help-26

[grim swift]

guys is this politically correct

[mint crescent]

uhhh

if you're asking if the two are equivalent

then no

[mellow venture]

isnt "politically correct" a dogwhistle

[drifting swift]

are you serious or was this an attempt at a joke

[grim swift]

w8

i wanted to say correct

but i add this

[drifting swift]

what's politics got to do with it

anyway no these two are not equal

[grim swift]

hmmm because it was part of inverse function operation

so i also got confused

as these 2 are not equal

w8 i put whole

🤔

this is solved task btw

Maybe like

[drifting swift]

...

"whole"...

[compact moss]

is the inverse

[drifting swift]

ok you've introduced like 5 symbols that we DON'T know the meanings of

[compact moss]

has nothing to do with being equal

[grim swift]

it dont matter just focus on x , p

those 2 are variables

like

in inverse

ye

from this context it would be correct?

[compact moss]

x=8/sqrt(p) isolate p

to find inverse

[grim swift]

hmmmmmmm

make senses now hmmm

[compact moss]

idk what youre saying is equal though

[grim swift]

🤔

[compact moss]

to the left of P

[grim swift]

very nice?

[compact moss]

very nice

[grim swift]

👨‍🏫🤝🕵️‍♂️👍👍👍

yo

sorry

@compact moss

this is not possible?

[compact moss]

if p=64 then x(64)=8/8=1

but p(1)=8/sqrt(1) is not 64

[grim swift]

😮

so every time when make inverse, it has to be clean of any roots or ^'s ?

[compact moss]

no

[grim swift]

🫠

[topaz sinew]

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[static jacinth]

I'm struggling with the pigeonhole principle formal definition in this example

[static jacinth]

Say I have the set S = {1,2, ... ,10}

And 5 subsets of S, each with 4 elements in it. Let's call them A_1, ... A_5. Let A = {A_1, ..., A_5}

Let R be a relation in AxS such that a set A_i is related to an element n if n is in A_i

R has cardinality 20 because each one of the five sets is related to four elements

Now, this inequality holds

There is some $A_i \in A$ such that $#{n\in S / A_i R n}\geq \frac{#R}{#A}$

[thorny flame]

casiel368

[static jacinth]

But #R/#A = 4

Oh hmm I see what is wrong

I want to prove that there is some number in two of the A_i

Using pigeonhole, because this is a toy problem for another one that can't be solved using the non-injectivity of a function

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[deep condor]

How many 8-card hands are possible if you want: Two different 3-of-a-kinds and one 2-of-a-kind?

Why is it 13choose2 * 4choose3 * 4choose3 for the first part? Why isn't it 13choose1 * 4choose3 * 13choose1 * 4choose3?

[worthy storm]

two different 3 of a kinds

if you had 13 choose 1 twice, that would mean that the two 3-of-a-kinds could be the same card value

[deep condor]

Oh, in that case could you do 13choose1 * 4choose3 * 12choose1 *4choose3?

[worthy storm]

in that case you would be saying that order matters

i.e. (three aces) and (three kings) would be treated as a separate case vs. (three kings) and (three aces)

[deep condor]

Okay, thank you so much!

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[silk pewter]

show that $\frac{\log 12}{\log 18}$ is irrational

[thorny flame]

bettimsucks

[topaz sinew]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[silk pewter]

1

[ocean mirage]

perhaps?

[neon iron]

irrationality proofs usually use contradiction

[silk pewter]

how does that help?

[silk pewter]

so i consider it to be rational

?

[neon iron]

yeah

suppose for a contradiction that 18^(p/q) = 12....

etc

[silk pewter]

so uh

let $\log_{18} 12$ be rational and be $=\frac{p}{q}$

[thorny flame]

bettimsucks

[silk pewter]

aaand $18^{\frac{p}{q}} = 12$

[thorny flame]

bettimsucks

[silk pewter]

how do i move forwards?

[neon iron]

clear denominators, then factor

[silk pewter]

huh

18^p = 12^q

?

[neon iron]

yes

[silk pewter]

then

:9

[neon iron]

write everything as powers of 2s and 3s

you should be able to get a contradiction from that

[silk pewter]

$2^{p-2q} = 3^{q-2p}$

?

[thorny flame]

bettimsucks

[silk pewter]

@neon iron

[neon iron]

yup

[silk pewter]

and then

lol i cant even get one step on my own im shit

[neon iron]

there is only one thing this can be equal to

(you can use fundamental theorem of arithmetic)

[silk pewter]

p-2q = q -2p =0
?

[neon iron]

yes

do you see the contradiction now?

[silk pewter]

yeaa

if i add those equations

i somehow get q=0

so it contradicts the fact that log thingy is rational

thus its irrational?]

