help-0

[wary fable]

ok u got this part?

[warm quarry]

no sadly not

[wary fable]

explain what u fail to understand

[warm quarry]

why that counts

[wary fable]

elaborate more

[warm quarry]

well uhm how can |x-1| / |x+1| <e <=> |x-1| < e . 2|x+1| cuz this is not always true right?

wait actually i think it is

[wary fable]

lmfao u trippin or what?

1/2 < 3 <=> 1 < 6 not always true?

[warm quarry]

ye okay

[wary fable]

try to understand why this is true

[warm quarry]

aaah okay now i get it

[lone heart]

@warm quarry Has your question been resolved?

[warm quarry]

<@&286206848099549185> can someone check if this is correct?

[solemn seal]

Why you have two accounts 😭

[warm quarry]

cuz when i am at uni i can quickly swap from devices

[solemn seal]

Are you trying to find radius of convergence 😐

[warm quarry]

the what what

[solemn seal]

Wait this doesn’t look like calc 😭

[warm quarry]

it is :-:

[solemn seal]

Oh sorry I’m not the right person idk what min is

[warm quarry]

oh okay no worries

[solemn seal]

What math is this

[warm quarry]

calculus

analysis

[solemn seal]

🗿

So real analysis

[warm quarry]

probably not

this is basis of math

[solemn seal]

I’m taking that later this yr

[warm quarry]

i am in my first year of uni

[solemn seal]

Oh im still in hs lol

[warm quarry]

aaah damn

alright

[solemn seal]

I’m on multivarible calc

I’ve never seen that kind of notation before 😭

[warm quarry]

that should be more complicated than this

[solemn seal]

Yeah

I think your in real analysis

[warm quarry]

noooo definitely not

[solemn seal]

I’ll be there later this year as I said 😭

[solemn seal]

Hmm odes then?

[warm quarry]

Idk it is called analyse in the netherlands

but i thought it is calculus

cuz we use a calculus book

[solemn seal]

🗿

I think your on complex analysis

Or real analysis

[warm quarry]

that cant beeeee

i just learned the definition of the limit and stuff

[solemn seal]

🗿

[solemn seal]

I don’t understand what the symbols mean

Min, etc.

[warm quarry]

min{1,e} means if e is greater than 1 then it will give the output 1 if e is smaller than 1 then it will give the output of e

[solemn seal]

Trex can you help him 😭

[warm quarry]

Aaah i think i am in analysis

not calculus

[solemn seal]

Yeah bro 😭

I’ll be able to help you in 6-8 months though 👍

[oak chasm]

Does your book ask you to prove theorems (not epsilon delta results for individual problems)?

Or does it just state the theorems?

[warm quarry]

we have to proof stuff

but not insanely difficult stuff

[oak chasm]

What stuff?

[warm quarry]

mostly with limits and continous and closed open intervals

[oak chasm]

OK, so your proofs are just proving the answers to particular limits?

[warm quarry]

but this cant be analyse already right

[warm quarry]

ye for now yes

[solemn seal]

What’s your class called?

“Math analysis”

What does it say on your syllabus

[warm quarry]

Analysis

[oak chasm]

Who is the author of your textbook?

[warm quarry]

stewart

[oak chasm]

Which of these: https://www.stewartcalculus.com/?

[solemn seal]

So it’s kinda like real analysis

[warm quarry]

no

i dont think so

[oak chasm]

OK, if it says early transcendentals, then it's probably a basic calculus course, not analysis.

[warm quarry]

no not that one

wait

[oak chasm]

Oh.

[warm quarry]

that was in first semester

wait it does say it

mb mb

but can someone check my answer :-:

[warm quarry]

.

[oak chasm]

Yeah, that's just for calculus, not analysis.

[warm quarry]

ye bro i am too stupid for analysis

[oak chasm]

It looks like your first line that starts with for all is wrong.

[warm quarry]

Oh huh why

[oak chasm]

It should be forall epsilon > 0 there exists delta > 0 such that 0 < |x - 3| < delta -> 0 < |f(x) - 15| < epsilon.

[warm quarry]

Huh

i am like 90% sure we always do it the other way around

[oak chasm]

(\forall \epsilon > 0 \ \exists \delta > 0: 0 < |x - 3| < \delta \to 0 < |f(x) - 15| < \epsilon)

[warm quarry]

look she does it like this

our prof

[ocean seal]

Chai T. Rex

[warm quarry]

ye wth

whattt

i am so confused

my professor been doing it wrong for 2 weeks now then

HUH

[oak chasm]

The comma there is strange as well.

Or the lack of one there.

[warm quarry]

Ye

my brain

[oak chasm]

You don't just write a proposition followed by another.

[warm quarry]

only if they are equivalent you are allowed to right

but this one isnt 💀

[oak chasm]

Oh, I get it.

It's sort of written in natural language rather than correctly.

[warm quarry]

so it is correct?

[oak chasm]

It looks like it's written in natural language.

Like this is true for all xs where ....

But that's just <-.

-> already handles the for all part in their statements.

[warm quarry]

aaah okay

[oak chasm]

If it's P -> Q, then if P is false, it doesn't count against it being true. So, P -> Q only looks at when P is true.

[warm quarry]

ye true

[oak chasm]

What does it say on the 6th line on your proof?

[warm quarry]

proof delta = {1,e} works for all x

[oak chasm]

No, I mean "<=> |x - 3| < epsilon" or something like that.

