help-43

[kind wind]

id slow

[paper anchor]

Okay, so the further you move away from the mean position (centre of earth), the slower you move, until you ultimately stop?

[kind wind]

yea

like i will oscillate

[paper anchor]

And can I take that motion to be acceleration?

[kind wind]

from the mean position

[upper bane]

friendly reminder

slowing down is also a form of acceleration

[paper anchor]

Therefore the further you move away from the centre, the more you lose acceleration you had from the momentum you had

[kind wind]

yea

[paper anchor]

Hence, as displacement increases from mean position, acceleration decreases

[paper anchor]

Exactly

[kind wind]

thus a prop -x how does that connect

??

[upper bane]

think there might be some mistake in this sentence

[paper anchor]

x is proportional to -a yeah?

[kind wind]

nooo

like

they said its not no ??

[upper bane]

the further you are from the center, the more you lose velocity, but gain acceleration in the other direction

because the increasing acceleration in the other direction is dragging your velocity down

[paper anchor]

yeah but you lose acceleration from your momentum and gain acceleration to the restoring force

[upper bane]

hm that might be important context for OP

[paper anchor]

I'm just making it simpler for him

[upper bane]

aye

sorry to interrupt

[kind wind]

wtf

[paper anchor]

True

[kind wind]

thats not simple bro

again

[paper anchor]

who said it's not

[kind wind]

wait didnt they

!

[paper anchor]

Think logically

[kind wind]

oh

a to x

[paper anchor]

You said that you'd oscillate

[kind wind]

she said

yea

[paper anchor]

and that's correct

[kind wind]

ohh

so it is correect

[paper anchor]

yes

[kind wind]

-a prop to x

yea

[paper anchor]

Yes!

[kind wind]

fr twin

k k

continue

[paper anchor]

just switch the minus sign, you get a is proportional to -x

[kind wind]

yea

[paper anchor]

add a proportionality constant, a = -kx

[kind wind]

which all together mean negative acceleration after the equality

am i correct

??

[paper anchor]

What do you mean by equality though?

[kind wind]

=

and after a constant k

[paper anchor]

Yeah yeah

you're correct

[kind wind]

it all together establishes that as x increases a increases opp to x

because

-k

[paper anchor]

k is a constant, I think you mean -x

[kind wind]

ahmmmmmm as -x increases ?? what like

ohhh

yea

yea

as x increases a increases and a's direction opp to x because of -x

[paper anchor]

switch the minus sign to a, that'll be easier for you to grasp

[kind wind]

innit

???

[paper anchor]

yeah, exactly

[kind wind]

k k

thanks mate

[paper anchor]

Anytime

[kind wind]

and the constant is omega

squared

[paper anchor]

Study well

[kind wind]

which i may not require to know

how it got derived just yet

[paper anchor]

That's from the graph

[kind wind]

might be

[paper anchor]

from the sine graph, you'll get an equation for SHM, there you'd understand omega

[kind wind]

k k

[paper anchor]

Alright then, study well and ask if you struggle to understand anything

[kind wind]

ok

[paper anchor]

you can close the channel if you don't have any more immediate questions

[kind wind]

.close

[compact pewter]

Channel closed

Closed by @kind wind

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[true lava]

,rccw

[true lava]

how am i suppose to draw a graph that has two points of inflection one absouloute maimum and no absouloute minimum

[kind viper]

well let's see

since there's only one absolute max, you'd want the curve to first go up, then go back down

and you want the curve to be concave-down near the maximum but concave-up both far to the left and far to the right, and not change concavity any more than that

[true lava]

huh

[kind viper]

if you know what a concave-up curve looks like without reaching any local minimum, this is kinda enough to fit together the basic shape of it

[true lava]

i lost myself reading that

[kind viper]

ok let me try going a bit slower

im gonna try to make you a sketch of the curve without the coordinate axes but like pretend they're there

[true lava]

ok

[kind viper]

so heres a portion of the curve with a maximum point highlighted in red

[true lava]

yeah

[kind viper]

ok

now we want the curve to change concavity twice, yeah?

cause of the two inflection pts

[true lava]

yeah but if we do that the

then we will have 2 miniumus

[kind viper]

or will we?

you are imagining something like this, yes?

[true lava]

yeah?

[true lava]

(insert vsauce intro here)

[kind viper]

but consider:

what if we just never let those concave-up pieces reach their minimum pts?

[true lava]

so it being disconnected?

[kind viper]

oh no, i was gonna join these up just now

but the point is that those minimum pts are not something that has to happen

[true lava]

ah so its aroc become 0 therfore not becoming minimums

[kind viper]

something like this

no, the idea's that the slopes never become 0

[true lava]

so before they become 0?

[kind viper]

like this

you can stretch these out as much as you want

[true lava]

yeah

[kind viper]

the purple bits are inflection pts

[true lava]

yeah so thats it?

[kind viper]

i.e. the places where a concave-up and a concave-down bit meet

and yeah that's p much it

make a sketch mimicking mine just about anywhere you want on the grid

[true lava]

yeah oki doki

i got another one

has one local maximum two global minima and 2 points of inflection

[kind viper]

try to think in terms of the curve pieces again

[kind viper]

actually look back at this picture

[true lava]

mustache shaped

[kind viper]

it's got just about everything you want, doesn't it?

(it just needs to be joined up again)

[true lava]

yeah

[kind viper]

and you'll want these minimum pts to be level with each other of course

[true lava]

yeah

and the longer bits on the side need to be higher then the maximum?

because its a local maximum

and not global]

[kind viper]

ehhhhh

well i guess yeah sure

you can even have them shoot off to +∞

[true lava]

yeah so just give em arrow

next one is : has one point of inflection, no relative extreama and no absouloute extream

im confused on the no relative extreama

idk what that means

@kind viper ?

