help-23

[safe radish]

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[wind cypress]

having some trouble understanding this question and how to approach it

[wind cypress]

So far I have that since $A$ is positive definite, then by Sylvester's theorem/law of inertia, then $A$ is similar to a matrix of positive eigenvalues, so then there exists some $C$ invertible such that $CAC^* = I$

[flat frigate]

TestTickler

[wind cypress]

but i dont really know how to show the next part and go forward

actually not similar but congruent*

[safe radish]

@wind cypress Has your question been resolved?

[safe radish]

@wind cypress Has your question been resolved?

[zinc hornet]

Just plug the right thing into f(X;A) = XAX* such that when you simplify, it reduces to I.

[wind cypress]

yeah i think i ended up getting it, since B is hermitian, then we get that $CBC^*$ is still Hermitian, so then by Spectral Theorem, hermitian matrix is diagonalizable by a unitary matrix $U$, so then we get that $U(CB(UC)^**$

[flat frigate]

TestTickler

[wind cypress]

i think i got it now

.close

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[vivid bison]

I tried to find MP using similar triangles but got nowhere, pls help

[vague phoenix]

What ratios did you get?

[vivid bison]

2:3

this problem was from my geometry test and it was poorly made so this could be unsolvable

[vague phoenix]

Have you tried to draw out the line ratios between triangle PNM and PBD?

[vivid bison]

no

i can say that

AP = 2x, PB = 3x and AM=MB=2y

[vague phoenix]

x and y are?

[vivid bison]

oh wait

forget about AM and MB

this is the ratio i can get

[vivid bison]

If AP = 2x, PB = 3x and AM=MB=y, then PM : PB = (3x-y) : (3x)

[split kayak]

Oh, okay, with this info the problem is "relatively" easy as far as i can tell.

I hope this makes some sense

We know that 2x * cosθ = 10 and that 3x * cosθ = 15; notably, we have that
x * cosθ = 5

[vivid bison]

Yeah

[split kayak]

Since the line AB = 5x

And M lies at the middle, its in 2.5x

Notably, the line i painted red has a length of 0.5x

[vivid bison]

So 0.5x × cos(theta) = 2.5

[split kayak]

and youre done

[vivid bison]

Am i?

[split kayak]

MN is what they asked you for, yea

[vivid bison]

Ohhh

[split kayak]

the triangle on top is just the rotated version of what its in the middle of the diagram.

Also, it makes sense that its 2.5, since we know that PDB has 6 times the length(s) of PNM

3x / 0.5x = 6
15/2.5 = 6

[vivid bison]

Yeah it does

[split kayak]

you were right in trying to approach this with the fact that these are all similar triangles.

[vivid bison]

Interesting

Well this was way easier

Thanks for help

[split kayak]

np

good luck

[vivid bison]

.close

[safe radish]

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[woeful shale]

How do I begin with 20th question

[nimble wyvern]

can you please send a clearer image

[upper lance]

Or write the question here

[nimble wyvern]

"Prove that no straight line can be drawn within a triangle which is greater than the greatest side"

i think

[safe radish]

@woeful shale Has your question been resolved?

[woeful shale]

Yes that is the question

I have a solution . But am nt able to understand that

[karmic merlin]

Show us so we can see where you're struggling to understand

[woeful shale]

1 min

[woeful shale]

Did u understand the solution?

I got stucked in diagram itself

[median sand]

,rcw

[woeful shale]

Understood?

[median sand]

I am reading.

[woeful shale]

OK

Forget abt this solution

Solve it ur own way and explain

[prisma karma]

u might want to clean ur camera lol, which question r u asking abt?

oh nvm u said 20th mb

[woeful shale]

20

[prisma karma]

do u want a way to do it or just to understand what the solution does

ig my method is the same as theirs ok

[woeful shale]

So ya I need explanation for diagram

[prisma karma]

Let P,Q be two points in the triangle and suppose for the sake of contradiction that PQ is somehow greater than the greatest side BC

Let us extend PQ to hit the sides of ABC and for a new line segment P'Q' then clearly, P'Q' will be longer than PQ right?

[woeful shale]

P q touches line of trainable or inside triangle?

[woeful shale]

Yes

[prisma karma]

PQ are inside the triangle (it could touch the edge of the triangle)

We are proving by contradiction

[woeful shale]

Then how will we extend it to touch the edge ?

[prisma karma]

basically suppose a line PQ exists greater than BC, we want to show someone is very wrong (so a contradiction occurs)

[prisma karma]

by extending I am including extending by 0 length

it is possible P'=P

good point, I should say P'Q' is at least as long as PQ

it is possible they are equal

[woeful shale]

Caus3 its already touching right ?

[woeful shale]

Wt

Can u share ur diagram 1st

[prisma karma]

PQ might be already touching: I am extending PQ such that they will be touching

[woeful shale]

K

[prisma karma]

i dont have a diagram yet just going off my head

[woeful shale]

So I will recreate other diagram for me . U say its right or wrong

[prisma karma]

i just drew it lol

[woeful shale]

Thank u

[prisma karma]

It is possible that P=P'

that's the case when P is on the edge of ABC

[woeful shale]

Ya may be

[woeful shale]

But most case it won't

[prisma karma]

but it doesn't matter, all I need is this fact:

If PQ is a counterexample (that PQ is longer than BC), then P'Q' will be too

[prisma karma]

yeah sure. We are doing a contradiciton tho we have to consider all cases

we suppose PQ exists (it can be in any shape or form) and trying to find a contradiciton

[woeful shale]

S

[prisma karma]

what does S mean

[woeful shale]

Yes

[prisma karma]

ok good

[woeful shale]

How to find contradiction?

