help-42

[glacial inlet]

now we just need to add them together

[tiny steppe]

98

[glacial inlet]

yep correct

[tiny steppe]

YESS

[glacial inlet]

that’s correct

[tiny steppe]

THANK YOU

[glacial inlet]

ywyw!

[tiny steppe]

.close

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[stone flame]

Just kinda need help understanding this. So g is from P(A) to P(B), defined by g(X)= B-X, for every subset X of B. Does that just mean g is from P(B) to P(B)?

[kind grotto]

that's kinda weird cause g should be defind for every value in P(A)

[stone flame]

And if f(X) is defined for every subset X of A and g is defined for every subset X of B, would that make (gof)(X) defined for every subset X of A?

[potent gazelle]

lucky who understand math

[stone flame]

So g should be for every subset X of A?

I think my prof might’ve made a typo in the question. Either that, or I’m just slow.

[kind grotto]

I guess, worth checking

because its weird if g is defind from P(A) it doesnt specify what will be the output of {1,2,3} for example

maybe im also missing something so maybe wait for someone else...

[stone flame]

I just thought g should be from P(B) to P(B) since g(X) is defined by every subset of X of B which is just the power set of B

So g(X) is defined for every element of P(B)

Which is why I’m confused as to why g is from P(A) to P(B)

[glass heart]

well you can still compute B-X even if X is not a subset of B

[stone flame]

How would I prove it’s one to one tho?

[glass heart]

given that P(A) has 16 elements, you could just write down what g(X) is for every subset of A

that will also help you better understand what is going on

[stone flame]

Why does it say g(X)=B-X for every subset X of B tho?

[glass heart]

oh I didnt see that

they fucked up

[stone flame]

Thats what I thought at first but I’m not smart enough to know for sure haha.

So should g be from P(B) to P(B)?

[glass heart]

I think they still mean P(A) to P(B)

but its impossible to tell. can you ask your teacher?

[stone flame]

Yea thats probably a good idea. Thanks for the help.

.close

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[vast parcel]

Can someone explain me this result in the comment of this video?
https://www.youtube.com/watch?v=qgvmJTmJIKs
(Sorry, the two images below are reversed, but I did not manage to put them back in order...)
Explanation high school level if possible... I understood the video, but not this comment with a lot of likes...

[cloud nexus]

first time ive seen someone ask for math help on a teded riddle

[vast parcel]

I understood the video, but not the comment...

<@&286206848099549185> ?

Tell me if you’re trying to find the answer, or I’ll close the discussion...

Nobody? OK, I'll retry tomorrow...

.close

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[copper hollow]

Alguien me puede resolver es trabajo practico de funcion cuadratica, lo ocupo para el jueves.

[calm coral]

Please don't occupy multiple help channels.

[copper hollow]

ok

Alguien me puede resolver es trabajo practico de funcion cuadratica, lo ocupo para el jueves.

[mellow crater]

Resolver no, pero podemos ayudarte mientras lo haces tu mismo 🙂

Tienes algún conocimiento de este tema?

[calm coral]

@copper hollow Has your question been resolved?

[copper hollow]

el problema es que tengo que entregarlo para el juebes y no se como resolverlo no entiendo nada

[calm coral]

@copper hollow Has your question been resolved?

[vital plume]

Ps es lo básico sobre funciones cuadráticas

Buscando información por internet supongo que en dos horas podrías aprenderlo por completo

[golden flare]

what r u ding or asking?

is this french

Idk what u r doing

[calm coral]

@copper hollow Has your question been resolved?

[vital plume]

He's asking for someone to do his homework

Also, this is Spanish

[golden flare]

k then but the ruls stat that we canot do their home work we can help tho

[vital plume]

Yeah, but you know how this world is

[golden flare]

I know

[calm coral]

@copper hollow Has your question been resolved?

[sinful pasture]

Mamma mia

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[eager token]

How would i do 64

[eager token]

Would it be 2logbase2of3 + logbase2of5?

[pure kayak]

that seems fine

[eager token]

Okk ty

What about for this

How would i rewrite logbase2of5 as logbase2of20

[pure kayak]

you wouldnt, but you can split it

[eager token]

?

[pure kayak]

split log_2(20) into two logs

[eager token]

Ohhh

Wait how

And why

[pure kayak]

you want things in terms of log(5), so get a log(5)

then see what you can do with the leftover

[eager token]

But wouldnt the leftover be log(4)

[pure kayak]

it would

[eager token]

How would u write that if u can only use log(3) and log(5)

[pure kayak]

log_2(4) evaluates quite nicely

to just 2

[eager token]

Oh

[dusky flax]

log_2 (20/9) = log_2 (2^2) + log_2(5/9)

[eager token]

So it would be {2 + log(5)}/2log(3)?

[dusky flax]

hm?

[pure kayak]

-, not /

[dusky flax]

how did you get that?

[eager token]

Ah

[dusky flax]

and no brackets

log_2 (2^2) + log_2(5/9) = log_2 (2^2) + log_2(5) - log_2 (9)

[eager token]

2 + log(5) - 2log(3)?

[dusky flax]

👍

[eager token]

TY BOTH :DDD

.close

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[left sorrel]

i can't find the length of this rectangle right?

