help-42

[hearty heart]

how do you get 2 max

out of this integral

do you get what they did here?

[calm coral]

@hearty heart Has your question been resolved?

[near agate]

first integral gives x_3 —> (max + max)

So 2max

[hearty heart]

but one says -2 max

[near agate]

Top-bottom

Bottom is negative

So it becomes +

[hearty heart]

ah ok intersting

[hearty heart]

when doing integrals like this it is always top - bottom?

[near agate]

Yes

For any integral

[hearty heart]

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[flint mist]

For this question I was thinking

$\int_{0}^{1} \int_{0}^{x_2} 3x_2 dx_1 dx_2$

but apparently thats not it

[potent lotus]

Calc III Victim

[flint mist]

It should be this?

But I dont understand why I have to split here

[lament ledge]

you have the condition that X1+X2 <=1

[flint mist]

right

oh so x_1 <= 1 - x_2 thats how they got the bound for the secondintegral

so the first one is for 0 < x_1 <x_2 < 1 region

and the second one is for x_1 + x_2 < 1?

[lament ledge]

(i also dont know the answer and im puzzling along with you)

[flint mist]

lmao

[lament ledge]

im a bit confused by the 1/2

[flint mist]

Im honestly not 100% sure if that is the answer

I got the second image from chatgpt because I couldnt find this question online

[lament ledge]

the 1/2 doesnt make much sense

oh...

!nogpt

[flint mist]

do you agree with this one?

[calm coral]

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

[lament ledge]

no

we have to look at our bounds

x1+x2<=1

0<x1<=x2

x1<=x2<=1

[flint mist]

so u would say

$\int_{0}^{1} \int_{0}^{(1-x_2)} 3x_2 dx_1 dx_2$

shit

[lament ledge]

almost

you forgot an _

*2 of them

[potent lotus]

Calc III Victim

[flint mist]

so u agree with this one?

[lament ledge]

no

[flint mist]

[lament ledge]

the inner integral is x1 right?

[flint mist]

ye

0 < x_1 < 1 - x_2

[lament ledge]

this condition says that x1<=1-x2

[lament ledge]

this condition says that x1<x2

[lament ledge]

this condition says x1>0

[flint mist]

so the inner integral bounds should b correct no?

we have 0 < x_1 <= 1 - x_2 from all of that

[lament ledge]

but what about x1<x2?

i think i see where the 1/2 is coming from

[flint mist]

so then x1 < x2 and x1 < 1-x2 should be two seperate integrals?

[lament ledge]

no, we should combine them

i think

[flint mist]

how will that work

[lament ledge]

x1 is at most 1/2

[flint mist]

will that mean the outer integral bounds will be diff?

[lament ledge]

since if x1 is greater than x1, x2 is greater than 1/2, and 1-x2 is smaller than 1/2, which is not allowed

[flint mist]

woah Im a bit confused

one min

[lament ledge]

i think i see it now

okay

[flint mist]

so we know

0 < x1 < x2 < 1
and
0 < x1 < 1-x2

but where did the 1/2 come from

oh

Im slow

[lament ledge]

its a bit hard to see

[flint mist]

I get it

[lament ledge]

but 1/2 is the turning point when it goes from one bound to the other

it comes from x2=1-x2

[flint mist]

I get it now

[flint mist]

so this should be correct

[lament ledge]

$x_2=1-x_2\iff 2x_2=1\iff x_2=\frac12$

[potent lotus]

Flappie

[lament ledge]

yes, i think so too

[flint mist]

tysm I get it now

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[lament ledge]

but its not so easy to see

[calm coral]

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[pearl jay]

how do proove that the series converges?

[pearl jay]

tried integral test, i got this

which diverges

[brittle isle]

i think you can show that it's telescoping

for k = 1, you get -4sqrt(2) + 2sqrt(3) + 2sqrt(1)
for k = 2, you get -4sqrt(3) + 2sqrt(4) + 2sqrt(2)
for k = 3, you get -4sqrt(4) + 2sqrt(5) + 2sqrt(3)

you might be able to continue this and find a pattern to what cancels, but i notice that the sqrt(3) term goes away, i'm sure other stuff does in the same way

[calm coral]

@pearl jay Has your question been resolved?

[pearl jay]

ok so this is the pattern i get, but how do i generalize to a formula

[brittle isle]

well everything cancels except for those three terms at the top left

like if you continued this

idk if you need a specific formula

now that you know it all cancels

[pearl jay]

wait, but wont we always have 3 terms at the bottom right remaining?

[brittle isle]

telescoping series always technically have some terms left at the bottom like this

but as you take them out to infinity you just consider the top terms

the stuff that doesn't eventually cancel

[pearl jay]

then i get this

but it still diverges

[pearl jay]

wolfram alpha says it is convergent

[brittle isle]

you're no longer evaluating the limit

the series is exactly the three terms that didn't cancel

[calm coral]

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[silver holly]

Edexcel A level maths resources to start from scratch (I’ve just finished year 11 and I wanna get ahead)

[kind quarry]

You could work through the Year 1 textbook yourself

[calm coral]

@silver holly Has your question been resolved?

