help-13

[viscid perch]

i know but how do i put it for undefined values

[cedar kiln]

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[upper abyss]

You sure you want to go this route? Bans happen fast here

<@&268886789983436800>
Spamming swastikas and deleting them

bruce is

Edgy.

[calm sierra]

.close

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[upper abyss]

Thx bae ilu

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[mystic wren]

i am completely lost on mclaurin and taylor series, can someone explain it to me along with this problem?

[cedar kiln]

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[silent viper]

Well, what is the Maclaurin series @mystic wren?

Do you know what the maclaurin series for sin(x) is?

[mystic wren]

No 😭

[silent viper]

Well a Maclaurin series looks like:

It is a taylor series centered at 0... if that makes sense

So, what is your function?

f(x)=sin(5x) correct?

So what would be the first four terms?

[mystic wren]

I would apply this to sin(5x) right?

[silent viper]

yes

[mystic wren]

I actually have to go right now but I’ll return in a few hours , should I close this chat and make a new one or am I allowed to keep it 🤔

[silent viper]

Eh, your choice. I don't think it necessarily hurts anything to leave it.

But I do know the bot might time it out anyways

So, you might just want to assume it won't be here when you come back.

Maybe screen shot our conversation.

If you need more help you can dm me and we can work on it

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[random urchin]

how would you go about this? doing year review but i forgot this process

[rain drift]

Remember that $\sqrt[n]{abc...z} = \sqrt[n]{a}\sqrt[n]{b}\sqrt[n]{c}...\sqrt[n]{z}$

[wraith dagger]

MellowDramaLlama

[cedar kiln]

@random urchin Has your question been resolved?

[random urchin]

back

okay so

in this case it would look like

on the top

im a bit confused cuz below that the answer has a sq root 5 in it

and im also wondering when do we get to simplify the exponent

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[celest otter]

Im confused at what to put in the white squares

If I use u = x^2 + 4

[cedar kiln]

Please read #❓how-to-get-help

[gaunt trench]

oh

sry

[obsidian coral]

What does the "this formula" link to?

[celest otter]

I see i will investigate this

honestly the formula just makes it more confusing :/

[obsidian coral]

This is what you use to get the final blank

The second to last blank, I think it just wants you to input the fraction from above, when you split it

[celest otter]

where does the 2/x come from?

[white spindle]

7/2In(u)

[obsidian coral]

The initial step you did

You split the 9x^2 -8x +8 fraction into two fractions, on the first line, then you split the 7x - 8 fraction on the second line, the third line just wants all the fractions you ahve

And the last line is the final answer

[white spindle]

Intergal part:[7/2d(u)]/u

[obsidian coral]

If you're referring to the integral blank, no

And

[cedar kiln]

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.

[white spindle]

ah

[celest otter]

I used the formula for the third integral to get the tan stuff

Did i mess up the second integral

nevermind i think i did it correct

thank you all

.close

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[woeful shale]

I need help with some desmos work with area shadding

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[jolly sable]

Does it kean obsolete ?

[jolly sable]

Here o?

[crimson delta]

are you asking what the small o notation means?

[tropic oxide]

no, the little o does not stand for obsolete.

[jolly sable]

Yes

[jolly sable]

Then?

Does it mean obsolete higher order terms can be ignored?

[crimson delta]

as a result it means that more or less, yes. not as a definition

as a definition it just means that those terms go to 0 faster than x^3

[jolly sable]

By how it goes to 0?

[crimson delta]

wdym

[jolly sable]

Terms which faster than x^3 why it goes to 0?

Are we ignoring by force? Or suitable to question

Denominator

[crimson delta]

those terms are of a higher order

x^5, x^7 and so on

if you take the quotients x^3/x^5, those go to 0

meaning that x^5 goes to 0 "faster" than x^3

you could also write those out in the question but you will notice that they make no difference in the end

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[wild spear]

How do i start proving this for positive real x, y, z?

[paper crown]

Perhaps harmonic progression could be used somewhere here?

[cedar kiln]

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[wild spear]

How can I use harmonic progression in here?

[paper crown]

Sorry, I meant harmonic mean

I am not sure about this, but usually questions like this in which we have to prove that LHS is less than equal to something use it

[wild spear]

The QM - AM - GM - HM inequality?

[paper crown]

yeah

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[prisma bluff]

Yo

[prisma bluff]

How

[paper crown]

What have you tried?

[cedar kiln]

@prisma bluff Has your question been resolved?

[cedar kiln]

@prisma bluff Has your question been resolved?

[paper crown]

.

[paper crown]

I can't help unless you answer my questions

I want to know what you understand from this question and what you have tried to solve it

why not?

that's okay. You can still try something

Start by asking yourself what the question ask you to do

yes

oh, is this question from an exam that is ongoing right now?

<@&268886789983436800>

[fallow steppe]

.close

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[mint raft]

Can someone please help me with these 🫡

[mint raft]

For question A I know m=4 is incorrect

But please help 🙏

[sly oxide]

What makes u think m=4 is incorrect

[mint raft]

Let me show you my steps

@sly oxide bro could you please help me with the other question please 🙏

Yesterday I asked for help and I got completely ignored -_-

[sly oxide]

Well u ady got m and c, so what's eqn of g(x)

[sly oxide]

And this is correct btw

[mint raft]

Oh so m is correct :)?