[untold compass]

I don't think you'll be able to solve for any length yea

you'll have to work entirely within areas which can be a little restrictive

[left sorrel]

what do you think is a good approach to solve the question?

[untold compass]

the only interesting thing is that the slope is constant and the spacing along AB is constant

your goal should be to find the area of the leftmost fifth piece, but to get you started on that I'd try to think about what the difference between the two shaded regions really looks like

[left sorrel]

Yeah the orange region is 3cm^2

I don't see how useful that'd prove

[untold compass]

yup, now think of the difference between the light-shaded and the white region

if you can relate the two shapes you'll have the entire area

this grid system might help, you don't know the length/width of the lines but it'll still be useful

[left sorrel]

right thanks, let me have a look

[left sorrel]

These two triangles are similar right?

hmm wait maybe i should use your grid system

[left sorrel]

okay basically fromt his i have that right + top = 3

to get the answer i need to find 2 * Right

I think i'm missing one little detail maybe

[untold compass]

wait what's right and top

[untold compass]

the red thing you shaded a few messages back is worth 3 grid rectangles, and the difference between white and light-shaded is 4 grid rectangles

[left sorrel]

yeah right + top = 3

[left sorrel]

how do you know the second part? 😭 i'm confused

This is my "naming" scheme btw

[left sorrel]

R + T = 3

i have to find R + R

[untold compass]

oh gotcha

try to see that T is half of R

using like triangle is 1/2base*height

like you can just see T is 2 triangles and R is 4

[left sorrel]

yeah i agree with you but like that's because of your grid system right?

I can intuitively also see that if i rotate T clockwise i can have the otherpiece of T

and T + T = R

[untold compass]

yea

since the slope and segments are constant you get to make this huge grid of similar triangles

[left sorrel]

oh i see it now

u mean like this right?

those two are congruent triangles

now rotate the upper one to the left or right

[untold compass]

does that rotation still work with a very skinny grid?

[left sorrel]

no i guess

[untold compass]

you might have to disect and shift things around instead of rotate

[left sorrel]

idk i wouldn't see the T = 1/2R tbh

your grid works but i don't see an actual justification

[untold compass]

like for the grid at all?

[left sorrel]

Nah like dismissing the grid entirely

[untold compass]

you're kind of inventing a variable x for between the segments and a variable y for slope*x

oh

yea there are cute ways to rearrange the triangles

[left sorrel]

I can see R +T = 3 but i can't make the connection for R + R

by deducing that T = 1/2 R (this part i can't really justify)

the rest i get now

[untold compass]

cut T vertically into T1 and T2

and you can build R with two T1's and two T2's

[left sorrel]

do we use a pictoral argument for this part?

and yeah i think i follow

it's the rotating thing again, i guess, right?

[untold compass]

well you're rearraning this time

[left sorrel]

Like my issue is when i'm right here

I don't see how i'd instantly go

Oh yeah 2*T1 + 2*T2 = R

If that makes sense

[untold compass]

actually I guess it's not quite that

more that a triangle is half the area of the rectangle it's in

[left sorrel]

oh i see and you're claiming that T is 1/4th of its rectangle?

whereas R = 1/2 its rectangle

[untold compass]

right

er idk what rectangle is here

[left sorrel]

That's fair yeah, thanks

[untold compass]

like a big one

[left sorrel]

I guess this rectangle?

flip this thing and it kinda is 1/4th of the

golden rectangle

[untold compass]

ah yea that's good

[left sorrel]

Oh yeah btw i had a different argument for this problem

I wanted you to confirm if this is valid

[untold compass]

sure

[left sorrel]

can i extend the left half like this?

and claim that extended triangle also has base n?

[untold compass]

no, I'm not sure how far out that would be

there's an answer I'm sure but it's not obvious from the problem setup

[left sorrel]

surprisingly that gets the answer btw

[untold compass]

oh boy

[left sorrel]

wait let me confirm what we get

it's

,calc (6 + 10 + 13)*2

[potent lotus]

Result:

58

[left sorrel]

wait lol what

hold on

oh i typo

,calc (6 + 10)*2 + 13

[potent lotus]

Result:

45

[untold compass]

there's like a sequence of areas where it goes down by a constant amount, like 10 light-shade, 6 white, 2 magic golden triangle you drew

[left sorrel]

yeah the triangle thing gets exactly 45

[untold compass]

so yea it's n

but from some strange way they picked the 10 and 13

[left sorrel]

so it's just luck then?

[untold compass]

yea

[left sorrel]

wow that's crazy

[untold compass]

if I said 11 and 17 it'd be like 1.75n or something dumb

amazing

[left sorrel]

wait i might be dumb

i don't really follow

are you talking about from the dark region?

[untold compass]

I'll just draw more random grids

this doesn't explain anything ngl

[left sorrel]

😭

i think i can follow actually

they're all just congruent triangles

but that's under the assumption that your height is also 2

unless that's by construction

[untold compass]

uh this is from that light-shade is 10 and the difference across areas is 4 because of that whole shaded-area logic

the way the problem is set up the length and width of the pentagon can be anything, but at a constant grid area

[left sorrel]

oh yeah so you drew rectangles of area 4

that's hard

[untold compass]

where is this problem from it's kind of wild

[left sorrel]

i stole it from someone here

it's an olympiad problem

😭

[untold compass]

_>

[left sorrel]

They have another one too

which i found interesting

[untold compass]

baiting me into olympiad geometry

[left sorrel]

i'm sorry

idk if it's olympiad

but i don't think it's regular geometry

at least imo

i don't do competition math either

i just thought this was approachable so i wanted to try it

[untold compass]

gotcha

[left sorrel]

okay so just to confirm, you basically do this whole thing with grid lines lol

at least to see the extended thing is "n"

[untold compass]

yea I started from that the slope is constant so there's all these equal area things

[left sorrel]

yeah fair enough, thanks!

