help-13

[upper laurel]

this can completely replace the b = a^n work earlier

[blazing zephyr]

question, a^2023/b being an integer doesent itnmean b | a^2023 or is it the same as b | a^2024

nvm

ok i fully underatand where you got (2,2^2024) to be the only sol

[upper laurel]

now other than that I dont really have any other ideas on what to do next

I think the k = a^2023/b + 1/a should have something to it

k - 1/a = a^2023/b
assuming that a > 1, k - 1/a isnt an integer
so a^2023/b isnt an integer
however b | a^2024, so a^2024/b is an integer
so b doesnt divide a^2023 but it does divide a^2024

[eager sapphire]

what is going on here lol

[upper laurel]

we gotta

but Im going in circles

[eager sapphire]

ah i see

i think this works to show (1,1) and (2,2^2023) are the only solns:

i think it's already been mentioned above that $a \mid b$ and $b \mid a^{2024}$

[wraith dagger]

AlphaNull

[eager sapphire]

then let $b = ja, a^{2024} = kb \implies j \cdot k = a^{2023}$, so WLOG let $j = a^L$

[wraith dagger]

AlphaNull

[eager sapphire]

then letting $a^{2024} + b = cab \implies a^{2024} + a^{L+1} = c \cdot a^{L+2}$ (subbing in $b = ja$

[wraith dagger]

AlphaNull

[upper laurel]

its also been established that if b is a power of a, then (2, 2^2024) is the only solution

[eager sapphire]

divide by $a^{L+1}$ on both sides to get $a^{2024 - L - 1} + 1 = c \cdot a$, which is not possible if $2024 - L - 1 > 0$

[wraith dagger]

AlphaNull

[eager sapphire]

pretty sure b has to be a power of a though

[upper laurel]

you already did that here

[eager sapphire]

i'm confused, then why isnt this thread closed

[upper laurel]

to be honest I was thinking that if a | b and b | a^2024 then theres still a chance that b isnt a power of a

I didnt notice that jk is a power of a which would force this to happen

[eager sapphire]

ah yeah

[upper laurel]

so the completed proof first had that a | b and b | a^2024

then what you did which shows that b is a power of a

then this here shows that the only solutions are (1, 1) and (2, 2^2024)

[eager sapphire]

oh dear wait i cant assume j is a power of a if a isnt prime

nvm i dont think what i have is complete then

[upper laurel]

damn I wanted to hope

[upper laurel]

we did find that b doesnt divide a^2023 but it does divide a^2024

(assuming a and b to be prime powers also didnt give any new solutions)

[pulsar scroll]

Is 2^2025|2^2024+2 true?

[blazing zephyr]

no

[blazing zephyr]

:((

[upper laurel]

2^2025 | 2^2024 + 2
2^2025 * k = 2^2024 + 2
2^2024 * k = 2^2023 + 1
2^2024 | 2^2023 + 1
even | odd
which is why it isnt true

[pulsar scroll]

Yeah so why is (2,2^2024) a solution?

[upper laurel]

you have the problem written down wrong

[blazing zephyr]

it implis 2^2025 | 2^2024+2^2024 is a solution (which it is)

[upper laurel]

its 2 * 2^2024 | 2^2024 + 2^2024

[blazing zephyr]

also literally 2^2025>2^2024+2

[upper laurel]

to be honest, I doubted that so much that I didnt do it

[cedar kiln]

@blazing zephyr Has your question been resolved?

[cedar kiln]

@blazing zephyr Has your question been resolved?

[upper laurel]

let b = b₂₀₂₄

ab | a²⁰²⁴ + b
ab₂₀₂₄ | a²⁰²⁴ + b₂₀₂₄
ab₂₀₂₄k = a²⁰²⁴ + b₂₀₂₄
ab₂₀₂₄k - a²⁰²⁴ = b₂₀₂₄

ab₂₀₂₄k - a²⁰²⁴ = b₂₀₂₄
ab₂₀₂₄ k - a²⁰²⁴ ≡ b₂₀₂₄ mod a
b₂₀₂₄ ≡ 0 mod a
let b₂₀₂₄ = ab₂₀₂₃
a(ab₂₀₂₃)k - a²⁰²⁴ = (ab₂₀₂₃)
ab₂₀₂₃k - a²⁰²³ = b₂₀₂₃

...

ab₁k - a¹ = b₁
ab₁k - a¹ ≡ b₁ mod a
b₁ ≡ 0 mod a
let b₁ = ab₀
a(ab₀)k - a¹ = ab₀
ab₀k - 1 = b₀
ab₀k - b₀ = 1
(ak - 1) b₀ = 1

b₀ = 1
b₁ = ab₀ = a
b₂ = ab₁ = a²
...
b₂₀₂₄ = a²⁰²⁴
b = a²⁰²⁴

abk = a²⁰²⁴ + b
aa²⁰²⁴k = a²⁰²⁴ + a²⁰²⁴
ak = 1 + 1
a | 2

a = 1, a = 2
(1, 1), (2, 2²⁰²⁴) are the only solutions

[blazing zephyr]

(ak - 1) b₂₀₂₄ ≡ a²⁰²⁴ mod a
b₂₀₂₄ ≡ 0 mod a
what happend here?

rhs makes sense but where did (ak-1) go

[upper laurel]

let me add in more guardrails knowing you immediately got stuck on that

[blazing zephyr]

