help-39

[pearl pond]

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[winter fox]

is this question asking me to use limits to find the derivative or the derivative rules?

[dense jasper]

"first principles" usually means the limit definition

[winter fox]

ok thanks

.close

[pearl pond]

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[dense jasper]

[pearl pond]

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[green meteor]

When I differentiate 5x + ln(2/(1-2x), i got 5 + 2/(1-2x) but the answer is 5-2/(2x-1).

[flint zenith]

Those are the same thing

[lilac jackal]

1-2x=-(2x-1)

[green meteor]

O

[flint zenith]

2/(1-2x) = ((-1)(2))/((-1)(2x-1))

And the -1/-1 = 1

[green meteor]

Alr

Thanks

🙏

.close

[pearl pond]

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[pearl pond]

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[sharp smelt]

So, I want $E( \mu^2+ \sigma_{0}^2)$

[jolly parrot]

wai

[sharp smelt]

Is there no "simple" way to find this

Var(X) =E(X^2)-E[X]^2

so $E[Var(X)]= E[E(X^2)]-\mu^2$

[jolly parrot]

wai

[sharp smelt]

so just $E(X^2)- \mu^2$?

[jolly parrot]

wai

[sharp smelt]

that gives me just $\sigma_0^2$

[jolly parrot]

wai

[west sapphire]

shouldn't it be a function of x1, ..., x_n?

[sharp smelt]

mhm, it should

[sharp smelt]

so I find the variance of that sample now, right

[west sapphire]

hm well how do you find umvu in general

[sharp smelt]

I thought I find plausible UEs first and the work with that

ah, order statistics

[west sapphire]

do you have a theorem like this?

[sharp smelt]

oh, I missed that

I suppose I'll do that first and then back to this

[sharp smelt]

sorry :(

.close

[pearl pond]

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[pearl pond]

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[grave parcel]

Can someone check my working please

[shell brook]

i recommend taking logs in cases like these, imo it greatly simplifies the process of taking derivatives

[grave parcel]

Agreed 😭 😭 this took so long

Can u explain how to do it by logs

U haven’t learnt it yet that way

[shell brook]

for example y = x(sqroot(x+1))

just take log on both sides

log y = log x + 1/2 log (x^2+1)

give it a try if you want to

[grave parcel]

would I be able to do this method without a calculator ?

[shell brook]

yeah

[grave parcel]

Oh really that’s cool

[shell brook]

yeah take natural logs

i had this course in my first semester

[grave parcel]

Is this the formula?

[grave parcel]

So this just differentiates it?

[shell brook]

no it basically relies on the fact that log(xy) = log (x) + log (y)

so you dont use tedious product and quotient rules

[shell brook]

my prof suggested it

[grave parcel]

If u don’t mind would you be able to show me the steps using the example I’ve given please

[grave parcel]

Exactly it’s so time consuming and many small mistakes can occur

😭

[grave parcel]

Does it work on any function basically? Or just the square root ones?

[pearl pond]

@grave parcel Has your question been resolved?

[grave parcel]

.close

[pearl pond]

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[pearl pond]

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[pearl pond]

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[obsidian wing]

oops

[pearl pond]

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[bronze atlas]

I am trying to understand this inductive proof by rewriting what I understood from it in my own words, filling in implicitly considered steps as shown in another image. Based on what I am doing here, I have some questions:

1) Is it true that it is tacitly implied that there is bijection between variables when reindexing is done as in this example so that we can rewrite the summation with different variables (indices)?

2) How can I move on from here to show that there is a bijective relation between them? There are "i" and "j" indices to consider so I need to phrase it correctly such as in "for a fixed i/j, then this..."

[pearl pond]

@bronze atlas Has your question been resolved?

[pearl pond]

@bronze atlas Has your question been resolved?

[pearl pond]

@bronze atlas Has your question been resolved?

[pearl pond]

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[pearl pond]

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[timber cape]

I need help on how to properly help people in the help channels

[rigid mist]

#helpers-info

[timber cape]

Oh

Ty

Mb

.close

[pearl pond]

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[rigid mist]

@bronze atlas pack it up

[bronze atlas]

I thought he freed the room and wanted to reopen my question I asked earlier.

[rigid mist]

you'll have to open a new channel/thread

[pearl pond]

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[zenith mural]

I don't understand how it weent from the second step to the third

[minor cloud]

$x^a = e^{a \ln(x)}$

if that's what you're confused about?

sorry if it isn't

and oops

[jolly parrot]

differential Towametry (Towa)

[zenith mural]

i got this yeah

[minor cloud]

oh I apologize then

[zenith mural]

thats fine i dont get how it went from here to the next line

[minor cloud]

you mean how all the e^{somethings} are combined?

[zenith mural]

yes

[minor cloud]

product rule of exponents

[zenith mural]

the logs and lnz vanished?

[minor cloud]

the common base is e, so just add all the exponents together

though they did the expansion individually first before adding the exponents together

[zenith mural]

oh i j realized

oml okay yeah

why is it log somewhere and ln somewhere? is it the same thing?

[minor cloud]

that is, they expanded, say, (ln y - ln z) ln x out

eh it's supposed to be ln really

[zenith mural]

yeah got it

[minor cloud]

not sure why they wrote log there because you can see in the next line they switched to ln

it's supposed to be ln all the way

[zenith mural]

yeah I was confused how it worked

alr thanks!

