help-23

[gentle hedge]

it's unituitive imo and it's not rigorous

[calm cloak]

how about using L'Hôpital's rule

[gentle hedge]

how though

some additional context is this is an exam question

[frigid locust]

try rewriting that to lnh / (1/h^2) and then read the logic they used

[gentle hedge]

and this part of the question is probably worht like

[plucky elk]

Why not

[gentle hedge]

2-3 marks

so i want a method that takes 2-3 minutes

[calm cloak]

$\lim_{h \to 0^+} h^2 \ln h = \lim_{h \to 0^+} \frac{\ln h}{1/h^2} = \lim_{h \to 0^+} \frac{1/h}{-2/h^3} = \lim_{h \to 0^+} \left(-\frac{h^2}{2}\right) = 0$

[flat frigate]

| caesar |

[gentle hedge]

i mean once i substitute h= e^n then i can see it

[calm cloak]

if you really wanna calculate the limit like this

[gentle hedge]

oh fair

that works i think

tysm

[calm cloak]

!done, if you dont have any more doubts

[safe radish]

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

[gentle hedge]

.close

[safe radish]

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[safe radish]

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[errant parcel]

https://www.analyzemath.com/stepbystep_mathworksheets/vectors/cross_product.html

On this website, the first diagram saying what the cross product of u and v are defined by contains a 3x3 matrix with i, j, and k on top.

What does the i, j, and k mean?

[simple gazelle]

the basis vectors

let me draw sum

[errant parcel]

ok, for your reference i have no background in linear algebra whatsoever

[arctic raven]

i is 1 step in positive x direction, j in y direction, k in z

[errant parcel]

this is my first time learning it

oh

so i is a unit vector in positive x, and j and z is that?

basically just converting the scalars to vectors along those axis right

[arctic raven]

or if you like, $\mathbf i=(1,0,0),\mathbf j=(0,1,0),\mathbf k=(0,0,1)$

[flat frigate]

Flip

[simple gazelle]

prob fucked up the coloring scheme

[errant parcel]

ok ok

got it guys

thank you

so how do i close this

[simple gazelle]

do .close

or sum

[arctic raven]

you type .close

[errant parcel]

.close

[safe radish]

Channel closed

Closed by @errant parcel

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[simple gazelle]

also remember that every vector is defined using basis vectors

[errant parcel]

yep, will do

[safe radish]

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[gentle hedge]

is it possible to do this question as is or is there a typo for the P(Y=y) function

[gentle hedge]

they're independent

[scenic dove]

there's not really a question in that screenshot is there

[gentle hedge]

mb

i dont want help with the question

i just need to know

if there's a typo

or if thats the actual probability fucntion

[scenic dove]

oh it should probably be y=2,3,6 instead

rather than x=

[gentle hedge]

ok ok

ty

.close

[safe radish]

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[lapis token]

I have a subset A of the real numbers where the elements are labeled using two different indexes from index sets
The sets are called I and J
My subset elements are labeled a_i,j

How do I interpret this subset and how do I prove this statement

[opaque fern]

it would be kind of like a matrix i assume?

[lapis token]

I thought of it as a table with I for the columns and J for the rows

[opaque fern]

so the middle part is the supremum of the row supremems

and the right part is the supremum of the column supremums

[opaque fern]

yeah u totally can

,, A = {a_{i,j}\where i\in \mathcal I, j \in \mathcal J}

[lapis token]

Why supremums
Why multiple

[opaque fern]

I dont understand

[lapis token]

Isn’t there one supremum for the columns and one for the rows

[opaque fern]

no

[split kayak]

I got baited so fucking hard mb

[opaque fern]

I mean there are ones for each row

And one for every column

[lapis token]

Huh
Did you get laplaced

[opaque fern]

Its like

If you have a matrix

[
[1, 5, 3],
[2, 4, 8]
]

You can consider maximums across each row and column individually

So if you srr considering it across the rows your set would be {5, 8}

[lapis token]

And 2,5,8 next

I think I get that part

Now we have a table

And a bunch of sups

What do we do from there

The order seems to be irrelevant
Whether we search horizontally or vertically first

But “just do whatever it’ll be fine“ doesn’t go well with math homework

[safe radish]

@lapis token Has your question been resolved?

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[safe radish]

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[graceful lichen]

How coordinates of Q' is x1 - y1

or is it x1,y1 but mis printed in the book

[tardy mango]

Double ordinates are perpendicular to the axis of the parabola. The axis of the parabola is the $x$-axis, so $QQ'$ is a segment that is perpendicular to the $x$-axis. This means that $Q$ and $Q'$ have the same $x$-coordinate. Since the parabola is symmetric across the $x$-axis, we know that $Q$ being on the curve $\implies$ its reflection across the $x$-axis is also on the curve. And of course, reflecting over the $x$-axis negates the $y$ coordinate, giving $Q'=(x_1,-y_1)$.

[flat frigate]

Civil Service Pigeon

[graceful lichen]

oh okay

thats the thing

in book if you see there is not comma

nvm its clear

[tardy mango]

[safe radish]

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

[graceful lichen]

also

which formula they used to find trisection?

of R and T

[tardy mango]

Is that not just the section formula?

[graceful lichen]

mb i forgot that fr

[tardy mango]

rip

[graceful lichen]

didn't look at it in a while

[safe radish]

@graceful lichen Has your question been resolved?

[safe radish]

Channel closed

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[safe radish]

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[crisp hull]

no idea how to do these

ik how to graph linear and quadratic functions btw

those were prev questions n now they jumped onto this

[calm cloak]

try squaring both sides and graphing the new function (but with a little constrant ofc which is x>0 for the first one)

[crisp hull]

is there a proper method?