[prisma karma]

so what the text you sent says something like "we might as well consider a counterexample where P, Q are touching the edge"

That is, we might as well use P' Q' as our counterexample

Now we delve into cases:

What if P' is on AB and Q' is on BC, and what if P' is on BC and Q' is on AB?

[woeful shale]

Didn't understand

[prisma karma]

which message?

[prisma karma]

^this?

[woeful shale]

This one

[prisma karma]

Basically if PQ is a counterexample then P'Q' is a counterexample. We might as well arrive at a counterexamle with P'Q' since it is easier to work with.

[woeful shale]

Wt does counter example mean ?

[prisma karma]

I mean a counterexample to what they are claiming

[prisma karma]

Suppose that there is a counterexample PQ (that is, PQ proves the claim wrong: PQ is longer than BC)

if this is confusing just ignore it. Suppose that PQ>AB with PQ inside ABC, we want to reach a contradiciton

Then we proved that length(P'Q')>length(BC)

we want to prove that length(P'Q')>length(BC) is impossible

Three cases:

1. the secment P'Q' lies on the sides AB, BC (doesnt matter if P' is on AB or Q', it is the same)

2. the segment P'Q' lies on the sides BC, CA

3. the segment P'Q' lies on the sides CA, AB

is this part clear?

[woeful shale]

Y

[woeful shale]

Yes

[prisma karma]

cuz that's our strategy remember? we are trying to reach a countradiction by assuming PQ is longer than BC

if we can prove that this assumption is always wrong

then it shows that PQ must be at most as long as BC

does this clear things up

[woeful shale]

Yessss
So if pq greater than bc then obv pq greater than ab ,ac as bc is largest side

[woeful shale]

Yes

[prisma karma]

so basically if we can prove it length(P'Q')<=length(BC) this shows that our original deduction of length(P'Q')>length(BC) is wrong, and thus our assumption of length(PQ)>length(BC) is wrong

oops i was using AB

[woeful shale]

Yess

Now I easily understood

[woeful shale]

Ya that made me be confused lil

[prisma karma]

mb

[prisma karma]

eitherway there are these three cases we consider

like there can only be three possible types of positions for the line segment P'Q'

[woeful shale]

Yes

[woeful shale]

Yes

[prisma karma]

Let me do the first case:

1. the segment P'Q' lies on the sides AB, BC (doesnt matter if P' is on AB or Q', it is the same)

i drew the graph for you

Now (I am using len to denote length now, cuz I am lazy)

len(P'Q')<= max(len(P'A), len(P'C))

And since len(P'A)<=len(BA) and len(P'C)<= max(len(AC),len(BC))=len(BC) (since BC is longest by assumption)

It follows that max(len(P'A),len(P'C))<=max(len(BA),len(BC))=len(BC)

Combining this inequality with the first inequality we obtain:

len(P'Q')<=len(BC)

lmk if there are any confusing bits

[woeful shale]

Y did 1st line of proof come?

And how can it even be write as bc is greatest side

So pq can't be less than anything write ?

[prisma karma]

there is a lemma about triangle that your solution used

r u not aware of that? i can write down the lemma

[woeful shale]

Lemme? Meaning

[prisma karma]

Lemma

a minor proven statement

[woeful shale]

Can u state that

[prisma karma]

Lemma:

Let ABC be any triangle and let D be any point on BC

Then len(AD)<=max(len(AB),len(AC))

[woeful shale]

In general terms?

[prisma karma]

wdym?

[woeful shale]

Like how do u state it

[prisma karma]

state what?

i just stated the lemma

u mean what you don't know lemma is?

[woeful shale]

Yes idk wt lemma is

[prisma karma]

it's just a noun we use to say "a statement that is used to prove the what we want"

[prisma karma]

this is a lemma

it is some statement that we can use in our question

[woeful shale]

Ok

Can u apply it in are question?

[prisma karma]

it is some random nouns we label out statements by: less important statement are called lemmas, more impotant ones are called proposition, and most impostant ones are theorems

[woeful shale]

Ok

[prisma karma]

but these are all rando math fact u don't need rn

[woeful shale]

Ok

[prisma karma]

^

this is the thing we need

have you seen the proof for this before?

[woeful shale]

Can u explain rhis lemma in our question

[woeful shale]

Noo

[prisma karma]

in the question it is a direct application. Using the triangle AP' C, and the point Q' on AC we get that len(P'Q')<= max(len(P'A), len(P'C))

[prisma karma]

oh lol, idk what ur lecturer is doing then, using these unproven facts

[woeful shale]

I jst started my 12th

[woeful shale]

Never heard this proof nor in my syllabus but still fine

[woeful shale]

Oh k

I get it

[prisma karma]

how it applies you get right?

Ill need to give you a proof of this fact, so that you don't have lingering doubts

[woeful shale]

OK till then I will be reading the proof u shared

[prisma karma]

yeah there are two times where I applied this lemma, be mindful

[woeful shale]

Oh I got it

Once I will solve myself again and let me see am I able to or nt
I will coneback to u in few mins

[prisma karma]

have you learned sin and cos stuff yet?

[prisma karma]

this is only one of the three cases

[woeful shale]

Yes

[prisma karma]

I invite you to try them yourself

[woeful shale]

Oh no

[prisma karma]

the other cases are quite similar

using the lemma

[prisma karma]

jave u learned that in a triangle ABC sin(angle A)/len(BC)= sin(angle B)/len(AC)?

[woeful shale]

Ya I will try and get back to u

[woeful shale]

Noooo

Isn't here only 1 case ?

[prisma karma]

have you learned pythagorean theorem?

[prisma karma]

it said in the last line "proceeding similarly for the other cases"

but u should try them yourself, helps to get somw practice in

[prisma karma]

I can prove the lemma with pythagorean theorem too

[woeful shale]

Yes

[woeful shale]

Oh k

But will that require those trigo things ?

