help-28

[proven lynx]

just dont go

[buoyant ravine]

i wont

[proven lynx]

this is how i notate it

[buoyant ravine]

what are the xs in the middle column?

[proven lynx]

Like to mark if theyre included

In the event

[buoyant ravine]

i see i see

[proven lynx]

but my question is

Do i include Et

or no

[robust slate]

orz u

[proven lynx]

because its "or"

[buoyant ravine]

well the people ordered by telephone, right?

[proven lynx]

WAIT

u include everything

YES

i just incude everything

that has the event

[buoyant ravine]

🙂‍↕️

[proven lynx]

Because its U

Holy shit

Man its so sad geniuenly tho

[buoyant ravine]

hello you also orz

[proven lynx]

I have to study 4 more topics

[buoyant ravine]

you got this!

[proven lynx]

I also have to do hypothesis tests, bernoulli,

[buoyant ravine]

you'll breeze through them, i'm sure :)

[proven lynx]

how do i do all of this

I hope so too man

[buoyant ravine]

oop cant help you with those, my math isnt good enough

alr well best of luck!

!done

[full forum]

If you are done with this channel, please mark your problem as solved by typing .close

[slate glacier]

,close (don't mind me)

[buoyant ravine]

.coose /j

.close

[full forum]

Channel closed

Closed by @buoyant ravine

Use .reopen if this was a mistake.

[buoyant ravine]

i love clopen perms

[full forum]

Available help channel!

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[proven lynx]

hello so in the b)

the event C

It says the customer either orders online or orders tickets for a sporting event

ive notated it like this but i checked and "So" shouldnt be ticked, why?

ojhh

because "Orders online U Tickets for a sporting event“

"So" is included in both

therefore it only appears once

no

i dont understabd

[glacial stream]

What does Xo and X+ mean? Is it online and offline?

[glacial stream]

Oh it's Xo and Xt

[proven lynx]

Wdym Xo and Xt

its online and telephone

[glacial stream]

Yea I just got that

[proven lynx]

The X mark if theyre in the evnt

[glacial stream]

Yea

[proven lynx]

so why shouldnt be „So“ be marked

[glacial stream]

Because it's either-or

[proven lynx]

Huh

[glacial stream]

Rather than either or, or both

either or is exclusive or

[proven lynx]

so he orders online or orders a ticket for sporting

[glacial stream]

Yep, but not both

[proven lynx]

so

Wtf i dont understand

[glacial stream]

Well this is a translation

[proven lynx]

its accurate

[glacial stream]

Yea which is why it matters

[proven lynx]

But it says

Look

[glacial stream]

In language a lot of the time, Either or is used as an exclusive or rather than just an or

Think about this

If you read the instructions manual, either add boiling water or vegetable stock

What does that read like?

[proven lynx]

one of both but not both

[glacial stream]

Bingo

[proven lynx]

holy shit

wait

[glacial stream]

In language either or is used when you want exclusive or

[proven lynx]

Why cant we mark "So" and not mark "St"

[glacial stream]

Because you either book online or you book sports event via telephone

The second part of the condition is directly St

[proven lynx]

Yes

Because all sports

But then Youd have to do Sports online too

WTF man

[glacial stream]

S = So U St

[proven lynx]

Because its "Online U Sports event"

[glacial stream]

Nope

[proven lynx]

Huh

[glacial stream]

I believe it would be

Online ∆ Sports

There is a reason why algebra pushed the field of math forward so far, and it was because it lowered ambiguities like these so much

[proven lynx]

man i fucking hate this subject

i hate every inch of this subject

not one thing did i like

hypothesis, bernoulli, nothing

[glacial stream]

We all have felt that way at least once in our life lol

[proven lynx]

i didnt understand it

because

[glacial stream]

Either or is being used here as symmetric difference or exclusive or

[proven lynx]

yeah i jsut didnt understand why ''So'' isnt being used

[glacial stream]

.

Either online or sports

O ∆ S

[proven lynx]

but sports can be online too?!

[glacial stream]

Yea but if it were online it would be both

Language my dear friend is complicated asf.

[proven lynx]

Wait so

Either online, or a sports event

But if its an online sports event that would mean that its not either anymore

holy hit

i think i understood it

because

If its ''So, Ko, Mo, Eo''

If youd have So then

its would be contradictory to C

Because the customer either orders online or a ticket for a sporting event

[glacial stream]

Ik you repeated the statement but hopefully you understood it lol

[proven lynx]

OMG

BECAUSE

[proven lynx]

if you would have ''So'', youd have both online and sporting event

done

i understood it

[glacial stream]

Yep

Good job

[proven lynx]

Okay we move on

[glacial stream]

lol

!done

[full forum]

If you are done with this channel, please mark your problem as solved by typing .close

[proven lynx]

no no

wait

ur staying here with me buddy

[glacial stream]

Ok lmao

[proven lynx]

,rotate

[proven lynx]

Okay so

D = A not U C not

that means that we basically do

A and C

just flip everything right

[glacial stream]

There's a bar over that as well

[proven lynx]

oh

so its

[glacial stream]

Also you can use latex

[proven lynx]

A not and B not

because the Bar means another ''not'' right?

[glacial stream]

Yea

That's what I would assume

$\overline{\bar A \cup \bar C}$

[glossy valve]

@glacial stream

[proven lynx]

whats the mathematicall term for and

the uspide down cup

[glacial stream]

The name of the operation is intersection

The operator is called cap

[proven lynx]

${\bar A\cap\Bar C}$

[glossy valve]

law

[glacial stream]

There would be double bar

[proven lynx]

huh

so the bar doesnt get removed?

[glacial stream]

$\bar{\bar A} \cap \bar{\bar C}$

[glossy valve]

@glacial stream

[proven lynx]

never seen that before

[glacial stream]

Oh not even $\bar{\bar A} = A$?

[glossy valve]

@glacial stream

[proven lynx]

i mean yes

that is what i did bro

wtf

[glacial stream]

No you used $\bar A$

[glossy valve]

@glacial stream

[proven lynx]

oh

But what would ''A not intersection C not'' be

[glacial stream]

not not A = A ≠ not A

[glacial stream]

It would not be A or C I tell you that much

[proven lynx]

i have to relearn these rules

[glacial stream]

This is a pun, it's "not (A or C)"

This is De Morgan's Law

[proven lynx]

yes

exaclty

i gotta rewatch those

[glacial stream]

K, cool

Anything more you want to ask your honor?

[proven lynx]

uhhh

[proven lynx]

,rotate

[proven lynx]

so i can do

0.55 = -4 * (3b) ...

and so on

right

thats 1 equation?

but how do i get the other 2

no we need 4

we have 4 unknowns?