[silver holly]

Thanks I found a free pdf

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[royal raft]

How can I integrate this

[strange lichen]

combine like terms

@royal raft

[royal raft]

oh thank you

i completely missed that

[strange lichen]

u got the answer?

[royal raft]

I have a question about something, am I allowed to do this? I know cos^2(x) + sin^2(x) = 1, but i’m not sure how it works when there is a constant with the x so I did u sub to make it just u, idk if I can do that tho

[strange lichen]

cos^2 (u) + sin^2 (u) just simplifies to 1 if thats what ur asking

as long as the arguments are the same

doesnt matter if its 10t

or 1111115t

or -238482934892t

as long as its in the form cos^2(k) + sin^2(k), it will equal 1

[royal raft]

oh ok thank you

got the correct answer thanks for the help

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[woeful ore]

I am working on a personal project and am running into the end of my knowledge of graph theory / group theory:
Is it possible to decompose the complete directed graph on 6 vertices into 5 Hamiltonian cycles?
More generally, for what number n is this possible with n vertices and n-1 cycles?

[woeful ore]

It is trivial to show that prime numbers of vertices are possible, and also I have brute-forced the case of 4 vertices to show that 4 is not possible.

[marsh summit]

...i don't think the numbers add up there?

you would need there to be 36 edges but there's only 30

[woeful ore]

I am ignoring degenerate edges

[marsh summit]

yeah so that gets you the numbers i said

a hamiltonian cycle has 6 edges in it, right?

and there's 6 of them

so 36

but there's only 30 nontrivial edges, 6*5

[woeful ore]

OOPS yes I typed my problem wrong

I did mean 5 cycles

[marsh summit]

ah ok that makes more sense

[woeful ore]

I have shown that these are the 14 cycles

And I know some properties about them. The problem is that most of those properties rely on their being arranged clockwise around this circle, when really any solution to this problem stays a solution under permutation of the points

But all my invariants rely on considering the chords as different from the minor diagonals as different from the major diagonals

Another direction I’ve tried is to consider it as a Latin square where the row tells you the source, the column tells you the destination, and the color tells you which of the 5 cycles it should belong to. And the game is to make sure you don’t accidentally make a 3-cycle or something

@marsh summit are you still here or should I ping for others or should I let it hang for a couple hours? I don’t know the etiquette

[lament ledge]

After 15 minutes you can ping helpers role

[marsh summit]

i'm still sort of here and just haven't made any progress and also got distracted by a different conversation

[woeful ore]

<@&286206848099549185>

[calm coral]

@woeful ore Has your question been resolved?

[calm coral]

@woeful ore Has your question been resolved?

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@woeful ore Has your question been resolved?

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[calm coral]

@ripe bison Has your question been resolved?

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@ripe bison Has your question been resolved?

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[austere topaz]

i got a question on elimination

[austere topaz]

I'm given a equation

-2x-9y=-25
-4x-9y=-23

i wanna solve for x

and since both coefficents are the same i subtract

but I can't because that does not get rid of the y

so what do i do

[warm warren]

Show me

How does it not get rid of the y

[austere topaz]

-9-9 is 18

[warm warren]

That’s -9 + -9

[austere topaz]

but don't i subtract

[warm warren]

But we’re subtracting the 2 equations

[austere topaz]

yeaa

[warm warren]

This is adding the equations together

[austere topaz]

well yea

but I can't add

[warm warren]

So minus them

That’ll be -9 - -9

[austere topaz]

oh

[warm warren]

Instead of -9 + -9

[austere topaz]

well

yea

yea that makes sense

https://tenor.com/view/giants-ny-obj-gif-19465394

ok

ty

[warm warren]

👍

[austere topaz]

ima finish the question rq

make sure i got it right

ok i did

ty

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[stark brook]

hello hello

[stark brook]

I have a quick question regarding my math work. I am identifying wheather the graph is quadratic, linear, and exponential.

so far

I put

a

exponential

since it doesnt touch the X axis

and I put d as linear since its straight and never ending

also I put e as quadratic

for question 2

for question 2 b would that be quadratic? because to me it seems like it could be either exponential or quadratic

[bright lagoon]

no, it kind of "hockey-sticks"

like an exponential

however, it should exist when x is negative

[stark brook]

hmmmm

[bright lagoon]

well

perhaps they are limiting the domains randomly

[stark brook]

hmmmm alright thank you so much

this has helped me a lot tysm

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[lapis bobcat]

Can i know the answer for 13th question

[blazing coyote]

,rotate

[calm coral]

@lapis bobcat Has your question been resolved?