[long orchid]

when plotting composed function

[calm cloak]

well squaring the first funtion gives you a quadratic and you said you can graph quadratics

[long orchid]

@calm cloak i don't agree

it'll be 3x

that does no help OP to graph its sqrt

[calm cloak]

y^2=3x

[crisp hull]

it gives me ysq = 3x

[long orchid]

how is taht useful

[crisp hull]

ye i dont get it

[calm cloak]

huh? OP said he can graph quadratics, this is quite literally a quadratic in y instead of a quadratic in x

[long orchid]

plotting 3x is elementary, it's a line

then how do you take the sqrt of that

[calm cloak]

but its not 3x right? its sqrt(3x)

[long orchid]

yeah, so square that

3x

[calm cloak]

isnt that what I have been telling?

[long orchid]

yeah

and then what ?

[calm cloak]

then plot the the new funtion but with x>0 in mind

[crisp hull]

wait so like first i draw a quadratic graph of y sq then a linear for 3x?

[calm cloak]

its not y=3x lol, its y^2=3x

[long orchid]

yeah sure

so you'll end up with a graph of y^2

then what ?

how do you plot the blue one from here ?

[calm cloak]

we will end up with a graph of x=ky^2, k being some constant

but with x>0

thats quite literally the blue line

[long orchid]

i'm sorry

i was mistaken

[calm cloak]

alright

[crisp hull]

alright now pls tell me in simple words

its my first time doin sq root graphs

[long orchid]

(yh mine too lol, it isn't taught like this here)

[calm cloak]

alright so can you plot the graph of y=x^2?

[long orchid]

basically you have to plot the graph of x = 1/3 y^2

so horizontally

(which you know how to graph because it's a quadratic)

and then just focus on the top bit

and thta's it

[calm cloak]

edit the equation and correct it

[crisp hull]

ye

[calm cloak]

and can you do the same for x=ky^2?

[crisp hull]

what k is here

[calm cloak]

any constant

[crisp hull]

ye i can

[calm cloak]

so if you square y=sqrt(3x), what do you get?

[crisp hull]

y sq = 3x

[calm cloak]

is that the form of x=ky^2?

[crisp hull]

ye

[calm cloak]

so you can plot it that right? but remember in the original question we had y=sqrt(3x)

so x>0,y>0

[crisp hull]

ye

ye

lemme try

[safe radish]

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[safe radish]

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[grizzled kindle]

I dont get what does the polynomial refer to in the next question, can anyone help?

the question reads "find all roots of the polynomial and then split them into linear factors: Z^3 - 1"

[grizzled kindle]

is Z^3 -1 the polynomial or Z?

and also isnt Z = a + bi?

[stoic dune]

z³ - 1 is a polynomial

z is "a complex number".
We may choose to represent any complex number as a + bi

[grizzled kindle]

right

[stoic dune]

Note that re^(it) is also a good representation of a complex number

[grizzled kindle]

Not sure we learned that yet

but also if i dont know what Z is how can i find its roots?

[stoic dune]

Fair enough! Yeah you probably want to use z = a + bi here

Z can be anything, it's a variable!

You plug it into the polynomial Z³ - 1 to get another complex number

The question is: for which values of Z do we get Z³ - 1 = 0

[grizzled kindle]

I think theres a confusion here on my side

When im told to find the root of Z^3 - 1, is the meaning not square root?

[stoic dune]

No. The roots of a polynomial are the values for which Z³ - 1 = 0

[grizzled kindle]

ohh

[stoic dune]

You can take the square root of a complex number, so if the question is asking for square roots specifically, we can talk about that too

[grizzled kindle]

No, im pretty sure thats what they mean.

But the thing is, i could expand Z using a + bi and compare it to 0 but i vaguely remember another way to do it

is comparing to 0 the right way?

[stoic dune]

There's a bunch of ways to do this problem! Expanding Z could work, but there's an easier way here

Know any tricks for factoring Z³ - 1?

[grizzled kindle]

Nah not really but i already expanded it lol

I remember one method but i forgot its name

not sure if its for roots or like nth roots

[grizzled kindle]

whats the easier way?

WAIT i think i remember

Sorry for the wait im trying something

[stoic dune]

All good! lmk if you need any help with it. Good luck!

[grizzled kindle]

Alrighty i think i know how to do this but i messed something up :<

i wrote the equation as Z^3 = i then i gotta turn "i" into its polar form

so R = 1 (sqrt(0 + 1)) and Theta = arctan(0) = 0

@stoic dune What am i missing in converting the imaginary number to polar form?

i actually got it, thx!

.close

[safe radish]

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[balmy sparrow]

Question: Prove that a zero vector 0 of a vector space V is unique.

Would someone be able to check over my proof, and let me know if its correct or if anything can be improved upon?

[brave wolf]

Looks good, btw there is a shorter, but similar proof which considers 01 + 02

[balmy sparrow]

Does that use the idea of commutativity? saying that 01+02=02+01?

[brave wolf]

either that, or the fact that you can cancel 0 both from left and from right

its basically that
01 + 02 = 01, because adding 02 doesnt change anything
but also
01 + 02 = 02, because adding 01 doesnt change anything

so 01 = 01 + 02 = 02

[balmy sparrow]

Might be a stupid question, but how would I get into a position where I can say that 01+02, or that it equals either of the two? I understand the relationship. Is this just by definition of the zero vector again and instead of v we just do 01+02=02 or 01+02=01?

[brave wolf]

yep, exactly

v + 0 = v holds for all vectors v, including v = 01 and v = 02

[balmy sparrow]

Gotcha. I appreciate you helping me out with this and checking over my work. Thank you!!