@glacial stream r u there bro

<@&286206848099549185>

hello anyone

[dire swallow]

If P(-2≤x≤2)=0.35, what can you deduce

[proven lynx]

uhh

that from -1 to 2

its 0.35

or wdym

[dire swallow]

Yup

[proven lynx]

so

how do i notate that

[lost sandal]

b+c+d-c+0.05=0.35 no?

jus add up the probabilities

[dire swallow]

Then you need one more equation

Here's a hint, think of what the sum of all probabilities have to equate to

[proven lynx]

1

[proven lynx]

wtf

[proven lynx]

yes cuz

but

wait

what about the 4th equation

[lost sandal]

why do you need four

there are only three variables

[proven lynx]

oh

im delusionla

sorry

yes tysm guys

i wish i was smart

il continue tomorrow

[dire swallow]

Just practise more and more and you'll get there eventually

[full forum]

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[outer thistle]

Im sure i use the 2nd formula

But I forget what does a equals to?

[lime trellis]

u have it written down

9.81

[outer thistle]

no i just wrote it down

but im still not sure

do i include the mass of the ball

[lime trellis]

no

a=9.81 ms^-2

depending on if u define down negative or down positive

[outer thistle]

Ah

my final answer comes out to 8.43

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[full forum]

@summer creek Has your question been resolved?

[ionic knoll]

Have you drawn a sketch?

[full forum]

@summer creek Has your question been resolved?

[cobalt summit]

Forget my abysmal drawing skills but this should be the figure.

why is it so dark in the picture.

😭

I found a scale so uh

Here.

[ionic knoll]

,rccw

[cobalt summit]

should I send a brighter screenshot or is it fine? 😭

[ionic knoll]

It's good

[cobalt summit]

kk.

[full forum]

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[ionic knoll]

tbh I have no idea best to wait for someone else

.reopen

[full forum]

✅ Original question: #help-28 message

[cobalt summit]

also what is a circumradius.

[ionic knoll]

I think thats the circle that goes through the 3 points

[cobalt summit]

circumcircle.

[ionic knoll]

Radius of it*

[cobalt summit]

radius of circumcircle. Fire. Lemme now try to find a way.

first thing I'm thinking of is like. To extend the perpendicular bisectors inwards because if we take PQ and PS as chords of the circle. Then the perpendicular bisectors meet at the center.

[ionic knoll]

Sorry I g2g (tbh I'm not good at geometry anyway, just tips and stuff, you can ping Helpers role if anyone else is online)

[warped frost]

try not to ping for OP though

cuz it's pointless to ping for OP when OP is MIA and now you have a horde of helpers waiting on OP

[full forum]

@summer creek Has your question been resolved?

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[undone radish]

!status

[full forum]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[undone radish]

2

[grand hatch]

Show your work

As a hint I would note that 8=2*2*2

[undone radish]

,, \begin{align}
x^9 -x &= x(x^8 -1)
\&= x\left((x^4)^2 -(1^4)^2\right)
\&=x\left((x^4+1)(x^4-1)\right)
\&=x\left((x^4+1)((x^2)^2 -(1^2)^2\right)
\end{align}

bruh

[onyx glen]

\\ for newlines

[grand hatch]

You are on the correct path

Continue

[glossy valve]

calvin

[onyx glen]

why are you wrapping (x^4-1)(x^4+1) in an extra set of brackets

[undone radish]

idk

[onyx glen]

are you an associativity hater

[undone radish]

whats that

[onyx glen]

keep going tho

[onyx glen]

associative law is the law that says (a×b)×c = a×(b×c)

[undone radish]

,,x(x^4 +1)(x^2 -1)(x^2 +1)

[glossy valve]

calvin

[undone radish]

,,x(x^4+1)(x+1)(x-1)(x^2+1)

is that it

[glossy valve]

calvin

[stone matrix]

Yes

[undone radish]

oh

thanks guyd

.close

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[undone radish]

.reopen

[full forum]

✅ Original question: #help-28 message

[undone radish]

Find two four digit numbers such whos product is $4^8 + 6^8 + 9^8$

[glossy valve]

calvin

[delicate torrent]

,w calc 4^8 + 6^8 + 9^8

[glossy valve]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by iaminfinityiq

[undone radish]

genius

no calculator btw

🌚

[delicate torrent]

Hm, notice you can write $4$ as $2 \cdot 2$, $6$ as $2 \cdot 3$ and $9$ as $3 \cdot 3$

[glossy valve]

1 divided by 0 equals Infinity

[undone radish]

yes

so its

,,(2 \cdot 2)^8 + (2 \cdot 3)^8 + (3\cdot 3)^8?

[glossy valve]

calvin

[delicate torrent]

Yep

Use some exponent rules

Aka if i let $a = 2^8$ and $b = 3^8$, what would the expression turn out to be

[glossy valve]

1 divided by 0 equals Infinity

[undone radish]

,,3a+ 3b

[glossy valve]

calvin

[delicate torrent]

How'd you get that

[undone radish]

simplifying

,, \begin{align}
3 \cdot 2^8 + 3 \cdot 3^8
\ &= 3a + 3b
\end{align}

[glossy valve]

calvin

[undone radish]

right?

[delicate torrent]

I don't get how you got $3 \cdot 2^8 + 3 \cdot 3^8$

[glossy valve]

1 divided by 0 equals Infinity

[onyx glen]

(2×3)^8 ≠ 2×3^8 btw

[undone radish]

oh im mistaken

[undone radish]

,,2^8 \cdot 2^8 + 2^8 \cdot 3^8 + 3^8 \cdot 3^8

[glossy valve]

calvin

[undone radish]

,,a^2 + ab + b^2

[glossy valve]

calvin

[delicate torrent]

Yep

[undone radish]

so now

[delicate torrent]

That should be your first instinct

[undone radish]

idk

[delicate torrent]

Now you want to factorize this

But this can't be neatly factorize

[undone radish]

,,(a+b)^2 -ab

[glossy valve]

calvin

[delicate torrent]

How are you supposed to factorize this

That's a problem

[undone radish]

wait ngm

nvm

idk how to factorise it

[delicate torrent]

Another approach is that you can let $a = 2^4$ and $b = 3^4$

[glossy valve]

1 divided by 0 equals Infinity

[sour shore]

,,(a+b-\sqrt{ab})(a+b+\sqrt{ab})

[glossy valve]

ch3rry

[sour shore]

Root shouldn't be a prob its even powers

[signal moat]

Hey does someone know how to do?