[kind viper]

it means you don't have any local max or min points

[true lava]

so nada max or mins oki doki

can i do a straight line?

diagonal line*

[kind viper]

that won't have any inflection pts

[true lava]

a slight curve that doesnt reach a aroc of 0

[kind viper]

mmmmy yeah sure

[true lava]

kinda like the one in the first 2 quadrant you drew

like the top left one

ms ann?

[kind viper]

the curve has to have a segment like the top left and one like the top right.

and a transition point between them in the form of an inflection point.

[true lava]

so it is correct right?

[kind viper]

im not sure i like any of your wording, but maybe if you show me what curve you've cooked up i could validate it

[true lava]

essentially that

[kind viper]

so far that's only the top-left quadrant in my breakdown, i.e. f' and f'' both > 0

so now you need to attach a segment to, say, the left of this bit which looks like my top-right thing (f' > 0, f'' < 0)

(ie still increasing but this time concave down)

[true lava]

huh\

[kind viper]

... idk how else to word this im sorry

[true lava]

hmm

can you sketch it?

ms ann?

@kind viper

[kind viper]

one moment

[true lava]

ah

that makes alot of sense

i got another one absouloute extremum,no points of inflection, and one local extremum

cant that be a vertical line because a local extremum and absouloute extrememum doesnt 1 extremum count for both?

@kind viper

[kind viper]

a vertical line is not the graph of a function at all

[true lava]

oh ok.

[kind viper]

refer to this shape chart again

also uh

absouloute?

[true lava]

so just a semi cruved line?

[kind viper]

did you mean absolute?

[true lava]

yeah

i cant spell

[kind viper]

don't know what a "semi-curved line" is

you can't have any inflection points so your curve has to be concave-down throughout or concave-up throughout

[true lava]

one that is sligtly curved yet no aroc of 0

like the image

like the one i did prior

[kind viper]

mmm

yeah ig you can try to make something similar to that

at this point though i have already told you so much im almost doing these things for you

[true lava]

yeah

but im not always 100% sure on these

or i have an idea for this one

like this?

@kind viper

[kind viper]

that has like 2 inflection pts...

[true lava]

that middle part isnt meant to be flat

and that bottom is meant to be a straight line

@kind viper

?

[kind viper]

i do not know what else to do or say, but what you just drew is incorrect.

[true lava]

oh ok

yeah well then i got no more ideas

i meant it more like this if that helps

or maybe a big concave up smiley face?

@kind viper

<@&286206848099549185> i need help with the question: sketch f( x) whioch has one absouloute extremum,no points of inflection, and one local extremum

[ebon pewter]

Just join

The two above graphs

[true lava]

huh

[ebon pewter]

Wait il draw it for you

Can you send me a blank canvas?

[true lava]

wdym send you a blank canvas?

[ebon pewter]

This without the drawing

[ebon pewter]

Would be something like this

[true lava]

really?

[ebon pewter]

Yes

See it for yourself

[true lava]

but that has points of inflections

[ebon pewter]

Oh wait

I read the question wrong

;-;

Wont

[true lava]

so? any ideas

[ebon pewter]

$f(x)=x^2 $

[true lava]

huj

[ebon pewter]

Wsit

,w plot y=x^2

[boreal girder]

Results provided by WolframAlpha

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

Requested by .evix.

[ebon pewter]

Wont this work

[true lava]

hmm

would one side be higher then the other then?

@ebon pewter

[ebon pewter]

I mean not really

[true lava]

oh ok

[ebon pewter]

Worry im kinda sick and sleepy

So i just die

In between

[true lava]

oh ok

[ebon pewter]

Yes

[true lava]

can you help me with this next question?

,rccw

[true lava]

@ebon pewter

[ebon pewter]

Well how do you approach the question

Tell me

[true lava]

well i look at the graph and cant figure out how positive and decreasing works when the whole graph is positive so then the second question doesnt make sense

wait

its talking abt roc

im dumb

4-7?

[ebon pewter]

Ok positive decreasing means

It is going up but its turning to the right

You get me?

It is increasing but the rate of increase is slowing down

[true lava]

yeah ?

at 4-7 right

[ebon pewter]

At 4 to 7 its strictly increasing

[true lava]

from -1-1?

[ebon pewter]

Have to find something like this

[ebon pewter]

Yes

Amazing

Bravo

[true lava]

next one is 1-4?

[ebon pewter]

Yes

You got this

Im gonna do sleep

[true lava]

can you just check these last 2 man please

4-7 for question 3

[ebon pewter]

And the last?

[true lava]

im confused

there isnt any spots where it can be

[ebon pewter]

Wdym?

[true lava]

well the only negative spot isnt increasing but decreasing

so thats why im confused

[ebon pewter]

Could you mark it on the graph perhaphs?

[true lava]

wdym

[ebon pewter]

Ok this about this

How do you express rate of change of a function

Mathematically

[true lava]

you used pointslop form?

or just m

[ebon pewter]

Do you know how diffrentiation works?

[true lava]

no?

[ebon pewter]

Oh well

Ok so

Do you know about tangents?

[true lava]

yes

[ebon pewter]

Ok so

[ebon pewter]

Try and draw a tangent on any point on this

And see how it goes

And you go upwards in that curve

[true lava]

the flatter it gets

would 1-3 work tho?