There's something under the epsilon.

[warm quarry]

oooh just a 1

so i know why it is |x-3| < epsilon

[oak chasm]

Sorry, I need to take a break.

[warm quarry]

no worries

[lone heart]

@warm quarry Has your question been resolved?

[lone heart]

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[limpid turret]

<@&268886789983436800>

[lone heart]

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[night geyser]

poor kid was asking about trig earlier today and got their account compromised

[winter bay]

Sad to see a fellow algebraic geometrician succumb to such things huh

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[main sun]

can someone try to find these answers

[shy cedar]

does any of the reason say "given"?

[main sun]

yep

[shy cedar]

ok, so what info is given to you?

[main sun]

that ac is perpendicular to bd

[shy cedar]

yeah thats probably the first box if its one of the options

but is anything else given?

[main sun]

its in the ss

[shy cedar]

you know what the tickmarks mean on ad and dc right

[main sun]

theyre congruent?

[shy cedar]

yeah, so thats another given

[main sun]

i just dont get like what step i'd put it on and everything

[shy cedar]

you put the given info on the top box

[main sun]

so those top two boxes i should put those two givens

[shy cedar]

yes

[shy cedar]

for the very first one, put that ad is congruent to dc

[main sun]

oh okay

done

[shy cedar]

yes, now what do you know about perpendicular segments

[main sun]

they make right angles???

[shy cedar]

yes, but what are the right angles of the triangle

[main sun]

adc bdc

[shy cedar]

bdc isa right angle but adc is a straight angle

if you know what linear pairs are, bdc and what other angle form a linear pair?

[main sun]

adc

i didnt mean adc my bad

adb

ive been meaning to say adb

[shy cedar]

Oh i see

[scenic lichen]

Is math induction important to study or not?

[shy cedar]

so you know that adb and bdc are right angles, so what is the reason

[main sun]

linear pair?

[scenic lichen]

anyone?

[shy cedar]

well i was using linear pair as an example, so that probaby isnt a choice

whats another reason?

[main sun]

perp lines make right angles

[shy cedar]

yep, or another way to say is "Definition of perpendicular lines"

so thats the third box

[main sun]

bet okay whats next

ur the goat

[shy cedar]

Ok, so lets think about abd and cbd as triangles, as they split them up. is there anything they have in common?

[main sun]

bd

do i reflexive bd

[shy cedar]

yep you got it

[main sun]

okay so last two boxes

[shy cedar]

ok, can i see the statement options for the box on the very right

[shy cedar]

those statements are using x and y? thats weird

[main sun]

well id plug in whatever letters

[shy cedar]

Oh i see

So, adb and cdb are congruent since they are both right angles, same measure

[main sun]

whats the reason

oh its gotta be all right angles are congruent right?

[shy cedar]

yeah

so what congruence postulate proves that the triangles are congruent

with the info we have

[main sun]

SAS??

[shy cedar]

yep

[main sun]

it says somethings wrong

[shy cedar]

wait

[shy cedar]

deos it say what

[main sun]

it just says ur wrong try again

[shy cedar]

the box on the righr isnt even necessary.. but i did forget something

its the HL theorem

[main sun]

so just HL instead of SAS and keep the box on the right

ik the box on the right feels redudant which makes me think its wrong

[shy cedar]

im not sure if you should keep it, or do something like adb and cdb are 90 degrees

you should prolly ask your teacher

cause that last box is just holding us back lol

[main sun]

my teacher never responds till like a week later

u seem smart tho could u help me with a proof that doesnt have a flow chart and is just wtv u want

[shy cedar]

oh yeah sure

i dont like flowcharts either, especially ones you have to use the boxes

[main sun]

i tried this for so long like genuinely hours

[shy cedar]

ok well if a segment bisects another segment, what does it do?

technically the figure is a parallelogram, so they may be called diagonals too

[main sun]

idek tbh

make sure u read the little fine print note

[shy cedar]

well do u know what segment bisectors are?

[main sun]

yes

[shy cedar]

ok, what is it

[main sun]

like when it splits it right in half

like ac to bd

[shy cedar]

yeah, a bisector splits either a segment or angle in half

so if ac bisects bd, then ?? is congruent to ??? and ??is congruent to ??_

[main sun]

ae ec and be ed

[shy cedar]

yep. so once we find that those statements are true, then we are done

[main sun]

but it isnt proven that their bisectors yet??

[shy cedar]

Well, if we do find out that ae is congruent to ec and be is congruent to ed, then we are done. thats our main objective. not sure if theres an option that says its not possible, but most of the time it is possible

[main sun]

its possible

[shy cedar]

ok, think about how we proved 2 triangles are congruent. if two triangles are congruent, their corresponding sides are congruent (CPCTC). go step by step in order and try to see if you can figure out if be is congruent to ed, and ae is congruent to ec. use paper if needed because you will need to mark yourfigure with tickmarks or the symbol that show that an angle is congruent to another angle

think about aas, asa, sas, etc

i gtg to sleep, but you can ask other ppl for elp

[main sun]

ok thank u

.close

[lone heart]

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[zenith flax]

whats the formula for that semi circle

I tried (1/2(12)(6) ) / 2

[charred summit]

you know you can calculate it using geometry right?

[zenith flax]

Yes, would that be two triangles

?