Btw can i ask the other question?

[untold compass]

sure

[left sorrel]

I'm guessing the other one is probably harder

It's this

first thought was finding the side length of the equilateral triangle

but that wasn't pretty so that wasn't the way to go i guess

using sqrt(3)/4 s^2 = 120 but yeah

through some geometry i did establish that BE: EC = 1:2

and also DF: FC = 1:1

[untold compass]

oh this one is pretty chill

you've got most of it just use AD:DC = 1:3

[left sorrel]

how do you know that?

[left sorrel]

oh lol

maybe i wasn't trying hard enough after that one

[untold compass]

like you can cut it up into 4 equal pieces and they all have the same height if you tilt your head so the top right is the base

height from AC to B

[left sorrel]

wait so you're claiming that BD = BF = height?

oh yeah

i mean pictorially it does look like it lol

[untold compass]

uh F isn't related

just that these 4 triangles have the same base/height at a certain angle so the way to draw BD is to go 1/4 along AC

I guess the problem doesn't say it but I saw it as drawing BD to cut in 4, DE, to cut in 3, and EF to cut in 2

[left sorrel]

but doesn't that change the height?

sorry i'm lost LOL

if it's at an angle then you can either have that angle influence the base or the height

[untold compass]

gotta rotate it and it looks normal

they all have the same height so to divide into 4 the base is 1/4 each

[left sorrel]

Oh i think i get you

do you mean something like this?

BD * AC = 240

BD * AD = 60

AC/AD = 4

so AC = 4AD

so DC is 3/4 of AC

and we had that DF = DC

[untold compass]

no clue how you're starting that

[left sorrel]

stating this?

[left sorrel]

or this?

[untold compass]

the 240 part

[left sorrel]

oh

1/2 (BD) (AC) = 120

[untold compass]

oh BD isn't a height

[left sorrel]

oh, it looks like it in the pic

wait

it still gives the same conclusion you had right

These questions are getting to me

[untold compass]

actually losing it at these alternative fake solutions that work

[left sorrel]

😭 rightt

[left sorrel]

I think i'm just generating all these fake solutions

😭 sorry

hilarious that it again works though

[left sorrel]

okay so i can't use this at all right?

btw before i making new fake solutions

can i even say h_A-h_D = h_F?

wait

this kinda looks like the other problem lol

[untold compass]

lol

nah F is like 3/8, D is 6/8, A is 8/8

[left sorrel]

ah yeah

this question's supposed to be chill? dang

[left sorrel]

wait another idea

the base AD generates ADB and that has area 30

the base DC and generates DCB which has area 90

AD/DC = 3 ?

Nah this one feels wrong

but it again just works

[untold compass]

uh the numbers are meaningless but that's sort of the reason AD/DC is 1/3 yea

[left sorrel]

okay what's your reasoning again?

[untold compass]

the picture splitting it into 4 above, each triangle has equal area and height

so 1/4 area means 1/4 base

[left sorrel]

oh yeah that picture but i don't get the "equal height" part

[left sorrel]

BD is not equal to this though?

do i have to believe the pictorial representation? 😭 is there a mathematical justification

I'm dumb 😭 to make pictorial arguments

[untold compass]

oh god yea I'm like full visual learner

[untold compass]

like the heights are clearly equal right

[faint ether]

Why AD/DC=1/3?

[left sorrel]

that's exactly what we're working on

well i am

[faint ether]

[left sorrel]

U gotta look at the rotated pic plurmorant posted i guess

[untold compass]

yea I'm at a loss of how to explain it with words/equations

[left sorrel]

OP said it was for grade 8

I'm in grade 11

Lol

[faint ether]

Is it ur assumption sort of thing

[faint ether]

Damn

[untold compass]

the problem says you draw BD to cut the triangle into 1/4 the full triangle area

so like clearly you do it 1/4 along the length

[faint ether]

Why

[untold compass]

[left sorrel]

[untold compass]

this #help-42 message and this #help-42 message

[faint ether]

It's doesn't necessarily have to be like that?

