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[knotty sierra]

hi

[knotty sierra]

my first approach to this problem was completing the square but idk how

[warm breach]

domain of the function would be [2,38]

[knotty sierra]

yeah

[small jackal]

Are you allowed to use calc

[knotty sierra]

no

it says in the question you are not allowed

u could just find minima with derivative

[warm lynx]

isnt it just 2 and 38

cause the maximum is 6root2

and the minimum is 6

[warm breach]

2 and 38 is domain

i thought that too

[warm lynx]

that is the answer just extreme cases

[knotty sierra]

how do u prove that

[warm breach]

ok i just confirmed from banished source

that when we have domain we could do like:
put extreme points and the symm. point i.e 20 ({38+2}/2)

[devout snow]

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@knotty sierra Has your question been resolved?

[somber ore]

Very hard algebra problem. Find E

[lavish prawn]

Sooo by shifting the function (x -> x + 20) and rescaling it (x -> x/18), this essentially asks for the minimum of sqrt(1 + x) + sqrt(1 - x)

If we square that, we get 2 + 2 sqrt(1 -x^2) I think? maybe?

And that function is a lot simpler and its minimum can be immediately read off

I'm leaving the details to you

[devout snow]

@knotty sierra Has your question been resolved?

[knotty sierra]

how would i explain it tho

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[tardy rose]

I need help understanding this

[devout snow]

@tardy rose Has your question been resolved?

[tardy rose]

<@&286206848099549185>

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@tardy rose Has your question been resolved?

[tardy rose]

scratch that

explain this

[bright scarab]

Hey

[tardy rose]

hey

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[limpid bridge]

i know the whole shift part and which part of the equation moves/makes the graph im stuck on this part which is doing it backwards starting from the graph and ending with a equation

[devout snow]

@limpid bridge Has your question been resolved?

[jade oak]

First identify what parent function you think can be transformed into the graph. Once you have the parent function, lets call it f(x), think about how you can replace a, b, c, and d in af(b(x-c))+d to make certain features match up. Like c and d always represent horizontal and vertical shifts, while a and b represent stretches/compressions (or for trig functions you may think of it as amplitude/period)

[frosty gyro]

^ and which trig function do each look like?

[limpid bridge]

Ok thanks Imma keep this open as I work thru it in case of extra questions

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[void knot]

how can i calculate $\int_1^2dx\int_0^1x^{y} \ln x dy$

[woven radish]

riyobi

[heavy current]

well, the inner integral is just going to be a number

so you can technically factor it out...

you should probably compute it's value first though

any thoughts?

[feral agate]

That dx looks like it’s in the wrong place

[heavy current]

it is not

this is common notation in, e.g physics

[feral agate]

Oh alright then

So it’s still just a normal double integral?

[final sand]

maybe solving the inner one with respect to y (not sure just throwing things out)

[heavy current]

yeah, I suspect this is a better idea

[feral agate]

Yeah

This

[heavy current]

it will turn out to just be a constant though

[final sand]

yes

[feral agate]

How?

[final sand]

then the outer with respect to x id think

[feral agate]

Wouldn’t it be x-1

[heavy current]

oh, hold on

am I being silly?

[feral agate]

I think you put the 0 and 1 in the base instead of in the exponent

[trim bronze]

you can factor the ln x to outer integral

the inner is just int x^y dy

you use an exponential trick to solve this

[heavy current]

I believe you are correct haha

[feral agate]

You can, but it’s actually better to keep it inside

[heavy current]

I am indeed being silly

[final sand]

i did it and yes it’s x-1

[heavy current]

yes

[trim bronze]

oh true

[heavy current]

my mistake

I'm too tired for this

[feral agate]

I’ve never seen this before though

What’s the point even

[heavy current]

to remind yourself what you are taking the integral wrt to

my physics lecturer did this, and said it's very common

[feral agate]

Huh, weird

[heavy current]

when you have dxdydz, ig it can be hard to remember at times

[void knot]

what's the trick

this?

[feral agate]

Remember that $\frac{\dd}{\dd{x}}\left(a^x\right)=a^x\log a$

[feral agate]

Yeah or this

[woven radish]

kheerii

[trim bronze]

yes

[void knot]

got it, thank you

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[trim bronze]

https://tenor.com/view/darkfantasykanyewest-gif-25617862

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[hot ferry]

Anna's Gedanken light clock has a height of 1m between the mirrors, and relative to Chloe her spaceship is travelling at 90% of the speed of light. One tick is the time for light to go from one mirror to the other.

[hot ferry]

As the light takes a zigzag path in her frame, Chloe sees the clock ticking at a slower rate, te.

d) What is the tick time of the clock in Chloe's frame?

I don't have any idea how to do this

(this is special relativity physics by the way)

i found the answer by dividing the lorentz factor by the speed of light but that literally was just me putting in random things until i found the answer and i need to actually understand and derive the calculation

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@hot ferry Has your question been resolved?

[hot ferry]

<@&286206848099549185>

[stark wigeon]

not many people can help here. try discord.gg/physics

[hot ferry]

i know, i hate to post physics here bc i feel bad

but that server has a lot slower response time, i've also posted there but this server tends to have a lot of people who with crossovers with maths and physics so i get responses faster here.

[stark wigeon]

oh yea makes sense then

[devout snow]

@hot ferry Has your question been resolved?