[sly oxide]

Ye

[mint raft]

Oh which one looks wrong oof go the 4 to you?

[sly oxide]

b should be wrong

See if u can find the mistake

[mint raft]

Okay let me show you what I said for b… in my calculations

…

[sly oxide]

Look at ur coordinates again

[mint raft]

(2;10)are the coordinates I’m working with

[sly oxide]

And what values of x and y did u sub in here

[mint raft]

Oh I what if I’m supposed to work with 2 then not the ten

Because wouldn’t the 10 be the x axis?

And 2 be the y axis?

Unless I’m mistaken

[sly oxide]

Wdym 10 be the x axis

As in x=10?

[mint raft]

No No like in terms of the coordinates…

Because when I was solving for b I used 10

I’m confused now 😭

Am I supposed to use the 10 or 2?

For solving for b?

[sly oxide]

Well ur suppose to use both

Coordinates r in the form (x,y), and in this case ur working with (2,10). That means x=2 and y=10

So ur suppose to sub x=2 and y=10

But u substituted x=1 and y=10 instead

[mint raft]

Oh I see now okay let me go work it out again 😂

Be right back

[sly oxide]

K

[mint raft]

Bro I’m getting two answers for b…

[sly oxide]

Ye, u shld

So u have to reject one of them

[mint raft]

Okay let me send my a pic of my calculations

[sly oxide]

Yep,

So which one do u think u will have to reject

[mint raft]

I’d say -3

[sly oxide]

And why

[mint raft]

Because of the way the graph is plotted it can’t be a negative

Wait-

[sly oxide]

Yep ur correct

Tho if the qn asks for an explanation, idt that'll be good enough

Either way, you'll just get b=3 as ur final ans

[mint raft]

Thanks man 🙏

[sly oxide]

Np

[mint raft]

Now can you help me with the rest 😭?

[sly oxide]

💀

[mint raft]

Like the other question

[sly oxide]

Aight I'll look at it

Well, like I said, u have m and c

So what's the equation of g(x)

[mint raft]

These are the ones I need help with will send now

[sly oxide]

Oh ok

[mint raft]

That one… blank 5

And this one please bro 💀😭🙏

[granite knoll]

what does "Blank 5" mean?

[mint raft]

It’s like a blank that I have to fill in

So it’s like a question

[granite knoll]

well just calculate g(-1) for part b)

[mint raft]

so can I please get help 😭🙏

[granite knoll]

and then calculate f(-1) for part c)

[mint raft]

For blank 5?

[sly oxide]

Ye

[mint raft]

So I’d literally just say f(-1) to get my answer for blank 5 y’all are confusing me

[sly oxide]

No

[mint raft]

Im not good with this part of math sorry 😭

[sly oxide]

For blank 5, u need the equation of g(x)

[granite knoll]

you substitute "-1" for x

[mint raft]

For blank 5?

[granite knoll]

if g(x) = 4x + 2, then you simply substitute a -1 anywhere you see "x"

[sly oxide]

Ye

[mint raft]

Oh okay let me go calculate now be right back

g(x)=mx+c though…

[granite knoll]

did you not find what m and c are?

[mint raft]

I did

Okay let me calculate again

I got -2

For blank 5 is that correct?

[sly oxide]

Ye

[granite knoll]

yes

for the next one, you just calculate f(-1)

[mint raft]

Okay let me go calculate again 🙏

[granite knoll]

and then for the last one, you just calculate the distance between the points you found in parts b) and c)

[mint raft]

I got 4 for blank 6

Is that correct?

[cedar kiln]

@mint raft Has your question been resolved?

[cedar kiln]

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[grim thistle]

help

[still vessel]

@grim thistle do you want the answer or the path

[grim thistle]

the path

[solemn torrent]

You dont give out the answers

[grim thistle]

i drew lines but im confused

[solemn torrent]

The explanation comes first

[still vessel]

you should use the parallel

[grim thistle]

i gave b the point (x,y)

and did the gradient from a

which equals to 1

[still vessel]

give it (a,a) since it passes through x=y

[grim thistle]

why

isnt a on that line too

[still vessel]

question says b lies on x=y

do you remember line equation, ax+b

[grim thistle]

mx+c?

[still vessel]

yeah

[grim thistle]

yes

[still vessel]

try to write ab and oc line's equation

do you see anything similar

[grim thistle]

they both parallel

oc= (t-6)/t x

[still vessel]

i just wanted to explain they are just same equations but ab is carried

also question says a and c is on same y level

[solemn torrent]

This may or may not help but given your information and a diagram you mightve expanded on, the shape is an isosceles trapezoid

[grim thistle]

oh i think i know

we know the distance

we plug b into the formula of distance

[still vessel]

try to solve

fully

[still vessel]

i just realized i solved wrong

[grim thistle]

how

[solemn torrent]

Shape OABC is an isosceles trapezium if closed line segment AC is parallel to the ordinate axis then segment AC is contained in the infinite straight line y=6

[cedar kiln]

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[limber snow]

am I doing this wrong?

[crimson sedge]

i dont think so

[limber snow]

so

would this b valid?

[crimson sedge]

i have not done many proofs,, cant really say

[limber snow]

ah shit aight

[granite knoll]

I don't really see how you're proving anything

[limber snow]

shit

for part a

can I say that

since 3, -4, 1/2 and 2/5 in Q and we can say F = Q and we know Q subset R and Q is a field

and also sqrt2 and pi not in Q

soomething like that for part a

[granite knoll]

but that doesn't say anything about F specifically, does it?