[minor cloud]

glad to help!

[zenith mural]

.close

[pearl pond]

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[pearl pond]

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[haughty wraith]

If E = mc4 then b = mc2

[tepid mauve]

Hello, first what do you exactly need help with, and what area of math is this?

[muted lotus]

[muted lotus]

if you were a square

youd be square root

[tepid mauve]

If you were a vector, you’d be pointless

[autumn fossil]

[latent quail]

What is this?

[autumn fossil]

Do you have any legitimate questions or are you just trolling?

[surreal siren]

Correct

• Ai

[haughty wraith]

I was just putting a small meme….

[timid lodge]

!redir

[pearl pond]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[autumn fossil]

I see, you can use #chill for that

help channels are for math help

[haughty wraith]

k

[autumn fossil]

I'll close this now

.close

[pearl pond]

Channel closed

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[surreal siren]

Kind of insane how people summon this bot like mahoraga instead of just typing it out

[pearl pond]

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[royal harness]

is this doable

[plush bramble]

,w int 0 to log(2) of x exp(exp(x))

[jolly parrot]

Results provided by WolframAlpha

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

Requested by rie.mann

[plush bramble]

Doesn't look like it

[surreal siren]

Some Feynman method might

[royal harness]

our teacher gave it to us as a "bonus exercise"

[surreal siren]

If ur teacher gave u then yes it must be

[royal harness]

that's what I tried but I didn't get nowhere near solving it

[royal harness]

he couldn't solve it himself tho 😭😭

[short lagoon]

anyone jee aspirant?

[surreal siren]

Got it maybe

Lemme see

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

@royal harness Has your question been resolved?

[surreal siren]

@royal harness

Can u integrate e^xlnx?

Do u know how to?

Cause if u do I can leave it to u

I’ll do it

[royal harness]

I don't think this integral is doable or is it ?

[surreal siren]

I’m using Feynman again

That’s right baby

[swift onyx]

What is going on here 🥹

[surreal siren]

?

[royal harness]

where are you going to introduce the parameter

[surreal siren]

I just substituted and used Feynman

[surreal siren]

Some random

I’m thinking h

[royal harness]

no like where

the ln ?

[surreal siren]

I integrated by parts

[swift onyx]

bro I thought I’m learning maths here what are these calculations

[surreal siren]

This

[surreal siren]

Oh yea ignore

[short lagoon]

yoo................................actually i joined this group cuz i needed guidance from you all guys as my maths is really weak and i have time till my april attempt ..............your advice is really appreciated

[surreal siren]

Then I got e^ay/a da

[surreal siren]

Not here

Pls go sm where else

[royal harness]

which is the Ei(x) function with an added parameter

[surreal siren]

Yep

But I’m not sure how to solve those

So I’m using Feynman again

I’ll show u

Oops a little goof up

[short lagoon]

how old u guys ?

[surreal siren]

Pls leave this chat and talk in #chill

[short lagoon]

bruh

[surreal siren]

@royal harness

[royal harness]

ohhh right

so i needed to do a double Feynman technique damn

I stopped trying after doing 1

[surreal siren]

Hm

So now im confused since im self taught on Feynman

Can u just put h as 1 here?

[royal harness]

so you replace h by 1 to recover the integral we had and then integrate with respect to y

[surreal siren]

Yep

[royal harness]

same

[royal harness]

ues

yes

[surreal siren]

Cool

Then we get

[royal harness]

yeah and now what's left is just easy

thanks so much for your help!

[surreal siren]

I hope this is correct

Do u want to stick around and get the answer?

[royal harness]

I mean it looks like it is

[royal harness]

if you want to sure !

[surreal siren]

Cool let’s do it

😂😂😂😂😂

Dude

U won’t believe this

Oh wait nvm

I thought we need Feynman a third time

[royal harness]

lmaooo

the result should be 1.2366

I HOPE THIS IS CORRECT COME ONNN

[surreal siren]

I(0)?

Uh oh

Not getting that

[royal harness]

I(1)

[surreal siren]

Ok ok

Oh wait

I messed up the sign

[royal harness]

it's fine lol it happens

[surreal siren]

What is ln2?

0.693?

[royal harness]

yup

[surreal siren]

LETS GO

[royal harness]

YOU GOT IT ???

[surreal siren]

Close enough

[royal harness]

😭😭

[surreal siren]

Lol

Acceptable or nah?

[royal harness]

did you put the values as they are or did u approximate

[surreal siren]

Approximate

[royal harness]

oh that might be why

[surreal siren]

Yea

[royal harness]

cuz 1.2366 is a small number

[surreal siren]

Hm

[royal harness]

it should be correct if it's close ig lol

damn GGS

[surreal siren]

Ggs bro

Holy

2 times Feynman then the crazy fucking

Omg

[royal harness]

I was not gonna do all of that in the 10min remaining on the exam lmao

our teacher is sadistic istg

[surreal siren]

Wait what

This was on ur exam?

Dang

[royal harness]

as a bonus exercise yes 😭😭

high school too

he just wanted to troll us ig

[surreal siren]

How old are u bro?

If u don’t mind me asking

[royal harness]

17

[surreal siren]

Same

All the best for college dude

[royal harness]

yeah but we should be fine dw 🔥🔥

[surreal siren]

Should we close this?

[royal harness]

yup thank you for you help have a good day !