[covert nexus]

I'm trying something that I don't remember much, what was the answer by the way?

[hot ferry]

7.6 x 10^-9 s

[covert nexus]

okay, give me a second

Something we can do in a simple way is calculate the ideal time of a tick and multiply it by the Lorentz factor to obtain the time from Chloe's perspective, since in this case we can use the Lorentz factor as a quantifier of the existing time difference.

[hot ferry]

???

[covert nexus]

I'm trying to translate, in general what we can show as equations is this
time of the tick = t1 = distance between mirrors/speed of light

To obtain the time it takes for the tick to occur at 90% of the speed of light, we use the Lorentz factor.

where v is the speed at 90% of the speed of light. If we calculate this factor it will be something like 2.3
If we calculate t1, it will be something like 3.33*10^-9 [s]
If we multiply this time t1 by the Lorentz factor we can obtain the time of a tick when being at 90% of the speed of light

I don't like to translate, but I don't know if you can understand what I'm trying to say.

[hot ferry]

what language do you speak?

i will try to do what you are saying and see if it works

[covert nexus]

i speak spanish

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[hot ferry]

it worked by the way!

[hot ferry]

thank you!

i have been learning spanish so it may have worked

[hot ferry]

i'd like to learn how to speak spanish with maths concepts

can i add you and possibly you could help me a little?

[covert nexus]

No problem, I will try to help as much as I can.

[hot ferry]

Gracias!

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[copper harbor]

Sketch the region corresponding to the system of constraints. Then find the minimum and
maximum values of the objective function (if possible) and the points where they occur,
subject to the constraints.
Objective function:
z = 4x + 5y
Constraints:
x ≥ 0
y ≥ 0
2x+3 y ≥ 6
3x - y ≤ 9
x +4y ≤ 16

[tidal basin]

did you sketch the region?

[copper harbor]

i'm doing it rn

nvm

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[hard oar]

Excuse me, is there a way to simplify this or is this it?

[cold bone]

I don't think you can simplify furthur

[hard oar]

Oh, alright thanks mater

[cold bone]

mater💀

[hard oar]

Mate

My bad

Typo

[cold bone]

nw

[hard oar]

Btw, is this the correct answer? Because I don't really know the concept of the question means.

Its on number 8

[cold bone]

there's a subtle mistake

in second last step

[left flower]

simplify sqrt(75) also sqrt(12)

[left flower]

$\sqrt{x^2 \times 3} = x\sqrt{3}$

[woven radish]

Newt

[left flower]

so for this

$\sqrt{4 \times 3} = \sqrt{2^2 \times 3} = 2\sqrt{3}$

[woven radish]

Newt

[hard oar]

So basically its like this? On number 8

Alright thx mate

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[quick peak]

What's the difference between moment of inertia and moment?

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[plush cobalt]

right or wrong

[mild sorrel]

,w a^n+1 + b^n+1/(a^n+b^n) = (a+b)/2

,w (a^n+1 + b^n)/(a^n+b^n) = (a+b)/2

[woven radish]

Results provided by WolframAlpha

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Requested by anshulyolo121

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[devout snow]

@plush cobalt Has your question been resolved?

[plush cobalt]

?

[mild sorrel]

.CLOSE

[plush cobalt]

this is mine

[mild sorrel]

oh sorry

.close

jk

[plush cobalt]

yo ^^

[mild sorrel]

[plush cobalt]

this one ?

[mild sorrel]

I am in class

[plush cobalt]

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[median pike]

Apparently my algebra is wrong here. I simply added cos in every term. Is this not permitted?

[mild sorrel]

,rotate

[median pike]

the answer is supposed to be x = 1 apparently

[mild sorrel]

nvm

[median pike]

I also tried this way, which is simpler, and I get the same result

[mild sorrel]

the original question is $4cos^-1 (x) - 2 * pi = 2 * cos^-1 (x)$

[woven radish]

Meolve

[mild sorrel]

how you add cos?

wdym 'add cos'?

[median pike]

sorry it's -

[mild sorrel]

,w 4cos^-1 (x) - 2 * pi = 2 * cos^-1 (x)

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[median pike]

In step 3 you could simply divide by 2 and then add COS in both sides

[mild sorrel]

you did a mistake

[median pike]

that gets you x = 1

however I don't understand why what I do gives me x = 1/2

[mild sorrel]

here

ohok

[median pike]

4cos^-1x - 2cos^-1x

[mild sorrel]

i gtg

[median pike]

Why do I get 2 different answers? What I am doing wrong in the first approach?

[restive mauve]

from step 1 to step 2, when u moved the 2 cos inverse x to the other side, it should be negative not postive

[median pike]

yes

My main issue with the last 2 pictures

[restive mauve]

ur last picture is correct

[median pike]

and why the one above is not?

I multplied everything by COS

wait

I think I know

[restive mauve]

cos ((cos inv x) + (cos inv x)) is not the same as x + x

[median pike]

yea

it's a cos function that's not how it works lol

thanks

[restive mauve]

mhm!

[mild sorrel]

you told the answer should be 1.

But it's -1

[mild sorrel]

.

[devout snow]

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[dusky sundial]

Help

[cold bone]

How you are supposed to do this?
Pure algebra or some fancy method?

