help-17

[hidden kelp]

All good

I really appreciate it

[dim quail]

yw

[hidden kelp]

This helps a lot

[jagged ibex]

if f(x)=x^(2)+px +1 where f(0) and f(1) have opposite signs so that f has a root in (1,0) ,then what is the value of p?

[hidden kelp]

.close

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[median harbor]

$h_1+h_2=1$

$\pi r_1^2 + 2\pi r_1h_1 + \pi r_2^2 + 2\pi r_2h_2 = 8\pi$

$\pi r_1^2h_1 + \pi r_2^2h_2 = 2\pi$

$\frac{r_1}{r_2} = \frac{h_1}{h_2}$

[twin meteor]

candies

[median harbor]

I KNOW THIS IS LIKE THE MOST CONFUSING NOTATION EVER

but i need to solve for all possible answer sets for r_1, r_2, h_1, and h_2

[boreal remnant]

wolfram alpha

[floral seal]

thats painful

but let h1, h2, r1, r2 be a b c d

[hushed pewter]

First question is why

Why do you need this?

[floral seal]

you may get a=c/(c+d) and b=d/(c+d)

then idk

[hushed pewter]

The solution is a cacophony of quadratics and cubics. Are you sure you need this?

https://xyproblem.info/

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[vivid sigil]

I have figured out how to get 2nd equation, but now I am struggling to invert it to get rho in terms of z

[vivid sigil]

any tips would be greatly appreciated

[vocal sleet]

@vivid sigil Has your question been resolved?

[vivid sigil]

<@&286206848099549185>

[regal bane]

Easier if it's:
(1 + ρ)²z = 4ρ?

Then just quadratic equation for ρ

[vivid sigil]

ah

idk why i didnt see that

thank you

i will try that

so im not seeing how this will actually yield this

[vivid sigil]

i thought this would work but for some reason it isn't giving me what i expect

the squre root will be in the numerator

[regal bane]

Must actually be the same number, but possible to algebraically rearrange

,w solve x = 4y / (1 + y)^2 for y

[twin meteor]

Results provided by WolframAlpha

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Requested by Kaynex#7520

[vivid sigil]

you see the difficulty, yes?

yeah ok they are equal

why am i not seeing the algebraic manipulation to get it into that form?

[vocal sleet]

@vivid sigil Has your question been resolved?

[indigo ledge]

,w solve x = (y/e)^y (2ky)^1/2 for y

[twin meteor]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by PrimeSparrow#4587

[indigo ledge]

,w solve x = (y/e)^y (2πy)^1/2 for y

[twin meteor]

Results provided by WolframAlpha

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Requested by PrimeSparrow#4587

[indigo ledge]

sus

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[inner lodge]

can someone help me make sin^2 theta in the middle

I thought of squarin gboth sides

but it gives 1 and 1 on each side

which dooesnt really make sense

[left crest]

ok, when we square inequalities we have to be careful

[inner lodge]

right

[tiny obsidian]

Begin by representing the inequality as $$ \mid \sin\theta \mid < 1 $$ Then proceed to squaring both sides, which would yield $$ \sin^2\theta < 1 $$ which can then be rewritten as $$ 0 \leq \sin^2\theta < 1 $$

[twin meteor]

waris

[inner lodge]

wait how do you go from the sescond line to third?

[tiny obsidian]

the argument is that the square of a real function cannot be smaller than 0; hence it is greater than or equal to 0 i.e. 0 <= f, where f is the concerned function

[inner lodge]

Okay would this also work for x^2 > 3 for example?

Or would you have to do that differentlky

I guess it woul dbe differently actually

ALright nvm

thanks for the help

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[tepid viper]

Hello could i have some help with c

i have the answers to a and b if it helps

I honestly dont even really know what it is asking me to do to be honest so im really stuck

<@&286206848099549185>

do i get another @ in another 15 minutes lol

[charred sierra]

what is eigenvalues

[tepid viper]

bro

i think i get another <@&286206848099549185> Please

[wicked sierra]

how many eigenvectors did u find @tepid viper

[tepid viper]

3

theyre parametric tho i had to do the whole let t be a real number and included a t in the vector components thing

[wicked sierra]

oh hmm

[tepid viper]

and you dont think i can just do [ a b c] = s[e1] + t [e2] + u [e3]

?

[wicked sierra]

thats what i was thinking of

[tepid viper]

yea

[wicked sierra]

but would it work for param is what im wondering

[tepid viper]

honestly it seems like itll be fine

well the parameter would just be adjusted

by the real multiple

honestly

ima just submit it like that i have a lot of work to do lmfao

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[wicked sierra]

alr lmao

[finite hatch]

hello!

can i have help finding angle abd

[median harbor]

do you know thale’s theorem?

