help-39

[little merlin]

a is any real number

which in our case happense to be |h(x)|-1

[royal galleon]

sorry I didn't see the a) part I thought that theorem 2.2 was if |ab| = |a||b|

[royal galleon]

but we had | |h(x) | - |1| |

[little merlin]

oh xd

[royal galleon]

because then it looks like the reverse triangle inequality is useless. Just go straight to c. for c we can set a to anything?

[little merlin]

good point

but

when you are proving |1/h(x)-1|< epsilon, the simplification leads us to an expression that looks like this:
(|h(x)-1|/|h(x)|)< epsilon

the denominator has |h(x)|

oh wow

these inequalities are really hard to read

i should have typed all of this in latex

[royal galleon]

this I follow

[pearl pond]

@royal galleon Has your question been resolved?

[pearl pond]

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[crimson haven]

Can someone help me

The answers suppose to be 8

[mortal flower]

you canceled out an extra factor of x^10

you said sqrt(2x^20)/sqrt(2x^10) was 1

[glacial charm]

instead of splitting x^20 in the denominator it will just cancel out with the numerator term which you factored out

[crimson haven]

For which part pls

[mortal flower]

the first step

or 2nd step sorry

[crimson haven]

Ohhh your right

Tysm

[glacial charm]

you can split the 18 in the denominator as 9*2 but keep it x^20 as it is it will cancel

also the 2

[crimson haven]

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[pearl pond]

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[cunning compass]

can someone explain in the simplest way possible why the y/x coordinates is sqrt2/2?

[edgy wren]

are you confident with solving quadratics and the pythogrean theorem?

[pearl pond]

@cunning compass Has your question been resolved?

[cunning compass]

yea

[edgy wren]

first off, we know that the triangle constructed in the picture is a right triangle correct ?

[cunning compass]

yes

hypo is 1

[edgy wren]

what about the other 2 sides?

[edgy wren]

hint: we know 2 of the interior angles of the triangle.
what is the value of the third angle?

[cunning compass]

oh

would the opposite just be 1sin(45)

[edgy wren]

yeah

[cunning compass]

then howd we end up with sqrt2/2

or does that just mean sqrt2/2

[edgy wren]

we'll get there

[edgy wren]

try answering this

[cunning compass]

uhhh

idk 💔

oh

180 - the value of the 2 interior angles

[edgy wren]

yes, and we know 2 angles already

[cunning compass]

wait huh

is the other angle 45 aswell?

[edgy wren]

yes

[cunning compass]

oh

[edgy wren]

now here comes the fun part,
there is a property that links equal angles in the triangle with its sides

[cunning compass]

yeah

[edgy wren]

if any two (or three) angles in a triangle are equal then the sides of opposite of those angles are equal

[cunning compass]

so both the oppo and ad are congruent?

[edgy wren]

yep we can denote them with any variable of our liking so we can go to the last step in the process

[cunning compass]

yes

[edgy wren]

and know we can use pythogrean theorem

$z^2 + z^2 = 1^2$

[jolly parrot]

qimmah

[edgy wren]

can you go on and solve this?

[cunning compass]

yea

+/- sqrt2/2

ohhhhh

[edgy wren]

only the positive since we are dealing with lengths and lengths can't be negative but yeah
since P(x,y) = (cos(t),sin(t)) being the angles
and we know the x , y being both sqrt(2)/2 then we basically concluded (derived) that sin(45)=cos(45)=sqrt(2)/2

[cunning compass]

alr

thanks g

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[pearl pond]

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✅ Original question: #help-39 message

[eager jewel]

i did d=lambda(x+2,6,-2)

then i did dot product a both sides

and i got the relation lambda(x+6) = 1

and i also calculated (axb).c = 20-9x

so i only need x value

how do i proceed

[uneven cypress]

You know lambda in term of x @eager jewel

[eager jewel]

yeah

now what to do

[lean star]

use a.d = 1 to find x

[eager jewel]

huh but didnt i just use that to get the relation

[eager jewel]

^

[lean star]

d is a unit vector

[eager jewel]

oh huh

alr ye i can use that

ty

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[tiny wadi]

how do they know a and p are orthogonal?

[slender stream]

they dont really do anything that need a and p to be orthogonal

[tiny wadi]

but like

they're writing in column form

a and p

dont u do it when they're like orthogonal

[slender stream]

they are writing AP and PB vector in column form and taking their dot product

[tiny wadi]

like if i write (4 )
( 2)

[slender stream]

oh yeah i get your point

[tiny wadi]

doesn't that mean go 4 units in one direction, then 2 units in the direction orthogonal to it

[slender stream]

oh yeah you are right, sorry cant think of any reason

[pearl pond]

@tiny wadi Has your question been resolved?

[pearl pond]

@tiny wadi Has your question been resolved?

[fierce sky]

@tiny wadi think of it like $$\left(\mathbf{-a} + \mathbf{p}\right) \cdot \left(-\mathbf{p} -\mathbf{a}\right)$$
then using the distributive property of dot products

[jolly parrot]

Roy

[fierce sky]

what they did with putting vectors as components of vectors I'd have to assume it's really sketchy

[tiny wadi]

yea

sketchy ass proof

[tiny wadi]

that's my whole issue with the solution

[dense jasper]

Define a new basis bla bla bla

[tiny wadi]

:/

.close

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[keen dock]

i am having trouble calculating the probability that the network is operational at any given time

ignore the annotations, that was me

[tulip ore]

are you open to brute-forcing?