[lapis bobcat]

.no

[worn belfry]

I would start by plotting those points on some graph paper

[lapis bobcat]

Not allowed to use graphs

Must find it through equations and gradient

If anyone explains me how to find d or c I could continue the question

[lapis bobcat]

Those stuff

[worn belfry]

That's absurd, but whatever floats their boat. Plotting the points will just give you an idea of how it's oriented. But perhaps you can picture in your head where points A and C are relative to each other.

From there, you know that side AB is gonna have an equation parallel to the line y = -3x

And side AB passes through A

[calm coral]

@lapis bobcat Has your question been resolved?

[lapis bobcat]

Oh ok

I’ll try

[calm coral]

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[storm birch]

can someone go through this working i dont understand it

[pallid halo]

what is Z?

[storm birch]

z score or z value

[pallid halo]

this?

[storm birch]

yeah

[pallid halo]

so in other words, the probability that a standard normal random variable is less than 1.2

i assume you have something called the error function, erf or maybe erfc?

[storm birch]

no

[pallid halo]

(assuming here that x is normal, is it?)

[storm birch]

what does that mean

im rly ass at this topic

@pallid halo

[kind quarry]

Seems like you're just meant to be use a calculator. Z typically denotes the standard normal distribution $Z \sim N(0, 1)$

[potent lotus]

StrangeQuarkAL

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@storm birch Has your question been resolved?

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[toxic haven]

can someone help me awnser this step by step

[toxic haven]

i found the derivative it is

,rotate

[lament ledge]

!occupied

[calm coral]

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

[worn belfry]

The derivative of velocity is acceleration. Since you've found the derivative, you now have the equation for acceleration. Do you know how to find the abs max/min of a parabola?

[toxic haven]

u just equal it to zero?

factor the common decide

factor

[toxic haven]

it seems very complication to use a quadratic function on my equation because there’s so many numbers

[worn belfry]

They gave you some pretty long decimals to work with, but that shouldn't make the math more complicated. It's just a parabola.

[toxic haven]

okie

so i should use quadratic form?

[worn belfry]

You're looking for the lower vertex of that parabola which is the absolute minimum that the acceleration is going to be. c - b^2/4a.

[toxic haven]

ok

so what would the next step be ?

sorry i’m a bit confused

[worn belfry]

Did you get the minimum?

[toxic haven]

Don’t you have to find the coordinates first and do the test point so right now I’m trying to find the coordinates by equalling the derivative to 0

[worn belfry]

You could do it that way yes. But the thing you want to keep in mind is that the equation you've been given on the original problem is Velocity. It's asking you for the minimum of the Acceleration.

Acceleration is the derivative of Velocity. So if you want the derivative of Acceleration, you need to get the second derivative.

[toxic haven]

min and max acceleration

oh

okok

i got 15.913

[worn belfry]

Alrighty, there's your minimum acceleration then

[toxic haven]

yess

okie

i got the max too

tysm

[worn belfry]

Now to find the maximum, evaluate t = 59.3, because that's when acceleration stops, since the boosters were removed.

Oh nice

[toxic haven]

.close

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[bronze adder]

I'm having a bit of trouble understanding how exactly this proves the limit looking back at the definition of a converging sequence

[remote mural]

what's troubling you in the proof exactly

as in, which part is tripping you up?

[bronze adder]

I get the up to the bottom of the first page

But after that it seems almost working backwards?

Okay are we basically "cheating"?

[remote mural]

that's often how we do it - because we want to find such an N in \mbb{N} which satisfies the convergence condition for n > N

[remote mural]

well, you could think of it like that

[bronze adder]

Right, I guess I'm just not used to it then

[remote mural]

what you're doing is

[remote mural]

that "N > 1/eps" is something that you wouldn't have necessarily figured out as you were writing the proof, as it's at the very start. you find that AFTER you've already done the algebra with the |a_n - a| < eps and figured out this value of N

notice how he does all the algebra before he formalizes the proof

[bronze adder]

So I assume every convergence proof boils down to finding a relation between epsilon and N and putting it back into the definition?

[remote mural]

that's because it's pretty much impossible to accurately guess the value of N by inspection

[bronze adder]

Right

[remote mural]

yeah pretty much that's what you're doing.

the definition is very eloquent and powerful and it's not possible to make a better one than that

so it's a matter of applying it

[bronze adder]

Alright, I appreciate the help, thanks :)

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[reef tulip]

anyone know what these type of grpahs are called

[reef tulip]

like sin-1(cosx)

[dreamy hull]

Do you mean the function or the graph "shape"?

About the shape, this is continuous, piecewise smooth, and periodic function.

