help-10

[obtuse pebble]

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[fickle surge]

why do the bounds change from 1 0

to pi/4 0?

[unreal musk]

Was the substitution u=arctan(x)? If so then when x=0, u=0, and when x=1, u=pi/4

Your original limits correspond to x values, so you need to change them to u values once you substitute

[fickle surge]

that always is the casey

?

so unless i’m using trig stuff then would my bounds usually stay the same

another question

[unreal musk]

No, they “usually” won’t be the same thing most of the time

[fickle surge]

for a question like this

i’m aware i have to split it

[unreal musk]

You’d have to check the limits in a similar manner ("when x='something', u=...")

[fickle surge]

but why should i split it?

i’ve been trying to find out how to easily find out if i should split an integral

ik i should do this

but idk how to easily spot it

[unreal musk]

Ah, I get you - basically, because both of those are "easy" to integrate

Not sure if I can say in a simple way how to easily spot it

[fickle surge]

so just split it if it’s easier to that way

i can do this for any function times another function?

[unreal musk]

But for me, looking at it, my thought process would be something like "hmm, I would really like the top to be 'almost' the derivative of the bottom, as I instantly know how to deal with that"

"but then it isn't, because of that +1, so what happens if I consider them 'separately'?"

"...Oh, that's nice actually, I know how to find the integral of both of them when I split the sums up, that makes my life so much better!"

[unreal musk]

Errmmm, not sure what you mean here?

Could you elaborate?

[fickle surge]

like if i had 2 different things multiplying each other

could i split those 2 different things and integrate them separately?

[unreal musk]

Ah, so do you mean like
$$
\int f(x) g(x) dx = \left( \int f(x) dx \right) \left( \int g(x) dx \right)
$$
for functions $f,g$?

[warm shale]

chartbit

[fickle surge]

yes

[unreal musk]

No, you can't, for one think of the "integration by parts" rule

[fickle surge]

we never learned that haha

he kinda just introduced it

as saying u could do this

[unreal musk]

But e.g. taking $f(x)=1$ and $g(x)=1$ gives
$$
\int f(x) g(x) dx = \int 1 dx = x
$$
yet
$$
\left( \int f(x) dx \right) \left( \int g(x) dx \right) = \left( \int 1 dx \right) \left( \int 1 dx \right) = x \cdot x = x^2
$$
with the latter not differentiating to 1.

[warm shale]

chartbit

[unreal musk]

I think maybe you might be thinking of whether you can do e.g.
$$
\frac{f(x) + g(x)}{h(x)} = \frac{f(x)}{h(x)} + \frac{g(x)}{h(x)}
$$
(you can), and
$$
\int f(x) + g(x) dx = \int f(x) dx + \int g(x) dx
$$
(you can also do that too!)

[warm shale]

chartbit

[obtuse pebble]

@fickle surge Has your question been resolved?

[fickle surge]

ahh so question

would it be possible to take the integral of both functions if they were multiplied

and then multiply them together after?

hey i has one more question

am i doing this right so far?

[unreal musk]

Could you write what you're thinking in terms of a function, just so I'm clear with what you're thinking of doing?

[fickle surge]

also now that i’m reading it i have one maybe dumb question

i’m not 100% sure that i’m right here but

if the final answers 9 couldnt i have just don’t -3*3 and done something else i’m not thinking of to get this answer instead of all that

[unreal musk]

Be careful of writing that integral sign there, as you already did the integration by then, but otherwise it looks right to me up until that point

[fickle surge]

alr sick

sorry for all the questions i’m just doing a final bit of review before my final

[unreal musk]

Well, the results from each are likely not the same, the top one isn't [necessarily] equal to the bottom one

[fickle surge]

for area between 2 curves

ik i usually set them equal to each other or something like that

but i’ve been stumped on this question

oops

would i set the y= to each other?

Or just x^2 = 25

and then since x =5 i would integrate x^2 + 25 with the bounds of 0 to 5?

[unreal musk]

Yea the result you get is right in the end

[unreal musk]

This one

[fickle surge]

ah ok thank u

[unreal musk]

No wait, wrong pic

[fickle surge]

for the latest question i just put (area between two curves)

[unreal musk]

The answer of the whole question is 9, but you could have written that a bit better

[fickle surge]

yea sorry haha just trying to speedrun them

[unreal musk]

Kinda - your "top curve" (well, it's a line) is y=25, and the "bottom curve" is y=x^2, so do the integral of 25 - x^2

Between those limits

[fickle surge]

so the integral would be instead this?

well would i just add a - sign to it

sorry i think it is yea

i had 2 more questions for now and i promise i’ll be out of ur hair for at least 5 minutes @unreal musk haha

i already have the correct answer allegedly for them just was confused on how they got it

[unreal musk]

Haha fair enough, send them through and I'll look at them!

