help-42

[upper nova]

yep AED seems right

[torpid dune]

so my final answer theoretically should be correct?

[upper nova]

So AED=CFD=pi/4-1/2

[torpid dune]

yeah

[upper nova]

And EDF=$\frac{1}{4}\pi(2-\sqrt{2})^2=\frac{3}{2}\pi-\sqrt{2}\pi$

[potent lotus]

Dri111

[torpid dune]

yeahyeah

[upper nova]

,w 3/2*pi-sqrt(2)pi+(pi/4-1/2)*2

[torpid dune]

uh what

[upper nova]

nvm

Then $\pi+\frac{3}{2}\pi-\sqrt{2}\pi+2\left({\pi \over 4} - {1 \over 2}\right)$ right?

[torpid dune]

yeah

[upper nova]

try again your calculation

[torpid dune]

i think the area includes circle ABCD, no?

correct me if im wrong

cuz i calculated the are of the entire figure

[upper nova]

I see

[torpid dune]

im not too sure if it includes ABCD now that u point it out

[potent lotus]

Dri111

[torpid dune]

yeah

[upper nova]

for ABCD circle

then I think you're right

[torpid dune]

yayy

now i have another question ||can u tell im bad at math olympiad?||

,rotate

[torpid dune]

ACD and ABC are similar

i think thats supposed to help me but idk how

wait i think i see something

[upper nova]

so is it good then?

[torpid dune]

im still trying to follow through but

for now i know that

AD is 4u, BD is 5u and AC is 6u

[hard kernel]

????

[torpid dune]

@hard kernel can u still assist?

<@&286206848099549185>

[calm coral]

@torpid dune Has your question been resolved?

[calm coral]

@torpid dune Has your question been resolved?

[torpid dune]

.close

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[fierce mulch]

I'm trying to find $\frac{\partial z}{\partial x}$ $z=(x^2 y+ xy^2) ^cos(xy)$. i took logs of both sides to get $\ln(z) = \cos(xy) \ln(x^2 y +xy^2)$, I used the product rule to get $\frac{1}{z} \frac{\partial z}{\partial x} = -x \sin(xy) \cdot \ln(x^2 y + x y^2) + \cos (uv) \cdot \frac{2uv + v^2}{u^2 v +uv^2}$ simplifying to $(x^2 y + xy^2)^{\cos(xy)-1} [\cos (xy) \cdot (2xy+y^2) - y\sin (xy) \cdot \ln (x^2y+yx^2) (x^2 y + y^2 x) ]$ i wasnt given a solution, so i checked it with wolfram, but its answer looked different to mine. im not sure if it factored it a different way, or if i have made a mistake. any help would be great. thanks

[potent lotus]

lewis_f04

[fierce mulch]

this was wolframs answer

[glass heart]

from a quick glance that looks like the same thing

well ok you have a small typo in your ln, you mean ln(x^2y+y^2x) but wrote ln(x^2y+yx^2)

(dont forget that you can also ask wolframalpha whether two expressions are the same)

[calm coral]

@fierce mulch Has your question been resolved?

[azure willow]

you are doing a partial derivative?

[calm coral]

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[fringe hamlet]

hi

[fringe hamlet]

\textbf{LEMMA 1.} A Cauchy sequence is bounded.

\textbf{Proof:} Let $\epsilon = 1$; by hypothesis there exists $v \in \mathbb{N}$ such that $|a_k - a_h| < 1$, $\forall h,k > v$.

Let's set an index $h_0 > v$. Then, by the properties of the absolute value, it follows:
[a_{h_0} - 1 < a_k < a_{h_0} + 1, \quad \forall k > v.]

Let $A = \min {a_1, \dots, a_v, a_{h_0} - 1}$ and $B = \max {a_1, \dots, a_v, a_{h_0} + 1}$; evidently it turns out:
[A \leq a_k \leq B, \quad \forall k \in \mathbb{N},]
and therefore the sequence is bounded.

[potent lotus]

alee

[fringe hamlet]

i dont understand this: Let's set an index $h_0 > v$. Then, by the properties of the absolute value, it follows:
[a_{h_0} - 1 < a_k < a_{h_0} + 1, \quad \forall k > v.]

[potent lotus]

alee

[glass heart]

$\abs{a_k-a_{h_0}} < 1$ means $-1 < a_k-a_{h_0} < 1$

[potent lotus]

Denascite

[fringe hamlet]

what is a_h_0?

[glass heart]

the h_0'th entry of the sequence

h_0 is some number, maybe 7340

then it would be a_7340

[fringe hamlet]

h_0 > v, right?

so the book takes a random number h, which is greater than v?

@glass heart

[glass heart]

yes

could be v+1

[fringe hamlet]

but is a_k or a_h_0 - 1 bigger?

[glass heart]

$a_{h_0} -1 < a_k < a_{h_0} + 1$

[potent lotus]

Denascite

[glass heart]

so a_k is bigger than a_h_0 - 1

and smaller than a_h_0 + 1

[fringe hamlet]

but

why the book put a_k in the min?

if a_k > a_h_0 - 1

[glass heart]

it didnt

it put into the min only up to a_v

[fringe hamlet]

ah okay

[calm coral]

@fringe hamlet Has your question been resolved?

[calm coral]

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[white atlas]

hi, how are you supposed to do this question?

[white atlas]

i'm just going to guess and say that the third pivot becomes 9

and the second blank is supposed to be 2

[calm coral]

@white atlas Has your question been resolved?

[faint charm]

I assume you are trying to diagonalize a square matriz using the pivot method ?

[white atlas]

i don't think so?

i don't know the context for this question either

i guess i'm right but idk what the rationale behind this question is

or how it was supposed to be done

[faint charm]

@white atlas I do not know the context either. It sounds to me like a matriz diagonalization by pivoting question, but I am just guessing by inference there. Besides, in that case, the question wouldn't make much sense without know the shape of the full matrix either.