[finite hatch]

i know

the one with right angle

[median harbor]

nice so where are you stuck

[finite hatch]

and the one with

um

double

[median harbor]

take a look at triangle ACB

you can find all 3 angles of it

[finite hatch]

this

[median harbor]

yes double angle theorem

[finite hatch]

ok

[median harbor]

^^

[median harbor]

^^^

[finite hatch]

yes

i figure that out to be 72 degrees

[median harbor]

nice so now take a look at triangle ADB

[median harbor]

nice

[median harbor]

you can also find all 3 angles

[finite hatch]

yes

[wicked sierra]

oops guys i accidentally opened it so ill close it @finite hatch just open it once again sorry abt the confusion

.close

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[median harbor]

??

whatever continue here

[finite hatch]

so for angle abd do i do 72 - 37

[median harbor]

i mean why not

💀

[finite hatch]

.reopen

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

call me a genius or whatever

[median harbor]

genius

[finite hatch]

thx

:D

ok bai

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[rigid carbon]

statistics

[rigid carbon]

hi can canyone tell me why the selected answer is wrong

oh nvm

0 included

woops

.close

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[long mortar]

if a set A={a,b,c}
how many equivalence relations can we make?

[dark kiln]

1+3+1?

[vocal sleet]

@long mortar Has your question been resolved?

[long mortar]

how

[dark kiln]

3 has three partitions, you get it like 2+1 or 1+1+1 or 3

so there's 3 ways to break a,b,c into 2+1

and 1 each for the others

and i don't know which part could be tricky to see

so i must be misunderstanding the question, i'm sorry

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@long mortar Has your question been resolved?

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[jade raptor]

If I have two sets of data. One has mean time of 484,21 and sd of 101,18. The second one has mean time of 569,14 and sd of 94,13. How do I compare their standard deviation? Do I just substract one from another?

[vast shale]

what does "compare" mean?

could you show the original question?

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[naive thistle]

Hello

[naive thistle]

Is this solution correct for this question

[merry python]

use

[naive thistle]

sin3x/sinx = 2 cos2x +1
cos3x/cosx = 2 cos2x -1
= 8 cos^2 x -4

is this correct?

[merry python]

yeah

that works too

[naive thistle]

which way is more effeciant

[merry python]

using this right

[rose mauve]

i have a doubt

could you please help me with that

[merry python]

#❓how-to-get-help

[naive thistle]

Whats your doubt my friend

[naive thistle]

@merry python yes

[rose mauve]

if 3sinx - 4sin power 3 x -k = 0 such that 0<k<1

what is the value of x

can i neglect k since its too small to affect the angle x

[naive thistle]

cotx - tanx = (cos^2 x - sin^2 x)/(sinx cosx)
= cos2x / (sinx cosx) = 2 cos2x/sin2x = 2 cot(2x)

@merry python is this correct

[rose mauve]

and then sin square x = 3/4 ---> sin x = root 3 / 2---> x = pi/3

[merry python]

please open another channel

[naive thistle]

@rose mauve are you helping me

Im confused

[merry python]

he is with his own doubt

not helping you

[naive thistle]

ok cool

[naive thistle]

can you refer to this please

[merry python]

yeah wait

[rose mauve]

after him/her please look into my problem

[naive thistle]

just open a new chat

[merry python]

yeah that is correct
just reverse the process

[naive thistle]

what do you ean by reversing the process

[merry python]

I mean start with 2cot(2x)
and go like 2cot(2x)=2cos(2x)/2sin(2x)

and so on

since you want to prove lhs=rhs not rhs=lhs

[naive thistle]

can you show me how you format it please

@merry python you still there

[merry python]

just reversed

[naive thistle]

can you show you reversing it

or is that the reverse version

[merry python]

$2\cot\left(2x\right)=\frac{2\cos\left(2x\right)}{\sin\left(2x\right)}=\frac{\cos\left(2x\right)}{\frac{\sin x\cos x}{2}}=\frac{2\cos\left(2x\right)}{\sin x\cos x}=\frac{\cos^{2}x-\sin^{2}x}{\sin x\cos x}=\cot x-\tan x$

[naive thistle]

ok cool

[merry python]

Yeah just reverse of what you did

[naive thistle]

are you able to show me how to format my previous question as well

[naive thistle]

this one

[merry python]

$2\cot\left(2x\right)=\frac{2\cos\left(2x\right)}{\sin\left(2x\right)}=\frac{2\cos\left(2x\right)}{2\sin x\cos x}=\frac{\cos\left(2x\right)}{\sin x\cos x}=\frac{\cos^{2}x-\sin^{2}x}{\sin x\cos x}=\cot x-\tan x$

[twin meteor]

B-eard

[merry python]

mb earlier one was wrong

[naive thistle]

nah bro ur good

I like the beard ngl

[merry python]

I used this

[naive thistle]

if that is you

So is this correct

or my version

[merry python]

Idk if there is any formula for sin3x/sinx so I'm not sure abt your process

You may be right

[naive thistle]

ok

so sin3x is the lhs and 3 sinx-3sin^3x is the rhs correct?