:3

[upper laurel]

what does that mean

[blazing zephyr]

a face

if it helps i understood everything except the dissapearance of (ak-1)

[upper laurel]

oh thank fuck

in any case the proof has been slightly shortened so this ak-1 business is avoided entirely

[blazing zephyr]

whar

oh

[upper laurel]

whoops the ak - 1 is still there in the bottom, let me erase that

[blazing zephyr]

oo thank you

[upper laurel]

np

[blazing zephyr]

tysm for staying patient with me (somehow)

[upper laurel]

the trick was to not teach you at all

(and also that I got stuck on this for long enough that when I found the right way, it turned out simple)

did the teacher give you a time limit to do these sorts of problems

[blazing zephyr]

err

not really they are just practice ones

[upper laurel]

thats fair

did you ever see this sort of trick done before

[blazing zephyr]

nope

[upper laurel]

yea even if I found it I wouldve taught it

aint no way youre just supposed to stumble across that out of all of the things you couldve done

I spent a good 5 minutes trying to see an easier way to write it than that

[blazing zephyr]

yea very wierd question

[upper laurel]

were the previous 8 questions like that

[blazing zephyr]

not even close

[upper laurel]

I was hoping for a yes thats nasty

I still think theres a shortcut to be found here that we missed

I was thinking that ak = 0 mod a

so ak - 1 isnt 0 mod a

so ak - 1 isnt 0 mod a^2024

[blazing zephyr]

#help-9 message
this was number 8

[upper laurel]

that burns to see

[blazing zephyr]

huh

alright im going to close this now, thank you very much

.close

[cedar kiln]

Channel closed

Closed by @blazing zephyr

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[cedar kiln]

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[subtle halo]

How can I do this?

[subtle halo]

<@&286206848099549185>

[livid hound]

Please don't ping Helpers before 15m have elapsed since asking your question.

[subtle halo]

oh i forgot that

sorry

[livid hound]

did you do part i)?

[subtle halo]

im doing ii)

and i did that

what to do now

wait

[outer stag]

Multiply both sides by 50/3

[subtle halo]

oh

then?

[outer stag]

Simplify

[subtle halo]

now what to do

[livid hound]

why are you indicating an intention to multiply both sides by 3/50

[subtle halo]

um

can someone show me the whole process?

im new to this

[livid hound]

there's no need to take major steps

you initially intended multiplication by 50 to both sides

start with that

[subtle halo]

okay

.close

[cedar kiln]

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[cedar kiln]

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[lost thunder]

tips to calc |A|?

[mental trail]

what I suggest: subtract x*(row 3) to row 4

then subtract x*(row 2) to row 3

etc...

[cedar kiln]

@lost thunder Has your question been resolved?

[cedar kiln]

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[crimson sedge]

Im supposed to be solving for these angles using the one I have but I don't even kno where to start I'm mad confused

[vernal basin]

the two horizontal lines are parallel right

[subtle halo]

you want to find the other angles?

[crimson sedge]

yup yup

[crimson sedge]

indeed

[subtle halo]

70 = Angle 3

[vernal basin]

don't give answers @subtle halo

that's like

[subtle halo]

Vertically opposite angles

[vernal basin]

the one rule for helping 😭

[subtle halo]

im trying to give him a hint

[crimson sedge]

so like

[subtle halo]

that's not an answer

[vernal basin]

you just gave the answer for the third angle but its ok

[crimson sedge]

is 5 = angle 3

[subtle halo]

you can take all out by knowing the angles

[vernal basin]

yes because of alternate interior angles

[subtle halo]

that's not really an anwer

*answer

[vernal basin]

do you know what we mean when we say vertical angles and alternate interior

[subtle halo]

because he needs to find all the angles

[crimson sedge]

its these terms man

they just look like words to me

[subtle halo]

then how can we help you

[crimson sedge]

what do they mean

[vernal basin]

take notes in class 👍

[crimson sedge]

i did

we don't keep the notes

[subtle halo]

ok so

[subtle halo]

these are vertical angles

[subtle halo]

do you know what is a linear pair?

[crimson sedge]

nah

[vernal basin]

[subtle halo]

Ang 1 + 70 = 180 degree bro

you don't know that

how can we help you then

look, we're not allowed to give direct answers

so yeah

[crimson sedge]

ahhh

makes sense

so then

[subtle halo]

180 - 70 = Angle 1

and also

you need to calculate the answer

i gave you the equation

[crimson sedge]

i think i cooked up

[vernal basin]

Yes good job

[crimson sedge]

bet bet bet

also

[subtle halo]

good

[vernal basin]

Just study up on the rules

[subtle halo]

This was correct

anyway

[vernal basin]

It’s a lot of memorization but it gets more intuitive when you think about it

[subtle halo]

You gotta focus on the classes

[crimson sedge]

theyre saying

[subtle halo]

Bruh

[crimson sedge]

x+4=60

[subtle halo]

another one?

[crimson sedge]

but wouldnt it equal 64

[subtle halo]

bro algebra

[subtle halo]

No

[crimson sedge]

cant even see the board fr fr i just be thugging it out

[subtle halo]

x + 4 = 60
x = 60 - 4

[crimson sedge]

ahhh

[subtle halo]

ask your teacher

Just learn to form equations

[subtle halo]

bro this isn't a joke place or smth

[crimson sedge]

im not joking

[subtle halo]

im not gonna play that

[crimson sedge]

"cant do dat one"

[subtle halo]

what

what are you doing now

[crimson sedge]

thats what he said in the video

[subtle halo]

ok

?

and?

