help-23

[thick dagger]

ok makes sense cuz i was doing that before 😭

[spark zinc]

yeah

[thick dagger]

took me like three tries to get the right answer for each qs

[spark zinc]

all good

[thick dagger]

tysm!!

mwah ❤️

[spark zinc]

done?

[thick dagger]

yup!

[spark zinc]

ok welcome bye

[thick dagger]

❤️

.close

[safe radish]

Channel closed

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[safe radish]

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[restive lance]

hello , can someone help me with question

[split ether]

Hint: ||Expand (A + 4I)(A - 5I)||

[restive lance]

you mean A + 2I ?

[split ether]

No, but you will need to expand the (A+2I)(A-3I) as well, yeah

[restive lance]

i did but I didnt know what to do next

[split ether]

Expanding (A+2I)(A-3I) in the given equation will tell you what A^2 - A is. The latter term also appears when you expand (A+4I)(A-5I)

[safe radish]

@restive lance Has your question been resolved?

[restive lance]

Im trying to know what to do next

[plucky elk]

The rest is basically algebra, multiplying by inverses, using the A^2 - A equation wherever you see an A^2 etc

There's a lot of ways to do it

[restive lance]

.close

[safe radish]

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[safe radish]

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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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[echo gazelle]

for 15b what do these mysterious symbols mean

[spark zinc]

have you tried pmod exam?

trinity is not good

[echo gazelle]

but I go to trinity

we use WAEP

there’s no point of me doing pmod

[vagrant ice]

you're stuck on part b right?

try translating that statement into words

[echo gazelle]

yea

for all integers x there belongs an integer

[spark zinc]

to answer your question....

Bracket 1: all values x are integers

Bracket 2: there exists a value y in the set of integers

Bracket 3: then just 2xy=24

[spark zinc]

Pmod is better

trust

[vagrant ice]

and it mentions counterexample so you just need to choose an integer value of x

[echo gazelle]

how is that going to benefit me

[vagrant ice]

such that brackets 2 and 3 don't follow

[spark zinc]

do practice exam from Pmod its the best school in perth

hardest exams are better

trust me

[echo gazelle]

pmod doesn’t use WAEP and im going to do a WAEP exam

[vagrant ice]

why are there so many WA people in chat

bum chicken is also from WA

[echo gazelle]

no it’s not i have like 4 friends that moved in like year 9

to like hale or ccgs

[spark zinc]

ccgs exams are good

trust me

[echo gazelle]

ccgs uses waep

[spark zinc]

for chem and physics especially

[echo gazelle]

oh

[spark zinc]

methods and spec idk

ccgs physics comes in clutch

do u have methods?

[echo gazelle]

I already did it

[spark zinc]

How did u go

[echo gazelle]

it was kinda easy except for that optimisation on non calc

I think calc was easier than non calc

[spark zinc]

the optimisation with the log?

not like logrithm but acutal wooden log

[echo gazelle]

there was more interesting questions compared to the old ones

[spark zinc]

did you get that one in your exam?

[echo gazelle]

yes

[spark zinc]

no fucking way

like the structural integrity of a log is measured by ayx^2 or something

I skipped it didnt know what the fuck to do

had like 4 variables

[echo gazelle]

yea same

i did the first part

and then skipped b and c

but i answered everything else

I reckon 90+

[spark zinc]

I know I got 75

My average is like 90 but I always fuck up the exams

[echo gazelle]

wait and that last probability question

[spark zinc]

idk why

[echo gazelle]

was a bit strange

[spark zinc]

they're literally easier than the tests

which one

[echo gazelle]

at tc the tests are harder than the exams

in year 10 we did year 12 spec

[spark zinc]

bro what?

[echo gazelle]

there was year 12 content in our year 10 exam

[spark zinc]

what was the last probability one?

[spark zinc]

lol

[echo gazelle]

the prerequisite for spec was 60%

[spark zinc]

yeah

I think I got like 84

you mean the probability for calc or calc free?

[echo gazelle]

calc

[spark zinc]

remind me...

[echo gazelle]

idk I forgot

[spark zinc]

oh ok

[echo gazelle]

i did it last week

[vagrant ice]

high chance that at least one of you is Asian

[spark zinc]

I did it yesterday hahahah

[spark zinc]

nah

[echo gazelle]

no tc is full of white people

[spark zinc]

tc?

[echo gazelle]

we have like 10 aboriginal kids for our indigenous program

[spark zinc]

oh trinity college

[echo gazelle]

and that’s it

oh and refugees

[vagrant ice]

damn

[echo gazelle]

refugee scholarship

[spark zinc]

I know one guy that went there idk if he's still there

you're in yr 11 right?

[echo gazelle]

yes

[spark zinc]

obviously wtf am i saying

Who is the sportiest kid in your year?

at tc

[vagrant ice]

jesus all this competition only for most people to get into a uni course that only needs 90 ATAR

[echo gazelle]

what sport

[spark zinc]

like athletics

[spark zinc]

what course

[echo gazelle]

our median was like 89 lol

or 88

[vagrant ice]

bachelor of arts or bachelor of science, like it can't be more than 90

[spark zinc]

I swear every course is 80 or under except for med and law

for art?

drawing shit requires 90+?

