪🇨🇪🇬$$
$$🇬🇶🇸🇻🇪🇷🇪🇪🇪🇹🇪🇺🇫🇰🇫🇴🇫🇯🇫🇮🇫🇷🇬🇫🇵🇫$$
$$🇹🇫🇬🇦🇬🇲🇬🇪🇩🇪🇬🇭🇬🇮🇬🇷🇬🇱🇬🇩🇬🇵🇬🇺🇬🇹$$
$$🇬🇬🇬🇳🇬🇼🇬🇾🇭🇹🇭🇳🇭🇰🇭🇺🇮🇸🇮🇳🇮🇩🇮🇷🇮🇶$$
$$🇮🇪🇮🇲🇮🇹🇯🇲🇯🇵🇯🇪🇯🇴🇰🇿🇰🇪🇰🇮🇽🇰🇰🇼🇰🇬$$
$$🇱🇦🇱🇻🇱🇧🇱🇸🇱🇷🇱🇾🇱🇮🇱🇹🇲🇴🇲🇰🇲🇬🇲🇼🇲🇾$$
$$🇲🇻🇲🇱🇲🇹🇲🇭🇲🇷🇲🇶🇲🇺🇾🇹🇲🇽🇫🇲🇲🇩🇲🇨🇲🇳$$
$$🇲🇪🇲🇸🇲🇦🇲🇿🇲🇲🇳🇦🇳🇷🇳🇵🇳🇱🇳🇨🇳🇿🇳🇮🇳🇪$$
$$🇳🇬🇳🇺🇳🇫🇰🇵🇲🇵🇳🇴🇴🇲🇵🇰🇵🇼🇵🇸🇵🇦🇵🇬🇵🇾$$
$$🇵🇪🇵🇭🇵🇳🇵🇱🇵🇹🇵🇷🇶🇦🇷🇪🇷🇴🇷🇼🇷🇺🇼🇸🇸🇹$$
$$🇸🇲🇸🇦🇸🇳🇷🇸🇸🇨🇸🇱🇸🇬🇸🇰🇸🇽🇸🇮🇬🇸🇸🇧🇸🇴$$
$$🇦🇼🇿🇦🇰🇷$$
$$ 🇺🇳 the world 🇺🇳 $$
doublerone
$$🇳🇬$$
v̵̋̎o̴̾͝ǐ̸̌d̸̙͘.̴̓͋
Racist.
$$ \frac{3^x }{3^4y} = (3^3) * (3^0.5) $$
$$ 3^(x) * 3^(-4y) = 3^3.5 $$
$$ 3^(x-4y) = 3^3.5 $$
$$ x-4y = 3.5 $$
$$ x - 3.5 = 4y $$
$$ \frac{x}{4} - \frac{7}{8} = y $$
duoblereno
breh
$$ \frac{3^x }{3^4y} = (3^3) * 3^(0.5) $$
$$ 3^(x) * 3^(-4y) = 3^(3.5) $$
$$ 3^(x-4y) = 3^(3.5) $$
$$ x-4y = 3.5 $$
$$ x - 3.5 = 4y $$
$$ \frac{x}{4} - \frac{7}{8} = y $$
duoblereno
-_-
$$ \frac{3^x }{3^4y} = (3^3) * 3(^0.5) $$
$$ 3^(x) * 3^(-4y) = 3^(3.5) $$
$$ 3^(x-4y) = 3^(3.5) $$
$$ x-4y = 3.5 $$
$$ x - 3.5 = 4y $$
$$ \frac{x}{4} - \frac{7}{8} = y $$
duoblereno
$$ \frac{3^x }{3^4y} = (3^3) * 3^0.5 $$
$$ 3^(x) * 3^(-4y) = 3^(3.5) $$
$$ 3^(x-4y) = 3^(3.5) $$
$$ x-4y = 3.5 $$
$$ x - 3.5 = 4y $$
$$ \frac{x}{4} - \frac{7}{8} = y $$
duoblereno
omg
$$ \frac{3^x }{3^{4y}} = (3^3) * 3^{0.5} $$
$$ 3^x * 3^{-4y} = 3^{3.5} $$
$$ 3^{x-4y} = 3^{3.5} $$
$$ x-4y = 3.5 $$
$$ x - 3.5 = 4y $$
$$ \frac{x}{4} - \frac{7}{8} = y $$
duoblereno
tf
Alt
how to make under root using this bot?
\sqrt
ty sahdoooooo bhai
Gave +1 rep to .shadoooo!
$\sqrt(4)$
Alt
who here got the allergy
$$ \int \frac{t}{2e} dt = \int \frac{1}{2e} t dt = \frac{1}{2e} \frac{t^2}{2} $$
Geagle
$$ \frac{a+b}{1} $$
dumbledore (apparently)
$$ \frac{\not{a+b}}{1} $$
dumbledore (apparently)
$$ \frac{\not{a} + \not{b}}{1} $$
dumbledore (apparently)
$$ \frac{\not{a} + \not{b}}{\not{a} + \not{b}} $$
dumbledore (apparently)
$$ \frac{\not{a} + \not{b}}{\not{abcdef}} $$
dumbledore (apparently)
$$ \frac{\not{a} + \not{b}}{\not{a}\not{b}\not{c}} $$
dumbledore (apparently)
$$ \frac{\frac{\not{a}}{\not{b}\not{c}d\not{e}}}{ \frac{\not{f}}{\not{g}\not{h}\not{i}}} $$
dumbledore (apparently)
lg
$$ f'(g(x))\cdot g'(x) $$
shingsuu (mimi hate club vp)
dumbledore (apparently)
:/
$$ [ (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots ]
For ((u+2)^{1/2}), we can rewrite it as:
[ (u+2)^{1/2} = 2^{1/2} \left(1 + \frac{u}{2}\right)^{1/2} ]
Using the binomial expansion for (\left(1 + \frac{u}{2}\right)^{1/2}):
[ \left(1 + \frac{u}{2}\right)^{1/2} = 1 + \frac{1}{2} \cdot \frac{u}{2} + \frac{\frac{1}{2} \left(\frac{1}{2} - 1\right)}{2!} \left(\frac{u}{2}\right)^2 + \frac{\frac{1}{2} \left(\frac{1}{2} - 1\right) \left(\frac{1}{2} - 2\right)}{3!} \left(\frac{u}{2}\right)^3 + \cdots ]
Simplifying the terms:
[ \left(1 + \frac{u}{2}\right)^{1/2} = 1 + \frac{u}{4} - \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{u^2}{8} + \frac{1 \cdot (-1) \cdot (-3)}{6} \cdot \frac{u^3}{48} + \cdots ]
[ = 1 + \frac{u}{4} - \frac{u^2}{32} + \frac{u^3}{128} + \cdots ]
Multiplying by (2^{1/2}):
[ (u+2)^{1/2} = \sqrt{2} \left(1 + \frac{u}{4} - \frac{u^2}{32} + \frac{u^3}{128} + \cdots\right) ]
Thus, the expansion of ((u+2)^{1/2}) is:
[ (u+2)^{1/2} = \sqrt{2} + \frac{\sqrt{2}u}{4} - \frac{\sqrt{2}u^2}{32} + \frac{\sqrt{2}u^3}{128} + \cdots ] $$
silver
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$$what the fuck how to you figure this out$$
Velpun
magic
Attention required <@&572709257036955649>
Reason provided:
scam link still remaining
thanks whoever it was (im too lazy to go check)
Twas me lol
thank you :>
Gave +1 rep to oriole1906!