[crimson sedge]

so

[subtle halo]

You were studying?

[crimson sedge]

if x is 60-4

[subtle halo]

Yeah?

[crimson sedge]

nd im solving for x

then that means

bet

[subtle halo]

bro

it's simple algebra

[crimson sedge]

i know that part

but these not easy man

it gets slightly less simple

[tall spindle]

dont be condescending they re learning

[subtle halo]

i know

[subtle halo]

im not trying to be mean to him

[crimson sedge]

i know that part

geometry the only thing ive ever struggled with in life

because wtf is this

[subtle halo]

i just studied it 6 days ago

i can give answer

[crimson sedge]

no

i need help not answers

[brazen dagger]

They assuming one angle as 5x and another as 5x+10

[subtle halo]

5x + 5x + 10 = 180 degree (Linear Pair)

[subtle halo]

yeah

[crimson sedge]

so then i guess i solve thus

wait

[brazen dagger]

First understand and remember those angles sums upto 180 because they are linear pair , their can be linear 3 angles or 4 angles as well..all sums up to 180

[crimson sedge]

ive read this a few times now

[subtle halo]

i think so

[brazen dagger]

Yep

x=17

[subtle halo]

yeah

[crimson sedge]

ok

[brazen dagger]

Original angles were
5x
5x+10
Substitute values of x in them and find the value

[crimson sedge]

makes sense

[brazen dagger]

Bro why are you solving

[subtle halo]

when did i?

im not solving

[brazen dagger]

Wth is this

[subtle halo]

x = 170/10

he was confused

[brazen dagger]

Your job was
5x+5x+10 = 180 that's it

[subtle halo]

i explained

[crimson sedge]

ngl this is way more helpful than a written explanation idk what hes doing wrong

[subtle halo]

no he was still not able to do that

[brazen dagger]

Happens, not everyone understands it that fast

[crimson sedge]

i was able to do that indeed

[brazen dagger]

Maybe he has started learning

[crimson sedge]

algebra is free

[subtle halo]

didn't you like that explanation

[brazen dagger]

[crimson sedge]

good stuff

[subtle halo]

i was not solving

i was showing him

i have also started learning

that's why i was solvin

im in 7th grade

[crimson sedge]

so 5x and 5x+10 ARE the angles

[brazen dagger]

That's great, create a separate channel for your understanding

[subtle halo]

yes

[subtle halo]

i know

[crimson sedge]

so im not solving for angles just X

[subtle halo]

i have got help from this server a lot of times

[subtle halo]

if you take out the value of x you'll know it automatically

[brazen dagger]

Once you find x then only you can find those angles values , that is 5x and 5x+10

[subtle halo]

yeah

[crimson sedge]

OHHHH

i AM solving for the angle

i just needed x first

i dont get it but i can do it

[brazen dagger]

If angles are given in terms of x, first job is find x. Then substitute x in those angle values

[crimson sedge]

thank you man

nd thank the other guy

nd the other guy

[brazen dagger]

No problem

[crimson sedge]

.close

[cedar kiln]

Channel closed

Closed by @red falcon

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[brazen dagger]

5x = 85 if x = 17

Not 95

Same for 5x+10

[cedar kiln]

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[dry fossil]

$(24+2a)x + (2b)y - 2b = 0$ for all $x$ and $y$.

[wraith dagger]

alee

alee

[dry fossil]

Could someone help me?

[brazen dagger]

Well just put random values of x and y , that would be easiest way

If the equation=0 for all x and y

You can put any values of x and y to get equations in terms of a and b

[dry fossil]

Without doing it this way, how can I do?

[upper ruin]

Better if you use easy values like "x=0, y free" and then "y=0, x free"

[dry fossil]

I had already done it this way

[brazen dagger]

Ohhh idk that sorry

[upper ruin]

Comparing the coefficients on the LHS with the ones on the RHS

[dry fossil]

I didn't specify this, my mistake

[upper ruin]

...x + ...y + ... = 0·x + 0·y + 0

[brazen dagger]

They don't want to put values it seems

They want another way

[upper ruin]

Yeah I was suggesting to compare coefficients of x and y

[brazen dagger]

Ohhhhhh

Yesss

[upper ruin]

But maybe there are other ways, as well

[dry fossil]

I should do it like this wait

[wraith dagger]

alee

[dry fossil]

But I didn't understand how to do it 😦

Did you understand?

[brazen dagger]

Its what Alberto was explaining

Compare the coefficient

LHS RHS

RHS all coefficient are 0 for x and y term

[dry fossil]

What is LHS and RHS?

[brazen dagger]

Left hand side , right hand side

[dry fossil]

@upper ruin Could you please explain to me what I should do?

[upper ruin]

Curiously, we don't use these names in Italy

[brazen dagger]

Oh wow

[upper ruin]

I learnt them in YT with BPRP and so on lol

[brazen dagger]

I have used them my whole life

[dry fossil]

...x + ...y + ... = 0·x + 0·y + 0

Do I have to do this?

[upper ruin]

Yes

[brazen dagger]

Yes

[dry fossil]

Oh ok thanks 🙂

.close

[cedar kiln]

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[cedar kiln]

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[cedar kiln]

Please don't occupy multiple help channels.