[neon iron]

yes correct

[silk pewter]

thnks

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[rare kindle]

does anyone undewrstand hwo to do this

[topaz sinew]

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[rare kindle]

.close

weird O_O

[strange whale]

you gotta go get another channel

[rare kindle]

ok

[strange whale]

if you delete your msg it will close no matter what

[rare kindle]

Okay

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[neon iron]

Is it possible to take the reciprocal? Or is that not a rule

[bold nebula]

it's a way

now * 4

[neon iron]

yeah i know how to solve the problem

just wondering if u can do that

thanks anyways

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[knotty ledge]

partial fraction decomposition

[topaz sinew]

@timid moth Has your question been resolved?

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[pseudo granite]

.open

[last ingot]

yes

<@&286206848099549185>

[timber basin]

what do u need help with

[last ingot]

my brain isnt braining

[last ingot]

everything

b

alr

np

19

oh

12

shouldnt we be doing caitlins one

hers is the tallest

16

the answer is a number

because it wants the frequency

16

alr

I have to answer both first

a and b

7+9+2 is 18

so its 19

thanks!

Yes!

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[last ingot]

cya!

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[polar blaze]

\begin{align*} P(z)&=z^2+1, \ Q(z)&=z^3+2, \ R(z)&=(z+1)^6 + P(z) \cdot Q(z). \end{align*}Note that $R(z)$ has degree $6$ and constant term $3,$ so it satisfies the conditions.

We need to find the solutions to\begin{align*} P(z) \cdot Q(z) &= (z+1)^6 + P(z) \cdot Q(z) \ 0 &= (z+1)^6. \end{align*}Clearly, the only distinct complex root is $-1,$ so our answer is $N=\boxed{\textbf{(B)} : 1}.$

can someone explain

[thorny flame]

auroraborealis13

[polar blaze]

https://artofproblemsolving.com/wiki/index.php/2021_Fall_AMC_12B_Problems/Problem_14

[thorny flame]

auroraborealis13

auroraborealis13

[polar blaze]

can someone explain the last part of the problem

[odd pagoda]

you need to be more specific

explain what about it

where are you stuck

[polar blaze]

i dont understand the solution

specifically the part where i asked

see the first msg

i dont understand the last 3 lines

[odd pagoda]

the question asks for solutions to P(z)*Q(z)=R(z)

[polar blaze]

yes

[odd pagoda]

if you plug the definition of R(z) in this is then P(z)Q(z)=(z+1)^6+P(z)Q(z)

the P(z)Q(z) on both sides cancels

leaving just 0=(z+1)^6

which only has one solution

z=-1

[polar blaze]

why only 1 solution?

how do you know there aren't any complex solutions

[odd pagoda]

its a product of terms that equals 0

[polar blaze]

oh wait

i see

magnitude has to =0

ok thx

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[topaz sinew]

Please don't occupy multiple help channels.

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[analog dome]

c

[analog dome]

how do i do this?

im supposed to expand it

[sweet shard]

,tex .log rules

[thorny flame]

rie.mann

[analog dome]

i know but

this one is a square root @sweet shard

which makes things awkward

if this was a regular natural log, this wouldnt be bad

like, the answer is there but idk how

[knotty ledge]

how do you write a square root as a power?

[analog dome]

the onl lead i have is that i should put a square root

square*

square to remove the square root

then youre left with ln Cx(axb)^2

wait a minute

howd the 2 become the denomiator then

[neon iron]

recall that $\sqrt{x}=x^{\frac{1}{2}}$

[thorny flame]

mist9912

[analog dome]

they didnt teach us that property in school

so this is new to me

[neon iron]

,tex .power rules

[thorny flame]

mist9912

[analog dome]

ill google it

[neon iron]

yea i forgot the command

search exponent/index rules

[analog dome]

i dont see the square root thing

wait

[neon iron]

see fractional exponent

[analog dome]

i think i might get it now

[analog dome]

sq root(axb) would be the same as 2sqroot a x b

wait no

what am i even saying

[neon iron]

np u shd try to apply the fractional exponent rule again

[analog dome]

would 'c' in this question be 'n' in the fractional exponent picture?

[neon iron]

hmm no

[analog dome]

so would 'n' be 1?

[neon iron]

notice how n in the picture is in superscript?

[analog dome]

since theres no number on it

so i think itd be 1

and since i add a sqare outside the bracket

[neon iron]

$\sqrt[2]{}$ means $\sqrt[]{}$

[thorny flame]

mist9912

[analog dome]

'2' would be 'm'

[neon iron]

no, u don't see the m here

m = 1

[neon iron]

take note of this

[analog dome]

oh

wait

[neon iron]

yea no one writes the 2 out

[analog dome]

so you shouldnt but the square symbol outside of the bracket to get rid of the sq root?