[dapper fable]

i think first you square the first equation

to find ab + bc + ac

[cold bone]

that's literally the pure algebra and most disgusting way too

[restive river]

nah

[potent nebula]

yeah but is there a better way

[dapper fable]

exactly

[potent nebula]

i’m interested in one

[warm breach]

uhm

[restive river]

(a+b+c)(a²+b²+c²-ab-ac-bc)+3abc = a³+b³+c³

you know abc from this

[warm breach]

you guys are gonna kill me, if you know what I thought

[restive river]

now (a²+b²+c²)² = a⁴+b⁴+c⁴+2(a²b²+b²c²+a²c²)

[warm breach]

is ans. 46

by any chance

[restive river]

(ab+ac+bc)² = a²b²+a²c²+b²c²+2abc(a+b+c)

[warm breach]

wait it's following a pattern

I am smart

[potent nebula]

if u find abc and ab + bc + ac we can use newton sums ong but idk if it’ll be any faster than the raw algebra but it could be useful for if you ever wanted to find like the 20th powers or some shit

[warm breach]

a^n + b^n + c^n = (n+1) + 2n

[cold bone]

that's definitely not gonna work

[warm breach]

wait no

I thought wait

yes nvm

I am wrong

[cold bone]

bro how to @ the OP

I can't find them\

[mild sorrel]

@dusky sundial

[warm breach]

.

[cold bone]

how did you do it?

[warm breach]

(Pinned msg)

[cold bone]

oh okok
I'm dumb

[cold bone]

it's not on my screen

[devout snow]

@dusky sundial Has your question been resolved?

[velvet rock]

it has to do some stuff with the Identities

[warm breach]

I think best way to simplify it

the old and scary one

[velvet rock]

why I feel an intuition that the binomial theorem has some relation with this question?

[cold bone]

trinomial you mean?

[velvet rock]

yeah, some what of that

[cold bone]

is there even a general formula for that?

never heard of any

[warm breach]

uh wait

[velvet rock]

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

[warm breach]

nono

[cold bone]

wtf

[warm breach]

wait I clicked

[velvet rock]

yeah, wikipedia

[chrome isle]

If a b c are integers there is no solution

[velvet rock]

I also never heard of it, new stuff to learn about

[velvet rock]

yo, how?

[chrome isle]

Ten as sum of 3 sqaures can be 1² + 3² + 0² and that satisfies the first eq but dosent satisfiy the 3rd

[warm breach]

yeah I thought of something but it's useless

[velvet rock]

ah, you are trying brute force...

but actually, yeah

shouldnt be valid

[chrome isle]

Not exactly I mean the only way to write ten as sum of 3 sqaures is this and it's quite easy to check

[velvet rock]

I mean yeah

but the question doesnt want us to do it this way

different methods run in different directions

[chrome isle]

So abc must be real Or even complex

[velvet rock]

yup

as per this pov

well then, this means

[velvet rock]

@dusky sundial , more context please

[chrome isle]

Question want s us to use a+b+c whole square and a+b+c whole cube

[velvet rock]

yeah, Identities, right?

[chrome isle]

Yup

[velvet rock]

means basically the trinomial expansion

here we go again

[chrome isle]

I mean if it was 28 and not 22 life would have been so good lol

[warm breach]

[velvet rock]

the textbook doesnt want the life to be good, it is its purpose

[chrome isle]

True lol

[velvet rock]

or be the internet

[devout snow]

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[north roost]

!help

[devout snow]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[north roost]

i dont know what i did wrong

🥺

[north roost]

<@&286206848099549185>

hullo?

[lost laurel]

What's the question?

[north roost]

Find ${\int_{C} (x^2+ y^2) dy }$ along the curve ${C}$ joining the points ${(0,0), (0,3), (3,3)}$ as shown in the diagram.

[woven radish]

:D

[devout snow]

@north roost Has your question been resolved?

[devout snow]

@north roost Has your question been resolved?

[long pasture]

Shouldn't it be
C_1=‹t,0›
C_2=‹3,t›
?

[north roost]

Yes

I think I should go to sleep 😭😭

Thanks a lot🙏🙏

.close

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[heady bane]

how is this wrong

nvm i got it

.close

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[last burrow]

solve 7x+12y = 220

[last burrow]

integal

integral

Please preserve the integrity

[tall knoll]

what do you mean "solve"

[devout snow]

@last burrow Has your question been resolved?

[mild sorrel]

in positive integers

divide the equation by 7

x+y + 5y/7 = 31 + 3/7
5y/7 - 3/7 must be an integer
multiply by 3 so that the difference is 1 less and you can separate so that y is left with coefficient 1.
2y + y/7 - 1 -2/7
(y-2)/7 must be an integer
y = 7t+2 (general solution)
substitute in the original equation to get x.
Find positive integers.

solutions

Oh nvm

Wait, jsut put t = 0 to get y = 2 and x = 28
so x = 28-12t

https://tenor.com/view/stupid-people-johncarlduck-johncarlroat-gif-1065725086417614870

Has your question been solved lil bro?