[ebon pewter]

Bobito

Im getting really sick rn

Il ping someone to take over ;-;

<@&286206848099549185>

Mind taking over in done for the day

[true lava]

oki doki

ok

[ebon pewter]

And dw you will get it!

[arctic portal]

which question?

[true lava]

27

[arctic portal]

what's your status

[true lava]

huh

im just stuck

i think its 1-3

but not sure

[arctic portal]

how come 1-3?

[true lava]

becuase its negative

and there is no where else where its negative

[arctic portal]

also i'm assuming that you're supposed to eyeball it? bc we have no equation for h(x)

[true lava]

yeah

[arctic portal]

but there's a second part to it

also, the slope btwn 3 and 4 is also negative

[true lava]

ik

but then the rate of change becomes 0

[arctic portal]

yes, but before that the slope is nonzero but still negative, correct?

[true lava]

yeah

[arctic portal]

so can you tell me what the interval is where the rate of change is negative?

just 1 criteria

[true lava]

[1-4]

[arctic portal]

that would be nonpositive, but it's the right idea

[true lava]

then 1-3?

[arctic portal]

because slope at 1 and 4 is 0

dyk the difference between [ and (?

in terms of interval notation

[true lava]

yes

the square bracket is x values for intervals

and the ( is used for coordinates

[arctic portal]

not quite, both are used in interval notation even without a graph

btw are you american? i dont want to be feeding you wrong info

[true lava]

i go to an american school

[arctic portal]

ok good

so we usually use ( when we don't want to include something, while [ is used for when we want to include something

[true lava]

union

[arctic portal]

so for instance the solution set for x>3 is (3,inf), while x>=3 is [3,inf)

and we use () for positive and negative infinity

[true lava]

but this thing doesnt have infinity

[arctic portal]

that was just an additional fact

if it confuses you, ignore it for now

[true lava]

it doesnt

[arctic portal]

if we have 3<x<5, solution set is (3,5), but 3<=x<=5 it's [3,5]

have you learnt that before?

[true lava]

yeah

[arctic portal]

ok good

[arctic portal]

so just change one little thing about this

[teal quartz]

you all are wrong

[arctic portal]

to make it the set of negatives, not nonpositive

[true lava]

fun times

[arctic portal]

@molten badger am i tripping or

[teal quartz]

if it happens then i will remove this question quickly

[molten badger]

Nahh, I think you're good

-# eventhough it's too long for me to read everything

[arctic portal]

buddy i dont think that's possible

[true lava]

im going to lose my shit soon

this shouldnt take this long man

[arctic portal]

did evix go over the other questions with you?

[true lava]

yes

[arctic portal]

they all follow a similar thought process

[true lava]

those ones are easy

but i cant find where its negative and the graph is growing sharper

i think im just blind too see

[arctic portal]

what'd you get for 24?

[true lava]

-1,1

[arctic portal]

if i were you i would eyeball where the derivative is at its minimum somehow

[true lava]

im tryna man

[arctic portal]

like literally eyeball it

[true lava]

we dont use derivatives in ap precalc

[arctic portal]

i dont think you can even show work

[true lava]

youre not meant to man

[arctic portal]

its another word for slope

[true lava]

oh ok

is it not 1-2?

[arctic portal]

wait what you get for 25

[true lava]

1,4

[arctic portal]

bc the answer for 27 is just the other part of the negative section

hmm idts but i do have to go 😭

[true lava]

do you not have an answer either?

[arctic portal]

<@&286206848099549185> i summon thee kek

[true lava]

ima tweak

yall passing the ball to eachother

[arctic portal]

i do, but i cant tell you directly per mathcord rules

[true lava]

like every 4 seconds

[covert crater]

what's the question?

[true lava]

number 27

and i need some checking for the other ones

[arctic portal]

mx check his answer for 25 too

[mild sky]

(1,4) @true lava can you see why?

[true lava]

for 27 right?

[mild sky]

yeah

[true lava]

ive said this like 14 times.

ima loose my shite

[covert crater]

where's that though

or what's his answer

[true lava]

but for 24 i got -1,1 and 25. i got 1,4 and for 26 i got 4,7

[arctic portal]

^^

ok imma dip

[true lava]

this 4 questions shouldnt take more then 40 minutes bro

[last fern]

what the fliz

[pearl burrow]

how did you get (1, 4) for 25? the derivative is not decreasing in the whole interval right?

[true lava]

1,3*

[pearl burrow]

its more like 1, 2 i think

[last fern]

^

[true lava]

oh ok

[pearl burrow]

and what do you think the answer to 27 is?

[true lava]

1,4

[last fern]

you can think of the rate of change increasing/decreasing as which way the function curves

[pearl burrow]

we said in (1, 2) the rate of change is decreasing. then how can it be increasing in the whole (1, 4)?

[true lava]

we said its negative an increasing

its going to 0

which is increasing

[pearl burrow]

increasing in (1, 2)? can you tell me how would you see if the rate of change is increasing in an interval?

[true lava]

at 1,4

not 1,2

[pearl burrow]

why?

[true lava]

because its going from negative and goes to positive therefore its increasing

[pearl burrow]

its not necessarily increasing, here it is decreasing in an interval and then increasing in another interval

even if its always negative

[true lava]

rewind and start from the begning bro im losing my brains

[pearl burrow]

In the interval (1, 4) the rate of change is negative right?

[true lava]

yes?

[pearl burrow]

And you know the rate of change at a point is the slope of the tangent passing through that point, right? Then how is the slope of the tangent changing as you go from x=1 to x+4?

[true lava]

its becoming flatter

[pearl burrow]

what do you mean?

[true lava]

the tangents

[pearl burrow]

are the slopes decreasing as you go from 1 to 2?