[delicate torrent]

!occupied

[full forum]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[delicate torrent]

I just realized that after i introduced a new variable declarations

[undone radish]

,,
2^8 \cdot 2^8 + 2^8 \cdot 3^8 + 3^8 \cdot 3^8 = \left(2^4\right)^2 \cdot \left(2^4\right)^2 + \left(2^4\right)^2 \cdot \left(3^4\right)^2 + \left(3^4\right)^2 \cdot \left(3^4\right)^2
\= a^2 \cdot a^2 + a^2\cdot b^2 + b^2 \cdot b^2 \= a^4 + a^2b^2 + b^4

bruh

[delicate torrent]

Miscalculations?

Should be $a^2 \cdot a^2 + a^2 \cdot b^2 + b^2 \cdot b^2$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

Then continue

[undone radish]

idk

[sour shore]

,,x^a\cdot x^b=x^{a+b}

[glossy valve]

ch3rry

[undone radish]

did i fuck up my exponents

[sour shore]

...

[undone radish]

what the hell am i doing sorryyy

[sour shore]

Nope there's no 2 here

[undone radish]

omh

[sour shore]

Neither should it be here

[glossy valve]

calvin

[undone radish]

im selling

[sour shore]

Why's arent u doing this

[undone radish]

can you teach me how to factorise like that

[undone radish]

,,(2^8 + 3^8 -2^4 \cdot 3^4)(2^8 + 3^8 +2^4 \cdot 3^4)

[glossy valve]

calvin

[sour shore]

,, x^2-y^2=(x+y)(x-y)

[glossy valve]

ch3rry

[undone radish]

yrh

yeah

[sour shore]

Yep
...now u gotta calc it i think :>

[undone radish]

no calc

😭

[sour shore]

,w 2⁸+3⁸-6⁴

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ._ch3rry_.7

[sour shore]

,w 2⁸+3⁸+6⁴

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ._ch3rry_.7

[sour shore]

,w 8113×5521

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ._ch3rry_.7

[undone radish]

oh its the answer

it was no calc

[delicate torrent]

||i accidentally introduced just to make factorizing neater||

[undone radish]

how do you get this

[sour shore]

Uhh try identifying what x and y are here

[undone radish]

wdym

oh its

ohhh

i basically helped with the answer

im so smart

😂😂😂

[delicate torrent]

Glad you said that

[undone radish]

@delicate torrent you should’ve pointed that out

🫥

[delicate torrent]

😭😭😭

[sour shore]

-#

[undone radish]

all good thanks for the help guys

[delicate torrent]

Tbh i couldn't see that factorization bc it ain't neat

[undone radish]

true

.close

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[delicate torrent]

I would want to work with neat expressions

[undone radish]

ye

[delicate torrent]

Because this has square roots which i didn't even think of

[undone radish]

yh true

[full forum]

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[sturdy vine]

$(1+i)^{2z}=n$

[glossy valve]

Goofy Joe

[sturdy vine]

Find z

[delicate torrent]

!status

[full forum]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[sturdy vine]

$\ln((1+i)^{2z})=\ln(n)$

[glossy valve]

Goofy Joe

[sturdy vine]

$z=\frac{\ln(n)}{2\ln(1+i)}$

[glossy valve]

Goofy Joe

[sturdy vine]

$\ln(1+i)=\ln(\sqrt{2})+i({\frac{\pi}{4}+2k\pi)$

[glossy valve]

Goofy Joe
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[sturdy vine]

$z=\frac{\ln(n)}{\ln(2)+i({\frac{\pi}{2}+4k\pi)}}$

[glossy valve]

Goofy Joe

[sturdy vine]

Right ?

[sturdy vine]

4

[queen flame]

right

[sturdy vine]

👍

Thanks

.close

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[somber pike]

correct as principal solution but not as general solution

not sure which one you're looking for

[sturdy vine]

.reopen

[full forum]

✅ Original question: #help-28 message

[sturdy vine]

Wdym

@somber pike

I put the periodicity

[queen flame]

why not general

[sturdy vine]

<@&286206848099549185>

[zealous stirrup]

Hi

[sturdy vine]

Can you check my work pls

[vapid shell]

Of what

[somber pike]

log is multivalued for complex numbers so for a general solution, you would need an extra +2piik term

in that step

[sturdy vine]

Where

[somber pike]

the one i replied to where you take the ln

your first step

[sturdy vine]

I just need to calculate the logarithm of n

$z=\frac{\ln|n|+i2k\pi}{\ln(2)+i({\frac{\pi}{2}+4k\pi)}}$

[glossy valve]

Goofy Joe

[sturdy vine]

So this

[somber pike]

What is the original question?

[sturdy vine]

$(1+i)^{2z}=n$

[glossy valve]

Goofy Joe

[somber pike]

Just find z or is there any other info given?

[sturdy vine]

Solve the equation

[somber pike]

If I were you I'd rewrite the left side like this first $(1+i)^{2z} = e^{2z\ln(1+i)}$

[glossy valve]

bartdestinkerd34

[somber pike]

Then take the multivalued ln

$e^w = n \Rightarrow w = \ln n + 2\pi i k,\quad k\in\mathbb{Z}$

[glossy valve]

bartdestinkerd34

[full forum]

@sturdy vine Has your question been resolved?

[sonic stratus]

also, ln n is meant to be ln|n|

[sturdy vine]

Thanks all

[full forum]

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[ember quartz]

Hello, could someone check if this proof looks good please?

\begin{Definition}[Divisibility]
A nonzero integer $a$ is said to \emph{divide} an integer $b$, written $a \mid b$, if $b = ak$ for some integer $k$.
\end{Definition}

% --------------------------------------------------------------------------------

\begin{Theorem}
Assume $a, b$ and $c$ are integers. If $abc$ is a multiple of 10, then at least one of $ab$, $ac$, or $bc$ is a multiple of 10.
\end{Theorem}

\begin{proof}
Let $a, b$ and $c$ be integers, and $abc$ a multiple of 10.
So, by the definition of divisibility, $abc = 10k = 2(5)(k)$ for some integer $k$.
This is means that $2 \mid abc$ and $5 \mid abc$.

Without loss of generality, assume that $2 \mid a$ and $5 \mid b$, which means that $a = 2l$ and $b = 5m$ for some integers $l$ and $m$. 
Then, $abc = 2l(5m)(k) = 10lmk$.
In this case, $ab$ is a multiple of 10.

A similar case can be constructed for $5 \mid a$ and $2 \mid 5$, and with either $a$ and $c$, or $b$ and $c$.
Then, at least one of the three is divisible by 2 and one of the two left are divisible by 5.
Therefore, at least one of $ab$, $ac$, or $bc$ is a multiple of 10.
\end{proof}

[glossy valve]

Mor Bras

[spiral vigil]

you can't assume 2 | a and 5 | b, for example if i pick a = 10 and b = c = 1

[marsh spruce]

looks alright but i think defining a := 2l and b := 5m seems redundant. you can just conclude that if 2 | ab and 5 | ab then 10 | ab. also 2 | 5 looks like a typo?

and yeah you're neglecting the possibility that 10 | a and not any other of b or c

[ember quartz]

Certainly, I'm fixing the proof accordingly to your comments

[full forum]

@ember quartz Has your question been resolved?