[neon iron]

it's equivalent

i mean $\sqrt[n]{a^m}=\sqrt[n]{a}^m$

[thorny flame]

mist9912

[analog dome]

ohh i get it now

[tall dirge]

u can write sqrt x as x^1/2

[analog dome]

but still

[analog dome]

howd the 2 go on the denominator

[tall dirge]

and then it would be like (a*b)^1/2

and bring out the 1/2 to the front

[neon iron]

look at power rule

[tall dirge]

just wondering how did u do the image embed code thingy

[thorny flame]

mist9912

[analog dome]

ive never understood it either

ok anwyays

so after the first step,

[neon iron]

it was the error code for my latex, you react to the code

[analog dome]

the next step should be ln c(axb)^1/2 right?

[neon iron]

press the emergency sign

[neon iron]

yea

[analog dome]

then c(1/2a x 1/2b) right?

[neon iron]

yes

do u see which log rule to apply?

[analog dome]

well, since all of it is bracketed, i understand that I have to do the product rule

but which rule do you use to get the 2 on the denominator?

[neon iron]

yes we shd now at least have $\ln{c}+\ln{a^{\frac{1}{2}}}+\ln{b^{\frac{1}{2}}}$

[thorny flame]

mist9912

[neon iron]

use power rule

that's the only rule you see a power

[analog dome]

yeah i was thinking of that actually

ln 1/2 a + ln 1/2 b

[neon iron]

uh

[analog dome]

since 1/2 is the exponent

im just talking about for a and b btw

[neon iron]

yea the order was probably reversed

maybe a typo?

[analog dome]

on the answer?

[neon iron]

no this

do u mean sth like in the ans in red?

(ln a)/2 + (ln b)/2?

[analog dome]

ik thats the answer but

since the exponent should go in front of the ln

i was wondering why it wouldnt be 1/2ln a + 1/2ln b

[neon iron]

yes it is (1/2)lna + (1/2) lnb

[analog dome]

waait a sec

OHHHHHHHHHHHHH

1/2 ln a is the same as lna/2 lol

i completely forgot

[neon iron]

yea i hope u see it now

[analog dome]

since taking the half of something means to divide it by 2

but would (1/2)lna + (1/2)lnb still be correct if I put it on an exam?

[neon iron]

yes

[analog dome]

ok cool

thanks

[neon iron]

np

[analog dome]

you were a big help

since im kinda new to logs and self learning it

[topaz sinew]

@analog dome Has your question been resolved?

[analog dome]

wait

1 lost question, if anyones still here

in this question, they used a square to get rid of the square root

but the fractional exponent property gave a different answer

wouldnt sqroot(x+2) be (x+2)^1/2?

[acoustic tangle]

Yes, you should end up with the same thing

[analog dome]

wait a sec, ill try it real quick

ok so

the 2nd step should be:

e^1 = x+2

[acoustic tangle]

e^1?

[analog dome]

i put it in exp form

a^c = b

[acoustic tangle]

Yes, but are you sure it's e^1

[analog dome]

yeah

[analog dome]

in this equation, 1 - c

x+2 = b

[acoustic tangle]

And what happened to the fractional power?

[analog dome]

and e is the base, which is 'a'

[analog dome]

in the next step

(e^1)^1/2 = (x+2)^1/2

but this is where it gets confusing..

the original way was to just use a square to get rid of the sqroot, but i have to understand the fractional way

[neon iron]

hmm let's start again with the fact that we know sqrt(x+2) = (x+2)^(1/2)

[analog dome]

ok

[neon iron]

so the original equation is $\ln{\sqrt{x+2}}=1$

[thorny flame]

mist9912

[analog dome]

yes

[neon iron]

using this, we have $\ln{(x+2)^{\frac{1}{2}}}=1$

[thorny flame]

mist9912

[neon iron]

do u see any log rule we can apply here?

[analog dome]

yes

power rule

[neon iron]

yes, after applying the power rule, we have $\frac{1}{2}\ln{(x+2)}=1$

[analog dome]

yeah

[thorny flame]

mist9912

[neon iron]

so we should now multiply both sides by 2 to have $\ln(x+2)=2$

do u see what to do next?

[thorny flame]

mist9912

[analog dome]

wait

just to clarify

we multiply both sides by 2 to get rid of the 1/2 behind the ln right?

[neon iron]

yes

[analog dome]

ok cool

i think ill probably but this in exponential form now

and try to solve from there

e^2 = x + 2

[neon iron]

yes exactly

[analog dome]

e^2 -2 = x

[neon iron]

so that's your answer here

[analog dome]

its the same answer!

[neon iron]

nice

[analog dome]

so it does work

ill do more practice with this so i can get better

thanks for the help once again

[neon iron]

np

[topaz sinew]

@analog dome Has your question been resolved?