[fickle surge]

ok so

it’s for part b and c for this

for part b the answer is

but im confused why it’s just squared

i get its a square so the sides are equal but i thought the shape itself wasn’t a square so why would that effect the answer

or is it because the “slice” is the shape of a square

[unreal musk]

Hmmm personally I'm not too sure about the last two parts, sorry

[fickle surge]

ahh ok

[unreal musk]

Haven't seen a layout like that before and I can't visualise in my head what they're looking for

[fickle surge]

i think they drew a different part too

does this help?

[unreal musk]

Hmmm I think I kind of see what they're doing, it looks similar to those solids of revolution integrals, but I don't want to mislead you and say something incorrect!

[fickle surge]

all good

for my final question

r u good with integrals and volume stuff then?

[unreal musk]

Errmm not like the ones there tbh

[fickle surge]

what about cones?

with r similar triangles and such

[unreal musk]

Hmm not sure 🤔 but I have to go unfortunately!

[fickle surge]

all good

[unreal musk]

Have to go out to get things

[fickle surge]

thanks for the help boss

[unreal musk]

Sorry 😭

[fickle surge]

just wondering for later

r u familiar with Work stuff

like W = Fd

[unreal musk]

It’s been ages since I’ve done anything like that myself, but I’m sure if you post it here, someone who is might help you!

It’d be like relearning it 😂

[obtuse pebble]

@fickle surge Has your question been resolved?

[obtuse pebble]

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[timid silo]

hello can you help out with this question: finding the value of x. f(x) = 10/ sqrt of x

[lean adder]

Unclear instructions? Or maybe i just dont know how to solve this XD

[timid silo]

i will take a picture

[thick oracle]

You can't find x lol

X can be any number

Any number>0

[lean adder]

yeah even <0 if you work with complex numbers

[thick oracle]

Translate

[timid silo]

It is basically saying you should answer with two decimals

[lean adder]

but there is no exact answer to this, but if they want to get the solution for x == 10 / sqrt(x) then its cbrt(100) ==(4.64)

[thick oracle]

no that's not the function

[lean adder]

yeah ik

[pure cobalt]

there must be a domain

else it shouldnt ask the value

of fn

[timid silo]

what do you mean with cbrt

[lean adder]

cuberoot

[thick oracle]

Yeah but what you wrote has nothing to do with the question

[lean adder]

i mean the question is unanswerable XD

[thick oracle]

Yes

And that's what he should write

[timid silo]

thanks for trying to help out.

.close

[obtuse pebble]

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[plucky pilot]

Hello im having a hard time simplifying algebraic expressions

[plucky pilot]

I have a final exam tomorrow

This is the only lesson im having a problem with

This is kinda urgent

Since its like 12 am for me

[silk flax]

you need to give us examples of problems you need help with so that we can help

[plucky pilot]

Sure

[silk flax]

and its probably more import that you sleep than cram last minute content but i can try to help

[plucky pilot]

At this point i dont care about sleep

I just wanna get a good grade

[silk flax]

okay so the thing with simplifying these expressions is you just need to combine the terms that look the same

often it is helpful to circle and underline them as you go

so if we look at part a

we have y^2 terms, y terms, and m^4 terms

so lets first combine the y^2 terms
we have 3y^2 - y^2 which is 2y^2

then lets combine the y terms
we have 2y + 8y which is 10y

and lastly we only have one m^4 term

so our final answer is 2y^2 + 10y + 7m^4

does that make sense?

[plucky pilot]

Yes

Thank you so much

I couldnt find something like this on youtube, khan academy

[silk flax]

oh okay

no problem

[torn wave]

Khan academy has a few videos on combining like terms

[obtuse pebble]

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[plucky pilot]

.reopen

[obtuse pebble]

✅

[plucky pilot]

Hey im so sorry

Just 1 more question

I dont know why but everytime i divide i get 1 number more

And i dont know how to solve it

[obtuse pebble]

@plucky pilot Has your question been resolved?

[obtuse pebble]

@plucky pilot Has your question been resolved?

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[granite zinc]

Hi, does anyone know how I would begin to solve this? It is pre-calculus, matrixes. I am being asked to make a matrix representing the system of equations.

[obtuse pebble]

@granite zinc Has your question been resolved?

[tardy epoch]

https://www.khanacademy.org/math/algebra-home/alg-matrices/alg-solving-equations-with-inverse-matrices/v/matrix-equations-systems

[obtuse pebble]

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[chrome arrow]

can someone help me understand how to do this

Determine how much is in the account on the basis of the indicated compounding after the specified years have passed; P is the initial principal, and r is the annual rate given as a percent. (Round your answer to the nearest cent.)
P = $6000, r = 4.3%, compounded annually for 1 year.

[chrome arrow]

this is what it says if you can’t read from the picture

[obtuse pebble]

@chrome arrow Has your question been resolved?

[chrome arrow]

<@&286206848099549185>

[obtuse pebble]

@chrome arrow Has your question been resolved?

[obtuse pebble]

@chrome arrow Has your question been resolved?

[obtuse pebble]

@chrome arrow Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[astral garnet]

how do do this problem

[obtuse pebble]

@astral garnet Has your question been resolved?