[merry python]

you straight up wrote sin3x/sinx=2cos2x+1

[merry python]

4sin^3x

[naive thistle]

ohh ok Ill see where I went wrong

sorry for the typo

but is that how it is

[merry python]

which one do you mean

Man I'm confusing myself

[naive thistle]

this

[merry python]

I did this

$\frac{\sin3x}{\sin x}=\frac{3\sin x-4\sin^{3}x}{\sin x}=3-4\sin^{2}x$

[twin meteor]

B-eard

[naive thistle]

but What Im trying to understand is how I would put this on a LS RS Chart

@merry python

[merry python]

similar for cos3x/cosx

[merry python]

That's just a part

let me frame

[naive thistle]

ok man much appreciated

whats 3-

[merry python]

$\frac{\sin3x}{\sin x}+\frac{\cos3x}{\cos x}=\frac{3\sin x-4\sin^{3}x}{\sin x}+\frac{4\cos^{3}x-3\cos x}{\cos x}$

[twin meteor]

B-eard

[merry python]

$\frac{3\sin x-4\sin^{3}x}{\sin x}+\frac{4\cos^{3}x-3\cos x}{\cos x}=3-4\sin^{2}x+4\cos^{2}x-3$

[twin meteor]

B-eard

[naive thistle]

which ones do i refer to

how many steps

[merry python]

wdym

there are 3 steps

last 2 ones are alr

[naive thistle]

ok thanks

can you attach or highlight all three steps

[merry python]

wait

$3-4\sin^{2}x+4\cos^{2}x-3=-4\left(1-\cos^{2}x\right)+4\cos^{2}x=8\cos^{2}x-4$

[twin meteor]

B-eard

[merry python]

Step 1

[naive thistle]

ok

[merry python]

2

[merry python]

3

[naive thistle]

alright bro thanks alot

would you mind doing the same for this question

[merry python]

yeah welcome

$2\cot\left(2x\right)=\frac{2\cos\left(2x\right)}{\sin\left(2x\right)}=\frac{2\cos\left(2x\right)}{2\sin x\cos x}=\frac{\cos\left(2x\right)}{\sin x\cos x}=\frac{\cos^{2}x-\sin^{2}x}{\sin x\cos x}=\cot x-\tan x$

Just this

[twin meteor]

B-eard

[naive thistle]

only one step?

[merry python]

well that's 5

in 1

[naive thistle]

so in both cases ls=rs right

it does thanks alot man

your literally amazing

would I be able to stay in contact with you for any future help

[merry python]

sure

[vocal sleet]

@naive thistle Has your question been resolved?

[naive thistle]

.close

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[vast shale]

What is point symmetry?

[vast shale]

in functions?

[vast gorge]

what

[vast shale]

what is point symmetry

[vast shale]

what's the what for

[vast gorge]

do you what's prabola?

[vast shale]

wait

this

right one

how do you define that

[vast gorge]

okay, correct do you know what's symmetry is?

[vast shale]

when something is identical?

[vast gorge]

yeah where left and right are same,equal. And the point where this left and right seperate is called point of symmetry

[vast gorge]

for example in your first example the point of symmetry is (0,0)

[vast shale]

I mean

the left one is symmetry to the y axis

so even if it were 0,1 it wouldn't matter

but the right one says "Urpsrung" which is 0,0

but what I mean is, what's "symmetrical" about the right one?

Like for the left one it's just that the right and left are the same if you turn them 180 degrees

flip them*

not rotate 180 degrees mb

[vast gorge]

this is called line of symmetry not point of symmetry
if you know what's vertex is on parabola that's the point of symmtery

[vast shale]

noted

but i think you misunderstood

me

[vast gorge]

yeah, sorry for that try pinging helpers gtg

[vast shale]

I'm just tryna understand why/how the right example is symmetrical?

[vast gorge]

oh, okay

[vast shale]

oh no worries

<@&286206848099549185>

[vocal sleet]

@vast shale Has your question been resolved?

[leaden wren]

are you familiar with odd functions ? if not then just know that when the graph of the function rotated by a 180 matches its original configuration or shape it is said to have a point of symmetry

in better sense

from the central/ origin point , the distance is same however in opposite / reflective direction

@vast shale

[leaden wren]

in this you can see that x^3 function has a point of symmetry

whereas x^2 is an even function it also has a symmetry in rotational symmetry , but that is only an identity symmetry

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[spice iris]

how do they get mEC= -5/3 without the y value of point E?

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[spice iris]

i got it nvm

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[vast shale]

.

[vocal sleet]

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[vast shale]

How can I do 480 times 0,25 the the fastes

[cerulean isle]

times .25?

480*0.25?

[vast shale]

Yes

[cerulean isle]

no

[wooden totem]

120

12

[vast shale]

Yes but how can I

Do it the fastes

[cerulean isle]

well .25 is one fourth of the number

which is one half of one half

what’s half of 480? then find half of that

[vast shale]

240

[cerulean isle]

now what’s half of that

[vast shale]

170

[cerulean isle]

no

what’s half of 24

[vast shale]

Sry

12

[cerulean isle]

it’s okay

yes

[vast shale]

So 120

[cerulean isle]

so what is half of 240

yes

[vast shale]

K

[cerulean isle]

.close

!close

[vast shale]

.close

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[drowsy ridge]

Hey! Have I solved this properly so far?