.close

[safe radish]

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[safe radish]

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[fringe dock]

regarding part b ii i am very confused

because i learnt that to find the range of values if it was concave is f'(x) < 0

or where the graph is increasing/ decreasing

now the worked solution has done f''(x) < 0

[tardy mango]

concae is not the same as increasing/decreasing

[fringe dock]

but

arent they literally the same thing

because

concave/convex is just a graph that increases or decreases

[karmic merlin]

I think you should check out the definitions of concave in like your books or anywhere on the internet

[fringe dock]

oh

dam

okay

[fringe dock]

to show that a graph is increasing/ decreasing do we look at f'(x)

[calm cloak]

yep

[fringe dock]

okay

thanks everyone

.solved

[safe radish]

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[shell mango]

i hv a question

[quiet plume]

[warm warren]

!da2a

[shell mango]

lmfao okay it was a dare

[safe radish]

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.

[shell mango]

to ask in a help channel if i can get a gf who is also in math 💀

anyone up 💀

sorry mods

[warm warren]

just .close

[cunning spear]

What?

[warm warren]

!done

[safe radish]

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

[shell mango]

faaaaa

[warm warren]

look its either this or mods.

your choice.

[golden nebula]

Wtf is this

[shell mango]

okay sorry

mb mb

.close

[safe radish]

Channel closed

Closed by @shell mango

Use .reopen if this was a mistake.

[rare hound]

No it fucking wasn’t a dare lmao

[upper forge]

Fr

[rare hound]

U did that all on ur own

[rare hound]

I feel like u were scared to reopen it for a sec

😭

But yeah bro did that willingly

[upper forge]

Tryna score baddies, when you need to score in exams

[rare hound]

@hollow canopy he did it

[warm warren]

Um its closed, can we stop.

[thin bridge]

Don't be creepy.

[golden nebula]

Why are the mods serialised?

*sexualized?

[thin bridge]

These aren't things to "joke" about.

[upbeat swan]

Wdym

[golden nebula]

Like the bot

[shell mango]

? i didnt ask creepily gng , i just wana sit with someone in vc while i solve math

[warm warren]

.

[golden nebula]

The bot is a "she/they"

[upbeat swan]

The pronouns?

Yes

[golden nebula]

Yes

And the other bot is a she/her i think

[upbeat swan]

It'a an attempt to spread awareness

[golden nebula]

But its not necessary is it?

[warm warren]

!redir

[safe radish]

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

[thin bridge]

Asking itself is creepy. take it as a warning and don't do it again.

[stoic dune]

It's not necessary to have a cookie, but I have a cookie anyway

[shell mango]

gng im so sorry

[hollow canopy]

REALLY?

[shell mango]

i diodnt mean to do it that way , sorry , it wont repeat

[warm warren]

peoples, can we PLEASE STOP

[upbeat swan]

Ah course its nona

Btw Annie when did you get Helpful

[hollow canopy]

??

[warm warren]

rn

[upbeat swan]

Yeah just noticed

Its cool

[hollow canopy]

Wdym off its nona what about me

[rare hound]

Yeah lmao

[upbeat swan]

You just seem like the type of person to do that

[hollow canopy]

Do what exactly

[rare hound]

Don’t talk to Nona like that

[golden nebula]

@warm warren getting so pissed looking at this

[rare hound]

Fuck u bitch

[warm warren]

!redir

[safe radish]

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

[warm warren]

OR MODS

[upbeat swan]

<@&268886789983436800>

[hollow canopy]

???

u started this btw

why were u talking about me like that

[rare hound]

Why were u saying that to Nona

[warm warren]

Can we all please stop and take this to #chill

[upbeat swan]

I didnt swear at you

You did

[hollow canopy]

tsk bye

[rare hound]

Yeah because ur starting shit with someone for no reason

[exotic anvil]

what is going on

[hollow canopy]

talking to u is useless

[rare hound]

Nona didn’t do anything to u

[proper cove]

I genuinely cant tell whats going on here, so im going to ask you all just close the channel and move on

[hollow canopy]

And this is a help channel

Just leave

[exotic anvil]

the channel is closed

lets all take a step back

[radiant isle]

go to #chill and resolve it yourselves

[lunar radish]

Damn why does everybody named larp suck

[radiant isle]

these are called conflict resolution skills

[upper forge]

Wtf guys. Stop discussing in this channel. Ts closed

[golden nebula]

@warm warren finally got mods

[rare hound]

.close

[upper forge]

Lock the channel mods

[exotic anvil]

okay I will start muting ppl shortly

[radiant isle]

its already closed

[upper forge]

Already closed man

[golden nebula]

Wait i have a genuine question

[thin bridge]

Going to start dishing out mutes to anyone that still sends messages 2 minutes from now

[rare hound]

Okay my bad

[upper forge]

I mean lock like lock the channel

Not close

[golden nebula]

How do you get the role that does nothing?

[radiant isle]

no.

[thin bridge]

.remindme 2minutes

[safe radish]

Created reminder #help-23 message for <t:1776613517:f>

[upper forge]

Open it afterwards?

[exotic anvil]

shhhhh

[golden nebula]

How do you get the role that does nothing?

[exotic anvil]

#help-23 message

[safe radish]

@thin bridge asked to be reminded <t:1776613397:R>,

[safe radish]

Available help channel!

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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.

[stray crown]

.close

[safe radish]

Channel closed

Closed by @stray crown

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[safe radish]

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[stray crown]

ok i actually need help

[stray crown]

<@&286206848099549185>

[outer path]

For the rhombus, you also need to show that opposite sides are equal since we are trying to prove that the shape is a rhombus

[stray crown]

wait

wrong image

sorry

[outer path]

From here I think you can state something similar to the sides, just that opposite angles are going to be congruent as well since it is a parallellogram.

After that, since opposite sides are congruent and opposite angles are congruent, that should be enough evidence to conclude that you have a rhombus.