$$ \frac{x(x+4)}{x^2+4} \not= \frac{x}{x^2+4} + \frac{x+4}{x^2+4} $$
Huang
uwu
$$\frac{x(x+4)}{x^2+4}=\frac{Ax+B}{x^2+4}+\frac{C}{x^2+4}$$
aaaaaaaaa
$$\frac{x(x+4)}{x^2+4}=\frac{x^2}{x^2+4}+\frac{4x}{x^2+4}$$
aaaaaaaaa
$$ \int \frac{1}{y} dy = \int dx $$
Huang
Given that $x - \frac{1}{x} = \sqrt{5}$, find the value of $\frac{x^{10} + x^6 + x^4 + 1}{1 + x^2 + x^8 + x^{10}}$
Saif
$$ \left(x - x^{-1} \right)^2 = x^2 - 2 + x^{-2} = \left(\sqrt{5} \right)^2 $$
$$ \fbox{$\therefore x^2 + x^{-2} = 7$} $$
$$ \left(x + x^{-1} \right)^2 = x^2 + 2 + x^{-2} = 7 + 2 = 9 $$
$$ \fbox{$\therefore x + x^{-1} = 3$} $$
$$ \left(x + x^{-1} \right)^3 = x^3 + 3x + 3x^{-1} + x^{-3} = 3^3 $$
$$ x^3 + x^{-3} + 3 \left( x + x^{-1} \right) = 27 \rightarrow x^3 + x^{-3} = 27 - 3(3) $$
$$ \fbox{$\therefore x^3 + x^{-3} = 18$} $$
$$ \left(x + x^{-1} \right)^4 = x^4 + 4x^2 + 6 + 4x^{-2} + x^{-4} = 3^4 $$
$$ x^4 + x^{-4} + 6 + 4 \left( x^2 + x^{-2} \right) = 81 \rightarrow x^4 + x^{-4} = 81 - 6 - 4(7) $$
$$ \fbox{$\therefore x^4 + x^{-4} = 47$} $$
$$ \left(x + x^{-1} \right)^5 = x^5 + 5x^3 + 10x + 10x^{-1} + 5x^{-3} + x^{-5} = 3^5 $$
$$ x^5 + x^{-5} + 5 \left( x^3 + x^{-3} \right) + 10 \left( x + x^{-1} \right) = 243 \rightarrow x^5 + x^{-5} = 243 - 5(18) - 10(3) $$
$$ \fbox{$\therefore x^5+ x^{-5} = 123$} $$
Saif
$$ \left(x - x^{-1} \right)^2 = x^2 - 2 + x^{-2} = \left(\sqrt{5} \right)^2 $$
$$ \fbox{$\therefore x^2 + x^{-2} = 7$} $$
$$ \left(x + x^{-1} \right)^2 = x^2 + 2 + x^{-2} = 7 + 2 = 9 $$
$$ \fbox{$\therefore x + x^{-1} = 3$} $$
$$ \left(x + x^{-1} \right)^3 = x^3 + 3x + 3x^{-1} + x^{-3} = 3^3 $$
$$ x^3 + x^{-3} + 3 \left( x + x^{-1} \right) = 27 \rightarrow x^3 + x^{-3} = 27 - 3(3) $$
$$ \fbox{$\therefore x^3 + x^{-3} = 18$} $$
$$ \left(x + x^{-1} \right)^4 = x^4 + 4x^2 + 6 + 4x^{-2} + x^{-4} = 3^4 $$
$$ x^4 + x^{-4} + 6 + 4 \left( x^2 + x^{-2} \right) = 81 \rightarrow x^4 + x^{-4} = 81 - 6 - 4(7) $$
$$ \fbox{$\therefore x^4 + x^{-4} = 47$} $$
$$ \left(x + x^{-1} \right)^5 = x^5 + 5x^3 + 10x + 10x^{-1} + 5x^{-3} + x^{-5} = 3^5 $$
$$ x^5 + x^{-5} + 5 \left( x^3 + x^{-3} \right) + 10 \left( x + x^{-1} \right) = 243 \rightarrow x^5 + x^{-5} = 243 - 5(18) - 10(3) $$
$$ \fbox{$\therefore x^5+ x^{-5} = 123$} $$
$$ \frac{x^{10} + x^6 + x^4 + 1}{1 + x^2 + x^8 + x^{10}} \rightarrow \frac{\not{x^5} \left( x^5 + x + x^{-1} + x^{-5} \right)}{\not{x^5} \left( x^{-5} + x^{-3} + x^3 + x^5 \right)} $$
$$ \frac{x^5 + x + x^{-1} + x^{-5}}{x^{-5} + x^{-3} + x^3 + x^5} \rightarrow \frac{x^5 + x^{-5} + x + x^{-1}}{x^{-5} + x^5 + x^{-3} + x^3} \rightarrow \frac{123 + 3}{123 + 18} $$
$$ \frac{x^{10} + x^6 + x^4 + 1}{1 + x^2 + x^8 + x^{10}} = \frac{126}{141} = \frac{42}{47} = \frac{m}{n} $$
$$ \fbox{$\therefore m + n = 89$} $$
Saif
$$ \left(x - x^{-1} \right)^2 = x^2 - 2 + x^{-2} = \left(\sqrt{5} \right)^2 $$
$$ \fbox{$\therefore x^2 + x^{-2} = 7$} $$
$$ \left(x + x^{-1} \right)^2 = x^2 + 2 + x^{-2} = 7 + 2 = 9 $$
$$ \fbox{$\therefore x + x^{-1} = 3$} $$
$$ \left(x + x^{-1} \right)^3 = x^3 + 3x + 3x^{-1} + x^{-3} = 3^3 $$
$$ x^3 + x^{-3} + 3 \left( x + x^{-1} \right) = 27 \rightarrow x^3 + x^{-3} = 27 - 3(3) $$
$$ \fbox{$\therefore x^3 + x^{-3} = 18$} $$
$$ \left(x + x^{-1} \right)^5 = x^5 + 5x^3 + 10x + 10x^{-1} + 5x^{-3} + x^{-5} = 3^5 $$
$$ x^5 + x^{-5} + 5 \left( x^3 + x^{-3} \right) + 10 \left( x + x^{-1} \right) = 243 \rightarrow x^5 + x^{-5} = 243 - 5(18) - 10(3) $$