[reef tulip]

like the function

cause imtrying to write a summary sheet for my test

[alpine stone]

I do not believe there is a standard name

[reef tulip]

ohh okok

[dreamy hull]

Yeah that does not have a name

[reef tulip]

on the obttom righ tcorner is that all of the veriations of these grpah

[dreamy hull]

You may denote it like e.g. s^(-1) c for shorthand

[dreamy hull]

For sin^(-1) (cos x)

[reef tulip]

okok

all good then

thank u

how do i end this channel

[dreamy hull]

You say .close

[reef tulip]

.close

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[upbeat hemlock]

Hello

[upbeat hemlock]

I have a algebra exam in 1 hour and I don’t know that much about it

I’m in grade 9

<@&286206848099549185>

I have a sheet the exam will be similar too

[ornate moth]

show me your money

[upbeat hemlock]

huh

[ornate moth]

it was a jk

[upbeat hemlock]

Oh lol

[ornate moth]

i have seen it int youtube so i thought americans liked it

[upbeat hemlock]

😂

[ornate moth]

soooo whats you question?

[upbeat hemlock]

I’m just tryna learn how to do these equations

I’d say this is better

[ornate moth]

have you trie simplyfi both sides?

@upbeat hemlock

[upbeat hemlock]

Nope

[ornate moth]

then dot it and after this try to put the x alone

x or etc

[upbeat hemlock]

Hm ok

[calm coral]

@upbeat hemlock Has your question been resolved?

[gleaming mason]

So to solve number 2, put the like terms on the same side

For number 3, do distribution method and if possible, like terms on the same side

[calm coral]

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[gaunt lichen]

Hi I'm a bit confused with how to approach 16b

I've done 16a

I know I have to set up an expansion for 16b

So what I did was nC0(p)^n + nC1((p)^n-1)(1-p) = 0.09478

and I know np= 3.5

But apparently that's wrong and I'm meant to reverse the 1-p and p terms?

I'm a bit confused with that

[calm coral]

@gaunt lichen Has your question been resolved?

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@gaunt lichen Has your question been resolved?

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[stoic oak]

Hi I need help for a

[stoic oak]

<@&286206848099549185>

[pine lynx]

Ta

[stoic oak]

Got anything?

a) Using a compass and ruler only. Construct a line parallel to GJ and passing through K to meet HJ produced at W

I'm not sure how to do this

[pine lynx]

Do you understand what it's asking you to achieve?

[fervent tundra]

can you rotate the first image?

[stoic oak]

Sure

[stoic oak]

@fervent tundra

[quick karma]

green line and blue line are parallel, pink line is hj produced to w

you need to find the length of hw

the pink line

segment

[stoic oak]

Ohhhh

Thank you so much for your time and teaching

.close

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[flint mist]

Theres something im confised abt

[flint mist]

so basically when finding the correlation coefficient

it takes two paramaters right

say X1 and X2

but on my exam it had something like

p(standard deviation of X / standard deviation of Y)

how does that work

thats missing another parameter right

[calm coral]

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[quaint sphinx]

@raw dirge i'd start with u = the whole thing

[quaint sphinx]

.close

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[raw dirge]

what

can you reopen this?

[quaint sphinx]

.reopen

[calm coral]

✅

[raw dirge]

alright

so u = the whole thing

,,du/dx = \frac{-cotx csc^2x}{\sqrt{cot^2x - 1}}

[potent lotus]

Wither

[raw dirge]

u^2 +2 = csc^2x

like that?

[quaint sphinx]

ye

[raw dirge]

it converts to

$\int{\frac{u^2}{\sqrt{u^2 + 1} (u^2 + 2)}du$

[potent lotus]

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

[quaint sphinx]

ye

after this i'd try to remove the square root

[raw dirge]

another substitution?

[quaint sphinx]

yeah
the one i would use here is ||u = sinh(v)||

[calm coral]

@quaint sphinx Has your question been resolved?

[brittle isle]

why not tan?

[calm coral]

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[normal wasp]

Indian tea is 5/4 times more expensive than American tea. What proportion do we need to mix the indian and American teas so that we get a tea that is 6/5 more expensive than American

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[quaint sphinx]

since it's worse

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[astral pawn]

help

[astral pawn]

I need to solve for the critical values of cos(2x)+sinx+1=f(x). I set the derivative equal to zero and got 0.25 and 2.9. However, there are two other critical values that I don't understand how to get mathematically.

[brittle isle]

what did you get for f'(x)?

also there's infinitely many critical values, are you looking for the critical values over a certain interval?

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[lost mural]

• I have to calculate the lenght of the line between 60 and 60 via the Pythagorean theorem.
• I am currently stuck at finding a way to do so.
• I'd be happy about a tip to nod me in the right direction, as the problem is simple enough
  (Ignore the lead (grey) pencil marks and numbers)

[lost mural]

Solved it

.close

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[remote mural]

for some reason

[remote mural]

With questions like these I can't visualize where I would put the lines to get the moments

like

for the clockwise, if I was tryna find moments from B

for the 60n force Im confused why I would be finding that horizontal distance

to get me 60 * 8sin70

[supple mural]

where have you set your point of reference?

[normal wasp]

.close

[supple mural]

?