[keen dock]

anything :')

[tulip ore]

lets go through each way that the servers are operational

lets say server B operates correctly

then whats the chance that server C operates correctly?

[keen dock]

lemme try

[tulip ore]

as a hint,

its in the question

and you even annotated it

but its not 70%

[keen dock]

30% then

oh yea

complement

right

but i thought that could be P(NOT B| NOT C)

hmm

is it because P(B) and P(C) failing would be
P(B intersection C)

tag me when you reply :)

[tulip ore]

@keen dock thats not exactly how you read this scenario

I said P(C | B), not P(!C | !B)

if server B operates correctly, whats the chance that server C operates correctly

[tulip ore]

thats not correct either

[bronze heath]

Seems correct

[tulip ore]

you need to learn bayes theorem

its instead true that $P(C\mid B)=1-P(\overline C\mid B)$

[jolly parrot]

mtt

[tulip ore]

that complement thing doesnt work for the "not-B" case

for example,

it tells you that P(C | not B) = .7

so we know P(not C | not B) = .3

[tulip ore]

this reads as "the chance of C, given that B isn't, is .7"

which matches the 70% annotation

"when server B fails, the probability that server C remains operational drops to 70%"

however, if server B doesnt fail,

[tulip ore]

that means this no longer applies

whats the probability that server C usually remains operational? and again, you annotated the number

[tulip ore]

theres only one other number besides 70% thats related to server C here, try and find it

[keen dock]

0.92 for C

[tulip ore]

yep

and to say that, we needed to know that server B didnt fail

[keen dock]

oww

[tulip ore]

(because if it did, it couldve been 0.7)

[keen dock]

soo

[tulip ore]

so P(C | B) = 0.92

[keen dock]

woah

i seeeeeeeee

i thought it was like this but i was completely wrong, however, P(A) is just 0.96 right

[tulip ore]

P(A) is straight 0.96, yes

[keen dock]

alright thank you thank you, you just opened my eyes :O

[tulip ore]

now as a reminder, again you cant say this

[keen dock]

yep!

[tulip ore]

to find the chance of C, youd need to know what B is

so rewrite that to have the necessary bit before it

P(C | B) = 0.92

,,P(\overline C\mid B)=0.08

[jolly parrot]

mtt

[tulip ore]

you also arent done yet

we next gotta figure out P(A and B and C) properly

how would you have gone about finding P(A and B and C)?

[keen dock]

hmm

a min

but we dont have P(B| !C) right? in order to P(AandBandC)

[tulip ore]

changing B to B | C doesnt really make sense

you know what server C is depending on server B
its not the other way around

[keen dock]

noted

[tulip ore]

go post another image with the facts again but with the | C thing fixed

[keen dock]

P( !B | C ) doesnt make sense right?

[tulip ore]

you forgot to erase the | C

I dont know why you were wondering to keep it around or not

given we had to change B | C to B to get that correct

[keen dock]

now we just left to do P(A and B and C) right?
is the P (!C | !B) ok?

[tulip ore]

yea thats correct

alr now we use this to try and figure out P(A and B and C)

[keen dock]

alr lemme try

btw

it cant be P(A and B and C) right?

[keen dock]

cuz at least 2 right

not all 3

right?

@tulip ore

[tulip ore]

I was just choosing an easy example but yea you have a point

that would mean something like
P(A and B or A and C or B and C)

looking at this I dont think its gonna be healthy to do all of that at once

[keen dock]

so like

[tulip ore]

itll be included in our calculations ultimately, go try this one first

[keen dock]

alr

[tulip ore]

we can reorganize the 2 out of 3 probability as: P((B and (A or C)) or ((not B) and A and C))

the P(B and (A or C)) is usually done as P(B and A or B and C) = P(B and A) + P(B and C) - P(B and A and C)

so we do need the P(A and B and C) probability as essentially the 'hardest' step we'll need to do for this portion of the calculation

[keen dock]

hmm

[tulip ore]

as a reminder, A is independent from B and from C

maybe that can make the A part of the probability easier to do

[keen dock]

hmm

ts hard

(0.96*(0.940.92)) + (0.96(0.940.08)) + (0.96(0.060.70)) + (0.04(0.94*0.92))

[tulip ore]

@keen dock ok Im back

oh wow what a coincidence that you happen to post it seconds ago

[keen dock]

(0.96 * (0.94 * 0.92)) + (0.96 * (0.94 * 0.08)) + (0.96 * (0.06 * 0.70)) + (0.04 * (0.94 * 0.92))

[keen dock]

hehe yea

[tulip ore]

thats also a lot cleaner then how I was going to present this

thats also correct

[keen dock]

i like this one better :')

[tulip ore]

yea Idk why I figured itd be better to do it the other way

you are implicitly using a fact here

youre having P(A and B and C) = 0.96 * 0.94 * 0.92

now the 0.96 *, thats always correct since A is independent of everything else here

P(A) P(B and C)

then P(B and C) = P(C | B) P(B)

P(B and C) = P(B) P(C | B) = 0.94 * 0.92

it sort of includes the "we know that B is true in this case, so C can assume that B is true"

similar details for the other three cases

@keen dock oh right, are you done with this then

[keen dock]

i think soo give me like 5 min

[tulip ore]

cool

[keen dock]

1.5 hours at it

not bad

just the concept is this right

i think i got it

my answer is slightly inaccurate ig

@tulip ore

[tulip ore]

@keen dock bro couldnt you just have asked me to check the answers again, youre resorting to AI?