[devout snow]

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[mild sorrel]

||Disclaimer: It was only my friend so don't worry||

[devout snow]

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[mild sorrel]

.close 💩 Bro atleast try once to solve it

[mild sorrel]

.close

[hearty sparrow]

hi! i just need someone to check my 7 questions to see if they are correct or not (:

[lunar harbor]

,w find t(58cos(19 deg)) if -1.9=(58sin(19 deg))t-(1/2)(9.81)t^2

[woven radish]

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[lunar harbor]

[lunar harbor]

,w x^2+y^2+7x-4y=0 polar form

[woven radish]

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[lunar harbor]

,w expand (x+4)^2+5(x+4)

[woven radish]

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[devout snow]

@hearty sparrow Has your question been resolved?

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[kindred belfry]

Please help me, i dont understand my own note

😭

I dont even know if its correct

[warm breach]

it's 1^infinity form

[dapper fable]

what is "tan x 1"?

[warm breach]

i suggest redoing it by taking log and then keeping log(x) when x ending to 0 as x-1

[kindred belfry]

TwT i dont know

[kindred belfry]

How..

[warm breach]

log y = cot 6x log(1+tan3x)
log y = cot 6x tan 3x
y = e ^ (cot6x tan 3x)
limit will be ending to 0 in exponent (whether I mention it or not)

(cos 6x/ sin 6x)(sin 3x/cos3x)

sin 6x can be written as 6x and sin 3x as 3x so which will give you 1/2 in exponent, as cos 0 is 1

[kindred belfry]

Ohh hmm

[kindred belfry]

I dont understand the log part

I dont know how do i do log

Im sorry

[warm breach]

why are you sorry lol

it's completely fine if you don't understand

[kindred belfry]

TwT because i make people busy

[glossy dew]

have you learnt about log?

[kindred belfry]

Thankyou

[kindred belfry]

Uhh, maybe but i forgot

[glossy dew]

do you know what $log(a^b)$ simplifies to?

[glossy dew]

to start with

[warm breach]

do you know taylor series?

crazeie

[kindred belfry]

I dont know T T

[kindred belfry]

Hmm i dont know

[warm breach]

uhm then I think you wouldn't understand what I did

[kindred belfry]

Yeah.. T T

[warm breach]

have you learnt something like

if x ending to 0 then you can replace log (1+x) as x

in limits?

[kindred belfry]

Havent..

[warm breach]

yeah then you wouldn't, we try diff approach i guess

L Hospital?

do yk it?

[kindred belfry]

Uh, no

[warm breach]

uhm

[kindred belfry]

uh

Hmmm

[warm breach]

i think wait for another helpers

as idk any diff approach

XD

[kindred belfry]

Yeah 😅

[warm breach]

btw never be sorry for things you don't understand

[kindred belfry]

My friend gave me this

[kindred belfry]

Thankyou

[kindred belfry]

But i still dont understand

[warm breach]

yeah he did same as me

[kindred belfry]

Ohh

[warm breach]

can you send me your notes page of standard limits?

[kindred belfry]

This?

[warm breach]

uhm

[kindred belfry]

They always gave example beside the question

[kindred belfry]

Uh

[warm breach]

OHH

ok

lmao they have directly given you

to learn some formd

without derivations

[kindred belfry]

I know they just erase details

[warm breach]

now idk how your friend brought it in form of
(1 + f(x))^1/f(x)

I mean he kept cot 6x as 1/tan 3x as it gives same value?

on that limit

[kindred belfry]

I dont know.. hm imma ask

[dapper fable]

(1 + f(x))^1/f(x) approaches e as x aproaches 0 i think

[warm breach]

they derived it from log

[dapper fable]

yeah

[warm breach]

and then applied taylor

and then asked their students to remember it

[dapper fable]

does your book provide a proof?

[warm breach]

never opened

btw my book doesn't have taylor series

or L Hospital

everything is easy in ut

i just learnt it in my coaching class

[kindred belfry]

So uh, my friend said
They use ^1/tan3x × tan3x just so they can get e

Because this is e

[cold bone]

this is not!

take limit as f(x) tends to zero

then it will be

[kindred belfry]

Right

[kindred belfry]

Wait, its not limit to infintiy?

[cold bone]

bruh

no

,w lim_x->infinity {(1+x)^{1/x}

[woven radish]

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[cold bone]

wtf

[cold bone]

no

[kindred belfry]

Oh dang

[cold bone]

see this

[kindred belfry]

💀omagad

[cold bone]

[kindred belfry]

Oh wait you are right

[kindred belfry]

I misread

[warm breach]

[kindred belfry]

[kindred belfry]

Right

[cold bone]

f(x) tends to zero

[kindred belfry]

[cold bone]

not x

[kindred belfry]

Oh wow

This time i didnt misread

[kindred belfry]

Thankyou for fixing

[warm breach]

[kindred belfry]

Thankyou y'all!

[warm breach]

ques is still left

[cold bone]

what question?

[kindred belfry]

[warm breach]

pinned message

[kindred belfry]

I think i start to understand it

[cold bone]

this?

[kindred belfry]

Yeah, its not right

[warm breach]

yh

[kindred belfry]

My friend gave me this

[warm breach]

Ik how to do it

[kindred belfry]

Is it correct now

[warm breach]

but my approach is diff

[kindred belfry]

I think i kinda understand it now tho

[kindred belfry]

Yea.. i kinda understand it now

[warm breach]

btw what book is that

[kindred belfry]

Its my school book

[warm breach]

kinda or completely

[kindred belfry]

Hm kinda

[warm breach]

I mean official book from gov?