[limber snow]

sorry I forgot to add F = Q

[granite knoll]

you have to justify whether you can say F = Q

[limber snow]

wait if a set A is a subset of another set B which is a field can we say that set A is a field as well?

[granite knoll]

I don't think you can necessarily say that. you would have to show that A is a field

[limber snow]

so I show F is a subset of Q and Q is a subset of F

[granite knoll]

yes but presumably that wouldn't be necessary for part a), since part b) is to prove that Q is a subset of F

[limber snow]

mhm

so what would b the appropriate approach for part a

[granite knoll]

I'm not sure. what class is it for?

[limber snow]

math: Introduction to proofs

[sacred grail]

use the fact that F is a field to show how you can reach those numbers shown in part a

[limber snow]

wth how would I do that

reach those numbers..?

[sacred grail]

F is a field, so it must contain certain elements

[limber snow]

shit im just gonna wait on my TA to solve it later today im lowkey lost

ty doe

.close

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[marsh pond]

sorry for blur, i can repost if difficult to read

i found the determinant of this 5x5 matrix using row operations to get a triangle matrix, then multiplied by the diagonal

but would it have been faster to use the cofactor method? (can't remember exact name)

[tropic oxide]

oh god

matrices written not on graph paper

[marsh pond]

yep

i should have used GoodNotes for this one

next time..

[tropic oxide]

$\bmqty{3&0&0&2&0 \ 3&0&-1&0&0 \ 0&3&0&0&-3 \ 0&0&0&-3&-3 \ 0&-1&1&0&0}$

[wraith dagger]

Ann (glomed)

[tropic oxide]

is this your matrix?

[marsh pond]

yes

[tropic oxide]

you've got a lot of zeros here so yeah i'd go with cofactors

it appears there's 3 zeros in every row and col so from that alone theyre all equally good

[marsh pond]

i just went with the first row

[tropic oxide]

i would extract factors of ±3 from col 1, row 3 and row 4

to get $-27 \vmqty{1 & & & 2 & \ 1 & & -1 & & \ & 1 & & & -1 \ & & & 1 & 1 \ & -1 & 1 & &}$

[wraith dagger]

Ann (glomed)

[marsh pond]

i'm not familiar with this method, did you apply a row operation of some sort to get these numbers?

[tropic oxide]

yeah

two row operations and a column operation

[marsh pond]

1/3 R -> R?

[tropic oxide]

this is not a method

[marsh pond]

but why is 2 not 2/3?

[tropic oxide]

i scaled the first COLUMN down by 3.

[marsh pond]

you can do column operations?

[tropic oxide]

yes

[marsh pond]

how have i not learned this yet lol

[tropic oxide]

they have the same effect on the det as would row operations of the same type

[marsh pond]

i have only learned about row operations so far

[tropic oxide]

pretty sure column operations are not a huge stretch of the imagination

[marsh pond]

i will probably avoid this method for the midterm just because we haven't learned it yet

if you see on the bottom of the screenshot i posted, i'm just a bit confused where to go next?

i wrote the scalars to the 4x4 matrix

[tropic oxide]

you got it down to two 4×4s

[marsh pond]

yes

[tropic oxide]

right ok so like

just keep going with cofactor expansions

you have ideal rows/cols to do that on in each

[marsh pond]

do i need to find the determinant of those 4x4 first? or not yet

[tropic oxide]

(ideal = has all but one entries zero)

[tropic oxide]

wdym first

[marsh pond]

like do i now multiply 3 with the 4x4?

3 is the scalar

or should it be 3[det(A)]

[tropic oxide]

you could multiply it into one of the rows if you wanted but that just makes your life difficult

3 det(A) is 3 det(A)

youre overthinking again

[marsh pond]

so now i do the same thing to the 4x4 matrices?

make them into 3x3

it will be 3 * (4x4 scalar) * [3x3 matrix]?

[tropic oxide]

theres no such thing as a four-by-four scalar

but like. yes. you keep going.

[marsh pond]

i mean the scalar from the 4x4 matrix

[tropic oxide]

the point of cofactor expansion is to reduce the computation of an n×n determinant to that of several (n-1)×(n-1) determinants.

if you want to do that method in full, you do it recursively.

[marsh pond]

i just can't remember if I need to find the determinant as the last step? like once it reaches 2x2 i can finally do ac-bc and multiply all the scalars?

like i did this one too many times and got a huuuge number, i must have multiplied the scalars by accident during the process

[tropic oxide]

you're overthinking it still

[marsh pond]

i should wait to multiply scalars as the last step, is what i am saying?

not overthinking, just not understanding

[tropic oxide]

theres no protocol for what NEEDS to be done at this stage or what NEEDS to be done at that stage and You Shall Not Deviate Even One Bit From It And Absolute One Hundred Percent Adherence To The Sacred List Of Steps Is Required At All Times

like

you've got an arithmetic expression

which involves determinants sure

you can do intermediate simplification if you want

like you can strike out those added and subtracted zeros youve got

if you go from 3 * det(something 4×4) to 3 * 7 * det(something 3×3) you can combine the 3 * 7 into 21

like

its just an arithmetic expression like any other

if you arent sure what to do with large dets then just dont touch them

[marsh pond]

so this would be correct for the next step?