[calm coral]

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[calm coral]

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[remote mural]

.reopen

[calm coral]

✅

[remote mural]

at B

idk what you mean point of reference

[ocean epoch]

First assume horizontal line abt the point ur taking the moment ( here along B)

To calc moment of force at A drop a perpendicular from A to the line

That's basically the distance we want

[remote mural]

yeah

doesn't explain why tho

like

we're taking a moment from B which is a point of pivot basically

the force is like directly from that point

[ocean epoch]

See in vector form moment = R x F

Cross product of r and f

So it's RFsintheta

Where theta is angle between them

[remote mural]

nope

no idea

what that is

im astronomically cooked

[ocean epoch]

Just use this R × F x sin theta

[ocean epoch]

Or use this method it's simpler

[ocean epoch]

R is the line joining points A and B

[remote mural]

this is what im seeing rn

in my head

the mg is a perpendicular distance away from B

[remote mural]

yes Im aware of that

but

I dont get why

I knew we need that height from the answers

[ocean epoch]

What's the formula for moment?

[remote mural]

distance x force

?

[ocean epoch]

Perpendicular distance × force

[remote mural]

yeah fairs

[ocean epoch]

Or distance × force × sin theta

[remote mural]

yes but that has no perpendiculr distance away

because the force is acting directly from that point

[vague path]

No they are using the force from A

which is solved in the previous part of the question

[remote mural]

oh yeah

I forgot about it

💀 ok that makes sense

Wait so I was right

there is no moment to be taken it just gets negated like the reaction at that point

[vague path]

Well yeah the force at B has no distance from B and so it creates no moment

[remote mural]

never do maths while licked it is not the method

[remote mural]

yeah that's what i've been saying the whole time

yh ok ok thanks

[calm coral]

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[gusty plinth]

Currently going through Rational Points on Elliptic Curves and I could use a hint for exercise 1.8.a. which asks to prove that, for every $k \geq 1$ the congruence $x^2 + 1 \equiv 0 \mod 5^k$ has a solution in $Z/5^kZ$. I'm trying to solve it by induction, but I'm struggling to find the connection between the k and k+1 case.

[potent lotus]

Corn

[graceful dust]

maybe you can try lifting with hensel’s lemma

[calm coral]

@gusty plinth Has your question been resolved?

[gusty plinth]

We are trying to prove a specific case of Hensel's lemma, so I think this would go against the goal

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[honest arch]

Can someone explain to me how to approach graphing autonomous differential equations such as the one above? I understand how to determine if a critical point is stable/unstable but have no idea how to graph it

[drifting ravine]

i think there is an atractor or repulsor and you look at the vector fields or something.

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[nimble badge]

If you can get the equilibria and the phase portraits the graphs are simple

As x gets larger what happens to a point in each of those intervals?

They get closer to stable equilibria

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[honest arch]

how do you determine what the value of the stable equilibria actually is?

[honest arch]

or is that just the critical point

[calm coral]

@honest arch Has your question been resolved?

[nimble badge]

You find the critical points and classify them. Remember if the derivative is zero, then it doesn’t change. Once we have the equilibria we can classify them by the sign above and below the point / the second derivative test

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[idle fractal]

The number of elements of order 8 in the group ℤ24 is
(a) 1 (b) 2 (c) 3 (d)4

[idle fractal]

So here 8a mod 24=0

So i am getting a={0,3,6,9,12,15,18,21}

[ocean cypress]

in the additive group?

or multiplicative group

[idle fractal]

I guess additive

[ocean cypress]

24 = 2 x 2 x 2 x 3

so any element divisible by 3 should have order AT MOST 8

now let's go through your list

0 has order 1, because it's the identity

3 has order 8, because 3 * 8 = 24
6 has order 4, because 6 * 4 = 24

keep going

@idle fractal

[idle fractal]

I see

8*6=48 is divisible by 24 no?

[ocean cypress]

define the word 'order' for me

in the context of an additive group, it is the least number a such that a * g = 0 for an element g in the group

8 x 6 = 48, but 8 x 3 = 24, and 3 is smaller than 6, so 8 has order 3

@idle fractal

[idle fractal]

I see

9*8

15*8

[ocean cypress]

use the prime factorizations

[idle fractal]

Prime?

I see

[ocean cypress]

24 = 2^3 x 3, so to determine the order of 9, for instance, use the fact that 9 = 3^2, so then 9 must have order 8 because you need 3 copies of 2 to reach a multiple of 24

[idle fractal]

Is this wrong?

[ocean cypress]

let's see

15 = 3 x 5

15 does not have any factors of 2, so we need 3 copies of 2 to reach a multiple of 24. so yeah it has order 8

[calm coral]

@idle fractal Has your question been resolved?

[idle fractal]

So here i got
3,9,15,21

Am I correct?

@ocean cypress

[ocean cypress]

yeah looks good buddy

[idle fractal]

What if the operation was multiplication?

[ocean cypress]

then you'd have to solve the equation a^8 = 1 mod 24

very difficult to do in general

[idle fractal]

i see

so in our addtive case answer should be 4?

[ocean cypress]

yep

[calm coral]

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[slate cipher]

where did i go wrong?