[remote crater]

i need to find a and got here

i mustve done smth wrong

or just be stupid

[dire geode]

Can you close your other channel

[remote crater]

yeah mb

[cedar kiln]

@remote crater Has your question been resolved?

[cedar kiln]

@remote crater Has your question been resolved?

[cedar kiln]

@remote crater Has your question been resolved?

[cedar kiln]

@remote crater Has your question been resolved?

[cedar kiln]

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[pale cloud]

specifically (0, 1/4)

.close

[cedar kiln]

Channel closed

Closed by @pale cloud

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[pale cloud]

.reopen

[cedar kiln]

✅

[pale cloud]

If x^2 + xy + y^3 = 1,
find the value of y''' at the point where x= 1.

[cedar kiln]

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[dire geode]

pain

[pale cloud]

indeed

[dire geode]

reopen a new available channel with your question since you deleted your original message

[pale cloud]

i did

[cedar kiln]

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[bold pond]

find lengh of curve $x^2 = (y-4)^3, P(1,5), Q(8,8)$

[wraith dagger]

wakamole

[dreamy void]

[bold pond]

is bounds from 5 - 8

or do i need to do distance formula

cause its dy

so can i just look at y

[dire geode]

exactly the same as you did before

[bold pond]

before wasn't 2 points

[dreamy void]

not exactly

[bold pond]

it was just 0 < x < .5

[dreamy void]

dont repeat your mistakes

else you'll be off by a minus again

[dire geode]

make the mistakes again here so you don't on a test

[bold pond]

yea that's true

[dreamy void]

superstition

[bold pond]

so it's distance formula

between 2 points then right?

$\sqrt{(x-x)^2 + (y-y)^2}}$

[dreamy void]

you do exactly like before

[wraith dagger]

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

[bold pond]

for the bounds

[dreamy void]

you wanna find the arc length again

[bold pond]

P(1,5), Q(8,8)

[dreamy void]

between 1 and 8

[bold pond]

but it's dy this time

so why not use y

[bold pond]

5 - 8

so u still use x bounds even tho it's dy?

[dreamy void]

oh right

mb

[bold pond]

eee

[dreamy void]

5 to 8

[bold pond]

yea ha told u

lol

jk

[dreamy void]

yeah h

you told me so hard

it's like i almost feel something

[bold pond]

i bet

how do i simplify this tho

cause x^2

[dreamy void]

sqrt

[bold pond]

yeah i tried that but i got dif answer

like if y = 8

[dreamy void]

then you get two versions

you need to take the positive

[bold pond]

but lets say y = 8 it doesn't work

[dreamy void]

because your points are positive

[bold pond]

cause like 8-4 = 4 => 4^3 = 64

x^2 = 64 => x = 8

oh wait nm

[dreamy void]

you see

[bold pond]

yeah

[dreamy void]

you basically jumped and fell

[bold pond]

so take sqrt of both sides

[dreamy void]

ye

[bold pond]

so x = (y-4)^2

[dreamy void]

no

the power is 3

the pain is real

^3/2

[bold pond]

my algebra skill need some dusting of

off

[dreamy void]

at least you admit

others are straight up

⛹🏻

[bold pond]

playin

[dreamy void]

😂

[wraith dagger]

bacc the sigma😔🤞

[dreamy void]

x'

[bold pond]

$\frac{dx}{dy}=\frac{3}{2}(y-4)^\frac{1}{2}$

oh

i forgot to put the exponent

[wraith dagger]

wakamole

[bold pond]

$\int_5^8\sqrt{1+(\frac{3(y-4)^\frac{1}{2}}{2})^2}dy$

[wraith dagger]

wakamole

[bold pond]

surprised i wrote that correctly the first time

@dreamy void do i distribute the numerator first

can i do $\sqrt{1+\frac{3\sqrt{y-4}}{2}^2}dy$

[wraith dagger]

wakamole

[dreamy void]

can you

[bold pond]

yea

i think that is easier

$\frac{9(y-4)}{4}$

[dreamy void]

ok bro

[wraith dagger]

wakamole

[dreamy void]

better

[bold pond]

$\int_5^8\sqrt{\frac{4}{4}+\frac{9y-36}{4}}dy$

$\int_5^8\sqrt{\frac{9y-32}{4}}dy$

[wraith dagger]

wakamole

wakamole

[bold pond]

split?

i did something wrong?

wait what happened

[bold pond]

is this one easier then the last?

[dreamy void]

ask your mirror

[bold pond]

i think.. am i doig it right

or did imess up

[dreamy void]

looks like it yea

[bold pond]

plz lmk

where?

[dreamy void]

nice

[bold pond]

was my initial step wrong

[dreamy void]

nice

[dreamy void]

nice

[dreamy void]

nice

[dreamy void]

nice

[bold pond]

so

$\int_5^8(\frac{9y-36}{4})^\frac{1}{2}dy$

[dreamy void]

damn

[bold pond]

haha

[dreamy void]

Substitute

[wraith dagger]

wakamole

[dreamy void]

,, u = \frac{9y-32}{4}

[wraith dagger]

bacc the sigma😔🤞

[bold pond]

ok

[dreamy void]

not 36

[wraith dagger]

bacc the sigma😔🤞

[bold pond]

quotient rule

o

[bold pond]

typo

[bold pond]

9y-32 / 4 u dont need quotient?

true

[dreamy void]