[calm coral]

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[zinc stirrup]

I'm unsure where to start on this question

[zinc stirrup]

The only thing I can think of is that x = u cosθ and y = v sinθ

[calm coral]

@zinc stirrup Has your question been resolved?

[calm coral]

@zinc stirrup Has your question been resolved?

[zinc stirrup]

<@&286206848099549185>

Please may I have a bit of support? I've now figured out x = u cos theta - v sin theta

and y = u sin theta + v cos theta

[calm coral]

@zinc stirrup Has your question been resolved?

[potent lotus]

spq_64_t

[calm coral]

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[mild jetty]

Im failing to see how they get from fig 1 to fig 2

This is euclids proof of pythagorean thm

oh bruh

how are you suppose to see it lmao

.close

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[simple loom]

You want to solve when the function is 0

I need help for question 2

[remote mural]

i don't think there's necessarily an analytical way to solve this

you have a polynomial and a trigonometric function

how will you find the roots?

[simple loom]

I don't know

but we were never taught any numerical methods to solve this

so I have no idea what to do

[leaden thunder]

Maybe try plugging in special angles

[simple loom]

I tried none work

[leaden thunder]

What did you try

[simple loom]

[-3, -2, -1, 0, 1, 2, 3], anything bigger than this isn't a solution

[leaden thunder]

,w solve 9cos^2(3x) - 6 cos(3x) -2(3x-2)(3x+1)=0

[potent lotus]

Results provided by WolframAlpha

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

Requested by rie.mann

[leaden thunder]

I see. Yea ask your teacher if there's a typo. This probably requires numerical methods

[simple loom]

ok thx

[calm coral]

@simple loom Has your question been resolved?

[calm coral]

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[warm warren]

so im looking at poisson processes, and im up to the proof of this theorem

[warm warren]

https://ocw.mit.edu/courses/6-262-discrete-stochastic-processes-spring-2011/3a19ce0e02d0008877351bfa24f3716a_MIT6_262S11_chap02.pdf page 73

i've got this as the given definition of memoryless

or at least how can i tell from just the information that X is a memoryless RV

i dont see how the memoryless condition helps on (2.7) here

[warm warren]

this should have X_1's everywhere

[marsh agate]

Write the conditional probability as per its definition

[warm warren]

yeah this one right?

[marsh agate]

Notice that X1 > z+t and X1 > t is just X1 > z+t

Hence it's got probability P(X1 > z) P(X1>t) / P(X1>t)

[warm warren]

wait what

[marsh agate]

Because z >= 0

[warm warren]

OH

[warm warren]

so this was not what i was meant to do

[warm warren]

i was meant to go to the definition of conditional probability from 2.6

[marsh agate]

Well at least that's one way

[warm warren]

don't see how that's using the memoryless property though

[marsh agate]

Here

[warm warren]

oh the second part

right

so i do $\text{P}(X_1>z+t|X_1>t)=\frac{\text{P}({X_1>z+t}\cap{X_1>t})}{\text{P}(X_1>t)}$ then realise on the top that 1 is a subset of the other so the intersect is just the bigger one, which is ${X_1>z+t}$

[potent lotus]

Frosst

[marsh agate]

That's this

[warm warren]

yeah

ok and then you just use the memoryless property and cancel the denominator

then tada just P(X_1>z) remains as shown

great

thanks!

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[oblique crypt]

(Read image first for problem description)
So

f(x) = {1/4800 for 10200 <= x <= 15000,
       {0          elsewhere 

So if I bid 13100, the probability I win is

(13100 - 10200) / 4800 = ~0.60

because 13100 beats all bids from 10200 to 13100. And we divide by 4800 (really we multiply by 1/4800) to find the area under the curve.
The area under the curve represents the probability that the competitor's bid will be less than 13100. Then, we find the profit we would expect if we did win. 16000 - 13100 = 2900. So we have a ~60% chance to profit 2900. So to calculate the expected profit, we multiply them together.

~0.60 * 2900 = ~1752.08

But that's wrong. What did I do wrong?

[calm coral]

Please don't occupy multiple help channels.

[oblique crypt]

I also tried

(13099 - 10200) / 4800 = ~0.60

since we may be able to assume that if I bid 13100, that doesn't beat the competitor's bid of 13100.
The profit if we win would still be 2900. Multiplying them gives
~0.60 * 2900 = 1751.48.
This answer is also wrong.

[amber bolt]

like the loss is part of the expected profit

[oblique crypt]

It doesn't cost anything if we lose the auction though.

[amber bolt]

I would think it's 0.60 (2900) + 0.40 (−13100 )

oh that makes sense

i get 1752.08

which is 1752.1

guess you rounded wrong

@oblique crypt

[oblique crypt]

Read the question again. It says round to 2 decimals, not 1.

[amber bolt]

i see

so there's a hidden 8 too

[oblique crypt]

Oh yeah, it just got cut off by the box.

Sorry for the confusion.

[amber bolt]

no idea then, my answer is the same

[oblique crypt]

<@&286206848099549185>

[calm coral]

@oblique crypt Has your question been resolved?

[oblique crypt]

<@&268886789983436800>

[calm coral]

@oblique crypt Has your question been resolved?

[calm coral]

@oblique crypt Has your question been resolved?

[calm coral]

@oblique crypt Has your question been resolved?

[calm coral]

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[ionic fog]

who an help with geometry

[calm coral]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[ionic fog]

Can I ask a question about geometry

[brazen elbow]

!da2a

[calm coral]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[ionic fog]

In a right triangle ABC ∠C=90° and the catheter AC is 12. The second catheter at point D is divided into 5 and 30. Find the distance between the radii of the inscribed circles of the triangles ACD and CBD.

[brazen elbow]

have you drawn out the shape?