[ember quartz]

Here's the fixed version:

\begin{Definition}[Divisibility]
A nonzero integer $a$ is said to \emph{divide} an integer $b$, written $a \mid b$, if $b = ak$ for some integer $k$.
\end{Definition}

\begin{Theorem}
Assume $a, b$ and $c$ are integers. If $abc$ is a multiple of 10, then at least one of $ab$, $ac$, or $bc$ is a multiple of 10.
\end{Theorem}

\begin{proof}
Let $a, b$ and $c$ be integers, and $abc$ a multiple of 10.
So, by the definition of divisibility, $abc = 10k = 2(5)(k)$ for some integer $k$.
This is means that $2 \mid abc$ and $5 \mid abc$.

Without loss of generality, assume that $2 \mid a$, which means that $a = 2l$ for some integer $l$. 
Then, $abc = 2lbc$.
Now, either $ 5 \mid l$ or $5 \cancel\mid l$.

If $5 \mid l$, then $ 10 \mid a$, so $10 \mid ab$ or $10 \mid ac$.
If $5 \ \cancel\mid \ l$, then, either $5 \mid b$ so $10 \mid ab$, or $5 \mid c$ so $10 \mid ac$.
In any case, $ab$ or $ac$ is a multiple of 10.

A similar case can be constructed for with either of the rest of the variables $b$ and $c$.
Then, at least one of the three is divisible by 2 and one of the two left are divisible by 5.
Therefore, at least one of $ab$, $ac$, or $bc$ is a multiple of 10.

\end{proof}

[glossy valve]

Mor Bras

[spiral vigil]

you already said WLOG so you don't really need that last paragraph. i'd phrase it like 2 | abc, so we know either 2|a, 2|b, or 2|c. WLOG, assume 2|a, which means...

[brittle steeple]

have some bad sols because they were too funny not to post here

[ember quartz]

Yes, I was thinking about the pingeonhole principle too

But with the boxes as 2 and 5, and the pigeons as a, b, and c, but that doesn't work

[brittle steeple]

boxes are ab, ac, bc, pigeons are 2 and 5

yeah saying WLOG is the same thing as saying the rest follow from a similar case

in this case it's probably obvious enough that you don't need to explain why beyond clumsy's suggestion

[ember quartz]

\begin{Definition}[Divisibility]
A nonzero integer $a$ is said to \emph{divide} an integer $b$, written $a \mid b$, if $b = ak$ for some integer $k$.
\end{Definition}

% --------------------------------------------------------------------------------

\begin{Theorem}
Assume $a, b$ and $c$ are integers. If $abc$ is a multiple of 10, then at least one of $ab$, $ac$, or $bc$ is a multiple of 10.
\end{Theorem}

\begin{proof}
Let $a, b$ and $c$ be integers, and $abc$ a multiple of 10.
So, by the definition of divisibility, $abc = 10k = 2(5)(k)$ for some integer $k$.
This is means that $2 \mid abc$ and $5 \mid abc$.
So we know that either $2 \mid a$, $2 \mid b$ or $2 \mid c$.

Without loss of generality, assume that $2 \mid a$, which means that $a = 2l$ for some integer $l$. 
Then, $abc = 2lbc$.
Now, either $ 5 \mid l$ or $5 \ \cancel\mid \ l$.

If $5 \mid l$, then $ 10 \mid a$, so $10 \mid ab$ or $10 \mid ac$.

If $5 \ \cancel\mid \ l$, then, either $5 \mid b$ so $10 \mid ab$, or $5 \mid c$ so $10 \mid ac$.
In any case, $ab$ or $ac$ is a multiple of 10.
The rest follow from a similar case.
% Then, at least one of the three variables is divisible by 2 and one of the two left are divisible by 5.
Therefore, at least one of $ab$, $ac$, or $bc$ is a multiple of 10.

\end{proof}

[glossy valve]

Mor Bras

[brittle steeple]

Looks good. The biggest "leap" here is saying that if 5 does not divide l, then it must divide b or c. If you're comfortable with why that is, then I'd say you're done

I would get rid of "The rest follow from a similar case". That's already covered when you assume 2|a WLOG.

[ember quartz]

10 must divide abc by assumption, and 2 divides a. If 5 doesn't divide a, then it must be that either 5 divides b or c, so to have 10 divides ab or ac, and so 10 divides abc. I may need to think in more about the detail of "if 5 does not divide l, then it must divide b or c"

[brittle steeple]

If you want a similar, slightly simpler step to check justification, why does 2|abc imply 2|a, 2|b, or 2|c ?

[ember quartz]

Because, in the case of 2 | abc implies 2 | a or 2 | bc, either

• 2 | a, then is given,
• or 2 doesn't divide a, then since 2 is prime, the gcd(2, a) = 1, and by bézout's identity, 2 must divide bc.

Reaplying this for bc, then 2 divides either b or c. Therefore, 2|abc imply 2|a, 2|b, or 2|c

Now I see, I'll correct this with a couple of lemmas.

[full forum]

@ember quartz Has your question been resolved?

[brittle steeple]

I don't mean to make you do everything in excruciating detail

[brittle steeple]

The justification for this is fairly similar, so I was checking if you knew it

[ember quartz]

It allows me to be more verbose, which is something I want to do, but I didn't know I could be more detailed in this, so thanks for pointing it out 👌

[brittle steeple]

Ah all right, nice

[ember quartz]

I'll close this for now, as I'm having troubles with latex

Thanks for your responses!

.close

[full forum]

Channel closed

Closed by @ember quartz

Use .reopen if this was a mistake.

[ember quartz]

.reopen

[full forum]

✅ Original question: #help-28 message

[ember quartz]

Ok, so here's the fix:

\begin{Definition}[Divisibility]
A nonzero integer $a$ is said to \emph{divide} an integer $b$, written $a \mid b$, if $b = ak$ for some integer $k$.
\end{Definition}

\begin{Lemma}
Let $a$, $b$, and $c$ be integers, and let $p$ be a prime.
\begin{enumerate}
\item\label{l1} If $p \ \cancel\mid \ a$, then $\gcd(p,a) = 1$.
\item\label{l2} If $a \mid bc$ and $\gcd(a, b) = 1$, then $a \mid c$.
\item\label{l3} If $p \mid bc$, then either $p \mid b$ or $p \mid c$.
\end{enumerate}
\end{Lemma}

% --------------------------------------------------------------------------------

\begin{Theorem}
Assume $a, b$ and $c$ are integers. If $abc$ is a multiple of 10, then at least one of $ab$, $ac$, or $bc$ is a multiple of 10.
\end{Theorem}

\begin{proof}
Let $a, b$ and $c$ be integers, and $abc$ a multiple of 10.
So, by the definition of divisibility, $abc = 10k = 2(5)(k)$ for some integer $k$.
This is means that $2 \mid abc$ and $5 \mid abc$.
So we know that, because 2 is prime, by lemma 2.3, either $2 \mid a$, $2 \mid b$ or $2 \mid c$.