[kindred belfry]

Yeah

[warm breach]

oh

[kindred belfry]

Maybe i need more practice

[warm breach]

yes true

[kindred belfry]

Imma close for now

Thankyou all!

.close

[devout snow]

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[cosmic junco]

I know that when there are different things shown such as ^2 its usually displayed on graph as bounced

but, the x

the multiplied x, does that just mean an extra one plotted at (0,0)??

Like, I understand (x-4) has no exponents so it goes straight through

(x+2) would be x = -2 which bounces off of it

I just don't get what the multiplication of x has to do with all this

[safe jasper]

yes it just means going straight through (0,0)

your reasoning for (x-4) is that it goes straight through 4

multiplying by x is the same as multiplying by (x-0)

no reason this case is any different

[cosmic junco]

what if it was like 2x or something else?

oh?

[safe jasper]

well 2x is the same thing as 2 (x-0) so you still have it passing straight through 0

[cosmic junco]

can you explain that a bit more, someone told me about this and I didn't understand it

[safe jasper]

all that the 2 does is vertically stretch your function

[safe jasper]

why did you say that the sketch would pass straight through 4?

based on the (x-4) term?

[cosmic junco]

yeah

[safe jasper]

and you're confused about the extra x being multiplied

[cosmic junco]

yes

[safe jasper]

you bounce off -2 since you have (x+2)^2

you go through 4 since you have (x-4)

so x is just (x-0)

or (x+0)

which means 0 is a root of your function with multiplicity 1

[cosmic junco]

oh

[safe jasper]

multiplicity is how many times the root shows up in your function

so (x+2)^2 has -2 as a root with multiplicity 2

(x-4) has 4 as a root with multiplicity 1

and x has 0 as a root with multiplicity 1

when there's an odd multiplicity you pass through the point, when it's even you bounce off

[cosmic junco]

ohhh, I see

wait so

[cosmic junco]

this right here

[safe jasper]

the 2 doesn't change anything, it just vertically stretches your function

[cosmic junco]

the 2(x - 0) thing would still have me at (0, 0)

[safe jasper]

for example, consider the function y = 2x vs y = x

both have roots of multiplicity 1 at 0

the only difference is 2x looks "steeper"

it's been stretched vertically

roots didn't change

[cosmic junco]

would it change how the graph looks?

[safe jasper]

just this

[cosmic junco]

I don't have a clue what that is either

[safe jasper]

imagine y = x^2

normal parabola right

[cosmic junco]

yeah

[safe jasper]

y = 2x^2 is just that same parabola but steeper

y = 2x vs y = x, it's the same line but steeper

[cosmic junco]

ohh

so I would graph it steeper?

[safe jasper]

y = sinx vs y = 2sinx are the same sine functions but one is twice as tall

etc

it's like graphing the original function on stretchy paper, and pulling at the top and bottom of the paper to stretch your function

[cosmic junco]

How steep it went after the (0, 0) point, is that significant?

[winter torrent]

it's helpful to play with this in desmos

[safe jasper]

since it's just a sketch it probably won't matter

main concerns are probably the zeroes and turning points

[cosmic junco]

I see, okay

[winter torrent]

https://www.desmos.com/calculator/5chfrrupig

[cosmic junco]

Thank you, I got it noted down

[cosmic junco]

I will check this out too

.close

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[turbid flame]

Hi there, can someone walk me through the logic of the solution please? I completely messed up this question. I did (2 x 7 x 1)^3 but I'm very certain the logic in which I got the (2 x 7 x 1) part was incorrect anyway...

[turbid flame]

forgot to give this

[winter patrol]

...

that's a big difference

[turbid flame]

..... ik

sorry 🙏

[fervent hornet]

lmao

[winter patrol]

why raise to the power of 3

[turbid flame]

Idk, I mistakenly calculated using the 6 as well (but you can't get greater than 600 so... I would have done to the power of 2 though)

[winter patrol]

how many possibilities are there for the
first digit
second digit
third digit

[turbid flame]

2, 7, 2(?)

[winter patrol]

why 2 for the third digit

[turbid flame]

but last one can only be 2 if it's not 400

[fervent hornet]

why 2 for the last digit?

[turbid flame]

0 or 5 (divisible by 5)

[winter patrol]

0 isn't in the list

[fervent hornet]

is there a 0 in the list

[turbid flame]

AH!!!

Blunder i say!

thanks! 😆

.close

[devout snow]

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[lost crag]

a > b, c > d
a/c > b/d
?

[winter patrol]

what's the question

[misty stream]

Put numbers and see if its true

[winter patrol]

are you being asked to prove/disprove?

[misty stream]

Probably to tell if its correct or not

[lost crag]

just did, damnit

[misty stream]

And

[lost crag]

a > b, c > d
ad > bc?

[glossy dew]

not necessarily

[cold bone]

if they are all positive then yes ig

[glossy dew]

ONLY if a, b, c, d are positive

[lost crag]

might have hit the jackpot then

[glossy dew]

lucky

[lost crag]

i thought of it at first but didnt know if it would hall cause of 0

but i realized im dealing with exponentials

.close

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[vivid wing]

This is my teachers working out. (Probability normal distribution)
How did she solve for M, what did she do to get to M? She missed a chunk of working out

[vivid wing]

It’s supposed to be 193, not 195
1.4051 = 193-M/150-M/-0.7722

I thought you would just do solve N, but it came up with no solutions

[devout snow]

@vivid wing Has your question been resolved?