[tropic oxide]

appears correct, but suboptimal enough to warrant a warning.

for $\vmqty{3&0&-1&0 \ 0&3&0&-3 \ 0&0&0&-3 \ 0&-1&1&0}$ i would expand along the first COLUMN rather than the first row, because the first column has all entries zero except one. so you will make less work for yourself

[wraith dagger]

Ann (glomed)

[crystal raptor]

Also avid if you're signifying a determinant use straight vertical lines like above^ otherwise you're going to end up adding together what you think are matrices like earlier

[tropic oxide]

also that

[marsh pond]

again, i don't think we learned column operations, only row operations, so i think I will avoid this at least for the time being

[tropic oxide]

im not TALKING about column operations.

im talking about laplace expansion along something other than the top row.

[marsh pond]

I think I can simplify it into this

[tropic oxide]

ugh.

whatever.

i mean like if you want to do things the painful way then sure who am i to stop you

[marsh pond]

we didn't learn laplace expansion. i specifically asked the professor about this and she said we do not cover it in this class

[tropic oxide]

what

then how tf are you doing this

this IS laplace expansion youre doing???

[marsh pond]

so i will go with what we learned, for the midterm

[tropic oxide]

https://en.wikipedia.org/wiki/Laplace_expansion

like this IS what you are doing...

[marsh pond]

i asked her and she didn't even know what i was talking about, she said it's not covered in this course

[tropic oxide]

even if it isnt called laplace expansion

cofactor expansion whatever

that shit can be done along any row not just the first
or along any column

[marsh pond]

yes, but it's a different method than laplace expansion

[tropic oxide]

WHAT ARE YOU TALKIGN ABOUT

THEYRE TWO DIFF NAMES FOR THE SAME SHIT

[marsh pond]

^

[tropic oxide]

what is this called

in your class

what is the name for this method i just gave a screenshot of

[marsh pond]

ugh.. now i'm confused

https://en.wikipedia.org/wiki/Laplace_expansion

Laplace expansion, named after Pierre-Simon Laplace, also called cofactor expansion

wth, i asked about Laplace Expansion and the professor said "not covered int his course"

[tropic oxide]

and you ahvent answered my question

ok

let me retract that

and instead present you with two other questions

[marsh pond]

wait i haven't found the determinant yet

[tropic oxide]

on this screenshot, a method of calculating determinants is presented.

1. is this method, taken as-is, without any reference to names, covered in your course?
2. if the answer to question 1 is "YES", what is the name for this method as used in your course?

these are simple questions and i want you to answer them

[marsh pond]

i am just wondering which method would be easier to find determinant of this 5x5 matrix. so far i am finding triangle matrix with row operations to be easiest

[tropic oxide]

if you find the triangle matrix thing easiest then thats whats easiest for you

[marsh pond]

ok

[tropic oxide]

you can do it every time

[marsh pond]

no need to yell

[granite knoll]

make sure you keep track of how row ops affect the determinant

[tropic oxide]

i cant see any other way of like

pulling you out of the overthink fog

except trying to dispel it by saying things in a way that best approximates LOUD AND CLEAR

[marsh pond]

not understanding = overthinking

[marsh pond]

oh, so i got lucky with the sign?

i didn't keep track of that

i did 4 row operations

if it was 3 row operations you are saying my determinant would have been 27 instead of -27?

[tropic oxide]

i think theres a difference between those

fuck i give up

[granite knoll]

no, adding a multiple of one row to another doesn't change the determinant

[tropic oxide]

my attempts at explanation fall on deaf ears or on mirages in your perception of complexity that isnt there

my attempts to present methods i think are good run into "this isnt covered in the course" or "i think each problem has only one path forward, and branches, and the concept that more than one plan of action may be valid, terrify me on a fundamental level"

my attempt to just do the problem myself in full is 99.99% likely to be met with confusion at my planning, execution, or both

like i have a feeling ive seen this happen at least a couple dozen times
you come here with a problem
the problem has, say, 2 different solution methods, or its solution may branch into two or more slightly divergent methods

and you just hopelessly buridan's-ass yourself over it

obsessing over which one is The Right One

and as of recently thats been manifesting as "which one would y'all prefer" / "which one is faster" etc

as if youre trying to formulate some universal choice resolution algorithm that tells you in each and every such branch case which branch is "the correct one"

its frankly exhausting trying to get past that

[marsh pond]

what row operations do i need to look out for? the ones that would effect the determinant

[tropic oxide]

• adding a multiple of one row to another has no effect on the determinant
• scaling a row multiplies the determinant by the same factor
• swapping two rows multiplies the determinant by -1

this is note-take-able

in fact i am certain this should have come up in class

and i dont want to hear a "not covered"

[marsh pond]

Alright. So here as the very first step I swapped 2 rows:
R2 <-> R3
R2 <-> R5

(-1)(-1) = 1

But I got very lucky. I need to keep track of this next time.

.close

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[lost fractal]

I see it, but cant understand. Help me plz!

the problem:
If a third of the number is divided by its seventeenth part, the remainder will be 100. Find this number.

An answer:

(x/3)/(x/17)=17/3
x/3=5*x/17+100
x/3-5*x/17=100
2*x/51=100
x=2550

Why (x/3)/(x/17) is equal to 17/3?

[crude gale]

do you know how to do (1/2)/(2/3) manually?

