help-0

[icy dagger]

the 1 in front the sqrt 2 doesnt need to be there right

since it being multiplied by 1

ahh i see

thanks for ur help guys i get it now

.close

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[velvet mango]

I could use some help on question 10

I get the set up involving

P(μ-κσ < X < μ+κσ) where κ = 1, 2, 3

I'm having trouble figuring out what to do when my probability goes beyond 1

i.e. for part a κ = 2

P( 1/2-(2sqrt(1/12)) < X < 1/2+(2sqrt(1/12)) )

[velvet mango]

mb pic sent rotated

i'm assuming i have to do something involving restraining the bounds but im at a lost for how to do it

[lone heart]

@velvet mango Has your question been resolved?

[velvet mango]

<@&286206848099549185>

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[blissful ether]

Not quite sure why Alpha is wrong

[gusty bronze]

Hi

I need help with solving rational inequalities

[blissful ether]

this channel is already taken

go to one of the available ones

[empty ridge]

How did you calculate alpha?

[blissful ether]

Expected Rate of Return - Required Rate of Return

[empty ridge]

Ok, how did you calculate your required return?

[blissful ether]

risk free rate+Beta*(Market rate-risk-free rate)

[empty ridge]

8 + 1.88 * (14-8) =/= 15.50

[blissful ether]

Market return is 12 not 14

I already tried 14 too

just in case

If it was 14 ^

[empty ridge]

... 14.75 - (8 + 1.88(14-8)) = -4.53, right?
E(r) = alpha + rf + beta*(ERP - Rf) :.
alpha = E(r) - (rf + beta*(ERP - Rf))

[blissful ether]

oh yea -4.53 is right

It's weird because I was following a couple cheggs solutions and neither had that step

trying to figure out what I missed

[empty ridge]

In general, for the CAPM: E(r) - rf = a + b (Rm - Rf) {This is what traditionaly we see on books}

Done?

[blissful ether]

Yea I just am trying to figure out where I went wrong

Ah I figured out why

I was calculating required rate of return wrong

Thanks so much for your help! 😄

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[sharp rock]

Hi! I am just looking for some intuition/hint (it is an assignment question so I really would not like an answer, just some idea) on how to rigorously write the answer to a question on Riemann Integration:
If f is decreasing and improperly integrable on [0,inf), show limit of f is 0.

Clearly if f's limit is not 0, you can find an infinite area but how does one actually write about an infinite area. I thought about using the upper/lower riemann sum but I'm not sure how to send it to infinity?

[olive sinew]

say f's limit is some number c > 0. Then f(x) >= c for all x > 0. So take a partition of [0, a], then the inf of each element of the partition is at least c.

(of course the goal here is to prove the contrapositive)

[lone heart]

@sharp rock Has your question been resolved?

[olive sinew]

By the above, the idea is that for each a, the lower integral has at least a rectangle with width a and height c

[vale sapphire]

so for any x, the integral of f over [0,x] is at least cx

which tends towards infinity

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[serene yacht]

hello may i get some help, second question asks a - b and first one asks how many different numbers can n take

[charred flint]

I'd need a translation for the 2nd pic, but for the first one you're mostly just counting how many 4's factor into 27!

which is just the number of 2's in the prime factorization /2

so like 2 gives one, 4 gives one, all the way up to 26 giving one

[alpine sable]

Your guestlist so far has at least 30 guests and you know for sure that 4 people are coming, including yourself. You have a budget with a maximum of $500 to spend. You spent about $200 so far and you figure it will cost about $10 per guest. What is the possible number of additional guests you can invite?