$$ \fbox{$\therefore x^5+ x^{-5} = 123$} $$
$$ \frac{x^{10} + x^6 + x^4 + 1}{1 + x^2 + x^8 + x^{10}} \rightarrow \frac{\not{x^5} \left( x^5 + x + x^{-1} + x^{-5} \right)}{\not{x^5} \left( x^{-5} + x^{-3} + x^3 + x^5 \right)} $$
$$ \frac{x^5 + x + x^{-1} + x^{-5}}{x^{-5} + x^{-3} + x^3 + x^5} \rightarrow \frac{\left(x^5 + x^{-5}\right) + \left(x + x^{-1}\right)}{\left(x^{-5} + x^5\right) + \left(x^{-3} + x^3\right)} \rightarrow \frac{123 + 3}{123 + 18} $$
$$ \frac{x^{10} + x^6 + x^4 + 1}{1 + x^2 + x^8 + x^{10}} = \frac{126}{141} = \frac{42}{47} = \frac{m}{n} $$
$$ \fbox{$\therefore m + n = 89$} $$
Saif
Given that $x - \frac{1}{x} = \sqrt{5}$, where $x$ is a positive rational number, find the value of $\frac{x^{10} + x^6 + x^4 + 1}{1 + x^2 + x^8 + x^{10}}$. Express your answer in the form $\frac{m}{n}$ where $(m, n) = 1$, and therefore find the value of $m + n$
Saif
@valid ice Do you think there could be a simpler solution to this?
Or is this the optimal way?
Nice solution. That’s probably the best approach.
I have a couple other qs I need your help with, lemme send
$(6m+7)(6n+7) - (5m+4)(5n+4) = 2024$ where $m, n$ are integers. Find $m + n$
Saif
,rotate
Q19
A and B take turns tossing a standard 6-sided dice. The first person who accumulates 6 or more dots wins. A starts first. If the probability for B to win by tossing the dice twice is $\frac{m}{n}$, and $(m, n) = 1$, find $m+n$
Saif
If there is only one element in set $A = { x \in \mathbb{Z} | x^2 - (a+2)x + 2 - a < 0 }$, and the range of a positive real number $a$ is $(m, n]$, find $m+n$.
Saif
Would be great if you could help with these at your convenience, no rush!
What I did for this was:
Irrespective of whether m, n are odd, the first term would always be odd. Therefore the second term needs to be odd too. For the second term to be odd, both m and n need to be odd. Therefore m and n are odd.
If m and n are odd, 5m and 5n would always end in 5, therefore 5m + 4 and 5n + 4 would always end in 9. Two numbers ending in 9, when multiplied, give a product that ends in 1. Therefore term two has last digit 1.
Since term two has last digit 1, term one must have last digit 5, so that they subtract to give 4. For the last digit of term one to be 5, at least one of 6m and 6n need to end in 8. The only last digit that fits both criteria is 3. Therefore at least one of m and n ends in 3.
Those have been my deductions so far.
$\sin(2x)=\frac{2tan(x)}{1+tan^2(x)}$\
$\cos(2x)=\frac{1-tan^2(x)}{1+tan^2(x)}$\
Huang
$$ (6m + 7)(6n + 7) = 2024 $$
$$ 36mn + 42(m + n) + 49 - 25mn - 20(m + n) - 16 = 2024 $$
$$ 11mn + 22(m + n) + 33 = 2024 $$
$$ mn + 2m + 2n + 3 = 184 $$
$$ mn + 2m + 2n + 4 = 185 $$
$$ m(n + 2) + 2(n + 2) = 185 $$
$$ (m + 2)(n + 2) = 185 $$
$$ 185 = 5 \cdot 37 \rightarrow 185 = (3 + 2)(35 + 2) $$
$$ \fbox{$\therefore m + n = 38$} $$
Saif
$$f^-1(x) = \frac{1}{f'(f^-1(x))}$$
Huang
$$f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}$$
Saif
thanks lol
Gave +1 rep to saifurrahman!
For Q19 above, let g(x) = f(x) - 2024x, => g(1)=g(2)=g(3)=g(4)=0 => (x-1)(x-2)(x-3)(x-4) | g(x).
will take look at the others if i have time
wall text
,help
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,help
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$$log_3(x^2+4x)>0$$
where $$x^2+4x>0$$
Huang
h
$$ h $$
doublerone
$heh$
doublerone
doublerone
x^2
$$test$$
doublerone
Compile Error! Click the
reaction for more information.