[quaint sphinx]

your answer is right

[slate cipher]

the correct answer is from the textbook

[quaint sphinx]

the textbook is wrong

[slate cipher]

really?

[quaint sphinx]

well you can differentiate their "answer" to check for yourself

[slate cipher]

might have to email the maker lol jk

lemme try that i always forget I can do that

oh i wus looking at the wrong part of the key in the textbook

myb

.close

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[remote mural]

What is meaning of travial solution?

[kind grotto]

usually the 0 solution but can you point where are you seeing it?

[remote mural]

@kind grotto summary

[remote mural]

0 solution or solution "0"?

[kind grotto]

for example in a matrix, the vector (0,0,...,0)

[remote mural]

Umm can u explain in terms of equations

[kind grotto]

its just that all of the variables are 0

[remote mural]

Cool
So it says if det is 0 it is 0,0,0 as solutions
Right?

But isn't it wrong

[kind grotto]

exactly

[remote mural]

Because... 1st

[kind grotto]

sorry sorry the other way around

[remote mural]

I didn't get

And here 1st is opposite

[kind grotto]

if det!=0 you have only one solution which is 0,0,0

[remote mural]

@kind grotto here in summary
If det is 0 and det 1 det 2 det 3 are non zero then there is NO solution

[kind grotto]

thats another condition

you need det=0 and also this thing

its important to say that when I say there is a solution its for all of the equations equal 0 at the RHS

[remote mural]

Ohh
So...

These are for RHS is 0

[remote mural]

And this is for RHS is d in general

Right?

@kind grotto

[kind grotto]

I think so

[remote mural]

Yessssssss
Thanks a lot man
Love u man

🫂

[kind grotto]

no problem

[remote mural]

Have a beautiful day ahead

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

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[idle fractal]

Hints

[blazing coyote]

consider two cases

as you approach 0 from the right

and from the left

I'd also divide the numerator and denominator by $e^{\frac{1}{x}}$

[potent lotus]

ƒ(Why am. I here)=I don't Know

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[idle fractal]

.reopen

@blazing coyote

[blazing coyote]

yes?

[idle fractal]

Is this correct sir?

What should I do next?

It seems like oscillator

[blazing coyote]

notice that e^{1/x} tends to zero as we approach it from the left

[blazing coyote]

on the other hand we approach it from the right, it tends to $\infty$

[potent lotus]

ƒ(Why am. I here)=I don't Know

[blazing coyote]

$e^{\frac{1}{x}}$. is what I'm talking about

[potent lotus]

ƒ(Why am. I here)=I don't Know

[blazing coyote]

so think about how you'd deal with that

[idle fractal]

Yes it is infinity that is why it is 0

[blazing coyote]

the left hand limit is zero IMO

,w lim of (e^{1/x} sin(1/x)/(1+e^{1/x}) x \rightarrow 0

[idle fractal]

What

[potent lotus]

Results provided by WolframAlpha

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

Requested by physicsrocks

[idle fractal]

It has to be 0+ 0-

[blazing coyote]

yes

wait a minute please

[idle fractal]

Yes sir

[blazing coyote]

what's the limit at 0^-?

[idle fractal]

Here limit denominator should be e^(-1/x) no?

[blazing coyote]

this is the limit, yes?

[idle fractal]

Yes this is same as i solved out

[blazing coyote]

yeah

so the left hand limit is what?

[idle fractal]

0

[blazing coyote]

and do you know how to find the right hand limit ?

[idle fractal]

Did you see this?

It is inderminante

[blazing coyote]

yes

[idle fractal]

And?

[idle fractal]

.

Is it d?

[blazing coyote]

yes

I think so

I'm not sure though

[idle fractal]

Me too

What is C?

[ancient thistle]

Using Rudin's definition of a discontinuity of the second kind for a function. f has a discontinuity of the second kind if either f(x+) or f(x−) does not exist.

The essential discontinuity of the second kind occurs when only the right- or the left-hand limit does not exist.

[blazing coyote]

oops, my source must be wrong then

sorry

[ancient thistle]

if you dont have a definition of what a "discontinuity of the second kind" is or what an "infinite discontinuity" is, nobody can help you

there does not seem to be a consensus on what the definition of a "discontinuity of the second kind" is

[calm coral]

@idle fractal Has your question been resolved?

[idle fractal]

Yes i googled and found essential discontinuity

[idle fractal]

.close

[calm coral]

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[ripe lintel]

got (-1/((x)(sqr(1-x^2))) when the answer is (-1/(sqr(1-x^2))) without the additional x in the denominator in my answer

[covert lark]

What is the problem?

[ripe lintel]

im using this formula btw

[covert lark]

Ok...

[ripe lintel]

when i use this formula i end up with an additional x in my denominator

[lament ledge]

what is the question???