[lunar saddle]

What's the area of a circle

Formula

[charred summit]

for the first part and third yeah you calculte the 2 triangles using geometry
but i was talking about the semi circle specifically

[zenith flax]

oh

well yes for geometry but i didnt know the formula to use

area of cirlce is pir^2

[charred summit]

yeah

[zenith flax]

I tried pi(12)^2/2

got a long ass number, wrong

[charred summit]

what is the radius of the circle in the graph

[zenith flax]

oh

its 6

[charred summit]

yeah

[zenith flax]

i always mix the two up

[charred summit]

also you need to see that the integration would be negative

as well

so it is negative the area of the semi-circle

[zenith flax]

erm

I did

-PI * (6)^2 / 2

[charred summit]

may be just right
-18pi

[zenith flax]

that was right

thats so stupid

lmfoa

one quick quick question

[charred summit]

LoL
they should at least put a margin of 1% error

[zenith flax]

I was breaking this up into shapes

[9,15], i made into a triangle
[15,21] i made into a triangle
[21,27] i made the lower half into a triangle
[ 21,27] i made the upper half a sqaure

[charred summit]

you realize that
integration from 0 to 21 = integration from 0 to 15 + integration from 15 to 21

oh wait

i saw that as 0 not 9

[zenith flax]

i was thinking the same thing

would make sense to do it that way, or write the question that way

but we gotta be difficult

[charred summit]

ok how did you get taht the area from 9 to 21
is -45

[zenith flax]

uno

[charred summit]

I think with just by looking you would realize it is 0

[zenith flax]

wow

oh my god

its 9 to 21

[charred summit]

yeah

it didn't mention 27

[zenith flax]

realizing that now

[charred summit]

anywhere

[zenith flax]

i got ahead of myself

last questiont hen im off

We've done step four before in class, I did a problem just like this one below this also

I've just never done it with a square root

[charred summit]

,rotate

[charred summit]

so you used the formula and it was wrong?

[zenith flax]

I'm unsure of how to get the right endpoints iwth this is all

This is the question I solved that was similar, but it doesnt ask for right end points

[charred summit]

isn't the ith end point here supposed to be the right endpoint?

[zenith flax]

So would i be 3 if its the end?

I'm referring to part 4

im a bit confused

[charred summit]

use i instead of 3

the second 3

[lone heart]

@zenith flax Has your question been resolved?

[zenith flax]

i did try that and it didnt work lol

[charred summit]

oh that is because in riemann sum you don't just sum the result of f(x) at the endpoints
you also multiply the width of each interval by the value inside

so it should be
$$\frac{2}{n} \sqrt{3+(1+\frac{2i}{n})^2}$$

[ocean seal]

Sherif Player

[zenith flax]

that is so damn stupid of me lmfao

I added the first two answers together, and got the area of the last shape (triangle).

36+(-18pi)+(1/2(3)(3))

[charred summit]

don't evaluate it numerically

[zenith flax]

ah

[charred summit]

just put it in terms of pi

[zenith flax]

thank you

[charred summit]

no problem

[zenith flax]

.close

[lone heart]

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[vital kindle]

Erm

[vital kindle]

Help

What does the set notation mean

Like the x>_ 4 thing

Move the lil line over a bit

[reef wren]

x is larger than or equal to 4

[vital kindle]

So it’s not true right cuz the graph goes into negative infinity

[reef wren]

it's not true yeah but what do you mean by the graph goes into -inf ?

[vital kindle]

Errrr

Like the arrow is pointing -inf

[reef wren]

yep but also in the +inf direction too

[vital kindle]

Yeah

[vital kindle]

It doesn’t mention the -inf part

Wait what

[reef wren]

yeah

[vital kindle]

Wait errr what is going on sob

[reef wren]

x>=4 is just like [4,inf)

[vital kindle]

Yeah

But doesn’t that mean the graph starts on the point 4

Not going (inf, inf)

Wait no

(-inf, inf)

[reef wren]

yeah the graph is defined for all x and maps all the y values

[vital kindle]

Ohh

So it’s the last one

[reef wren]

nod

[vital kindle]

Okayy lol thanks

.close

[lone heart]

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[quasi vector]

<@&268886789983436800>

[lone heart]

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[solemn juniper]

nobody gets a fair trial anymore

[quasi vector]

Sending the link to an nsfw server in a math server is.. a choice

[north rover]

what happened

[lone heart]

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[raw jetty]

suppose $x_1,x_2,x_3,\dots,x_{2025}$ are positive integers such that $x_1+x_2+x_3+\dots+x_{2025}=x_1x_2x_3\dots x_{2025}$, find the maximum of $x_1+x_2+\dots+x_{2025}$

[ocean seal]

skissue.in.a.teacup

[raw jetty]

by logic most of x would be 1, as else the rhs would be much much higher

yeah that was my only logic :shrug: guessing around gave me x_(1 to 2023)=1, x_2024=2, x_2025=2025

[mortal trellis]

I'm not convinced that there arent for example two of the numbers somewhere around sqrt(2025)=45 so that the sum is still near 2025 but the product also is roughly 2025

or similar for cube or bigger roots

tho I suppose that wont be relevant if you only want the max of the sum

[vocal barn]

Didn't this question end

I can't even see the question

[raw jetty]

what

[vocal barn]

Nothing I am just new in this channel

Solving question entertain me so I joined the channel 😺

[raw jetty]

but like, if so lhs would be like around 2025+90 but my guessing strat got 4050

[mortal trellis]

which is why I wrote this

[raw jetty]

i mean the max of the sum is still equal to the product so you can reword it as the max of the product

[little drum]

=> x_1 x_2 ... x_2024 ≤ 2025 ≤ x_2 ... x_2025
This sets a restriction for x_i's```

[raw jetty]

i dont get how you get x_1x_2...x_2024<=2025

[lone heart]

@raw jetty Has your question been resolved?