[topaz sinew]

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[topaz sinew]

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[odd pagoda]

<@&268886789983436800>

[analog dome]

yeah mb

i was just stuck on a question but

when i posted it here, thats when i figured it out

feel free to close the channel

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[night slate]

How would you do question 7a

[night slate]

I tried splitting the numerator into x^n-1 and x in order to get the In-2

But there is a log that messes the entire thing up

I also tried by parts with 1 and x^n/(1+x^2)

But then a tan inverse falls out

<@&286206848099549185>

<@&286206848099549185>

[thick musk]

Add and subtract x^n-2 from x^n (add 0)

[night slate]

How does that help

[thick musk]

You separate the fraction into 2 fractions the x^n + x^n-2 part and the -x^n-2 part

[night slate]

Oh

[thick musk]

The -x^n-2 part becomes the -I_n-2

The other part factors and cancels

[night slate]

How did you see that

[thick musk]

I tried to make the -I_n-2 part first

[night slate]

Is your first reaction for these questions to immediately apply by parts

Oh wait the question asked for algebraic manipulation

Thanks for the help

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[night slate]

.reopen

[topaz sinew]

✅

[night slate]

Sorry @thick musk but how would you simplify the x^n and x^n-2 terms

[thick musk]

Factor x^n-2 out of it

The x^2 + 1 cancels

[night slate]

Then you are left with x^n-2

But that is not x^n-1/(n-1)

[thick musk]

Integrate

[night slate]

Oh yeah

That makes sense

Thanks

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[neon iron]

is the determinat e of something transposed the same as if it were not trasnsposed for matrixs

[keen venture]

Yes

[neon iron]

ok thanks

[keen venture]

|A^T| = |A|

[neon iron]

thank you

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[vale heath]

how does simpson's 1/3 rule consider more than 3 points?

[vale heath]

it doesnt seem ok to assume they lie on a parabola

[odd pagoda]

what exactly do you mean? do you mean the composite version? you split up the interval [a,b] into several smaller ones and then use simpsons on each of those, then add stuff up

[vale heath]

so each three consecutive terms get to be their own parabola?

like y0 y1 y2 and then y3 y4 y5

are two parabolas

[odd pagoda]

basically

[vale heath]

ok thanks

[odd pagoda]

although you would let them overlap

[vale heath]

hmm?

[odd pagoda]

y0 y1 y2 is the first parabola and then y2 y3 y4 the next

[vale heath]

more like y0 y1 y2 and then y2 y3 y4

great

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[topaz sinew]

@neon iron Has your question been resolved?

[fiery saddle]

y=x-2

it is very simple

you draw a line between D and D'

and the perpendicular bisector of DD' is the line of reflection

close enough

with acurate measures

you can see that this line should have points (0,-2) and (2,0) on it

therefore the equation is y=x-2

first draw a line connecting D and D'

Then bisect this line perpendicularly

and then you get the equation

bisect

great

when a line is bisected

it means to split it into two half with the same measure

for exxample when you have a 10 meter line

to bisect it

you have to cut it into two 5 meter parts

yep

i think so

unless your teacher wants you to put your work in words

otherwise i think you are fine

yep

that is the line of reflection

Pythagorean theorem?

this question doesn't need that

it just asks you to find the equation that represents the line

long

the line of reflection

that is the long one

yep

sorry quick question

what grade are you?

oh me too

I am becomig 10th this year

extra studies + pain

good luck

you are welcome

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[fiery saddle]

have a good day

bye

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[frosty thorn]

how can you find the time, when you are only given displacement equal to 250.8 m and a acceleration of 9.81meter per second squared?

[drifting swift]

show entire problem?

sounds fishy / insufficient-info as stated

[frosty thorn]

hold up

this question is quite confusing?

help pls?

[drifting swift]

,rccw

[drifting swift]

ah

so the object drops from rest. that's important info.

btw you should not have put those vector-arrows there

tbh just d = at^2/2

[frosty thorn]

so that would be my equation?

i'll remove the vector arrows

and how did get that equation?

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[latent salmon]

I just had two quick questions about the problem

1. what does (n,m) = 1 mean?
2. what does the dot next to the n mean?

[neon iron]

(n,m) probably means gcd. So (n,m)=1 means n,m coprime

dunno what the dot means (probably some sort of typo)

[latent salmon]

okok

ty

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[stuck jay]

Why is the „3” equal to 1/3? https://media.discordapp.net/attachments/903430332324405288/1126138890688929854/307914411648090113.png

[stuck jay]

And how would you eliminate the 3?

You need to take the number 27 outside of the sqrt?

[eternal wing]

cube root of 27 equals 3

not 1/3

oh are you asking why

[wintry phoenix]

I am not sure what he is asking really tbh.

[eternal wing]

$\sqrt[3]{27} = 27^{\frac{1}{3}}$

[thorny flame]

redstoneplayz09

[wintry phoenix]

oh, that's what's he is asking xD

[eternal wing]

are you asking why these are equal?

yeah probably

[topaz sinew]

@stuck jay Has your question been resolved?