[true lava]

yez

[pearl burrow]

and what are they doing from 2 to 4?

i tried drawing the tangents if it helps

[true lava]

yeah from 1-2 its getting sharper which is increasing

and its negative

so 27. is 1-2

[pearl burrow]

sharper but more negative so they are decreasing

[true lava]

yeah but were talking about rate of change?

[pearl burrow]

yeah

[true lava]

if its sharper the rate of change is increasing

and the slope is negative

[pearl burrow]

the rate of change is decreasing if the slopes are getting more negative

you need to consider whether the y values increase or decrease

[true lava]

really u sure ?

that doesnt make sense

[pearl burrow]

yeah

what makes you say that

[true lava]

well the negative change is increasing

becoming sharper

rate of change is talking about slope

meaning the slope is increasing more negativly

[pearl burrow]

increasing more negatively means decreasing. You can have negative slopes and slopes that get more negative, and that is what i mean by decreasing roc

[true lava]

oh ok

so i have 0 clues to what 27 is then

[pearl burrow]

lets step back a bit, what is the answer to 25)? negative and decreasing?

if you know this by exclusion we can get 27

[true lava]

1-2?

[pearl burrow]

so 27 asks about negative and **increasing **roc. given the roc is negative only in (1, 4), and in (1, 2) is decreasing, where can the rate of change be increasing?

[true lava]

1-3?

[pearl burrow]

so are you saying the roc is both increasing and decreasing in (1, 2)? how come

[true lava]

its only talking about that said section

not anything after it

[pearl burrow]

ok, my question is what makes you say it is increasing in (1, 3)?

[true lava]

idk anymore man

ive spent over an hour on this question

ive said all possible combinations

and somehow nobody said any are righjt

what am i suppose to do now.

[last fern]

yo so

from a to b, the slope of f is negative right

and decreasing

(for most of it)

[true lava]

yeah

[last fern]

alright

so we can use what the function looks like to gather the intervals

[true lava]

yeah

[last fern]

at this interval, you would say it looks like its negative and decreasing

right

[true lava]

yes

[last fern]

so that interval cant be negative and increasing

[ebon pewter]

Yaal still stuck

[last fern]

so you shouldnt include it in your answer

[true lava]

yeah

[last fern]

for 27

[true lava]

ion got an answer for 27

[ebon pewter]

Well tree got it

So i dont think im needed

[last fern]

but u can rule that out

[true lava]

does that mean its 2-3

[ebon pewter]

Thank you tree! :D

[last fern]

hm

whats the functions derivative doing here then?

[last fern]

ts has been going on for too long 😭

[true lava]

its negative and its going to 0

[true lava]

an hout and 20 minutes btw

[last fern]

hence, increasing?

[true lava]

yeah

[last fern]

so shouldnt that interval be your answer for 27

[ebon pewter]

https://tenor.com/view/kermit-falling-building-fall-woodz-gif-24715383

[true lava]

so it is 2-3

2-4*

[last fern]

yes

[true lava]

and just to double check

24 is -1,1 25 is 1,2 abd 26 is 4,7

[last fern]

hold on let me resend

[last fern]

that all sounds good!!

[true lava]

FINNALY

im still not free.

tree please save me man

[last fern]

💀

i wanted to add

u can use sin(x) as a reference

when its above the x axis, the rate of change of the slope is always decreasing

and when its below the x axis, the rate of change of the slope is always increasing

[true lava]

oki doki

please help me with one more question

[last fern]

using just the shape of the curve for reference u can quickly do problems like these

[last fern]

sure !!

[true lava]

,rccw

[true lava]

sorry for the blurry tings

[last fern]

um apparently these are supposed to be limited to one question?? but

[true lava]

they are?

ive used this wrong for months then

[last fern]

what do you think?

if no one complained about it ur probably fine

im kinda new to this

[true lava]

ah ok

3-4 looks like 27

so is it negative and decreasing

[last fern]

decreasing?

27 was negative and increasing

[true lava]

oops wrong one

[last fern]

all good

[true lava]

thats what i meant tho

[last fern]

awesome

what about the rest?

[true lava]

b i beleive is positive and decreasing?

because its going to 0 from a positive number

[last fern]

i agree with that

[true lava]

8-9 looks like negative and decreasing

and last one is posotive and increasing?

[last fern]

alright, all those answers sound good!

out of curiosity, did you use your answers for the first problem as reference?

[true lava]

yes.

ive learnt from my hour and a half of mistakes

[last fern]

😭

i mean

its a valid method

[true lava]

i mean if i suffered

ima use what i learnt

[last fern]

yea but you cant really reference that on a test

[true lava]

well i remember it now

so therefore i can

[last fern]

oh awesome

just checking

[true lava]

bad experiences is what you remember

therefore learning badly is like revision

guess what my teacher just posted the answer key

[last fern]

💀

[true lava]

so i did this for nothing.

[last fern]

well, that certainly was an experience..

[true lava]

an hour that i wonr get back fun time

nvm she forgot the one page i need to do

fun time

i want to finish it yet i dont at the same time.

you want to help me speedrun 4 more questions?

.close

[compact pewter]

Channel closed

Closed by @true lava

Use .reopen if this was a mistake.

[compact pewter]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[old rivet]

Let ac(A) be the acculumation points of A. Usually perfect sets are defined as ac(A) = A and closed but my book defined them as just ac(A) = A

[old rivet]

Does ac(A) = A imply A closed?

The answer is surely no, right?

Otherwise we would usually just define perfect sets like that

[dry canopy]

There are quite a few ways to define perfect sets

But yes, ac(A) = A implies that A is closed and has no isolated points (i.e. is perfect).