[little drum]

In the first place, you do not necessarily have to bug with such a terrifying looking problem:

=> x_2012 ... x_2024 ≥ 2^{11} = 2048 
=> x_1 ... x_2024 ≥ 2048 which does not satisfy the restriction```

So essentially, this problem simplifies by plugging x_1 = x_2 = ... = x_2012 = 1

You can similarly do for:

=> x_2013 till x_2018 are ≤ 2```

[raw jetty]

going higher doesent seem to help

i cant think of how the idea goes with this

[little drum]

You don't need to go higher actually :p Assume all 6 of them are 2s

=> (2^6k - 1)x_2025 = 2012 + 12 + ... + x_2024 ≤ 2024 + 8x_2025
=> (64k - 9)x_2025 ≤ 2024
=> x_2025 ≤ 2024/(64k - 9), where k = x_2019 * x_2020 * ... * x_2024 ≥ 2^6
so not all of them can be 2s either```

Can you use the same idea and show none of them can in fact be 2s?

[little drum]

Another hint :p you can totally use that x_2025 = 2025, x_2024 = 2, all other x_is = 1 is a solution

[little drum]

Here's how I'd proceed.

=> x_2025(2x_2024 - 1) = 2022 + 2 + x_2024 ≤ 2024 + x_2025
=> x_2025(x_2024 - 1) ≤ 1012, where x_2024 ≥ 2
=> x_2025 ≤ 1012
=> (x_1 + x_2 + ... + x_2024 + x_2025)_{max} = 2022 + 2 + 1012 + 1012 = 4048
Since this falls shorts of 4050, x_2023 must also be 1```

the rest can be solved as such: 2024 = (a - 1)(b - 1) => (a + b)max happens at a = 2025, b = 2
=> 4050 is indeed the best maximum we can find

[raw jetty]

wait so the max is 4050?

[little drum]

It'd appear so

[raw jetty]

sorry toom a shower

[little drum]

toom

[raw jetty]

alright

.close thankyoh again arya! <3

[lone heart]

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[lone heart]

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[dreamy echo]

Let f:ℝ→ℝ, f(x)=ax+b, a,b∈ℝ, a≠0.
Show that there is an x₀∈ℝ\ℚ such that f(x₀)∈ℝ\ℚ

[carmine reef]

are you allowed to use arguments about cardinality?

[lone heart]

@dreamy echo Has your question been resolved?

[lone heart]

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[gaunt turret]

can someone help me construct the pmf on the letter B

[lone heart]

@gaunt turret Has your question been resolved?

[gaunt turret]

<@&286206848099549185>

https://tenor.com/view/dono-wall-gif-24048347

why does nobody wants to help me

T_T

[gaunt turret]

<@&286206848099549185>

[safe meadow]

What does the sum of the seven spot mean

1+6 or you get 7 different face on the 2 dice

[runic glacier]

The former. So 1+6, 2+5 etc

[safe meadow]

So like you roll both dice and get 7

I don't know about pmf

But the probability of getting 7 by rolling 2 individual dice is 1/6

6/36

[gaunt turret]

probability mass functionn

it's like the table

x and P(X=x)

[safe meadow]

The way to construct them that I don't understand

[runic glacier]

@gaunt turret first think of the values that x can take

[dreamy echo]

yes. any idea?

[runic glacier]

Go to a new channel

[dreamy echo]

ok

[runic glacier]

@gaunt turret

[gaunt turret]

1-6?

[lone heart]

@gaunt turret Has your question been resolved?

[lone heart]

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[vital ocean]

So how do I go about solving this equation

[mental python]

find the first few terms?

thats where i would start

[vital ocean]

I have a exam on it in 3 weeks and I have no idea how to do it at all I was out

[little drum]

start with plugging u_n = v_n + k, k constant

after rewriting in terms of v_n, you can choose the value of "k" such that the equation has no constant terms

[vital ocean]

sorry u_n=v_n + k ?

[carmine reef]

yes make that substitution

so everywhere you see u of something replace it with v of that same thing and then add k

[vital ocean]

Vn+k - 2Vn+k - 35Vn+k

sorry i havent dont any of these yet 😭

[little drum]

$u_n = v_n + k$

[ocean seal]

Arya

[little drum]

Substitute this into your eqn.

[vital ocean]

and what to do if its Un+2

do i make it Vn+2+k ?

$u_n+2 = v_n+2 + k$

[little drum]

Yes

because v_n is also a function of u_n, so yes... if you change n on the LHS, n on RHS changes as well

[vital ocean]

sorry but isnt this just a quadratic?