Please anyone

This is hw

And I can’t figure it out

[serene yacht]

@charred flint second one says if xyxy2 is a five-digit natural and if the xy is a two digit natural number then find a - b

[charred flint]

@alpine sable repost this in an available channel like #help-16

[alpine sable]

Kk

[charred flint]

hmm, I'm not familiar with those lines in the pic 2 diagram

[serene yacht]

its division

lines

xyxy2 = xy . a + b if that helps

[charred flint]

cool yeah that makes sense

for pic 1 are there any restrictions on A? I get something like 12

[serene yacht]

@charred flint both n and a are positive

and integers

[charred flint]

hmm oh well I'll try to help with that anyway even though I get the wrong answer

so you're looking for the total number of 2's in the prime factorization

the set {2,4,6,...26} each give you a 2 in 27!, since they're even

[serene yacht]

o so i do 27/2

[charred flint]

but this doesn't use that 4 has two 2's, so you need to count the factors of 4 too

so {2,4,6,...26} + {4,8,12,...24}

and again this doesn't count all the 2's in 8's factorization

if you keep going with this, the number of 2's is the size of {2,4,6...}+{4,8,12....}+{8,16,24} + {16}

does that make sense?

[serene yacht]

yea

[charred flint]

cool, so that's 13+6+3+1=23

so the number of 4's in 27! are 11, half that (rounding down)

oh nevermind that's an answer nice haha

[serene yacht]

lol

cool

[charred flint]

kinda confused about the 2nd pic, you have xy*1010+2, and apparently dividing by xy gives a with remainder b

not sure how a-b only has one possible answer, does the question specify anything else?

[serene yacht]

nope

[charred flint]

ummm, I mean one strategy is to just pick random numbers because the answer should be the same for anything

[serene yacht]

also b < xy is a rule

[charred flint]

I meeeean, just use x=1, y=1

11112 divided by 11 is 1010 with remainder 1

so the answer is 1009

feels super cheaty lmao

[serene yacht]

oo

i have never seen sb solve this like that

but this way is much better ngl

[charred flint]

oops

I guess I see the actual way now

[charred flint]

made a mistake, should be remainder 2 for 1008

but like, we already said xyxy2 is xy*1010+2

so like obviously dividing by xy gives 1010 plus 2 left over

[serene yacht]

got it

[lone heart]

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[small pine]

Hello fellow people

[small pine]

Can someone explain me how to solve this step by step?

[charred flint]

what are you trying to do with that expression, I don't see what's there to solve

[small pine]

oh i forgot to say the most important thing
...well I need to simplify it

[charred flint]

ah sure

so you want everything to be in a big fraction divided by x-1, but right now there's 1- on the outside

1 can be turned into a fraction though, it's just (x-1)/(x-1)

now you can put the two fractions together and solve it

[small pine]

what would be the final result?

[wary stream]

That's your job to find

[small pine]

like i can find it but im not sure, im here to check

its not homework btw, its some practice im doing

[wary stream]

What did you get as a final answer? Post your work

[small pine]

ill do it, gimme a sec

[novel pond]

can someone explain why E[X -E(X)]^2= E[X2] -[E(X)]^2 it is somehting about expect value

[wary stream]

Don't ask in an occupied channel

Read #❓how-to-get-help

[novel pond]

sorry

[small pine]

xd

not sure

[lone heart]

@small pine Has your question been resolved?

[alpine sable]

Uhhh

This is 1 - x^2 + x
———-
X-1

So

[tight locust]

don't write like that lmao

[alpine sable]

2 - x ^2 + x + 1
___________
X??

Idk

[tight locust]

stop it

just use parentheses or latex like a sane individual

[late parcel]

Is this channel open

[blissful ether]

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

just to help w/parentheses

[ocean seal]

@hasty hound

[hasty hound]

@small pine

[small pine]

hmm ye, im here

yes

[blissful ether]

yes

[small pine]

can someone just solve it on paper and send the photo?

wait now that i check what i did, its all wrong. Excuse me, my neurons just shut down xdd

lmao

.close

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[solemn schooner]

hi I don't know how to solve this the anwer is sqaure root of 16 (4) but idk how to get there

[glass lichen]

expand and simplify

[solemn schooner]

idk how

no one teached me that and my exam is tmrw

[raw shard]

@solemn schooner do you know that, for example, (a/b)^2 = (a^2)/(b^2)?

[solemn schooner]

nope

didnt know that

its the square roots thats fucking me up

[raw shard]

so you’re trying to get ((sqrt31)+1)^2

there’s a formula that can help with this too

(x+y)^2 = x^2+2xy+y^2

[solemn schooner]

did the sqrt cancelled the ^2?