[old rivet]

Alright, thanks

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[south sage]

Is there an actual algorithm to determine if a (potentially multivariate) polynomial is prime?

[viscid canopy]

do you mean irreducible polynomials?

could you define what a prime polynomial is

[south sage]

Not factorable

[viscid canopy]

Over real / complex coefficients, there are only a subclass of cases where this happens

With complex coefficients: Only linear polynomials cant be factored
With real coefficients: Only linear or quadratics with imaginary discriminant cant be factored

[south sage]

Well like yeah, because C is the splitting field. But let's say we're working with integer coefficients

[viscid canopy]

I think the fastest way to tell is to factor over Q and see if there are any rational roots

"A corollary of Gauss's lemma, sometimes also called Gauss's lemma, is that a primitive polynomial is irreducible over the integers if and only if it is irreducible over the rational numbers."

https://en.wikipedia.org/wiki/Gauss's_lemma_(polynomials)

[south sage]

For more context, if it's helpful, I was working with a student on Factoring by grouping (all multivariate polynomials with 4 terms) and it's surprisingly difficult to explain wh:
x^2y + y^2x + x + y can be factored into (xy+1)(x+y), but x^2y + y^2x + x + 1 cannot

And I was wondering if there's an easier way to determine primality other than brute force testing in these cases

[signal willow]

Easiest way I have is: one can think of it as a polynomial in y with C[x] coefficients, check that quadratic factor doesnt work (degree and gcd considerations) so you are left with
(Ay+B)(Cy+D), so you want BD=x+1 and AC=x, these are obviously prime and you have finitely many cases to consider

Okay maybe C[y][x] is easier because you instantly have
(Ax+1)(Bx+1) which splits into only two cases to check

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[sullen canopy]

,rccw

[sullen canopy]

.close

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[molten badger]

[sullen canopy]

if you didnt notice the package was fragile

[dark merlin]

@sullen canopy

can u help me

[sullen canopy]

heyyyyyyyyyyyyy

open a channel

[sullen canopy]

#help-45

just join and type anything

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[young flame]

Hello

[young flame]

How do I know if A is true or not

[shell tendon]

What is this? What is happening here?

[molten badger]

Could you translate it

[compact pewter]

@young flame Has your question been resolved?

[chilly basalt]

There are three non-zero vectors a, b, c.
If (x, y, z) = (3, 2, 1) is the solution of the set
-# that set shown in the question

If d = (1, 1, 1) = 2a + 3b -c, which of the following statements are true

A) The set
-# bla bla bla
has infinite solutions.

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[lost hamlet]

How do I find the individual number of terms in this expression

[compact pewter]

@lost hamlet Has your question been resolved?

[molten badger]

Wdym by "individual number of terms"

[lost hamlet]

total number of induvidual terms*

sorry lmao

[glass swallow]

Just to understand you correctly, if it was (x^2+1)^3, the answer would be 4, because (x^2+1)^3 = x^6 + 3x^4 + 3x^2 + 1 having 4 different powers of x.

[lost hamlet]

yeah exactly

[glass swallow]

So in the last expression, we can skip the denominator, because it just shifts all powers of x by -15, right?

[lost hamlet]

all three are the same expression

[glass swallow]

Yeah I know, but in it's last form, it is easier to see this.

[lost hamlet]

im trying to find the total number of possible scores in a test of 300 marks and 75 quesions

each correct question gives you +4 , wrong question gives you -1 and unattended question gives 0 marks

[lost hamlet]

okay

[lost hamlet]

yeah

[glass swallow]

If we multiply (x^5+x+1) 15 times, the highest power of x is reached, when multiplying x^5 15 times -> x^75

[lost hamlet]

okay

[glass swallow]

If we take x^5 only 14 times, we reach x^70, but have on times "1" or "x", so we can reach x^70 and x^71

No, the numerator in the last line is (x^5 + x + 1)

[lost hamlet]

okay im sorry

[lost hamlet]

yeah

[glass swallow]

Now we take only 13 times x^5, then we can have 1, x or x^2 for the last 2 factors, so we reach x^65, x^66, and x^67

two more steps and we can reach all other powers of x down to x^0=1

[lost hamlet]

2 more?

arent there like 12 more?

[glass swallow]

12 times x^5 and 11 times x^5, then you will see, that you can reach everthing between the steps of x^5.

[lost hamlet]

omg

yeah

so i just gotta calculate the power of x then

[glass swallow]

An do not forget x^0

[lost hamlet]

it does nothing right

[glass swallow]

It is 1^15 = 1 = x^0

[lost hamlet]

oo

okay

thanks @glass swallow

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[tropic agate]

I know that if $x_n\to x$ and $y_n\to y$, then $\langle x_n,y_n\rangle\to \langle x,y\rangle$. Now, the author of my book claims that it follows from this result and the linearity of the inner product on a Hilbert space $H$ that $E^\perp={x\in H:\langle x,y\rangle=0\text{ for all }y\in E}$ is a closed subspace of $H$. What does linearity of the inner product have to do with anything?

[boreal girder]

psie

[vivid breach]

dont know much about this, but were you able to prove this without using linearity?

[tropic agate]

no 😅

[vivid breach]

at least for a vector space youd need it, since
<ax1 + bx2, y> = a <x1, y> + b <x2, y> = 0

maybe its a similar deal here

[tropic agate]

ah yeah, the linearity probably refers to the subspace thing. I was constantly thinking about the fact that E^perp is claimed to be closed. Do you understand why it is closed?