Without loss of generality, assume that $2 \mid a$, which means that $a = 2l$ for some integer $l$. 
Then, $abc = 2lbc$.
Now, either $ 5 \mid l$ or $5 \ \cancel\mid \ l$.

If $5 \mid l$, then $ 10 \mid a$, so $10 \mid ab$ or $10 \mid ac$.

If $5 \ \cancel\mid \ l$, then it must be that $5 \mid bc$, and, since 5 is prime, by lemma 2.3, either $5 \mid b$ so $10 \mid ab$, or $5 \mid c$ so $10 \mid ac$.

In any case, $ab$ or $ac$ is a multiple of 10.
The rest follow from a similar case.
Therefore, at least one of $ab$, $ac$, or $bc$ is a multiple of 10.

\end{proof}

[glossy valve]

Mor Bras

[delicate torrent]

im impressed with the proof you got there

[delicate torrent]

did you learn congruency yet?

[ember quartz]

No, I'm learning how to write proofs, right now I'm with direct proofs, and divisibility is one of the topics the book covers for this kind of proofs

[delicate torrent]

hm

then i think you would be interested in congruency

in number theory ofc

[ember quartz]

Certainly

[delicate torrent]

definition: $a \equiv b \mod{m}$ for $a, b, m \in \mathbb Z$ when $m \mid a - b$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

by words, $a$ and $b$ has the same remainder when dividing by $m$

[glossy valve]

1 divided by 0 equals Infinity

[ember quartz]

Ah yes, the book touches this is some exercises

[delicate torrent]

cool

try to prove some basic properties of it first

for example:

if $\begin{cases} a \equiv b \mod{m} \ c \equiv d \mod{m} \end{cases}$ then $a + c \equiv b + d \mod{m}$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

or

if $\begin{cases} a \equiv b \mod{m} \ c \equiv d \mod{m} \end{cases}$ then $a - c \equiv b - d \mod{m}$

[glossy valve]

1 divided by 0 equals Infinity

[ember quartz]

Of course

[delicate torrent]

or

if $\begin{cases} a \equiv b \mod{m} \ c \equiv d \mod{m} \end{cases}$ then $ac \equiv bd \mod{m}$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

these only need some algebraic manipulations

[ember quartz]

For sure

[delicate torrent]

aight

let's see how you would do these

[ember quartz]

<@&286206848099549185> Hello, could someone else check if this proof looks good?

[delicate torrent]

hello

[ember quartz]

Hi

[delicate torrent]

send in the proof

[ember quartz]

Yes

[delicate torrent]

or is it still this one

[ember quartz]

\begin{Definition}[Divisibility]
A nonzero integer $a$ is said to \emph{divide} an integer $b$, written $a \mid b$, if $b = ak$ for some integer $k$.
\end{Definition}

\begin{Lemma}
Let $a$, $b$, and $c$ be integers, and let $p$ be a prime.
\begin{enumerate}
\item\label{l1} If $p \ \cancel\mid \ a$, then $\gcd(p,a) = 1$.
\item\label{l2} If $a \mid bc$ and $\gcd(a, b) = 1$, then $a \mid c$.
\item\label{l3} If $p \mid bc$, then either $p \mid b$ or $p \mid c$.
\end{enumerate}
\end{Lemma}

% --------------------------------------------------------------------------------

\begin{Theorem}
Assume $a, b$ and $c$ are integers. If $abc$ is a multiple of 10, then at least one of $ab$, $ac$, or $bc$ is a multiple of 10.
\end{Theorem}

\begin{proof}
Let $a, b$ and $c$ be integers, and $abc$ a multiple of 10.
So, by the definition of divisibility, $abc = 10k = 2(5)(k)$ for some integer $k$.
This means that $2 \mid abc$ and $5 \mid abc$.
So we know that, because 2 is prime, by lemma 2.3, either $2 \mid a$, $2 \mid b$ or $2 \mid c$.

Without loss of generality, assume that $2 \mid a$, which means that $a = 2l$ for some integer $l$. 
Then, $abc = 2lbc$.
Now, either $ 5 \mid l$ or $5 \ \cancel\mid \ l$.

If $5 \mid l$, then $ 10 \mid a$, so $10 \mid ab$ or $10 \mid ac$.

If $5 \ \cancel\mid \ l$, then it must be that $5 \mid bc$, and, since 5 is prime, by lemma 2.3, either $5 \mid b$ so $10 \mid ab$, or $5 \mid c$ so $10 \mid ac$.

In any case, $ab$ or $ac$ is a multiple of 10.
The rest follow from a similar case.
Therefore, at least one of $ab$, $ac$, or $bc$ is a multiple of 10.

\end{proof}

[delicate torrent]

so it's still the old one

grammar error

[ember quartz]

Yes, the whole thread is for this proof

[delicate torrent]

[glossy valve]

Mor Bras

[delicate torrent]

i think the proof is okay

do you need to prove the lemmas btw

[turbid badge]

great proof taught me some latex 👍👍

[delicate torrent]

$\LaTeX$

[glossy valve]

1 divided by 0 equals Infinity

[turbid badge]

thank you for teaching me more latex

[ember quartz]

It's true, but then the proof would be quite lengthy, and because I think the topic is "basic", I just proof the theorem, and the rest are definitions and lemmas

[turbid badge]

now all that I need to know is if it is lah-tek or lay-tek

[delicate torrent]

what's $\pi \left( \frac{1}{2} - \frac{1}{4} - \frac{1}{8} - \frac{1}{16} - \frac{1}{32} - \frac{1}{32} \right)$ btw?

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

lah-tek

@turbid badge

[turbid badge]

hold on

0?

or am i trippin

[delicate torrent]

🔥

yessir

[turbid badge]

okay we've sidetracked too much mb I'll take my leave

[delicate torrent]

specifically 0 and some latex

[turbid badge]

night

[delicate torrent]

anyways

whenever you finished perfecting your proof

you can

!done

[full forum]

If you are done with this channel, please mark your problem as solved by typing .close

[ember quartz]

!done

[full forum]

If you are done with this channel, please mark your problem as solved by typing .close

[ember quartz]

Ok

I'll leave it at that, thanks for your responses!