[tulip ore]

yes this is true

[keen dock]

:')

[keen dock]

sorry

[tulip ore]

the question doesnt really specify enough to even say this is wrong unfortunately

"C works 92% of the time" could mean "C normally works 92% of the time (which is what we did)" or "C ultimately works 92% of the time (which is what ChatGPT did)"

[keen dock]

i see

[tulip ore]

to be honest I dont know which to go for here

[keen dock]

hmm

its fine i'll just keep my answer as is, hope that's okay

.close

[pearl pond]

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[tulip ore]

hopefully np about this

though if they go with P(C) = 0.92 it would make sense

[pearl pond]

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[royal galleon]

for part a I can see how this can work if epsilon <= |x| and delta <= pi/2. but how to extend that for 2pi?

[royal galleon]

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[void estuary]

how to solve this?

[naive zinc]

Use cross product

[pearl pond]

@void estuary Has your question been resolved?

[naive zinc]

I use “times” to represent cross product
viewing O as original, A,B,C as vectors, you have aA+bB+cC=0 and a+b+c=1
(Your case a,b,c=3/6,2/6,1/6, but you don’t have to keep these numbers in mind for now)
Like 2 area of triangle OAB is +/- A times B right? But A times B is O-A times O-B= ((a-1)A+bB+cC) times (aA+(b-1)B+cC). You will find it equals c times something
Similar for B times C and C times A, you will find out they are also a times something and b times something
In the end you will obtain the ratio of areas of OBC, OCA, OAB

[pearl pond]

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[pearl pond]

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[long garden]

$\lim_{\epsilon^{+} \to 0} \int_0^{\tan\left(\frac{\pi}{8} - \epsilon\right)} \frac{\ln (2t)}{t^2 +2t - 1} , \mathrm{d}t$

[toxic lichen]

!msgdel

[pearl pond]

The original post of this help channel has been deleted, and it will abruptly close and possibly lock. (This is irreversible.) Please claim a new channel, and don't delete the first message of any future channel you claim.

[jolly parrot]

impract1cal

[long garden]

ok

[pearl pond]

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[wary vault]

,tex
Hello. So, we are prepping for exams and our professor sent us a bunch of practice exercises, but im not sure how much rigor the proofs need, so i need some educated guesses. heres one of the exercises: find the supremum and infimum of $$A= { 1+\left( -1\right) ^n +\frac{\left( -1\right) ^n}{n} , \ n\in\mathbb{N}} $$
For this, i split it into two sequences, $ a_n = A_k | k=2n $ and $b_n = $ same thing with odds. Is it good enough to just state that since $a_n \geq b_n $ we know that the supremum lies on a and the infimum on b? i struggle with these types of exercises because im never sure what the professor expects as proof. of course ill go ask him too, but id appreciate some advice from here too. thank you

[jolly parrot]

fijokazż

[wary vault]

.close

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[nocturne grail]

that looked easy, do you have a brain? are you lobotomized? i dont think math is for you. quit.

[wary vault]

im on my 37th lobotomy

[slow oak]

I’m gonna assume this is an in joke but for the sake of not giving me a heart attack let’s maybe not make jokes like that in the help channels, I thought youd randomly become evil

[nocturne grail]

noooo its my friend! okay, shall do! sorry.

[slow oak]

Nw I just got a fright 😭

[pearl pond]

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[violet prism]

why is b a counterexample? isnt b increasing for all x < 0?

[proper nova]

b is not increasing for all x < 0

well

example: x = -1

and x decreases to -2

WHO GOT A NITRO HERE

[shut elm]

More important is how you came up with that premise

[violet prism]

well

[bronze heath]

,w plot f(x) = -|x|

[jolly parrot]

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[violet prism]

yeah

that

[plucky oriole]

it is

[violet prism]

i thought it was a reflection of the modulus graph

in the x axis

[bronze heath]

it is

[proper nova]

i don't see it's increasing here

[shut elm]

Are your eyes okay

[shut elm]

I feel like answer choice a was a great answer

Maybe they just input it wrong when doing the questions

[bronze heath]

-x^3 is increasing isn't it?

nah

mb

[shut elm]

,w plot f(x) = -x^3

[violet prism]

,w plot -x^3

[jolly parrot]

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[violet prism]

hmm

im trying to figure out how likely it is that the website is wrong

eh whatever

thanks guys

.close

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[toxic lichen]

how the fuck does this website unironically write In(x)

what a load of shit

[pearl pond]

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[oak quiver]

what have you tried?

[inland laurel]

so i can calculate the wavelength using

[inland laurel]

i have made a triangle

whose h = 25m

base = 120m

and assumed at one vertex there is source and at other is observer

[oak quiver]

correct

can you share a pic of the thriangle?

[inland laurel]

from reflection from roof the angle of incidence = angle of reflection

[inland laurel]

yes

[oak quiver]

yes thats correct

[inland laurel]

i am having issue to calculate what is the path difference now

[oak quiver]

what will be the hypotenus of the trig?

[inland laurel]

65

[oak quiver]

good

so poath diff is just difference between the paths

so 130 - 120 in this case

[inland laurel]

what is 130

[oak quiver]

65 * 2

because you go up then down

[inland laurel]

oh

[oak quiver]

does this make sense?

[inland laurel]

yes

getting 20,20/3,20/5

[oak quiver]

does the comma mean decimal point?