[drowsy ridge]

I’m also a bit confused on how to solve -1(x-1)(x+1), is it multiplying only (x-1) or?

[peak matrix]

doesnt matter. -1(x-1)(x+1) = (-(x-1))(x+1) = (1-x)(x+1) = (x-1)(-x-1) = (x-1)(-(x+1)) etc

[drowsy ridge]

alright. Have I solved it properly though?

[peak matrix]

,rccw

[peak matrix]

so far it seems alright

instead of -1(x-1)(x+1) you can just write -(x-1)(x+1)

[drowsy ridge]

that’s just the same as (x+1)(x-1), right?

[peak matrix]

-(x-1)(x+1) isnt same as (x+1)(x-1), but 1(x+1)(x-1) is same as that

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@drowsy ridge Has your question been resolved?

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[tawdry haven]

What's a topology in math?

[sudden compass]

Branch of mathematics afaik

[tawdry haven]

What comes in it?

[sudden compass]

Related to 3d polygons I think

[tawdry haven]

Okk

Thanks

[sudden compass]

Yeah properties of geometric objects under various condition

[tawdry haven]

Ok

.close

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[strange pier]

out of these graphs

why does only A don't have an inverse function

shouldnt B not have an inverse function as well? how does B even have an inverse function

[karmic imp]

Because its inverse isn't a function

[karmic imp]

The inverse of B is a function

[solemn oyster]

A is x to multiple y

[strange pier]

ah i see

so in this sense B isnt a function?

[karmic imp]

Yes, in its current form, B is not a function

But the inverse will be

[strange pier]

alr thx for the help

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[vast shale]

How to solve y^2<=1

[vast shale]

<@&286206848099549185>

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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.

[pallid zenith]

whatd you try

[vast shale]

Ok

Wait

[timber orchid]

This step is incorrect

[vast shale]

Please correct me

[timber orchid]

y^2 <= 1

means -1 <= y <= 1

y can't be too big in magnitude, in other words

[vast shale]

I don't understand 😭 plz help

Why can't we take union of both cases

[hushed pewter]

There are a few things wrong with it. First, your wrote $y^2\le1$ followed by $y^2\le\pm\sqrt{1}$. I think you meant for that $y^2$ to just be $y$.

[twin meteor]

SWR

[hushed pewter]

Second, if you did mean $y\le\pm\sqrt{1}$, that's just $y\le1$ and $y\le-1$

[twin meteor]

SWR

[hushed pewter]

But consider $y\le-1$, we could choose $y=-2$ since it is less than $-1$. But $(-2)^2=4$ and that is NOT less than or equal to $1$. That is, it violates $y^2\le1$

[twin meteor]

SWR

[vast shale]

So then how can I solve this from starting

[hushed pewter]

The solution to $y^2\le a$ is $-\sqrt{a}\le y \le +\sqrt{a}$

[twin meteor]

SWR

[timber orchid]

For some a >= 0

[hushed pewter]

yes

[vast shale]

Ok man/woman thank you for helping

[hushed pewter]

Conversely, the solution to $y^2\ge a$ is the union of $y\le-\sqrt{a}$ and $y\ge+\sqrt{a}$

[twin meteor]

SWR

[vast shale]

Ok

Thnx for help

.close

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[vocal sleet]

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[worldly marsh]

There is a pair (a, b) of coprime positive integers (a, b > 1). In one operation you can change (a, b) to (a-b, 2b) if a >= b or (2a, b-a) if b >= a. Show that after a finite number of operations, you can get a pair (x, y) where min(x, y) < min(a, b)

I don't even know where to start on this problem...

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[gentle totem]

$\int_0^n \lfloor x^2 \rfloor x dx = ?$

[twin meteor]

Oğuzhan

[gentle totem]

What should be the first step to solve this?

[hard atlas]

probably try to separate the interval into pieces over which the floor is constant

[plain minnow]

Wouldn't an integration by parts work?

[gentle totem]

I tried seperating like this:
$\sum_{i=0}^{n-1} \int_i^{i+1} \lfloor x^2 \rfloor x dx$

[twin meteor]

Oğuzhan

[gentle totem]

But it's not always constant

For example integral from 2 to 3's square's floor will be 4 to 9

[gentle totem]

How so?

[plain minnow]

Consider $\frac{d}{dx} \floor{x^{2}} = 0$

[twin meteor]

Mikkel

[gentle totem]

Yes

Oh

[gentle totem]

So this becomes $\lfloor x^2 \rfloor \frac{x^2}{2}$ right?

[twin meteor]

Oğuzhan

[gentle totem]

From the integration by parts

[plain minnow]

Without evaluating the bounds, yes

[gentle totem]

Obviously x replaced with n

Alright this helped, thanks!

.close

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[plain minnow]

[gentle totem]

.reopen

[vocal sleet]

✅

[gentle totem]

I just re-thought about it, what about the x positions like 1 or 2?

Are the derivatives defined at those values?

[gentle totem]

I checked the answer and it was wrong

[plain minnow]

[gentle totem]

I checked the graph*

It's 3 at x = 2

From 0 to 2

[gentle totem]

I guess the issue is this? It probably doesn't have derivatives at those points

[plain minnow]

The left derivatives and right derivatives at those points are both 0, so I guess it can be concluded the derivative exists and is 0

[plain minnow]

What did the answer say?