[high lily]

you have two points the tangent line passes through

f'(8) will be the slope of the tangent line at the specified point (8,10) on the curve

and can be determined from slope formula

[obtuse pebble]

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[obtuse pebble]

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[hot dock]

How would i find primitive roots of unity in a certain space?

[hot dock]

for example in Z_29

is it eulers totient phi(phi(29))?

but that would be phi(28) = 7^1 2^2 = 6(2)=12 nvm i answered my question

.close

[obtuse pebble]

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[hot dock]

.reopen

ok

[obtuse pebble]

✅

[hot dock]

So how would i find

$log_3(5)\text{ in}\mathbb{Z}_7$

[warm shale]

Splash

[zenith raft]

primitive roots of unity?

[hot dock]

yes

[zenith raft]

hm ok

[hot dock]

then also is

$\sqrt{3}\text{ in }\mathbb{Z}_11$

[warm shale]

Splash

[hot dock]

is this {5,6}

since 3^{(11+1)/4}=3^3=3^1 3^2 = 3* 9 = 27 mod 11 = 5 mod 11, but the inverse is also true so it would be 5,-5 mod 11

[kind hawk]

well if you want sqrt to be a function then you have to choose one of them, it can't be both

over the reals there is an easy canonical choice (take the positive one) but over Z_n there isn't

especially if you wanna have rules like sqrt(ab) = sqrt(a) sqrt(b)

[hot dock]

what

i dont see how it cant be both

[kind hawk]

"if you want sqrt to be a function"

a function has one output

[hot dock]

well im not looking for a function

but yeah, sqrt x = something mod p

[kind hawk]

then maybe your notation is not the best cause the sqrt notation as given is usually only used for functions

maybe better to write ${x: x^2=3 \mod 11} = {5, 6}$

[warm shale]

Denascite

[hot dock]

good point

on my test last month it was literally written as sqrt 3 in z_11

thats why i wrote it as such

[obtuse pebble]

@hot dock Has your question been resolved?

[obtuse pebble]

@hot dock Has your question been resolved?

[obtuse pebble]

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[steel marsh]

Can one of you guys help explain this solution to me? I don’t get how my prof got the two orbits or why it uses the points (1,1) & (1,2) for the stabilizers.

[stable rain]

#❓how-to-get-help

[timid silo]

#❓how-to-get-help

[stable rain]

...

[steel marsh]

What is happening

[wild swallow]

someone probably posted a q

[steel marsh]

I saw that lol but what is this interaction

[wild swallow]

[steel marsh]

Snow are you familiar with group theory?

[wild swallow]

maybe

[steel marsh]

I’m having trouble understanding orbits and stabilizers. Or rather, how to calculate them for different groups

[steel marsh]

Re:

[wild swallow]

well

think about what the orbits could possibly be

[wild swallow]

so both elements must remain the same

but π could send 1 anywhere

[steel marsh]

Ok that makes sense for the first orbit actually

And using the stabilizer to find the order is clear following that as well

But what about the second orbit?

[wild swallow]

well say you have $x = \tuple {1, 2}$

[steel marsh]

This is an arbitrary point correct?

[wild swallow]

well im just picking it because 1 isnt 2

[steel marsh]

So these are the only distinct orbits?

[wild swallow]

yes because they partition X

[steel marsh]

I see now

So for calculating the stabilizer of the second orbit, why is Stab(1,2) congruent to S_m-2 ?

Is this in some theorem for stabilizers?

[steel marsh]

Hmmmm

[steel marsh]

My apologies but I’m lost with that statement

Is it possible to explain in a different way? Or perhaps provide an example so I can see it hands-on?

[wild swallow]

like

if you fix the first 2 elements of the permutation

then you are free to permute the remaining n-2

that gives you (n-2)! possibilities

[steel marsh]

Oh okay okay I see

No that makes sense now

I was imagining that the value of j was impacting the remaining permutations and got confused

[wild swallow]

oh

thats a typo

[wild swallow]

here i meant to type σ(i-2)

[steel marsh]

Ahhhhh okay okay

Thank you for your help

.close

[obtuse pebble]

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[timid silo]

Pentagon $\Rightarrow$ $5$ sides.\
Perimeter $\Rightarrow$ sum of side lengths.\
$\frac{\textup{Perimeter}}{\textup{Number of sides}}\Rightarrow$ side length.\
Side length $\Rightarrow$ diameter of the semicircles.

I think you've got it.

Jeez, it's broken.

[warm shale]

3317

[trail cloak]

Hmm

[obtuse pebble]

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[lost tree]

and why are u stuck

[drifting wraith]

it means this

that would be the largest portion

[lost tree]

yeah

u know the total area of the square

u can use that to find the length of a side

192 is the area of a square

196* actually

if the square has an area of 196, then what must the length of a side of the square be?

u must find that before u find the answer to ur problem

yes

and what is that equal to?

yes

so if 14 is the side length of the square

[lost tree]

what is the diameter of the circle u just drew

what is the diameter equal to?

if u know this value, u know the radius of the circle

and hence its area

can u mark the diameter of the circle on ur picture?

and also mark the length of a side of the square on ur picture

?

wdym theres nothing

[obtuse pebble]

@timid silo Has your question been resolved?