[faint ether]

Ok lemme see

[untold compass]

yea I show that the heights are all the same

[faint ether]

If it was given like this then Ad/dc would be 1/3
But here BD isn't = ED ryt

Ok well

Maybe it's right

[untold compass]

idk what E does here

the problem starts with all the triangles are 1/4 the total area

[left sorrel]

in these questions u can do randoms tuffs

[faint ether]

Is EF perpendicular to FC

[left sorrel]

and it somehow works too

it's all stupid

[left sorrel]

no

[faint ether]

Grade 8 yea we are supposed to do this stuff ig

[left sorrel]

well yeah i guess it should be

i used that idea to show that DF = FC

[left sorrel]

that's why i asked

this is too hard 😭

The other question was crazy too

ugh

[faint ether]

Ok I will read all of them now

[untold compass]

unlucky

[left sorrel]

wait

this is your labeling btw?

just so we're on the same page

[untold compass]

no F

maybe it's easier to see it as you cut in half at h', then cut in half at D

maybe it's about a perspective on how the problem is constrained

you can't figure out how to make triangle EFC until you've drawn everything else, since E and F are free to move along their sides

but ABD is fixed because A and B can't move

[left sorrel]

I think i can see your pov

[faint ether]

It's all so stupid

[left sorrel]

we conjecture that green is the height of ABC

[faint ether]

I didn't do 8 grade now that I think

[left sorrel]

orange also has height green has height

so (BH)(AC) = 240 and (BH)(AD) = 60

divide both the equations and you have AC/ AD = 4

[left sorrel]

frfr 😭

this is insane

holy smokes

[faint ether]

good luck

[left sorrel]

lol u gave up?

u should solve it too

this isn't my question

[faint ether]

Yeah I can't even come up with some fake soln

[left sorrel]

😭

[faint ether]

Plurmorant will give u some soln ik

[untold compass]

oh I thought it was solved like an hour ago

[left sorrel]

whatttt

so we're assuming that EF is perpendicular to DC right?

oh yeah it is because DF = DC so it is a perp bisector

[left sorrel]

yeah i think solving this part is trivial

the annoying bit was the altitude thing

nice nice

thankss

[left sorrel]

wait just so we're clear

you find FE right with pythag

that isn't even nice

[untold compass]

question asks for FC

[left sorrel]

oh bruh

whoops

i forgot what the question was

[calm coral]

@left sorrel Has your question been resolved?

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[cerulean compass]

hey everyone

[cerulean compass]

If I am trying to prove that given f is continuous, then so is |f| (via epsilon delta definition of continuity) could I show via alternate triangle inequality that $||f(x)| - |f(x_{0})|| >= |f(x)| - |f(x_{0})|$ and $|f(x) - f(x_{0})| >= |f(x)| - |f(x_{0})|$

[potent lotus]

Zouni

[cerulean compass]

all smaller than some epsilon

is that sufficient given f is assumed to be continuous?

[calm coral]

@cerulean compass Has your question been resolved?

[cerulean compass]

<@&286206848099549185> ?

can you @ helpers is that fine

[modern peak]

yea that's alright

as long as it happens at least 15m after post, so yea ur good

[calm coral]

@cerulean compass Has your question been resolved?

[cerulean compass]

.close

[calm coral]

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[rich ore]

I need help with this i feel like in overcomplicating it. question d

[calm coral]

@rich ore Has your question been resolved?

[rich ore]

<@&286206848099549185>

[polar jetty]

Try changing everything in terms of cos(x) and sin(x).

Basically, get rid of tan(x) and cot(x).

[rich ore]

I tried that in last step but i think thats overxomplicating it

[polar jetty]

No. You need to do that at the beginning.

Basically, adding the two terms together is already a misstep(in the sense that it makes it unnecessarily convoluted).

[rich ore]

Should i do that in all such probs?

[polar jetty]

Well, some people do that. It tends to make things simpler if you don't remember all the identities.
It's not really the case that you should always do it though. It's about trying things and learning by experience. After enough practice, you would be able to see multiple ways of doing such questions in your head.

[calm coral]

@rich ore Has your question been resolved?

[calm coral]

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[safe moss]

hey uhm can someone correct me here if im wrong?

the question states:

Upper and Lower Bounds Show that the given values for a and b are lower and upper bounds for the real zeros of the polynomial.

and for the first example we got $$2x^3+5x^2+x-2$$ with a = -3 and b = 1

[potent lotus]

ates

[safe moss]

so i tried b = 1 but clearly it is not a zero of this polynomial

-1 is

as i know we divide x-b (for b>0) with the polynomial

so we divide x-1 with the polynomial

however, i used synthetic division and when we use this way we take -1 and turn it into postivie ( as i know we are doing it the same with negatives to positive ) because clearly for x-1 x = 1 is the zero

my question is am i wrong at stating that -1 is the zero and not 1 ?

well, i tried b = -3 as well and thats not a solution either

i think i know upper and lower bounds theory incorrect i just checked an example given by the book and it makes no sense tbh.

<@&286206848099549185>

i realized i know this theorem wrong, so got one question now

could u tell me if a zero can be an upper/lower bound?

[calm coral]

@safe moss Has your question been resolved?

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[eternal laurel]

I am new to rigorous math, my question is that what is the relationship between The Field Axioms and other axioms like ZFC, PA etc…

[eternal laurel]

I know that you can prove PA using ZFC, but I can’t find anything that relates field axioms to other axioms, is it really axioms or just a list of rules?

<@&286206848099549185>

[remote mural]

while axioms are indeed fundamental rules or properties assumed to be true without proof they are not typically derived from other axioms within the same system instead it provides the framework upon which mathematical structures and theories are built

[eternal laurel]

So what is Field axioms really?

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[unkempt drift]

Yeah?

[remote mural]

so why can we drop it here exactly?