$\frac{4}{9} \int_{\frac{13}{4}}^{10} \sqrt{u} : \dd u$

[wraith dagger]

bacc the sigma😔🤞

[dreamy void]

split it

9y/4 + constant

[bold pond]

ok lmc

[dreamy void]

9/4

[bold pond]

32 / 4 = 8

oh ur sying dy/dx 9y/4 = 9/4

lol

[dreamy void]

yea trivially

constant is 0

and y is linear

[bold pond]

yep

[dreamy void]

so 9/4

then du = 9/4 dx solve for dx = 4/9 du

plug in the old bounds into the sub

[dreamy void]

this

[bold pond]

y not just keep bounds and put the term back in u

just wondering

doesnt matter ig uess

[dreamy void]

you can do that too

doesnt really matter

bro

sorry

you have a hard time

you need support

[bold pond]

$\frac{8u^\frac{3}{2}}{27}$ bounds are 5 to 8

[wraith dagger]

wakamole

[bold pond]

my teachers just never explained any of which is what

[dreamy void]

i always hear that

are teacher so badly failing

[bold pond]

$\frac{72y-256}{108}$ bounds 5 to 8

[wraith dagger]

wakamole

[bold pond]

they are stressed out

and rushed

and most are new

[dreamy void]

did you sub back?

,w (728-256)/109 - (725-256)/109

nope

[bold pond]

yea $\frac{\frac{8(9y-32)}{4}}{27}$

[wraith dagger]

wakamole

[dreamy void]

,w 8((98-32)/4)^(3/2)/27 - 8((95-32)/4)^(3/2)/27

nani

[bold pond]

wut is that

[dreamy void]

bro you forgot the powers

[bold pond]

where

omg

3/2

[dreamy void]

that's it

[bold pond]

right now im at $\frac{\frac{8(9y-32)^\frac{3}{2}}{4}}{27}$ bounds 5 to 8

[wraith dagger]

wakamole

[dreamy void]

$\frac{8}{27} \cdot \left ( \frac{9y-32}{4} \right )^{\frac{3}{2}} \bigg |_5^8$

[wraith dagger]

bacc the sigma😔🤞

[dreamy void]

that

[bold pond]

oh

makes sense

in some way

[dreamy void]

u^(3/2)

[bold pond]

yah

ok makes perfect sense

[dreamy void]

u = (9y-32)/4

before you continue

just check the answer

so you dont waste time

[bold pond]

ok but i can do $\frac{8}{27}\frac{\sqrt(9y-32)^3}{4}}$

[wraith dagger]

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

[bold pond]

how do i chek the answer

[dreamy void]

yea dont

i thought your book

[bold pond]

y its just to omuch work

[dreamy void]

had solutions

[bold pond]

oh

this doesnt hav it's even number

[dreamy void]

supposed to be this

[bold pond]

only odds

ok lets see

[dreamy void]

makes sense

if you do pythagoras it's approximately really close

[bold pond]

i know pythagorean theorem

[dreamy void]

the blue

[bold pond]

cool

[wraith dagger]

Results provided by WolframAlpha

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

Requested by adoniszgks

[dreamy void]

we have a tiny bit more arc length of course

makes sense since it's arc

i gotta go now

good luck on your exams

and good doing

[bold pond]

nice i got it

ok thank you, thanks for you helping me out that was cool

.close

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[noble dock]

Hey, if we have a “for all x, y \in R if P(x) then Q(x)” Is it equivalent to “if x, y \in R then if P(x) then Q(x)”

[noble dock]

what does the emoji/reation mean?

because i dont qutie udnerstand how those two are equivalent

because if we take the ocntrpaostive of both

we get for all x y in R if not Q(x) then not P(x)

and for the second statement

if P(x) and not Q(x) then x, y not in R

[cedar kiln]

@noble dock Has your question been resolved?

[mild citrus]

Is there an easy way to solve this integral?

[cedar kiln]

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[blazing scaffold]

for each real number p, points (ap + b, p) lie on the graph of each equation in the xy plane for the given system, where a and b are constants, what is the value of a + b
5x+6y = 8
20x + 24y = 32

[blazing scaffold]

i've plugged in the points to get p(5a+6) + 5b = 8, but i do not know how to proceed from there

also what exactly are constants in math? how are you supposed to interact with them?

[slate lintel]

"constants" are i guess numbers that don't change

some of them come from physical constants, such as the gravitational pull of the earth, or the mass of the sun

in this case, they're essentially just variables -- they have some value, we're not telling you what that value is, but in this case we're giving you enough information to figure that value out

[blazing scaffold]

ok got it

the question i am asking wants to know what a and b is

and uh that is all i know

[cedar kiln]

@blazing scaffold Has your question been resolved?

[blazing scaffold]

.close

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[cedar kiln]

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[crimson sedge]

can anyone help me? What is 2+2?

[nova snow]

6

[radiant scaffold]

3

[valid dome]

I need help for this equation pls. I'm stuck in my reasoning.
2+2=4 @crimson sedge

[crimson sedge]

bro legit u do math solv for 2+2

[radiant scaffold]

Weird q asked in this channel, imma still need you to get a new one sorry

[crimson sedge]

u are weak too jk

[dusk finch]

<@&268886789983436800>

[crimson sedge]

weak too jk

[crimson sedge]

oh i forgot its my2nd thime 💀

[nova snow]

https://tenor.com/view/case-caseoh-your-banned-you'e-banned-get-out-gif-7698576230662898937

[livid hound]

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[pearl junco]

hey guys, I just can't solve for 'a' and 'b' in this equation:
2a + b = 30
(its for triangle, I am really dumb or its just hard)

[dreamy void]

If you bring 2a to the right you solved for b

[pearl junco]

I have only this:
a = 15 - 1/2*b

[crimson delta]

from 2a+b=30 you cant uniquely solve for a and b

!original

[cedar kiln]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[pearl junco]

2a + b = 30
I get to this:
a = 15 - b/2

and when I calc for number its:
a = 10

but its isosceles triangle

so this cant be it

[pearl junco]

why tho?

the question requires to have set numbers

[brazen dagger]

What is a and b first of all?