[vivid breach]

gonna be honest, I dont know what a hilbert space is

[hushed magnet]

closed just means the limit of the sequence is in the space. which is easy to show with the property you wrote down

linearity is just for it being a subspace as you noted

[tropic agate]

Ok. Makes sense. I think I get it now. 🙂 Thanks.

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[cedar hawk]

im confused on this question, it seems easy but its not and i dont know where to begin because ive tried simplifying it but ended up with x = y which cant be

[chilly basalt]

you’re extremely close to the question

[cedar hawk]

?

[chilly basalt]

If x ≠ y, how about x = -1/y

[cedar hawk]

but how would that come to be

[chilly basalt]

Could you show your work?

[cedar hawk]

wait hold on

i think i read the question wrong

[ripe saffron]

minus

[verbal rover]

thats actually solvable

[chilly basalt]

Mb

Fixed

[verbal rover]

simply multiply both sides by xy and simplify

[cedar hawk]

ive gotten to $x^2y - y = xy^2 - x$

[boreal girder]

Marco-Polo20

[spark folio]

group them all, and factorize

[cedar hawk]

but then doesnt it just end up as x = y

[verbal rover]

throw the -y to the other side and bring xy2

[cedar hawk]

wait nvm nvm

mb

is it $xy = (-x + y) / (x-y)$

[boreal girder]

Marco-Polo20

[verbal rover]

yeits -1

[cedar hawk]

yeah it was -1

but how do you get -1 from that equation

i mean i get it

but shouldnt it be -2

oh wait nvm

you just times all of the numerator by - 1

[verbal rover]

bro get enough sleep

[chilly basalt]

You already arrived at this through simplification.

[verbal rover]

its good for your health

[cedar hawk]

good idea

but how do you know i need more

this question was actually easy

idk why they put the difficulty of it as 4

[verbal rover]

its clear for me based on how your attention is bad

[cedar hawk]

its because i ask before thinking it through properly

[verbal rover]

you dont need to think if you had good sleep

[cedar hawk]

bruh

thanks for the help guys

.close

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[chilly basalt]

.close

[winged lion]

that thing doesn't pop up for you Swaamii?

[chilly basalt]

It didn’t show up to me

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[hazy obsidian]

Hey! im stuck on a geometry question, one second

[hazy obsidian]

wait bro

[boreal girder]

Copter

[hazy obsidian]

i honestly dont know where to start

wait R isnt even visible

(uh R doesnt always lie on Omega_A by the way, its just a coincidence)

status 1

[compact pewter]

@hazy obsidian Has your question been resolved?

[hazy obsidian]

<@&286206848099549185>

i can sense nobodies answering this

[compact pewter]

@hazy obsidian Has your question been resolved?

[compact pewter]

@hazy obsidian Has your question been resolved?

[hazy obsidian]

alright man

😭

[compact pewter]

@hazy obsidian Has your question been resolved?

[calm mist]

Hey

Whats up

[calm mist]

Ohkay!

I can tell why nobody has rushed to answer this 😭🙏

Gimmie 10 and ill try something

[calm mist]

What on earth

Ok lets see

Im gonna think out loud here

Given

• ABC is triangle with circumcircle ω
• Ω_A is the A-excircle of ABC
• Y and X are the intersection points of ω and Ω_A
• P and Q are projections of A onto the tangent lines to Ω_A at X and Y respectively
• The tangent line at P to the circumcircle of the triangle APX intersefts the tangent line at Q to the circumcircle of the triangle AQY at point R

So P and Q are foots of perpendiculars from A to the tangent lines to Ω_A at X and Y

Since P and Q are projections as stated in the problem, $AP \perp$ tangent at X and $AQ \perp$ tangent at Y

[boreal girder]

i need $10

[calm mist]

So the tangent lines at P and Q to the circumcirlces of APX and AQY are the polars of P and Q with respect to these circles

The point R lies on both tangent lines, so it lies on the radical axis of the two circumcircles

We know the radical axis is perpendicular to the line joining the centers of the two circumcircles

BC is the chord of the circumcircle ω, and the A-excircle Ω_A is tangent to BC externally

This means that the line AR is perpendicular to BC as R lies on the radical axis of the two circumcirlces, which is perpendicular to BC

Thus we can conclude that $AR \perp BC$

[boreal girder]

i need $10

[calm mist]

Im too lazy to put the line on top

But i think thats right

@hazy obsidian

[compact pewter]

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[placid spruce]

i'm trying to solve this on desmos but i'm not getting anywhere

[twin anchor]

try writing a parabola in vertex form

[placid spruce]

i did

[twin anchor]

what would the equation look like?

[placid spruce]

y=a(x-8)^2 + 16

[twin anchor]

perfect. can you write b and c in terms of a now?

[placid spruce]

what equation would i use to do that

[twin anchor]

just expanding that expression

we know y = ax^2 + bx + c and y = a(x-8)^2 + 16

just match coefficients

[placid spruce]

y=ax^2-16x+80 ?

[twin anchor]

you dropped a's

[placid spruce]

oh is it (ax-8a)^2

[twin anchor]

not exactly

[native shard]

how can you write a + b + c in terms of f(x)?

[twin anchor]

try expanding the (x-8)^2 first

and then multiply that by a

[placid spruce]

ax^2-16ax+64a

[native shard]

no do not expand

no

[twin anchor]

that works too

[native shard]

^

remember f(x) = ax^2 + bx + c

what is a + b + c?

what do we need to let x =

[placid spruce]

should i divide by the xs?