[lost fractal]

well

I think so

[crude gale]

how would you do it manually

[lost fractal]

what do you mean?

[crude gale]

without a calculator, how would you solve (1/2)/(2/3)

[lost fractal]

(1/2)/(2/3)
(1/2)*(3/2)
// 1*2 and 3*2
3/4

[crude gale]

right because theres this rule where dividing by a fraction is same as multiplying by the reciprocal of that fraction

try to do the same thing to (x/3)/(x/17)

[lost fractal]

sec

aaaa

x/3 * 17/x = 17/3

[crude gale]

yep cause now the x in the denominator and numerator cancel out

[lost fractal]

🤦🏻‍♂️

Thanks. How do I close the thread?

[cedar kiln]

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[pine minnow]

In the first image (Şekil 1) the water level is the same with the bottom cone.
The same cylinder is being flipped upside down and it is shown as (Şekil 2)

What is the ratio between the heights of the both cones?

[pine minnow]

I tried to solve with namin cone heights as h1 and h2

I found the water volume

tho I couldn't do the same in the flipped version

oh yea btw in the second image, water level is h2+h1/2

thats the information from the question

[cedar kiln]

@pine minnow Has your question been resolved?

[umbral lodge]

@pine minnow

[pine minnow]

ypu

[umbral lodge]

do you know solid angles?

[pine minnow]

uh like 3 4 5 ?

[umbral lodge]

no not that

ok so w8

the answer is pretty complicated

@pine minnow

is this the answer?

[pine minnow]

no its with normal numbers

wait

these are the answers

[cedar kiln]

@pine minnow Has your question been resolved?

[pine minnow]

I solved it myself

[cedar kiln]

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[light pine]

in this question, in the mark scheme, they say at y = 0, t = 0

but surely isnt it y = 1000, at t = 0?

they later say this

but cant u just say y = 1000, at t = 0

[dawn junco]

you do you

you can choose your own reference altitude no problem

[light pine]

ight kl

[dawn junco]

you should get the same results in the end

[light pine]

wat would u do just curious

[dawn junco]

starting at y=1km sounds nicer visually

maybe the computations suck more idk

[light pine]

ight kl

i got the same answer anyway

thanks pal

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[shy monolith]

how to find interior, closure and boundary of sets made of functions?

[shy monolith]

you need to post your question in one of the available help forums

[leaden coral]

Oop arry wrong chat

[cerulean sail]

is there any restriction on a for the first set?

[shy monolith]

just that a<b

and that a and b are real

[cerulean sail]

Well for a start, for set A, there's a minimum value of a such that taking that or below, your life becomes somewhat easy...

Otherwise, it might be good to try sketching out those sets maybe?

[shy monolith]

yup i was thinking that for set A the interior would be all values greater than "a" and "a" is greater than -5

would that be correct?

[cerulean sail]

Well a can take any value you want, but "something special" happens when a is equal to or less than -5 sure

And given an a greater than -5, then there's also a "less than" to consider too

[shy monolith]

im still a bit confused, might need a minute to think this through

wait so would the interior be x^2-5>a AND y cannot be a?

imconfused because the interior doesnt include all points of the set (such as the outer limits)

but in this case what are those outer limit numbers?

[cerulean sail]

Ah, think I misunderstood what you meant by "all values" in the previous message: $y$ can be anything, there's no restriction on it, but yes you would want $x^2 - 5 > a$ (provided that as per before $a$ is strictly greater than -5), which gets you that $x > \sqrt{a + 5}$ or $x < -\sqrt{a + 5}$

[wraith dagger]

@cerulean sail

[shy monolith]

could you just leave it as x > sqrt(a+5)

or does the - sign indicate something?

[cerulean sail]

Yes, because

(hmmm I don't think I remember how to plot it)

But like you'll basically have a "gap" in the middle of the xy plane of the values that you can't take, but the other "sides" are fine

Let me see if I can get it plotted

Something like this here

[shy monolith]

Ohhhhh, ok i think it makes a bit more sense

comparing it makes sense

how would you do the boundary part for set A though?

[cerulean sail]

Boundary is pretty much what the dotted lines are in that plot

[shy monolith]

although here, isnt x^2-5>a reffering to y axis intervals so shouldnt the coloured part be horizontal?

[cerulean sail]

Hmm, what do you mean by y axis intervals?

x^2 - 5 > a restricts your x values

[shy monolith]

Like this

[cerulean sail]

Ahh, not quite, basically

the x and y values here aren't "linked"

So you can choose the y value to be whatever you want, but you have to have x such that x^2 - 5 >= a

[cerulean sail]

While a graph like that can help you find the values such that the inequality holds, it's slightly different to the restrictions on x and y in R^2

[shy monolith]

Wouldn't the graph be wrong though because X and Y are not linked?

[cerulean sail]

This one here is fine and correct

[shy monolith]

Yup mb i just thought you meant my one 😅

Any tips for set B though?

[cerulean sail]

Sketch it too

It would make life easier(!) cause you can make similar arguments

[shy monolith]

Alright I'll try it and see

Until then I'll free up this channel for others

.close

[cedar kiln]

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[warm tree]

stuck on this one

it's not -0.058

but i'm not sure how to find the answer with just the graph

[mighty drift]

Using the given tangent, it's easy to estimate that the derivative is just a bit below less (in absolute value) than -0.1

So -0.092 makes sense

[warm tree]

can you elaborate how you do that, I'm just starting this topic today

[mighty drift]

Do you know which dashed line is the tangent ?