.close

[pearl pond]

Channel closed

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[pearl pond]

Available help channel!

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[long nova]

can i solve this by factoring out n under each radical instead of using the sum formula?

[long nova]

i tried it earlier but i got like infinity

[plush bramble]

!show

[pearl pond]

Show your work, and if possible, explain where you are stuck.

[solar ember]

I doubt it.

[hazy pilot]

why not just use the formula?

[solar ember]

The issue here is the term in the parentheses. As $n \to \infty$, the number of terms inside that sum is also growing. You aren't only factoring $n$ out of a fixed number of terms; you are factoring it out of $n$ terms.

[jolly parrot]

Ajay

[long nova]

kinda like this

[long nova]

idk i wanted to try something new

[hazy pilot]

fun!

[long nova]

cuz like we know like 1/n or 4/n or 5/n is equal to 0

any real number over infinity is just 0

[hazy pilot]

well this doesn't really work logically I think?

[solar ember]

@long nova You can read what I mentioned above

That line of reasoning doesn't work in this case

[long nova]

oh i see

[solar ember]

If you want to try something different, how about taking the average of the terms inside the radicals?

[long nova]

oh well i guess this method aint gonna work

[solar ember]

Something like this:

For the first radical there are $n$ terms. The average of any linear sequence is just the average of the first and last terms, so we write: $\frac{1 + n}{2}$. Therefore, the sum is the number of terms times the average = $n \cdot \frac{n+1}{2}$. As $n \to \infty$, this is essentially $\sqrt{\frac{n^2}{2}}$.Similar working for the second radical also gives $\sqrt{\frac{n^2}{2}}$.

[jolly parrot]

Ajay

[solar ember]

This is essentially the motivation for the sum formula, however, we are not explicitly using it

Do you understand so far @long nova ?

[long nova]

yeah i think so

so ur doing like n/2 times (n + 1) where n is the last term and ur just taking the average

[solar ember]

Yes

Now i'm going to take a bit of a leap off into a different direction

[long nova]

lets hear it

[long nova]

holy

[solar ember]

If you look at your first radical, this is $\sqrt{T_n}$. Geometrically, this is like finding the side length of the n-th triangle.

[jolly parrot]

Ajay

[solar ember]

Similarly, for your second radical, this is $\sqrt{T_{n-1}}$.

[jolly parrot]

Ajay

[solar ember]

So if you take the area of the triangle for the first radical, you get: $1/2 n^2$

[jolly parrot]

Ajay

[long nova]

ya

[solar ember]

And for the second radical: $\frac{1}{2}(n-1)^2$

[jolly parrot]

Ajay

[solar ember]

Take the square root of both and you end up with $\frac{n}{\sqrt{2}}$ and $\frac{n-1}{\sqrt{2}}$

[jolly parrot]

Ajay

[long nova]

then what

we must factor or multiply by something ya

?

[solar ember]

So your limit now transforms to $$\lim_{n \to \infty} \frac{n}{\sqrt{2}} - \frac{n-1}{\sqrt{2}} = \frac{1}{\sqrt{2}}$$

[jolly parrot]

Ajay

[long nova]

ahh

thats a very interesting way of doing it

[solar ember]

I actually learned about it from my niece who's in the 8th grade 🙂

[long nova]

:0

smartness runs in the bloodline eh?

thanks for the solution though

.close

[pearl pond]

Channel closed

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[solar ember]

The key thing to remember is: The total number of dots ($T_n$) is the "area" of that triangle.

[jolly parrot]

Ajay

[long nova]

yup

[solar ember]

This can be universally applied to a lot of things

In this case, sequences

[feral sedge]

calc or precalc?

[long nova]

calc but technically precal

[feral sedge]

eh whatever there are other ways to illustrate

$\underbrace[test]{abcd}$

[jolly parrot]

Dreyuk

[feral sedge]

$\lim_{x\to\infty}1 = \lim_{x\to\infty}\underbrace{\frac1x+\cdots+\frac1x}_{\text{x times}$

$=\underbrace{\lim_{x\to\infty}{\frac1x}+\cdots+\lim_{x\to\infty}{\frac1x}}_{\text{x times}$

[jolly parrot]

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

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

[feral sedge]

$=\underbrace{0+\cdots+0}_{\text{x times}}=0$

[jolly parrot]

Dreyuk

[feral sedge]

its like this lol

troll math

[pearl pond]

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[urban jacinth]

is there a way to get angle OBA easily when theta = 90 degrees?

nvm im an idiot LOL

.close

[pearl pond]

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[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[ocean galleon]

Hepl

[minor cloud]

hi, please send your question!

[feral olive]

!da2a

[pearl pond]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[ocean galleon]

Can I get help on AP m mechanics problem

[minor cloud]

please don't ask to ask. just send the question!

[pearl pond]

@ocean galleon Has your question been resolved?

[ocean galleon]

Sorry if my handwriting is messy. I don't need the answer straight help me approach the problem

A is a trolley

[ivory basin]

Well, you want the friction to cancel out the weight

Calculate those and set them equal

[ocean galleon]

Im not English so please be more umm spesifik

🙏

Okay okay understand

[full current]

isnt this like physics

not maths?

[ocean galleon]

No it's actually maths

It's part of simplified rb motion

Pls help what should I do next

[pearl pond]

Channel closed

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[ocean galleon]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[ocean galleon]

@ivory basin pls help

[pearl pond]

@ocean galleon Has your question been resolved?