Un+2=x^2
Un+1=x
Un=1

[little drum]

No. This is inhomogenous

Quadratic happens for Homogenous

[vital ocean]

ahh ok

[little drum]

You're in the process of homogenizing it

[vital ocean]

thankss alright ima do what i can and send a pic of where im at

[little drum]

,texsp $v_{n+2} - 2v_{n+1} - 35v_n = 144 + 36k = 0$ for $k = -4$

[ocean seal]

Arya

[little drum]

let me know when you have this

[vital ocean]

$v{n+2}+k - 2v{n+1}+k - 35v_n+k = 144$

[ocean seal]

maT

[vital ocean]

i did it wrong

[little drum]

Yes you did

$u_{n+1} = v_{n+1} + k \implies - 2u_{n+1} = -2v_{n+1} - 2k$

[ocean seal]

Arya

[vital ocean]

im charging my phone so Ill be able to send pics but I dk how to use the formatting for the maths thing

so the -2 affects the k aswell?

[little drum]

Yes it does. it is multiplied to u_{n+1}

so when substituting, it multiplies with everything u_{n+1} is

[vital ocean]

ohhh so also for the -35?

and a + 1

yess i got it now

and remind me why do we let it equal to 0 and solve for k to get -4

[little drum]

scratch your second line, that's bs

[vital ocean]

yup done

[little drum]

you need to delete the constant term from the fourth line so you have a homogenous equation that you're familiar with

plug k = -4 to get rid of 144 + 36k. Can you solve the rest homogenous eqn?

[vital ocean]

ima try

i think i have it

nooo 😭

[lone heart]

@vital ocean Has your question been resolved?

[vital ocean]

I mixed up the 7 and -5 however I still dont know what to do after that

[lone heart]

@vital ocean Has your question been resolved?

[vital ocean]

.close

[lone heart]

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[river yacht]

The question Down and first this queston is <0 right?

[river yacht]

Or not 🤔 it's >5

[patent vale]

yes its smaller than 0

[river yacht]

Yup I'm right ty

I can't get do questions like Multiplication of fractions but i can do logarithmic and trigonometry lol :@@

[river yacht]

Tysm

Hell nah wtf questions coming😭

[wide quest]

nice theyre teaching you what they taught me last week

[river yacht]

What about this down 3😭

[hardy path]

Which language is this?

[river yacht]

Georgia

[hardy path]

I see

[river yacht]

😅😆

[patent vale]

down 3?

[river yacht]

Yup

The larges

How tf i can multiply it 💀

[reef wren]

what is the question even asking ?

[river yacht]

Compare the value to 0.

[tardy stag]

that should probably be $\log_{\blue{a}^2 + 2}(a^2 + 1)$

[ocean seal]

hayley, who shakes the world

[river yacht]

Yup

[lone heart]

@river yacht Has your question been resolved?

[lone heart]

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[spiral dirge]

someone solve the integral please

[junior kraken]

Why wouldn't someone not like to solve this cute integral?

[tight pier]

Consider || z^2 = 1-sqrt(x) ||

[spiral dirge]

cool thank you

[lone heart]

@spiral dirge Has your question been resolved?

[lone heart]

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[kindred obsidian]

why is it wrong i put it in a graph and its wrong😭

[sour verge]

!xy

[lone heart]

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.

[kindred obsidian]

i made this problem so this is it

[sour verge]

It's just more efficient if we know what you're trying to do

Instead of piecing it together from your work

[kindred obsidian]

i want a quadratic function (opening downwards) that has the tangent 2 (so going through P(10|5) and Q(11|7)) and i want it to be 25 up

[little drum]

A quadratic function has variable tangent.

[kindred obsidian]

so this

sorry its flipped

[little drum]

Do you mean it must be passing through some point and has a tangent 2 at that point?

[kindred obsidian]

yeah

or m

[little drum]

what point is that

do you have its coordinates given?

[kindred obsidian]

is tangent the same as gradiant?

[kindred obsidian]

yeah

[little drum]

what are the coordinates

[kindred obsidian]

P(10|5) and Q(11|7)

[little drum]

just so you know, if the curve for the quadratic function passes through 2 points, the line joining them is not tangent to the curve

[kindred obsidian]

ah

thats ok

[little drum]

so if you want the function to pass through these two points, it'll simply be a quadratic function that passes through two given points

[kindred obsidian]

yes

[kindred obsidian]

.

[little drum]

the way to find this is to write $\frac{1}{a} = \frac{(x - h)^2}{f(x) - 25}$

[ocean seal]

Arya

[little drum]

Passing it through the two points gives you

$\frac{(10 - h)^2}{5 - 25} = \frac{(11 - h)^2}{7 - 25}$

[ocean seal]

Arya

[little drum]

did you solve this?

[kindred obsidian]

uhh

no

had bottom and top switched

[little drum]

looks like you did.

[kindred obsidian]

oh

i guess so😭

[little drum]

Let me verify your roots rq

[kindred obsidian]

okok

[little drum]

,w 10(11-x)^2 - 9(10 - x)^2 = 0

[ocean seal]

Results provided by WolframAlpha

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Requested by ansh_3502

[little drum]

You got one of them right, you missed h_2 = 20 - √90

[kindred obsidian]

huh

i had h2

[little drum]

right, I had it cropped so I didn't notice

[kindred obsidian]

youre good

[little drum]

let me verify your "a"

[little drum]

You're bad ._>

You missed the exponent in the denominator

[kindred obsidian]

oh

[little drum]

a_1 = -20/(-10 - √90)² ; a_2 = -20/(-10 + √90)²

Do that and you're good

[kindred obsidian]

hm

[little drum]

btw you might want to rationalize your denominators, plugging that ugly "a" into your quadratic form would displease whoever you're trying to show this equation

[kindred obsidian]

OK

thanks

.close

[lone heart]

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[sick tangle]

ok so

[sick tangle]

can i just flip all the fractions

[limpid turret]

No

[sick tangle]

am i correct

[reef wren]

it works yea

[patent vale]

a little overcomplicated but it works

[sick tangle]

but its still wrong did i type it wrong

aaaaaaaaaaaaaaaaaaaaaaa

[reef wren]

put a bracket in 1/(4a)

[sick tangle]

yes ty!