[lost tree]

hey sorry my computer just started lagging

im back

so i guess u get it now?

[obtuse pebble]

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[ancient acorn]

I want an algebraic proof of the FTA. The one I've been taught uses a bit of analysis because it needs to prove odd degree polynomials have at least one real root using the Intermediate Value Theorem.

Is there an algebraic way to prove an odd degree polynomial has at least a single real or complex root without the FTA and without using analysis?

[ancient acorn]

Wikipedia says you need to but I'm not sure of how correct that is

[royal basin]

you cannot divorce yourself entirely from analysis or topology for FTA.

[ancient acorn]

Aw

[ancient acorn]

Any good resources on this?

Like, is it that it's not been found, or is there some metamathematical hard proof

[royal basin]

i do not know of any "metamathematical hard proof", no.

[ancient acorn]

Alright, thanks

.close

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[azure gazelle]

@ancient acornyou could read this https://arxiv.org/pdf/1504.05609.pdf

[azure gazelle]

.close

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[silver ore]

it will be e^-3t

the eigenvector is $\begin{bmatrix}-1\ 1\end{bmatrix}$

[warm shale]

arrow891

[silver ore]

it says that the answer is going to be

but i don't see how

not funny

[obtuse pebble]

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[silver ore]

<@&286206848099549185>

.close

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[agile scaffold]

how would you integrate that as the 8x will be 8x^-1 wont it? just started learning integration

[cedar lichen]

Using the power rule, you do get division by 0, yes

However, the integral of 1/x is a famous one

∫ 1/x dx = ln|x| + C

[agile scaffold]

so 8ln|x|?

[cedar lichen]

Yes

[obtuse pebble]

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[true lily]

hello, how can I solve this?

[patent jetty]

Similarity

[true lily]

this is what I tried so far

[patent jetty]

Or trig

[true lily]

yes

[patent jetty]

Correct

Cross multiply

[true lily]

I don't know how to do that

[patent jetty]

7(x+8) = 13x

You might wanna watch a lecture series on basic math as it will help in long term

[patent jetty]

Just multiply it like a CROSS '×'

[true lily]

do I really need to know how to cross multiply in order to solve this?

I never used that method

[patent jetty]

No

Do u know trigonometry

Or like just try trial and error

[patent jetty]

It's best if u cross multiply... It is a very simple thing to do

[true lily]

I tried solving using algebra and got 4/3 but that was wrong

[patent jetty]

What did u do

[true lily]

ok could you explain to me how I can cross multiply?

[patent jetty]

a c
_ = _ is ad=bc
b d

[patent jetty]

@true lily this is it

[true lily]

ok so should I do x + 8 * 7 and x times 13

[patent jetty]

Yup

[true lily]

I don't understand it. what should I do after i multiply

[patent jetty]

Solve for x

[true lily]

7x + 56

how do I solve for x

[patent jetty]

Ok so that's on the left side

And there's 13x

[patent jetty]

x times 13 is on the right

[true lily]

left side? you never explained how I should set it up

could you explain cross multiplying

[patent jetty]

Look at this setup

[true lily]

oh okay so I should make it an equation

[patent jetty]

Ye

[obtuse pebble]

@true lily Has your question been resolved?

[obtuse pebble]

@true lily Has your question been resolved?

[true lily]

so I got my answer to x = 56/6

how can I solve from there?

<@&286206848099549185> could someone help me with this please

[slim basalt]

yeah x = 56/6 or 28/3

[leaden lion]

This is the correct answer

Just reduce the fraction

[true lily]

okay thanks for the help all

.close

[obtuse pebble]

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[obtuse pebble]

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[wooden cipher]

no dx 💀

[thick oracle]

Lol

[final hornet]

no dx no proof

[copper spade]

No proof no question

[slender nebula]

heres what I did

I get there's a mistake in the question, it wd b very helpful if u cd overlook that and assume that the dx is there

[final hornet]

how did you get the first equality

sin becomes cos ?

[wooden cipher]

cofunction

[slender nebula]

if you replace x by pi/2 - x

its a property of definite integrals

[final hornet]

ah I see

[slender nebula]

or if u see the sin and cos graphs

[final hornet]

yeah true

[slender nebula]

between 0 and pi/2

theyre the same, just in reverse

[final hornet]

yep

change of variable u = sinx does it lead somewhere?

[lost tree]

i guess integration by parts works

[slender nebula]

maybe

[slender nebula]

it a function of a function

[final hornet]

yeah probably

[slender nebula]

by parts works for product of functions right?

[lost tree]

$\int_0^\frac{\pi}{2}\log\left(\sin x\right),dx=\int_0^\frac{\pi}{2}1\times\log\left(\sin x\right),dx$

[warm shale]

SilverSoldier

[lost tree]

treat 1 as a function

[slender nebula]

ah ok lemme try

[final hornet]

yeah this shoud be straightforward

we get rid of the log function

[lost tree]

yeah ig ud get to integrate x*tan x

x*cot x rather

[slender nebula]

is that correct?