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hmm $x^2$
$$test$$
doublerone
oo
Finding X intercepts,
For equation 1, $y = 16 - 2x$
Let y = 0
$$0 = 16 - 2x$$
$$x = \frac{-16}{-2}$$
$$x = 8$$
Hence ;
$$(x-8)(R) = x^3 - 15x^2 + 66x - 80$$
Finding First term of R
$$xR - 8R = x^3 -15x^2 + 66x - 80$$
Here only xR can lead to $x^3$, so it is assumed the first term of R is equal to $x^2$
Finding Second term of R
$$xR - 8R = -15x^2$$
Since the first term of R is $x^2$,
$$xR - 8x^2 = -15x^2$$
$$xR = 8x^2 - 15x^2$$
$$xR = -7x^2$$
$$R = \frac{-7x^2}{x}$$
$$R = -7x$$
$$(x-8)(x^2 - 7x + c) = x^3 - 15x^2 + 66x -80$$
Expanding the first equation to find c
$$x(x^2 - 7x + c) - 8(x^2 - 7x + c) = x^3 - 15x^2 + 66x -80$$
$$x^3 - 7x^2 + cx - 8x^2 + 56x -8c = x^3 - 15x^2 + 66x - 80$$
$$Simplify$$
$$(c+56)x - 8c = 66x - 80$$
$$c+56 = 66$$
$$c = 10$$
$$R = x^2 - 7x + 10$$
Simplifying R:
$$R = x^2 - 2x - 5x + 10$$
$$R = x(x-2) - 5(x-2)$$
$$R = (x-5)(x-2)$$
equation:
$$y = (x-8)(x-5)(x-2)$$
X intercepts: (8,0) (5,0) (2,0)
doublerone
Finding Intercepts between line and curve:
$$x^3 - 15x^2 + 66x - 80 = 16 - 2x$$
$$x^3 - 15x^2 + 68x - 96 = 0$$
Since both graphs intersect at (8,0)
$$(x-8)(R) = x^3 - 15x^2 + 68x - 96$$
We know that the first and second term of R are $(x^2 - 7x)$
So $-8R = -96$
$$R = \frac{-96}{-8}$$
$$R = 12$$
$$(x - 8)(x^2 - 7x + 12)$$
simplifying R:
$$R = x^2 - 7x + 12$$
$$R = x^2 -3x -4x + 12$$
$$R = x(x-3) - 4(x-3)$$
$$R = (x-4)(x-3)$$
$$y = (x-8)(x-4)(x-3)$$
$$y = 16-2x$$
$$y = 16 - 2(4)$$
$$y = 8$$
$$y = 16-2(3)$$
$$y = 10$$
Graph intercepts: (8,0 (4,8) (3,10)
doublerone
finding area under graph by Integrating $x^3 - 15x^2 + 66x - 80$
$$\int x^3 - 15x^2 + 66x - 80 dx$$
$$ = \frac{x^4}{4} -5x^3 + 33x^2 - 80x$$
area from x = 3 to x = 4
$$\left(\frac{4^4}{4} -5(4)^3 + 33(4)^2 - 80(4)\right) - \left(\frac{3^4}{4} -5(3)^3 + 33(3)^2 - 80(3)\right)$$
$$ = 9.75$$
area from x = 4 to x = 5
$$\left(\frac{5^4}{4} -5(5)^3 + 33(5)^2 - 80(5)\right) - \left(\frac{4^4}{4} -5(4)^3 + 33(4)^2 - 80(4)\right)$$
$$ = 4.25$$
Area from x = 4 to x = 8
$$= \frac{1}{2}(4)(8) - 4.25$$
$$= 11.75$$
Area from x = 3 to x = 4
$$= 9.75 - \frac{1}{2} (8+10)$$
$$= 0.75$$
Shaded area $= 11.75 + 0.75 = 12.5$ Ans.
doublerone
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,list
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$$ X_x$$
doublerone
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Render LaTeX code and configure rendering options.
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$$[x]^{31291381}_{342852}$$
doublerone
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znc grow
This command can only be used in bot-stuff channels.
omg
$\int_{a}^{b} f(t) dt=\int_{a}^{b} f(a+b-t) dt$
Huang
doublerone
If the set of vectors $v_1,v_2,v_3$ are not linearly independent, $\forall c_1,c_2,c_3\quad c_1\textbf{v}_1 + c_2\textbf{v}_2 + c_3 = 0$
Huang
If the set of vectors $v_1,v_2,v_3$ are not linearly independent, $\forall c_1,c_2,c_3\quad c_1\textbf{v}_1 + c_2\textbf{v}_2 + c_3\textbf{v}_3 = 0$
Huang
$ \\To print NREC \
\document class [ac, paper] {article}
\begin {document}
NREC
\end {document} $
$
\begin{aligned}
\frac{\partial(r,\theta)}{\partial(x,y)}
&=
\begin{matrix}
{\frac{x}{r} }& {\frac{y}{r} }
-\frac{y}{x^{2}+y^{2}} & \frac{x}{x^{2}+y^{2}}
\end{matrix}
\frac{x^{2}+y^{2}}{r(x^{2}+y^{2})}
= \frac{1}{r}.
\end{aligned}
$
Kurosaki Ichigo
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
$f(x) = 3^{x+2}$
$y = 3^{x+2}$
$x = 3^{y+2}$
$\log_3 (x) = y-2$
$y = \log_3 (x) + 2$
$f^{-1} (x) = \log_3 (x) + 2$
Muhammad Hasan
$\alpha = 38.68 \text{ (taking it positive)}$
$\text{ Since sinx was negative, we'll look at quadrants where sin is negative in CAST diagram and it'll be third and fourth quad}$
$x = 180 + 38.68, 360 - 38.68$
Muhammad Hasan
$2c+30 = \frac{9c}{5} + 32$
$2c - 2 = \frac{9c}{5}$
$10c-10 = 9c$
$\boxed{c = 10}$
Muhammad Hasan
\begin{tikzpicture}
\draw (0,1) to[bend left=10] (0,6);
\draw (0,1) to[bend right=5] (0,0);
\draw (0,0) to[bend right=10] (2,0);
\draw (2,0) to[bend right=5] (2.25,2);
\draw (2.25,2) to[bend right=10] (3,2);
\draw (1.75,2.125) to[bend right=8] (2.25,2);
\draw (3,2) to[bend right=5] (3,-0.25);
\draw (3,-0.25) to[bend right=10] (5,-0.25);
\draw (5,-0.25) to[bend right=5] (5.25,7.25);
\draw (5.25,7.25) to[bend right=30] (3,10);
\draw (3,10) to[bend right=15] (2,10);
\draw (2,10) to[bend right=30] (0.5,8.28);
\draw (5.25,7.25) to[bend left=5] (6.25,7.15);
\draw (6.25,7.15) to[bend left=5] (6.25,2.5);
\draw (6.25,2.5) to[bend left=20] (5.25,2.25);
\draw (-1,7.08) to[bend right=90] (3.25,7.08);
\draw (-1,7.08) to[bend left=90] (3.25,7.08);
\end{tikzpicture}
\begin{tikzpicture}
\draw (0,1) to[bend left=10] (0,6);
\draw (0,1) to[bend right=5] (0,0);
\draw (0,0) to[bend right=10] (2,0);
\draw (2,0) to[bend right=5] (2.25,2);
\draw (2.25,2) to[bend right=10] (3,2);
\draw (1.75,2.125) to[bend right=8] (2.25,2);
\draw (3,2) to[bend right=5] (3,-0.25);
\draw (3,-0.25) to[bend right=10] (5,-0.25);
\draw (5,-0.25) to[bend right=5] (5.25,7.25);
\draw (5.25,7.25) to[bend right=30] (3,10);
\draw (3,10) to[bend right=15] (2,10);
\draw (2,10) to[bend right=30] (0.5,8.28);
\draw (5.25,7.25) to[bend left=5] (6.25,7.15);
\draw (6.25,7.15) to[bend left=5] (6.25,2.5);
\draw (6.25,2.5) to[bend left=20] (5.25,2.25);
\draw (-1,7.08) to[bend right=90] (3.25,7.08);
\draw (-1,7.08) to[bend left=90] (3.25,7.08);
\end{tikzpicture}'
kayla
ohhh
What's this
latex
Nuh uh
yah
Nah
..