I cant lie 90 is easy

lie 67% average

[vagrant ice]

yeah I mean 90 or below

[spark zinc]

My predicted is like just over 99

I have 3x bonus

hahahahah!

actually it might be like 99.5 because of scaling

[echo gazelle]

dawg what

[spark zinc]

thank god I didnt chose apps or some of those bs subjects

[echo gazelle]

mine is like 94

[spark zinc]

spec methods language

[echo gazelle]

u go to PSA?

[vagrant ice]

like psychology and physio usually require higher

maybe low 90s depending on demand

[spark zinc]

where is psa?

nah I dont mean like some rural school where they add like 5 atar

[echo gazelle]

like trinity hale ccgs aquinas

guildford

[spark zinc]

but I get 10% bonuses for methods, spec and second language studies

[vagrant ice]

but you can just choose a uni that is out of state and it will be back to like 80 or 85

[echo gazelle]

scotch

[spark zinc]

scotch has good chem exams

but no

[echo gazelle]

oh

[vagrant ice]

I think the highest I've seen for something not crazy like dentistry or advanced or double degree in the degree title

bachelor of mathematical sciences at ANU is 95

[spark zinc]

Or be smart and not go to uni because 90% of people don't know what they want to do with their lives

:)

are u all in yr 11?

[vagrant ice]

a lot of the humanities and arts are at 80 actually

[spark zinc]

still light work

[vagrant ice]

just figured you might not be interested in those

[vagrant ice]

lmao I switched from maths to sociology at uni

[spark zinc]

I'm doing mech engineering bachelor with bachelor of cs probably

something like that

probs UWA

[vagrant ice]

ah okay cool

[spark zinc]

u at curtin ?

[vagrant ice]

I don't know about Scotch in WA but in Melb your parents can sign you up onto the waiting list as soon as you're born

[echo gazelle]

i found our y10 exam omd

[vagrant ice]

no not a WA person

[echo gazelle]

i got like 73 on this 😭

[spark zinc]

oh ok

[vagrant ice]

but yes Aussie ofc

[spark zinc]

oooh yr 10?

I was gonna say this is piss easy

there was some shit like this in the methods remember?

[echo gazelle]

i think this was y12 spec

or methods

[spark zinc]

it was like if x^2x-2=13 what is x^2+1

I think that was exactly it

[vagrant ice]

that's actually normal for year 10 advanced

[spark zinc]

what did you get for that?

[spark zinc]

its not advanced

[echo gazelle]

mine had different numbers

[vagrant ice]

exactly

[spark zinc]

its achademic extesnion

[echo gazelle]

it was easy though

it was like if 5^(n-1) = 15

what is the value of 5^(n+1)

[spark zinc]

15 is div by 5 tho

so easy

mine was 13 and 2 and I was like wtf is this

[echo gazelle]

calc has u1 spec

[vagrant ice]

it doesn't even matter you have to multiply by 5^2

15 could be anything

[spark zinc]

yeah, had some matricies though I swear

wait

u mean methods my bad

[echo gazelle]

methods is really easy i feel like

because its just everything we learnt from y7-10

but i cant lie i kinda struggle with spec

[spark zinc]

not in yr 12

[echo gazelle]

everything is new

[spark zinc]

yr 12 is all calculus

ALL calculus

[echo gazelle]

oh

and probability

[vagrant ice]

https://pastpapers.papacambridge.com/viewer/caie/igcse-mathematics-0606-2024-may-june-0606-s24-qp-21-pdf

if you do maths enrichment in an international school you get this

[spark zinc]

I got 100 on my calculus test and I knew I was going to cook for year 12, comes very easily to me

[vagrant ice]

for year 10

[spark zinc]

@echo gazelle do you do chem or phy?

[echo gazelle]

yea

i do both

[spark zinc]

uh oh

[vagrant ice]

don't worry it's 52/80 for an A on this paper

to get A* they combine marks from both papers so you'd need like 83% on average

[spark zinc]

I'd probably only get like 60 rn

imagine yr 10

no knowledge of calculus

id get 30/80

[echo gazelle]

damn its 75% for one at my skl

[vagrant ice]

a lot of people struggle at that level but it's good prep for y11 and y12

[spark zinc]

have you done the exams?

[echo gazelle]

yea

[spark zinc]

both?

[echo gazelle]

yes

[spark zinc]

I have mine monday and wed

interesting

[echo gazelle]

have u done spec

[spark zinc]

easy?

[spark zinc]

yes

i have chem and physics to go

[vagrant ice]

yeah that means it must be curved pretty badly as well

cause I checked, for WACE 75% and higher is excellent

as in you can't get a higher markband

[safe radish]

@echo gazelle Has your question been resolved?

[safe radish]

Channel closed

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[edgy harbor]

<@&286206848099549185> I don’t know how to find the total length of the path
A dump truck starts to slide down a serpentine to the bottom of a paraboloid-shaped quarry. The helical pitch of the road is given - this is the ratio of the quarry radius (60 m) to the quarry height (40 m). The dump truck entered the road at a speed of 16 km/h and began to slow down with uniform acceleration at |a| = 2 km/h/s. Determine the final speed of the dump truck if, after driving uniformly at a speed of 1 km/h for 7.5 seconds, the dump truck reached the bottom of the quarry with uniform acceleration in the same time that it had already spent on the descent as a whole. Give your answer in km/h and round to the nearest tenth.

[hard crest]

use Pythagoras

[edgy harbor]

here paraboloid

what pythagorean here, where

[hard crest]

yeah I'm not exactly sure what they mean by that

[edgy harbor]

Can I throw a photo, how should the road chart look?