[stray crown]

it says that i need to prove AB = BC

@outer path

[outer path]

True, since all 4 sides need to be the same.

I am writing it out now. I will come back when I have a better solution

[stray crown]

thank you

[outer path]

You should be able to prove that triange DEC is congruent to triangle BCE

THat would then prove that DC = BC

That would also really be the only reason we needed to know that CE is a bisector is so we can use that angle in some way.

Also, Knowing that <ADE is congruent to <ABE & <ABC is congruent to <ADC, this means that <CDE is congruent to <CBE
That means you have enough info for AAS triangle congruency (CE = CE)
From there it is enough evidence to state that DC= BC.

[stray crown]

alright thanks

.close

[safe radish]

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[lusty spear]

I don't know what to do with this exercise, can someone help?
The task is to find for U_1 the so called "annulator" (U_1)^0.
So for a subspace U of V this U^0 is defined as shown ("für alle" = "for all")
lin(...) is the same as span(...)

Idk if this is common notation: V* is the set of all linear maps from V to the field F over which V is a vectorspace.

So for any f in (U_1)^0
0 = f( a (1 2) + b (4 3)) = a f(1 2) + b f(4 3)

Now I don't see how I could be any more specific about f

[lusty spear]

So I feel like I should treat f just as a "regular" vector conceptually

and formulate a basis

but idk cant really put the pieces together in my head

[stoic dune]

That's exactly it. What do you need me here for lol

[lusty spear]

lol

Idk

[stoic dune]

One way we can discuss elements of V* is with row vectors. For example, [1 0] will multiply (4 3) and give back 4

[lusty spear]

I was confused as to how to do this

[stoic dune]

Where I'll use [] for a row vector, why not

Actually, even before that, there's easier useful details we can iron out. What's the dimension of U^0 here?

[lusty spear]

that is a good question

which I need to think about for a second

my gut feeling tells me it should be dim(U)

but no real argument

that I can think of rn

[lusty spear]

I dont see why one would start thinking of this row vector here

[stoic dune]

You don't NEED to represent these with row vectors, but it's a pretty natural choice.
I chose [1 0] just as an example, I could have also chose [2 2]

[2 2] acts on (4 3) by doing 2*4 + 2*3 = 14

I am implying with this choice that the set of all linear functionals on R^2 is a vector space with dimension 2. Which, it is

However some don't map U1 to 0, so the space we're looking for must have lower than dim 2

[lusty spear]

ok makes sense to me

[stoic dune]

Actually, an easier question I could have asked first, what's the dimension of U1 itself?

[lusty spear]

didnt check whether the two vectors a linearly independent

but I'd say they are

so 2

[stoic dune]

They definitely are! U1 is just R2 itself

[lusty spear]

ok I can see that

with (0 1) and (1 0) we can generate this basis

[stoic dune]

We can figure this out mechanically by multiplying
[a b] (any choice of vector) = 0

For two choices of vectors

[lusty spear]

wait if I remember correctly

if U is subspace of V and dim U = dim V then U = V ?

yeah I am pretty sure this is true

we have an isomorphism from V to F and from F to U

wait

oh ok I think it should make sense since U is subspace of V

anyways

[stoic dune]

This is a really good strategy. What are a and b in [a b] here?

[lusty spear]

they differ for each vector, dont they?

for (1 2) I'd have 2(0 1) + 1(1 0)

for (4 3) it is 3(0 1) + 4(1 0)

actually should I not want to generate (0 1) and (1 0) from (1 2) and (4 3)?

so the other way around

[stoic dune]

[1 2] (0 1) = 2
[4 3] (1 0) = 4
You can think of these like a dot product. The important part is they map a vector to a real number

[lusty spear]

because obviously I can generate any v in R^2 from the canonical basis

[lusty spear]

I dont understand why we'd do this now

I understand the computation

but not why

[stoic dune]

These are just examples of how these row vectors would be used

The thing you want to find now, is an a,b such that:
[a b] (any vector in U1) = 0

We can actually solve this as a system of equations by choosing two vectors to plug in. You had a good idea with (1 0) and (0 1)

[lusty spear]

ah ok I think I get it now somehow

because

I was also thinking about the system of linear equations thing

but I didnt really know how to do it

ok so I will try it again from here

thanks a lot!

[stoic dune]

Okay! Feel free to ask if you'd like any help with it

[lusty spear]

thank you!! thats really kind, I will probably do that xD

.close

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[safe radish]

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[quiet plume]

So what's your feeling about a) after those few examples you computed? (Note that you've only shown that 4 doesn't satisfy a), not that composites don't satisfy a), although the argument is similar).

Do you think it's a definition of prime numbers? In other words, do you think it is it true that a natural p satisfies a) iff p is prime?

• If so, can you prove it?
• If not, can you provide an example that exhibits that it's not the case?

[spiral saddle]

idk how to continue

[quiet plume]

Can you answer those questions?

[spiral saddle]

im not sure

[quiet plume]

Well you've tested two primes and they seem to satisfy a), and you've tested a composite and it doesn't seem to satisfy a).
Do you suppose that it's a definition of prime numbers or not?

[spiral saddle]

well

sometimes a conjecture works for some examples but not in general

is hard to tell really

[spiral saddle]

so far the composite number made a) be false

[quiet plume]

The point is that those examples, if they don't yield counterexamples, should provide some amount of intuition about whether you think the statement is true.

[spiral saddle]

im not sure what should I expect out of the quasi definition if I use a p that is composite

you understand?

[quiet plume]

Does the fact that 4 does not satisfy a) invalidate the statement "p satisfies (a) iff p is prime"?

[spiral saddle]

unsure

I tried multiple possible values for p, being prime and composite

im not sure

should I have done only cases when p is prime and avoid p composite? @quiet plume

[quiet plume]

I am asking you specifically for p = 4. Does the fact that a) does not hold for p=4 mean that "p satisfies (a) iff p is prime" is false?