[ionic fog]

not yet

[brazen elbow]

i assume catheter here is cathetus

[brazen elbow]

well, what are you waiting for

[ionic fog]

yes

[ionic fog]

Im waiting for solution

I cant do that

[brazen elbow]

im here to help, not to dispense solution

[ionic fog]

ok

so

[brazen elbow]

draw out the shape

follow their instruction

[ionic fog]

ok

got it

[brazen elbow]

also, what do you mean by this?

The second catheter at point D is divided into 5 and 30

does it mean the point D belongs to the second cathetus and creates 2 segments of measurements 5 and 30?

[ionic fog]

like AB is divided by D on 30 and 5

[ionic fog]

yep

[brazen elbow]

if so, CBD aint a triangle

[ionic fog]

its a triangle

draw it

u will see

[brazen elbow]

yes i did and no it isnt

why dont you draw it out and see it for yourself

[ionic fog]

i draw it

a and c is near

b is on top

[brazen elbow]

show your work ™️

[ionic fog]

one sec

[brazen elbow]

well, that wasnt what the question meant

but i assume it is a typo

[ionic fog]

?

wdym

[brazen elbow]

it asks that the point D divides the second cathetus into 2 segments

but what you drew seems correct

[ionic fog]

but

hypotenuse

[brazen elbow]

then it is a typo

[ionic fog]

i diveded the hypotenuse

well

mb

A is 90

[brazen elbow]

then it still doesnt make sense

if C is 90 and D belongs to the hypotenuse, then the question is coherent

[ionic fog]

wait

I know

Like this

have any ideas?

[brazen elbow]

its basically asking for the distance of 2 incenters

[ionic fog]

yes

idk how to do it

[calm coral]

@ionic fog Has your question been resolved?

[calm coral]

@ionic fog Has your question been resolved?

[calm coral]

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[scarlet nova]

I am unable to transform it to a form where I can apply telescope method. Is there any other method? If not please help me how to telescope this

[glad plover]

is this the original question

[scarlet nova]

Idk my teacher gave this to me

[glad plover]

hmm

it feels like an integral somehow

[spare beacon]

I don't know why your teacher would ask this. There is some context missing here for why you are even doing this to begin with

[glad plover]

there might be a trick

it feels like the sort of thing with a trick

[scarlet nova]

What?

[spare beacon]

I'm asking why you are trying to solve this to begin with. Is this an assignment? Extracurricular thing?

[glad plover]

are you sure it's (2k-1)pi on the top and then (k-1)pi on the bottom?

that's a bit of a bugger

[scarlet nova]

That's the question I was sent

[glad plover]

hrm

[scarlet nova]

If this feels incorrect I will ask my teacher tomorrow

[scarlet nova]

To get good marks in college entrance exams ig?

[glad plover]

maybe it's an angle addition formula thing

[scarlet nova]

I tried that, by splitting sin but can't get it in a telescopable form

[spare beacon]

Here is a hint:

$\sin\left(\frac{(2k-1)\pi}{2n}\right) = \frac{\cos^2\left(\frac{(k-1)\pi}{2n}\right)-\cos^2\left(\frac{k\pi}{2n}\right)}{\sin\left(\frac{\pi}{2n}\right)}$

[potent lotus]

JessicaK

[scarlet nova]

Ok I got it now

Thank you🙏

.close

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[potent smelt]

@hard forge topic multidimensional fourier transform

[potent smelt]

So a standard fourier transform is given by:

[
\int_{-\infty}^{\infty} f(t) \mathrm{e}^{-i \omega t} , \dd{t}
]

[potent lotus]

全能の存在

[hard forge]

Yes

[potent smelt]

For the multidimensional case, we want two frequencies

so it's just:

[
\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(s, t) \mathrm{e}^{-i \psi s - i \omega t} , \dd{t} \dd{s}
]

[potent lotus]

全能の存在

[potent smelt]

where $\psi$ and $\omega$ are your frequency variables in the s direction and t direction respectively

[potent lotus]

全能の存在

[potent smelt]

and when you do the inverse transform you need to use a leading factor of 1/(2pi)^2 instead of 1/(2pi)

@hard forge ^ does this all make sense?

[hard forge]

Yes

[potent smelt]

So the discrete case will be

[hard forge]

Just one thing

[potent smelt]

what's up?

[hard forge]

What is f(s,t)

S and t are the pixel location

[potent smelt]

your two dimensional input signal.

yeah, in your case s and t can be your x, y coordinate

and since you have discrete data, you will have:

[hard forge]

Then f of that is the intensity

[hard forge]

Of that pixel value

[potent smelt]

[
F(v, w) = \sum_{x=0}^{m} \sum_{y=0}^{n} f(x, y) \mathrm{e}^{-ivx - iwy}
]

[potent lotus]

全能の存在

[potent smelt]

Yeah, and if you have multiple channels, you'll need to handle them separately

If you watched the videos I sent, you can consider converting to YUV instead of using RGB

[hard forge]

From what I’ve watched grayscale is better

[potent smelt]

it's certainly simpler, only one channel.

[hard forge]

RGB is how many channels

3?

[potent smelt]

yeah, and so is YUV

[hard forge]

Then I’ll stick with grayscale

[potent smelt]

👍 you can consider using FFT as well

but if you just need to demo a little bit of math by hand

[hard forge]

That would help yes

[potent smelt]

then it might be easiest to calculate a random frequency pair using the above.

[hard forge]

Since we’re doing grayscale f(x,y) is intensity as we discussed right

[potent smelt]

yup

[hard forge]

Brightness

[potent smelt]

f(x,y) is the brightness at that pixel value.

[hard forge]

I’ve seen people say they do it row by row

And then column by column

Don’t know tho

Is that a thing?