[raw shard]

no

[small pine]

ye

oh this is the actual steps taking place

[solemn schooner]

idk anything abt sqrt rool thats the thing

[raw shard]

wait i didn’t know which power of 2 you were talking about

[raw shard]

yes it canceled the power of 2 for the c

i just sent 2 formulas that can help you

(a/b)^2 = (a^2)/(b^2) and (x+y)^2 = x^2+2xy+y^2

yes i know there are lots of letters

they’re just placeholders

[solemn schooner]

i learned this one (x+y)^2 = x^2+2xy+y^2

[raw shard]

so you want to get (((sqrt31)+1)/2)^2

[solemn schooner]

could you send a pic of you doing a really detailed step by step?

im really tired i cant think properly

[raw shard]

i can just explain it step by step

so what i just sent is equal to (1/4)*(1+sqrt31)^2

i rewrote it a bit so i don’t need as many parentheses

(1+sqrt31)^2 = 1+2sqrt31+31

[solemn schooner]

i got it thank you

this channel is now free

[raw shard]

ok, do .close

to close the channel

[solemn schooner]

,close

.close

[lone heart]

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[charred flint]

the -2 comes from x^2-2x+1

[torn bridge]

Hey is anyone here that can finish a grade 8 work just need an answer

[charred flint]

the (x-1)^2

[silver viper]

shouldnt it be 2(x-1) + 2(y-1)y' = 0

is y a function of x?

[raw shard]

not sure why it wouldn’t be in this case

[silver viper]

in that case derivate of (y-1)^2 should be 2(y-1)y' by chain rule

[lone heart]

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[halcyon plaza]

8. The burning time for a certain type of low-energy lamps in street lamps is stated to be 10,000 hours with a standard deviation of 550 hours. The municipality had 913 lamps of this type. How many of these lamps need to be replaced before they burn 8900 hours?

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@halcyon plaza Has your question been resolved?

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[wicked breach]

Y is directly proportional to x, and y=216 when x=2.
a)Find y when x=7?
b)Find x when y=540?

[misty bridge]

I cannot finish this proof

it says the last line is wrong

[wicked breach]

this channel is occupied

[misty bridge]

oh

[wicked breach]

go to the available channels

[misty bridge]

WAIT, I SEE HOW THIS WORKS, I AM SO SORRY

[wicked breach]

👍

[buoyant edge]

You still need help @wicked breach ?

[wicked breach]

yea

[buoyant edge]

Alright. Do you know what direct proportionality means?

[wicked breach]

ye

im just stuck here

[buoyant edge]

What do you mean

[wicked breach]

216 = k x 2

so k=216/2=108 right?

[buoyant edge]

Yep!

[wicked breach]

alright

i just needed clarification

[buoyant edge]

Sounds good.

[wicked breach]

this is 50% of my grade

.close

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[compact lynx]

https://gyazo.com/e2e58734342322e9f195079e25e4a3ba
i thought this would be P(exactly 5 spades) + P(at most 1 king) - (P(4 spades and king of spades) + P(5 spades no king) + (P(5 spades 1 king of non-spades)). Can someone tell me why we can just ignore the last case?

[compact lynx]

oh wait the calculation being shown is P(A not B)

nevermind. lol

.close

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[compact lynx]

.reopen

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✅

[compact lynx]

.close

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[molten iris]

Hi

[molten iris]

i need help with implicit differntiation

for Calc AB

i dont really understand how dy/dxworks

like in the first question

i thought it wud be y=x^5 which wud equal dy/dx = x^5 which wud be equal to

5x^4 but then why does my teacher multiply it by 1

[polar barn]

pretty sure 5x^4 . 1 is still the same

[lone heart]

@molten iris Has your question been resolved?

[molten iris]

yea i know

its the same but in a different problem

its not gonna be the same

[rough compass]

@molten iris just illustrating that chain rule was used on x

d/dx of x is just 1 tho which is why ur teacher multiplied by 1

[polar barn]

this

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[coral inlet]

Is (logx)^3m-1 equal to (logx)^1-m

[lone heart]

@coral inlet Has your question been resolved?

[vale wigeon]

@coral inlet no, not in general. why would you think these are equal?