[ocean galleon]

Nope

[minor cloud]

react to the bot. the bot cannot read text replies to that prompt.

[ivory basin]

What is your native language, just in case it's one I speak

[feral olive]

Do you know how friction works?

As in the formula for it

[ocean galleon]

I speak sinhala

[ocean galleon]

F=uR

What can I do with it

[feral olive]

If the trolly is accelerating against the object its applying a force to it

So the reaction force is equal to that but in the opposite direction

[pearl pond]

@ocean galleon Has your question been resolved?

[pearl pond]

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[pearl pond]

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[chilly jay]

in this question, since the vectors are linearly independent, we can write A as PDP^-1

here, P = {(1, 1, 0), (2, 1, 1), (3, 0, 1)} and D = {(-1, 0, 0), (0, 2, 0), (0, 0, 2)}

is there a way to solve this further without having to explicitly calculate P^-1 and then multiplying all three matrices?

when i tried to do that, my resulting matrix was coming out to be {{-8.5, 10.5, -4.5}, {-3, 5, -3}, {-4.5, 4.5, -2.5}}

which is obviously not correct

[pearl pond]

@chilly jay Has your question been resolved?

[daring charm]

Hello

yeah just use
$$A P = P D$$

so solve

$$
A [v_1\ v_2\ v_3] = [-v_1\ 2v_2\ 2v_3]
$$

[jolly parrot]

Ezra

[daring charm]

directly for the entries of $$(A)$$

[jolly parrot]

Ezra

[daring charm]

avoid $$(P^{-1})$$ entirely

[jolly parrot]

Ezra

[chilly jay]

why does it work?

oh

got it! thank you

.close

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[daring charm]

sure

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[flint basalt]

could someone check my answers and steps to make sure i did this correctly?

[flint basalt]

please ping me if you respond to this^

[uncut vapor]

you got Q1 -B wrong

it should equal to zero

@flint basalt

[steep saddle]

that is a lot of work vs simple parameterisatiom

[uncut vapor]

i hate line integrals ngl

[pearl pond]

@flint basalt Has your question been resolved?

[flint basalt]

got it i'll take a look in a bit, thank you for checking

[pearl pond]

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[flint basalt]

oops

.reopen

[pearl pond]

✅ Original question: #help-39 message

[flint basalt]

you mean i shouldn't be doing it this way?

[uncut vapor]

nope its correct way to do it like that

but i think u missed smth with the integral that gave you 1/2

it should equal zero

u are integrating three odd fucntions over a period of -1 to 1 they all should equal to zero

[steep saddle]

you turned x^4/4 from -1 to 1 into 1/2?

[uncut vapor]

i guess thats what he did

it should equal to zero

[pearl pond]

@flint basalt Has your question been resolved?

[flint basalt]

ah i see it's -1/4 + 1/4 i think i just missed a sign somwhere in my mental math

so then it just turns everything into zero that makes sense

thanks for your help, is the rest all good?

[uncut vapor]

didnt really check

the rest

i got smth rn

[pearl pond]

@flint basalt Has your question been resolved?

[flint basalt]

thats fine i can go through them myself then thanks

.close

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[exotic shore]

Hello, me again. I understand the meaning of coterminal, however, with it being in this form it confuses me a little on how exactly to begin

[dense jasper]

$360^{\circ}=2\pi$ rad

[jolly parrot]

Civil Service Pigeon

[dense jasper]

[exotic shore]

that's really awesome ty

Knowing this, I would do 3pie/4 - 2pie/1 right?

[dense jasper]

pi, not pie, but yes.

[exotic shore]

my bad lol

so, would my answers ne -5pi/4 and 11pi/4 ?

nvm not a negative

5pi/4 and 11pi/4 ?

[dense jasper]

What

-5pi/4 is negative

It legit has a negative sign on it lol

[dense jasper]

yes

[exotic shore]

well I did the 3pi/4 - 2pi/1 and I should have done 2pi/1 - 3pi/4

[dense jasper]

what

[exotic shore]

I'm just saying my answer shouldn't have been a negative

[dense jasper]

This would give you that 3pi/4 and 5pi/4 are coterminal

But these aren’t even the same quadrant

,w graph angle of 3pi/4

[jolly parrot]

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[dense jasper]

welp worth a shot

[exotic shore]

Lol

I'm saying uh, in general to find the coterminal it'snt it 360+## and 360-## ? does it not matter if it's actually ##-360 ?

does that make sense what I'm asking?

[dense jasper]

$360-x$ and $x-360$ negatives of each other, and $-\theta$ definitely isn’t always coterminal to $\theta$

[jolly parrot]

Civil Service Pigeon

[dense jasper]

In general, the angles coterminal to $\theta$ are all described by $\theta+2\pi k$

[jolly parrot]

Civil Service Pigeon

[dense jasper]

(where k is an integer)

Because you’re “adding on” full rotations

In a rough sense, something like $2\pi-\theta$ would be reflecting $\theta$ over the $x$-axis to go from $\theta$ to $-\theta$, then adding on a full rotation to go from $-\theta$ to $2\pi-\theta$

[jolly parrot]

Civil Service Pigeon

[exotic shore]

that makes sense

I am a little overthinking lol But I understand

[dense jasper]

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[exotic shore]

.close

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[random lintel]

Can anyone help me with B

[random lintel]

I can do a closer up photo of the graph if you’d like

[rotund ferry]

you can split the area into a rectangle and a triangle

[minor cloud]

I can give you a hint

actually nvm Matcha alrd gave you one. I'll step back

[random lintel]

Sorry I keep cropping it by mistake

[rotund ferry]

okay nicee

[minor cloud]

one heads-up: be careful about the sign of the final answer!

right back out now

[random lintel]

I got -40.5 as my final answer

[rotund ferry]

why negative?