[potent smelt]

There might be an equivalent representation. But I don't know it off of the top of my head

[hard forge]

I think this is it

Because the sums are from pixel 0 to 16 in my case

[potent smelt]

yeah, that makes sense.

though this is column by column then row by row, the order of summation can easily be switched

[hard forge]

The videos I watched said it didn’t make a difference

So thanks for this

[potent smelt]

no worries.

.close

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[remote mural]

Guys 1/cos x should be differentiated as sec x or should we differentiate like 1/-sin x or like - sin x/ (cos^2 x)

[celest stream]

use 1/cosx and u/v rule

[remote mural]

But it makes it very lengthy

[glad plover]

what are you doing

[celest stream]

its noot that lenghty tbhh

[glad plover]

1/cos(x) is not 1/(-sin(x))

[celest stream]

yea i assumed that to be a typo

[remote mural]

And I'm getting it different if I do u/v rule and sec x way

What 💀

[glad plover]

#❓how-to-get-help

#❓how-to-get-help

[glad plover]

what do you get if you do u/v

what do you get if you do sec

[remote mural]

-(tan x/ cos x)

[glad plover]

sure

[remote mural]

And only sec x tan x, no minus, if i do with sec x

[glad plover]

#❓how-to-get-help

[glad plover]

i see the problem

you should get -(-sin)/cos^2 = sin/cos^2 = sec tan

[remote mural]

Oh yeah you're right! Thanks!

[glad plover]

poste em um dos canais disponíveis como #help-35 ou #help-44｜stanley-🌲-v2-dans. este está sendo usado por outra pessoa

[glad plover]

note that since the top is 1, you don't even have to use u/v

you can just treat it as (cos(x))^-1

and do chain rule instead

which is a little faster

[remote mural]

Exactly, amazing

Appreciate the help g

[calm coral]

@remote mural Has your question been resolved?

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[calm coral]

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[vivid barn]

10

[obsidian bridge]

Don't need the answer

[vivid barn]

whats your question then

[spice mason]

x-1/x+2 < 1

[obsidian bridge]

How to get it

[unique jackal]

note that sqrt(ab) = sqrt(a)sqrt(b)

[spice mason]

answer anyonee??'

[obsidian bridge]

Bro

[spice mason]

watt

[obsidian bridge]

This channel is taken

[spice mason]

can u just help me

[obsidian bridge]

No

[spice mason]

I need the answer quick

[obsidian bridge]

Get your own channel

[spice mason]

then free me dummy

[obsidian bridge]

Stfu

[spice mason]

Pls who can answer the question

[spice mason]

Tell that to ur father

[gloomy badge]

<@&268886789983436800>

[obsidian bridge]

Nice English

[spice mason]

Yhh
Thanks

[obsidian bridge]

So sqrt 100 = sqrt a sqrt b?

[spice mason]

Can anyone pls answerr plss

[obsidian bridge]

And a is 5 b is 20

[oblique current]

@spice mason read #❓how-to-get-help , open your own channel and dont type in this one

[spice mason]

I did that no one is responding

I just want help

[glad plover]

they did

[obsidian bridge]

Bro mute this kid

[spice mason]

They didn;t

[oblique current]

well then wait patiently and quit invading other peoples channels

[glad plover]

they did

[obsidian bridge]

He's prob cheating anyway

[spice mason]

bro leave mea lone

[glad plover]

they asked you what you'd done already

[spice mason]

okk

[obsidian bridge]

Gtfo of my channel then damn it

[spice mason]

Just dumb

[obsidian bridge]

Yeah I know you are

[glad plover]

can you smite this guy already

[obsidian bridge]

Fr

Banish em

[spice mason]

Maybe ur talking to urself

[unique jackal]

sqrt(5)*sqrt(20) = sqrt(5*20) yes

[obsidian bridge]

2nd grader ass insult

[spice mason]

Okay i have made a channel can someone answer me now?\

[spice mason]

U done?

[glad plover]

<@&268886789983436800>

[obsidian bridge]

Bro <@&268886789983436800>

[oblique current]

stop interacting with eachother jfc

[gloomy badge]

@spice mason go to help-19, someone is trying to help you there

[obsidian bridge]

Maybe mute him

[glad plover]

agreed

[spice mason]

fr thanks so much

[obsidian bridge]

Ight imma send my shit again cuz of this mf

[leaden thunder]

.

find a and b

[obsidian bridge]

So the answer is 100?

Radical 100

[rich carbon]

which is

[obsidian bridge]

10

[unique jackal]

there you go!

[obsidian bridge]

Ayy

Thanks

Got some more if that's fine?

[spice mason]

The answer is 168√30

[obsidian bridge]

Need the process

[blazing coyote]

$a\sqrt{x}b\sqrt{y} = ab\sqrt{xy}$

[spice mason]

Ok

[potent lotus]

Why am. I here

[obsidian bridge]

I don't get this

[blazing coyote]

$a\sqrt{x} \cross b\sqrt{y} = ab\sqrt{xy}$

[potent lotus]

Why am. I here

[oblique current]

you first wanna simplify the roots, try finding the biggest square number that is a factor of a number under a root

[obsidian bridge]

So 4 radical 24 multiplied by 3 radical 245 is equal to 4×3 radical 24×245?