[dull wagon]

can you show your work

[ripe lintel]

i have to find the derivative of this

my bad for not clarifying

[covert lark]

You can use chain rule here

So you replace x with u = 1/x and calculate chain rule

[swift sapphire]

bro simply replace, x with cos theta

1/cos theta = sec theta

arcsec(sec theta) = theta

x = cos theta

then, theta = arccos(x)

then simply differentiate cos inverse x and you have your answer

substitution will be wayyyyy faster in this method

than doing this algebraically

[ripe lintel]

thanks

.close

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[dreamy crest]

 Sketch what the distribution of possible observed values could be for the following cases:

    The number of people visiting your local cafe assuming Poisson distribution

[covert lark]

Do you know what Poisson distribution looks like?

Or do you know it's functional form

[dreamy crest]

Yes, I plotted it with the Python.

Could you please confirm whether I am right or wrong?

[dreamy crest]

import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import poisson

# Define the lambda (mean number of visitors)
lambd = 20

# Define the range for K values (from 0 to a bit more than the lambda to capture the tail of the distribution)
k_axis = np.arange(0, 40)

# Compute the Poisson PMF for each K value
distribution = poisson.pmf(k_axis, lambd)

# Plot the PMF using a bar chart
plt.bar(k_axis, distribution, color='skyblue')
plt.title('The number of people visiting your local cafe')
plt.xlabel('Number of Visitors (K)')
plt.ylabel('Probability')
plt.grid(True)
plt.show()

[covert lark]

Yeah looks good to me

[dreamy crest]

Yay!

.close

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[blazing coyote]

so I feel L would be $^{p+q}C_p$

[potent lotus]

ƒ(Why am. I here)=I don't Know

[blazing coyote]

M=$^2C_1 ^{p+q}C_p$

[potent lotus]

ƒ(Why am. I here)=I don't Know

[blazing coyote]

is that right so far?

[fathom shuttle]

the N in this question is entirely useless

yeah thats right so far

[blazing coyote]

N is incomplete

it should be p books and q books

so that would be $^{p+q}C_q$

[potent lotus]

ƒ(Why am. I here)=I don't Know

[fathom shuttle]

oh ok

[blazing coyote]

is this all right so far

but this means $^{P+q}C_q =^{p+q}C_p$

[potent lotus]

ƒ(Why am. I here)=I don't Know

[blazing coyote]

[fathom shuttle]

whats wrong with that

[blazing coyote]

right

nothing

That means p=q

[fathom shuttle]

what

[blazing coyote]

doesn't it

[fathom shuttle]

whats 6C2 and 6C4?

[blazing coyote]

oo

right

I forgot that property

sorry

thanks a lot

.close

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[remote mural]

A YouTuber said set the parenthesis to equal 0 and solve to get the phase shift

Is that the same thing as factoring

If there was a number in front of x

I would need to factor or set it to 0 to find the period and phase shift?

Then in this form I don’t need to do anything

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[remote mural]

Can someone explain how logarithms work?

[mystic scaffold]

What part do u need help with

[minor gust]

https://www.tiktok.com/@onlocklearning/video/7335141960241401120?lang=en

[remote mural]

Thanks guys

[remote mural]

Im on my way home, I just didnt understand well what the teacher said

[mystic scaffold]

Ah okay

[remote mural]

How do I close the thread

[median oyster]

do .close

[remote mural]

.close

.close

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[static walrus]

i have 0 knowledge

need help

[coarse minnow]

where could the point be for the triangle to be obtuse and/or acute respectively?

[calm coral]

@static walrus Has your question been resolved?

[static walrus]

for this specific rectangle, there is a certain height where it begins being obtuse

the central angle is like 126

if you put it in a cirlce

[coarse minnow]

what portion of all points is that?

[static walrus]

no idea

[coarse minnow]

there should exist some arc within the rectangle on which it's all right triangles (by Thales's theorem)

the arc in question being the semicircle over AD

so obtuse is anything UNDER that arc (but still inside the rectangle)

find the area of said arc enclosed in the rectangle, then find the area of the rectangle in total

that's it basically step-by-step

then round to the nearest X/400

[static walrus]

why is it a semicricle over AD

the point on the very bottom of the rectangle is 90 deg

if its a semicricle

@coarse minnow

or do we discount the 100

and just 314

[static walrus]

@coarse minnow

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[static walrus]

.close

.reopen

[calm coral]

✅

[static walrus]

.close

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[reef cipher]

stop you're about to get banned

sigma

[carmine sorrel]

!15min

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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.

[fierce smelt]

Az

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[coral dew]

I don’t see how to do this pls help

[coral dew]

Co interior??

Same side interior?

Ah ok

Yes yes but both have an X in them

So I can’t say they equal to 180 right

Bc both have X

To find

So I can do x + 3x + 20 = 180?