[inland laurel]

no

disticnt values for distict n

[inland laurel]

here

[oak quiver]

oh yeah

20, 20/3, 20/5

[inland laurel]

yeah

thank you

[oak quiver]

yes that correct

nice

[inland laurel]

.close

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[inland laurel]

btw

.reopen

[pearl pond]

✅ Original question: #help-39 message

[inland laurel]

what is it mean by component waves?

like i have another question

The ratio of max to min intensity due to the superposition of two waves is 49/9. Then ratio of the intensity of component waves is

[oak quiver]

superposition or component of waves?

[inland laurel]

component waves

[oak quiver]

lets say you have wave1 and wave2

their interference will cause the intensity to inc or dec

the question wants you to calc the ratio of intensity of wave1 and wave2

they mean wave1 and wave2 by component waves

[inland laurel]

oh to calculae the original intensities from max and min intensisites

right?

[oak quiver]

yess

[inland laurel]

i am getting it as 9/49

beta = 10 log_ 10 (I_max/I'_O)
beta = 10 log_10 (I_min/I_O)

divided both

getting 9/49

[oak quiver]

can you send a pic of what you did?

[inland laurel]

[
\beta = 10 \log_{10}!\left(\frac{I_{\max}}{I_0'}\right)
]

[
\beta = 10 \log_{10}!\left(\frac{I_{\min}}{I_0}\right)
]

Dividing both equations,

[
1 = \frac{I_{\max}}{I_{\min}} \cdot \frac{I_0}{I_0'}
]

Given,
[
\frac{I_{\max}}{I_{\min}} = \frac{49}{9}
]

[
1 = \frac{49}{9}\cdot \frac{I_0}{I_0'}
]

[
\frac{I_0}{I_0'} = \frac{9}{49}
]

[jolly parrot]

BlackidoZΣ

[inland laurel]

using loudness formula

[oak quiver]

but we are talking about intensity?

we dont need loudness

[inland laurel]

true

[oak quiver]

you are supposed to use
$I_max = [\root(I_1) + \root(I_2)]^2$

[inland laurel]

yes

[
\frac{I_{\max}}{I_{\min}}

\frac{\left(\sqrt{I_1}+\sqrt{I_2}\right)^2}
{\left(\sqrt{I_1}-\sqrt{I_2}\right)^2}
]

[jolly parrot]

BlackidoZΣ

[inland laurel]

this?

[oak quiver]

correct

[inland laurel]

what are you getting

i am getting I_1/I_2 = 25/4

oh yeah thats the anser

answer

tha nks

[oak quiver]

yes thats correct

nice

[inland laurel]

dude do you have any tip

i am in so much pressure right now

[oak quiver]

use grammarly lmao

[inland laurel]

eh

i am typing informally

[oak quiver]

im sorry

[inland laurel]

no

[oak quiver]

that was a joke

[inland laurel]

i have to complete homework of physics, chemistry and math(2 chapters) and i have only 2 hours remaining for school to start

lol

[oak quiver]

that is a lot

are you supposed to submit all of it today?

[inland laurel]

kind of yes, altho they don't worry wheter i do it or not. but if i skip something i have to spent extra time on it later, and by today they might give extra homework of today

;-;

should i try each problem 2 min only and look at solutions quickly?

[oak quiver]

are these chapters really important for your exam?

if yes then you should give time to them

[inland laurel]

yes

they are coming in upcoming exams

[oak quiver]

hmm

[oak quiver]

you can try this

but if you are unable to do a bunch of questions from a sub topic, then mark it and make time for it

the easier sub-topics you can cover quicly

btw cbse?

[pearl pond]

@inland laurel Has your question been resolved?

[pearl pond]

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[late mist]

can anyone help me in solving this

[late mist]

i am not getting were to start from

[sullen cedar]

I'd start by expanding the second term on the LHS.

[late mist]

i was trying to reduce the inequality to one variable

like then proving that inequality of one variable

<@&286206848099549185> help me

can anyone tell me what to substitute so that i can get all the inequaltiy in one variable

I am getting this inequality

But what to substitute so that I will get a single variable inequality equation

[lean star]

just expand and go sicko mode

maybe let b-a = k to make things simpler

[jolly parrot]

Herzog

[sullen cedar]

Can you finish it now?

[plucky oriole]

what in da world 💀

[plucky oriole]

you can expand second term here

put b-a to some constant k

[sullen cedar]

.

[plucky oriole]

ah nvm

[sullen cedar]

It should be quite easy from here.

[plucky oriole]

just get rid of shitty integral symbol, put I there

[pearl pond]

@late mist Has your question been resolved?

[late mist]

oh yes i can resolve it now

[sullen cedar]

Use 0<= I <= (b-a)

[late mist]

though i got soluiton by sustitution m= I/k

[late mist]

ok

[sullen cedar]

0<=m <=1 and you have to prove m+(1-m)^3<=1

[late mist]

yes yes

now its easy to solve that inequation

[sullen cedar]

Sure.

[late mist]

ok thanks i may try your method

[sullen cedar]

Type .close if your question is resolved.

[sullen cedar]

It's the same as that.