[spiral saddle]

unsure

care to give a hint

[quiet plume]

"p satisfies (a) iff p is prime" is the same as

• if p is prime, then p satisfies (a), and
• if p satisfies (a), then p is prime.
  Things that would invalidate this are : an example of a prime number which does not satisfy (a), OR an example of a number which satisfies (a) but is not prime.

Does the fact that 4 does not satisfy (a) fall into one of those two cases?

Similarly, does the fact that 2 satisfies (a) fall into one of those two cases?

And does the fact that 3 satisfies (a) fall into one of those two cases?

[spiral saddle]

nothing falls under

those two cases

[quiet plume]

Then you've tried three numbers and have failed to produce a counterexample. You can either keep trying to find more in hopes of finding a counterexample, or try to prove that the statement is actually true at this point.

[spiral saddle]

how to prove

[spiral saddle]

how many examples should I try before I give up?

because it might be just luck that any p in that range works but p outside that range doesnt @quiet plume

[quiet plume]

In evaluating (a) for some numbers, primes and composites alike, you should have an inkling of an intuition as to whether it should be true in the general case or not.

If you try to prove it and realize it doesn't work, then chances are you'll be in a better position to go back to the drawing board and pick a relevant counterexample.

[spiral saddle]

ok how to prove

[quiet plume]

What do you want to prove?

[trim mauve]

can anyone help me please

[quiet plume]

This channel is currently occupied. Refer to #❓how-to-get-help for details on how to find available channels to post your question in.

[spiral saddle]

that a) is a characterization of prime p > 1

[quiet plume]

What is the statement you're trying to prove explicitly? (What does it mean that a) is a characterization of prime numbers?)

[spiral saddle]

a)

[spiral saddle]

i meant to say def sorry

[quiet plume]

It doesn't change what I'm asking. What is the statement you're trying to prove explicitly? It's a double implication.

[spiral saddle]

a) holds for prime p <=> p is prime

its an if and only if

@quiet plume

[quiet plume]

a) holds for p <=> p is prime

[spiral saddle]

p is prime when a) holds?

or is it composite?

@quiet plume

[spiral saddle]

why

[quiet plume]

You're trying to show that a number is prime iff it satisfies a).

If you assume p is prime of course p is prime.

[spiral saddle]

yes

[spiral saddle]

can you help me with left direction

[quiet plume]

What have you done so far?

[spiral saddle]

i did right direction

[quiet plume]

Ok, and how far along are you with the left direction? (I assume you mean proving that if p is prime then it satisfies a)?)

[spiral saddle]

no

I want to prove that if a) => p is prime

[quiet plume]

Ok.

Since p satisfies a), what divisors does p have?

Or rather, what does a) say about the divisors of p which are less than p?

[spiral saddle]

a) is forall d in N, (d < p and d | p) => d = 1

[spiral saddle]

unsure

they are 1

[safe radish]

@spiral saddle Has your question been resolved?

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[elfin pawn]

Hey

[elfin pawn]

I need help

[plucky elk]

Yes ask

[plucky elk]

And follow the rest of this

[elfin pawn]

I’m try to solve it first

Should be easy

[elfin pawn]

The answer is 70?

[grand basin]

yeah

howd you know?

[elfin pawn]

Because it’s pretty simple I just need someone to double check my work if that’s cool

For this I got 62

[grand basin]

howd you figure that

[elfin pawn]

Because I read it ?

[grand basin]

Read it where?

[elfin pawn]

Bro

[vague phoenix]

Yes that is correct

[elfin pawn]

Can u just tell me if im wrong or correct so I can correct my answer

[elfin pawn]

Thanks

For this I got A

4.1 and -2.2

[grand basin]

Right

[rotund epoch]

[elfin pawn]

Thanks

I got 90

[vague phoenix]

How come?

[rotund epoch]

?

[vague phoenix]

Note the word "average"

[elfin pawn]

Did I get it incorrect?

[rotund epoch]

-# I think it would be a range(average), like the B grade - 80-89

[elfin pawn]

O

So like

I grab the scores and see what’s the average

Oh yea your right

Let me re do it

So this is what

Add the scores: 89 + 73 + 75 + 83 + 72 + 78 = 470
Divide by 6 → 470 ÷ 6 = 78.3

So your average is 78.3%

That’s the grade

And I need __ to reach 80

Are sure it’s not 90?

[rotund epoch]

Not 90

[elfin pawn]

O

What is it then???

I’m so lost

Did u get the answer?

[rotund epoch]

Maybe try 80-78.3 lolz

[elfin pawn]

Wait whatttt

[rotund epoch]

since you need at least an 80 for a B

[elfin pawn]

Yeaa

[rotund epoch]

If 78.3 is your score

and you need at least an 80

boom

[elfin pawn]

To get 80%

Not 90?

So the answer is 80?

[rotund epoch]

It does not say to get an A//next grade up? Unless there is more to that question

[elfin pawn]

Well I feel like 80 is not going to bring up your grade by 2 percent

I have 2 attempts I’ll let you know if your right

Can you help me on this?

I got -15.9

No change that I got -17.4

[rotund epoch]

on a second thought, it might be 90, because you need
89 + 73 + 75 + 83 + 72 + 78 = 470
with 78.3 as the mean
and to make the mean a 90,
(89 + 73 + 75 + 83 + 72 + 78 +** 90**) / 7
~_~ sorry!

[elfin pawn]

Noooo it’s totally fine

It’s okay

[elfin pawn]

Could u help me with this

I’m not the brightest at math

[prisma wren]

Try to write out the mean as an expression of the known numbers, and the missing one

[elfin pawn]

Did I get it incorrect stitches ?