[warm tree]

the one that fits against the solid line where c is

[mighty drift]

So the "middle" one

[warm tree]

yea

[mighty drift]

What is its slope ?

[warm tree]

14.625/5

[mighty drift]

That's the rate of change you want

[mighty drift]

How so ?

[warm tree]

wait, I think my brain is burnt out. can you explain how to get the slope

it's ΔY/ΔX

[mighty drift]

Yes

[warm tree]

so how do i do that though with one point

[mighty drift]

Find another point on the line

Perhaps approximate if you can't find an exact one

[warm tree]

so when I used point B i got -0.058

[mighty drift]

Is B on the line ?

[warm tree]

on the solid line?

[mighty drift]

Have an educated guess

[warm tree]

none of the points are on the dashed line but C

[mighty drift]

Find one

There aren't only 3 points in a plane

[warm tree]

that goes against a tangent line, but youre saying I can use an approximate point close to c and use ΔY/ΔX to get answer?

[mighty drift]

Doesn't have to be close to C

Can be anywhere on the tangent line

By virtue of it being a line, you'll get the same answer, so take a convenient point

[warm tree]

so if i use the point at 6, I get -0.095

ok

[mighty drift]

Sounds close to -0.092 to me

[warm tree]

yea kinda makes sense now

lol

alright thanks

.close

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[errant ether]

I kind of understand this but im having a little trouble finding the integral

[dire geode]

Show what you attempted

[errant ether]

I know it’s wrong

I just don’t know how to do it

[magic solar]

Find the interesection

Wait no shells method

[cedar kiln]

@errant ether Has your question been resolved?

[hollow trail]

we're looking at the area between y=10√x, the y-axis, and y=5, which is then rotated about the x-axis

[magic solar]

Not sure if this is the method you want but it would be the integral from 0 to the xintersection of 25 pi- the integral of the same thing but for the root equation squared

[cedar kiln]

@errant ether Has your question been resolved?

[errant ether]

nvm i solved it

can yall help me with another one

[cedar kiln]

@errant ether Has your question been resolved?

[cedar kiln]

@errant ether Has your question been resolved?

[bold vine]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

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[limber snow]

how tf would I do something like this

[short blade]

what is g

[bold vine]

That’s up for us to decide

[limber snow]

o apologies

lemme send

lmaoo no sorry

[violet flume]

g(x,y) = x-y

[limber snow]

here

thats g(x)

x + y / 2

[short blade]

x+y = x - y + y + y

use that then split the fraction

[limber snow]

$\frac{x - y + y + y}{2} \implies \frac{x - y}{2} + \frac{y + y}{2}$

[wraith dagger]

Calc II Victim

[limber snow]

and

[short blade]

what can you say about x-y and y + y

[limber snow]

I can rewrite x - y as

2k

so 2k/2 = k where k is an integer

right

[short blade]

yes

and what can you say about (y+y)/2

[limber snow]

2y /2

and y is an integer

so k + y in Z

[short blade]

next time you’re stumped like this just try to get the assumption into the expression you’re working with
like we found a way to put a -y in the expression this time

[limber snow]

tysm

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[short blade]

have you seen the graphs for each?

[violet flume]

theyre equivalent up to a translation so its really for convenience more than anything

[short blade]

other way around

and that should be your intuition for which to use

e.g. if you start at (0,0), probably easier to use sin

or more generally, if you start at a peak/trough, probably want to use cosine

but like jan niku said, they are equivalent up to a translation left/right

so you can use either

same thing

look at the behavior at x = 0

[cedar kiln]

@crimson sedge Has your question been resolved?

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[oblique prawn]

yo

[oblique prawn]

if you have a angle from standard position and has a negative measure do you take the smallest positive coterminal angle for finding the 6 trig values

[short blade]

well you could just use the angle given

do you have a specific example in mind

[oblique prawn]

so like im given a model of an angle theta with measure -(21pi)/4

i kept adding 2pi to get (3pi/4)

[short blade]

that works

[oblique prawn]

so like how would i find tan(theta) using the negative angle

[short blade]

well say the angle was -pi/4

[oblique prawn]

Ok

[short blade]

you could use the fact that cos(-x) = cos(x)

[oblique prawn]

should i convert it to degrees

[short blade]

and sin(-x) = -sin(x)

[oblique prawn]

so cos(theta)=cos(pi/4)

and sin(theta)=-sin(pi/4)

[short blade]

yes

and the other 4 you can get from that

[oblique prawn]

how can i do it graphically

[short blade]

what do you mean

[oblique prawn]

oh wait nvm i got it

.close

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[violet pagoda]

Stuck on 1: c, d

[violet pagoda]

This is my working. Answers are both wrong

Got c) YAHOO!

still stuck on d tho

[short blade]

@violet pagoda

you subtracted 4 from both sides on the last step of d

you should have added 4 to both sides

[violet pagoda]

lemme try that

ayo lets go

thank u

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[crimson sedge]

can any1 help me w/ part b

[crimson sedge]

i said it doesnt work bc it doesnt approximate to cube root of 9

[crimson delta]

note that 1-1/9 = 8/9 and that cuberoot(8)=2

[crimson sedge]

so what do i do

it said substituet x=1/9

so i assumed it meant to sub in 1/9 for x in part a

[crimson delta]

yes

but it didnt say that you would be done after that

[crimson sedge]

so what do i do after that

[crimson delta]

by subbing in x=1/9 you find an approximation for cuberoot(8/9)

[crimson sedge]

ok

[crimson delta]

then you rearrange to get an approx for cuberoot(9)

[crimson sedge]

wdym by rearrange

what does this symbol mean??