[lost tree]

whers the 2x coming from

[slender nebula]

log(sin 2x)

[lost tree]

oh u r continuing from here

[final hornet]

x log(sin2x) should be evaluated at pi/2

[lost tree]

u cud have directly gone from here

[slender nebula]

ahh right

Ill try that

[lost tree]

okay im not sure if this is gonna work

idk how to integrate xcotx

[slender nebula]

so by parts wont work?

[lost tree]

wait

idk if u can find out a way to integrate xcotx u can use it

[rugged shoal]

Go by parts it'll work

[lost tree]

again by parts?

wont that lead us back to where we started

[slender nebula]

yep doesnt work

[rugged shoal]

I don't know but to integrate xcotx you can use it

[lost tree]

we wanted to integrate ln(sinx)

[slender nebula]

I just tried it

[slender nebula]

yeah thats what v get back

[lost tree]

and we turned it to an integral of xcotx

[rugged shoal]

Oh

[lost tree]

maybe try using the pi/2-x thing here

for the xcotx

[slender nebula]

is there any way of converting the first integral to the question

[slender nebula]

cuz then it cd b expressed as I

and the second integral is half of the result v want

in the last step

[slender nebula]

doing that

.close

[obtuse pebble]

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[rugged shoal]

Hey I got it

[obtuse pebble]

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[slender nebula]

woah can u explain it

[rugged shoal]

It's in my notes lol

I'll send you a pic

[slender nebula]

thanks

[timid silo]

Yoo

[rugged shoal]

@slender nebula

[lost tree]

oooh

[slender nebula]

ahh understood, thank you very much

[lost tree]

again sorry for taking ur time

sparkle

[slender nebula]

no probs, thnx for trying

[lost tree]

maybe im not the best person to ask for integration help lol

[rugged shoal]

I forgot too but I knew I did something like that earlier

[slender nebula]

dw these questions arent too simple

at least for me

[slender nebula]

i see

thanks again

[rugged shoal]

your welcome

[slender nebula]

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[old oak]

.reopen

<@&286206848099549185>

[old oak]

it’s sending

can u plz help

<@&286206848099549185>

[obtuse pebble]

@old oak Has your question been resolved?

[old oak]

<@&286206848099549185>

[obtuse pebble]

@old oak Has your question been resolved?

[obtuse pebble]

@old oak Has your question been resolved?

[obtuse pebble]

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[hexed lance]

Looking for the name of this notation. I know I have seen it before in my math class but it's escaping me.

[tardy epoch]

can you show the whole page

looks like "angle of negative arctan of omega"

[obtuse pebble]

@hexed lance Has your question been resolved?

[hexed lance]

this would be it, I just plug in the values for omega and calculate the tan inverse which gives me the angle. And the angle symbol doesn't mean anything additional happens to it right?

[tardy epoch]

maybe? you should look up the definition of that symbol in your book

[hexed lance]

umm how do I say it? It's an engineering course and the textbook is very limited with definitions? here is the image of the whole example given (no comments or text is included)

[tardy epoch]

yea that should be fine

[obtuse pebble]

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[pine onyx]

anyone know what test to find if this series is diverging or converging?

I know I can't use limit or integral test for this since it has negative values

Divergence test is also hard idk how to take the limit of this

[tardy epoch]

limit test

also called divergence test. it is the way to do this problem

[pine onyx]

Idk how to take the limit of this🥲

[tardy epoch]

neither term converges

periodic functions with nonzero amplitude don't converge

[pine onyx]

How can I show that?

[tardy epoch]

if you haven't done eps-N proofs, maybe you have to do a different test.

,w int sin(x) tan^2(x)

[warm shale]

Results provided by WolframAlpha

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

Requested by riemann#2529

[tardy epoch]

ah haha integral test seems easier

[novel knoll]

You can’t use that. Check assumptions

[tardy epoch]

ty scapeprof

root test

[obtuse pebble]

@pine onyx Has your question been resolved?

[pine onyx]

Im having a hard time understanding the epsilon proof🥲

[tardy epoch]

.

don't suppose ratio test is any better?

[obtuse pebble]

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[pine onyx]

Ill try ratio test

[pine onyx]

Yeah it's too hard😭

I give up

[brisk matrix]

the first test you can always conduct is that $\lim_{n\to\infty}a_n\ne 0\implies\sum_{i=1}^\infty a_n$ diverges

[warm shale]

maximo

[brisk matrix]

i don’t know if you really need an epsilon delta proof to claim tan^2n * sinn does not converge to 0, assuming this is a calc class

[obtuse pebble]

@pine onyx Has your question been resolved?

[pine onyx]

Oh how do I show that the given series diverge tho🥲

[brisk matrix]

i just explained that

[pine onyx]

Oh idk how to get the limit of this function

[obtuse pebble]

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[pine onyx]

. reopen

[obtuse pebble]

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[woeful epoch]

need help on this theory

[obtuse pebble]

@woeful epoch Has your question been resolved?