[blazing coyote]

yup

[obsidian bridge]

12 radical 5880

[blazing coyote]

just a suggestion, in the future please use latex to render your working

[obsidian bridge]

I have no idea on how to use it

[blazing coyote]

https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference

read this when you're free

not evrything, just the introductory sections

or got to #latex-help

[obsidian bridge]

Will do

But how would I find the square root of 5880

Can't use calculators

[blazing coyote]

there's an algorithm

https://www.wikihow.com/Calculate-a-Square-Root-by-Hand

see this

[limpid flame]

i think they mean in radical form

[blazing coyote]

ah

[limpid flame]

doing actual roots by hand is hardly ever a thing

[blazing coyote]

Depends on the region

[obsidian bridge]

Yeah I don't think we're supposed to do it by hand

My school is cruel but not that cruel

[limpid flame]

you can seperate 5880 into 580 x 10

[blazing coyote]

ok, express 5880 as the product of a perfect square and an imperfect square

[limpid flame]

by repeating this over and over you can simplify

[blazing coyote]

so $58 \cross 10^2$

[potent lotus]

Why am. I here

[glad plover]

no

5880, not 5800

[limpid flame]

typo mb

u get the idea

[glad plover]

yeah you just factorise and then look for squares

[blazing coyote]

,w prime factorise 5880

[potent lotus]

Results provided by WolframAlpha

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

Requested by physicsrocks

[blazing coyote]

ok, so that's $3 \cross 5 \cross 2 \cross(14)^2$

[potent lotus]

Why am. I here

[blazing coyote]

does that help

[obsidian bridge]

Yeah but I'm kinda stuck on the factorization part

Since it's such a big number

[blazing coyote]

you have to do that manually

look for prime factors and proceed

[obsidian bridge]

Think I did it

[blazing coyote]

yup

you should be able to do it from here

[oblique current]

You should have done this for each individually before multiplying the roots together

Way easier imo

[obsidian bridge]

12 radical 2 cubed×3×5×7 squared

[obsidian bridge]

Can you write it and send?

[oblique current]

Just factor 24 and 245 first

[obsidian bridge]

Wait can u help me in this rq?

The answers there but idk how

One sec imma go in 30

.close

[calm coral]

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[calm coral]

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

how do i prove that cosh >= 1 for all x

[glad plover]

do you know how to define cosh

[remote crown]

yes

[glad plover]

define cosh

[remote crown]

(exp(x) + exp(-x))/2

[glad plover]

ok

so you just have to show that this is always >= 1

[remote crown]

yes

without using derivatives

[glad plover]

yeah

i mean ok here's a hint

let y = e^x

then it's just (y + y^-1)/2 >= 1

you should be able to do this

[remote crown]

ill give it a try

so far so good?

@glad plover

[glad plover]

not the way i would have done it but it definitely works

[remote crown]

how would you have done it?

[glad plover]

multiply through by y and solve the quadratic

or just complete the square and show it's >= 1

but this works too

[remote crown]

at which step would u have done it

[civic dirge]

If
(y + 1/y)/2 = 1
Then
y² + 1 = 2y

This is a quadratic in y, and can tell us about important points

But AM-GM is slicker. Our way is just the mechanical way to do it

[remote crown]

am-gm seems to easy tho

pretty slick tbh

[remote crown]

and this is now always 1

if you solve for y

[remote crown]

would this be the way

and then just solve for y

and just imagine it is = 0

.close

[calm coral]

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[spiral ruin]

Hello, looking for some help with this question, not really with the multiplying of matrices but just checking if each one possible and why

So far I've gotten:

a) not possible, A and B are not the same size
b) not possible, A has more scalars per row than B has scalars per column
c) doable, spits out a 2 x 3 matrix
d) not possible, A and B are different sizes and AC and BC are different sizes
e) not possible, A and BC are different sizes
f) doable, CB is a 3x3 matrix
g) I think... no, because BCB would make a 2x3 matrix and there's no 2x3 identity matrix? I'm really not positive on this one
h) Seems fine, A^2 and 3A are both 3x3
i) same issue with g?

[upbeat venture]

all correct :)

[spiral ruin]

Awesome, thank you!

.close

[calm coral]

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[remote mural]

need some help

[hasty fiber]

!status

[calm coral]

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

[red moat]

what exactly do you need help with

[red moat]

Do you know natural logarithm ?

[remote mural]

am i supposed to give the x values, because the question is marked as wrong

[remote mural]

no

wait yea

[red moat]

and do you know the formula for continuous compouding ?

[hasty fiber]

this equation isn't what you want to be solving

[remote mural]

i dont

[remote mural]

oh

[red moat]

ah well i guess you can't use it then in solving this

[remote mural]

do you mind showing how youd set up the equation perhaps?

[red moat]

i'd use the formula for continuous compouding ?

[hasty fiber]

yeah ideally "continuous compounding" is something you should have already been given a formula/calculator thing to compute

[red moat]

replace with the known values

and solve for t

[remote mural]

ah alright

thanks guys

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

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[sudden nebula]

So

Are the coordinates for A <s/c , 0 >?

[calm coral]

@sudden nebula Has your question been resolved?

[calm coral]

@sudden nebula Has your question been resolved?

[calm coral]

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[undone wing]

Let x, y, z be three different real numbers.
Prove that -1/3 < x, y, z < 1.
How do I begin tackling this question?

[undone wing]

I'd like not the exact answer, but some guidelines on how to solve questions similar to this.

[calm coral]

@undone wing Has your question been resolved?

[undone wing]

<@&286206848099549185>

[calm coral]

@undone wing Has your question been resolved?

[calm coral]

@undone wing Has your question been resolved?

[calm coral]

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[fresh halo]

Sorry, this is lower end math, but I’m studying for the final and this came up?? I don’t even remember how to do it anymore

[fresh halo]

Just the top question that I tried already

[remote mural]

try the pythagorean theorem (you want #1 right)

[fresh halo]

Yes

Thanks

.close

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[elfin tusk]

yo guys I am back I gotta ask if the first diff of a sequence is there than the formula to write a rule would be a_n=a_1+(n-1)d but what about the second diff and so and so forth? Thanks in advnace 🙂

[elfin tusk]

@pure kayak

[pure kayak]

ello

[elfin tusk]

ello

quesiton is above

[pure kayak]

if theres a constant second difference then its a quadratic sequence

which has its own method

ive never gone beyond that to a constant 3rd difference

[elfin tusk]

so what would the arthimetric rule be for it then?