[prisma karma]

nah

but it requires, cases...

draw a perpendicular from A to BC that meet the line BC at H

two things will happen:

1. H is outside of the line segment BC

2. H is inside

[prisma karma]

in the fist case, we may as well assuming H is on the right of C. (if it were on the left of B then the proof is exactly the same, just with B and C swapped)

Now AB^2=BH^2+AH^2>=DH^2+AH^2=DA^2 Thus since length are positive, AB>=DA and so AD<= max(AB,AC)

[prisma karma]

in the second case, we might as well assume D is on BH (if D is on CH th proof is the same just with B,C swapped)

Now AB^2=BH^2+AH^2>=DH^2+AH^2=DA^2 Thus since length are positive, AB>=DA and so AD<= max(AB,AC)

Thus we have shown that AD<=max(AB,AC)

i stopped writing len(AB) and just wrote AB (mainly cuz i wanna go to bed soon)

Anyways this is the proof to the Lemma, and you can use it to solve the original question

[safe radish]

@woeful shale Has your question been resolved?

[woeful shale]

Yes

[woeful shale]

Where did our pq go ?

[safe radish]

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[barren fjord]

Can someone help me with the last part I’ve done the first two bits

[honest perch]

which question

q4 or q5

[barren fjord]

4 srry

[stone shale]

Bro

Trig

[barren fjord]

?

What

[stone shale]

Which class is it?

*grade

[barren fjord]

Er

Im not American

[stone shale]

I'm not too

[stone shale]

Wdym Er

[barren fjord]

12

[stone shale]

Oh alr

[barren fjord]

Can you help?

[honest perch]

what have u tried

for the last part

[barren fjord]

I thought I could use roots of polynomials

[stone shale]

Sry bro, but idk that lvl trig

[dense lark]

engineering

[stone shale]

Basics

[dense lark]

maybe i can

[stone shale]

💀

[dense lark]

aur bhyi prithvi ke haal chaal

[stone shale]

Badhiya mittar

[barren fjord]

Can you talk elsewhere pls

[dense lark]

last part of which question

[stone shale]

Ale

[honest perch]

so first i think u wanna substitute theta = p/12, 5pi/12, etc into part ii

[barren fjord]

Question 4

[dense lark]

alr where r u getting stuck

[honest perch]

what

[dense lark]

can you send till where you have done?

[barren fjord]

I did the first two parts

I and ii

[barren fjord]

Won’t that just be 0

[dense lark]

so you're not at all able to do the third part yea?

[stone shale]

Yes

[honest perch]

the lhs would be 0 yes

[dense lark]

alr solving it rn for you, I'll send the pic once im done

[barren fjord]

Won’t both sides be 0

[stone shale]

How bro?

[honest perch]

rhs = lhs so yes ofc

but then u got a polynomial u can solve for cos((2n+1)pi/12)

[dense lark]

hey @barren fjord

[barren fjord]

Hm

[stone shale]

How (iii) justifies our answer?

[loud wren]

you're not supposed to solve it for them, please don't send the picture

[honest perch]

!nosols

[safe radish]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[dense lark]

did you take in account in which quadrant the cos is lying, so acc to that their values turn positive nd negative?

[dense lark]

ah alr

[barren fjord]

I still don’t get it

[dense lark]

alr alr mb

[dense lark]

have you taken the quadrants in account?

[barren fjord]

Yh

[dense lark]

alr

so 1/12 pi is 15 degree

[stone shale]

Oo

[dense lark]

nd 5/12 pi is like 5*15

which is 75 degree

then 7/12 pi is 105

[honest perch]

why are you converting them to degrees

[dense lark]

and finally 11/12 pi is 55 degree

[dense lark]

let me cook chat

[honest perch]

?

[dense lark]

shush bbg

[honest perch]

watch your mouth

xD

[stone shale]

Cos 75 • cos 105 • cos 55?

[dense lark]

which are easy

[loud wren]

wut

its 165 is it not

[dense lark]

uh 1/12 pi is 180/12 which is 15

right?

[honest perch]

looks at ii. the rhs is a polynomial in cos(theta) and the lhs will be 0

[loud wren]

and 11/12pi=(pi-1/12pi)

[dense lark]

same logic, but you multiply 11 in the numerator

[loud wren]

15*11 is not 55

[dense lark]

shoot

mb

[barren fjord]

So I set 2c^2-1= 0 and the quartic to 0 and solve ?

[loud wren]

^^ this is also something i'd suggest to reduce the amount of things you'd have to calculate

cos(pi-x)=-cos(x)

[honest perch]

yeah so u can find all roots of the polynomial

explicitly

[loud wren]

hmm. what solutions would those yield?

[barren fjord]

Ohhhhh

[loud wren]

hint: ||chose the correct polynomial on the RHS to use when solving||

[barren fjord]

Pi/4 3pi/4 between 0 and pi for the quadratic

I didn’t get exact values for the quartic

[loud wren]

what's one way you could solve the quartic?

[honest perch]

sure you could do this but the question wants you to use the previous parts

[dense lark]

ah

ah ah

[barren fjord]

Let x=y^2

[loud wren]

well, do those look like pi/12, 5pi/12, etc.?

[barren fjord]

Solve as a quadratic

(Root6+root2)/4

As one of my solutions

[loud wren]

k

which theta value did it correspond to?