[random lintel]

I made it negative because it is below the x line

[rotund ferry]

oops mb

i misread the integral as 8 to 2

ur good now

[minor cloud]

anything else, OP?

[random lintel]

No sorry

.close

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[inner jacinth]

this isnt right guys?

[inner jacinth]

my professor did this but

he kept doing it as its the right way, so i dont understand

isnt it supposed to be 2/3q vector

[plush bramble]

What part do you think is wrong

[inner jacinth]

isnt this missing?

i dont understand the logic behind just removing 4/9 completely

bcs if we were to mutliply |q| by 2/3 we would just get 2/3|q|

and that is not original statement

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[green meteor]

why is 5pi/12 not an answer even tho its in the range

[green meteor]

also is there another way i can find nature without using d2y/dx2 cuz im lazy and its too long

[solar ember]

look at the denominator

The function is undefined whenever $\sin 3x = 0$. In your range, this happens at $3x = \pi$, which means $x = \frac{\pi}{3}$. Since $\frac{\pi}{12} < \frac{\pi}{3} < \frac{5\pi}{12}$, there is a vertical asymptote at $x = \frac{\pi}{3}$.

[jolly parrot]

Ajay

[green meteor]

why does the asymptote affect anything

is x=5pi/12 not a stationary point?

[solar ember]

Look at the domain of your curve and check the continuity

[green meteor]

what does continuity mean i dont think my textbook uses it

[solar ember]

In simple terms, continuity means you can draw the entire graph without lifting your pen from the paper.

[green meteor]

But why doesnt it count as stationary

And ehat do i say for my reasoning for rejection

[sharp vigil]

can you show the answer key? i'm not really sure why it isn't being counted as stationary either

[green meteor]

It just says x=pi/12 min/max wtv it was

[solar ember]

It is a stationary point

[sharp vigil]

maybe it was a mistake of the question writers where they intended to make a domain that would only include the stationary point at pi/12 and accidentally did not

[green meteor]

Hm

There are a lot of mistakes in cambridge

Also how do i find nature without using product plus quotient rule for d2y

Or is that the only way

[sharp vigil]

the alternative is to find the sign of the derivative on each side of the stationary point

[green meteor]

How do i do that

Do i just use angles close to stationary points

[solar ember]

I don't think it is a mistake

[sharp vigil]

as long as the angles are between the stationary point and any other stationary points or discontinuities, that would work

[solar ember]

While $x = \frac{5\pi}{12}$ is in the range, it lies beyond the vertical asymptote at $x = \frac{\pi}{3}$, so the stationary point for the principal curve is $x = \frac{\pi}{12}$.

[jolly parrot]

Ajay

[green meteor]

Ok

[sharp vigil]

it seems to me like if they wanted to only include the "principal curve" they would have restricted the domain appropriately

[green meteor]

Why would there be a domain if its just the principal curve

[solar ember]

The question asks for the exact value

So I assume it owuld be a point within the principal curve

[sharp vigil]

i would think "exact value" means "give the answer as a fraction of pi rather than a decimal approximation"

[green meteor]

Hm

Ill ask my teacher

[solar ember]

But if they wanted (or at least expected) two values wouldn't they have written exact value(s)?

I'm also saying coz of the asymptote at x = pi/3

[green meteor]

But similar questions to this they included stationary points after asymptote

[solar ember]

I think that this is just a poorly written question

I agree it is a stationary point

[sharp vigil]

i agree that they expected only one answer, but that means they should have restricted the domain appropriately. it's not reasonable to expect people to only look at the "principal curve" (and who says 0 < x < pi/3 is principal rather than pi/3 < x < pi/2? the choice would be arbitrary)

[solar ember]

fair enough

[green meteor]

Alr

Ill double check with my teacher

Sometimes cambridge has weird wording and stuff

But thanks guys

🙏

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[pearl pond]

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[feral sedge]

real

[pearl pond]

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[fathom sequoia]

can someone help me figure out how i can make this specific arch using exponential/logarithmic functions

[fathom sequoia]

in desmos

[dreamy stratus]

idk

[fathom sequoia]

man

[fathom sequoia]

ill close this channel actually im gonna do this later

.close

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[shell brook]

try using math

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[rare holly]

am i allowed to do thi

[minor cloud]

no

you'll need an extra 3

[rare holly]

alr

ill need to factor out the 3?

[minor cloud]

and it will become [3(x+4)] [3(x-4)]

hence 9(x+4)(x-4)

[rare holly]

like this?

[minor cloud]

yes if that center x represents multiplication

[rare holly]

yh

thanks

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[normal magnet]

Let a, b, c, k and x be real numbers. Let a, k and x be all not equal to 0.
c^2 - ab = kx^2
a^2 - bc = kx
b^2 - ac = k
Find the value of x.

[normal magnet]

I have nother question.