[pure kayak]

its not arithmetic
the nth term would look like an^2+bn+c

the second difference = 2a, get a from there
the difference between the first two terms is 3a+b, get b from there
the first term is a+b+c, get c from there

[elfin tusk]

so if the second diff lets say would be 6 then a would be 3?

[pure kayak]

yup

[elfin tusk]

how tf do u know this I have not learnt this and I was told to guess by doing n^2 or (n-2)n shit like that

and then I have a test tmr and my teach asks hard af questions

[pure kayak]

1,7,19,37
6,12,18
6 6
2a=6 a=3
3a+b=6 9+b=6 b=-3
a+b+c=1 c=1
3n^2-3n+1

n=1 1
n=2 12-6+1=7 etc

simple example

[pure kayak]

learnt it in highschool yonks ago

[elfin tusk]

lol my alg2 class is teacing me sigma noation like wtf

I gotta lock in

.close

[calm coral]

Channel closed

Closed by @elfin tusk

Use .reopen if this was a mistake.

[elfin tusk]

godspeed

.reopen

shit

[calm coral]

✅

[elfin tusk]

how would I find the sequence if the denominators are not the same ie 2/3, 2/6, 2/9, 2/12 ?

@pure kayak

lol

[pure kayak]

ello again

[elfin tusk]

ello

[pure kayak]

hm, so it looks like the sum from n=1 of (2/3n)

thats the nth term 2/(3n)

[elfin tusk]

wdym

[pure kayak]

ignore the sum, dk why i jumped to that

[elfin tusk]

lol

allg

[pure kayak]

the denominator follows an arithmetic progression

3,6,9,...

so just express that as its arithmetic general term, here 3n

[elfin tusk]

wait idu

[pure kayak]

then the sequence here has nth term 2/(3n)

[elfin tusk]

?

[pure kayak]

hmm, which parts causing confusion

[elfin tusk]

when we started

like how to find the diff

between terms

[pure kayak]

there isnt a constant diff between terms

[elfin tusk]

2nd diff?

[pure kayak]

this isnt one of those sequences

no

[elfin tusk]

oh

wait

I get it

do u mean like
2/3(n1), 1/3(n2), 2/9(n3), 1/6(n4)

[pure kayak]

its odd to write n1,n2 etc
we just say n=1,2,...

but yeah, you get the idea

2/(3n)

[elfin tusk]

I found the ans key but I idk what its sayin:

thats a clean screenshot I took right there

[pure kayak]

theyre just saying the denominator is an arithmetic progression really

[elfin tusk]

oh

[pure kayak]

you can see the rule

[elfin tusk]

ohhhhhhhhhhhhhhhhhhhhhhhhh

damn

imma leave this thread open cuz I may have a bit more questions 😉

[pure kayak]

no worries lol

[elfin tusk]

out of curiousity this may come in my test but what if the common diff is not a diff but multiplication or division?

how would I approach it then?

@pure kayak

sry for @tting u

[pure kayak]

ello, that would be called a geometric sequence

the multi is called the common ratio

[elfin tusk]

oh ya

I remeber now

thanks xd

.close

[calm coral]

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[calm coral]

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[elfin tusk]

how to make a rule for -3 + 4-5 + 6 - 7 @pure kayak

[pure kayak]

ello

[elfin tusk]

ello again

[pure kayak]

how could i express -1,1,-1,1,...

[elfin tusk]

idk u tell me lol

wait dont tell me

lemme try

shit idk

tell me now

[pure kayak]

(-1)^n 🙂

[elfin tusk]

lmao

oh'

bruh this was on my other tab

[pure kayak]

lmao

[elfin tusk]

🙈🙈🙈

wait is that the rule

[pure kayak]

half

[elfin tusk]

?

[pure kayak]

how would i express 3,4,5,...

[elfin tusk]

1, 1, 1

n + 2

😉

[pure kayak]

so the sequence is (-1)^n * (n+2)

[elfin tusk]

how am I suppose to remeber this in my test

[pure kayak]

if there are alternating signs, then there will be a (-1)^n

[elfin tusk]

would it be better if you could explain this so I could make it on the go

[pure kayak]

just slap it on

[elfin tusk]

lmao

but how do I know I am supposed to append n+2 to -1^n etc?

wrong timing mb

aight

[pure kayak]

because you have the sequence n+2... just with alternating signs lol

[elfin tusk]

https://tenor.com/view/you-didnt-have-to-cut-me-off-meme-you-didnt-have-to-cut-me-off-big-brain-big-brain-meme-gif-26965490

Oh

ic thanks

https://tenor.com/view/you-didnt-have-to-cut-me-off-meme-you-didnt-have-to-cut-me-off-big-brain-big-brain-meme-gif-26965490https://tenor.com/view/you-didnt-have-to-cut-me-off-meme-you-didnt-have-to-cut-me-off-big-brain-big-brain-meme-gif-26965490

this looks so cool

...

[pure kayak]

bunch of random equations

classic

[elfin tusk]

lmao fax

[pure kayak]

wonder where they get them

[elfin tusk]

when u look at those equations u laugh

[elfin tusk]

from me

hahahaha

[pure kayak]

https://tenor.com/view/calculation-math-hangover-allen-zach-galifianakis-gif-6219070

[elfin tusk]

I found sigma noation, eulers identity, abs

intergral

[pure kayak]

lmao, random fourier transform

[elfin tusk]

fax idk what that is

searched what it is and it looks complicated af

https://tenor.com/view/doug-maclean-nhl-hockey-stats-statistics-gif-13408391465163304912

if u see what hes writing on the board its bull LMAO

anyways break time over back to grindin

https://tenor.com/view/deez-gif-26390778

if I wanted to find the sum would I just manipluate the formaula to be 2(n(n+1)/2)

wish I knew how to use LaTeX

@pure kayak

[pure kayak]

2(n(n+1)/2 ?