[hard crest]

photo would be really helpful. for a normal helix you're looking for the blue length

[edgy harbor]

there is one nuance

[edgy harbor]

Do you want me to multiply the number of such possible turns per resulting blue line?

[hard crest]

I'm trying to figure out what's going on with the quarry, i guess it's a bowl shape

[edgy harbor]

Paraboloid. just look on the internet

[hard crest]

something like this I think?

[edgy harbor]

yes, such a black line. and now I think how to make an equation to find it

[hard crest]

figure out what a is based on the initial radius of the quarry (60m)

[edgy harbor]

and height of the quarry 40

[hard crest]

yeah

oh that integral isn't right, we need path length

$L = \int_0^{40} \sqrt{(2\pi r)²+1²}\dd{y}$ i think??

[flat frigate]

hayley is stateside!!

[edgy harbor]

but we need to count the integral for the general case, suddenly the numbers are different

[edgy harbor]

.

[hard crest]

well yes, but figure out what the relationship is between r and y first

[edgy harbor]

Wolfram Alpha doesn't want to solve this

[edgy harbor]

it counted

thank you

[frozen marlin]

hayley did you purge

[hard crest]

isn't it just 60 = a √40 ?

[edgy harbor]

a - is this a helical stroke?

[hard crest]

does this equation make sense to you, given the diagram?

[edgy harbor]

Okay, let's go through this equation, but I don't know what we're going to do with A next

[hard crest]

well, we'll need it to calculate that integral

[edgy harbor]

I would change this.. from the parametric equations, we replace r with a coefficient u, which is already equal to a*sqrt(ф) where a is the helical pitch and ф is the angle, which was achieved through the parameterization of the circle.

[hard crest]

oh yeah that might be simpler

[edgy harbor]

it is necessary to make parametric equations of a paraboloid, I think

and there..to get it somehow through integration, I'm still thinking

dy/dx is a fraction, but 1/∞ isn't .. oh limits

[frozen marlin]

inf isn't a number (in most sets)

[edgy harbor]

tell me how to solve my problem further..

[frozen marlin]

frankly i have no idea what your problem even is

[edgy harbor]

the length of the helix on the paraboloid

[edgy harbor]

this is used in the task, I can't output the formula

@hard crest Didn't you decide further?

This channel will not close..

[frozen marlin]

the cmd is .close

[edgy harbor]

I know

[safe radish]

@edgy harbor Has your question been resolved?

[safe radish]

@edgy harbor Has your question been resolved?

[safe radish]

@edgy harbor Has your question been resolved?

[safe radish]

@edgy harbor Has your question been resolved?

[edgy harbor]

<@&286206848099549185>

[lean otter]

?

[edgy harbor]

I couldn't figure out an equation for the length of the paraboloid's spiral. how to do this?

[lean otter]

are u a girl

[edgy harbor]

not

[edgy harbor]

@lean otter

[lean otter]

yes

dm me ill help u

[edgy harbor]

Do you know about the length of the helical spiral of a paraloid?

[lean otter]

i said dm me or no help

[cedar owl]

💀

[lean otter]

Banana who goes to the navy to fail?

[cedar owl]

alr bro

[lean otter]

tell me

[edgy harbor]

.close

[safe radish]

Channel closed

Closed by @edgy harbor

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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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• Be polite and have a nice day!

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[modest holly]

help?

\frac{8}{x+5}-\frac{3}{x}=5

[marsh walrus]

$\frac{8}{x+5}-\frac{3}{x}=5$

[flat frigate]

jan Niku

[modest holly]

thanks

[marsh walrus]

where'd you get stuck

[modest holly]

i dont know what the common denominator i can multply is

[marsh walrus]

just multiply by every denominator

and they will all cancel

[modest holly]

what?

[marsh walrus]

$\frac{8}{x+5} - \frac 3x = \frac 5 1$

[flat frigate]

jan Niku

[marsh walrus]

multiply by $(x+5)x$

[flat frigate]

jan Niku

[marsh walrus]

(the 1 is there, its implied)

[modest holly]

that works for all?

[marsh walrus]

well the (x+5) factor will cancel in the first fraction

and the (x) factor will cancel in the second

you may imagine the 1 factor cancelling in the third, if it makes you happier

and now we have no fractions left

[modest holly]

now im getting confused

we still need to multply numerator

and denominator

if i have 8/x+5

and i multiply both sides by x(x+5)

i get 8x^2+40x

as numerator

and denominator is

[marsh walrus]

you do not

lets do it slowly

[modest holly]

alr

[marsh walrus]

will you give me a moment to type

[modest holly]

sure

[marsh walrus]

it might take me a few tries to get it right

[modest holly]

np

[marsh walrus]

$\qty( x(x+5) ) \qty( \frac{8}{x+5} - \frac 3x ) = \qty( x(x+5) ) 5$

[flat frigate]

jan Niku

[marsh walrus]

here's step 1

we multiplied both sides by x(x+5)

[modest holly]

ok i see

[marsh walrus]

do you see why I'm using x(x+5)?

[modest holly]

yea

[marsh walrus]

its nothing special, just used every denominator multiplied together

alright

now we can distribute

$\frac{ 8x(x+5) }{ x+5 } - \frac{3x(x+5) }{ x } = 5x(x+5)$

[flat frigate]

jan Niku

[modest holly]

ok so we only distrbute with numerators

oh the x+5 cancel

the x cancel

8x- ?