[autumn fossil]

What's the context?

do we know a,b,c,k?

[normal magnet]

Nope.

[autumn fossil]

so then its 3 equations 5 unknowns

i doubt that's solvable

[verbal whale]

!xyp

[pearl pond]

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.

[normal magnet]

The possibility is higher, given the condition that a, k, x ≠ 0

[autumn fossil]

https://www.desmos.com/3d/myjjhuygs1

X, K can be anything (almost)

[pearl pond]

@normal magnet Has your question been resolved?

[pearl pond]

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[normal magnet]

Can you give a five-ple of values of a, b, c, k and x with with every criteria satisfied with distinct x's

[autumn fossil]

hm

,w c^2 - ab = 1
a^2 - bc = 1
b^2 - ac = 1

[jolly parrot]

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[autumn fossil]

here you go, k = x = 1

[normal magnet]

So x = 1 is the answer to the original question?

If not, then I would like another set of values with x not equal to 1.

[autumn fossil]

sure

,w c^2 - ab = 4,
a^2 - bc = 2,
b^2 - ac = 1

k = 1, x = 2

[jolly parrot]

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[normal magnet]

a = 0

[autumn fossil]

you'll almost always find a sol

for almost any x

[normal magnet]

which dissatisfies that a is not equal to 0

[autumn fossil]

oh okay ic

,w c^2 - ab = 27,
a^2 - bc = 9,
b^2 - ac = 3

hmm interesting

[normal magnet]

lol

[jolly parrot]

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[autumn fossil]

i wonder what forces a = 0

<@&268886789983436800>

[autumn fossil]

resolved, it just got deleted, sorry for the ping

oh wait

i think i got it

c^2 - ab = (b^2 - ac)x^2
a^2 - bc = (b^2 - ac)x

oh nvm, doesnt really work

but at least there is one less variable now

now we can square the second one and divide them

(c^2 - ab)(b^2-ac) = (a^2 - bc)^2

[normal magnet]

hm

[autumn fossil]

now x = 1 iff a^2 - bc = b^2 - ac I think?

[warm tiger]

wouldnt it be (a^2-bc)^2 in RHS

[autumn fossil]

true

,w expand (c^2 - ab)(b^2-ac) - (a^2 - bc)^2

oh and its sol is a union of planes

that changes things

[jolly parrot]

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[autumn fossil]

yeah okay so we factor out a

[warm tiger]

we get a+b+c = 0 or a = b = c

[autumn fossil]

a(-a^3 + 3abc - b^3 - c^3) = 0

[autumn fossil]

You sure?

wouldnt it be a = 0 or a^3 + b^3 + c^3 = 3abc?

the second one will probably simplify a bit more

[warm tiger]

if u factor the second one

then u get those 2 conditions

since a != 0 here

[autumn fossil]

oh so a = 0 or a=b=c or a+b+c=0

where the first one gets crossed out technically

okay great

[warm tiger]

a=b=c wont be possible coz that would mean either of k or x has to be zero

[autumn fossil]

now its just casework i suppose

so now a + b + c = 0

[warm tiger]

x = (a^2-bc)/(b^2-ac)

ig put c = -(a+b) here

[autumn fossil]

$x = \frac{a^{2}+ab+b^{2}}{b^{2}+a^{2}+ab} = 1$

done

[jolly parrot]

MathIsAlwaysRight

[pearl pond]

@normal magnet Has your question been resolved?

[normal magnet]

ok thanks wow

[pearl pond]

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[spare carbon]

this is the way professor solved it what i did was simplify it at the beginning and got Is this legal?

[minor cloud]

wait which is which?

[spare carbon]

First is professors

[minor cloud]

the written one is yours?

[spare carbon]

No

[minor cloud]

oh ok, and the typed one is yours

[spare carbon]

mine is screen shots

Yea

[timber cape]

Both are correct

You just rationalised yours

[spare carbon]

Okai

ty

[minor cloud]

yeah, ok, both are right, agreed

but you can further simplify yours

[spare carbon]

how so?

[timber cape]

Uhm no

[spare carbon]

simplfying root 14?

[minor cloud]

the 4 and 14 have common factors

[minor cloud]

what?

[timber cape]

It is preferred to not have an irrational number in the denominator*

[minor cloud]

and OP didn't? did you see pic 3?

[spare carbon]

Mhm

[timber cape]

Oh there are 3 picks

[minor cloud]

no, the 4 against the 14

the root 14 is done

[timber cape]

God bless phones 🙏

[spare carbon]

by 2

Right?

[minor cloud]

yup

[spare carbon]

yea i think we can do that

ty missed it

[minor cloud]

nps. anything else?

[spare carbon]

nope i will post if any other doubts do come

[minor cloud]

sure thing. you can close the channel if you're done with .close. welcome to the server and enjoy your stay!

[spare carbon]

.close

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[normal magnet]

Let BD = DC.
Let AF/FD = 2/3
Find BF/FE

[frank violet]

wut

#help-36 message

duplicated?

[still hamlet]

yeah same question

[normal magnet]

this came in my exam today

[still hamlet]

start by drawing a line from D parallel to BE

mark the point where it intersects AC

[normal magnet]

I have found the value of AE/EC

[still hamlet]

well what is it and how did you get to it

[normal magnet]

DG || BE
Hence GE = GC
now i used the BPT in the triangle ADG to get the ratio of AE and EG, and hence the ratio of AE and (2EG) = AE and (EG + EG) = AE and (EG + GC) = AE and EC

[still hamlet]

what's G?