[elfin tusk]

yep

same thing

right?

[pure kayak]

you mean (n(n+1))/2

[elfin tusk]

the OG formula is n(n+1)/2

which is for the rule i

[pure kayak]

ah okay i see what you did

[elfin tusk]

however if it is 2i would I just mult by 2?

[pure kayak]

yeah that works

[elfin tusk]

k thanks

would this be the same if the rule were 5n -2 hence the rule would be 5(formula)-2?

@pure kayak

[pure kayak]

nearly

[elfin tusk]

damn

[pure kayak]

it would be 5(formula) - 2n

[elfin tusk]

why

[pure kayak]

the -2 is still in the sum for all n indexes

so its there n times

-2n

[elfin tusk]

ohhhhh

ic

thanks

[pure kayak]

no worries

[elfin tusk]

I have to ask smth, how are u this active?

on discord?

[pure kayak]

just having an all nighter

,ti

[potent lotus]

The current time for aldrnari_ is 04:48 AM (GMT) on Thu, 15/02/2024.

[elfin tusk]

wtf

how old are u

[pure kayak]

average uni student

20

[elfin tusk]

u dont get paid for this tho

lmao

[pure kayak]

just a bit of fun

[elfin tusk]

I thought u were a:

https://tenor.com/view/skragzire-gif-19722425

LMAO

[pure kayak]

woah

[elfin tusk]

lol

[pure kayak]

hell no

[elfin tusk]

u sound like u go to the gym

that was random

[pure kayak]

i used to, but the gym at my uni is too expensive for what it offers

so im slacking

[elfin tusk]

lol why tf is uni gym paid

lol

[pure kayak]

(idk man)

[elfin tusk]

thats cheap of them

bro is gonna lose all his gains

😦

[pure kayak]

definitely weaker, but physically i just becomes more athletic

[elfin tusk]

gyatt

haha

I am sry for being a immature 14 y/old

[pure kayak]

all good

i was once one, long ago

[elfin tusk]

lul

good ol days for u ig

[pure kayak]

yeah im old enough to say that now

[elfin tusk]

fr

the moment I found out that i have a week off next week I was so estactic, I guess you never rlly appreciate holidays before u are in HS and above

tmr imma get bombarded with my FR test and math test and my cooking lab

thursdays suckass for me

anyways gotta get back in the grind

https://tenor.com/view/locked-in-gif-17282991172107551550

[calm coral]

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[minor gust]

.close

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[calm coral]

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[main phoenix]

Please help to find focii and directrices of (x-1)(y-2) = 2

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[past schooner]

Question 11

[calm coral]

Please don't occupy multiple help channels.

[leaden thunder]

!1c

[calm coral]

Please stick to your channel.

[potent igloo]

.close

[calm coral]

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[steel marlin]

Can someone help me with this question pls

[calm coral]

Please don't occupy multiple help channels.

[steel marlin]

I know it’s speed, distance and time

[celest stream]

using thee speed and time ava takes to travel from e to g

find the distance EG

and then use speed =distance/time for ben

[steel marlin]

Distance between G to F is 16 miles

If we assume that the distance between E to F is “d”

…the distance between E to G is (d-16) miles

Am I right

[celest stream]

mb

[steel marlin]

What do I do next

?

<@&286206848099549185>

[harsh siren]

do you have the answer by any chance

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@steel marlin Has your question been resolved?

[calm coral]

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[steel marlin]

.reopen

[calm coral]

✅

[steel marlin]

No

<@&286206848099549185>

What do I do next

[calm coral]

@steel marlin Has your question been resolved?

[calm coral]

@steel marlin Has your question been resolved?

[steel marlin]

<@&286206848099549185>

[vivid vine]

can you show your work?

[vivid vine]

start with writing equations of displacement (x(t)) for ava and ben

and also write equations for |EG| and |GF| in terms of velocities and times spent

[remote mural]

Let EG = x, FG = y. Suppose Ava's speed is v and she drove GF in time t1, and Ben's speed is u and he drove GE in time t2. Then we have:
y = vt1
x = ut2
However, we also know that Ava drove EG and Ben drove GF in the same time. So:
x/v = y/u
So, we have a system of three equations:
y = vt1
x = ut2
x/v = y/u
You know v, t1 and t2. You need to find u. I think you can proceed

@steel marlin

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@steel marlin Has your question been resolved?

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[spark adder]

How should I start with this integral

[leaden thunder]

,w int 0 to (pi/2)^(1/3) x^8 cos^6(x) dx

[potent lotus]

Results provided by WolframAlpha

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

Requested by rie.mann

[leaden thunder]

yea i guess have fun using power reduction trig identities and integrating by parts 8 times

[dense rapids]

💀

[spark adder]

😢

bruh i hoped that something ez will come out of this

like it being odd with a -a limits and being 0

anyways thanks

[calm coral]

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[fringe hamlet]

$$\int_{0}^{1} \int_{0}^{1} \tanh^{-1}\left(\frac{x}{y} + \frac{y}{x}\right) ,dx,dy$$

[potent lotus]

alee

[fringe hamlet]

i need help 😦

[wraith robin]

Maybe write it as a log and then simplify? Not sure tbh

Think that works actually

Split the log into two and then integrate

[fringe hamlet]

like this?