[devout snow]

@vivid wing Has your question been resolved?

[upper schooner]

Do you have the original question? Presumably you have somewhere else that also gave you -0.7722 = (150 - mu)/sigma, at which point rearrange that one and solve for mu, hence sigma

[devout snow]

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[upper schooner]

E.g. from her second line, you can then rearrange that to get 1.4051(150 - mu) = -0.7722(193 - mu), expand both sides, rearrange and find mu that way

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[limpid sentinel]

i tried taylor but i got stuck on ln(x)

[devout snow]

@limpid sentinel Has your question been resolved?

[faint gorge]

Maybe it's important to note from which side we are approaching too

From the right side it just seems to be [0 - 0²ln(1)]

but from the left we don't even get to approach x = -1 because we would have ln(0)

[finite briar]

its a divergent limit

[limpid sentinel]

i mean its from a calc book and they got 1/2

[faint gorge]

lol

[limpid sentinel]

they specifically asked to use taylor

[faint gorge]

,w plot y=x-x^2ln(1+1/x) between -2 and 2

[woven radish]

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[limpid sentinel]

strange

welp

[faint gorge]

1/2 nice

[limpid sentinel]

yea thats weird but that can happen

ig

[faint gorge]

,w 1/2 = x-x^2ln(1+1/x)

[woven radish]

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[limpid sentinel]

no wonder

[faint gorge]

Wait

[limpid sentinel]

maybe its to inf

typo maybe

[faint gorge]

,w Limit[x-x^2ln(1+1/x), {x-> infinity}]

[woven radish]

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[limpid sentinel]

yea

[faint gorge]

Yep

[limpid sentinel]

thats what i tought

th

[faint gorge]

ahhhhhhh

[limpid sentinel]

lmao

i get itnow

[limpid sentinel]

i use taylor of ln(1+x) with 1/x

simplifies nicely

[faint gorge]

ye

[devout snow]

@limpid sentinel Has your question been resolved?

[devout snow]

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[sullen stone]

Hello I need help figuring out the following:

I observed that parabolae translate in what seemed to be a parabolic trajectory when modifying only one of the parameters and leaving the other fixed like so: (x+a)(x+b) | a=0, b is variable.

I could confirm this by finding the roots for the derivative of p(x)=ax(x+b) when equal to 0.

This yielded the x translation in terms of b:

$\frac{-b}{2a}$

This allowed me to obtain the y translation in terms of b:

$\frac{-b^2}{4a}$

Which made possible to obtain the y translation in terms of x, where x is a function of b (x=-b/2a):

$-ax^2$

This confirmed the conjecture that the vertex translates in a parabolic path.

I later checked this also works for cubics’ extrema, and seems to hold for linearly-decomposable polynomials in general, though not when all fixed parameters are 0, seemingly.

I attach one link for [the parabola example](Parabola translations | Desmos)

And another for the cubic example

My questions are:

• Why do polynomials translate their extrema in parabolic trajectories in this way?
• Do other functions also do this, perhaps following other trajectories also defined by a function?
• Could you please point me to reading material about this, if you know of any?

Thank you!

[woven radish]

Jake The Human

[devout snow]

@sullen stone Has your question been resolved?

[sullen stone]

<@&286206848099549185>

[devout snow]

@sullen stone Has your question been resolved?

[devout snow]

@sullen stone Has your question been resolved?

[devout snow]

@sullen stone Has your question been resolved?

[vivid wing]

THANK YOU!! 🙏🙏🙏

[devout snow]

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[lethal vault]

when i solve for x i get 5/4 but when i put the value and solve then its wrong. if i solved correctly for x then why when i substitute the value of x in the equation it is not satisfied

[cold bone]

because it doesn't have any solution

at least not any real one

[lethal vault]

but when i solve i get 5/4

[cold bone]

To solve it you must have squared it

[lethal vault]

yes

[cold bone]

and whenever we square in any equation
the possibility of unnecessary solution arises

[lethal vault]

huh

[cold bone]

like x=2, if we square it we get x^2 = 4
implies x^2 - 4 =0
solving further by factoring (x-2)(x+2)=0
from this we get x= 2 and -2

hence whenever we square in any equation we must check if the final answer satisfies the original equation or not

[lethal vault]

both 2 and -2 satisfy the eq

[cold bone]

no

x=2 is our real equation

it is only satisfied by 2

[lethal vault]

oh

[cold bone]

got it?

[faint gorge]

When you square a square root outside you have to consider the domain

By squaring you solve the equatipn but you have to keep the domain restricted to depending on your previous terms

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

That only holds as long as x or rather the argument is non negative

,w plot sqrt(x)^2

wolfram smh

[lethal vault]

domain?

[faint gorge]

yes

If one part is not satisfied for one solution then the solution is simply invalid

[lethal vault]

these conditions are not given in the question. how do u know these need to be there

[faint gorge]

You have to define a common domain where $$x + 1 \geq 0 \textbf{ and } x - 1 \geq 0 \textbf{ and } 4x - 1 \geq 0$$ are all satisfied.

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

Is the square root of a negative number defined?