[gentle totem]

So what's the problem

There is no reference

I made the question

F(1) = 0
F(2) = 3
F(3) = 18
F(4) = 60

[gentle totem]

This doesn't work for any of the inputs

And this is the graph

Of the integral*

Even wolfram alpha seems to be clueless

Probably because it has floor

<@&286206848099549185>

[gentle totem]

The original question ^

[plain minnow]

I don't know why I said the derivative of the floor function exists everywhere, when it is obviously discontinuous at the integers

[gentle totem]

Oh so that's why it didn't work

Are there methods to solve floor integrals like this?

Or is this just a non-elementary integral

[plain minnow]

Maybe there is a closed form for it, but it's probably hard to find

The function itself seems to slant itself in a line over a lot of intervals. And those intervals seem to atleast have a pattern in them

From 0 to 1 the function is split into 1 piece, from 1 to 2 it is split into 3 pieces, from 2 to 3 it is split into 5 pieces and so forth

[gentle totem]

Hmm

[plain minnow]

So it seems the amount of pieces the function is split into from n - 1 to n is the n'th odd number

[gentle totem]

Hmm

Are you sure

I think it should be the perfect-squares between those values?

[plain minnow]

Well, yeah, it can also be said to be the differences between the square of n and n - 1. But this is just n² - (n - 1)² = n² - (n² - 2n + 1) = 2n - 1

[gentle totem]

Yeah

[plain minnow]

Maybe if it's possible to write up the pattern that determines the horizontal sizes of these pieces depending on which interval you look at, and then use that to find the area under

It's probably gonna look ugly, but maybe it'll work

[gentle totem]

I'm gonna sleep in a few minutes, should I leave this channel as it is right now; or should I close it?

[gentle totem]

Oh

[plain minnow]

It's gonna close automatically anyways

[gentle totem]

Oh, okay then

[vocal sleet]

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[crystal vessel]

how to go from line 1 to 2

[flat whale]

draw a picture

[crystal vessel]

ok i got it
wanted to know if there was some other way of breaking down the integral that would make sense within the context of the question

[flat whale]

what do you mean "other way"

what way do you know

[crystal vessel]

the graphical representation:
instead of the whole area between 1 to infinity we summed up the small bits bounded between n and n+1

right?

[flat whale]

yes

that's the best way to understand the equation

[crystal vessel]

👍

thanks

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[opal flare]

Hey guys, trying to prove that $\langle S\rangle = \langle gcd(S)\rangle$
Trying to prove each set subset the other but im getting confused.
Any ideas?

[twin meteor]

meitar5674

[opal flare]

Using this definition

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@opal flare Has your question been resolved?

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[balmy pike]

basic integral question i keep getting wrong

[ornate ember]

hi do you want to show your work? We can prob tell you where it went wrong

[balmy pike]

ok

thats my working out

[ornate ember]

,rotate

[ornate ember]

ah okay yeah you were on the right track but your distribution is a bit off.

[balmy pike]

did i go wrong when expanding bracket

[ornate ember]

sorry one moment

[balmy pike]

ok

[ornate ember]

$\left(t + \frac{2}{\sqrt{t}}\right)^2$
$\\ = \left(t + \frac{2}{\sqrt{t}}\right)\left(t + \frac{2}{\sqrt{t}}\right)$
$\\ = t^2 + t\frac{2}{\sqrt{t}} + t\frac{2}{\sqrt{t}} + \left(\frac{2}{\sqrt{t}}\right)^2$
$\\ = t^2 + \frac{4t}{\sqrt{t}} + \frac{4}{t}$
$\\ = t^2 + 4t^\frac{1}{2} + 4t^{-1}$

[twin meteor]

MellowDramaLlama

[ornate ember]

so the only thing you had wrong was the middle part

it's 4t^1/2

also remember that $\int x^{-1} = ln(|x|) + C$

[twin meteor]

MellowDramaLlama

[balmy pike]

so it will be 4log + C

[ornate ember]

specifically ln

but yeah

4ln(|t|) + C

for that part of the integral

[balmy pike]

for this part how did u break it down

to 4/t

did u seperatly expand that bracket

[ornate ember]

$\left(\frac{a}{b}\right)^c = \frac{a^c}{b^c}$

[twin meteor]

MellowDramaLlama

[ornate ember]

I was just wrapping it in parentheses to make it easier

or more clear what was going on

[balmy pike]

ok i see i see

thanks for helping me with that

[ornate ember]

yep np!

[balmy pike]

alright have a good day

[ornate ember]

you as well

[vocal sleet]

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[amber hound]

dy = (22 m/s) (sin63°)/ 2 (9.8 m/s)

after I calculated the sin theta of 63°, I got 0.89100652419

should I round it off to 0.891?

[worthy citrus]

When calculating dy? No

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[fringe pawn]

I don't know the basics. Tried finding but couldn't make a logic.