[unkempt drift]

Cause a multiple of y1 (t) gives 0 when you sub it in

Since t^(-1) is a solution to the DE, the question says

As in $(k y_1 (t))' = k (y_1 (t))'$ and so on, so you are multiplying everything by k, but that's just k * 0

[potent lotus]

south

[remote mural]

ah i see

fair enough thanks

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[pallid wing]

Given:

AD is parallel to BC
ADC = 120°
BAD = 3CAD
E is a point on AC
ABE = 2CBE
On the line AC, a point P is taken such that ABP = DCA

Find the ratio of CBP to ABP.

[pallid wing]

how to do this guys

[winter elbow]

!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

[pallid wing]

1

i have no idea with that

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[remote mural]

For [
y'' +3y' +2y = e^t(t^2 + 1)\sin(2t) +3e^{-t} \6\cos t + 4e^t
]
What would be a suitable form for a particular solution? Please note I am not meant to be solving this; just need to find a somewhat reasonable form to put down

[remote mural]

So i guess the last two terms are pretty direct

but im unsure how to handle the first

[quaint sphinx]

basically you combine everything

[remote mural]

im trying to think of it logically so like i dont get extra unnecessary terms that will eventually cancel

even though that would still be correct

the polynomial seems like it will always be a power of two if u differentiate the first term twice

okay

so i guess it really is something like

,, \8{Ae^t}\8{Bt^2 + Ct + D}\8{E\6\sin{2t} + F\6\cos{2t}}

[remote mural]

hmmm

seems sus

[quaint sphinx]

it’s sorta close

[remote mural]

maybe like

,,e^t\6\cos t\8{A_0t^2 + A_1t + A_2} + e^t\6\sin t\8{B_0t^2 + B_1t + B_2}

[remote mural]

i was thinking of putting a coefficient in front of the e^t's but thats redundant

[calm coral]

@remote mural Has your question been resolved?

[quaint sphinx]

yeah that’s pretty good

[remote mural]

aight

ty

[quaint sphinx]

though the insides of the cos/sine terms are 2t but i think that was just an incidental mistake

[remote mural]

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[remote mural]

oh yeah

righty thanks

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[feral thistle]

Im trying to solve this problem about finding the volume of a general spherical cone. My answer seems to be missing a factor of 1/3 though. Does anyone know where I made a mistake? Thanks!

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[feral thistle]

Nvm im stupid

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[deep crescent]

Can this be integrated without using Ln 2+tanx? I haven't tried it myself yet

[drifting seal]

why

this is a clear du/u

[civic dirge]

I'm sure you could find some strange way to write it in terms of other functions

But the answer is literally ln(2 + tan(x)) + C and I don't personally know why you'd want to complicate that further

[deep crescent]

Yeah was just curious thanks

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[sharp arrow]

Hey, been stuck on for ages, meant to use integration via substitution apparently, thanks 🙂

[civic dirge]

This is a special kind of substitution called a trig sub. Seen much of that?

[sharp arrow]

a bit, not given what the sub was so not sure

ik tan^2 + 1 is sec ^2 so might try that

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[trim dust]

hello

[trim dust]

can this also be
5t/26
(-25/26 )*sint
thats what i got
idk how tf i got that b4

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[wild nymph]

8y'=y^2-2y-15, y(0)=-2

[wild nymph]

how to find the explicit solution y(x)

[calm coral]

@wild nymph Has your question been resolved?

[wild nymph]

<@&286206848099549185>

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@wild nymph Has your question been resolved?

[wild nymph]

<@&286206848099549185>

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@wild nymph Has your question been resolved?

[tulip veldt]

Hi @wild nymph I believe we have a solution for this question. Start by factoring the RHS and then this shows us that this is a separable equation.

[wild nymph]

yeah I did it

[tulip veldt]

ok sweet!

we got this is you want to compare

remember to .close when youre done!

[wild nymph]

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[spiral forum]

16. E)

[spiral forum]

.close

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[remote mural]

Can someone explain to me how this came into being?

[remote mural]

(xiii)

[brazen elbow]

chain rule

[exotic falcon]

⛓️

it actually is not even true as written but we ball anyways

[shy light]

where did they change the variables?

[exotic falcon]

who wrote this book

[brazen elbow]

yeah its supposed to be dX not dx on the first equal

[remote mural]

But there is not need to. f(ax+b) = f(X)

Wouldn’t the derivative just be ‘a’

[brazen elbow]

d/dx f(X) makes no sense

the variable needs to match with the differential

[exotic falcon]

sir

it does make sense

X=g(x)

but

it just isn't true because this implies that d/dx f(ax+b) = a^(j) d/dx f(ax+b) for all a for all natural j which clearly isn't true

[brazen elbow]

yes

pardon my mistake kind sir

[remote mural]

Where did you get j from? I don’t understand

[exotic falcon]

the formula is written in your book is wrong

it's stating that

d/dx f(ax+b) = a d/dx f(ax+b)

but this implies that

d/dx f(ax+b) = a d/dx f(ax+b) = a^2 d/dx f(ax+b) = a^3 d/dx f(ax+b) = ...

[remote mural]

Ohhhh

By induction right?

[exotic falcon]

sure

or just by multiplying the original formula by

a

and then comparing

and repeat

[shy light]

so then a==1 which is not always true?

[exotic falcon]

yes

[remote mural]

Yeah the formula would have to be incorrect then

[shy light]

then why is this even valid?

[exotic falcon]

it isn;t

it's just written incorrectly

they're trying to show you the chain rule

[remote mural]

What does a== 1 mean?