I got -17.4

[rotund epoch]

Close

[rotund epoch]

[elfin pawn]

Oh yeaaa thanks

30.9?

[prisma wren]

show work

[elfin pawn]

I only have to show work for the last questions

Like 2 of dem

[rotund epoch]

-# Still, good habit + it will help us help you

[prisma wren]

presumably you got your answer of 30.9 by doing some work. show that work so we can verify/tell you where you went wrong (if you did)

unless 30.9 fell to you from the heavens

[elfin pawn]

LOL

Sure

I’m

My hand writing is pretty bad

If u don’t mind

Skipped some steps Becuase I can do some in my head

[rotund epoch]

You only used 3 of the initial digits, the last 2 weren't subtracted from the mean I guess I was wrong again

And were not squared

[prisma wren]

Yes, 30.9 is correct, assuming population. presumably you used symmetry for 11

[elfin pawn]

Kk

Thanks so much

I’ll Make it easier for u guys to read for the next question

Just give me a second to finish this

The boxed is my answer

This is for part a

Sorry my hand writing is so baddd

My first language is Arabic so bear with me

[rotund epoch]

[rotund epoch]

If population, then it's correct

[elfin pawn]

So I submitted it

And the only wrong answer I got

Was the 30.9

Can we do that one again?

[rotund epoch]

I thought so..

[elfin pawn]

Yea I don’t know where I went wrong

Stitches said it was right

But it’s my fault

[rotund epoch]

Did it say the answer?

for a sample, it would have been 37.1

[elfin pawn]

Noo sadly it didn’t say the answer

Let me try to solve it again

Yes I think it is 37.1

Just give me a second

[rotund epoch]

you do

|| Just like the steps of standard deviation ----> (1 - 7.5)^2 + (4- 7.5)^2+ (5- 7.5)^2+ (6- 7.5)^2+ (11- 7.5)^2+ (18- 7.5)^2||
||OVER||
||The amount of data ---> 6 minus 1||
||185.5/5||
= ||final answer of 37.1||
and that is the variance for the sample

[safe radish]

@elfin pawn Has your question been resolved?

[safe radish]

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[worldly lantern]

why is this legal

[prisma wren]

Be more specific. (ii) in your photo is sort of like the chain rule https://en.wikipedia.org/wiki/Chain_rule . Implicit differentiation works cause of the chain rule being implicitly applied when necessary.

That (-) shouldnt be there

[worldly lantern]

its not chain rule

its just

differentiation of implicit function

let f(x,y) be given
now to find dy/dx we do partial derivative of f(x,y) wrt x and wrt y

this is what i mean

[golden nebula]

I don't think this formula is correct

Oh wait never mind

While not being the best explanation, I assume that its because every term that has a "dy/dx" must have a differentiated function with respect to y along with it. Everything without dy/dx is either a constant or is a differentiated function of x. Also, for a product of a function of y and a function of x, it splits into 2 (by product rule I think, I always forget the rule name). This is the best explanation I can do. If you want a clearer explanation, you can try googling or waiting for someone else to respond

[median vigil]

you can't just ``cancel'' the $\partial f$s the way you would with ordinary derivatives

[flat frigate]

clôud

[golden nebula]

https://math.stackexchange.com/questions/94570/implicit-differentiation-proof might help you

[worldly lantern]

why the negative tho

[median vigil]

well you have to approach it from the multivariable chain rule

if you have a curve $f(x,y) = C$ and you differentiate in such a way that $y = g(x)$ is an implicit function defined by that curve, then you essentially differentiate ``along the curve'', so since $f(x,y)$ is constant along the curve:
[
\odv fx = \pdv fx + \pdv fy \odv yx = 0 ]
then just rearrange for $\odv yx$

[flat frigate]

clôud

[worldly lantern]

damn

thanks

.close

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[honest peak]

How do I proceed from here, I know it must convert into an integral but how

[left gyro]

have you seen riemann sums before

[honest peak]

Nope
But I know limits of a sum

[left gyro]

riemann sums look like:

,,\lim_{n\to\infty}\sum_{k=1}^n\frac1nf(\lambda)=\int_0^1f(\lambda)d\lambda

[flat frigate]

mtt

[left gyro]

also, k isn defined when you move it outside of the ∑

[honest peak]

Oh
But how is 1/n = dλ

[left gyro]

its not a literal limit

[honest peak]

Oh yeah

[left gyro]

you cant just replace a ∑ with a ∫

[left gyro]

instead, you replace this ∑ f(k/n)/n with a ∫ f(x) dx

the idea for this is partly due to what an integral is

have you done integrals before?

[honest peak]

Yes

[left gyro]

what do integrals look like, on a graph?

[honest peak]

Summing slivers to get area under the curve

Ohhh

I get it

[left gyro]

this I think is partly the definition of such an integral

rigorously it can boil down to this, for continuous functions at least

[honest peak]

1/n can be thought of as a nudge in the x, a dx

[left gyro]

youre not using the word nudge correctly
nudge = movement

here, 1/n is the width of the rectangle instead

1/n width, f(k/n) height

[honest peak]

Got it

Thanks!

[left gyro]

np

[honest peak]

Yes thank you very much :>

.close

[safe radish]

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[left gyro]

I still need to check your work though

nvm there isnt anything wrong with it other than moving that k out

[meager ravine]

@plush ravine please help me w my maths hw😔

@plush ravine

[safe radish]

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[craggy tree]

question: why does g(x) = x|x| at x = 0 not serve as a counterexample to (c)?

[craggy tree]

oh wait i misinterpreted "all its derivatives exist" as "its first derivative exists at all points"

oops

ok nevermind

.close

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[burnt nymph]

3x²+2√6+2 find roots

[normal moss]

Is there supposed to be a linear term or no

[burnt nymph]

?