$$\int_{0}^{1} \int_{0}^{1} \tanh^{-1}\left(\frac{x}{y} + \frac{y}{x}\right) ,dx,dy = \int_{0}^{1} \int_{0}^{1} \tanh^{-1}\left(\frac{x^2 + y^2}{xy}\right) ,dx,dy$$

$$\int_{0}^{1} \int_{0}^{1} \frac{1}{2} \ln\left[\frac{1 + \frac{x^2 + y^2}{xy}}{1 - \frac{x^2 + y^2}{xy}}\right] ,dx,dy$$

$$\int_{0}^{1} \frac{1}{2} \int_{0}^{1} \ln\left[\frac{xy + x^2 + y^2}{xy - (x^2 + y^2)}\right] ,dx,dy$$

$$\frac{1}{2} \int_{0}^{1} \int_{0}^{1} \ln\left[\frac{x^2 + yx + y^2}{-(x^2 - yx + y^2)}\right] ,dx,dy$$

$$\frac{1}{2} \int_{0}^{1} \int_{0}^{1} {\ln(x^2 + yx + y^2) ,dx,dy - \ln(-1)\left[(x^2 - yx + y^2)\right]}$$

$$\frac{1}{2} \int_{0}^{1} \int_{0}^{1} \ln(x^2 + yx + y^2) - \ln(-1) - \ln(x^2 - yx + y^2) ,dx ,dy$$

$$\frac{1}{2} \int_{0}^{1} \int_{0}^{1} \left[\ln(x^2 + yx + y^2) - \ln(x^2 - yx + y^2) - \ln(e^{i\pi})\right] ,dx,dy$$

$$\frac{1}{2} \int_{0}^{1} \left[\int_{0}^{1} [\ln(x^2 + yx + y^2) - \ln(x^2 - yx + y^2)] ,dx - i\pi \int_{0}^{1} 1 ,dx\right] ,dy$$

$$\frac{1}{2} \int_{0}^{1} \left[\int_{0}^{1} \ln(x^2 + yx + y^2) ,dx - \int_{0}^{1} \ln(x^2 - yx + y^2) ,dx - i\pi \big[x\big]_{0}^{1}\right] ,dy$$

$$\frac{1}{2} \int_{0}^{1} \left[ \ln(1 + y + y^2) - \int_{0}^{1} \frac{2x^2 + yx}{x^2 + yx + y^2} ,dx - \ln(1 - y + y^2) + \int_{0}^{1} \frac{2x^2 - yx}{x^2 - yx + y^2} ,dx - ix \right],dy$$

[potent lotus]

alee

[fringe hamlet]

can you check this

[calm coral]

@fringe hamlet Has your question been resolved?

[wraith robin]

Not sure whether you should have combined fractions

Honestly I’m not sure tbh

[calm coral]

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[spark adder]

guys is the volume of this pi times integral of 3^2 - (2+sinx)

[spark adder]

with bounds 0 and pi/2

[calm coral]

@spark adder Has your question been resolved?

[glacial inlet]

(2+sin x)^2

[spark adder]

Right thanks

.close

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[alpine jetty]

Could some help explain to me where the power of 7 in the first line disappears to

[remote mural]

,rotate

[jolly flax]

its taken out as common

[alpine jetty]

What does that mean?

[jolly flax]

the term $(3x^2 + 4x + 2)$ was present in both term s in step 1, so it was taken out in the square bracket

[potent lotus]

Air-istotle

[alpine jetty]

I’m stupid I see what u mean but then how does the power of 6 still stay

[jolly flax]

because power 6 was taken out

power 7 is present in the first term

[alpine jetty]

Ok thank you

.close

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[unique dome]

Hello, I have a problem.
'Find a change of variables that will reduce the quadratic form ... to a sum of squares, and express the quadratic form in the terms of the new variables'

[unique dome]

$x_{1}^2+18x_1 x_2 +18x_2 x_3 - x_3^2$

[potent lotus]

CharlesWorthingtonShire

[unique dome]

https://math.stackexchange.com/questions/727030/find-a-change-in-variable-that-will-reduce-the-quadratic-form-to-a-sum-of-square
I think I am meant to do it via the method of the top comment here. But i am still confused

Please make a post in an unoccupied channel @upbeat breach

[upbeat breach]

ohhh okay my apologies

[unique dome]

I know that my matrix, A, would be

And I think this is P

<@&286206848099549185> any idea?

[tawny idol]

what’s P

[unique dome]

From the site I linked

but I am not used to using transposes in this situation so i am not sure what it is doing

[unique dome]

I can confirm that P^t A P does give a diagonal matrix but the numbers are nasty and I don't know that is actually an answer

and I am struggling to work out how to verify if it is

[calm coral]

@unique dome Has your question been resolved?

[unique dome]

.close

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[calm coral]

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[abstract basin]

i forget which formula im supposedto use

[full niche]

Work = Fs cos theta?

[abstract basin]

what does s represent

[full niche]

force * displacement

f *s

wait but the distance isnt mentioned here

wait wait

is it answer d

[full niche]

I think thats the formula u use yeah

[abstract basin]

.close

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[abstract basin]

thenks

[calm coral]

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[strong grotto]

could someone help me with this

.close

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[formal finch]

Determining the zero:

f(x) = x^2 - x

I'm confused as to how we start this equation, i tried using completing the square method but im somehow getting a different answer

[dull wagon]

simplest approach here would be to factorise followed by zero product/null factor
show yuor work for completing the square

[formal finch]

oh, we can factor it even if there's no coefficient?

i mean there's a 1 before x, so thats what we factor?

[dull wagon]

yeh

,rotate

[formal finch]

ohhhhh

i thought we can't do that

[dull wagon]

you didn't complete the square correctly

why are you choosing to add 2/4 to both sides

[formal finch]

i thought we add it to both sides?

[dull wagon]

adding the same value to both sides maintains equality
the addition of 2/4 itself is valid
but the step afterwards of converting x^2 - x + 2/4 to (x+1/2)^2 is not

because 2/4 isn't the correct value to use
hence why i was asking why you used it (how you got it)

[formal finch]

so we get x^2 - x + 2/4 then solve it normally?

[dull wagon]

no

[formal finch]

oh-

[dull wagon]

to complete the square, you want to introduce a value that actually completes the square

2/4 is not that value, how are you getting it?

[formal finch]

(b/2)^2