[shy light]

oh

[shy light]

a is always 1

[remote mural]

Probably a typo

Oh oh

Okay that makes sense

[shy light]

dw we use it in programming to assign variables

[remote mural]

Thank you so much guys:)

I appreciate it a lot

Have a good one!

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[slow trout]

How did this became 2/14? Dont you need to add 3/14 and 7/14 since those were the pages?

[slow trout]

Since its P(A) + P(B)

Oh so thats why its equal to 1 instead of those odd numbers?

Oh i get it now ive been so confused

I was doing multiple solutions earlier cannot understand why i havent got the same exact answer

It was just some simple logic

Thank you!

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[indigo cipher]

Sorry for asking such a dumb question, but my friend has been bothering me with it for two days. He wrote " -1 = i x i = √-1 x √-1 = √(-1)(-1) (under the same root) = √1 = 1" thus proving -1 = 1. Can someone please explain where is the error, I don't get it.

[pliant hamlet]

Iirc sqrt(a) * sqrt(b) = sqrt(ab) only holds where the square root of a and b are real numbers

[indigo cipher]

Thanks, i've been telling him this he keeps on gaslighting me you can... Thanks again!

[pliant hamlet]

Np

[indigo cipher]

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[slow trout]

How did this became a non mutually?

[slow trout]

Ik it shows the non mutually formula but how did it became one?

[half sparrow]

what is the non mutually formula

[slow trout]

This

[half sparrow]

ok

[slow trout]

How?

[half sparrow]

So I think 'either one showing a 4'
means that we accept one of the dice showing 4 but not both at the same time

...wait

[slow trout]

Suree

[half sparrow]

hmm i think this question is a bit poorly worded if that is the solution

If my interpretation was right
We should not count the case where both dices show 4
And the probability is 10/36

[slow trout]

How did it became 10/36?

[half sparrow]

There are 36 possibilities with 11 having at least one 4 showing

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)
(1,4),(2,4),(3,4),(5,4),(6,4)

exclude the case (4,4)
then it would be 10/36

[slow trout]

Ohh got itt thank you!!!

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[vital lotus]

Why is one of the solutions, x=pi, for the below equation not graphed on desmos nor geogebra?

[obtuse pendant]

I am not sure; maybe it's something with Desmos's precision. I tried 0.9999999 instead of 1, and it works.

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[ebon marlin]

how to prove $\lim_{x\to \infty}\frac1x = 0$ using $(\varepsilon, \delta)$ definition of limit

[potent lotus]

babario

[remote mural]

Can anyone once again explain where and how multisets are used?

[remote mural]

Oh shit

Wrong place

[ebon marlin]

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[lyric snow]

We can use |1/x-0|<epsilon
And |x|>delta
Then find connection between epsilon and delta

[calm coral]

@ebon marlin Has your question been resolved?

[ebon marlin]

shouldnt it be $|x-\infty|<\delta$?

[potent lotus]

babario

[lyric snow]

No x is increasing and going to infinity so we use |x|>delta

[lyric snow]

X-infinty is -inf and by mid it's +inf so doesn't make sense

[ebon marlin]

i dont quite understand

[ebon marlin]

what do you mean "by mid"?

[lyric snow]

Sorry mod

Modules

[surreal stratus]

[ebon marlin]

so like x is closer to infinity than delta is so |x|>delta and then |1/x|<1/delta so as long as i choose delta=1/epsilon the limit holds true?

[lyric snow]

Ya

[ebon marlin]

how to verify it tho?

say $\delta=\frac1\epsilon$, we have $|x|>\frac1\epsilon$.
from $\bigg|\frac1x\bigg|<\epsilon$, taking the reciprocal we also get $|x|>\frac1\epsilon$
which checks out?

[potent lotus]

babario

[ebon marlin]

erm <@&286206848099549185> can anyone pls help validate my ans?

<@&286206848099549185>

[midnight wedge]

Hi

[ebon marlin]

<@&286206848099549185>

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[formal kettle]

so i start of with x - x/1 =7

[strange lichen]

no

[formal kettle]

or should it be x/1 - x

[strange lichen]

no

whats the reciprocal of x

[formal kettle]

i meant x - 1/x = 7

[strange lichen]

yes

[formal kettle]

right

can i try to set the equation to 0?

or what should i be looking to do

[strange lichen]

write out, as an expression, what exactly ur question wants u to calculate

[unkempt drift]

Yes
It's a quadratic in disguise

But this time you're not looking to find the roots

[formal kettle]

x^2 + (1/x)^2 = ?

[formal kettle]

so no reason to do that then

i mean the roots would be x

[unkempt drift]

Yeah so don't find the roots (unless you want more steps)

[formal kettle]

and with x i can find the sum

[unkempt drift]

Nope, the sum of their squares

[formal kettle]

oh true ture

x^2 + (1/x)^2 = ?

ok so im practicing for a timed test

so i definately do not want more steps

if i shouldnt look for roots

what am i doing

system?

[unkempt drift]

Ah wait so yeah there's a way to get from x - 1/x = 7

To x^2 + 1/x^2

Can you see what that might be given the ^2?

[formal kettle]

squaring both sides?