[sick tangle]

can u tell me the quicker way?

[patent vale]

b = -4ac/(3*(4*a - 1))
is the cleaner way to say it

[sick tangle]

ty

[patent vale]

they are the same mathematically

[sick tangle]

there has to be a rule where i can flip everything

or im not allowed to

[tardy stag]

you can, if you wish, replace each side with 1 / that side

the result would be $p = \f1{\f1q + \f1r}$

[ocean seal]

hayley, who shakes the world

[sick tangle]

i see

[tardy stag]

i don't think you want to do that

[lone heart]

@sick tangle Has your question been resolved?

[lone heart]

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[tacit arch]

Double angle formula

[mint orchid]

okay i just googled what that was and that would change the numerator into 2sinxcosx right?

it would turn into this

is the next step to distribute? then integrate? or is 5/4 a constant that i can pull our before i integrate?

[tacit arch]

Cancel

Yes you can pull out 5/4 as well

[mint orchid]

cancel the sinx from top and bottom?

[north rover]

yes

[patent vale]

wait a minute thats illegal

[north rover]

[mint orchid]

wait what?

[north rover]

me confused now

😔

i dont see anything wrong with cancelling

[little drum]

$\int \frac{5\sin 2x}{4\sin x} \dd x \neq \int \frac{5\cancel{\sin} 2\cancel{x}}{4\cancel{\sin x}} \dd x$

[ocean seal]

Arya

[patent vale]

wait nvm you can

i lost a plot a little

mb

[vocal dirge]

kill it with fire

[north rover]

[north rover]

proceed with what you were doing

[little drum]

XD
Anyways, yes. Double angle formula of sin 2x and cancel the sin x

[mint orchid]

is this right or not?

[north rover]

wha

[little drum]

No no, the sin x got cancelled, why is that still there

[north rover]

"cancelling" meaning that you remove from both the numerator and denominator

[mint orchid]

oops

better?

[north rover]

yeah, better

[mint orchid]

okay now i integrate?

[north rover]

$\frac54\cdot 2=\frac{5}{\cancel{4}^2}\cdot\cancel{2}$

[ocean seal]

;(

[north rover]

yep

[north rover]

this is not 4^2 btw it means 4 reduces to 2

[mint orchid]

okay so i take out the 2 before integrating

[north rover]

yes

[north rover]

gg ez

[mint orchid]

yay thank u!

[north rover]

yw

[mint orchid]

WAIT!

it says that its wrong

[north rover]

methinks

meconfused?

meknows its right...

[north rover]

check for any mispellings

[north rover]

hell, it took 3hrs for me to realize a mistake on the other thread

alright let me work it out

$\int\frac{5\sin(2x)}{4\sin(x)}dx=\frac54\int\frac{2\sin(x)\cos(x)}{\sin(x)}dx=\frac52\int\cos(x)dx=-\frac52\sin(x)+C$

[ocean seal]

;(

[north rover]

??

bro im not doing this shit again

[mint orchid]

umm i can type it into symbolab

[north rover]

im just traumatized from the other thread

[patent vale]

how did bro get -

cosx integrates to sinx

not -sinx

[mint orchid]

oooh silly mistake

thank u!

[ripe perch]

yes

double angle

[north rover]

yeah im going to sleep

[ripe perch]

$\int \frac{5 sin(2x)}{4 sin(x)} dx = \frac{5}{4} \int \frac{2sin(x)cos(x)}{sinx} dx = \frac{5}{2} \int cos(x) dx = \frac{5}{2} sin(x) + C$

[ocean seal]

kronium_

[mint orchid]

.close

[lone heart]

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Closed by @mint orchid

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[lament wraith]

SOMEONE PLS

due IN 5 MINS

[serene scarab]

dont ask for help on a test

[lament wraith]

its not a test

dawg

💀

its homework

fuck makes u think its a test

[limpid jolt]

do you think cursing at somebody makes them inclined to help

[north rover]

please dont...beg.

[lament wraith]

what makes anyone think thats a test

[north rover]

so...do you wanna show your work, or...?

[lament wraith]

ok

1 sec

my phgone died can i just explain

what i did

[north rover]

yeah ok

so just to let you know you're not getting help in that 5 mins

[little drum]

,w (63.8 *(sin 58.5°))/(sin 99.5°)

[ocean seal]

Results provided by WolframAlpha

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

Requested by ansh_3502

[lament wraith]

thats fine

[little drum]

calculator error :p

[lament wraith]

sin99.5
∘

63.8
​

sin58.8
∘

b
​

is how i solve for b

[little drum]

Yes that's right

so it's calculator error

[lament wraith]

what

so whats the correct value for b

[lament wraith]

this ?

sin99.5
∘

63.8
​

sin22
∘

c
​

and for c is this

[little drum]

b = (63.8)(sin 58.5)/sin(99.5) right?