[marsh walrus]

yea, and this step, you can do more slowly, if you want

[modest holly]

3x+15?

[marsh walrus]

to say only with numerators makes me nervous

[modest holly]

8x-3x+15=5x^2+25?

[modest holly]

wdym?

[marsh walrus]

we aren't doing any special multiplication

we are just doing multiplication

[modest holly]

in 3x(x+5)/x

x cancel out

with multiplication?

[marsh walrus]

it does, but do you understand how the distribution happened here

[modest holly]

yea

[marsh walrus]

i mean, i just want to be sure

okay

if you feel you could duplicate that step then you are good

[marsh walrus]

it does, by design!

[modest holly]

yea

[marsh walrus]

since we mutiplied by the product of denominator

[modest holly]

so 3x+15 ?

[marsh walrus]

they will all cancel with part of what we multiplied by

[modest holly]

after the x cancels out

[marsh walrus]

yes

[modest holly]

so 8x-3x+15 ?

=5x^2+25?

[marsh walrus]

no

but youre close

[modest holly]

what am i doing wrong?

[marsh walrus]

you made a small sign error

[modest holly]

-15?

[marsh walrus]

it should be $\frac{8x}{1} - \frac{3(x+5)}{1} = 5x(x+5)$

[flat frigate]

jan Niku

[marsh walrus]

so $8x -(3x+15) = 5x(x+5)$

[flat frigate]

jan Niku

[modest holly]

oh so -15

ok so now i can solve the quadratic, right?

[marsh walrus]

yea, easy mistake to make, you hae to be careful

[modest holly]

@marsh walrus 2 more quick questions

pls

?

not problems

[marsh walrus]

whats up

[modest holly]

does this strategy work for all problems like this one?

where i multiply by both terms?

[marsh walrus]

the thing we multiplied by

[modest holly]

yea

[marsh walrus]

we got it by taking the product of every denominator

that will always work

[modest holly]

but we would use different terms because diff denominators?

ok so i can use that in other types of problems

[marsh walrus]

sometimes it will create more work for you

but, as a benefit, it will always work

[modest holly]

ok and one more thing

do you know where i can learn probability and statistics?

like a guide out there

[marsh walrus]

thats a tough question i would say you should take a class or find a tutor

[modest holly]

i only need it for alg 2

alg 2 probability and statistics

and i dont have a class or a tutor

[marsh walrus]

well thats a pretty broad topic

you could try an online course, something on youtube

there are free courses out there

[modest holly]

can i show u a preview of what im looking for?

[marsh walrus]

okay

[modest holly]

like a set of problems

alr

@marsh walrus for stats https://www.jmap.org/Worksheets/S.ID.A.4.NormalDistributions1.pdf for prob https://www.jmap.org/Worksheets/6.SP.C.7.ExperimentalProbability.pdf

[marsh walrus]

these problems are what would be covered in a normal introductory probability and stats course

im not sure what to say other than that

[modest holly]

should i try khan academy?

[marsh walrus]

the material is generic; if you find some intro prob stats course, this will be covered

lets see

[modest holly]

https://www.khanacademy.org/math/statistics-probability

unit 7 and unit 4?

[marsh walrus]

looks up up to unit 12

[marsh walrus]

i would say all but unit 13 onwards

[modest holly]

really?

[marsh walrus]

but again the material is generic

you can jump around

to other sources and such

[modest holly]

yea thanks tho jan

i got these finals i have to take

[lean otter]

who needs help with math?

[marsh walrus]

it normally takes people a while to like

[lean otter]

im intelligent and smart and really good at math

trust

[marsh walrus]

internalize the normal distribution stuff

[lean otter]

who needs help

[marsh walrus]

probability i think is more straightforward, for people

[modest holly]

i have some background so its not completely new

probability i find harder

[marsh walrus]

ah

[modest holly]

but its all good

[marsh walrus]

you're backwards

[modest holly]

haha

[marsh walrus]

well if you have funn prob problems i can help with

i like prob

stats is spooky to me

[modest holly]

thanks jan

[marsh walrus]

np

[modest holly]

if u want a preview tho

[marsh walrus]

good luck

[modest holly]

https://www.nysedregents.org/algebratwo/622/algtwo62022-exam.pdf

here

[lean otter]

WHO NEEDS HELP

I ASKED U QUESTIONS

WHO NEEDS HELP

I CAN ANSWER UR PROBLEMS

[marsh walrus]

see u can use like

lets see principle inclusion exclusion

[lean otter]

Do u need help sir?

why are u chatting here if u dont need help..

[modest holly]

huh?

[lean otter]

Do u need help sir

with math

[modest holly]

yea

[lean otter]

Then dm me i can help

[marsh walrus]

thats honestly almost all i see

@lean otter usually we dont do stuff in dms

people have their own channels

[modest holly]

i dont know wthat that is tho

[marsh walrus]

like this, you draw a diagram

[modest holly]

oh yea for those ones

[marsh walrus]

this one is just a definition

[modest holly]

yea its not that bad

pretty simple once u get the hang

[marsh walrus]

you just gotta practice

but fun exam

it looks harder than any algebra class i ever took

[modest holly]

what? no way

[marsh walrus]

my algebra classes were easy it was the later classes that were murder lol

im sure youll do great

[modest holly]

oh yea lol its all good tho

take it easy jan

[marsh walrus]

you too

[safe radish]

@modest holly Has your question been resolved?