[normal magnet]

A point on AC such that DG || BE

I drew another line through F parallel to DC intersecting AC at K. Then I found the value of EK/KC, and hence, EF/BF.

I would like a simpler solution.

I believe my solution is unnecessarily complex.

[still hamlet]

i dont know if my solution is much simpler

[normal magnet]

what is it

[still hamlet]

also are you sure AE=2EG?

[normal magnet]

no, that is not true

i did not say that, did i?

I just said that I can find the value of AE/2EG whatever it be.

[still hamlet]

from DG, derive the ratios AE/EG and EF/,GD and in ∆CBE, derive the ratio DG/BE and i think you'll see what to do from here

[timber cape]

-# my head is scratching with vectors

[normal magnet]

that is indeed simpler

[frank violet]

you can solve this in 1 line with vector yeah

[still hamlet]

you specifically typed (2EG)=AE

[timber cape]

Exactly

[still hamlet]

unfortunately they don't have that for grade 10 🥲

[normal magnet]

sorry for the confusion, but i meant to mean =>, not =

[frank violet]

really? they taught me vector in 10th grade lmao

[timber cape]

well

[still hamlet]

-# in CBSE, vectors are in grade 11 and 12

[timber cape]

There are basic vectors and then there are 'vectors'

[normal magnet]

wdym really. different systems have different syllabi

[timber cape]

well yeah

Caution: this shifted from helping to chilling

[frank violet]

-# idk, just surprised that you have to wait till 11th

[timber cape]

12th actually

Unless physics

[normal magnet]

ok

thank you for your help

i have another question

[timber cape]

ask away 🍃

wind noises

[normal magnet]

in this channel?

[timber cape]

Yeah

Even the channel named after you

Respect 🫡

[normal magnet]

I need help with (i).

[still hamlet]

what were you able to get?

also idk why the first image does not look like a tabletop

[normal magnet]

Yeah, it is quite vague

[still hamlet]

I'm thinking it's height of the solid base + diam of the sphere given the second image

[normal magnet]

oh

shouldn't it be the vertical distance between where the water emerges and where it lands

[still hamlet]

yeah not too sure about that

[pearl pond]

@normal magnet Has your question been resolved?

[normal magnet]

thank you

[pearl pond]

Channel closed

Closed by @normal magnet

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[pearl pond]

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[eager jewel]

i dont know how to start

[buoyant panther]

Well, I suggest simplifying the fraction there

Maybe PFD/factoring/long div

[eager jewel]

(n-1)^2 -2

is the best i can do

uhh i wouldnt know how to apply any of those here

[buoyant panther]

Mhm, it'd be better if we got (n+1)^2

(n+1)^2 = n^2 + 2n + 1, so we would have to subtract 4n + 2

(n+1)^2 - (4n +2)

so the whole fraction becomes

[(n+1)^2 - (4n +2)]/(n^2(n+1)^2) = 1/n^2 - 2(2n+1)/(n^2(n+1)^2)

Which is already nice

But we can go even further

[eager jewel]

oh ok how

[buoyant panther]

hint: 2n + 1 = 2(n+1) - 1

[eager jewel]

ohk yeah

we can break into more fractions

[buoyant panther]

and (n+1)^2 - n^2 = (n+1+n)(n+1-n) = (2n+1)(1) = 2n + 1 which is very helpful here

[eager jewel]

ahhhhh

[glossy tapir]

wait can i check the answer with someone

[eager jewel]

the whole thing simplifies to 2/(n+1)^2 - 1/n^2?

[eager jewel]

answer is 2

[glossy tapir]

okay thanks

[buoyant panther]

yes

I think so

So now

[eager jewel]

yea

[jolly parrot]

Modus

[eager jewel]

👍

does it telescope or smth

[buoyant panther]

Mhmm

We can transfrom it a little bit

to obtain the expression under the sum

Think about that

[eager jewel]

Multiple by 2

And shift 2/(1+n)^2 to other side

Alr I gtg to sleep now

I'll try again tomorrow

Thank u

.close

[pearl pond]

Channel closed

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[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
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[quiet sage]

I cannot figure out what I’m doing wrong here

[plush bramble]

how did you get your f(x)

was your parent function cot(x) ?

[quiet sage]

I originally was using tan(x) because of the shape and direction but could get it with that and switched then got the f(x) where it crossed at -2

My original answer was -tan(pi/6 x)-2

[plush bramble]

yea that looks more correct

[quiet sage]

I’m still getting it wrong tho and not sure why

[plush bramble]

show the error

[quiet sage]

What do you mean? It just says it’s wrong doesn’t tell me why so I’m
Not sure my error

[plush bramble]

just sounds like you're entering something wrong

[glossy tapir]

is this a multiple choice

[quiet sage]

No

[plush bramble]

what happens if you try entering just the function and not the f(x) = part

[quiet sage]

It tells me it’s a syntax error: this is not an equation

[plush bramble]

oh i missed the constant in front

f(x) = -Atan(pi/6 x) - 2 for some constant A>0

i had to assume that intersection point with the x-axis was (-1.5, 0)

[quiet sage]

Awesome. Thank you for the help. I have one more if that’s okay

[plush bramble]

yea looks similar except instead of A=2, it's probably 6/4 or 4/6

rude that they don't tell you exact points and just make you guess based on window size

[plush bramble]

A =2 here because 12/6=2

[quiet sage]

Yeah it makes it really challenging. I really appreciate the help. Thank you

.close

[pearl pond]

Channel closed

Closed by @quiet sage

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[pearl pond]

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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[naive sparrow]

hi

[minor cloud]

hi. what question do you have this time?