[lethal vault]

no

[faint gorge]

Thats how you know it

[lethal vault]

imaginary numbers

[faint gorge]

x+1, x-1 and 4x-1 are all linear functions that can become negative

[lethal vault]

cant the answer be an imaginary number

then x-1 could be negative

[faint gorge]

unless you are specifically asked, no

[lethal vault]

ok

i had another question

how to solve these exponent questions

[dapper fable]

i'm assuming lambert w

[cold bone]

why do you know about it

[lethal vault]

whats lambert w

[dapper fable]

W(ae^a) = a

[dapper fable]

?

[lethal vault]

i dont understand

[cold bone]

bruh what's happening in your school

what are you being taught in 7th-8th

[dapper fable]

rn prisms and stuff😭

[dapper fable]

W(x) is kinda like sqrt(x) where you dont exactly need to know what it is, but it's acceptable

[lethal vault]

dispersion

[dapper fable]

like sqrt(2)

[lethal vault]

u dont need to know but its acceptable?

what dont u need to know

[dapper fable]

if you put it in a caculator, it will tell you

[lethal vault]

what

[dapper fable]

well x = 1/4 also works for this

[lethal vault]

how did u get that

[dapper fable]

for most like these you have to guess

[lethal vault]

i dont know what function is

[dapper fable]

mostly from the fact that sqrt(256) = 16

[lethal vault]

i also did guess but isnt there a way to get that answer without guessing

[dapper fable]

not really

[lethal vault]

what

[dapper fable]

but an easy way to know where to guess is trying to estimate

like from the question you could assum x <1

[lethal vault]

so we cant algebraically solve that

[lethal pollen]

Most square roots are approximate

[dapper fable]

(yes but it takes way higher level stuff)

[lethal pollen]

Because the only real way you can solve them is by a good guessing method

[lethal vault]

ok

prove this

[lethal pollen]

Nuh uh

!noans

[devout snow]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[dapper fable]

real

[lethal vault]

bro these trignometry questions

how do i know what to do

[dapper fable]

trig questions mostly just use experience

do a lot of them

[lethal vault]

these 10 questions made my life hard

[cold bone]

why this seems like some NCERT crap

[lethal vault]

how do i know when i have to take tan as sin/cos and when 1/cot and what not

[lethal vault]

it is

[dapper fable]

yo guys i suck at trig help me

[cold bone]

I am out then, I literally hate those textbooks

[lethal vault]

why

[dapper fable]

depends on the question really😭

[lethal vault]

hmm

[cold bone]

I almost failed my schools once after being dependent on them

[dapper fable]

what's NCERT?

[lethal vault]

only way that i can solve these are by memorising the steps

[lethal vault]

where u from

[lethal pollen]

cot a = 1/(tan a)
And
(a - b) (a+b) =a^2-b^2
Is the hint

[lethal vault]

how would i get to know that i have to take cot as 1/tan and not cos/sin

[dapper fable]

isnt it a^2 - b^2?

[lethal pollen]

Mb

Finger slip totally

[dapper fable]

probably when the question has tan(a) and sin(a) you need to take as sin/cos

[lethal vault]

do i solve lhs and rhs seperately

[dapper fable]

for most proofs you do that

[lethal vault]

my school says just take one side and solve it to get the other

they wont accept when i solve both the sides and show that they are equal

[cold bone]

here, try to go from left hand side to right one

it will be more simpler to see what to do next

[visual hazel]

left side is two simple identities, right side is simplification

[pure topaz]

for this you could write it as
(1/2)^(8x) = x
raise both sides by power of (1/2)
(1/2)^(4x) = x^(1/2)
(1/2)^(2x) = x^(1/4)
(1/4)^x = x^(1/4)
x = 1/4

the reason for raising both sides by 1/2 is to try to balance it

[lethal vault]

ok

[pure topaz]

could also use tan^2 + 1 = sec^2

right

[lethal vault]

yes

so is sec = 1-tan

[pure topaz]

that's also what I did

[lethal vault]

its just better to solve both sides seperately

when i solve LHS i get tan + sec

[devout snow]

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[humble sierra]

ez

[faint gorge]

Hypergeometrische Verteilung

in einer Stichprobe zufällig
n Elemente ohne Zurücklegen entnommen. Die hypergeometrische Verteilung gibt dann Auskunft darüber, mit welcher Wahrscheinlichkeit in der Stichprobe eine bestimmte Anzahl von Elementen vorkommt, die die gewünschte Eigenschaft haben. Bedeutung kommt dieser Verteilung daher etwa bei Qualitätskontrollen zu.

Ich hab zu erst an Binomial gedacht

Aber das ist eig mit ZK was hier wenig sinn macht

Weil man kauft ja und behält es ja

Deswegen ohne ZK

außerdem gehts um erfolg/misserfolg

deswegen bam

Ja bei mehrmalligem miss/erfolg

Übrigens wirst du eine Summenformel brauchen

Am besten gegenwahrscheinlichkeit wieder nehmen

Übrigens

Wie würdest du hie die Zv definieren

,,\mathbb P (X>8940) = 1 - \mathbb P(X\leq 8940) = 1 - \sum_{k=0}^{8940} \mathbb P(X=k)

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

Musst halt P(X=k) herausfinden also formel benutzen

[trim bronze]

pretty sure this is binomial

chance of success is always .1

[faint gorge]

binomial means with replacement tho

[trim bronze]

binomial is n experiments

where each probability of success remains the same

in that sense it is with replacement because chance of success doesnt change

it is the same case here

when 1 flower germinates

it doesnt affect others

it says that in the question

[faint gorge]

if you buy a flower whether germinated or not you dont give it back?