[normal moss]

Well, as written I assume you mean solving

$3x^2 + 2\sqrt{6} + 2 = 0$

[normal moss]

This is a quadratic in the form $ax^2 + c = 0$

[normal moss]

c is 2sqrt(6) + 2

[burnt nymph]

Yes

[normal moss]

There's no bx, linear term

Okay

[burnt nymph]

ax²+bx+c

[normal moss]

Yes

But there's no bx

That's why I am asking

[burnt nymph]

Oh wair

[normal moss]

If it's supposed to be that

[burnt nymph]

Wait*

[normal moss]

Or is there an x missing

At 2sqrt(6)

[burnt nymph]

It's 2√6x

[normal moss]

Right

$3x^2 + 2\sqrt{6}x + 2 = 0$

[burnt nymph]

Yeah

[normal moss]

Anyway

!status

[safe radish]

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

[burnt nymph]

2

[normal moss]

Can I see your work

[burnt nymph]

First i thought a lot that how to split 2√6x but I couldn't find it

Then I did
(√3x+√2)²

[normal moss]

Are you supposed to solve by factoring?

[burnt nymph]

Yeah or what I have done

[normal moss]

Yes, that's factoring

Though in this case it can be slightly tricky to find it

Though you did

Correctly

$\left(\sqrt{3}x + \sqrt{2}\right)^2 = 0$

[normal moss]

Now what

[burnt nymph]

To get zeros

[normal moss]

How

[burnt nymph]

Idk that's why I came here

[normal moss]

I see

Well

We have an equation in the form of

m^2 = 0

When would this be 0

[burnt nymph]

√3x=√-2

X=√-2/√3

[normal moss]

Uh

Not exactly

[burnt nymph]

Then rationalisation

?

[normal moss]

Why minus inside the sqrt

[burnt nymph]

-√2

[normal moss]

Yes

When moving, you change the sign of the entire expression

sqrt(2) is the expression here

[burnt nymph]

-√6/3

[normal moss]

Hence - sqrt(2)

[burnt nymph]

Root

[normal moss]

Yes

[burnt nymph]

But there is only one root

[normal moss]

Yes

[burnt nymph]

But it's a quadratic

[normal moss]

Well technically

It's not one root

There's 2

But they are the same

[normal moss]

Look at this

It's squared

That means the root repeats

Because you can write the square as a product right

[burnt nymph]

Oh ok

[normal moss]

And you get the same root from both factors

Have you learned about the discriminant yet?

[burnt nymph]

No

[normal moss]

Oops

Auto correct

[burnt nymph]

What is discriminant

Btw

I have one more question

[normal moss]

Well it tells you what kind of solutions the quadratic equation has

Simply put

Sure

If it's positive, there's 2 unique real solutions. If it's negative, there 2 unique complex solutions (conjugates). If it's 0, there's 1 repeated root

[burnt nymph]

6x²-7x-3 find alpha⁶+beta⁶ and 1/alpha³+1/beta³

[normal moss]

Here the discriminant is 0

[bitter sparrow]

just find alpha beta and solve

[normal moss]

Well

There's 2 ways of doing this

You can find alpha and beta and compute

[burnt nymph]

No

Using identity

[normal moss]

Which here is simple

Otherwise

You need to know a bit about a^6 + b^6

[bitter sparrow]

Lmao why complicate it so much

[burnt nymph]

. close

.close

[clever tinsel]

no spaces.

[safe radish]

Channel closed

Closed by @burnt nymph

Use .reopen if this was a mistake.

[normal moss]

Saying "thank you", or "bye" at least is really not that difficult

😒

[near sky]

Thank you for your effort

Have a nice day

[astral glacier]

-# now kiss

[near sky]

Kiss xavier

[safe radish]

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[merry lintel]

I have 2 uhh functions?
$$f(t) = 5900 * 1.002^t$$
And
$$g(t) = 8000 * 1.012^t$$
f(t) is rent
g(t) is salary
The time unit is months
The question is
After how many months will the rent be 60 percent of the salary?

[flat frigate]

Kaz

[main mural]

it's \% for percentage signs in latex fyi

[merry lintel]

i see
thanks :)

[main mural]

how do you calculate 60% of something

or actually where are you stuck

[safe radish]

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

[merry lintel]

I found it to be 9833.33

60% of that is 5900 i think

[main mural]

found what out

can you show your steps

[merry lintel]

one sec

I tried drawing it but
typing was faster so

[buoyant shadow]

just write the correct equation

literally

you;re thinking too much

[merry lintel]

whats the correct equation

[main mural]

no

you threw away the exponential information

[merry lintel]

I tried 9833.33 = 8000 * 1.012^t

[main mural]

just solve $f(t) = 0.6 \cdot g(t)$

[flat frigate]

artemetra

[buoyant shadow]

5900 * 1.002^t rent
= is
0.6 (8000 * 1.012^t ) sixty percent of the salary

[merry lintel]

oh

I see

thanks for the help :)
thats it

[buoyant shadow]

sorry made a mistake

[merry lintel]

all good

thank you both

[main mural]

[merry lintel]

.close

[safe radish]

Channel closed

Closed by @merry lintel

Use .reopen if this was a mistake.

[safe radish]

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[karmic merlin]

What have you tried so far?

[near sky]

so we were doing this yesterday,
with angle AMC = 90, M is on a sphere with diameter AC
A point K on plane perpendicular to AB forms 90 angle for ABK so M lies on that plane too, locus is the intersection of the plane with the sphere

AM traces a cone like shape

@naive dragon

nvm

<@&286206848099549185>

[fleet willow]

!15m

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[near sky]

bruh its been 15 mins

[fleet willow]

since someone last answered

[near sky]

its since then technically 🥀

but i get it

[safe radish]

@near sky Has your question been resolved?