[barren fjord]

Ah pi/12

[dense lark]

so 1/12 pi is 15 degree--(1)
nd 5/12 pi is like 75 --(2)
then 7/12 pi is 105--(3)
and finally 11/12 pi is 165 degree--(4)

now we know some certain values of cos
now we split the degree values INTO the degree values of cos we are already familiar with by basic arithimatic addition and subtraction
for (1)
we do cos15= cos(60-45) ----(a)
for (2)
we do cos 75 = cos(60+15) [which we got from (a)]
for (3)
we do cos105=cos(60+45)
and lastly for (4)
we do cos165=cos(180-15) [which we got from (a)]

Lastly while putting the values we will take quadrants into account (which I leave for you to do) and jst multiple them alll

finally

[barren fjord]

5pi/12 7pi/12 11pi/12 there we go

So it is just roots of polynomials then

Multiplying all the roots gets u 1/16

[dense lark]

ah is it cz it says 'Hence find the value'

so you gotta take previous two subparts into account?

[barren fjord]

Yh

[dense lark]

hmmmmmmmmmmmmmmmm

so my method is a waste here 🥀

[small epoch]

what you did here was another way which also leads to the correct answer

[barren fjord]

Do u mean pi/2 or pi/12

Where’s that from

[dense lark]

90 degree

[small epoch]

oh wait mb

[small epoch]

yeah, it should be θ = x * π/12 (where x is odd here)

[barren fjord]

The issue is

U don’t get exact values

[small epoch]

you actually do, because that quartic is disguised as a quadratic

[barren fjord]

Thanks

.close

[safe radish]

Channel closed

Closed by @barren fjord

Use .reopen if this was a mistake.

[safe radish]

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• Show your work, and if possible, explain where you are stuck.
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• Be polite and have a nice day!

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[shadow moat]

Basically I want to calculate a daily buy in price that I have a 75% chance of actually hitting per month and then a buy out price which is like 90% chance of exit in 3 day timespan, what maths do I need to learn?

[outer pollen]

probability and statistics, it sounds like.

[shadow moat]

Are there specific types or equations I have essentially no maths skills

[main mural]

there's an entire field called financial mathematics

[shadow moat]

What's the difference between probability and statistics

[honest perch]

just buy and hold bro

[main mural]

just buy low sell high bro

[shadow moat]

That's what I'm trying to do

[outer pollen]

in very simplified terms, statistics attempt to derive information and build a model from existing data. probability is more or less the opposite - given a model, predict data.

[honest perch]

don't even sell

[main mural]

TO THE MOOOOOOOOOONNN 🚀

[shadow moat]

They're short term treasury bonds just an alternative to holding cash

[shadow moat]

So my math question tries to do both?

How long does it take to learn this

[outer pollen]

that's a very subjective question.

[errant ravine]

-# also not to discourage you but there are many many who try to do this and make some sort of pos winning trading models.. and its alot more complicated to realibly turn a profit since if u had such a model it would be worth billions ;3

[outer pollen]

I'll let other helpers with more experience answer that one.

[shadow moat]

I'm not attempting to compete in quant trading I know I will lose. I just want an extra fraction of a % on holding bonds vs cash and trade slightly higher gains for reduced liquidity

I basically plan to just hold a portion of bonds to reduce volatility in a mostly etf portfolio

And then go 100% equities during the next financial crash

So basically looking to buy regularly sell once

[honest perch]

the probabilities here depends on ur model

like, there are no inherent probabilities in the financial market

[shadow moat]

And it's a good opportunity to learn a bit by trying to optimise entry and exit

Any good resources or books on this?

I'll just eyeball my entries for now

[main mural]

-# >pre-uni

[shadow moat]

Any good resources for learning the maths?

[split kayak]

just in advance, youre about to get feet first into one of the most involved topics, but id say you can kinda start by picking up any basic quant trade book, youll probably see a few concepts that you dont understand, and from there you can search up on them.

Notice, since there are a lot of companies worth billions that rely on it, this is one of the fields that has the most activity

[errant ravine]

-# just because someone is preuni doesnt mean they shouldnt try to or learn to do something especially if they are willing to go through the effort of learning to do it! (but i alaso do agree its gonna pretty hard for them)

[shadow moat]

Nice

[split kayak]

Youll probably want some knowledge on at least calculus and basic notions of real analysis.

Apart from probability and statistics.

Which all eventually combine into stochastic analysis

[shadow moat]

I taught myself a language and do dense philosophy books for fun so I have confidence I can learn maths too

[errant ravine]

-# also just as a small tip when making ur models and stuff it might be helpfull to use past data that you know the outcomes of (but obv dont plug it into ur model) before testing it on live data

[errant ravine]

well if u dont have knowledge of calculus and basics in analysis it will take some time- and dedication but i wish u best of luck!

[main mural]

(fyi this is called backtesting if you want to read up on that)

[river pollen]

arguably it's not as simple as this, but it would be a good start yes

[errant ravine]

-# i actually did not know that thanks i wil look it up!

[split kayak]

quant is built upon on the combination of pattern recognition algorithms and stochastic analysis as far as i know, which are two huge topics.

[shadow moat]

Is it simpler if I'm not trying to calculate a future price but instead just the sell price that I have a 1/3 daily chance of hitting today

[river pollen]

overall, the shorter the time period, the less factors come into play, and thus the more you can be sure of your model's prediction

so simpler... I guess so?

[split kayak]

you run into the slight problem of never hitting the sell out price and always falling onto your stop loss.

[shadow moat]

I won't have a stop loss

[split kayak]

I cant word how bad of an idea that is

[shadow moat]

It's a fund composed of short duration us treasury bills less than 3 months and no leverage

Extremely low voloatility

[split kayak]

If i might add, outside of the whole math thing, if you want to pick up inversion, go for safe long-term markets

[errant ravine]

atp it might be better to gamble...