[crimson delta]

no clue. just classic teacher stuff. ask them

[crimson delta]

well you know, rearranging an equality

adding/subtracting/multiplying/... the same stuff on both sides

[crimson sedge]

i think im overthinking it

oh ok

i get it

[muted bear]

looks like gamma, but i dont think it means anything in context

[crimson sedge]

ok thanks guys

$\gamma$

[crimson sedge]

$\Gamma$

[crimson sedge]

is this the symbol for a photon?

[tropic oxide]

you might be better off asking the physics discord

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

.close

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[buoyant latch]

,rotate

[buoyant latch]

Have I done something stupid because the answer doesn’t have a 2 in it

In fact it should be a 3

Ok I’m really dumb I see the mistake

.close

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[grizzled spade]

hi, i have this questions:
prove/disprove: there is a f(z) entire function that holds:
|f(z)|>= |sin(z)| + |cos(z)| for every z in C

[grizzled spade]

I understand the proof but my question is:
how is it possible that cos z and sin z are the functions that their slope is the highest in the complex area

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[cedar kiln]

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[wild spear]

Given a sequence $(a_n){n\geq0}$ with $a_0=1221$ and $$a{n+1}=a_n+\frac{2}{a_n}$$ for every $n\geq0$. Prove that the sequence does not contain the square of a rational number.

[wraith dagger]

$nichoals$

[wild spear]

Will trying to prove this by contradiction work? So maybe assuming there exists p, q coprime such that a_n = (p/q)^2

[cedar kiln]

@wild spear Has your question been resolved?

[grizzled spade]

no

[cedar kiln]

@wild spear Has your question been resolved?

[wild spear]

<@&286206848099549185>

[wild spear]

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[brittle cipher]

So the area of the curved surface of a cylinder formula is PI2r*h and the measurements of mine are 5 and 12 and i get 376.99 for the answer and my teacher says it’s wrong

[fair mortar]

measurements of what

[brittle cipher]

cylinder

the curved surface

[fair mortar]

what is 5 and what is 12

[brittle cipher]

the formula is Pi times 2 times the radius ajd height

[brittle cipher]

radius is 5 height is 12

[fair mortar]

,calc 3.143512

[wraith dagger]

Result:

1318.8

[fair mortar]

wait n

,calc 23.155*12

[wraith dagger]

Result:

378

[brittle cipher]

see

[fair mortar]

seems right

[brittle cipher]

i got that too 378.99

but my teacher said it’s wrong

[fair mortar]

idk then

maybe he meant total surface area

[brittle cipher]

it didn’t say total

i’ll talk to him about it

when i get my test back

thank you

.close

[fair mortar]

yw

[cedar kiln]

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[faint tendon]

help

without lopithals rule

i did it with

[cedar kiln]

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[faint tendon]

<@&286206848099549185>

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[median valve]

calculate X

[median valve]

calculate x

[valid yacht]

do it!

[vague rapids]

you'll need a construction to solve this. Think what could it be?

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[eternal seal]

The last question part two, i understood the first part, which was that tan2x = square root 2, would be tan4x = square root 2. but the second part its just tan2x = square root 2. but why

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[shell arrow]

i don't understand your question

what exactly is your question

[shell arrow]

In the second part of (b), you want to find the argmin{3 - 2f(x)} = argmin{-2f(x)} = argmax{f(x)}. We know that the maximum of f(x)'s x-coordinate is a solution to tan(2x) = sqrt(2).

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[hollow tulip]

I am confused as to how this is possible, Can someone please give me a tip?:

If a function A(z) is divisible by (z-a) then find the remainder from dividing A(z) by (z-a)(z-b)

[hollow tulip]

I know how to do it If I know something about b, then I can find the remainder without dividing even

[royal loom]

what does it mean for A(z) to be divisible by (z-a)

[hollow tulip]

That the remainder is 0, meaning A(z)/(z-a) has a remainder of 0

[royal loom]

so, A(z)=k(z-a). Right?

(for k an integer)

[hollow tulip]

yes

no I mean

k can be anything

[foggy merlin]

Huh, not always an integer

[hollow tulip]

k can be another function

[royal loom]

o right

[foggy merlin]

a polynomial, not any other function

[hollow tulip]

Yes, my bad

[cedar kiln]

@hollow tulip Has your question been resolved?

[hollow tulip]

<@&286206848099549185> Can someone please push me in the right direction

[hollow tulip]

No luck huh?

[cedar kiln]

@hollow tulip Has your question been resolved?

[hollow tulip]

I will be waiting, and thank you if someone answered

[dire geode]

Not enough information given

Screenshot or take a picture of the original question

[hollow tulip]

I have translated the question from dutch and yes I also find that there is not enough information given. But this is the exact question.

Translated by deepl.com: "If A(z ) is divisible by z-a, determine the remainder of the division of A(z ) by (z a)(z b)."

What if we can just crazily assume the remainder of A(z)/(z-b) as a constant and just write this problem out as if you have information about two different divisions and their respective remainders.