[native shard]

😭

brother

f(what) = a + b + c

[placid spruce]

oh 1

[native shard]

yes

so we’re trying to find f(1) is f has a vertex at (8, 16) with two zeros

since f has two zeros and it’s vertex is above the x axis it means f is concave down so f(x) has a maximum at the vertex

hence f(x) <= 16 for all x

which leaves 15 as the only possibility

[native shard]

since a + b + c = f(1) <= 16

[placid spruce]

ohhh

[native shard]

i’ve seen this sort of question like twice now here, get used to it

seems to be a common sat problem

[placid spruce]

yeah i'm trying to drill the method of solving

[twin anchor]

thats kind of a crazy sat problem lowkey

[native shard]

eh

harder than solve for x if 7x + 3 = 2x + 13 but not too difficult

[twin anchor]

i mean nothing on the sat is too difficult but i think thats pretty noticeable

[native shard]

if you understand parabolas this is simple

[placid spruce]

its not difficult tbh its just learning how to approach

[native shard]

always try thinking of a graph and determine if it has to be concave up or concave down

[placid spruce]

i heart desmos

[native shard]

and relate it to the number of zeros/where the vertex is (above or below the x axis)

[placid spruce]

ty

.close

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[pulsar isle]

Hi, I have a question about the trick for solving this limit. I recreated it from a handwritten textbook, so the red or orange lines are significant. The method might be an unconventional one made by a professor

[frail bane]

whats your question?

[pulsar isle]

the trick or method used to solve this ?

[frail bane]

when (x,y) -> (3,-2) you have 2x + 3y -> 0 and because if you dont factor anything before doing the limit you will get 0/0 which doesnt help you, so you factor out 2x + 3y in the top and bottom to try and get rid of the 0/0

[covert spear]

how does (4x^2/2x + 6y^2/3y) = (2x-2y)?

[frail bane]

there must be a sign error

[covert spear]

yeah

makes no sense

[frail bane]

lol, its just a flipped sign

not that crazy of a mistake

[covert spear]

good point

[compact pewter]

@pulsar isle Has your question been resolved?

[pulsar isle]

there might be a typo. Thanks, I understand the problem is a 0/0 indeterminate form, but that doesn't seem to directly help with factoring out the term (2x+3y). Simply guessing how to factor this seems difficult. There might be a trick or a method for this simplification that is more systematic than guessing, which I want to know. , Also, some steps are skipped, which might be related to the orange lines and the numbers associated with them.

[frail bane]

it helps because 2x + 3y -> 0, its not really a guess, its just that if i have (x,y) -> (a,b), then (bx - ay) -> 0, so you factor that

its very systematic, not a guess

[pulsar isle]

Finding the factors of 2x + 3y is a systematic process, not a guess. The problem is how they factor it neatly without eliminating fractions , And what do the orange lines do? , However, if you think it’s taking too long and consuming others’ rights, I can close it.

[frail bane]

nono, take all the time you need to understand dont worry :)
I dont understand what you mean by "without eliminating fractions", they just factor it normally from the top and bottom
the orange lines are I think just to show that if you take 3y and 2x then it goes to 0, its a sort of "cross product" if you know what i mean, its just to visualize how to find the factor, not very important

[compact pewter]

@pulsar isle Has your question been resolved?

[compact pewter]

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[pulsar isle]

.reopen

[compact pewter]

✅

[pulsar isle]

So, there's no shorter way than 6x^2+(29/3)xy+y^2 => 3(6x^2+(29/3)xy+y^2) = 18x^2+29xy+3y^2 => (2x+3y) (ax+by) = 2ax^2+(2b+3a)xy+3by^2 => a=9 b=1 => (2x+3y) ((9x+y)/3) => (2x+3y) (3x+y/3) to factor (2x+3y) from denominator for example? , and , As you said, if we have a(x, y) → (a, b) that makes the limit zero, then the linear term (bx - ay) will be a factor. I would like to use this as a general method. If you don't mind, I'd be happy to understand or have a reference that proves why knowing a '0/0 point' (a point where the limit vanish) helps us find a factor so easily and in which contexts its valid ?

[ripe ether]

the linear term might not be a factor. for example, x^2-y when (x,y)->(1,1) is 0. However, x-y does not divide x^2-y

[upbeat gorge]

geometrically speaking, when you have more than one variable there is more than one dimension to approach in. so you can approach the limit in an infinite number of ways: linear paths, quadratic paths, spiral paths, etc.

and each of these paths, heuristically speaking, corresponds to some factor to take out

in this specific case, the linear factor worked out, but you could maybe construct a different quadratic path that -> 0 as (x,y) -> (3, -2)

this is the intuition, anyways

[compact pewter]

@pulsar isle Has your question been resolved?

[pulsar isle]

.close

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[dusky quartz]

12 13 16 14 15 36 16 17 __ find the next number in this sequence

[ashen lava]

64 ez

[dusky quartz]

that's crazy

but is it really

[ashen lava]

its a prediction tho

[dusky quartz]

you can make an infinite amount of sequences that passes through those points, and still produce infinitely different answers

[humble mortar]

2

[upbeat gorge]

you can, so is there any other context to this problem? or maybe what topic it originates from?

[humble mortar]

the next number is 2

[upbeat gorge]

actually it's 2

[dusky quartz]

its just the equivalent of a facebook "guess the next number" quiz

[humble mortar]

because the sequence is { 12 13 16 14 15 36 16 17 2 ... } repeating

jokes aside, hopefully this has demonstrated the problem with "find the next number" questions.
are there any other properties the sequence should have?

[sharp owl]

yo can someone help with Calc AB

[arctic portal]

see #❓how-to-get-help :)

welcome to the mathcord!