Ohh ok

Thxxx

Got it

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[rocky vapor]

are you talking about part b?

are you familiar with the limit definition of the derivative?

ah okay

sorry if that wasn't clear

so what part is confusing you?

ah ok

so first just for refrence purposes

$\frac{\mathrm{d}}{\mathrm{d}x}\left[f(x)\right] = \lim\limits_{h\to 0} \frac{f(x+h)-f(x)}{h}$

[potent lotus]

Mar the Marey

[crystal forge]

in your case the "f(x)" is s(t), does it help if you think of it as:
$\lim_{h\to0}\frac{s(t+h)-s(t)}{h}$

[potent lotus]

∫oosh (lemonsaurus appreciator)

[rocky vapor]

yea

[crystal forge]

what's s(t+h) ?

it's basically just plugging in t + h wherever t appears in the formula

[rocky vapor]

so it looks like you went a little wrong somewhere in evaluating your limit

so

in your second step you went from

$(t+h)^2$ to $(t^2+h^2)$

[potent lotus]

Mar the Marey

[rocky vapor]

however this isn't true

were you able to correctly evaluate the limt now?

[crystal forge]

yep, you actually solved for the generic derivative formula of 2t - t^2, which is 2 - 2t and yeah now you can plug in any t there to get the instant velocity for some specific t

once you learn the shortcuts for taking derivatives (probably next class) you can be slightly upset with your teacher for a second for making you solve this using the limit definition :p

well this is a 1 dimensional motion type situation where you have position (s) in terms of time (t) along a line basically, so yeah negative velocity would basically mean moving towards the negative direction on the line and positive velocity would mean moving toward the positive direction

seems right yeah

it's not conflicting pieces of information

here is a sneak preview of your next lesson power rule for derivatives

so you can see you went from s(t) = 2t - t^2 to v(t) = t - 2t, if you think of each of those terms individually you can see how easy the velocity formula is to find with the power rule shortcut

2t becomes 2 and t^2 becomes 2t

check the formula above and see how that works

basically the exponent goes down by 1 and also you multiply by the original exponent in front as a constant

-inf is right yeah

[rocky vapor]

just to add on to this rq, it might be a bit easier to think of it as left and right (if you werent already) where negative velocity is moving left and postive velocity is moving right

or if you think of it as driving a car is positive velocity and reversing the car is negative velocity

yea of course

if you don't already have a textbook to work out of

i reccomend using pauls online notes

https://tutorial.math.lamar.edu/

this is a college professor who posted notes for calc 1-3 online and each section has in depth explanations and solution

its how i personally taught myself calc so i vouch lol

[calm coral]

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[stable hollow]

okay so

[stable hollow]

if you had

(-6)^2

it would be 36 right

but if it was not in brackets it would be -36 correct?

[dull wagon]

yes

[stable hollow]

ty

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[dry shale]

why

why can x be in the range of -39 and 39

where did 39 come from

[dull wagon]

5 * 39 = 195 < 200

[dry shale]

huh?

[dull wagon]

5 * 40 = 200 right?

[dry shale]

yes

[dull wagon]

and |k| < 200 means
-200 < k < 200

if x = 40,
k = -12/40 - 200 < -200
which you don't want

[dry shale]

yes

[dull wagon]

it's fine to have x = 39 which gets
k = -195 - 12/39

same for the negative case

[dry shale]

but why do u assume that x = 40

[dull wagon]

because that's the dominant term

[dry shale]

why cant u assume x to be something above 40

[dull wagon]

200/5 = 40

[dry shale]

why do u divide it by 5

not 12

[dull wagon]

because you have 5x

[dry shale]

(sorry if im being annoying here)

[dull wagon]

for x > 12, 0 < 12/x < 1

which isn't really significant

[dry shale]

so 5 was used cuz rational numbers are harder?

[dull wagon]

no

because 12/x will be less than 1 for the larger values you're considering

[dry shale]

ohh

[dull wagon]

and you can approximate
k as -5x

[dry shale]

how did we get 39+39 in the end

[dull wagon]

39 integers from 1 to 39 inclusive

[dry shale]

and 39 from -39 to 0?

[dull wagon]

from -39 to -1 inclusive

[dry shale]

oh

why are they added tho

[dull wagon]

because you want the total

[dry shale]

oh

[dull wagon]

they all work

[dry shale]

in another solution of the same problem

why is -n the integer solution

[dull wagon]

from (x + n) = 0

[dry shale]

ohhh

[dry shale]

how do u get k=12/n+5n

@dull wagon(sorry but can please answer quickly cuz i have to go after 5 minutes)

[dull wagon]

expand out the ()

[dry shale]

so the answer to the first bracket in rhs is k?

[dull wagon]

no

[dry shale]

then?

[dull wagon]

expand (5x +12/n)(x+n)
and what do you get

[dry shale]

the original quadratic equation

with k replaced with n

[dull wagon]

I'm not looking for word answers

I want you to actually expand that out

[dry shale]

5x^2+nx+12=0

[dull wagon]

no

[dry shale]

its wrong?

[dull wagon]

still no

[dry shale]

why

[dull wagon]

because you just looked at the original and replaced k with n

don't take shortcuts

[dry shale]

oh