[late mist]

yes

thanks

.close

[pearl pond]

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[frank violet]

Let $f(x)$ be a cubic function with leading coefficient of 1 and $f(0)=\frac{-1}{2}$. Let $a$ be a positive number, we consider $g(x)=\int_0^x f'(t+a)\cdot f'(t-a) \mathrm{d}t$. It's known that $g(x)$ has 2 extrema at $x=\frac{1}{2}$ and $x=\frac{13}{2}$, compute $f(1)$

[jolly parrot]

Fionna The Unemployed

[frank violet]

$g'(x)=f'(x+a)\cdot f'(x-a) - f'(a)\cdot f'(-a)$

[jolly parrot]

Fionna The Unemployed

[frank violet]

$f'(\frac12+a)\cdot f'(\frac12-a)=g'(\frac12)=f'(\frac{13}{2}+a)\cdot f'(\frac{13}{2}-a)=f'(+a)\cdot f'(-a)$

okay I got stuck here

huh wiat

[naive zinc]

You sure this is correct?

[jolly parrot]

Fionna The Unemployed

[frank violet]

no

[naive zinc]

To me looks like g’(x) is simply f’(x+a)f’(x-a)

[frank violet]

what why?

[naive zinc]

$\frac{d}{dx}(\int_{m}^{x}h(t)dt)=h(x)$

[jolly parrot]

Cogwheels of the mind

[naive zinc]

I am drunk right now so can’t point the further direction toward a proof. I can only point out this one mistake.

But when you have a proof I am still capable of a proof checking I think

[frank violet]

oh yeah

F*ck

I get it now

[naive zinc]

It happens

[frank violet]

$\int_{m}^{x}f(t)dt)=F(x)-F(m)$

[jolly parrot]

Fionna The Unemployed

[frank violet]

F(m) is just a constant

Thank you

[frank violet]

.close

[pearl pond]

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[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

Channel closed

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as this action is irreversible, and this channel may abruptly lock.

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[fossil fern]

[bronze heath]

<@&268886789983436800>

[dim linden]

hello, please dont open a help channel unless you need help

[fluid oar]

I was here first

[dim linden]

do i .close it or something

[unborn abyss]

ye

[fluid oar]

Yeah

[dim linden]

.close

[pearl pond]

Channel closed

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[slow oak]

[eager jewel]

omg ur gay LOL 😂

[slow oak]

Do not use gay in a derogatory fashion.

[toxic lichen]

ok cishet

[pearl pond]

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[bronze heath]

No

[bronze heath]

But ts help

nvm .close

.close

[pearl pond]

Channel closed

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[proper nova]

!redir

[pearl pond]

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

[proper nova]

faka u

[bronze heath]

Oppenheimer was no gay

or he was

Truth died with him

[proper nova]

😭

[proper nova]

this person did not even hear the warning lol

[eager jewel]

?

[proper nova]

.

[eager jewel]

its not my channel

[proper nova]

well it WAS your channel and you occupied without a question

[eager jewel]

bro

are u tryna ragebait

[proper nova]

YOU are trying to ragebait

[proper nova]

as shown in this picture

[eager jewel]

bro LOOK A THE PINNED MSG

he opened the channel

[proper nova]

wait hold up my bad

[proper nova]

this guy opened first

im so sorry for the inconvienience

[eager jewel]

;/

[pulsar flax]

.close

[bronze heath]

fake green

[pulsar flax]

bot is dead

oh

[pearl pond]

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[abstract pagoda]

Is there a difference between bounded and finite set?

[abstract pagoda]

Does
Finite set <=> bounded set?

[pulsar flax]

No

[0,1] is bounded but infinite

[abstract pagoda]

Then
Finite => bounded?

A finite set that contains its infimum but not its supremum.

[pulsar flax]

That is true

[abstract pagoda]

Is this a correct statement

?

[pulsar flax]

A finite set of real numbers includes both its infinimum and supremum

[abstract pagoda]

Obviously no

But how we can prove it

[abstract pagoda]

Is there a prove

[pulsar flax]

Induction can work

[abstract pagoda]

Give me a hint to do that

[pulsar flax]

But it should be very intuitive to see why it works

[abstract pagoda]

Hmm should i use cardinality ?

For a set A contains one element |A|=1

MaxA=supa=element

Mina=infa=element

Suppose there exist an set A ‘with k elements (finite) |A’|=k
And this set has sup and inf

Hmm

Now we should prove the statement for a set with cardinality k+1

[toxic lichen]

[0,1] is bounded but is not finite

so yes there is

[abstract pagoda]

Thanks

[abstract pagoda]

Suppose B is the union of A’ and set with one element

A is any set with k element then we can do that

And the set with one element is any set with one element

MaxB=max(A’ union the set with one element)

[toxic lichen]

suppose we already know every finite set of k elements has a supremum and an infimum.
then let B = A ∪ {x} as per your notation, where x is the new element to be added.

x is exactly one of the following:

• x < inf(A)
• inf(A) ≤ x ≤ sup(A)
• x > sup(A)

[abstract pagoda]

That’s make maxB exist

Then supB=maxB

SupB exist

The same with infB

By induction every finite set has sup and inf

[pearl pond]

@abstract pagoda Has your question been resolved?

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[inland sequoia]

Again I came up with my question

[inland sequoia]

Question no. 26

Hint me

[west vault]

Do you have a diagram?

[inland sequoia]

Yup I have drawn it

[west vault]

Right so can we see it?

[inland sequoia]

Ofc

[left steppe]

what are we trying to do?

[inland sequoia]

AB is tower and point B is foot of the tower

[inland sequoia]

Check this

[left steppe]

ok thanks

[west vault]

Right, so what ratio can you use the relate the height of the tower (the perpendicular here) to the bases of the triangles?