.close

[full forum]

Channel closed

Closed by @ember quartz

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
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• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[proven lynx]

hello,

[proven lynx]

do i have to do -1 *(b+c)+2 times0,5 = 0.35

or is it jsut added

[full forum]

@proven lynx Has your question been resolved?

[robust slate]

Not sure what's going on here. The only possible values of $X$ in the interval $[-2,2]$ are $-1$, $0$, and $2$. So,
\begin{align*}
P(-2 \leq X \leq 2) &= P(X=-1)+P(X=0)+P(X=2) \
&= (b+c)+(d-c)+0.05
\end{align*}

gtg, but hope this helps!

[glossy valve]

Civil Service Pigeon

[proven lynx]

yes i understand i was confused

i thought i have to multiply the -1 and 0 and 2

with the corresponding P(x)

[robust slate]

That's when you find expected value

but this is a probability

[proven lynx]

oh yeah okay

[full forum]

@proven lynx Has your question been resolved?

[glossy valve]

LUC1DV1B3S_010

$\LaTeX$

[full forum]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[full forum]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cold sluice]

Hi, I'm trying to do this problem in this geometry video in Khan Academy before it's shown in the video how to do it and was wondering if someone could nudge me in some direction but without any hints.

[cold sluice]

This is my work so far

I'm thinking maybe I can construct a line between point E and C to make a statement that triangle ECG is congruent to triangle EBG because of side angle side triangle congruency and that might possibly help me with figuring out the measure of angle BED?

Then the sum of angle BEG and GEC and CED equals the measure of angle BED.

[sonic stratus]

Quick question, im allowed to refer to the unit circle?

[cold sluice]

I haven't learned it yet, it's in Algebra 2 in the trigonometry unit in Khan Academy.

I'm learning high school geometry on Khan Academy.

[sonic stratus]

Okay i know the angle that forms but i have 0 clue about how the proof for it being that angle is carried

lemme try to think a bit ig.

[cold sluice]

by that angle you mean angle BED right?

[cold sluice]

Take your time!! 🩷 ty for helping me <33

[sonic stratus]

once you find any of the useful angles the problem basically falls off as trivial tbh

[cold sluice]

Hmmm

I also know that if we continue line DE then the sum of angles (let's call the new point that forms point Z) then the sum of angle ZEB, BEG, GEC, and CED have to equal 180 degrees, does that help me?

Can you continue lines? is that a thing?

[sonic stratus]

Yeah you can continue lines

[cold sluice]

Nice

[sonic stratus]

You can anything basically, but im not entirely sure that helps

Since we dont know / cant prove where Z is

[cold sluice]

Right

[sonic stratus]

First of all, lets assume that AB is = 1

[cold sluice]

But I think maybe if we do algebra then that'd help immensely, like if we know 3 of the angles for example then the 4th has to be known with algebra.

[sonic stratus]

it just makes everything easier

[cold sluice]

Alright

that's fair

Unit square? is that an appropriate term?

[sonic stratus]

In reality i did it because of the unit circle, but yeah

the same idea applies

[cold sluice]

Nice

[sonic stratus]

Oh goddam im so dumb

pythagoras

BE is a radius

so its = 1

[cold sluice]

I also know that BE is the radius of the circle, because the center of it is at b and one line segment connects it to the circle so that has to be a radius right?

[sonic stratus]

BG = 1/2

[cold sluice]

How do we know that BG = 1/2?

i dont see it

Ohhhhhhhhhhhhhhhhhhhhhhhhhhh

Nevermind I don't see it

Can you show me?

[sonic stratus]

If AB = 1

All the sides = 1

BG = GC
BG + GC = BC = 1

[cold sluice]

Oh cool! Because a^2+b^2=1 a and b have to be 1

Yeah

[sonic stratus]

BG = GC = 1/2

have you learnt cosine law already?

[cold sluice]

Nope

[cold sluice]

This makes sense to me intuitively though

[sonic stratus]

okay, theres another way to show it anyways.

[cold sluice]

The distance between BG and BC is 1/2 the total line right?

[sonic stratus]

Notice, the triangle BEG is just a reflection of triangle CEG

[cold sluice]

Right!

[sonic stratus]

Since they are reflections

[cold sluice]

Yeah

[sonic stratus]

What is the length CE?

[cold sluice]

It's 1

[sonic stratus]

join the two triangles

[cold sluice]

Because it's congruent to BE

[cold sluice]

How do you do that?

Like add the area?

[sonic stratus]

Mb took a sec

[cold sluice]

NP

[sonic stratus]

I mean, we are looking at two triangles with a shared side

BEG and CEG

Ignore G and join them into a bigger triangle

CEB

[cold sluice]

Also I realized now that the radius BE is the same as the length BA and BC right? Or does that need to be proved?

[cold sluice]

Ah!

Okay

Omg

[sonic stratus]

It does not need to be proved, no

Check the sides of the triangle now

[cold sluice]

Now their measure has to be 90 degrees, which means that the measure of BED - 90 degrees = the unknown angle, and added together we get BED

[sonic stratus]

What are the lengths BE, BC and EC?

[cold sluice]

Ah now it's an isosceles triangle right?

[cold sluice]

1, 2, and 1

Right?

[sonic stratus]

BC = 2?*

[cold sluice]

No 2

oh it's 1

Because it's the same size as the radius

Mb

[sonic stratus]

What are the internal angles of any equilateral

[cold sluice]

60?

[sonic stratus]

Yep

But you can also find another useful triangle now

[cold sluice]

Why isn't angle BEC = 90?

[cold sluice]

The mirror of triangle DFE from the bottom left corner?

I suppose the reason that earlier we can pick radius 1 for segment BE is because whatever the true number is the proportions stay the same?

So if instead of 1 the radius was actually 4 then the logic still stays the same?

That's really sick

[sonic stratus]

Try to construct the Triangle BEC and check if its looks like a 90º angle

[cold sluice]

Also I really appreciate the ancient greeks figuring out SAS and other triangle congruencies because it makes it way easier to prove congruency of certain triangles.

[forest hawk]

I've got the answer
Sharing in some time

[sonic stratus]

To be honest, theres a point of no return in trigonometry where you learn that trig is in reality based on circles

And past that you become a lazy fuck like me

[cold sluice]

Logically it has to though? Because 90+2x=180 and so x has to equal 45 and BEG+GEC = 90?

[sonic stratus]

angle BEG there is 30º

angle BGE is 90º

[cold sluice]

I'm not looking for it, and that's not what this server is for. Please don't share it without guiding me properly.