[pearl junco]

a is side of a isosceles triangle, so 2a are 2 sides

b is the shortest one

looks smth like this

[brazen dagger]

Perimeter is 30?

[pearl junco]

yes

[crimson delta]

just from this information its just not enough to find the triangle. you are probably overlooking something in the problem statement

can you please post the problem statement

[pearl junco]

no, thats all

[plucky owl]

Then send an image of it

[pearl junco]

they are saying that perimeter is 30 and its isosceles triangle

[plucky owl]

Send an image of the full problem

[pearl junco]

its not in english tho

[plucky owl]

That's fine

Still send it

[undone halo]

Can you try translating it to the best of your abilities

[raven shard]

notice you can form a relationship between a and b

[pearl junco]

what releationship?

well

I can try making equation from this:

3b + 4a = x

where x is perimeter of that shape

b is shorter

and a is longer side

[crimson delta]

look at the sides of the middle triangle

[pearl junco]

yeah?

they are touching one point of other triangles

[crimson delta]

and two sides of other triangles

[pearl junco]

what other triangle?

the middle one?

or?

the one at the right

or left

[brazen dagger]

Instead of giving one liners , maybe help them understand what do you want to say

[crimson delta]

https://i.imgur.com/Z5VfsWe.png

[brazen dagger]

Ohhhh

[pearl junco]

they divide the side into two?

so b = a/2?

[crimson delta]

yes

[pearl junco]

oooh

so

[crimson delta]

and now you have two equations and two variables

[pearl junco]

2a + b = 30 can be
2a + a/2 = 30?

[crimson delta]

yes

[pearl junco]

ok thanks

so....

that means

5a = 60?

a = 12?

[crimson delta]

yes

[pearl junco]

ok thanks

.close

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[raven harbor]

Find the set S of all points x ∈ R where the function g(x) is continuous

[raven harbor]

I’m not really sure how to approach the question. Just by intuition it’s going to be oscillating between 1 and e^x, and I would guess that only at the point x=0 will it be continuous since e^0 = 1 so the limit from 0+ and 0- will approach 1

But how do I prove this?

[idle tusk]

epsilon delta definition will work here

[raven harbor]

With two cases for x is rational and irrational?

[idle tusk]

yep

[raven harbor]

.close

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[calm summit]

How do you find s4?

[calm summit]

The answer is 2 but idk how to find it

[vast pike]

A B and C are known

[calm summit]

Wait lemme process im not used to this format lol

[vast pike]

https://en.wikipedia.org/wiki/Vieta's_formulas

you can use Vietas's formulas to find A B and C

[calm summit]

Is it the S2=S1^2 -2 (Sum a b y)

[vast pike]

so now you only need to find the first number that is S_2 (squared)...but I leave it to you, as it is much simpler to compute than S_4

[calm summit]

Mhm

Imma try lol

[vast pike]

yes ... in your bracket seems that there is B

[calm summit]

Ya it’s B

I did that but I think my S2 still wrong

I substituted B with 1 and A with 2

A is my S1

Ohhh wait, should I use the S1 from my equation(y^3-2y^2+y-1) or should I use the ones from the original?

Because the S1 of the Y equation is technically S2 right?

[vast pike]

you are bloody right!

[calm summit]

Ohhhh ok ok i think I get it🙏

thank you!

[vast pike]

thank you too for that observation

[calm summit]

Have a good day

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[zinc crypt]

I dont understand what I'm doing wrong?

nvm figured it out

.close

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[copper geyser]

Q2 a)

[floral arrow]

,rccw

[floral arrow]

What have you tried?

[copper geyser]

I don’t get how do I do this

I’m new to math

[floral arrow]

Someone "new to math" wouldn't do that kind of exercise...
I'm sure you can at least expand the numerator

[crimson sedge]

if youre going to proceed with calculus, you need to know a decent amount of trigonometry

also parabolas, circles and stuff for integrals

[floral arrow]

$q = \frac{(v+2)^3-8}{v}, (v \neq 0)$

[wraith dagger]

Nel

[floral arrow]

Expand (v+2)^3 - 8 to remove the parentheses

[copper geyser]

Okay

And then?

[floral arrow]

Tell us your result

[copper geyser]

Is it
v^2+3v^2•2+ 3v2^2

[floral arrow]

No

[copper geyser]

What is it

Bruh I really dk math

[floral arrow]

Did you mean v^3 at the beginning?

[copper geyser]

No

[floral arrow]

you need to do some exercises on basic algebra, distributivity, factorization, ...

I personally can't help you with limits if you aren't comfortable with these things first

Maybe other people can

[copper geyser]

Okay

Thanks

[copper geyser]

I’ll try solving this once again

And get back

@floral arrow

Is it this

[floral arrow]

,w expand ((v+2)^3-8)/v

[wraith dagger]

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Upgrade to WolframAlpha Pro!