[shadow moat]

I mostly want to buy SP500 and similar products but concerned we're close to the end of a bubble so I want to hold bonds for 40-60% of my portfolio though I am looking into a little value investing for fun long term

The only way I'd get ruined holding short duration US treasury bonds is if Russia turns America into glass

For added context this is the asset

[safe radish]

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[safe radish]

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[wise hill]

why did my professor say that the smaller size N is the smaller the number of errors? What counts as a small sample size N?

[safe radish]

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[lean otter]

Ive been trying to Prove D lies on the circle but I am stuck for over a month.

[prisma karma]

stuck for a month is crazy

[pallid magnet]

what do u have so far

[lean otter]

MN is diameter

And M,N,P,H_a concyclic

[lean otter]

D is bothering me so much

[prisma karma]

give me some time to actually do the q lol ill assist in a moment

like 10 min

[lean otter]

Sure

[ionic blaze]

-# thought I'd seen this before, turns out it was overly simplified

[pallid magnet]

haven't solved it yet but for some reason i feel like there's a quick way to do it with inversion, just because there's so many circles

directly trying to prove it is umm... kinda hard

[prisma karma]

ok i got it

[prisma karma]

you did it by proving that $\angle NPM=90^\circ=\angle NH_a M$ right?

[flat frigate]

qwertytrewq

[lean otter]

Yes

[prisma karma]

ok so we only need to prove that $\angle NDM=90^\circ$

[flat frigate]

qwertytrewq

[prisma karma]

there is a pretty weird hidden similar trangles: $\triangle NED\sim \triangle DFM$

[flat frigate]

qwertytrewq

[prisma karma]

Can you show that they are indeed similar and try to go from there using the angle property you get

i think this might be the missing piece for u

<@&268886789983436800>

lmk if u r/where u r stuck

[verbal cobalt]

Anybody here familiar with Gina Wilson all things algebra ? If so please dm

[prisma karma]

i think you should start a separate help channel (and only one)

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

I GOT IT, $DF=CE=NE$ and $FM=FB=ED$, so $NED$ is congruent to $DFM$. $\angle{FDM}+\angle{EDN}=\angle{FDM}+\angle{FMD}=180^\circ-\angle{DFM}=180^\circ-90^\circ-\angle{DFB}=90^\circ-\angleCAB=90^\circ-\angle{EDF}$. Finally

[flat frigate]

Ihhi123
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lean otter]

Thx

[safe radish]

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[prisma karma]

fire

[safe radish]

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[carmine steeple]

How will I do this question? This is like impossible

[outer pollen]

hint: if you let the original width of the cardboard be, say, x, then you can express everything in this problem in terms of x and work out the value of x this way.

[carmine steeple]

I did that

[outer pollen]

then please show your work.

[carmine steeple]

I made the lenght (2x-8) and (x-8)

ik the height of the box is going to be 4

but idk what i need to do next

[outer pollen]

you know the volume of the box.

[carmine steeple]

oh yeah

wai

so i can put it as

2400 = (2x-8)(x-8)(4)

right?

then i solve for x

[outer pollen]

correct.

[carmine steeple]

how would I solve this

do i use the quad formula

[outer pollen]

you can.

[carmine steeple]

ur intimidating me

but alr

give me a second

[outer pollen]

if so, I'll step back, and I apologize for doing so. all the best.

[carmine steeple]

lol its all g the periods r just scaring me

seems like ur pissed at me for being dumb lol

[outer pollen]

I don't think I would be helping if I did, but either way I'll step back to avoid further misunderstandings.

[carmine steeple]

? ok

[prisma karma]

u got ur work written down somewhere? could u send a pic

so I can continue where u were left off

[carmine steeple]

uhh i coul copy and paste it for u

I made the lenght (2x-8) and (x-8)
ik the height of the box is going to be 4
Volume = 2400cm^2
2400 = (2x-8)(x-8)(4)

thats all

[prisma karma]

and just to confiem (x-8) is the width right?

[carmine steeple]

yes

[prisma karma]

hell ye

so we just gotta solve for x

[carmine steeple]

yes

[prisma karma]

well there are a couple things u can do first

u can divide both sides by some constant

so the numbers are smaller and easier to work with

like both sides are multiples of 4 arent it

[carmine steeple]

i was gonna devide both sides by like 4

oh yeah

read my mind

[prisma karma]

well it could be 8

cuz 2x-8 has a 2

[candid wraith]

you can take out the 2 from (2x-8)

[carmine steeple]

wait hwat how

mb guys math is my worst subject

[candid wraith]

2x-8

is basically

2(x-4)

cus that expands to 2x-8

[carmine steeple]

right

[prisma karma]

(2x-8)(x-8)4=(x-4)(x-8)8

[candid wraith]

so 2(x-4)(x-8)(4)=8(300)

[carmine steeple]

right

[carmine steeple]

huh

wait im confused now

[candid wraith]

sooo

lets replace (2x-8) with 2(x-4)

[carmine steeple]

yes

i understand urs

[candid wraith]

2(4)(x-4)(x-8)

[carmine steeple]

i dont realllt get qwerty

[candid wraith]

2*4=8

[carmine steeple]

right

[candid wraith]

so we can replace 2*4 with 8

[carmine steeple]

ik taht

[candid wraith]

8(x-4)(x-8)

[prisma karma]

oh i just multiplied 2 with 4 and put it to the right

so i got (x-4)(x-8)8

[carmine steeple]

right

[carmine steeple]

yes

[prisma karma]

so what is the result after factoring out the 8 from the original equation u had?

[carmine steeple]

2400 = 8(x-4)(x-8) (4)

?

[prisma karma]

there seems to be a 8 and a 4 on the right hand side

is that a typo?

[carmine steeple]

uh

no

im confused

bec

the origional thing

2400 = (2x-8)(x-8)(4)

is this no?