[lament wraith]

yea

im p sure

[little drum]

put that in your calc

what do ya get

[lament wraith]

does mathway work

[little drum]

sure

[lament wraith]

i got 55.15

[little drum]

,w (63.8)(sin 58.5°)/sin(99.5°)

[lament wraith]

wtf

[ocean seal]

Results provided by WolframAlpha

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

Requested by ansh_3502

[little drum]

then why'd you write 54.8 in your hw lol

[lament wraith]

idfk

LOL

[little drum]

solve similarly for c and lmk what you get

[lament wraith]

wait so how do i calculate c

what equation

(63.8)(sin 58.5°)/sin(99.5°)

this?

[little drum]

this one that you did

bring c to the left and every other thing to the right

[lament wraith]

i got 55.15 again

huh

[lament wraith]

iu put this in

to the caluclator

and got 55.15

[little drum]

why'd you put that.. that was the equation for "b"

[lament wraith]

o shi

[little drum]

To get "c" you solve the equation for "c"

[little drum]

.

what'd you get >_>

[lament wraith]

is this the equation

[little drum]

yes

that's the sine law

,w (63.8)(sin 22°)/sin(99.5°)

[ocean seal]

Results provided by WolframAlpha

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

Requested by ansh_3502

[lament wraith]

ok

i will assume thats right

cause theres no degree sin in mathway

[little drum]

Lol aren't you past your 5 min deadline

[lament wraith]

yes

its 24.23

OMG ITS RIGHT

OMG

TYSM ILY

can u help me wit one more

[little drum]

No Peter ._> Ask Louis

[lament wraith]

what

[little drum]

jk go ahead

[little drum]

solve

[lament wraith]

i dont understand

[little drum]

$\cos \angle A = \frac{AB^2 + AC^2 - BC^2}{2\cdot AB \cdot AC}$

[lament wraith]

c
2
=a
2
+b
2
−2ab⋅cos(C

[ocean seal]

Arya

[lament wraith]

ik thats the law

[little drum]

plug everything you have into this to get BC

[lament wraith]

ok gimme a min to do that

[little drum]

does solving triangle means you also ought to find the remaining angles?

[lament wraith]

uh

le mme see

im p sure we only need one row

how do i plug this in mathway is not working

[little drum]

use wolframalpha

[lament wraith]

oh type shit

[little drum]

:p

[lament wraith]

what

.

whats bc

then

[little drum]

Lmao gotta trust your hands on this one >.<

[lament wraith]

??

bruh i litr js needa solve this then im done

ok 23.75

for bc

[little drum]

No =_=

use the formula and check for your own

[lament wraith]

how do i plug it in

idk how to use

wolframalpha

[little drum]

$BC = \sqrt{AB^2 + AC^2 - 2\cdot AB \cdot AC \cos \angle A}$

[lament wraith]

idk how to do this either

[ocean seal]

Arya

[little drum]

that's cosine law

just need to plug your values here in wolframalpha

[lament wraith]

YO I GOT IT

BC=
34
2
+27
2
−2⋅34⋅27⋅cos(44
∘
)
​

ok

[little drum]

yeah

plug that in any calc

[little drum]

you missed a square root

the thing you wrote on the right side is BC² not BC

[lament wraith]

this the right equation

i dont udnerstand this bruh pls

[safe meadow]

Which part you don't understand

[lament wraith]

everything

[lament wraith]

its not solving for bc

only b

or c

[little drum]

you wrote the cosine law right

[little drum]

you need to replace "bc" with a variable

the calc is reading B, C as seperate variables probably

[safe meadow]

Bc is not b*c it's is the length from point b to point c

[lament wraith]

HOLY SHIT I GOT IT

23.75

[little drum]

is why I used BC = x

[lament wraith]

bc=23.75

okok

now what

is this the next equation for b

[little drum]

cosine law only gives the value for a side, in this case BC, given that you have 2 sides and including angle

[safe meadow]

Angle b or line BC

[little drum]

Now you gotta use sine law for the remaining angles b and c

[lament wraith]

angle b

[safe meadow]

Then the formula slightly changes but still the same

[lament wraith]

b is 19.25

when i plugged it in

[little drum]

huh? plugged what in T-T

what's your cosine law

[safe meadow]

Cos(a) = (BA^2+CA^2-BC^2)/2BA*CA

[lament wraith]

brah

[safe meadow]

I dont think this is the right formula because we use side a,b,c not point a,b,c

Let me revise it

[lament wraith]

how do i rearrange it

to find b and c

[safe meadow]

Just input the length

[lament wraith]

wdym

[safe meadow]

For example we want to find angle c

[lament wraith]

yes

[safe meadow]

Notice the 2 line adjacent to poin t c

Point

[lament wraith]

yes

[safe meadow]

We add the square of them

[lament wraith]

34 squred plus 27 squred?

squared

[safe meadow]

And subtract the square of opposite side

So we have BC^2+AC^2-AB^2

[lament wraith]

yeah

so

thats how we find c?

[safe meadow]

Then divide it by the the 2 adjacent side times the other adjacent side

That is the cos of the angle we want to find

[safe meadow]

Sorry my internet was bad

[lament wraith]

i dont understand

still

what

[safe meadow]

The angle

Use trigonometry

Or calculator

Some angle have easier value

Like cos 45 = √2/2

While cos 50

,calc cos50

[ocean seal]

The following error occured while calculating:
Error: Undefined symbol cos50

[lament wraith]

i got the answer

tysm

[safe meadow]

How do use Wolfram

Nice

[lament wraith]

nah i used triangle calculator

xd

tysm tho

appreciate the time

.close

[lone heart]

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Closed by @lament wraith

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[safe meadow]

Yeah it's fine

[lone heart]

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[hybrid basin]

I am confused about the last step in this example, I understand the solution to be something like x-y=0, but I'm not sure how or why it corresponds to the general vector of the form x2[1, 1].