That is the only possibility I can fathom

[stuck walrus]

eait

I'm confused

why can the remainder be another polynomial

I'm pretty sure it must be a constant?

@hollow tulip

deg remainder < deg divisor right?

and deg devisor is 1... so deg remainder must be 0? aka a constant?

[hollow tulip]

I am stupid, I was making things complicated for nothing

Thank you kind sir!

.close

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[plain badge]

this is my work for part A

for part A section b how can i write the Nullspace of the matrix

I wrote Null(M) = 2b + bx for some b in F but this would be the nullspace of T right

for Null(M) is there a specific way to write it

[granite knoll]

factor out b

[plain badge]

Null(M) = b(2+x) ?

[granite knoll]

no? the nullspace will be a span of some vector(s)

[plain badge]

okay that's waht i was confuse don

this is because we are taking Nullspace of the matrix right

if it was null of T it would be 2b + bx

[granite knoll]

it's the same thing as far as I know

[plain badge]

oh because for null of T I write it as 2b + bx

but it's my first time using nullspace of matrix

anyways

[granite knoll]

I guess in terms of the basis then yea

but you're asked for the nullspace of M

[plain badge]

how about Null(M) = {2b
b for some b in F}

since the first row corresponds to P0(F)

and the second to P1(F)

[granite knoll]

I would write it as $\text{span}\left{\begin{pmatrix} 2 \ 1 \ 0 \end{pmatrix} \right}$

[wraith dagger]

cwatson

[plain badge]

I see that would work too

thanks sir

[plain badge]

also one more question sir

for section c

when it asks for all polynomials that satisfy that equation

is it not the same as asking for nullspace of T

cause when (x+2)P'(X) = P(X)

[granite knoll]

yes that's how I'd interpret it

[plain badge]

it means (x+2)P'(X) - P(X) = 0 right

[granite knoll]

ya

[plain badge]

how does this look

[granite knoll]

that's fine but you did all the work in part b) already

[plain badge]

true

[cedar kiln]

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[crimson sedge]

yo

[crimson sedge]

I need help

so in khan academy

they claim that at a sharp turn, a function's differentiability is well none

because as you approach from either side, you get tangents with opposite signs

however, it's not the same for a curve

[faint dirge]

…. that is wrong

[crimson sedge]

damn fr

[faint dirge]

take x^2

Before x=0 its derivative 2x is negative

After x=0 its derivative 2x is positive

[fair geyser]

you don't get tangents with opposite signs

[crimson sedge]

yeah that is the point I am trying to make

[faint dirge]

It’s derivative at x=0 also exists, and is 0

[crimson sedge]

in a curve it isn't the case

but at a sharp turn according to them it is

[fair geyser]

there's a 0 in the middle

and like there's no 0 if there's a sharp turn

[faint dirge]

Oh you mean for like abs(x)?

|x| *

[crimson sedge]

sorry I don't quite understand

[fair geyser]

wel i don't know frankly

i suck at calc

[crimson sedge]

https://youtu.be/pwh1dK3vTkM?t=173

i added timestamp

if i wasn't clear about what i meant

[fair geyser]

a sharp turn has two wildly different tangents at the same time is what they mean

[crimson sedge]

wdym

[faint dirge]

Oh yeah by sharp turn they mean graphs like |x|

[fair geyser]

a curve you showed has one tangent equals 0

no conflict

[crimson sedge]

like at the same value of x

like take absolute(x)

wouldn't the tangent at the bottom just be 0

[obsidian coral]

That function doesn't have a unique tangent at x = 0

[faint dirge]

Which are not usually real continuous functions but piecewise function sets pretending to be one function

[crimson sedge]

why is that

[obsidian coral]

Because there are multiple tangents

[crimson sedge]

can't it just by y = 0

or well rather isn't it just y = 0

i don't see what makes this scenario different to x^2 for x = 0

[fair geyser]

yeah, not even 2 tangents

but at least 2

[obsidian coral]

Because for x^2, there is only 1 tangent point at x = 0

[civic eagle]

kinks in a curve are not differentiable

[obsidian coral]

|x| has multiple tangents at x = 0

[civic eagle]

in |x| case you can't have only one tangent, so its not differentiable.

[faint dirge]

|x| looks like a single function but it is not, its the combination of -x for x < 0 and x for x>0, 0 at x=0

A piecewise function.

[fair geyser]

these 4 are all tangents

[crimson sedge]

ah gotcha

[fair geyser]

i assume they meant specifically blue and green

[crimson sedge]

right but can't you do something similar for x^2

[civic eagle]

no

[fair geyser]

no, you just have one horizontal

[obsidian coral]

Because for x^2, there is only 1 tangent point at x = 0

[civic eagle]

because if you zoom in as close as you can to x^2 at x = 0, you'll see the line becomes flatter and flatter

eventually it'll look like a straight line

[faint dirge]

That’s a bad way to describe it

x^2 is s single continuous function

[civic eagle]

you can't do that trick with |x| or any kink in a curve

[crimson sedge]

yeah but what I don't understand is why that has one tangent and absolute x has infinite

[faint dirge]

As stated here |x| is two lines and a point pretending to be a single continuous function

[obsidian coral]

These are some of the tangents for |x|

[civic eagle]

it is continuous though

[crimson sedge]

okay I see, so are all sharp turns, piecewise functions

[faint dirge]

And that disembodied point at 0 is tangent to any line passing through it