[topaz sinew]

@stuck jay Has your question been resolved?

[topaz sinew]

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[tender tiger]

Is this correct?

is a= e/9 and r=1/3

[sweet shard]

nah

use exponent rules

,tex .exp rules

[thorny flame]

rie.mann

[tender tiger]

oh alright

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[weak estuary]

hiii sorry for the mess, I just started trig today and was given this question to do, I could solve the single triangle ones but this question has two triangles and I'm so lost and have no idea what i'm doing, can someone explain or guide me? I'm basically bullshitting my way through this question I have no idea what's going on

[noble laurel]

start by finding angle RPS

[weak estuary]

my goal is to find distance between p and q, I was going to use the adj of the entire triangle, and find the hyp, then find the opp of the entire triangle then subtract it by r-p to find p-q

[noble laurel]

based on already having the other two angles, and knowing the angles of a triangle sum to 180

[weak estuary]

okay :)

[weak estuary]

we have 31 already and the triangle is a right angle so 31+90 = 121-180 and I get 59

[noble laurel]

yes angle RPS is 59

good work

now find angle SPQ

[weak estuary]

180 - 59 = 121 for spq

[noble laurel]

good

now angle PQS should follow quickly

[weak estuary]

47

how do these angles help me in trig?

[noble laurel]

you already have the length of the hypotenuse, and now all 3 angles in the triangle

the other two side lengths should follow from law of sines / law of cosines

[weak estuary]

these angles are what I put next to the sin cos tan stuff? in the calculator?

[noble laurel]

not exactly

[weak estuary]

I don't understand how the inside of the triangle/angles are supposed to help me find the length of the triangle outside

I'm like so clueless

[noble laurel]

apply this law, to find the side length PQ

[weak estuary]

how do i find b and a ?

[noble laurel]

$\frac{11.45}{\sin{(125)}} = \frac{PQ}{\sin{(12)}}$

[thorny flame]

austinu

[noble laurel]

see how that works from your triangle?

then multiply both sides by sin(12)

and you will find the length of PQ

[weak estuary]

Where do I get 125 from ?

and also if I divide 11.45 by answer of sin (125)

how does that help me with anything for Pq/sin(12)?

I've never done this before

[topaz sinew]

@weak estuary Has your question been resolved?

[weak estuary]

I just tried again and got p-q = 2cm

[topaz sinew]

@weak estuary Has your question been resolved?

[topaz sinew]

@weak estuary Has your question been resolved?

[topaz sinew]

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[snow spindle]

Rest of the problems looks complicated ;-;

[craggy haven]

mostly just angle chasing

remember things like the sum of all angles of a quadrilateral is 360°, sum of angles of a triangle is 180°, straight line is 180°, right angle is 90°

[snow spindle]

Am trying to do problem e right now

When I tried to find x

I got 90

But then the answer page saids it’s 95

Why is that?

I’m soo confused

Okay so I found the answer for problem e

Just by imagining them as being co-interior angles

(Although it doesn’t seem like that)

[snow spindle]

How to do problem F exactly?

That shape looks like a triangle

Wait nvm

It’s not

<@&286206848099549185>

I’m not sure what my working out should look like

[bleak mirage]

hi

ok

so what would angle x be, based on the 180 rule

[snow spindle]

110?

[bleak mirage]

not quite

[snow spindle]

90?

[bleak mirage]

no

remember

[snow spindle]

150?

[bleak mirage]

yep

[snow spindle]

Wait

[bleak mirage]

yes?

[snow spindle]

Is it because

That 30 degree is co-interior?

Or?

Is it external?

[bleak mirage]

dont overcomplicate it

remember

[snow spindle]

Okay ;-;

[bleak mirage]

all angles tht mke up a line adds up to 180 degrees

[snow spindle]

So like

Even 30 degree?

[bleak mirage]

yea

the 30 degree angle and the y degree angle = the line = 180 degree

make sense?

so 30 + y = 180

and y = 150

make sense?

[snow spindle]

Apparently y = 50 though

[bleak mirage]

oops

sorry

i meant x

[snow spindle]

Oh right right

[bleak mirage]

whenever i said y, i meant x

my bd

bad

[snow spindle]

All good

[bleak mirage]

anyway, that do the angles in a quadrilateral add up to

[snow spindle]

So basically

360 -

[bleak mirage]

yea

[snow spindle]

The entire angles I have?

[bleak mirage]

yea

[snow spindle]

Got it

[bleak mirage]

the angles in the quadrilateral u have are 150, 70, 90, and y

make sense?

[snow spindle]

Makes sense

What do I do for problem g?

[bleak mirage]

r u done with f?

[snow spindle]

Mhm

[bleak mirage]

ok

do u know about vertical angles

[snow spindle]

Vertical angles

I don’t think I have learnt that yet

Other than

Knowing what vert. opp. Means

(Aka vertically opposite)

[bleak mirage]

ok

well

just know they are congruent

using that

what would angle y be

[snow spindle]

70?

no?