[unkempt drift]

Yep, exactly

[formal kettle]

45

[unkempt drift]

Nope

[formal kettle]

49

i mean

[unkempt drift]

You're moving too quickly

What's the left hand side squared?

[strange lichen]

inb4 freshdream

[unkempt drift]

kekw.

[strange lichen]

ok nvm

[formal kettle]

x^2 - 1/x^2 = 49

[unkempt drift]

Jesus Christ

[strange lichen]

im him

[unkempt drift]

$(a + b)^2 \ne a^2 + b^2$

[potent lotus]

south

[formal kettle]

oh shit

[unkempt drift]

mhm

[formal kettle]

it happens

whatever rman

[strange lichen]

@formal kettle btw this was me predicting u were about to do that

[formal kettle]

damn

bro im sick and tired asf

give me a break

but yes

[strange lichen]

nah we just playing dw abt it

[formal kettle]

yeah

ok so (x - 1/x)(x-1/x) = 49

[strange lichen]

right, now distribute

[formal kettle]

(x^2 - 2 + 1/x^2) = 49

[strange lichen]

ye

[strange lichen]

now do this

[formal kettle]

ohhhhh

alright

hella embarrased rn

but thanks guys

.close

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[formal kettle]

a, b, c are real numbers. If then which of the following may be true

[vivid vine]

prolly you need to move everything to one side

to get 0 on the other

[formal kettle]

i found that B and E might true

but idk

[vivid vine]

give an example

for both

[formal kettle]

b) 1/(3-2..5) is greater than 1

[left sorrel]

"may be true"?

[formal kettle]

e) 1^2 is equal to or lesser than 1*2

[vivid vine]

ok i need some more context

[formal kettle]

yes

[left sorrel]

b) might be true? consider a - b > 1

[vivid vine]

are there some restriction to a,b,c?

[left sorrel]

none of that is true

even the "may be true" just makes it subjective

[vivid vine]

ok

[formal kettle]

i typed it myself cus its wrongly formatted

[vivid vine]

im guessing they ask for all a,b,c

[formal kettle]

but that is the original question

and you can only pick one

[vivid vine]

ok

[formal kettle]

i think these quesitons are just badly designed

there isnt even a c

[left sorrel]

but none of this is true though lol

[formal kettle]

how

is what i said wrong?

[left sorrel]

what did you say again?

[formal kettle]

.

for example

[vivid vine]

ok maybe start one by one

[left sorrel]

for b) you consider any case where a- b > 1

and that isn't true

is it?

a = 3 and b = 1 for example

[formal kettle]

it can be true

[left sorrel]

in that case everything can be true

[vivid vine]

no

look at a

[formal kettle]

yeah

[vivid vine]

in a there is no example

that it is true

[left sorrel]

sure

[formal kettle]

yes

i think the question is just made wrong

idk

[left sorrel]

i was thinking a = b = 0

[violet prism]

Is it me or I see no c here

[left sorrel]

then you have undefined for A)

[formal kettle]

thats what im saying

[violet prism]

anyways

[left sorrel]

which you can't compare

but okay

[violet prism]

A is pretty straightforward
B can be
C can be
D straightforward
E can be....?

It's a bit too generic of a question tbh

[left sorrel]

there's no correct answer

[formal kettle]

yeah

[left sorrel]

[formal kettle]

i think ima just leave it like it is

[violet prism]

Like, 3 correct, 2 wrong

[left sorrel]

they're all false

[formal kettle]

and can only pick one

[left sorrel]

none of them is true

[violet prism]

it really depends here

We don't really know if any of them is true

But 2 are certainly false

[left sorrel]

if it depends then everything is true

[formal kettle]

if that were they case there usually is a "none" option

[left sorrel]

for a) consider b^2 = 0 so you have 1 > 1 which is false

[formal kettle]

same with all

alr guys i think its clear it doesnt make sense

[violet prism]

A and D are 100% false

[formal kettle]

just leave it here

.close

[calm coral]

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[left sorrel]

choose C

if u had to choose

I'm p sure that's what they want you to choose

[formal kettle]

alr thanks

[left sorrel]

that almost always works

actually always does

[formal kettle]

hahah alr

[left sorrel]

unless a = b = 0

[calm coral]

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[topaz niche]

shouldn't this be non differentiable in 3 points?

do the end points of the given codomain also count?

[calm coral]

@topaz niche Has your question been resolved?

[indigo knot]

I think its only not differentiable at x=0, x=-1, and x=1.

all points are solid implying they exist within f(x) too, (including the codomain).

thus there should be no other reason why they are not differentiable

[topaz niche]

points being solid implies continous

but it cant be differentiable if there is a sharp turn

[calm coral]

@topaz niche Has your question been resolved?

[calm coral]

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[wise pine]

I got it wrong but I want to know how to do it and what I did wrong, but basically just need someone to explain it to me

[remote mural]

is that a + or a -

in between?

[remote mural]

ok so

suppose you

have

-(a+b)

after you remove parantheses what will you have

note: - can be treated as a -1

[wise pine]

-1 + a + b ?