$\begin{pmatrix}\alpha_{11} & \cdots & \alpha_{1m} \\vdots & \ddots & \vdots \\alpha_{m1} & \cdots & \alpha_{mm}\end{pmatrix};\otimes;\begin{pmatrix}\beta_{11} & \cdots & \beta_{1k} \\vdots & \ddots & \vdots \\beta_{k1} & \cdots & \beta_{kk}\end{pmatrix}=\begin{pmatrix}\alpha_{11}\beta_{11} & \cdots & \alpha_{11}\beta_{1k} & \cdots & \alpha_{1m}\beta_{11} & \cdots & \alpha_{1m}\beta_{1k} \\vdots & \ddots & \vdots & \ddots & \vdots & \ddots & \vdots \\alpha_{11}\beta_{k1} & \cdots & \alpha_{11}\beta_{kk} & \cdots & \alpha_{1m}\beta_{k1} & \cdots & \alpha_{1m}\beta_{kk} \\vdots & \ddots & \vdots & \ddots & \vdots & \ddots & \vdots \\alpha_{m1}\beta_{11} & \cdots & \alpha_{m1}\beta_{1k} & \cdots & \alpha_{mm}\beta_{11} & \cdots & \alpha_{mm}\beta_{1k} \\vdots & \ddots & \vdots & \ddots & \vdots & \ddots & \vdots \\alpha_{m1}\beta_{k1} & \cdots & \alpha_{m1}\beta_{kk} & \cdots & \alpha_{mm}\beta_{k1} & \cdots & \alpha_{mm}\beta_{kk}\end{pmatrix} $\end{document}
attacker2007 🎀
finally
$\mathcal{U} =\begin{pmatrix}\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & 0 & 0 & \frac{1}{2} \\frac{1}{2} & \frac{i}{2} & -\frac{1}{2} & 0 & 0 & -\frac{i}{2} \\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & 0 & 0 & -\frac{1}{2} \0 & 0 & 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\frac{1}{2} & -\frac{i}{2} & -\frac{1}{2} & 0 & 0 & \frac{i}{2} \0 & 0 & 0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} &0\end{pmatrix}$\end{document}
$H \otimes \mathcal{I} = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \0 & 1 \end{pmatrix}=\begin{pmatrix} \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} & 0 \0 & \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2} & 0 \0 & \frac{1}{\sqrt{2}} & 0 & -\frac{1}\sqrt{2} \end{pmatrix} \end{document}
attacker2007 🎀
$H \otimes 1 = \begin{pmatrix} \frac{1}{\sqrt&{2}} & \frac{1}{\sqrt&{2}} \\frac{1}{\sqrt&{2}} & -\frac{1}{\sqrt&{2}} \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \0 & 1 \end{pmatrix}=\begin{pmatrix} \frac{1}{\sqrt&{2}} & 0 & \frac{1}{\sqrt&{2}} & 0 \0 & \frac{1}{\sqrt&{2}} & 0 & \frac{1}{\sqrt&{2}} \\frac{1}{\sqrt&{2}} & 0 & -\frac{1}{\sqrt&{2}} & 0 \0 & \frac{1}{\sqrt&{2}} & 0 & -\frac{1}{\sqrt&{2}} \end{pmatrix} \end{document}
attacker2007 🎀
$H \otimes 1 = \begin{pmatrix} \frac{1}{\sqrt&{2}} & \frac{1}{\sqrt&{2}} \\\frac{1}{\sqrt&{2}} & -\frac{1}{\sqrt&{2}} \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix}=\begin{pmatrix} \frac{1}{\sqrt&{2}} & 0 & \frac{1}{\sqrt&{2}} & 0 \\0 & \frac{1}{\sqrt&{2}} & 0 & \frac{1}{\sqrt&{2}} \\\frac{1}{\sqrt&{2}} & 0 & -\frac{1}{\sqrt&{2}} & 0 \\0 & \frac{1}{\sqrt&{2}} & 0 & -\frac{1}{\sqrt&{2}} \end{pmatrix} \end{document}
```Compilation error:```! Missing { inserted.
<to be read again>
&
l.49 ...es 1 = \begin{pmatrix} \frac{1}{\sqrt&{2}}
& \frac{1}{\sqrt&{2}} \\\...
A left brace was mandatory here, so I've put one in.
You might want to delete and/or insert some corrections
so that I will find a matching right brace soon.
(If you're confused by all this, try typing `I}' now.)```
...
the root twos broke >.<
[
H \otimes \mathbb{I}
\begin{pmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \
\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}
\end{pmatrix}
\otimes
\begin{pmatrix}
1 & 0 \
0 & 1
\end{pmatrix}
\begin{pmatrix}
\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} & 0 \
0 & \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \
\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} & 0 \
0 & \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}}
\end{pmatrix}
]
attacker2007 🎀
that is crazy
It's not that bad
It's basically applying a hadamard gate on a qubit and doing nothing on the other qubit
So you can get the tensor product between them if you wanted to
$x^2$
Ved Karjagi
Ved Karjagi
$$ x^2 $$
Ved Karjagi
$$
\center{x^2}
$$
Ved Karjagi
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
$$
\text{Fick's First Law of Diffusion states that: }
\mathbb{J} = -\mathbb{D}\frac{d\phi}{dx}
$$
attacker2007 🎀
$$
\text{Consequently, Fick's Second Law of Diffusion states that: }
\frac{\delta\phi}{\delta t} = \mathbb{D}\frac{\delta^2\phi}{\delta x^2}
$$
attacker2007 🎀
$$
\text{If we wish to compute in 2 or more dimensions, the Laplacian(}
\Delta = \nabla^2
\text{)}
\text{ must be used, which results in: }
\frac{\delta\phi}{\delta t} = \mathbb{D}\delta_{\phi}
$$
$$
\frac{\delta\phi}{\delta t} = \mathbb{D}\delta_{\phi}
$$
attacker2007 🎀
attacker2007 🎀
$$
\text{If we wish to compute in 2 or more dimensions, the Laplacian(}
\Delta = \nabla^2
\text{)}
\text{ must be used, which results in: }
\frac{\delta\phi}{\delta t} = \mathbb{D}\delta_{\phi}
$$
```Compilation error:```! Missing $ inserted.