[safe radish]

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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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• Be polite and have a nice day!

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[outer compass]

can i please get help on number 13🥹🥹

ive been trying to solve it for almost an hour now😂😢

[stoic saddle]

it feels like the area should depend on the radius of the circle...

and not be constant

cus i think i got a formula for the thing but its a wierd formula in r

Let AM = r and MB = x

and PB = t

we know r + x = 2

also BM.(2AM+BM) = PB^2

=> x.(x+2r) = t^2

therefore 4-r^2 = t^2 (as r+x = 2)

now, angle apb = 90

so area of triangle apb = 1/2 * r * t

and area of sector AMP = pi r^2 . theta/ 2pi

where theta is angle PAM in radians. but sin theta = t/2

thus the area of the shaded region is the triangle area - the sector area

$= \frac{1}{2}\cdot rt - \frac{1}{2}\cdot r^{2}\theta$

[flat frigate]

CherryMan

[stoic saddle]

$= \frac{1}{2}(\sqrt{4-r^{2}} - r^{2} \arcsin(\frac{\sqrt{4-r^2}}{2}))$

[flat frigate]

CherryMan

[stoic saddle]

by substituting in the vlaue of t as sqrt(4-r^2) and theta as arcsin(t/2)

and this doesnt look constant

@outer compass pls check if ive made a mistake in the process

$= \frac{1}{2}(r\sqrt{4-r^{2}} - r^{2} \arcsin(\frac{\sqrt{4-r^2}}{2}))$

[flat frigate]

CherryMan

[stoic saddle]

small correction

wake up @outer compass

[outer compass]

okay wait lemme check

thank you! ill close it now

.close

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[hasty narwhal]

I have a quick question about matrix calculation with unknowns, for example, what techniques are there to determine what values for a give what solutions?

[hasty narwhal]

I can use gauss-jordans and then see what values a might have to get trivial solutions for example

[light shoal]

well it could be anything

what is the question asking?

like a particular solution?

[hasty narwhal]

But the solution states that since 3-3a =/= 0, so a =/= 1

How do they find this "3-3a"

[light shoal]

!original

[safe radish]

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.

[hasty narwhal]

For all values of a, make a base for this matrix

so consider all cases

[light shoal]

do you mean a basis?

basis of what, the column space?

[frozen marlin]

yoooo bungo
been a while

[light shoal]

hey ren

[hasty narwhal]

the picture space, I am translating directly

[light shoal]

i guess it means image?

[hasty narwhal]

Probably

[light shoal]

ok, do you have any thoughts about how to do that?

[hasty narwhal]

Gauss Jordan, see what values of "a" would create leading rows of 1s

or 0s

and then the solution would be different cases of different values of "a"

[light shoal]

ok

and did you try that?

[hasty narwhal]

Yeah

I think it works, but I am more so curious what the solution is pointing towards

I could not find this when doing gauss jordan

I would expect to find it there, so either I did it wrong or they're using another technique that I am not aware of

[light shoal]

well do you see that if a=1 then you get 0 0 0 for your last row when you do the gauss jordan?

[hasty narwhal]

Yepp, I agree

I guess I am asking if there's a more direct way to find this

Because the solution seems to allude to this

[light shoal]

well, if you happen notice that when a=1 then row 1 + row 3 = row 2, that could be a shortcut

but in general it's not easy to notice such things

if you know about determinants, you could use that

[hasty narwhal]

I am aware about how to calculate them but not sure what they mean. I should read more about determinants, thanks

[hasty narwhal]

Yes that's fair

[light shoal]

short story is, if the determinant is 0 then the matrix has linearly dependent columns

so you could find what value of "a" makes the determinant 0

[hasty narwhal]

Ooh, this is where it's from then

Surely

I bet if I calculate the determinant it will end up as 3-3a

[light shoal]

possibly, give it a try!

[hasty narwhal]

awesome thanks!

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[outer river]

hello, I need help with the following: suppose $a_n$ is bounded and $\lim_{n\to\infty}(a_n-a_{n-1}-a_{n-2}) = 0$, does that imply $a_n \rightarrow 0$?

[flat frigate]

ω

[outer river]

I think it does

and since $a_n$ is bounded there exist $\limsup a_n$ and $\liminf a_n$.

[flat frigate]

ω

[outer river]

$|a_n-a_{n-1}-a_{n-2}| < \epsilon$

[flat frigate]

ω

[honest perch]

okay...

limsup and liminf are not linear btw

eg limsup(an + bn) != limsup(an) + limsup(bn)

note that if a_n has a limit then it must be 0

[outer river]

I know that

[outer river]

oh yes I see from the expression

makes sense

[honest perch]

so if u prove a_n has a limit ur done

[outer river]

yeah

[honest perch]

have u tried to think of a counterexample?

[outer river]

yeah, but I couldnt find one

[honest perch]

let's say u believe the statement is true

try to find a counterexample and see what goes wrong

and try to find it properly

the counterexample must be a bounded sequence which does not converge

what's the simplest such sequence?