[gaunt elk]

Let's define coordinates of $N$ : $(x_N,y_N,z_N)$.
Can you find an equation verified by those coordinates ?

[flat frigate]

Lin Xia

[fair hound]

I think there is a problem with the exercise, you might have miswrote something

this lead to a problem of definition for M

[gaunt elk]

why ?

[fair hound]

can't find such a point M, this def implies M is on the sphere of diameter AC and M is on the plane going through B and perpendicular to (AB)

[gaunt elk]

yes there re several solutions

[fair hound]

there aren't

[gaunt elk]

it's aa circle

[fair hound]

good luck to find a circle consisting of 0 point

[near sky]

oh

[gaunt elk]

hum you mean with those coordinates

[near sky]

must AMB then

[naive dragon]

you're goddamn right

someone insert heisenberg gif pls

[near sky]

https://tenor.com/view/damn-right-breaking-bad-gif-14029087

[naive dragon]

there are no such M

[gaunt elk]

unlucky

[naive dragon]

ty for trying that problem xD

[near sky]

:)

[naive dragon]

Actually someone said it does not exist but I just kinda, ehhh anyway

It's still a fun problem

[near sky]

yes

ty

[fair hound]

probably, yeah

[near sky]

.close

[safe radish]

Channel closed

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[safe radish]

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[timber karma]

Let $\ell$ be a line in the plane, and $90^{\circ} < \theta < 180^{\circ}$. Consider any three distinct points $P,Q,R$ satisfying the following: $\ i) P$ lies on $\ell$ and $PQ$ is perpendicular to $\ell \ ii) R$ is on the same side as $Q$ relative to $\ell$, but not on $\ell$. $\ iii)$ For any $A,B$ on $\ell$ such that $\angle ARB = \theta$, then $\angle AQB \geq \theta$. \ Determine the minimum of $\angle PQR$

[flat frigate]

Copter

[timber karma]

I have something that looks like this

but im kind of confused, cant we pick B=P and A large enough such that <ARP is theta, then <AQP is less than theta?

or am i misinterpreting the question

[golden nebula]

If the question states that angle AQB is greater than or equal to theta, I assume that your take would be invalid

[safe radish]

@timber karma Has your question been resolved?

[timber karma]

oh

why is it wrong then?

<@&286206848099549185>

[golden nebula]

It would be invalid because angle AQB would be less than theta

[timber karma]

well, couldnt i just pick the same thing regardless of where R is?

[golden nebula]

What do you mean?

[timber karma]

could i not do something similar to this

or am i trolling

[golden nebula]

If you pick B=P, then angle AQP is less than theta right? You said it yourself. So I think it might not me valid. Or I misunderstood the question

[timber karma]

but ARP is theta

[golden nebula]

Oh my bad. I meant that it would be invalid because theta is between 90 and 180. so if P coincides with B, theta woudl be less that 90

Sorry for the confusion

[timber karma]

oh wait

yeah i see

sorry mb

[golden nebula]

Also @timber karma, I recall you sent something related to inversion related to a circle. What does that mean?

(This is completely unrelated to the current question btw)

[small rampart]

For ANY A,B

[timber karma]

type of geometric transformation, look it up

[timber karma]

yeah i see my mistake now mb

.close

[safe radish]

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[safe radish]

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[copper shadow]

For 27 d I have that f(x)= x for 0<x<1 and for x>1 f(x)=e^x+c. How can I go about finding c? I know he gives f(0)=0. But e^x+c is for x>1 so I don't see how I can use both information in conjunction

[verbal cloud]

remember it's f(log)

[copper shadow]

Yes but I can just sub in x instead of logx

[verbal cloud]

yes

[copper shadow]

Sorry I'm not seeing why f(log) is important

[verbal cloud]

e^x+c is for x>1

so when x = 1 what happens

it's not f(1)

I think you should try to do the substitution formally

[copper shadow]

It is going to be handled by the other case

[copper shadow]

What do you mean by this? I set u =logx. And then found e^u. What is the method you are suggesting?

[verbal cloud]

not really since you want the function to be continuous, so both cases should handle it

[verbal cloud]

the domains for u and x are different

x > 1 iff u > ...

[copper shadow]

Wait why do we want it continuous?. Don't we just want a function f.

Oh wait

It is differentiable

[verbal cloud]

f is differentiable thus continuous

[copper shadow]

U greater than 0

[fathom jewel]

How?

[verbal cloud]

I think so

[copper shadow]

Well we would have $\lim_{h\to0} \frac{f(x+h)-f(x)}{h}=f'(x)$

[flat frigate]

BigBen

[copper shadow]

So multiple both sides by $\lim_{h\to0}h$

[flat frigate]

BigBen

[verbal cloud]

did you manage to connect the dots

[copper shadow]

Well then for e^u+c we have it defined for u>0. But that isn't 0

[verbal cloud]

what isn't 0

[copper shadow]

It isnt defined for 0

Only for u>0

[verbal cloud]

well it's continuous so it's essentially the same

[copper shadow]

I'm not following

[verbal cloud]

if you take the limit as u goes to 0 you have the same equality as if it was defined at 0

[fathom jewel]

Hmm I am not at home so I can't formulate my thoughts but I feel somethings off

[restive niche]

no, its correct

a function being differentiable at a point means it is continuous there

although you could phrase your proof a bit better

[copper shadow]

Ok so since our function is continuous the limit at a point is equal to the value of the function at the point

[copper shadow]

What do you suggest?

[verbal cloud]

yes

[copper shadow]

Ok f(0)= 0 if x is 0 and f(0) =0 if e^x+c. So c=-1

[verbal cloud]

yep