[naive sparrow]

no just saying hi

[minor cloud]

!redir?

[pearl pond]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[minor cloud]

don't think help channels are for greetings mate.

[cinder flower]

lol

[naive sparrow]

ok bye im goin to #latex-testing

[minor cloud]

cool

!done then

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[naive sparrow]

yall can be the owner of thise serve

.close

[pearl pond]

Channel closed

Closed by @naive sparrow

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[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[leaden iris]

hi guys

[leaden iris]

is there any method to solve ts?

[proper nova]

oof

[shell brook]

oof indeed

[cinder flower]

yes

[shell brook]

ig

[leaden iris]

well?

[shell brook]

theres one method

[leaden iris]

which is?

[shell brook]

which i cant recall rn

[proper nova]

haven't met these kinds of questions before

[shell brook]

😞

[leaden iris]

oh

well i have a few other doubtss

[paper compass]

I join this help channel and just see everyone saying oof 😭

[proper nova]

[leaden iris]

??

[cinder flower]

https://proofwiki.org/wiki/Product_of_Divisors i’m just gonna put this here

[paper compass]

Do you want me to help?

[leaden iris]

yes pls

[proper nova]

is this SAT or smth?

[paper compass]

Check the link they shared for the equation.

[tight cobalt]

oof

[proper nova]

oh i have 2 factoids for ya

!nosols

[pearl pond]

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.

[leaden iris]

nope 10th olympiad (ncert chapters)

[proper nova]

!noai

[pearl pond]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[shell brook]

oh my bad

[leaden iris]

how-??

[full frost]

45?

[tight cobalt]

it is clearly none of these

[shell brook]

idk man

[proper nova]

And no solutions too

[shell brook]

thats my bad as well

[leaden iris]

i'm sorry, this is of a higher level than my current grade... i'd prefer an easier method

[full frost]

7056=2^4 * 3^2 * 7 ^ 2

[worldly jewel]

it's very easy

[paper compass]

If you have no clue. Then check the proofwiki someone linked above.

[worldly jewel]

you just pair up the divisors as d and n/d

[full frost]

you get (4x3x7)^2

[proper nova]

I can just google ts up and get an answer

[worldly jewel]

if there are k divisors then there are k/2 pairs (assuming n is not a square)

that's where the n^(k/2) formula arises

[leaden iris]

okkk

[cinder flower]

amen 0lante

[full frost]

which is 84^2

[worldly jewel]

awomen slayla

[proper nova]

Oh yeah, the d and n/d cancels to n

[leaden iris]

what-

[full frost]

and you count the number of divisors (4+1)(2+1)(2+1)

[leaden iris]

huh-

[worldly jewel]

suppose n=12, then its divisors as 1, 2, 3, 4, 6, and 12

you pair 1 with 12, 2 with 6, 3 with 4

[leaden iris]

yes

mhm

[worldly jewel]

each pair multiplies to 12

[full frost]

and yeah then by the pairing thing you get (84^2)^45/2=84^45

[leaden iris]

yeah

[worldly jewel]

therefore the product of all divisors is 12^3 since there are 6/2 = 3 pairs

[leaden iris]

okk

well

[proper nova]

How would you do sum of divisors then

[leaden iris]

so r v jst supposed to find all the pairs for the divisors of 7056

isnt that very legnthy

?

[worldly jewel]

when n is a square number, you have to be careful with the square root since it cannot be paired up with anything else

[leaden iris]

mhm

[full frost]

not really since there is a way to count divisors

[leaden iris]

which is..?

[full frost]

if you split a number n into its prime factorisation

[proper nova]

Prime factorization

And then use the powers

[full frost]

for ex 24=2^3 x 3^1

[proper nova]

So the amount of factors is (3+1)(1+1)

[leaden iris]

does denote all it's prime factors not all it's factors

[leaden iris]

i dont get uT_T

[full frost]

we take the exponents and make it (exp1+1)x(exp2+1)x...

because

[proper nova]

Adds each power with 1 and times all of them

[full frost]

if there a divisor

[leaden iris]

wait

[full frost]

it is this form because each of those primes to some exponent

but the exponent could be 0

so we add 1 to the exponent that divides

[leaden iris]

like u said

3x2^3=24

since the power of 3 is 1 and the power of 2 is 3

[full frost]

but in the divisor it could be 0

so we need to account for that

[leaden iris]

(exponent of 3+1)(exponent of 2 +1)

[full frost]

adding 1 to each of those exponents a nd multyplying'

[full frost]

yup

[leaden iris]

wat-

[full frost]

this gives the number

[full frost]

we have 2

which is the divisor of 24

[leaden iris]

yep

[full frost]

but it is also the form 2^a x 3^b

but a=1 b=0

[leaden iris]

T_T

[full frost]

cause any natural number ^0 is 1

[leaden iris]

b=0 is always possible

[full frost]

exactly

[leaden iris]

ok?