[woeful epoch]

<@&286206848099549185>

[obtuse pebble]

@woeful epoch Has your question been resolved?

[woeful epoch]

<@&286206848099549185>

[obtuse pebble]

@woeful epoch Has your question been resolved?

[woeful epoch]

.close

[obtuse pebble]

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[honest snow]

mines just a basic question since i forgot about it and my final is tomorrow...

when doing u-substitution, what would x^u become? assuming that it hasn't been integrated yet

[honest snow]

example like this

[dark mango]

When you u-sub you want your integral in terms of u

not x and u

[honest snow]

oh yeah sorry i was just trying to make an example

what if it was something like pi^u

or is that just not a possible outcome

[dark mango]

Do you have an example question?

[honest snow]

no, im just trying to refresh my brain on some more uncommon things just in-case

[dark mango]

the integral of pi^u would be like pi^u/ln(pi) + C but I don't think thats what you mean

[honest snow]

so if would be something like this

i just treat u substitution the same as i would treat it like (number)^x

?

[dark mango]

Yes but after you integrate sub back in x

Or change the bounds of your integral

and leave it as u

[honest snow]

isnt that only for definite integrals that i change the bounds?

[dark mango]

Yeah

[honest snow]

yeah ok

thanks, just wanted to make sure

.close

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[obtuse pebble]

@safe plank Has your question been resolved?

[safe plank]

<@&286206848099549185>

[solar hornet]

@safe plank what's your question?

[obtuse pebble]

@safe plank Has your question been resolved?

[safe plank]

What did I do wrong?

[lost tree]

theres a 9

that u have missed

7 i mean

ur 7 looks like a 9

[safe plank]

Oh thanks

.close

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[dull delta]

Hey, how do I go about solving this problem?

[obtuse pebble]

@dull delta Has your question been resolved?

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[winter sun]

Why does an odd degree function have to have at least 1 zero

[gilded needle]

you mean polynomial ?

[winter sun]

Yea

[gilded needle]

look at the limit as x-> infinity and the limit as x-> -infinity

what can you say about those limits?

[winter sun]

They don’t exist

[gilded needle]

be more specific

[hybrid gull]

Because f(x) = -f(-x), and continuity means that f(x) has to be both + and -

[gilded needle]

ah that's nicer

[winter sun]

Uh

[zenith raft]

eh odd degree polynomial functions don't need to be odd functions

[gilded needle]

wait

[winter sun]

I think I meant odd degree though

Not odd function

[winter sun]

What were you alluding to here

[gilded needle]

as x-> infinity, f either goes to +infinity or to -infinity

as x -> -infinity, the opposite happens

so in particular there's some x with f(x) > 0 and some y with f(y) < 0

[winter sun]

So I understand thus

[gilded needle]

now use continuity and the IVT

[winter sun]

But

Why does this happen

That the opposite inifinity limit is the opposite sign

Like what’s the reasoning behind that

[gilded needle]

it's as follows

[winter sun]

I understand what u mean but I don’t; really see the logic

Yet

[gilded needle]

the leading term is of the form cx^n where n is odd

and c is nonzero

now if c is positive, then for large x, we have x^n positive and c positive, so cx^n positive

and for large negative x, we have x^n negative and c positive, so cx^n negative

if c is negative, the opposite occurs

[winter sun]

Ok lemme draw this out

[gilded needle]

where we use oddness of n is the fact that (-1)^n = -1 when n is odd

[winter sun]

Wouldn’t it be negative c tho

Since that’s the coefficient

[gilded needle]

if c is negative then f(x) -> -inf as x-> inf and f(x) -> inf as x -> -inf

if c is positive then the opposite occurs

c is what it is, it can be either positive or negative

for example if f(x) = x^3 then c is 1

if f(x) = -x^3 then c is -1

c is the coefficient of the leading term

[winter sun]

He

He

But what part of -x^3 gaur3ntees that as x->+infinity that y does the same

Wait sorry ignore that

What part of -x^3 gaurentees that as x approaches +infinity, y approaches -infnininty

[gilded needle]

what's y?

oh you mean y = -x^3

if x is any positive number then x^3 is positive, so -x^3 is negative

and the absolute value of x^3 goes to infinity as x goes to infinity

(because, for example, x^3 > x when x > 1)

[obtuse pebble]

@winter sun Has your question been resolved?

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[lone snow]

Why is it decreasing even if I plug the numbers and they are positive?

For example, plugging -1 yields a positive number

Why would it be decreasing then?

(on interval -sqrt(2), 0)

[timid silo]

a function can both be decreasing and positive?

[lone snow]

How is that the case?

Our professor didnt explain it that way

[timid silo]

if a function is decreasing in an interval, then as x increases, the function decreases - so for any interval where the gradient is negative, the function is decreasing

[lone snow]

Makes sense

Thank you

.close

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[edgy heron]

https://tenor.com/view/rickroll-roll-rick-never-gonna-give-you-up-never-gonna-gif-22954713

[broken bobcat]

could someone walk me through the step for factoring $27x^{2}-30x+3$

[warm shale]

Astassa

[wild swallow]

.close

[obtuse pebble]

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[pastel galleon]

in this problem, why $alpha$ must lie in the interval [n/k,(n+1)/k)?