[bleak mirage]

not quite

remember

vertical angles are across from each other

[snow spindle]

do they add up to 180 degree?

[bleak mirage]

no

they are equal

[snow spindle]

110 then?

[bleak mirage]

yep

then what would x be

[snow spindle]

70

[bleak mirage]

yep

good job

[snow spindle]

should I say the reason as

vertically opposite

or?

what should I write it as?

[bleak mirage]

vertical angles are congruent

thats what we used

[snow spindle]

okay

[bleak mirage]

is thst all?

[snow spindle]

still have problem h and i

[bleak mirage]

alr

[snow spindle]

am ready

for h

[bleak mirage]

alr

this will be harder

but

u got this

[snow spindle]

;-;

[bleak mirage]

what type of triangle is that

scalene

equalatiral

or isoceles

[snow spindle]

isoceles

[bleak mirage]

so

the idea is that base angles are always congruent for these types of triangle

got it?

[snow spindle]

got it

[bleak mirage]

good

can u repost the pic so its easier for me

[snow spindle]

or do you want me to reply to that post

so that you can go onto it

[snow spindle]

second image

[bleak mirage]

yea

csan u just repost it please

it would be much easier

[snow spindle]

sure

[bleak mirage]

thanks so much

[bleak mirage]

thx

so what are the three angles in that triangle

[snow spindle]

50

x

and y

[bleak mirage]

in the triangle

hint: base angles are congurnet

[snow spindle]

65 degree?

[bleak mirage]

not really

what are the two base angles there

oh wait

nvm

did u say x = 65

[snow spindle]

mhm

[bleak mirage]

good

now

do you know the corresponding angle theorem

[snow spindle]

I

suppose so

just minusing 180 by 65 right?

[bleak mirage]

not quite

oh wait

nvm lol

u got it

wow

nice

[snow spindle]

seems like

the more I understand

the better I get hang of it

anyways

[bleak mirage]

yeah thats kinda how it works

[snow spindle]

now since h is done

the final problem i

[bleak mirage]

yep

can u repost the pic

[snow spindle]

thought you might ask

here

[valid marsh]

use the fact that the triangle is isosceles

[bleak mirage]

ok

so what would be the non-base angle

remember 180 rule

[snow spindle]

3x = 180

divide both by 3

x = 60

what's the 138 degree for?

[valid marsh]

the triangle is not equilateral

[snow spindle]

oh shoot

right

my bad

[bleak mirage]

even order group => solvable, i assume ur gonna take over

bye serial

[snow spindle]

see you

hmmm

the angle next to 138 is 42 degrees right?

if that's the case...

2x + 42 = 180

[valid marsh]

you could also use exterior angles to get 2x = 138 directly

[craggy haven]

that's definitely not how i'd do it lol

[snow spindle]

he got a point though

since 180 - 42 is 138 anyway

[craggy haven]

if you want to do it that way go ahead yea

[snow spindle]

and then I just had to divided that by 2

which is

69

x = 69

:D

me finished now

thank you for the help!

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[weak estuary]

hello! sorry for the bother, can I get a second opinion on this, I'm really paranoid that this might be wrong because I have no idea what I'm doing, so the question is that I have to determine which pet of Hilary's is closer to catch, and my answer was dog, thank you very much : )

[valid marsh]

the dog is easier to catch

because the angle is larger

[bright crypt]

Greater angle=shorter distance

[weak estuary]

wait HWAT>???

[valid marsh]

only in this case

[bright crypt]

i mean im pretty sure it always applies if the angle is 90 or lower

for both sides I mean

[weak estuary]

looks right ?

okay thank you both very very much

hope you guys have a great night

appreciate the help : D

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[sturdy jasper]

I need help showing that the set of all functions f from N to N such that f is increasing is uncountable. I’ve made a list of functions f1, f2, … fn and i’ve listed out fi(0) fi(1) … fi(n) for all my functions. Is it sufficient to sum the diagonal (f0(0) + f1(1) + f2(2) … fn(n)) and call it a function f and say that it’s not in the list?

[craggy haven]

you seem to have a good idea but I can't quite tell what you mean. What would g(2) be?

[sturdy jasper]

i don’t have g?

[craggy haven]

g being the new function, sorry

[sturdy jasper]

oh it would just be f0(2) + f1(2) + f2(2) which isn’t listed?

[craggy haven]

so your g(0) would equal f0(0)?

you need to show that it's different than all of the f's

[sturdy jasper]

@craggy haven

[craggy haven]

@sturdy jasper

oh you edited

[sturdy jasper]

yeah

like that

[craggy haven]

that seems like an okay pattern except again, what's g(0)?

[sturdy jasper]

uhh it would be f0(0)

oh that’s in the listing tho

hi?