<inserted text>
$
l.53
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.```
$1+1$
hersheys
helpy/femdevil/takenbydemonlord😝
helpy/femdevil/takenbydemonlord😝
$$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$
helpy/femdevil/takenbydemonlord😝
$$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$
```Compilation error:```! Display math should end with $$.
<to be read again>
l.49 $$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$
The `$' that I just saw supposedly matches a previous `$$'.
So I shall assume that you typed `$$' both times.
Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2023/texmf-var/fonts/map/pdftex/updmap/pdftex.map}{/usr/l
ocal/texlive/2023/texmf-dist/fonts/enc/dvips/lm/lm-mathit.enc}{/usr/local/texli```
helpy/femdevil/takenbydemonlord😝
$
$\text {if $ b^2 - 4ac < 0 $, quadratic has no real roots}$
how does text work huh
uh
i see
helpy/femdevil/takenbydemonlord😝
ooo
$\text meow $ a-b $$
k
$$ ax^2 + bx + c = 0,
\text if $ b^2 - 4ac > 0, x \in \mathbb{R} $$
$R=mgcos{\theta}$
helpy/femdevil/takenbydemonlord😝
helpy/femdevil/takenbydemonlord😝
$$ T_A = \frac{9mg}{4} - mgcos{\theta} $$
$$ \newline T_B = 4T_A = 9mg - 4mgcos{\theta} $$
$$ 9mg - 4mgcos{\theta} = 4mg + \frac{9}{2}mg $$
helpy/femdevil/takenbydemonlord😝
$$ T_A = \frac{9mg}{4} - mgcos{\theta} \
T_B = 4T_A = 9mg - 4mgcos{\theta}
9mg - 4mgcos{\theta} = 4mgcos{\theta} + \frac{9}{2}mg
8mgcos{\theta} = \frac{9}{2}mg
cos{\theta} = \frac{9}{2*8} $$
ok
$$help
erm
helpy/femdevil/takenbydemonlord😝
,help
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@unkempt horizon any clue why it fails every time? havent been able to get members of a role in forever
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$$ T_A = \frac{9mg}{4} - mgcos{\theta}
T_B = 4T_A = 9mg - 4mgcos{\theta}
9mg - 4mgcos{\theta} = 4mgcos{\theta} + \frac{9}{2}mg
8mgcos{\theta} = \frac{9}{2}mg
cos{\theta} = \frac{9}{2*8} $$
helpy/femdevil/takenbydemonlord😝
$$ T_A = \frac{9mg}{4} - mgcos{\theta}
T_B = 4T_A = 9mg - 4mgcos{\theta}
9mg - 4mgcos{\theta} = 4mgcos{\theta} + \frac{9}{2}mg
8mgcos{\theta} = \frac{9}{2}mg
cos{\theta} = \frac{9}{2*8} $$
```Compilation error:```! Missing $ inserted.
<inserted text>
$
l.50
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.```
$$ T_A = \frac{9mg}{4} - mgcos{\theta} $$
$$ T_B = 4T_A = 9mg - 4mgcos{\theta} $$
$$ 9mg - 4mgcos{\theta} = 4mgcos{\theta} + \frac{9}{2}mg $$
$$ 8mgcos{\theta} = \frac{9}{2}mg $$
$$ cos{\theta} = \frac{9}{2*8} $$
helpy/femdevil/takenbydemonlord😝
$\int\frac{1}{x} dx = \ln{x} + C$
helpy/femdevil/takenbydemonlord😝
$$\int\frac{1}{x}dx$$
helpy/femdevil/takenbydemonlord😝
\hi\
$ \text{The mass-energy equivalence is described by the famous equation}
[E=mc^2]
\text{discovered in 1905 by Albert Einstein.
In natural units ($c$ = 1), the formula expresses the identity}
[E=m]
$
$\int u\frac{dv}{dx}dx = uv - \int v\frac{du}{dx}dx$
helpy/femdevil/takenbydemonlord😝
$\int f'(x) dx = f(x) + c$
helpy/femdevil/takenbydemonlord😝
$\lambda\begin{pmatrix} 1\ 2\ 3 \end{pmatrix}$
helpy/femdevil/takenbydemonlord😝
$cos{\theta} = \frac{\begin{pmatrix} a_1\ a_2\ a_3 \end{pmatrix} \cdot \begin{pmatrix} b_1\ b_2\ b_3 \end{pmatrix}}{\sqrt{a_1^2 + a_2^2 + a_3^2}\sqrt{b_1^2 + b_2^2 + b_3^3}}$
helpy/femdevil/takenbydemonlord😝
no clue fr
$$\int\frac{4}{x}dx = 4\int\frac{1}{x}dx = 4\ln{x} + C$$
helpy/femdevil/takenbydemonlord😝
$$\beginpmatrix 5 \ 4 \ 3 \endpmatrix$$
$(a+b)^{n} = \sum_{k=0}^{n} a^{n-k} b^{k}$
hersheys
$$tan{r+1} = \frac{tan{r}+tan{1}}{1-tan{r}{tan{1}}$$
$$=\frac{sin{r}cos{1}+cos{r}sin{1}}{1-\frac{sin{r}sin{1}}{cos{r}cos{1}}}$$
helpy/femdevil/takenbydemonlord😝
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
helpy/femdevil/takenbydemonlord😝
$$cos(r+1)=cos(r)cos(1)-sin(r)sin(1)$$
helpy/femdevil/takenbydemonlord😝
$$\frac{sin(r+1)}{cos(r+1)} = \frac{sin(r)cos(1)+cos(r)sin(1)}{cos(r+1)}$$
helpy/femdevil/takenbydemonlord😝
$$\frac{sin(r)cos(1)+cos(r)sin(1)}{cos(r+1)} - \frac{sin(r)}{cos(r)}$$
helpy/femdevil/takenbydemonlord😝
$$\frac{sin(r)cos(1)cos(r)+cos^2(r)sin(1)}{cos(r+1)cos(r)}$$