[remote mural]

nope

so when you have parantheses

it means the terms are locked there

and all you can do is multiply them 1 by 1 out of the parantheses

so it will be -1(a+b)

take them one by one

-1 * a + -1*b

[wise pine]

Oo so basically multiplying them by -1

[remote mural]

if you have -

if you have + multiply by +1

[wise pine]

Ok so there’s always a 1 there even if it’s adding or subtracting

[remote mural]

yes

[wise pine]

And then once you do that you just add like terms and solve like normal

[remote mural]

yes

[wise pine]

Hol up le me see if I get this right

[remote mural]

sure

[wise pine]

Ok I think I got it

Is it 2x

@remote mural

😄

Thank you for the help

.close

[calm coral]

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[calm coral]

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[rough ferry]

how to find radians

[rough ferry]

I answered it as $\frac{-4\pi}{6}$ but that's wrong.

[potent lotus]

rynite

[rough ferry]

should be $\frac{-4\pi}{3}$

[potent lotus]

rynite

[rough ferry]

but I don't know why

[civic dirge]

Remember the full circle would be 2π

But this is 4/6 of a whole circle

[rough ferry]

yeah but im still not sure why it's -4pi/3

im new to this

[civic dirge]

I'll ask a completely different question. It does relate to the original though:

Let's say there's 2π pizza. It's been cut into 6 slices. You took 4 of those slices. How much pizza did you get?

@rough ferry

[calm coral]

@rough ferry Has your question been resolved?

[calm coral]

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[vast acorn]

can anyone explain this

[vast acorn]

I thought it was e^x*cosx

or I it may have click

when u derrive something out it doesnt change the inside

[glacial inlet]

well you have a product of two functions

so you have to use product rule

[vast acorn]

which is

[glacial inlet]

$\frac{d}{dx}(f(x)\cdot g(x))=f’(x)g(x)+g’(x)f(x)$

[potent lotus]

y0shi

[vast acorn]

Ive never fully understood d/dx , I only understood derrive, but some have dv/dt

[glacial inlet]

well think of d/dx as just a notation for taking the derivative

[vast acorn]

dr/dt

[glacial inlet]

basically the left side says take the derivative of f(x) times g(x)

[vast acorn]

ye ok, ty. You may help me even understand Dv/dt= dV/dh*dh/dt

anyway your message is noted

very good rule infact

[calm coral]

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[calm coral]

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[neon valley]

I need help with a math induction problem

[cloud nexus]

sure

[neon valley]

https://mathb.in/78099

Here it is

I'm stuck there

[cloud nexus]

Prove using mathematical induction: $\sum_{k=1}^n\frac{1}{k^2} \le 2-\frac{1}{n}, n \in \mathbb{N}.$

The inequality is true for $n=1$ and $n=2$:

$$\sum_{k=1}^1\frac{1}{k^2}=\frac{1}{1^2}\le2-\frac{1}{1}=1,$$

$$\sum_{k=1}^2\frac{1}{k^2}=\frac{1}{2^2}\le2-\frac{1}{2}=\frac{3}{2}.$$

The hypothesis part:
$$\sum_{k=1}^n\frac{1}{k^2}\le2-\frac{1}{n}\implies \sum_{k=1}^{n+1}\frac{1}{k^2}\le2-\frac{1}{n+1}$$

This is the part where I'm stuck:

I know that
$$ \sum_{k=1}^{n+1}\frac{1}{k^2}=\sum_{k=1}^n\frac{1}{k^2}+\frac{1}{(n+1)^2},$$

so I can write
$$\sum_{k=1}^{n+1}\frac{1}{k^2}=1+\frac{1}{4}+ \cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}\le2-\frac{1}{n}+\frac{1}{(n+1)^2}$$

[potent lotus]

hired

[cloud nexus]

anyway

so whats wrong with just doing sum of n+1 is sum of n + 1/n^2

and going from there

[warped rampart]

subtract $\sum_{k=1}^n\f1{k^2}$ on both sides, then you can use the assumption which will cause the inequality to hold and then from there you can just simplify to get something which is obviously true

[potent lotus]

Duh Hello

[warped rampart]

or wait maybe its not that simple since the negative of the inequality wont quite work like that, hmm

ah but you can still use that [0\leq2-\f1n-\sum_{k=1}^n\f1{k^2}]

[potent lotus]

Duh Hello

[warped rampart]

i think, shouldve perhaps done the problem before responding

[neon valley]

Seriously, idk what to do with my life

I just want to know how to solve this

What

[neon valley]

Do you mean this? $\sum_{k=1}^{n+1}\frac{1}{k^2}=\sum_{k=1}^n\frac{1}{k^2}+\frac{1}{(n+1)^2}$

[potent lotus]

Mithridates

[cloud nexus]

yes

[neon valley]

Then what?

I can't say $\sum_{k=1}^{n}\frac{1}{k^2}=2-\frac{1}{n}$

[potent lotus]

Mithridates

[cloud nexus]

wait

we can do something even simpler

so the entire sum never gets greater than pi^2/6

so $\sum_{k=1}^{n}\frac{1}{k^2}<\frac{\pi^2}{6}$

[potent lotus]

hired

[cloud nexus]

for n < infinity

and so we need to test that this is true for 1, 2, and 3, because after that 2-1/n>pi^2/6

[neon valley]

Is this the only way to go?

[cloud nexus]

theres probably a way with induction

but this is the easiest way

[neon valley]

But I have to solve this using induction

[calm coral]

@neon valley Has your question been resolved?

[calm coral]

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[quick tiger]

any idea of how to calculate this