Dont' share the answer please

[forest hawk]

I'm solving it
On my own

[cold sluice]

Nice

[cold sluice]

How? every triangle has to add to 180 degrees.

[sonic stratus]

Angle EBG is 60º

30+60+90 = 180

[cold sluice]

You can have a right triangle with angles like that?

I thought that it was fixed

[sonic stratus]

Right angles always have a 90º angle

[cold sluice]

Yep

[sonic stratus]

theres nothing specified on any of the other two

[cold sluice]

This is insane, I didn't know that!

[sonic stratus]

Well, we have this now

[cold sluice]

How do I find this out though?

[sonic stratus]

Again, we stated that every equilateral has 60º angles

[cold sluice]

Yeah

Ah okay!

[sonic stratus]

and you were the one that found that all sides had length 1

[cold sluice]

Because EC and BE are equal and are 1

Yep

This helps a lot because all that's left is this triangle now and we have to find the angle.

[sonic stratus]

Yep

[cold sluice]

60+ that angle = BED

[sonic stratus]

This triangle as an important fact

[cold sluice]

That makes sense

[sonic stratus]

Try to find the two larger lengths

one you already did, its CE

[cold sluice]

Yep

[sonic stratus]

you also want CD too.

[cold sluice]

Also, we have a transversal right?

[sonic stratus]

The transversal doesnt really matter

[cold sluice]

Oh ok

[sonic stratus]

Try to check the two blue lengths there

and find the value

[cold sluice]

The middle one has length 1, and the bottom one has to also be one because it's congruent to BA because it's a square.

Am i right?

i mean does that prove it?

[sonic stratus]

Yep, so its an isosceles

You also technically know the bottom-right angle of that triangle.

[forest hawk]

We can use area of sector also

[cold sluice]

I don't know that

[forest hawk]

If we see the figure ACB is a quadrant

[cold sluice]

I don't see how

[forest hawk]

Ar of sector AEB + Ar of sector BEC= ar of quadrant ACB

[sonic stratus]

You know that the corner is a 90º, as of all squares
and you know the inner angle of the equilateral thats just above our triangle of interest.

[cold sluice]

i don't know what you're talking about, i'm gonna listen to @sonic stratus for now if that's ok

[forest hawk]

In which grade are you?

[cold sluice]

I won't share that

[sonic stratus]

this is completely unnecesary tbh, you can do this only by appealing to a really basic trig identity.

[cold sluice]

That's right!

So 90-60=30

[sonic stratus]

Now, isosceles have the nice property of sharing 2 equal angles apart from 2 equal sides.

[cold sluice]

And since we know that it's an equilateral the two opposite sides have to each be 75

Right?

[sonic stratus]

not equilateral, but yeah, the idea is right

[cold sluice]

Sorry I mean isosceles

[sonic stratus]

yeah, and you now already have BED

[cold sluice]

Omg tysm!!! 60+30=90=BED

sorry 75

[sonic stratus]

Make sure to draw over your problem to calculate the whole thing

[cold sluice]

60+75=135=BED!

[sonic stratus]

yea

[cold sluice]

https://tenor.com/view/cute-bunny-bed-gif-19199485

Tyyyyy!!!!

🫶

[sonic stratus]

Ill just spoil you like 2 years of study

[cold sluice]

Yeah?

[sonic stratus]

the thing i mentioned at the start where i already knew the important angle

[cold sluice]

That's just a trig identity?

[sonic stratus]

not really
is because its a result that theyll engrave into your brain, lmao

[cold sluice]

Oohhh XD

that's cool!!!

I really appreciate it!!!

@forest hawk Thank you too!!!

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[hot kernel]

$$x = \frac{2y^2+6}{3y-3}$$

[glossy valve]

hoax

[hot kernel]

Find dx/dy

my approach plz tell me what im doing wrong

lemme type it in latex rq

[delicate torrent]

Let's see how you cooked this

[hot kernel]

$$ x(3y-3) = 2y^2 + 6$$
$$\frac{d}{dy}(x(3y-3)) = \frac{d}{dy}(2y^2 + 6).... \text{differentiating both sides with respect to y}$$

[glossy valve]

hoax

[onyx glen]

overcomp

are you not able to just do quotient rule?

or better yet, do some long division and then differentiate

[hot kernel]

nah imma just do product rule instead

[delicate torrent]

Same thing

[hot kernel]

cuz it seeme way simpler that way

look

[delicate torrent]

Well uhm, this is complicated already

[hot kernel]

$$\frac{dx}{dy}(3y-3) + 3x = 4y+6$$

[onyx glen]

it's not

[glossy valve]

hoax

[hot kernel]

ain't this it/

f'g + fg'

[delicate torrent]

It's just complication

[onyx glen]

you will get something implicit when there is no good reason for it

[hot kernel]

and then from here i can just isolate the dx/dy

[delicate torrent]

Quotient rule and product rule are a mess to deal with if differentiated directly from this

I'd rather do long division

[hot kernel]

also there shouldn't be plus 6

i got the expression x(3y-3) on the LHS, and 2y^2 + 6 on the RHS

differentiating both sides with respect to y

we apply product rule on the LHS

[delicate torrent]

What's the purpose of this anyway

[hot kernel]

i am doing questions rn

for

math exam soon

[delicate torrent]

You're just overcomplicating yourself rn

[hot kernel]

it's from this if ur wondering anyway

[sour shore]

-# why u hate quotient rule

[delicate torrent]

I meant what's the purpose of cross multiplying here

[delicate torrent]

-# or long division

[hot kernel]

quotient rule ugly

u see the thing is

when i try my approach

[delicate torrent]

Then long division

[hot kernel]

i get (4y-3x)(3y-3)

[delicate torrent]

Long division not rlly complicated

[hot kernel]

dx/dy

[sour shore]

-# its p quick in this case

[hot kernel]

but

uh this is the answer

[sour shore]

-# quotient rule yay

[hot kernel]

like what am i doing wrong/

[delicate torrent]

-# i agree that quotient rule is ugly and i can't sven remember ts, but i'd rather do long division here

[hot kernel]

also why is ur text so tiny

[delicate torrent]

What even IS the purpose of cross multiplying in your case

[hot kernel]

if u look up and check my working

much easier to deal with

unless im doing something wrong

[delicate torrent]

How are you going to get $x'$ alone on 1 side

[glossy valve]

1 divided by 0 equals Infinity

[hot kernel]

and i also asked "what am i doing wrong here??"