helpy/femdevil/takenbydemonlord😝
$$\frac{\frac{sin(2r)}{2}cos(1)+cos^2(r)sin(1)}{cos(r+1)cos(r)}$$
helpy/femdevil/takenbydemonlord😝
$$\frac{cos(r)[sin(r)cos(1)+cos(r)sin(1)]}{cos(r+1)cos(r)}$$
helpy/femdevil/takenbydemonlord😝
$$\frac{sin(r)cos(1)+cos(r)sin(1)}{cos(r+1)}$$
helpy/femdevil/takenbydemonlord😝
$$\frac{sin(r)cos(1)cos(r)+cos^2(r)sin(1) - sin(r)[cos(r)cos(1)-sin(r)sin(1)]}{cos(r+1)cos(r)}$$
helpy/femdevil/takenbydemonlord😝
$$cos^2(r)sin(1)+sin^2(r)sin(1)$$
helpy/femdevil/takenbydemonlord😝
$$= sin(1)[cos^2(r)+sin^2(r)]$$
$$cos^2(r) + sin^2(r) = 1 \therefore sin(1)[cos^2(r)+sin^2(r)] = sin(1)$$
YES
helpy/femdevil/takenbydemonlord😝
$$\frac{sin(r+1)}{cos(r+1)} = \frac{sin(r)cos(1)+cos(r)sin(1)}{cos(r+1)}$$
$$\frac{sin(r)cos(1)+cos(r)sin(1)}{cos(r+1)} - \frac{sin(r)}{cos(r)}$$
$$\frac{sin(r)cos(1)cos(r)+cos^2(r)sin(1) - sin(r)[cos(r)cos(1)-sin(r)sin(1)]}{cos(r+1)cos(r)}$$
$$\frac{cos^2(r)sin(1) + sin(r)cos(r)cos(1)- sin(r)cos(r)cos(1)-sin^2(r)sin(1)}{cos(r+1)cos(r)}$$
$$=\frac{sin(1)[cos^2(r)+sin^2(r)]}{cos(r+1)cos(r)}$$
helpy/femdevil/takenbydemonlord😝
helpy/femdevil/takenbydemonlord😝
helpy/femdevil/takenbydemonlord😝
Write the equations in a standard form, and align the variables
$\qquad 2x + 3y = 21$
$\qquad x - y = -2$
Multiply one or both equations to make the coefficients of one of the variables the same with opposite signs
$\qquad 2x + 3y = 21$
$\qquad(x - y = -2) \times 3$
$\qquad3x - 3y = -6$
Cancel out one of the variables and solve for the other one
$\qquad \quad \phantom{3}\ 2x + \cancel{3y} = 21$
$\qquad +\quad 3x - \cancel{3y} = -6$
$\qquad \rule{3.5cm}{0.4pt}$
$\qquad \qquad 5x \phantom{+ \cancel{3y}}= 15$
$\qquad x = 3$
Solve for the variable that we cancelled out earlier
$\qquad x - y = -2$
$\qquad y = x + 2$
$\qquad y = (3) + 2 = 5$
Ved Karjagi
Yum
Write the equations in a standard form, and align the variables
$\qquad 2x + 3y = 21$
$\qquad x - y = -2$
Multiply one or both equations to make the coefficients of one of the variables the same with opposite signs
$\qquad 2x + 3y = 21$
$\qquad(x - y = -2) \times 3$
$\qquad3x - 3y = -6$
Cancel out one of the variables and solve for the other one
$\qquad \quad \phantom{3}\ 2x + \cancel{3y} = 21$
$\qquad +\quad 3x - \cancel{3y} = -6$
$\qquad \rule{3.5cm}{0.4pt}$
$\qquad \qquad 5x \phantom{+ \cancel{3y}}= 15$
$\qquad x = 3$
Solve for the variable that we cancelled out earlier
$\qquad x - y = -2$
$\qquad y = x + 2$
$\qquad y = (3) + 2 = 5$
t0ken
copy
whats 9 + 10
21
gotta put $ before and after
$2+2$
Ayesha
cute
!!
$=5$
himalayancat
$○ ○$
$ ^ $
himalayancat
$○ ○$
$ ^ $
```Compilation error:```! LaTeX Error: Unicode character ○ (U+25CB)
not set up for use with LaTeX.
See the LaTeX manual or LaTeX Companion for explanation.
Type H <return> for immediate help.
...
l.49 $○
○$
You may provide a definition with
\DeclareUnicodeCharacter```
$' '$
$ - $
himalayancat
$' '$
$ - $
```Compilation error:```! Double superscript.
<recently read> ^
l.49 $' '
$
I treat `x^1^2' essentially like `x^1{}^2'.
Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2023/texmf-var/fonts/map/pdftex/updmap/pdftex.map}{/usr/l
ocal/texlive/2023/texmf-dist/fonts/enc/dvips/lm/lm-mathsy.enc}]
(./705488212965064777.aux)```
himalayancat
$1234
56789$
radiohead ref
close enuf
$$\ln\left(\frac{p}{q}\right) = a, \ln\left(q^2p\right) = b$$
$$\frac{p}{q} = e^a, q^2p=e^b$$
helpy/femdevil/takenbydemonlord
$$hi$$
Raiyan
$$ln(5/3)$$
Raiyan
$$6x + 2y =7$$
$$2x + 4y = 8 $$
$$solve for x and y$$
Raiyan
/sum
$$/frac{0.17}{2.67} x 100$$
Raiyan
Raiyan
Raiyan
$$\frac{0.17}{2.67}$$
Raiyan
$$D = \frac{1}{no. of lines per mllimeter x 10^{3}}$$
Raiyan
$$D = \frac{1}{no.\ of\ lines\ per\ mllimeter\ x\ 10^{3}}$$
Raiyan
$$D = \frac{1}{no.\ of\ lines\ per\ millimeter\ x\ 10^{3}}$$
Raiyan
$$D = \frac{1}{no.\ of\ lines\ per\ millimeter\ \times\ 10^{3}}$$
Raiyan
$$+$$
Raiyan
$$_$$
Raiyan
$$_$$
Raiyan
$$hi\ ben$$
Raiyan
$\mathcal{L}$
Pugs
$\mathcal{LMAO}$
Pugs
$\mathcal{GAY}$
hayzy 月下獣
Sexy
$\mathcal{Sexy}$
EminemSomething
$\mathcal{SEXY}$
EminemSomething
Is this how they propose in roman?