you keep it

[trim bronze]

well just take some examples

P(X=0) should be .1^10000 right

the chance of success doesnt change

just try examples

[faint gorge]

hmm ok

[trim bronze]

P(X=1) is just 10000 choices for the germinate * .1^9999 * .9

and so on

[faint gorge]

Ok I understand thanks damn

[trim bronze]

np 👍

[faint gorge]

Ok it makes perfect sense

[trim bronze]

germinating flowers

lol

says at most

so its P(X<=8940)

you prob need to write

it as summation expression

for the answer

yr

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[left ember]

Consider a truth table with A, B and A->B
F F T
F T T
T F F
T T T

I want to know if my rational checks out.

Consider the case F T T (= not A and B, then A->B is true)

I define not A as R. So R and B are true. Still A can result in B. (If I buy a lottery ticket then I win the lottery. But you still can win the lottery if you find the ticket and not buy it)

Is this a good rational

Sorry for my english

[left ember]

<@&286206848099549185>

[devout snow]

@left ember Has your question been resolved?

[devout snow]

@left ember Has your question been resolved?

[devout snow]

@left ember Has your question been resolved?

[tender cobalt]

its a bit convoluted but it works
you can shorten your rationale to "if B is true, then A can directly -> B, but something different can also -> B, so either way it would be true"

[humble sierra]

help needed

find sum of nth term whose nth term is 3^n - 2^n

[devout snow]

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[tender cobalt]

you need to ask the question in an available help channel instead of someone else's

[devout snow]

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[gloomy gazelle]

how do i do a and b

[gloomy gazelle]

how do 2 lines that are skew not intersect?

[thin fern]

that's what skew means

[hushed wraith]

to show they dont intersect u can show that there do not exist s and t for which they point to the same point

[gloomy gazelle]

oh

[gloomy gazelle]

how do i do that and if they dont intersect and arent parallel how do i calculate the angle?

[hushed wraith]

its not quite clear what they mean by angle between the lines...they probably mean the angle between the directions?

[hushed wraith]

if they did intersect, how would u have looked for the point of intersection?

[gloomy gazelle]

hmm

im not sure

[gloomy gazelle]

so dot product would work

[hushed wraith]

yes the dot product between the direction vectors of the lines

[gloomy gazelle]

yes

[hushed wraith]

and just to help u see, these are two skew lines that do not intersect

[gloomy gazelle]

basically they go "behind" or "in front"

[hushed wraith]

yeah u can think that

[gloomy gazelle]

aight

[hushed wraith]

so u got r = (i - 2j + 3k) + (4i - j)s

do u see that by putting different values for s, u will get an arrow that points to different points in space?

as u vary s thru different values, u get a vector that traces out a straight line

thats clear right?

[gloomy gazelle]

yes

i have an idea

[hushed wraith]

and similarly for the other line

[gloomy gazelle]

so

[hushed wraith]

yeah?

[gloomy gazelle]

there are 3 axes x y and z right

for each value of x for one line i can equate that to the x value of the other line

and do the same for y or z

and solve them simutaneously

[hushed wraith]

yes. that way u can find the intersection if there exists one

[gloomy gazelle]

and substitue those values of t and s in the axis that i didnt use

[hushed wraith]

if u can show that no such solution exists, then uve shown that theres no intersection

[gloomy gazelle]

and if the values arent equal i can show no intersectio

aight

[hushed wraith]

yes

[gloomy gazelle]

are there any other diff techniques?

[hushed wraith]

im not aware of any others...

[gloomy gazelle]

aight

thank you

.close

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[sturdy swallow]

How did we get (-1) here?

[olive snow]

e^(-x) --> -e^(-x) while derivating

[restive river]

aka chain rule

[olive snow]

ye e^u --> u'e^u

[sturdy swallow]

oh, so if it were e^12x it would be derived to 12e^12x?

These are the rules out professor taught us, and I dont see a constant in e^x

[umbral shuttle]

yes

[sturdy swallow]

is that because of the rule that f(g(x)) --'--> f'(g(x)*g'(x)?

[umbral shuttle]

yep

[sturdy swallow]

perfect, thank you 🙂

[umbral shuttle]

where e^x = f(x) and -x is g(x)

[sturdy swallow]

can I ask a sepperate question here also?

.close

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[umbral shuttle]

yeah

ofc

welp

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[latent lance]

how do i solve this

[tall knoll]

i'd start by just computing the given product

[boreal vine]

Plug in -25b into the funct, plug in 5b into the funct. You should now have 2 expression involving only b, then multiply them together and maximize the product wrt b

[devout snow]

@latent lance Has your question been resolved?

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[viral lynx]

Im really not sure what to do here

I set f_a(1) = g_a(1) but I got e^0 = e^(1-1a)

And I dont really know what I could do with that

[dense jay]

You verified they both have the point (1,1) regardless of a

You havent shown its a right angle yet though

[viral lynx]

Did I? 🙃

Accidental

I had no idea what I was doing

[dense jay]

f_a(1)=e( (1-1)/a)=e^0=1
g_a(1)=e^(1-1^a)=e^0=1

Youve done that so far

[viral lynx]

e^(1-1^a) = e^0 ?