[delicate torrent]

That's the problem

[sour shore]

6 constant

[delicate torrent]

@hot kernel try to answer this

[hot kernel]

i thought it'd just be derivative of x with respect to x multiplied by derivative of y with respect tox

unless im thinking about the chain rule incorrectly

[delicate torrent]

Can you fact-check this? (Not the replied msg, the msg that was recently sent)

[sour shore]

Which msg what

[delicate torrent]

This one

[frigid carbon]

Why not just use quotient rule

[delicate torrent]

Ugly (that's the whole reason)

[frigid carbon]

It's better

Than this

[delicate torrent]

Not my opinion tho

Im not even crazy enough to cross multiply all of this

[hot kernel]

i have the expression x(3y-3)

[hot kernel]

dude it's not hard to cross multiply x(3y-3) what?

[frigid carbon]

^

[hot kernel]

so i have the expresssion x(3y-3)

if i decide to differentiate this expression with respect to y

applying the product rule, won't we just get (dx/dy)(3y-3) + x(3)

[delicate torrent]

How are you differentiating $x(3y - 3)$ with respect to $y$

[glossy valve]

1 divided by 0 equals Infinity

[hot kernel]

can u not/

?

[delicate torrent]

$x'(3y - 3) + 3x$ you mean?

[glossy valve]

1 divided by 0 equals Infinity

[sour shore]

Yh

[onyx glen]

but you will get dx/dy in terms of x and y

and you have to put x=(original fraction) back in

you don't actually save any effort in this way

[delicate torrent]

Which is more complicated

[hot kernel]

oh right

[sour shore]

Also d( 2y²+6)/dy = 4y

[hot kernel]

i get something implcit i see

[sour shore]

Diff of 6 is 0

[delicate torrent]

If you think quotient rule is ugly, try long division

[hot kernel]

no u see i am just used to implicit differentiation that i did not see we get dx/dy explicitly

[onyx glen]

that's called tunnel vision

[delicate torrent]

$\frac{2y^2 + 6}{3y - 3} = \frac{2(y - 1)^2 + 4(y - 1) + 12}{3(y - 1)}$

[glossy valve]

1 divided by 0 equals Infinity

[delicate torrent]

This is much easier to differentiate (thank god i did my calculations correctly)

[onyx glen]

i will do you one better

[sour shore]

-# poor quotient rule getting hate for no rsn

[frigid carbon]

$$x = (2y^2+6)*(3y-3^{-1}$$ product rule this if you not comfortable with quotient tbh, this would be more ugly

[glossy valve]

heisenberg

[onyx glen]

$\frac{2y^2+6}{3y-3}=\frac23\cdot\frac{y^2+3}{y-1}= \frac23\paren{y+1+\frac{4}{y-1}}$

[glossy valve]

Ann

[delicate torrent]

Or this

[hot kernel]

aight thx

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[undone radish]

can someone explain the solution please

!status

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7. None of the above

[undone radish]

1

i dont understand the explanation of the solution

AP means the addition principle btw

[bright wagon]

S1

[onyx glen]

x^2+y^2<=5 means x^2+y^2 is precisely any of 0,1,2,3,4,5

@undone radish they are counting up how many pairs (x,y) satisfy x^2+y^2=0, then x^2+y^2=1, etc up to 5. then they add those counts up.

[undone radish]

ohh okay

is the only way trial and error

[rare dock]

no

[onyx glen]

you can make do with some symmetry arguments i guess

[sour shore]

Does stars and bars work here?

[onyx glen]

but some case checking will be required

[undone radish]

hmm okay

thank you guys

[rare dock]

for numbers this small you might as well “trial and error”

[undone radish]

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[brittle steeple]

for large enough values of 5 it'll be more efficient to go row by row

[sour shore]

7C2...

[brittle steeple]

Like, try each x-value one by one and count number of y-values

[onyx glen]

probably just coincidence?

[rare dock]

you could do it more dynamically/recursively for larger numbers than 5

[brittle steeple]

Well we already see that for radius 2 the value is 13

that isn't nCr for any n≤13

[sour shore]

Yeah

[onyx glen]

it's 13C1

you meant n<13

[brittle steeple]

Oh yeah lol

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[brittle steeple]

I would start by writing out the definitions

[lean rock]

monic gf:
∀uv, g∘f∘u=g∘f∘v → u=v

[brittle steeple]

tfw no monic gf

[brittle steeple]

Ok, let's assume this is true and true to prove that f is monic

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@lean rock Has your question been resolved?

[lean rock]

Look I proved it with tableux!

tree closes, so it's logical truth ∎

[brittle steeple]

Wow

Looks interesting

[lean rock]

Sorry it's not entirely accurate with the variable because I left sketches, I can edit it, one sec

[brittle steeple]

Assume gf is monic. Then, we need to show for all u,v that fu=fv => u=v

fu = fv (assumed premise)
gfu = gfv (left-compose by g)
u=v (gf is monic)
if you want a more readable proof

[lean rock]

Thank you so much!

[brittle steeple]

np!

[lean rock]

the argument is
P
∴ Q

write premises with the conclusion negated in a tree:
P
¬Q

this means P∧¬Q

and show if premises are true the conclusion cannot be false,
this tree closes with the conclusion negated, so if the premises are true the conclusion cannot be false

but I maybe am making mistakes as I am learning this now

yours is more elegant 🌟

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[tropic jolt]

how do I do this?

[tropic jolt]

what I tried was, to map out the possibilities
(5, 2) (1/3, 1/4)
(2, 5) (1/6, 1/4)
(4, 3) (1/3, 1/4)
(3, 4) (1/6, ,1/4)
and then calc
3/12 / 1 = 1/4

[signal wasp]

its given that sum can also be more than 7, so you would need other cases like (5,3), (5,4), (5,5) etc

[dire swallow]

Try calculating probability that the sum is less than 7, then take its complement, it might be easier

[queen flame]

do we need to consider that the sectors are not equal?

[tropic jolt]

yeah

[tropic jolt]

OH

OML

i thought exactly 7 omg

i keep messing up cus i don't read the question

[tropic jolt]

yeah lemme try it out

okay well still seems wrong

(2, 2), (3, 3), (2, 3), (3, 2), (4, 2), (2, 4)

(1/6*1/4) x 4 + (1/3x1/4) x 2 = 1/3

where'd I go wrong this time?

[signal wasp]

(2,2), (2,3), (2,4), (3,2), (3,3) have probability 1/6*1/4

(4,2) has 1/3*1/4

[tropic jolt]

right

[signal wasp]

(1/6*1/4) x 5 + (1/3x1/4) x 1

[tropic jolt]

ohhhhhhhhhhhhhhhhh

oops

1-7/24 0 . 70

thank you oml

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[native furnace]

Hey how do I do number 7

✌🏼

[sonic stratus]

rare instance where we assume perfect friction

[native furnace]

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[delicate torrent]

!status

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2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
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[torn jolt]

1

[stiff pier]

What are you trying to do with this expression?

[torn jolt]

I’m tryna prove it u feel