$\mathfrak{NO}$
Pugs
AlexV
$\mathfrak{NOU}$
$\mathfrak{UVW}$
Boeing-747
$\mathcal{UVW}$
$\mathcal{GAY}$
Broski 6w9
$\mathcal{abcdeABCDE}$
AlexV
AlexV
$\textgoth{ABCDE}$
AlexV
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LOL
Apparently with the right package it can display blackletter font
$\textbf{PUGSISGAY}$
Pugs
Ohh
Yea
Ohh
$amazing$
ripilicious
Amazing
$/textit{WHO WAS IN PARIS}$
hayzy 月下獣
Wait lemme try sth
Well
Forward slash my lad
Fuck
Let's test it's limit
$$F^{'}{\mu \nu} = \partial{\mu}(A_{\nu} + \partial_{\nu}f) - \partial_{\nu}(A_{\mu} + \partial_{\mu}f)$$
$$F^{'}{\mu \nu} = \partial{\mu}A_{\nu} + \partial_{\mu}\partial_{\nu}f - \partial_{\nu}A_{\mu} - \partial_{\nu}\partial_{\mu}f$$
Pugs
$\textit{who was in paris}
u need to end it w a $
$\textit{who was in paris}$
hayzy 月下獣
IK I FORGOT
$who was in paris$
Pugs
Ah the spacing
Try a continued fraction
$F^{'}{\mu \nu} = \partial{\mu}(A{\nu} + \partial{\nu}f) - \partial{\nu}(A{\mu} + \partial{\mu}f) \rightarrow F^{'}{\mu \nu} = \partial{\mu}A{\nu} + \partial{\mu}\partial{\nu}f - \partial{\nu}A{\mu} - \partial{\nu}\partial{\mu}f$
rip
Pugs
Ah right
Ye
Hmm
$\textit{yall need to} \mathbf{CALM DOWN}$
TicTac
no space TT
Put a space
$\frac 1 {1+\frac 1{1+\frac 1 {1+\frac 1{1+\cdots}}}}$
$\mathfrak{NAH}$
Pugs
Wow
Wtf
What'd you try?
AlexV
nais
$\underline{white girls be like}$
hayzy 月下獣
Bruh
$1+\frac 1 {1+\frac 1{1+\frac 1 {1+\frac 1{1+\cdots}}}}=\varphi$
AlexV
$\aquarius$
AlexV
Nice
wha??
$\aries \taurus \gemini$
i was just about to do that xd
AlexV
XD
$\Re (z) + i\Im (z)$
AlexV
$\Re$
AlexV
AlexV
$\tilde \approx \aquarius$
AlexV
Nice
$\amongus
For small $x, (1+x)^n \aquarius 1+ nx$.
$\amogus
AlexV
For small $x, (1+x)^n \aquarius 1+ nx$
Broski 6w9
For small $ok
Wait what
Wat
$visibleconfusion$
ripilicious
,tex $$
\begin{tikzpicture}
\draw (0,1) to[bend left=10] (0,6);
\draw (0,1) to[bend right=5] (0,0);
\draw (0,0) to[bend right=10] (2,0);
\draw (2,0) to[bend right=5] (2.25,2);
\draw (2.25,2) to[bend right=10] (3,2);
\draw (1.75,2.125) to[bend right=8] (2.25,2);
\draw (3,2) to[bend right=5] (3,-0.25);
\draw (3,-0.25) to[bend right=10] (5,-0.25);
\draw (5,-0.25) to[bend right=5] (5.25,7.25);
\draw (5.25,7.25) to[bend right=30] (3,10);
\draw (3,10) to[bend right=15] (2,10);
\draw (2,10) to[bend right=30] (0.5,8.28);
\draw (5.25,7.25) to[bend left=5] (6.25,7.15);
\draw (6.25,7.15) to[bend left=5] (6.25,2.5);
\draw (6.25,2.5) to[bend left=20] (5.25,2.25);
\draw (-1,7.08) to[bend right=90] (3.25,7.08);
\draw (-1,7.08) to[bend left=90] (3.25,7.08);
\end{tikzpicture}
$$
alex or cal [she/they]
Thanks lads
Yes Pierrot?
sus
$\int \taurus^2 d \taurus$
The πerrot [inactive/NieR]
yes
a taurus integral
amazing
$\int \taurus^2 d \taurus = \frac{\taurus^3}{3}+\aries$
The πerrot [inactive/NieR]
where aries is a constant of integration
What the fuck
What the fuck
,calc 9+10
Result:
19
Result:
1.2247448713916
wow, just wow
:D
$:D$
ripilicious
$0_0$
ripilicious
$0_0$
$0_{0_{0_{0}^{0}}^{0_{0}^{0}}}^{0_{0_{0}^{0}}^{0_{0}^{0}}}$
The πerrot [inactive/NieR]
$0_{0_{0_{0}^{0}}^{0_{0}^{0}}}^{0_{0_{0}^{0}}^{0_{0}^{0}}}$
$\large\smile$
AlexV
$\large\smile$
Kamala Harris
why is there a channel called latex
who has a latex fetish
...oh its for a bot.

isomorphic to pi_1(D^2,0)
I do
$\LaTeX$
turtius - رمضان كريم
$\YoUr MoM$
turtius - رمضان كريم
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turtius - رمضان كريم
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bruh calm
everyone b goin wild with the bot fr
and this is the day where the ZNotes community learned how to use LaTeX
Lol
AlexV
Bruh-
😊
bad apple latex edition?
Thought other people would be experimenting with different colours :)
$\color{red}{\sum_{n=1}^{\infty}\frac 1{n^2}= \frac {\pi ^2}6}$
AlexV
So one weakness about the bot us that you can’t delete what has been typeset after some time.
$\Sigma \text{balls}$
The πerrot [inactive/NieR]
$\color{orange}{\mathfrak{Hello Everybody!}}$
AlexV
AlexV
$\color{red}{\int} \color{orange}{\int} \color{yellow}{\int} \color{green}{\int} \color{blue}{\int} \color{purple}{\int}$
AlexV
$\color{red}{G} \color{yellow}{a} \color{green}{y}$
The πerrot [inactive/NieR]
$\color{orange}{\mathfrak{i failed my gcse maths}}$
astro
oh shit
but you get an A for typesetting in LateX :)
$\color{purple}{\mathfrak{hello?}}$
mauve
$\color{green}{\mathfrak{1234567890}}$
AlexV
$\color{pink}{\mathfrak{swag}}$
alex or cal [she/they]
satan [in/jail]
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it ried
isomorphic to pi_1(D^2,0)
$\color{orange}{\mathfrak{greetings comrade}$
turtius - رمضان كريم
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