<@&286206848099549185>

[median harbor]

do you know what the angle of the two opposite-corner things’ intersection is?

[fringe pawn]

90?

[median harbor]

yes

now, just use < sum of triangle and alt angles to find x

[fringe pawn]

Will x be 50?

<@&286206848099549185> <@&286206848099549185> <@&286206848099549185> <@&286206848099549185> <@&286206848099549185> <@&286206848099549185>

thats a lot of waiting

[floral seal]

hello

[floral seal]

thats right

[fringe pawn]

Okay

.close

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[supple fractal]

How can I solve this problem
What are the prime numbers a and b such a>b and an a²+b²+ab is a complete square

[vocal sleet]

@supple fractal Has your question been resolved?

[supple fractal]

<@&286206848099549185>

[vast shale]

!status

[vocal sleet]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[supple fractal]

2

[vast shale]

!show

[vocal sleet]

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

[supple fractal]

It's written in Arabic

[vast shale]

then give a rough translation of the ideas you have and what you are stuck on

[supple fractal]

I have that for b=2 there is no solution over the natural numbers. I'm stuck on getting a solution for b being odd

[vast shale]

try completing the square

[supple fractal]

I did

[vast shale]

what did you get?

[supple fractal]

Nothing

[vast shale]

you just said you have completed the square

[supple fractal]

Actually since a²+b²+ab is odd can't I write it as (n+1)²

[supple fractal]

I meant I got nowhere from there

[vast shale]

it should help

once you complete the square, you can compare prime factors

[supple fractal]

a²+b²+ab=(a+b)²-ab

[vast shale]

suppose for a contradiction that it is equal to a square n^2

then rewrite as (a+b)^2 - n^2 = ab

you can factor and you should be able to find a contradiction

[supple fractal]

I factored it what do I do now

Never mind

[vast shale]

what are the possible factors of ab

check all of them

[supple fractal]

Thanks

[vast shale]

np

[supple fractal]

.close

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[finite hatch]

hi can someone explain this to me

[finite hatch]

so far i have a is equal to 180 - 75

which is 105 degrees

how do i find b and c?

[timid shell]

Maybe using the fact that 2 lines a parallel here

[little token]

the total is 360 degress is it not or is that a different square

[finite hatch]

idk

[timid shell]

Parallel lines.......

[finite hatch]

hi

[little token]

yes and there is also 2 paralel lines

[floral seal]

i remember for the polygon like this inscribed in th circle

[finite hatch]

what does inscribed mean people keep saying it to me

[floral seal]

the opposite angles add up will eb 180

[finite hatch]

yer

[timid shell]

Inside a circle where polygon's vertices touches the circle

[finite hatch]

ohh

[floral seal]

if you understand it then basically done

[finite hatch]

with that i got angle a

but idk how to get the othjers

[proven needle]

yes

[floral seal]

you know internal angles

[finite hatch]

cointerior?

[floral seal]

Of

parallel lines

[finite hatch]

ok we're skipping that question

now we're onto this

i have a question

[floral seal]

what the - that question is simple bro

[finite hatch]

oh.... nvm...

ur right

ok i might be back in 3 minutes

[floral seal]

why would you skip that

[timid shell]

Maybe he realised that was a dumb question

[floral seal]

well

[finite hatch]

is this whole angle 62 degrees

[floral seal]

no

opposite angle

[little token]

i learnt co interiors in year 7

[finite hatch]

lol im in year 10 pre calculus

[little token]

lol

[finite hatch]

i am going to fail

[finite hatch]

omg.............................................................................

thx

[floral seal]

dont worry

i dont study in uk but

[finite hatch]

im australian

[little token]

same

[floral seal]

Anyway

[finite hatch]

omg hi

[little token]

hii

[finite hatch]

did u study maths methods/extension 1 or 2?

[finite hatch]

yes

[floral seal]

How old are you?

[timid shell]

Btw he is a guy

[finite hatch]

15

[floral seal]

great

[finite hatch]

thx

[floral seal]

im 16

[finite hatch]

oh

[floral seal]

good luck for calculus

[finite hatch]

thx

idk if i want to pick it

loks too hard

[floral seal]

Australia also have math extension

[finite hatch]

where r u from

[floral seal]

hong kong

[finite hatch]

oh.... do u find my maths easy

[floral seal]

yes

[finite hatch]

thx a lot

[floral seal]

i study m2

[finite hatch]

im in the highest maths class in my school too 😭

what is m2

[floral seal]

Math extension module 2

[finite hatch]

oh

do uwanna see one of my tests i aced

can u tell me if its easy for hong kong

[floral seal]

Yes

ok

honestly

[finite hatch]

hi

[floral seal]

its not hard

[little token]

im in year 8 and doomed for the future........

[floral seal]

how long time for you to do this

[finite hatch]

for the firs tone

they give us 20 minutes

for the second one we get 25 minutes

[floral seal]

bruh

[finite hatch]

what

[floral seal]

so long

[finite hatch]

are u serious i was fighting for my life

[floral seal]

we usually only have about 15 mins for this kind of quiz

The second one may only have 10 mins tho

[finite hatch]

oh......

year 8 was calms

dont worry its not hard just word problems

[finite hatch]

thx im never going to a hong kong school

[floral seal]

Good choice

algebra question

others are all stats and pretty easy

[finite hatch]

omg...................................................