[naive valley]

x-y = 0 is the same as x=y

so any vector of the form (c,c) where x=y=c
(with c nonzero since eigenvectors are required to be nonzero)

[hybrid basin]

ok so x=y=x2, which is why x2[1,1] is the same as x2+x2 = x+y?

[naive valley]

yea a general vector with x=y looks like [1,1] times some constant

[hybrid basin]

that makes sense, but I'm used to solutions being equal to something, why isn't it written like x2[1,1] = 0?

[naive valley]

well it's not equal to zero, so =0 would not make sense

you could write [x,y] = [c,c] if you want it in that form

where c is any nonzero number

[hybrid basin]

ok, I understand now, thanks.

.close

[lone heart]

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[zinc haven]

i struggle to prove the other direction

essentially i need to find some linear map T such that for all linear S, ST = TS

i tried maybe finding some map S that would give me some clues, but i couldn't

[keen plinth]

you will probably need to pick a basis

[zinc haven]

okay, let e_1, ..., e_n be a basis of V

now do i just examine where T(e_i) sends to?

[keen plinth]

no yes

examine the commutativity of $T$ with $S_{ij}$, defined [ S_{ij} e_i = e_j, \quad S_{ij} e_{i'} = 0, i' \ne i ]

[keen plinth]

actually also yes

but in conjunction with what i wrote above

[zinc haven]

this sounds like a pain to prove it's linear

is S_ij even well defined?

[keen plinth]

not really

it's directly by the linear map lemma

[zinc haven]

linear map lemma only ensures S_ij(e_i) = e_j for 0 <= i, j <= n, what about the second clause when S_ij(e_i') = 0?

[keen plinth]

theres a different S_ij for every pair i, j

there are n^2 different maps you should be considering

for example, $S_{69}$ is defined
[ S_{69} e_6 = e_9, \quad S_{69} e_i = 0, i \ne 6 ]

[zinc haven]

yeah i understand, but i don't think that answered my question

how does linear map lemma tells us S_69 sends e_i to 0? and the fact that S_69 is unique makes it even less plausible

[keen plinth]

let us observe the statement

here, the T in the picture is our S_69

V = W for us as well

the basis is e_1, ..., e_n

ive defined

1. S_69 e_1 = 0
2. S_69 e_2 = 0
   ...
3. S_69 e_5 = 0
4. S_69 e_6 = e_9
5. S_69 e_7 = 0
   ...
   n. S_69 e_n = 0

so here, w_1 = 0, w_2 = 0, ..., w_5 = 0, w_6 = e_9, w_7 = 0, ..., w_n = 0

[zinc haven]

ahhh, i see

i was under the assumption w1, ..., wn is the list e1, ..., en or its alternate order thereof

that makes sense now

[keen plinth]

you now should look at how this commutes with T

what does that tell you about T?

in fact, you could even just try proving it for T : R^2 -> R^2

[zinc haven]

alright, one sec

[zinc haven]

for arbitrary vectors v or for just the basis vectors?

[keen plinth]

what does T look like?

[zinc haven]

er, a map sends elements from a vector space to the same space?

[keen plinth]

im not sure if you know this answer actually, since its not explicitly mentioned in axler

[keen plinth]

yes, but theres a way to write it out explicitly in some sense

ill show it, and you can tell me if it makes any sense

,align \MoveEqLeft[6] T(x_1 e_1 + \dots + x_n e_n) \
& = (A_{1, 1} x_1 + \dots + A_{1, n} x_n) e_1 + \dots + (A_{n, 1} x_1 + \dots + A_{n, n} x_n) e_n

[zinc haven]

epic overfull hbox moment

[zinc haven]

looks familiar

[keen plinth]

dont worry i have plenty of hbox

yes because its this exercise that you did

but in a more general setting where we replace F^n and F^m by V

and use bases

[zinc haven]

oh yes i can see the resemblance now

are you suggesting i examine how S_ijT commutes for e_1, ..., e_n, then extend it using this form?

[keen plinth]

i think you'll find that you can compute what A_i,j are equal to using the fact that S_i,j and T commute

[zinc haven]

alright, brb

[keen plinth]

birb indeed

[zinc haven]

[zinc haven]

how would i invoke this in the proof? since i only proved this for L(F^n, F^m)

[keen plinth]

the proof should be no different really

for example, by the linear map lemma, it suffices to determine what Te_i is for each i

then by the fact that you have a basis, you can write Te_i = A_1,i e_1 + ... + A_n,i e_n

and actually thats the entire proof

[zinc haven]

oh lol

so maybe put this in a separate lemma?

[keen plinth]

sure why not

you don't need to write the full thing tbh

[keen plinth]

you only need this bit

[lone heart]

@zinc haven Has your question been resolved?

[zinc haven]

yeah uh, i'm lost

[quasi vector]

oh hey we're doing the same chapter

[zinc haven]

[quasi vector]

Now use the linear independence of all the e_i

[zinc haven]

ok so that means A_jj = A_ii