[prisma karma]

yeah

[carmine steeple]

and htne

when we get the 8

[prisma karma]

we then factored out a 2

and we multiplied the 2 with the 4 we had to get the 8

So (2x-8)(x-8)(4)=(2)(x-4)(x-8)(4)=(8)(x-4)(x-8)

i think you might've forgotten to remove the 4 after combining it with the 2 to get 8

[carmine steeple]

OHHHHHHHHHHHHH

no wonder why

ok

so after that

2400 = 8(x-4)(x-8)

its just this

[prisma karma]

yeah u can divide both sides by 8

[carmine steeple]

but im gonna get a fraction

[prisma karma]

cuz it doesn't change nothin but makin our equation simpler

[carmine steeple]

no?

[prisma karma]

no? 2400/8=300

[carmine steeple]

no like

4/8

[prisma karma]

u mightve misremembered lol

[prisma karma]

its an 8 here

8/8=1

[carmine steeple]

no like the 4 inside (x-4)

[prisma karma]

$\frac{8(x-4)(x-8)}{8}=(x-4)(x-8)$ right?

[flat frigate]

qwertytrewq

[prisma karma]

we are not dividing every term by 8

we are dividing the right side as a whole by 8

for example if i write 64=8*4*2

[carmine steeple]

so would i just cancel out

the 8 infront of the thing

when i divide

[candid wraith]

ye

[prisma karma]

and i say "divide the equation by 8"

if you divide every tem by 8 you'd get 8=1*1/2*1/4

which would be a pretty wrong equation lol

[carmine steeple]

wait so

[prisma karma]

so you should really divide both side as a whole by 8

[carmine steeple]

300= (x-4)(x-8)

would be the thing

[prisma karma]

I dont quite understand

oh nv

m

typo i guess

[carmine steeple]

?

[prisma karma]

hell ye

[carmine steeple]

its right

?

[prisma karma]

you wrote 300/(x-4)(x-8) and i got confused lol

[carmine steeple]

nonono

[prisma karma]

but ig its just a typo

anyways this is right

[carmine steeple]

OMDS YES BRO

ok

now what do i do

[prisma karma]

300=(x-4)(x-8) is right

[carmine steeple]

yes

now do i like expand it

[prisma karma]

quadratic formula bruv its kinda annoying to do

but u can use a calculator right?

then u should be bing chilling

[carmine steeple]

yeah

okok

[carmine steeple]

oh wait what

i can do that?

300=(x-4)(x-8)

[prisma karma]

just expand, put it into the form of ax^2+bx+c=0

[carmine steeple]

how would i use quad

OH

[candid wraith]

ye

[carmine steeple]

okok thanks

that makes sense

[prisma karma]

hell ye just dont make any computational mistake lol

[carmine steeple]

okokok

thanks guys

.close

[safe radish]

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[prisma karma]

glad to be of help

[safe radish]

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[elder lagoon]

idk how to prove 3..
cuz if i sub the point A coordinates in my P2 eqn it doesnt satisfy.

(P2 : 4x + y -2z = 5)

[rancid lion]

How does it not satisfy?

[elder lagoon]

sub A point.

4(3) + 2 - 2(7) = 5.
12 + 2 -14 = 5
0 != 5.

[rancid lion]

there exists only one plane that is the perpendicular bisector plane of segment BC

[elder lagoon]

ok..

[rancid lion]

A is not related to plane P2

[elder lagoon]

if P2 intersects AB then A must satisfy the P2 equation.

right?

[rancid lion]

no

[elder lagoon]

A & B.

[elder lagoon]

what?

[rancid lion]

If A satisfies the equation for P2 then A has to lie on P2

[elder lagoon]

yea, by intersecting it lies on p2 right?

[rancid lion]

no

[elder lagoon]

huh?

[elder lagoon]

are you telling me to find cross prodct and see if the normal of the plane and it match?

[rancid lion]

lemme ask you

how did you find the answer for (2)

[elder lagoon]

i didnt find the ans

for 2 or 3.

i went from Back

[rancid lion]

???

[elder lagoon]

i checked from last option man

[rancid lion]

chịu

[elder lagoon]

...?

[rancid lion]

what statements have you answered

@elder lagoon

[elder lagoon]

1,4

[rancid lion]

ok so how do you know a line segment intersects a plane or not

[elder lagoon]

dk

wait dont tell me it shows sign change when u sub the points

in

the plane eqn

[rancid lion]

ye you got the idea

[elder lagoon]

bruh.

i didnt know tht welp, thanks!

i saw it from lens and thought it was wrong mb

.close

[safe radish]

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[patent citrus]

i will tell what im doing

can anyone correct me

[outer pollen]

for future helpers: please show your work then.

[patent citrus]

well i havent really done anything as of now

i just had an idea

like

umm

its wrong

very wrong but

ill show give me a min

[buoyant shadow]

i don;t see any way but a whole lot of cases

[outer pollen]

I'll let frowny take over then. all the best OP.

[buoyant shadow]

well 2 cases, we can take 3 3 2 2 or 4 2 2 2 balls

[patent citrus]

cant we merge them all in 1 like this

[buoyant shadow]

hm

[patent citrus]

tho the answer is only in 4 digits

thats why i said

its very wrong

[buoyant shadow]

uh

it should overcount not undercount

strange

[patent citrus]

yeah

[buoyant shadow]

anyway there's 6 ways to take 2 balls from a box, like you found

how many ways to take 3

[patent citrus]

um

[buoyant shadow]

3×2 if we take 2 red, and 3 if we take 2 blue, so total 9

,calc (9)(9)(6)(6)(6)

[flat frigate]

Result:

17496

[patent citrus]

why '6' 3 times

[buoyant shadow]

another times 6 because we choose 2 boxes where we take 3 balls

4c2

[patent citrus]

ah

yeah

[buoyant shadow]

how does this count so few lmao

[patent citrus]

so we had to make cases and solve

...