[warm shale]

Pacomodo

[pastel galleon]

this problem is relavant to dirichlet's theorem, but i don't know how to apply it in this problem

[inland mulch]

That is wrong

[pastel galleon]

wrong for what? first image? or second image?

[inland mulch]

The interval they gave is wrong

[pastel galleon]

you mean this solution is wrong?

[inland mulch]

Also uh

Ah I should do the question to know if it's true or false

It sounds wrong too

Say alpha is 10

And n is 1

There is no integer k that works

[pastel galleon]

this problem is from elementary number theory 6th edition, rosen

from ch 1-1, 33/

[inland mulch]

Ah is it (1/2) * k or 1/(2k)

[pastel galleon]

i am confused too

in solution, i think 1/(2k) is right, but then question is wrong

[inland mulch]

The question works if you swap k and n

So you are given k and you need to find n

[pastel galleon]

oh

wait

$$ if \alpha is a real number and k is a positive integer, then \exists n \in \mathbb{Z} s.t. | \alpha - k/n| \leq 1/(2n) $$

this is right problem??

[inland mulch]

Don't swap them in both halves

Just in the first half and not in the maths part

Do you have a screen cap of the question in the book

[pastel galleon]

wait a minute

[inland mulch]

Cursed

If alpha is a real number and k a positive integer then there is an integer n such that |alpha - n/k| <= 1/(2k)

[pastel galleon]

oh

really

thank you very much

really appreciate

👍

.closed

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[obtuse pebble]

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[sudden hearth]

6 question

Its Q in place for A

<@&286206848099549185>

<@&286206848099549185>

Ah help 🥲

. close

.close

[obtuse pebble]

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[sudden hearth]

.reopen

[obtuse pebble]

✅

[sudden hearth]

6th question

<@&286206848099549185>

<@&286206848099549185>

[obtuse pebble]

@sudden hearth Has your question been resolved?

[obtuse pebble]

@sudden hearth Has your question been resolved?

[solar hornet]

@sudden hearth you might know what x^2+y^2=1 is a circle

while y=m(x+2) is a line, moved two places to the left

example

[obtuse pebble]

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[solar hornet]

.reopen

[obtuse pebble]

✅

[solar hornet]

but then I'm not sure what Q is in this context, so it's harder for me to progress ..

[sudden hearth]

Ah A is Q

Its mis typed

Can u show me the solution?

I don't even know where to start

[solar hornet]

but on that spreadsheet they call for locus midpoint of PQ

well, assuming Q is meant to be A, then you need to find the perpendicular line to the one in the problem

[sudden hearth]

Yes yes

[solar hornet]

wait no, lol, locus of midpoint.. and it's a circle too..

[sudden hearth]

Yes

Mid point of chord

From the secant

Above mentioned

[solar hornet]

oh I get it. that's interesting

[trail cloak]

Wait wait

What is point Q?

[sudden hearth]

Can u pls solve it i am stuck it like 3-4 hrs

[sudden hearth]

Its A

Its mistyped

[trail cloak]

Ooo

Ok so

Since the slope is positive

[sudden hearth]

Slope isnt mentioned

He only said slope

[trail cloak]

Alrighty

[sudden hearth]

m

It fking didn't even give solution
That damn book

[solar hornet]

slope could be negative

it just says m

[trail cloak]

Yeah just realized

[solar hornet]

but we can notice that the line has equation y=mx + m, which is a funny way to put it

[trail cloak]

Anyways you can make the line and the half circles

Equal

$m(x+2) = \pm \sqrt{1-x^2}$

[warm shale]

VulcanOne

[trail cloak]

You should get the coordinate for x

I got a pretty big expression for x

[sudden hearth]

I did and it gave a big expression

[trail cloak]

Yep

[sudden hearth]

With m

[trail cloak]

$x = \frac{-4m^2 \pm \sqrt{16m^4 - 4(m^2 +1)(4m^2 - 1)}}{2m^2 + 2}$

[warm shale]

VulcanOne

[trail cloak]

Ok this method is hard

There has to be another way

[sudden hearth]

Yes there has to be

But i can't find it on web

Neither thr book given its solution

[wild swallow]

[sudden hearth]

Oh wait

We have to find range of m also

[solar hornet]

it's probably answer a)

why? well

[sudden hearth]

Its is answer a

[solar hornet]

just from looking at it I guess

but how to show it on paper..

basically for m=0 the midpoint is the middle of the circle

[trail cloak]

Oooo

I realized that

When y = 0, x =-2

For the line

Then we find the points of tangency between the line and the circle

[sudden hearth]

Oh m range is (-1/√3,1/√3)

[trail cloak]

Yep

[sudden hearth]

Well m off now if u can pls send the solution

[trail cloak]

The points of tangency and the center of x^2 + y^2 = 1 should be part of a circle

[wild swallow]

this is like

simple geometry problem

lemme draw

by construction CX = OX = OP = OQ

and M is midpoint of PQ

[sudden hearth]

How?