Also it's given that x and y are complimentary, so what other restriction does that give us?

[inland sequoia]

Idk

[west vault]

Opposite/Adjacent?

[inland sequoia]

Idk
Check what is given in the question

[west vault]

I'm asking you

That question is Socratic

[inland sequoia]

I know here we have to use
Tanx=p/b

[west vault]

Yep

That's correct

[inland sequoia]

b is given
But p and angles are not given
Angles are given in the condition of complementary

[west vault]

Right, we need to find p

That is the height of the tower

What does it mean for angles to be complementary?

[inland sequoia]

Yup it's the main question of answer

[inland sequoia]

if those angles are complementary,
x+y=90°

[west vault]

Yes

Now, what are the expressions for tan x and tan y?

[inland sequoia]

It's means
x= 90-y
And, y= 90-x

There are many possibilities to be x and y

[inland sequoia]

Wait

[west vault]

Yes

[inland sequoia]

Next?

[west vault]

Use the fact that x + y is 90 degrees

[inland sequoia]

In which way?

[west vault]

You can say that y = 90 - x

[inland sequoia]

But how do I calculate value of tan (90-x)?

[west vault]

$\tan (90^\circ - x) = \cot x = \frac 1{\tan x}$

[jolly parrot]

jewels!

[inland sequoia]

tanx + tany = AB/121 + AB/144 ???

Or
tan(x+y) = AB/121 + AB/144 ??????

[west vault]

In general, tan x + tan y is not the same as tan(x + y)

[inland sequoia]

Ooooooooooooooo

[inland sequoia]

Is it correct?

[west vault]

Neckbreaker

,rotate

[west vault]

Yes

[inland sequoia]

What is meant by Neckbreaker?

[west vault]

I need to turn my head like an owl

[inland sequoia]

How do you know that much at the age of 19?

[west vault]

um

[inland sequoia]

Hein! Why?

[west vault]

Your image was not upright

[inland sequoia]

Oh,

I have one more question

A ladder 10 m long resting 10 m below the roof of a house. If the angle of elevation of the roof of the house from the foot of the ladder is 60°. Find the height of the house.

I'm unable to draw th diagram

[icy wyvern]

They grow up so fast

🤧

[inland sequoia]

Hello me

[inland sequoia]

..

[quasi estuary]

do u mean ladder 10m long ??

@inland sequoia

[inland sequoia]

Yes

[left steppe]

it says its resting, resting on what

[inland sequoia]

I just need a diagram

[quasi estuary]

Can u send a photo or smth

[inland sequoia]

Wait

[quasi estuary]

If u can send a pic of the ques

[left steppe]

you dont need the upper triangle

[quasi estuary]

why not ??

[inland sequoia]

Question no. 26

[left steppe]

ohh

wait

nvm

you do need it

[west vault]

What kind of text even is this

[inland sequoia]

Devanagri Script (Nepali Language)

[left steppe]

they do have english under it

[quasi estuary]

Are u familiar with trigonometric ratio's ??

[west vault]

I'm not talking about the language, I'm talking about the amalgamation of topics

[inland sequoia]

What is the ladder and which line refers to the house?

[west vault]

The diagonal "10" is the ladder, the entire long left perpendicular is the house, and the section labelled 10 is the section between the top of the ladder and the roof

[inland sequoia]

Oh sorry my thoughts only to my next neurons

[inland sequoia]

Ok

[quasi estuary]

@inland sequoia

[inland sequoia]

Yup

[quasi estuary]

Use sin theta

If u could send the diagram with labelled points I could give u a hint

[left steppe]

wouldnt you use tan theta

[quasi estuary]

I dont think so

sin gives the answer directly

[left steppe]

cuz we need to find the height of the house

[inland sequoia]

Wait guys first tell me which angle is 60°?

[quasi estuary]

could u send a labled diagram

so i could name the angles and then help u

[quasi estuary]

angle ACB is 60

then u let angle(ACP) = x

and then tell me what does 2 sides of a triangle being equal tell u

@inland sequoia

[inland sequoia]

We can't use it here bcz the given side is hypotenuse not base.

[inland sequoia]

Ok wait

[left steppe]

do you know about exact values

[inland sequoia]

Sorry I was talking about PC but here we need AC

[left steppe]

why?

i guess laws of sines but i dont think its possible here

[inland sequoia]

The given angle is angle BCA

[left steppe]

yes

im not sure how you would find ac though

idk i did it another way

[inland sequoia]

AC = (20+2PB)/√3

[west vault]

I think you just start quantifying everything; for starters, AP = PC which means that angles A and PCA must be equal

Eventually you'll get a relation that'll help you get what you need

[inland sequoia]

If AP=PC
Then how
Angle A = angle PCA?
Explain me

[west vault]

PCA is an isosceles triangle

[left steppe]

isosceles

[inland sequoia]

Yesssss

[left steppe]

then what would you do with this information

[inland sequoia]

Just wait

I think I can solve

[inland sequoia]

Is this correct

[inland sequoia]

See what I did

[left steppe]

yeah

[inland sequoia]

Are you also in school?

[left steppe]

yeah

[inland sequoia]

Which grade?

[midnight haven]

any1 doing probability

[left steppe]

11th

[midnight haven]

jee level

[inland sequoia]

Oh I'm 10th grader

[left steppe]

nice

[inland sequoia]

Sorry bhai ham wo nahi kar sakte

[left steppe]

wrong channel

[midnight haven]

default channel?