[patent citrus]

no clue

lol

but it looks pretty right to me

idk why

since all balls are distinct

[buoyant shadow]

oh because it's not ×4 it's 4th power

6×6×6×6 obv

[patent citrus]

oh

shit

[buoyant shadow]

now how many ways to take 4 balls

2+3=5

we leave any one ball, 5 ways

,calc 17496 + (5)(6)(6)(6)(4)

[flat frigate]

Result:

21816

[patent citrus]

yeah

thats the final ans

ill go through it once more then close

if that works

[patent citrus]

ight tysm man

.close

[safe radish]

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[spiral saddle]

I proved that if a in N satisfies a = 3 (mod 4) then there exists a prime p with p = 3 (mod 4)

such that p | a

[dusk halo]

so you need help with the infinitely many part?

[spiral saddle]

the Hint has two parts we need to prove, I only proved the first part of the hint

[dusk halo]

ok so you can prove there are infinite primes that are congruent to 3(mod4) by contradiction

[loud wren]

how do you think you should approach the 2nd part

[spiral saddle]

unsure

i can show what i did for first part of the hint

[dusk halo]

try assuming that there are finite primes, from $p_1$ to $p_n$, and see if you can end up contradicting yourself

[flat frigate]

AMysteriousStranger

[loud wren]

what can you clearly notice when you take a mod 4

[spiral saddle]

a = 3 (mod 4)

[loud wren]

k so what can you imply from that

specifically regarding the primes dividing a

[spiral saddle]

there exists a prime p = 3 (mod 4)

[loud wren]

and therefore by our assumption?

[spiral saddle]

is false

[loud wren]

how

[spiral saddle]

is false that a is not divisible by any prime congruent to 3 (mod 4)

right?

[loud wren]

wut

it only has to be one of the prime divisors of a that is congruent 3 mod 4 for a to be congruent 3 mod 4

[loud wren]

ok why

[spiral saddle]

because p is congruent 3 mod 4

[loud wren]

hmm

[loud wren]

well that's not really where you're supposed to prove the contradiction

you're supposed to prove the "finite primes" part is false

[loud wren]

what you're meant to do here is to show that the p here is not actually a prime

hmm

[spiral saddle]

wut

[loud wren]

do you want to try to prove it directly instead, i think i took a longer route

[spiral saddle]

yes

[loud wren]

k

convert the statement "then a... congruent to 3 mod 4" into a mathematical statement

[spiral saddle]

wut

[loud wren]

into like a mod statement regarding a and the p_i

[spiral saddle]

if there were only finely many primes congruent to 3(mod 4), say p1,p2,...,pn then a = -1 + 4prod(1,n) pi would be greater than 1 and not divisible by any prime congruent to 3(mod 4)

[loud wren]

k

what are those "primes congruent to 3 mod 4" in the last part of your statement

[spiral saddle]

p | a

[loud wren]

k

[spiral saddle]

pi | a

[loud wren]

so what do we want to show

[spiral saddle]

there are no finely many primes cong to 3 m 4

[loud wren]

that's the whole problem statement, but what about for this little snippet

[loud wren]

you got close here but you have to show something slightly different

[spiral saddle]

-1 mod a

[loud wren]

closer

but uh we can't rly take anything mod a right

what can you take instead

[spiral saddle]

who tf knows at this point

[loud wren]

uhh

you did get rly close i can say that

but if $a=-1+4\prod_{i=1}^np_i$ then $a\equiv -1\mod (?)$

[flat frigate]

;(

[loud wren]

what's the ? here

[spiral saddle]

can't be 4

[loud wren]

so what could it be

[spiral saddle]

p

[loud wren]

ye

$a\equiv -1\mod p_i$

[flat frigate]

;(

[loud wren]

we have this directly right

[loud wren]

so is this statement verified or not

[spiral saddle]

there is no index for i

[loud wren]

hmm

why

[spiral saddle]

why

[loud wren]

bc a=-1+4p1p2p3...pn

so a mod p1 = (-1+4p1p2p3...pn) mod p1 = -1+(4p2p3...pn)p1 mod p1 = -1

etc.

[spiral saddle]

care to elaborate

[loud wren]

its the statement we were trying to prove

i gtg do smth sry

[safe radish]

@spiral saddle Has your question been resolved?

[safe radish]

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[desert pasture]

Determine if $y^5-x^2y+x$ in $F(x)[y]$ where $F$ is a field is irreducible

[flat frigate]

what a wonderful world(wai)

[desert pasture]

My first thought was einestien

My only thought wwould be to consider some prime ideal, and proceed

perhaps the converse?

if $f(x)$ is reducible in $R[x]$ then $a_0$ is an element of $P^2$ or $a_{n-1},\dots, a_0$ are all elements of $P$?

Does that sound right

[flat frigate]

what a wonderful world(wai)

[desert pasture]

but that doesn't help us does it

[unique thicket]

F(x) is a field

[desert pasture]

yea

oh

right

[unique thicket]

instead you should think about rational root theorem

[desert pasture]

yes, just realised

Thanks!

[desert pasture]

that would just tell me if it has a "rational" root

wouldn't it

[unique thicket]

uhhh yeah it wouldn't rule out 3 + 2 = 5

i'm actually not quite sure how to deal with 3 + 2

maybe you could consider a quadratic extension

[unique thicket]

oh sorry i misinterpreted what you meant. no this is fine since F(x) is the fraction field of F[x] which has unique factorisation

the problem is that this only detects linear factors

[desert pasture]

mhm, I should have worded it better yea

I'll look at some extensions I suppose

I'll close this for now

thanks!