[wild swallow]

then COM is a right triangle and so XM = XC = XO

[sudden hearth]

But who said x is mid point

[wild swallow]

your question

[sudden hearth]

Chords's midpoint

Here M will be mid point not X

[wild swallow]

yeah ?

XM is constant and equal to XC = XO

so the locus of M is a circle centred at X

[sudden hearth]

Nah xm isn't constant

[wild swallow]

[sudden hearth]

Even cp is variable

Its slope and intercept are variable

[wild swallow]

im talking about the length

[sudden hearth]

Then pls send me solution

I will check it later bz now

[solar hornet]

asking for solutions is taboo round these parts

[wild swallow]

this is the solution

[trail cloak]

Damn my geometry toolbox is dead 💀💀💀

But yes it makes sense

C is (-2,0)

[wild swallow]

-2

[trail cloak]

And O is (0,0)

And CX = 1, XO = 1

And r of the circle = 1

[warm shale]

ameo

[trail cloak]

So OP = OA

= 1

As we change the slope, Point M will go from the tangent line to the midpoint and then to the other tangent line

In an arc

[sudden hearth]

I still didn't get it

[obtuse pebble]

@sudden hearth Has your question been resolved?

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[fierce niche]

Find the functions f + g, f(g(x)) and g(f(x)).

[fierce niche]

My brain is boggled i dont get functions

[slim basalt]

Well, f + g means that you simply add them

f(g(x)) means that you insert g(x) into f(x) ---> f(g(x)) = 6 * (2x^2+ 3) + 5

g(f(x)) means you do it the other way around

[fierce niche]

so 6x + 2x^2+8 for the first one

and f(g(x))

u put g into x?

[slim basalt]

yep

[slim basalt]

yes exactly, you normally write it like this 2x^2 + 6x + 8 but its right

[fierce niche]

so its 6(2x^2+3) + 5

i see

thankyou 🙂

i can get that now

[slim basalt]

yep and you can also simplify that one

[fierce niche]

so 12x^2 + 18

• 5

12x^2 + 23

and for g(f(x))

would it be 2(6x+5)^2 +3

so 24x^2 + 53

also what about

Find the inverse function f^−1(x).

[gleaming ridge]

Do you know the definition of inverse function?

[fierce niche]

i do not

[gleaming ridge]

Basically if you let y = f(x), f^{-1}(y) = x

Although there are conditions, inverse function is only defined when f is 1-1 and onto

Aside that computation-wise

write x = f(y) and then solve for y

[fierce niche]

ok

so

[bold bane]

$y = f(x)\
f^{-1}(y) = y$

[warm shale]

Kookiemon

[fierce niche]

f-1(x) = x-5/6

[obtuse pebble]

@fierce niche Has your question been resolved?

Channel closed

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[obtuse pebble]

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

Triangle

[stone flint]

📐

[timid silo]

What?

[stone flint]

If I know all 3 angle values in a triangle

Could I tell size of it's sides

Sides

[timid silo]

No

[high lily]

from angles alone, no

[timid silo]

there are infinitely many possibilities

[stone flint]

Not even via trigonometry

[timid silo]

No, you need to know at least one side

[high lily]

you could determine the ratio of the sides though

[stone flint]

Ah, i see. Thanks for clearing that up.

[wild swallow]

note that scaling a triangle will not change its angles

[high lily]

enlarge the triangle until it reaches Pluto

[stone flint]

😂

.close

[obtuse pebble]

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[cursive tundra]

Hello i have a question regarding linear functions

[cursive tundra]

So i want to express that to gain 1 unit of Y I need 3 inputs of X yeah

So its y = 3x

But when I try to put it on desmos

Why is it oriented as if x is the dependent/response variable

And y is the independent

Y seem to be on the horizontal axis and not vertical

Ohhhh yeah!

So the rise/run in this case 3x becomes 3 Y’s for 1X

Ok i get it now hahaha

Thank u!!!

Would it be correct to intepret the slope as

Rise = to gain y,run = to input x?

[obtuse pebble]

@cursive tundra Has your question been resolved?

[obtuse pebble]

@cursive tundra Has your question been resolved?

[obtuse pebble]

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[silver plover]

is my part i correct?

[timid silo]

looks fine

,w plot sign(1-x)

[warm shale]

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Requested by Toby#2275

[silver plover]

could u help me with ii plz

lost on what to for that tbh

[timid silo]

I guess you know thats the symbol for floor?

[silver plover]

yep

[timid silo]

try substitute some numbers in and see if you can spot a pattern

hint: ||the range is [0,1)||

[silver plover]

so if u let x be any whole number

would it always be 0?

[timid silo]

yup

if x is an integer, its floor is equal to x

[silver plover]

i see

and why is the range [0,1)

i assume that means between 0 and 1 but not include 1

[timid silo]

yeah

this function gives us the fractional part of x