[little token]

fr im in advanced and im doing factorising

[finite hatch]

i love factorising :D

[little token]

i got 1 day to learn it then it was the test i struggled but understood it 5 minutes before the test lmfao its fun for me now

btw i got all the factorising questions correct

[finite hatch]

year 9 was kinda hard

[floral seal]

oh god you think factorising equations are fun

[finite hatch]

yeah

[little token]

hell ye im pretty sure im doing easy shit atm though

[finite hatch]

the hardest factorising thing ive done is like

completing the square idk

[floral seal]

huh

[little token]

whats the square

[finite hatch]

this stuff

[little token]

o i had that in my test but it was just perimetere

easy peasy but i read area

so i did the area

and got the easiest question in the test wrong

[little token]

i feel dumb in this server

[floral seal]

Everyone does

except those pros

[vocal sleet]

@finite hatch Has your question been resolved?

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[tawdry wedge]

How do you solve this pair of equations : (x/a) + (y/b) = a + b and (x/a^2) + (y/a^2) = 2

[lime gorge]

Solve?

You have 4 unknowns and 2 equations so

Hmm well

[tawdry wedge]

Thats literally what the worksheet asks me to do

The previous questions in the same exercise contain only 2 variables. Only this one looks strange

Wut?

oh well, thats too aggressive

[lime gorge]

Thats definitely not appropriate

<@&268886789983436800>

[tawdry wedge]

That is the craziest thing i've ever heard

Anyways, is there any ways to solve the above equations

[sterile forum]

We can probably assume that a and b are constants and y and y are the unknowns we want to solve for.

[tawdry wedge]

Probably yes, but the worksheet gives no instuctions

But if that is the case, then i can solve it by myself

[sterile forum]

It's what would make most sense to me. As Stephen implies, when you have more equations than you have variables, you can't really hope to get concrete values out of solving, and the best you can achieve in general is to express some of the variables as functions of the rest.

[tawdry wedge]

Well then i have to assume those as constants. Thank you for your help

.close

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[proud summit]

In the paper SHARPER CHANGES IN TOPOLOGIES (https://www.ams.org/journals/proc/1999-127-01/S0002-9939-99-04498-6/S0002-9939-99-04498-6.pdf), why does this underlined equality hold?

[proud summit]

the relevant definitions:

Polish: separable complete metric space

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[atomic hamlet]

How to proof that this diverges to inf?

[soft walrus]

is this a series? $\sum_{n=a}^{\infty}(-1)^n\cdot n+38n, a\in\mathbb{Z}$?

[twin meteor]

XxMrFancyu2xX

[atomic hamlet]

It’s a sequence

[civic pendant]

38 n + (-1)^n *n >= 37 n 🙂

[peak matrix]

that's because (-1)^n in the worst (smallest) case is -1. So it would be 38n + (-1) * n which is just 37n

[median harbor]

isnt that just alternating between 37n and 39n

[peak matrix]

||yes||

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[south token]

F(x) is wider

[vocal sleet]

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✅

[south token]

4 left 5 up i think

1a 2c 3b

[south token]

I think it’s step 2

[vocal sleet]

@vast shale Has your question been resolved?

[south token]

I’m not quite sure what the question is asking

[meager escarp]

What you can do is substract the second equality from the first, find x by substracting the second equality from the first and find y by inserting x.

So it's (-4; 2).

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✅

[regal slate]

test?

that looks like a test

set the equations equal to each other

from there it’s just pure algebra

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@vast shale Has your question been resolved?

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@vast shale Has your question been resolved?

[steep ocean]

Why does it appear you're just asking for the answers to the questions?

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[upbeat chasm]

Would this be a valid argument? I feel like my teacher over complicated it since this is much shorter, cant even remember her argument lol

[silver plinth]

@everyone you all love bbc

[upbeat chasm]

[vocal sleet]

@upbeat chasm Has your question been resolved?

[tall reef]

can you say what have writte after "p,q share no common factors"?

@upbeat chasm

[upbeat chasm]

oh yeah sorry

p^2 has an even number of prime factors, but 7q^2 has an odd amount of factors of 7, hence a contradiction so root7 is irrational

[tall reef]

I'm afraid that is not going to work here

You have to show the details - first is to show that 7 is a factor of p, then using that you have to derive that 7 is also a factor of q- which give sthe contradiction

[civic pendant]

But this is a different argument

What's wrong with the argument?

[upbeat chasm]

7 isnt necessarily a factor of p of q though

[civic pendant]

You can see the classical proof of sqrt{2} here https://en.wikipedia.org/wiki/Square_root_of_2

You could repeat it with sqrt{7}

[tall reef]

p/q=√7
=>p²/7=q²
and the rule is if a prime number k divides p², then it also divides p, and vice versa is also true. This implies that since 7 is a factor of p2 then it will also be the factor of p.