[sturdy jasper]

how would i avoid this

same pattern but add 1 at the end?

[craggy haven]

simpler than yours

actually wait yeah fn(n) + 1 wouldn't always be increasing

because my first function could be f0(0) = graham's number
and my second function could be f1(1) = 1

[sturdy jasper]

but f1(1) has to be greater than f0(0) no? so that’s not possible

[craggy haven]

no, my second function is the identity function, which is increasing

[sturdy jasper]

hmm what should i do then

[craggy haven]

I think something like what you had with the f0(n) + f1(n) + f2(n) + ... + fn(n)

but add 1 to it?

[sturdy jasper]

okay how do i show that it’s increasing and not in the list tho

[craggy haven]

well not in the list is pretty straightforward

increasing you'll need to use the fact that each of the input functions are increasing

also, don't be afraid to dream big, if it's convenient to multiply things by 1000, or take their factorial, or whatever, go ahead and do that - there are plenty of natural numbers to work with

[sturdy jasper]

Okay got it how do i go about showing that the set of decreasing functions N->N is countable

cuz you sound like a bot lol

[craggy haven]

why does this feel so much harder than the increasing one

i can't multiply literally everything together

oh countable

[sturdy jasper]

yeah countable

i think it has something to do with it being finite

[craggy haven]

what does "decreasing" mean here? strictly or less-than-or-equal?

[sturdy jasper]

<=

[craggy haven]

ok because if it was strictly it'd be very very finite lmao

[sturdy jasper]

How would i go about this

[craggy haven]

essentially you need to make a list of all of them

i would start by deciding whether 0 is a natural number or not

[sturdy jasper]

it is in my class

[craggy haven]

ok

if you kinda think about what it means to be a decreasing function on N, it's pretty restrictive, right?

[sturdy jasper]

yeah

for f(0) it must contain the largest value right?

[craggy haven]

yeah f(0) would be the largest value

[sturdy jasper]

so the range is kinda finite

[craggy haven]

yep
but, that doesn't mean that it'll always be 0 at a certain point, we could have f(x) = 3 for instance

[sturdy jasper]

hmm how would we go about listing

[waxen flame]

can't you ditch the increasing without loss of generality by using the cumulative sum of a sequence

[craggy haven]

we're on decreasing now

[craggy haven]

there's one function in particular that's especially stupid interesting

which is where i'd start my list

[sturdy jasper]

would it be f(x) = largest value

maybe set largest value to a variable

f(x)=k

[craggy haven]

well, there's no largest natural number

[sturdy jasper]

i’m not sure then

[craggy haven]

what's the simplest function that would satisfy this?

[sturdy jasper]

what are you referring to by “this”

[craggy haven]

oh, that would be from N to N and decreasing

[sturdy jasper]

f(x)=-x

[craggy haven]

natural numbers aren't negative

[sturdy jasper]

hmmm f(x)=0 then

[craggy haven]

yeah, that one's kinda... baseline? like everything else seems to fall down to it

[sturdy jasper]

yeah that’s true

[craggy haven]

what i was thinking about was like, how does the choice of f(0) influence the rest of the function?

so in this case if f(0) = 0 how many decreasing functions are there?

[sturdy jasper]

0?

i mean 1

[craggy haven]

yeah that's neat

what about if f(0) = 1?

[sturdy jasper]

2

[craggy haven]

think there are slightly more than that

[sturdy jasper]

hold up let me think

[craggy haven]

yep i'm planning to let you go for at least like ten minutes on that

[sturdy jasper]

is this this with the function f(x)=1

[craggy haven]

that would be an example of a decreasing function that satisfies f(0) = 1

there are others though

[sturdy jasper]

wouldn’t the only 3 be f(x)=1 and f(x)=0 and f(x)=1-x

[waxen flame]

I guess they're allowed to be zero/one after a certain point?

Otherwise you have to change decreasing to nonincreasing, and that means there aren't a finite number

[sturdy jasper]

it is non increasing

[waxen flame]

the set of nonincreasing functions from N to N seems uncountable to me, but not sure

[sturdy jasper]

it says to prove that it’s countable so i think it is

[waxen flame]

hmm maybe it is countable

[sturdy jasper]

not sure how to deduce that tho

@craggy haven helpppp

[waxen flame]

ah yes

[craggy haven]

i thought so at first as well

[craggy haven]

f(x) = 0 doesn't satisfy f(0) = 1 though

[waxen flame]

it does feel very similar to real numbers with terminating decimal representations though

[craggy haven]

think about it this way:

   0    1    2    3    4    5    6 ...
--------------------------------------
f  1

how to fill in the rest of this table?

[sturdy jasper]

ahh damn there’s prob a bunch of functions tho

has to be less than or equal to 1

[craggy haven]

don't worry about formulas and stuff, a function is just the set of its input-output pairs

[sturdy jasper]

right?