[left steppe]

i did it another way if you want to see that

[left steppe]

i mean i guess #discussion

[midnight haven]

oh ok ty

[left steppe]

or #math-discussion

[left steppe]

@inland sequoia

[inland sequoia]

Go on help 8,21, ... And post your question
@midnight haven

[inland sequoia]

Sure

@left steppe

[left steppe]

never mind

i mean i could write it down

but it will take some time

[inland sequoia]

Dm me your steps

Tysm for helping guys

.close

[pearl pond]

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[low matrix]

Confused about the last part

[low matrix]

How am I supposed to estimate

I calculated the percentage change from monday to sunday (7days) = (45-31)/45 = -31%

Then I applied this to the 73 visitors of this friday:
73 - 0.31(73) = 50

Is this it?

[cursive wraith]

I'm not sure, maybe they were talking about the growth rate from Monday to Friday

[low matrix]

its the growth rate per week

[cursive wraith]

and not from monday to sunday

[oak quiver]

the q makes no sense honestly

[cursive wraith]

since they're talking about "assuming the same growth rate"

[low matrix]

but friday to friday means the growth over seven days

monday to friday would be only 5 days

[oak quiver]

the asked you to calculate the percentage inc in the prev part

so it makes sense to use it in the next part

[cursive wraith]

monday to friday is 4 days

monday to sunday is 6 days

friday to friday is 7 days

[low matrix]

but using growth rate of 4 days for calculating growth after 6 days is not enuff

[cursive wraith]

so you run into a problem whether one or the other

[low matrix]

what

oh well

you are right

what should i do

[cursive wraith]

calculate the growth rate from the percentage increase in 5)

[low matrix]

i think mon-fri it is then

[low matrix]

it is +62%

[rough python]

i dislike the question

[cursive wraith]

percentage increase is 62%

[low matrix]

so js 73+0.62(73)?

[cursive wraith]

me too, it's too vague honestly

[low matrix]

its for data analytics

[cursive wraith]

this is the growth rate per week?

[low matrix]

from monday to friday

[cursive wraith]

so it's the growth rate per 4 days

[cursive wraith]

so you can't do that

[low matrix]

yeah'

but that is what i calculated in 5

[cursive wraith]

Yeah but it's the growth over a duration of 4 days

you need the growth over a duration of 7 days

so, rule of three

[low matrix]

wats that

[cursive wraith]

a/b = c/d

how do I express one in terms of the other three

62% is the growth rate over 4 days

we want x% the growth rate over 7 days

62/4 = x/7

[low matrix]

i got 108.5

so is it gonna be 73 + 1.08(73)

,w 73 + 1.08(73)

[jolly parrot]

Results provided by WolframAlpha

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

Requested by sean4245322_63090

[cursive wraith]

looks good to me, maybe it's what the question wanted

in any case the question is not great

[pearl pond]

@low matrix Has your question been resolved?

[low matrix]

.close

[pearl pond]

Channel closed

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[pearl pond]

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[eager jewel]

how to find maximum and minimum value of this function

[warm current]

Find critical points

[eager jewel]

meaning? i need to differentiate that and equate to 0?

[warm current]

Or find where derivative does not exist, or endpoints of your domain

[eager jewel]

i cant differentiate that shit bruh

too time consuming

[warm current]

Too bad

[eager jewel]

"too bad"?

[waxen agate]

too bad

[warm current]

As in you know what you need to do, but you are choosing not to do it.

[eager jewel]

no is there any other way?

[warm current]

Nope

[eager jewel]

but this was given in circles chapter

they wouldnt have given it to differentiate

[toxic lichen]

this expression sounds like the squared distance from (4,7) to a point on the graph of y=sqrt(8x-x^2-12)

[iron basin]

yeah i think that

second eqn is a circle

[toxic lichen]

and y=sqrt(8x - x^2 - 12) sounds like it could be massaged into looking like a circle equation

it's the upper half of a circle

[iron basin]

||y^2 + (x-4)^2 = 4||

[eager jewel]

oh ok after i convert that into a circle then what do i do?

[toxic lichen]

think geometrically

[eager jewel]

distance between the 2 circles?

[toxic lichen]

your function is equal to AP^2, where A is fixed at (4,7) and P moves along that arc

maybe make a picture for this

[eager jewel]

ohhhh

ohh alr alr got it

max is 41

and min is 9

thanks a lot

[eager jewel]

"nope"

.close

[pearl pond]

Channel closed

Closed by @eager jewel

Use .reopen if this was a mistake.

[warm current]

My bad. Wasn't thinking geometrically

[eager jewel]

no thats fine but i felt u were being a bit rude

[warm current]

I was. I don't care to hear "I ain't doing that shit" to help i offer. You could have been clearer that this was geometry from the start

But you are still right that i was rude. I'll work on that

[pearl pond]

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[pastel totem]

i understand what im doing but i need help distributing sqrt3 into (x-1)^3

[pastel totem]

the 2 blue are 3’s to see better because they were written badly

[supple shoal]

thats right

[pastel totem]

ya so far

im just stuck on what to do next

[supple shoal]

no like thats it

[pastel totem]

ik that x stays x ONLY because of past examples but idk how its actually x still

[supple shoal]

oh

ok so

[pastel totem]

can you elaborate how so ik aswell :)

[supple shoal]

yea sure one sec

[pastel totem]

ofcofc tyyy

[vocal olive]

Hey

[pastel totem]

hi

[supple shoal]

oh wait

i just realised

u cant

nvm

ur a bit off

depending on how the ans is wanting to be in

u can just leave it as

cuberoot((x-1)^3)+1)

[pastel totem]

wait ill send the question

[slender stream]

,tex $h(x)=\sqrt[3]{x+1}$ right?