@dusky quartz apologies for the ping, but are you still here?

[dusky quartz]

not dtated

stated

should probably just be 64 ig

.close

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[strange pendant]

the equivalence relation is given in words

why do they mention cardinality?

[random path]

The equivalence class is given by: two sets in P(A) are equivalent if they have the same cardinality. How many distinct equivalence classes are there? Find a representative for each equivalence class

[strange pendant]

oh ok

[x] = {C ~ D | |C| = |D|, C in P(A), D in P(A) }

[random path]

yup

[strange pendant]

I am having trouble definiing the equivalence class in set builder notation

[x] = {y in P(A) | y ~ x} subseteq P(A)

[random path]

Yeah that works!

Don't need the "subseteq P(A)" part to define it either, the text just says it to emphasize that [x] a subset of some set.

[strange pendant]

help please

[random path]

You can just say " [x] = {y in P(A) | y ~ x}"

[strange pendant]

no

[random path]

"y in P(A)" already tells you that all elements are of P(A)

[strange pendant]

X ~ Y <=> |X| = |Y|
X,Y ∈ P(A)
A = {1,2,3,4,5,6,7,8,9,10}

[random path]

Oh right, just write the cardinality condition for x~y instead

[strange pendant]

X,Y ∈ P(A)
A = {1,2,3,4,5,6,7,8,9,10}
X ~ Y <=> |X| = |Y|

[upbeat gorge]

!occupied

[compact pewter]

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).

[strange pendant]

A = {1,2,3,4,5,6,7,8,9,10}
X,Y ∈ P(A)
X ~ Y <=> |X| = |Y|

X, Y ⊂ A

[upbeat gorge]

do you have to define the equivalence class, or could you just define each partition?

[strange pendant]

help me write the set builder

[random path]

I think he just needs a representative for the exercise

[strange pendant]

this is the condition for equivalence

[random path]

But he wants to practice the set builder anyhow

[strange pendant]

first we need to find the quotient set or each of the equivalence classes

no?

[random path]

No it just says find the number of distinct equivalence classes I think

[strange pendant]

X R Y <=> |X| = |Y|

X,Y ⊂ {1,2,3,4,5,6,7,8,9,10}

[random path]

And then it also asks for a representative

[strange pendant]

the cardinality of the qoutient set bro

no?

first, how do we find the equivalence classes?

[random path]

Yeah

[random path]

Well, there's one for each possible cardinality that a subset of A can have.

[strange pendant]

nice dude

[random path]

And A is finite so that's easy to count

[strange pendant]

|AxA|?

[strange pendant]

whats the cardinality of P(A)?

[random path]

No just subsets of A. Like {1} has cardinality 1 for instance

[random path]

2^|A| but you don't need that for this problem

[strange pendant]

ok we are fucked

we need to sum the individual cardinality of the 2^n sets or something

[strange pendant]

how many are distinct? we are fucked

[random path]

Well, there's only a few possible cardinalities the subsets of A can have

Like {1} has 1

[strange pendant]

0,1,2,3,4,5,6,7,8,9,10?

[random path]

Yup!

[strange pendant]

0 for the empty set which is in P(A) jaajaj

jajajaj

so that's 11 distinct

equivalence classes

[random path]

yup!

[strange pendant]

the cardinality of the quotient set is 11

[random path]

yeah

[strange pendant]

I don't get it

[somber quail]

@proper portal

[strange pendant]

ø,1,2,3,4,5,6,7,8,9,10 are the representatives?

[random path]

No, it would be sets of those sizes

Subsets of A of those sizes

[strange pendant]

what?

[random path]

1 isn't a subset of A

[strange pendant]

the equivalence classes are 0,1,...,10

[random path]

Sooo the sets have cardinality of those sizes in each equivalence class

But the representatives are members of P(A)

Which are subsets of A

So for example for the equivalence class with subsets of cardinality 5 you could pick {1,2,3,4,5}

[strange pendant]

[{1}] = {{1},{2},...,{10}} is the equivalence class of {1}

[random path]

yup

For your sanity I wouldn't list out all the objects, some of those classes can get pretty big

[strange pendant]

it says list out a representative of each equivalence class

[random path]

like the equivalence class with subsets of cardinality 5 will have 10C5 subsets

Yeah just a representative

,w 10C5

[boreal girder]

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[strange pendant]

just the representatives are

[random path]

Like a single element from the class

[strange pendant]

{1},{1,2},{1,2,3},{1,2,3,4},{1,2,3,4,5},{1,2,3,4,5,6},{1,2,3,4,5,6,7},{1,2,3,4,5,6,7,8},{1,2,3,4,5,6,7,8,9}, {1,2,3,4,5,6,7,8,9,10}

also one for the equivalence class of [{}]

[random path]

Yup

[strange pendant]

{},{1},{1,2},{1,2,3},{1,2,3,4},{1,2,3,4,5},{1,2,3,4,5,6},{1,2,3,4,5,6,7},{1,2,3,4,5,6,7,8},{1,2,3,4,5,6,7,8,9}, {1,2,3,4,5,6,7,8,9,10}

[random path]

[strange pendant]

was it hard or meh?

[random path]

Difficulty is subjective

[full moat]

Got here a little late, but in case it helps you with equivalence relations and some intuition of them

Equivalence relations are anything that can partition a set, each class will be a slice of your original set, and a representative will simply be a randomly chosen element of your nonempty slice

While lastly the quotient set is the sliced up set(note X/~⊆P(X))

[strange pendant]

how did we found the equivalence classes

the equivalence relation is such that X,Y ⊂ A
X ~ Y <=> |X|=|Y|