[craggy haven]

ye correct

so what are your choices for f(1)?

[sturdy jasper]

either 1 or 0

and for f(2) if f(1) is 0 then the only choice is 0

[waxen flame]

okay if it's nonincreasing, I think I have a sketchy ass argument for it:
encode the function as a sequence of binary digits in the following manner:
start with m = 0 (which function value we're currently encoding)

1. Add f(m) 1s to the binary sequence
2. Add a 0 and increment m.

this thing is guaranteed to eventually repeat all 0s or all 1s, and then you can show that this set is contained within the rational numbers

[sturdy jasper]

uhh i think the answer involves showing that it’s part of a countable set or a union of countable sets

without binary digits

[waxen flame]

nah this is a common technique for various arguments

the binary thing is just an injection

decreasing, with the exception of zero is easier than nonincreasing, because you can guarantee that there are a finite number of functions with f(0) = n

but if it's nonincreasing, then it could go on forever

[craggy haven]

i'd stay away from the complexity of binary here

[sturdy jasper]

is there a way to do it without binary

[craggy haven]

yeah

[sturdy jasper]

[craggy haven]

so just limiting ourselves to the case where f(1) = 1, can you make a list of decreasing functions?

   0 1 2 3 4 5 6 ...
----------------------
f  1 0 0 0 0 0 0  ... (you said this one earlier)
f  1 1 ? ?

[waxen flame]

but this only works if the problem asks about decreasing functions

if it's nonincreasing, then you can't say the list of functions corresponding to f(0) = n is finite

[craggy haven]

i can and i will

[sturdy jasper]

it can keep going on forever

[craggy haven]

oh not finite no but countable

[sturdy jasper]

you can but isn’t it infinite?

[craggy haven]

there are no strictly decreasing functions from N to N

[craggy haven]

yeah, so? as long as it's countably infinite we're good

[waxen flame]

yeah but maybe they put in an exception for 0 idk

yeah if it's countable you're okay

just appeal to the countability of N x N

[sturdy jasper]

but how would i list that 💀

[craggy haven]

you'd list a pattern

like how you can list the natural numbers 0, 1, 2, 3, 4, 5, ...

[sturdy jasper]

well there’s 1000000000000
11000000000000000000
111000000000000 and so on but is there a way to generalize it

[craggy haven]

i mean, you just said "and so on" so it sounds like you think so

[sturdy jasper]

well the case with the largest possibilities will have a ton of possible values

[waxen flame]

listing is kinda weird, but you can show that all of the sequences will eventually end up repeating zeroes

[craggy haven]

except for one of them 👀

[sturdy jasper]

which one 🤯

[craggy haven]

the constant function that always returns 1

[waxen flame]

oh we were talking about just where f(0) = 1 lol

[craggy haven]

yeah

[sturdy jasper]

that’s just one function tho 🥲

have to list a generalized version of all ig

[waxen flame]

basically the idea is the same though

eventually the sequence of numbers you write has to repeat something

[sturdy jasper]

it’ll start at some f(0)=m and then keep decreasing from there

[waxen flame]

I was thinking of listing out the differences, but you could just list out the function values and it's the same thing

[craggy haven]

showing this as a first result is a good first step in the proof

[sturdy jasper]

by the well ordering principle there’s a smallest number so it must repeat eventually?

[craggy haven]

what dimension did you just pull that out of, yes that's right
you'll want to clearly state the set on which you're applying the WOP

[waxen flame]

that was honestly kinda unexpected

[sturdy jasper]

the set N

wdym by dimension tho

[craggy haven]

i mean i didn't expect you to know that theorem

[sturdy jasper]

😐

[craggy haven]

the set N does have a smallest element, it's 0
but functions don't have to repeat 0, they can repeat any other number

we already saw one that repeated 1

[sturdy jasper]

it’s a subset of N which is the range of values the function can output

[craggy haven]

yeah there you go

[sturdy jasper]

what should the next step be

[craggy haven]

(there are a few ways you could go with it)

[pseudo jetty]

||run length encoding||

[sturdy jasper]

if we let f(0)=k then there’s k-(smallest number) values before it repeats right

[sturdy jasper]

what does that mean

[craggy haven]

well, no, it could output k for the first 1000, then k-1 for the next million, then k-2 for like a billion

[waxen flame]

I advocated for something resembling decoding that, but it's kinda weird

[sturdy jasper]

mmm

must create a bijection somehow

[craggy haven]

you don't need a bijection, injection/surjection (depending on your function direction) is fine

[pseudo jetty]

there is no weirdness, ||each function starting at n is encoded by an n-tuple of run lengths, allowing for infinity (and zeroes afterwards possibly), which shows that the set is a countable union of of countable sets||

[craggy haven]

saccharine originally had some binary thing that I couldn't tell what it was doing and seemed overly complicated