[outer river]

$(-1)^n$ comes to mind first

[flat frigate]

ω

[honest perch]

yep

check if it works

oh wait

nvm i somehow thought that works but it doesn't

[outer river]

yeah it doesnt work

cause the limit does not exist

[honest perch]

it's becaue an - an-1 - an-2 doesn't converge to 0

[outer river]

oh sorry I worded it poorly thats what I meant

[honest perch]

maybe try a limsup liminf argument like u were suggesting

use limsup(an + bn) <= limsup(an) + limsup(bn)

gtg

[outer river]

alright

[honest perch]

to show that limsup an = liminf an maybe

[outer river]

yeah that would do it

[safe radish]

@outer river Has your question been resolved?

[outer river]

using Cauchy's criterion

I got that $|a_{n+1}-a_{n-2}| < \epsilon$

[flat frigate]

ω

[safe radish]

@outer river Has your question been resolved?

[potent jolt]

can't do just do it on contradiction? suppose $a_n \to k \neq 0$, then $\lim_{n \to \infty} a_n - a_{n-1} - a_{n-2} = \lim_{n \to \infty} a_n - \lim_{n \to \infty} a_{n-1} - \lim_{n \to \infty} a_{n-2} = k - k - k = -k \neq 0$

[flat frigate]

Sires

[potent jolt]

@outer river

oooh my bad could be that a_n doesn't converge at all

[outer river]

we already established that if its convergent then it is convergent to 0

[potent jolt]

yeah but I was trying to do contrapositive assuming it already converges

I believe the following is a counterexample:
$a_0 = 1, a_1 = 1, a_2 = 0, a_{n+2} = a_n - a_{n+1}$:
1, 1, 0, 1, -1, 2, -3, 5, -8, 13, -21, ....

[flat frigate]

Sires

[potent jolt]

probably easy to prove that from a_3 onwards its an alternated fibonacci sequence so it does not converge

[outer river]

is $a_n$ bounded?

[flat frigate]

ω

[potent jolt]

oh my bad

forgot about that

oh ok

but that can still be helpful

because that is the only way to construct a sequence that obeys the hypthosis given the first terms

nvm I'm halucinating

its just the only way to make one where the sum of 3 successive terms is always 0

[outer river]

yeah

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[jagged mica]

How do you find W, W^2, W^3 and W^4 for a.ii

[fathom adder]

A is the distance from center

Pi/4 is the orientation angle

So if a =

2

Since pi/4 represent the diagonal in the first quadrant of the unit circle

You can represent w

And for w^2, you take 2e^i*pi/4 and you put a square

This gives 4e^i*pi/2

And same you can draw it

[jagged mica]

Yes

I already got the stuff for W3 and W4

Its just im unsure what to do next for the drawing

Like im not sure how to get here

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[weary obsidian]

Does this proof relies on the existence of f^-1, that is, is it necessary that inverse function exists, or inverse image existing is sufficient

[peak estuary]

they used only the preimage

[weary obsidian]

ok thx for clarification

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[vague pewter]

ive tried taking logs on both sides, but im not sure how to proceed from there

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[sturdy stone]

Prove the inequality 10(x^2 + y^2) + 3(x + y) - 8xy <= 18 for all numbers x, y ∈ [0, 1]. Hello, please help me solve this inequality, I had many attempts to solve it, but unfortunately nothing worked out for me, I will be grateful for your help

[lament steeple]

try taking the function is equal to 18 and apply implicit differentiation to show dy/dx is 0?

not sure if this works but seems to be worth a try

[lament steeple]

ok i correct my sefl

[sturdy stone]

Thanks for the answer, but is it possible somehow without the derivative?

[lament steeple]

err

its like 10(x^2 + y^2) + 3(x + y) - 8xy=k is a relation

you can either implicit differentiate which yields 20x+3-8y=0 at a max point, take this back to the original function so its all x or y and then it'll be easy to find the max

or you can try to transform the relation into a function of y= something x with a k which could still probably yield the same result

[sturdy stone]

Why is there a maximum here?

[lament steeple]

because the question asked to prove a max?

[sturdy stone]

no

[lament steeple]

i dont know. derivatives are the first thing to come to when its max and min questions

[solar hazel]

probably

[sturdy stone]

you need to prove that the inequality is true when x,y belongs is from 0 to 1

[sturdy stone]

This is an inequality with 2 variables, why introduce a derivative here?

[lament steeple]

ok ok i get it

you can look at this function like 2 relations

10y^2+3y-8xy and 10x^2+3x-8xy

we wanna make both as large as possible so the whole thing is largest

we look at them separate, so in the first part only y is variable and in the second part only x is variable

you would yield -5y=3-8x and -5x=3-8y when both of these are in maximum

take this simultaneous equation and you find there is actually a solution x=1 y=1 that falls wthin the domain

take that back and the function yields 18 which is max

[solar hazel]

i am skeptical of the -8xy here

[safe radish]

@sturdy stone Has your question been resolved?

[sturdy stone]

I don’t understand what you want to do, why don’t you just show it???

[queen saddle]

you can view the left hand part as a parabola in x on interval x∈[0;1]. Its branches go upwards, so, the maximum value is reached only with x=0 or x=1. So you plug these into initial equation and find the maximum w.r.t. y. which is always <=18

for the same reason it is enough to check only y=0 and y=1.

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[full bobcat]

straight to the point: Find a cubic polynomial equation with roots -2, 2, and -4., i have absolutely no clue how to do this

[brave wolf]

How would the factored form of such a cubic equation look like?

[full bobcat]

p(x): ax^3 + bx^2 + cx + d

[kind forge]

that's not factored

[brave wolf]

Factored form is e.g. something like this:

(x-2)(x+3) = 0

[full bobcat]

oh factored

wait

(x+2)(x−2)(x+4)=0?