[tall reef]

how can you tell psquare has an even numbe rof prime factors? and q has odd amoount of prime factors?

[civic pendant]

It's not the argument

the argument is that for every number z, z^2 has even number of prime factors

Obviously

[upbeat chasm]

yeah thats basically the argument my teacher gave

[civic pendant]

So the left side of the equation has odd number of prime factors, and the right side has even number of prime factors

[upbeat chasm]

oh mine is here

[civic pendant]

@upbeat chasm I don't see a flaw in your argument, it's just using more advanced ideas

About prime factorization

Yea, cool

[tall reef]

ohhhhhhhhhhhhhh

[civic pendant]

I liked your argument

[tall reef]

got it

[upbeat chasm]

oh alright great, she just seemed confused when i presented it to her lol

[civic pendant]

New to me

[upbeat chasm]

:) awesome

thanks so much !

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[torn tapir]

hey

[torn tapir]

can someone help check if this is right

[worn bobcat]

checked, you did it right

[edgy matrix]

yes

[empty frigate]

yeah that looks good

[torn tapir]

thank youu

thats all i needed

.close

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[strange pier]

hi just curious what would your method of doing this would be

i could do it with cosine rule but im just wondering is there an alternate more elegant way

cuz with cosine rule, my answer leaves me with cos(pi - 2 ø) in one of the parts

[midnight rapids]

Not sure if this is better but maybe draw a perpendicular from O to PR

Let it meet PR at S

a cos(phi) = y/2

[strange pier]

yea which would get a cos (ø)

yep definitely better. cuz with cosine rule its pretty messy and im not sure if the mark scheme even accepts

[midnight rapids]

Angle PRQ is 2phi

[strange pier]

yep i know the rest thx

[midnight rapids]

So area of unshaded region is phi y^2

Great

[strange pier]

was just wondering could the y part be improved

thanks for your time!

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[humble carbon]

sat problem

[humble carbon]

how to do i turn 4 sqrt 9^3 into 3 sqrt 3

[lime gorge]

$9^{\frac 34} = (9^{\frac 14})^3$

[twin meteor]

Stephen

[lime gorge]

From there

U go

[humble carbon]

literally dont know what happened

i just know the radical/sqrt rules

[lime gorge]

This is not even true btw, ur trying to get from $9^{\frac 34} \rightarrow \sqrt [3] {3}$?

[twin meteor]

Stephen

[lime gorge]

O wait is the 3 not an index?

[humble carbon]

correct answer for this problem is

3 sqrt 3

Wait

ITS NOT AN INDEX

Oops

[lime gorge]

lol

[humble carbon]

So how do i do it

how do i get the answer from 4 sqrt 9^3

[lime gorge]

$9^{\frac 34} = ((9^{\frac 12})^{\frac 12})^3$

[twin meteor]

Stephen

[lime gorge]

Can u evaluate the inner term

9^(1/2)

[humble carbon]

yes its sqrt 9 right

[lime gorge]

Yes

[humble carbon]

ok i still dont get it

[lime gorge]

What don’t u get

sat math requires that you have knowledge of exponent rules

[humble carbon]

nvm i fet it

.close

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[humble carbon]

ty

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[supple fractal]

How can I prove the following inequality
a+b+c=1
Prove that
a²/(a+b) +b²/(b+c) +c²/(a+c)>=1/2

[vocal sleet]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[supple fractal]

1

[vast salmon]

Have you tried taking the lcm

[supple fractal]

The what?

[sterile geode]

this one please

this one please

[vast salmon]

Make the LHS denominator same

!help

[vocal sleet]

Please read #❓how-to-get-help

[supple fractal]

Yes

[vast salmon]

@sterile geode

[vast salmon]

What did you get

[supple fractal]

Forgive my ignorance but I really want to know the proof not how to get it

[sterile geode]

i did

[supple fractal]

<@&286206848099549185>

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[fresh crane]

i’m struggling with this question

i hope someone can explain the steps and why they are done

[flat whale]

where do you get lost first

[fresh crane]

okay let me give a little context

i tried to do the question

and i had the problem of not knowing the limits@

so i looked at the solution

i don’t understand why there is a +c in the 2 equations at the start

once x=4 is found

i see that 4 and 12 are the limits

and the difference in x between the 2 points of the curve used are 8

but i don’t understand why 4 is the first limit

[floral seal]

i guess you have to slove x when y=11

as the height is 11m

[fresh crane]

i don’t think so

the solution creates 2 simultaneous equations

for the 2 different heights

to get rid of the constant

and i don’t understand why the constant is there

because like you said, i would probably just solve for x

[floral seal]

the y axis of the 2 points are 11 and 5

which means you have to find 2 points on the graph of the equation taht

y axis of 2 point have a difference of 6

X axis of 2 points have a difference of 8

through the equation you know what the graph look like

than just let the first point be (x,y) and the second one be (x+8,y-6)

something like that

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[deft arch]

can we use row and column transformation simultanously in determinants?

[deft arch]

hloooooooooooo

[sly sierra]

can you show an example of what you mean?

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[astral apex]

n