.close

[safe radish]

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[nova plover]

can anyone solve this?

[safe radish]

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@nova plover Has your question been resolved?

[crude bear]

There is the solution on the second page

Also most of these questions are covered in the first course on a topic

[safe radish]

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[ember lark]

how do i find an inverse matrice with gausian method

[ember lark]

<@&286206848099549185>

[vagrant hill]

dont ping immediately

wait some time

[ember lark]

i'm sorry man i just didn't get my other questions answered and i have to hurry

[vagrant hill]

okay its fine

just dont immediately ping

[stiff snow]

^

[ember lark]

Yes okay i'm sorry

[vagrant hill]

thanks!

[hidden gyro]

Write this
[ A | I] where I is the identity matrix and A is your original matrix
Try to reduce A into [[1,0,0],[0,1,0],[0,0,1]
And match whatever operations you did onto the identity matrix

[ember lark]

okayy

[hidden gyro]

https://youtu.be/Fg7_mv3izR0?si=MYHiOmHUlD0dl54H

[ember lark]

but i need to use gausian method specificly

or whatever

[hidden gyro]

That’s what you will do

[ember lark]

but is that all i just reduce it to get the numbers?

[prisma elk]

yes

But it is slow at least I think

[ember lark]

wait but how did my friend get the fractions on the side

i don't think it's correct but i think there is another step or sum

[prisma elk]

you will get fractions

wait

[hidden gyro]

It’s about matching the operation
If I did row1 -> 2*row1
The operation must match in the in the identity matrix

[ember lark]

no i get that but the fractions just confuse me

like i am seriosly so lost

[prisma elk]

https://www.youtube.com/watch?v=uBmLelnJgvI

[ember lark]

OH WAIT

THE DET A AND ADJ A THING RIGHT?

[prisma elk]

Yea, you just split your 3x3 matrix into a bunch of 2x2 and find its determinant

[ember lark]

i see it now,but this is the gausian method still right?

[prisma elk]

after finding its determinant, apply the sign matrix

[prisma elk]

no gaussian method is when you are doing gaussian reduction

to your matrix

to make the left side look like the right side

[ember lark]

but the thing said to use Gauss Jordan method to solve it

[prisma elk]

yea then ig you have to first do that

and check with the other method that I sent

[ember lark]

I think the equation is set up wrong or sum

[prisma elk]

sent

[ember lark]

cs idk how to do it with just elimination

[ember lark]

It says with Gauss Jordan method find inverse matrice of the matrice D....

[prisma elk]

So what is your question sorry?

like how you do it or what?

[ember lark]

Yeah

like isn't gausian method just elimination?

[prisma elk]

yes

[ember lark]

like don't i need some sort of answer on the right to get any value on x,y,z

[prisma elk]

you will get something that

since you have inverse

the inverse will be on the right side

and the identity on the left

You are just trying to the left side the identity matrix and right side becomes your inverse

does that make sense?

[ember lark]

i understand yes

but

can this only be done with the adj a and det a thing

[prisma elk]

ye

[ember lark]

is that what the gauss jordan method is here

like this is the part that's confusing me

just that

[ember lark]

allright

[ember lark]

thank you for your help btw

[prisma elk]

this is gaus jordan elimination

[ember lark]

i get it now

.close

[safe radish]

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[prisma elk]

happy to help

[safe radish]

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[safe radish]

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[lilac relic]

Hello I'm attempting to find the estimated mean gall diameter of flowers, with a fall thickenss of 6.98mm

[lilac relic]

What am I doing wrong here?

I always get this problem, the object gall not found etc etc

[median sand]

Is this like a Google Sheets problem?

[lilac relic]

No it involves R studio

Which is basically a fancy caluclator

[median sand]

Interesting.

[lilac relic]

Would u know how to find the predictor value?

[median sand]

This is out of my scope sorry, but what happen if you apply it on Sheets, or on Excel?

They do have mean calculations.

[lilac relic]

I've never tried it

[opaque fern]

Its been ages since I used R but

Dont you have it mixed up on the lm function

Like

It needs to be lm(gall ~ wall) not the other way around

[lilac relic]

REALLY

LOL

Lemme try that

[lilac relic]

Now its giving the same error but for "wall"

So that wasn't the answer hahah

[opaque fern]

Wwll I still think thats a correct change

[lilac relic]

fsdofOwawjdf

[opaque fern]

Btw

Rename total1 to wall?

total1 doesnt exist in ur tsble

[lilac relic]

lezzgo

Thank you

My problems have been fixed

[lilac relic]

Now I suppose the answer would be 20.42832

[opaque fern]

Just for future reference you are better off joining a server that can help u with this sort of stuff other than this one

Tech support here is...slow lol

[lilac relic]

I have, the other server I'm on is even slower

they only get online past my deadline to hand in

[opaque fern]

Unfortunate

[lilac relic]

Recommend any good ones?

[opaque fern]

Nope

Im clueless about this sort of stuff

I did like one course a century ago

[lilac relic]

Not as clueless as me though haha

[lilac relic]

alright thanks again. I'll no doubt have more questions eventually.

is it all right to ask you?

[opaque fern]

If im available yea I can help u, but dont go out of your to DM or ping me since it is seen as bad etiquette here

[lilac relic]

okie then

.close

[safe radish]

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[safe radish]

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[lilac relic]

Are these two questions asking for the same thing?

[lilac relic]

I think they are

[safe radish]

@lilac relic Has your question been resolved?

[median sand]

I do not think they are the same. But a bit uncertain.

[lilac relic]

sFDishdfsidjf

.close

[safe radish]

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[safe radish]

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[shut gust]

<@&286206848099549185> pls fast exam in 5 min

cant wait too long

how solve these kinda problems

<@&286206848099549185>