[brave wolf]

now you can either expand it if you want it in the standard form, or keep it like this

[full bobcat]

how can i expand it?

[brave wolf]

FOIL principle or anything like that

just multiply it out

[full bobcat]

oke lemme try

i only know how to do foil on 2 of these ()

lol XD

[brave wolf]

fortunately, multiplication is associative

(x+2)(x−2)(x+4) = ((x+2)(x−2)) * (x+4)

so you can start by doing this: (x+2)(x−2)

and then take the result and multiply it by (x+4)

[full bobcat]

i solved it thanks!

x^3+4x^2-4x-16

thanks @brave wolf

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[hollow zenith]

.reopen

Hello can you help me with my problem involving theorems on circle?

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Please don't occupy multiple help channels.

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@hollow zenith Has your question been resolved?

[hollow zenith]

yes but still confused just got the answer

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[indigo seal]

did I do anything wrong

bc the answer shoudl've been

a = 2

b = 3/4

c = 41/8

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[halcyon light]

Bit lost here

Thinking of trying to rearrange the cos^3 theta such that i can replace the 12 sin theta in y with x

its not working out all too well for me

[long copper]

well i guess you could probably just use the generic equation that $$\int y \dd x = \int y \frac{\dd x}{\dd \theta} \dd \theta$$

[flat frigate]

I can't believe you've done this

[halcyon light]

oh idt i've come onto that yet

oh ok so ig i try differentiate cos^3 theta

oh right the d theta cancels out

nice

thank you

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[runic rampart]

I posted first 😤

[lean otter]

<@&286206848099549185>

[runic rampart]

go into another channel ya goof

[cosmic grove]

yea what is wrong ?

[runic rampart]

So do you have sin and cos in your x equation because the roots are imaginary?

[cosmic grove]

well yea

[runic rampart]

would it be c1e^rt+c2e^rt otherwise

with your two r values

[cosmic grove]

in general it's c1 exp(r_1 t) + c2 exp(r_2 t)

but since the roots are complex, cos and sin appear

[runic rampart]

hmm

there's also a formula omega=sqrt(k/m)

I think that's phase shift?

does that go into the formula anywhere?

because I thought you would have cos(omega*t) and sin(omega*t) but that doesnt seem to be the case

[cosmic grove]

for a damped harmonic oscillator, the differential equation is :
x''(t) + c/m x'(t) + k/m x(t) = 0
with m the mass, c the damping coefficient, k the spring constant

k/m = omega²

omega would appear indeed in the solution if you solve it

[runic rampart]

wait, can you also write it as mx''(t)+cx'(t)+kx(t)=0?

because thats usually how i see it

[cosmic grove]

you can

i just divided both side by m

[runic rampart]

ah

yeah that clears it up, thx

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[frosty stag]

Im so lost, idk where to even start, like their are 2 variables in this, how am i suppossed solve this?

[lean otter]

do you know what a quadratic whose only solution is 4 would look like?

or whose only solution is 3, or 2, or 1

or 0

[frosty stag]

uhh, nope

[lean otter]

if a quadratic has only one solution, then b^2 - 4ac = 0

[lucid trout]

don't you agree that if a polynomial $x^2 + mx + n$ has roots $r_1$ and $r_2$ then it's gonna be equal to $(x - r_1)(x-r_2)$?

[flat frigate]

combinatorics hater

[frosty stag]

yes

[winged flare]

substitute x = 4

oh wait

sorry

[frosty stag]

yeah thats what i did lol

[winged flare]

since x is the only solution

both roots are equal

sum of roots is 4 + 4 = 8

product is 4 * 4 = 16

now the reason we are taking it as two roots instead of a single 4 is that in a quadratic equation with all real coefficients the roots can only occur in two multiples

which is either none or two

here there is one

so only possibility is it to have two roots

and from here maybe @lucid trout can continue?

[lucid trout]

nop

[winged flare]

oh

😔

well theres nothing more anyway

just that sum of zeroes is -b/a and product is c/a

for ax^2 + bx + c = 0

i thought you were about to derive it

[frosty stag]

ty

Tho

[winged flare]

so here can you find what b and a are?

[frosty stag]

yea

[winged flare]

by comparing

[frosty stag]

b= -8

[winged flare]

yea

a = 1

[frosty stag]

yep

[winged flare]

so b = m = -8

[frosty stag]

when is says, "only x value" it means the only root is 4?

[winged flare]

yes

[frosty stag]

ight ty

.close

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[frosty stag]

.reopen

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✅

[frosty stag]

I feel like i should know this but, sm how I'm not getting the correct answer.
My thought process:

I got 8 for how many combinations of shirts and pants
1S 2S 3S 4S
1P 1P 1P 1P

1S 2S 3S 4S
2P 2P 2P 2P
P = Pant
S = Shirt
Then i js used the formule
*formula
8 options(n)
2 to chose from(r)
8!/6!
= 54
becuase it is a permuation + no repeats
I'm thinking that either, I didnt understand the problem and it is actally Combinations + no repeats, or I messed up with the shirt pant combos

[lavish rune]

hello

[frosty stag]

hello

[winged flare]

im guessing a pair of pants means a single pant?

according to language convention

[frosty stag]

uhh

yea

so 2 total pants

[winged flare]

slight inconvenience

you may have choosed two combinations with 1P

or 2P

in one of those

so instead

you can just calculate for shirts

and then multiply by 2

as there is only two ways you can display the pants

[frosty stag]

wdym

like i counted a combo twice?