[jolly parrot]

Yash_AR

[pastel totem]

yes

[supple shoal]

yea so expand out (x-1)^3

[pastel totem]

what do you mean by expand out?

[supple shoal]

(x-1)^3 = (x-1) (x-1) (x-1)

or a^2 = a * a

[feral leaf]

You don't need to expand

The problem is finding the simplified expression of h(g(x))

[slender stream]

,tex $\sqrt[3]{x+1}$ not equal to $\sqrt[3]{x}+1 $ your question has the later

[supple shoal]

ohr

[jolly parrot]

Yash_AR

[pastel totem]

wait im confused

[supple shoal]

x+1 is its own term u cant cube root the seperate terms

[feral leaf]

How you wrote it, the + 1 is under the cube root

It's on the outside

[supple shoal]

like is sqrt(5 + 4) the same as sqrt(5) + sqrt(4)?

[feral leaf]

$\sqrt[3]{x+1}$ is not the same as $\sqrt[3]{x}+1 $

[jolly parrot]

CaptainNova22

[pastel totem]

wait wdmy its not?

[supple shoal]

@pastel totem

[pastel totem]

i dont think so?

[supple shoal]

ur correct but why do u not think so

[feral leaf]

Plug in a value for x and you'll see

[polar sierra]

sprt(25) + sprt(16) = 9

sprt(25+16) = 6.403

[pastel totem]

bc they are different processes

meaning different answers

[supple shoal]

kind of

[polar sierra]

if you ever have ideas, try it with familar numbers.

[supple shoal]

because sqrt(5+4) = sqrt(9) = 3

but the other one

[pastel totem]

wait

[supple shoal]

is sqrt(5) + sqrt(4) = sqrt(5) + 2

sqrt(5) + 2 does not equal 3

[polar sierra]

you start with the brackets eg PEDMAS or BIDMAS. then you do the sqrt

[pastel totem]

im just more confused 😭

[polar sierra]

yh it does.

sqrt of 4 is 2

[pastel totem]

im not asking of mu answer is right or wrong

im just confused on what process i need to do

[supple shoal]

ok but ru familiar w BIDMAS

order of operations

[feral leaf]

Apply your exponent rules

[pastel totem]

nope

[polar sierra]

i mention both due to international.

[pastel totem]

what does this mean exactly

[polar sierra]

@pastel totem whats the orginal question?

[feral leaf]

These

[pastel totem]

this is where im stuck

ignore the = x because im trying to figure out how i got = x

[pastel totem]

yes i kinda understand

[feral leaf]

So your process is correct, with doing h(g(x)) = $\sqrt[3]{(x-1)^3} + 1$

[pastel totem]

right

[feral leaf]

It's just simplifing it now

[pastel totem]

dont i distribute sqrt3 into (x-)^3

[jolly parrot]

CaptainNova22

[polar sierra]

is it just x?

[feral leaf]

You don't need to

[polar sierra]

sqrt^3(A) is the same as A^1/3.

[feral leaf]

You can rewrite the $\sqrt[3]{x}$

[jolly parrot]

CaptainNova22

[pastel totem]

but can you explain why i can rewrite it as that?

[feral leaf]

Can you rewrite that with a fractional exponent?

Using the exponent rules

[polar sierra]

how do u write in this format?

[feral leaf]

It's just a property of exponents

[pastel totem]

i get what ur asking but i dont know how to rewrite it as that no

[feral leaf]

Like sqrt(x) is the same as x^(1/2)

[pastel totem]

wait so

[feral leaf]

Look at the table I sent

[polar sierra]

sqrt^3( (x-1)^3 ) === ( (x-1)^3 ) ^1/3

Then follow the power rule

[feral leaf]

^

[pastel totem]

ya i did im just confused how i would rewrite this into that

[pastel totem]

i think $ before n after

[feral leaf]

That's not helping, I am trying to walk them through the process

[feral leaf]

Start with $\sqrt[3]{x}$

[pastel totem]

are u saying that the sqrt3 is cancelled out by ^3

[jolly parrot]

CaptainNova22

[pastel totem]

therefore left with x?

[pastel totem]

okok

[feral leaf]

Can you rewrite that with an exponent?

[pastel totem]

yes

wait

sqrt3x^2

like that?

[feral leaf]

No

Do you understand the last rule in this table?

[pastel totem]

i dont understand the a^x/y

but after the = i do understand

[feral leaf]

a is the base, and x/y is the exponent and root

[feral leaf]

You can match it up

So you have $\sqrt[3]{x}$

[jolly parrot]

CaptainNova22

[pastel totem]

how would i go about matching it up?

okok

[feral leaf]

What would a be?

[pastel totem]

x?

[feral leaf]

Yes

What would x be?

[pastel totem]

1

[feral leaf]

And y?

[pastel totem]

right?

OHHH

[feral leaf]

Yes

[pastel totem]

so itcowuld be 3?

[feral leaf]

Yes y is 3

[pastel totem]

okok so wait

[feral leaf]

So can you rewrite $\sqrt[3]{x}$ with an exponent now?

[jolly parrot]

CaptainNova22

[pastel totem]

x^1/3 = sqrt3x ?

like that?