Requested by nelfeal

[floral arrow]

Yeah

[copper geyser]

Okay

Now?

[floral arrow]

Now the limit as v -> 0 should be obvious

[copper geyser]

Alright

[cedar kiln]

@copper geyser Has your question been resolved?

[cedar kiln]

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[compact gulch]

I want to know how to play with the limits

[nimble mountain]

play with?

[compact gulch]

As in

How do I change the limits to solve it

here's another example

[bitter plinth]

#❓how-to-get-help

[compact gulch]

?

This is an available channel

@nimble mountain idk how to explain

[nimble mountain]

this is the same problem?

[compact gulch]

nope, a different one

no need to solve

I want to know how to change the limits to solve

Idk

[nimble mountain]

u mean what would happen to the stuff in the summand if we replaced N with 2q for instance?

[compact gulch]

yeahh

[nimble mountain]

yea well u replace each instance of it exactly

[compact gulch]

but how?

[nimble mountain]

u just replace each instance if N with whatever you want

i do believe it is normally different if the top of the summand is not based on N as well

[compact gulch]

no it can't be whatever I want

[nimble mountain]

why

[compact gulch]

but idk why

[nimble mountain]

the left has terms from
N+1 to N^2

the right has terms
(1 to N^2) - (1 to N)

which is the same

[compact gulch]

but how?

is there a yt video for this

[nimble mountain]

try replacing them with numbers and writing it out

[compact gulch]

okay

[nimble mountain]

taking N=2

the left has terms from
N+1 to N^2

3+4

the right has terms
(1 to N^2) - (1 to N)

(1+2+3+4)-(1+2)

assuming the function is x

[compact gulch]

wait can you show me how you did this

okay wait nvm

3, and 4 are the only ones

[nimble mountain]

yes

[compact gulch]

It works

okay yes

But then how do you quikly determine this

[nimble mountain]

yay

[compact gulch]

?

Because rn I got this from an answer 😭

[nimble mountain]

whats the context

[compact gulch]

wdym?

no like

[nimble mountain]

whats this answer

[compact gulch]

no it's not the answer, that black image. It's like how to do it right

[nimble mountain]

honestly i assumed it to be true and worked backwards

😉

[compact gulch]

same

But in exam

they won't give us 😭

So how do I get this

[nimble mountain]

shouldn't your question be how to solve it

[compact gulch]

yes, I know how to tbh

It's just that

[compact gulch]

idk how to get this

like that black image

[nimble mountain]

then aren't you good

[compact gulch]

no wait

okay let's do another example pls

[compact gulch]

here

[nimble mountain]

in general

[compact gulch]

how do I create like a system to solve it

[nimble mountain]

1+2+...+n=
(1+2+...+k) + ((k+1)+(k+2)+...+n)

where k<n

thats all they did

[compact gulch]

I see

[nimble mountain]

well more specifically replacing

1+2+.. with f(1)+f(2)+..

[compact gulch]

I see but

[compact gulch]

in this case

this doesn't work

[nimble mountain]

it starts with k+1

[nimble mountain]

this had N+1

[compact gulch]

yea

but like how do I change this and stuff

[nimble mountain]

put a +1 to the n on the bottom left

[compact gulch]

why

you mean here?

[nimble mountain]

yes

[compact gulch]

but why

[compact gulch]

the question doesn't have +1

[nimble mountain]

same reason we put k+1 here

so the term doesnt overlap

[compact gulch]

I'm so conused 😭

[nimble mountain]

lets start from basics

we can split two series of additions right

like

1+2+3 =(1) + (2+3)

[compact gulch]

yeah

[nimble mountain]

so what we are doing here is splitting the series right?

if u notice, we stop the series first series at 1. and we start the next series at 2 (which is 1+1).

another example would be

1+2+3+4+5 = (1+2+3) + (4+5)

we end the first series at 3 and we start the next series at 4 (4=3+1)

thus the need for the +1 to denote we start at the next term

[compact gulch]

ohhh I see okay thanks @nimble mountain

[nimble mountain]

https://tenor.com/view/pokedance-pokemon-dance-bulbasaur-charmander-squirtle-gif-8027890098923710900

[compact gulch]

🫶

.close

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[ancient shadow]

I have the formula for a

confused how to apply it

formula being it's going to be 0, (Rsina)/a

[violet flume]

interesting problem

I'm trying to think of the formulation of equilibrium that is helpful here

maybe we want to define the angle between some reference point on the shape and the point of contact

then just work with the two centers of mass

and the centerline of the shape by dropping a perpendicular through the shape from the floor

[ancient shadow]

i see

[violet flume]

I don't know that thats it

just in my head this makes sense

[ancient shadow]

ah whatever i'll play w the problem more with this idea

thx

.close

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[atomic siren]

Kinda stuck on №10 of a problemset titled 'XOR and binary representation'. It goes something like this (paraphrasing due to translation):

"Consider planar graph G where every face is a triangle. Prove that the chromatic number of the graph is 4 if and only if the edges can be painted in 3 colors, such that no face has 2 edges of the same color."