[safe radish]

@frosty stag Has your question been resolved?

[winged flare]

i dont think you can choose the same pant twice

[safe radish]

@frosty stag Has your question been resolved?

[safe radish]

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[maiden sentinel]

i feel so silly for asking this but i literally forgot how to divide by a decimal

[narrow ridge]

represent the decimal as a fraction

dividing by a fraction is the same as multiplying by its reciprocal

[maiden sentinel]

thank you

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[dense sphinx]

could someone explain?

this is the definition of lim in sign function

epsilon > 2

therefore, limit doesn't exist

okay but why tho

could someone explain the definition of limits in simple terms?

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[dense sphinx]

<@&286206848099549185>

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[dense sphinx]

=( Anyone?

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[plucky elk]

It negates the definition of limit existing

You have to first understand the eps delta definition of limit it looks like

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[lofty prism]

Having trouble with convergence and divergence. Trying to solve 75

[plucky python]

the directions are cut off a bit

[lofty prism]

Give me a sec

[plucky python]

have you found the general term for a_n?

[lofty prism]

No how would I go about finding it? Would it be sn?

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<@&286206848099549185>

[lofty prism]

.close

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[lofty prism]

.reopen

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✅

[lofty prism]

.close

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[versed wave]

what is the formula for angular velocity?

one sec

[buoyant shadow]

the height of the tower doesn't matter but the location does

[versed wave]

yeah thats the part im trying to think of

unless youre a flat earther or something

this is not the formula i really want though

we dont know the tangential speed of the eiffel tower

[devout oracle]

What info are you given?

[versed wave]

what you want is omega = 2pi*f

[devout oracle]

oh wack

[buoyant shadow]

0.8552 rad latitude

[versed wave]

frequency

whoever gave you this question expecting you to know the exact latitude of the tower is evil

but also this is 3 dimensional so i dont know what to do here

[buoyant shadow]

oh wait does it matter?

it's just one circle per day

,calc 360 / 24 /3600

[flat frigate]

Result:

0.0041666666666667

[buoyant shadow]

hm not right

no clue

i would answer c and be like really confident

there's no way the yearly motion changes the answer enough to matter

and latitude doesn't do anything

yes, it's just the same thing as rpm, except unit

how many circles per second

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[buoyant shadow]

" The angular velocity of the Earth is approximately 7.272×10−5 radians per second."

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✅

[buoyant shadow]

ok yeah, degrees vs rad

they did something wrong

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[regal halo]

Could someone tell me why my particular solution won’t work.

[regal halo]

I guess I'm supposed to do Axsin(x)+Asin(x)+Bxsin(x)+Bsin(x)

Not sure why though

[eternal carbon]

roughly speaking, Axsinx + Bxcosx generates some sinx and cosx terms that cannot fully be accounted for by the Axsinx + Bxcosx term itself

in general, when you have xsinx or xcosx on the rhs and you wish to use undetermined coefficients and you confirmed the homogeneous solution doesn't have any overlapping terms, your guess should look like (Ax + B)sinx + (Cx + d)cosx

[regal halo]

Strange, this is the first example I've come across that has required that. Even if my homogenous solution has c1cos(x)+c2sin(x) I've always been okay with a particular solution of Axsinx+Bxsinx

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[proper axle]

Wait I need help how is this wrong 😭

[versed wave]

the average velocity between 2 points is the slope of the points' secant

[vagrant ice]

the average velocity only depends on the start and the finish

[versed wave]

it doesnt matter however the graph behaves; as long as they have the same endpoints, the average velocity is still the same

[vagrant ice]

it's the slope between the endpoints

[proper axle]

Ohh I see

[mystic bear]

Tru

[proper axle]

I thought I could use any point from the graph

[mystic bear]

Also if distance is 0 till 10 secs and then suddenly distance reaches 100m then also avg is same bcz avg is total distance/total time taken

[proper axle]

oh I see

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[upbeat basin]

wHy doesn't the answer include f(a,b,c) in the tangent plane equation?

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[brisk turret]

Im not able to understand how to prove q2

[flat frigate]

Blockydude

[brisk turret]

I got question 1 by doing the above part

[covert yoke]

Is this a misprint? It seems surprising to me that the centroid about x and the centroid about y would be exactly the same no matter the shape of f.

It's been a long time since I've done multivariable calculus seriously though.

[brisk turret]

yk it actually might be, i've spent nearly an hour on a whiteboard and its constantly showing x^2 f(x) instead of xf(x)^2 and chat gpt agrees

[brisk turret]

its not even multi-variate calc im in calc 1

our prof just likes making our life harder

[covert yoke]

ChatGPT agreeing is not worth much in general

[brisk turret]

true

its not a typo turns out, my prof has a published list of typos and its not in it

now im even more confused

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[deft pivot]

What is the probability of four of a kind in poker ?

[deft pivot]

I saw a video where it is 13C2 x 2C1 x 4C1 and the "story" behind this is that first we are choosing out of 13 cards 2 , say A and K , then we are choosing which one will be 4 jn number and the we are choosing the colour

[fleet condor]

what kind of poker? do you mean what is the probability of a 4 of a kind in a single draw 5 card hand?