For the direct proof, I represented the 4 colors of the vertices as 00, 01, 10 and 11, and painted edges connecting numbers differing in only the right bit color A, only the left bit color B and both - color C. The proof of the fact that no edge has 2 same-colored edges is pretty straightforward (2 number can't be the same, so if we want 2 edges of the same color we'd have to use all 4 colors in the 3 vertices)

I've also proven that if there's a coloring in 4 colors of the vertices, we can swap the colors so that in any vertice a selected color is chosen, but thats kinda obvious.

To prove the inverse statement, I tried to assign a random vertice the color 00 and stating that all the other vertice colors can only be restored one way using the edge colors, but I can't seem to prove we don't run into a contradiction.

[cedar kiln]

@atomic siren Has your question been resolved?

[crimson sedge]

@atomic siren This is a hard one, but I think I figured it out (might've goofed up though, will see)

Consider how a contradiction would occur

As in, like, describe it, in text, right here

[atomic siren]

Like if we go one way, we get that a vertice must be one color, but when we choose another path, it has to be another color

Can't list an actual example on account of it most likely not existing under the given task

[crimson sedge]

@atomic siren Sorry for the late reply

I had the same train of thought

But you can reduce this problem down

[atomic siren]

Oh, it's fine, really

[atomic siren]

wdym?

[crimson sedge]

The issue occurs when a vertex is in 2 different triangles, both with the 2 other vertices already decided, and which give a different value for the vertex in question

[crimson sedge]

@atomic siren Do you understand and agree with this?

[atomic siren]

understand
Not really, no. Doesn't have to come down to the last triangle. Maybe I can sketch smth

[crimson sedge]

Is it true that if every triangle a vertex is a part of agrees about that vertex's value then that there's no contradiction about that vertex's value?
And is it true that if there are 2 triangles which a vertex is a part of that disagree on that vertex's value then there is a contradiction about that vertex's value?

[atomic siren]

I don't really get how a triangle defines a point, if all it takes is the color of an adgacent vertice and the edge between them

[crimson sedge]

Oh right

Yes, good thing to point out

Let's consider a single triangle instead then

Say you know the values of 1 vertex and every edge

Can you show there will be no contradiction within that triangle?

[past wave]

Hello Mike Hermantraut (hope I spelled it right)

[atomic siren]

Yeah.
If both edges are 1 bit flippers, the results differ by 2 bits, so the other edge must be the both-bit switch.
If one is a 1-bit flipper and the other is a 2-bit switch, the results only differ by one bit, and since there are only 3 vertices, there are only 3 colors, which is less than the allowed 4

[crimson sedge]

@atomic siren After some more thinking I finally found a goof with my reasoning

dang

[atomic siren]

sadge

[crimson sedge]

@atomic siren Then you should try it with this

and this

and show how in all of these cases a contradiction can't arise

Then the idea was to say that if for every vertex, considering all of the triangles around it (which are all you need to determine whether there's a contradiction involving that vertex's value or not), you wouldn't arrive at a contradiction, then this algorithm would never arriver at a contradiction at all

Because no matter what new vertex you checked you'd know that all the already known vertices and edges around it would agree to its value

but then this happened

oh wait

yeah nah

Anyways I'm pretty sure I was on the right track there but I don't know how to proceed

oh wait

[crimson sedge]

Assuming we have a vertex that connects to 4 other ones like this in a graph as described in the problem, it's not possible for any of the 2 left vertices to connect to any of the 2 right vertices through any path

Because then we'd have a non-triangular face

(also the outside is technically a face but it's pretty clear we're not considering it)

So it would split the graph into 2 parts and you could just do 1 of them and then the other one

And this generalizes to any number of parts

@atomic siren So this should fix the goof in question

If you have any questions, please ping me

But I think this should be all you need to construct a proof

Also I may have just given too much information, oops...

[atomic siren]

Probably not

It basically boils down to why there can't be a loop that doesn't result in no changes, where for size 3 it's basically given

[crimson sedge]

So, do you understand what I said and are you convinced there are no more things to sort out?

I'm guessing not

[crimson sedge]

I forgot the part where you generalize this to any number of adjacent triangles

[crimson sedge]

Oops, thought what I said implied that they weren't directly connected as shown in the image

But I re-read what I wrote and nowhere did I imply this

my bad

Correction: They can't connect together through any path other than directly

@atomic siren With that clarification out of the way, do you see any remaining issues or unsolved questions?

[atomic siren]

While that may've cleared out the case where jt's connected to 2 independent triangles, I don't really see how it applies to bigger cases

Like this

[crimson sedge]

Have you considered the cases with 2 and 3 adjacent triangles first?

[atomic siren]

Only one way to restore with 3 (aside from permutations), many to restore with 2

[crimson sedge]

...Huh?

We're assuming every edge's value is already set

And all that remains is deciding the vertices

And that we have 1 vertex decided

[atomic siren]

Well, no cases have contradictions depth 1

[crimson sedge]

[atomic siren]

Probably going to have to ponder about it later. Could I PM you with a thanks when I understand?

[crimson sedge]

Sure

Does that imply you won't PM me for, like, the rest of the explanation?

Like I know what I said earlier isn't crystal clear but I can clarify anything

And I'm pretty sure I have all I need to solve this

@atomic siren Anyways, if you're gonna stop using this channel right now you can .close it

[atomic siren]

Eh, I'll try to figure some stuff out m'self

.Have a good one

.close

[cedar kiln]

Channel closed

Closed by @atomic siren

Use .reopen if this was a mistake.

[cedar kiln]

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[crimson sedge]

how do i go about this

i know taht u.v = 0 when the orthogonal as