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[coral inlet]

Wait sorru

It was

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[lone heart]

@alpine sable Has your question been resolved?

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@white meteor Has your question been resolved?

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[oak ferry]

I used the rule for it to get two possible values for n one value as an integer which is the answer and another value as a decimal which is refused value

But looking for another way to solve it

<@&286206848099549185>

[fringe ember]

Expand both the binomials, using the binomial theorem

subtract the two

[oak ferry]

But how will I know how many terms will cancel

[fringe ember]

i think you'll be left with only even terms

[oak ferry]

It depends whether n is even or odd

No

[fringe ember]

all right

[oak ferry]

A question from egyptian book

[fringe ember]

find the general formula for both

[oak ferry]

2(Odd terms)

[fringe ember]

in general, $(a+b)^{n} - (a-b)^{n}$ contains n/2 terms, rounded down

[ocean seal]

the.dorkest.knight

[oak ferry]

This (a+b)^n ++++++ (a-b)^n tho

Not ------

[fringe ember]

oops sorry my bad

will correct

how many terms does (a+b)^3+(a-b)^3 contain

[oak ferry]

2 terms

Because it equals 2(sum of odd terms of (a+b)^3

?

U r with me?

<@&286206848099549185>

[alpine sable]

@oak ferry I would first find the number of terms for n=1, n=2, and n=3, and then find a pattern for the number of terms for each n

Then I'd generalize this pattern into a formula and then apply that formula to find the n value for 11 terms

[oak ferry]

What I used was like this

In case of (a+b)^n +(a-b)^n
Number of terms if n is even
=n/2 +1
If n is odd = (n+1)/2

I solved for the two assumptions

[alpine sable]

Ah that works too

[oak ferry]

In case of 2n is odd n was 10.5

But I was looking for another way other than the rule

The problem is knowing if n is even or odd

[alpine sable]

hmm I'm assuming that n can only be integers

[oak ferry]

There is another way but also considering n two times one as odd and the other as even

(a+b)^n + (a-b)^n
Equals 2(sum of odd terms)

So in our case number of terms of first Equals 2n+1 if n is odd then 2n+1 is odd too
So we have 2(T1+T3+T5+.....+T(2n+1))

I m consfused

If we were counting from T1,T2,T3 till T(2n+1) we get 2n+1 terms
We started with odd and ended with odd so number of odd bigger by 1 from number of even

If we have 1,2,3,4,5,6,7 number of odd
= 4 =(7+1)/2

So number of odd = (2n+2)/2

=n+1=11

N=10

Yh that's the answer

But this answer confuses me

What always confuses me is when to add 1 or talk half directly and these things

[alpine sable]

I think n has to be an integer in this case, so whichever method gives an integer n for this question should be correct

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[wispy olive]

Did they mean 7! here since "CARAVAN" is a 7 letter word?

[small bear]

If the question is as simple as find the permutations of Caravan, then yes. Author was probably really tired when writing that lol

[wispy olive]

Lol.

Yeah it was that.

Cause like every other word's permutation seems to be the number of letters in that word.

[small bear]

Just write an additional s in the book, make it plural and problem fixed :)

[wispy olive]

xD.

Nice way.

Thanks.

.close

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[alpine sable]

set theory problem here

[alpine sable]

my answer came out to be 24
but the answerkey shows it to be 60
what are your views frnds?

[vale wigeon]

the text in your image is a bit too small for me to comfortably read

so we have three classes: math, physics and chemistry

and the data are

Math 100
Phys 70
Chem 40
Math & Phys 30
Math & Chem 28
Phys & Chem 23
All three 18

@alpine sable i take it you have a venn diagram you could show us?

[lone heart]

@alpine sable Has your question been resolved?

[vale wigeon]

bruh

[alpine sable]

hey unfortunately i am on pc and doing problem in notebook so can't take a pic and show it sorry for inconvenience 😦

altho i did made venn diagram

[vale wigeon]

do you have any image editing software on your pc?

such as MS paint?

or alternatively

you could tell me what the numbers are in your venn diagram

can you tell me the values of a through g in your venn diagram? @alpine sable

[alpine sable]

yeahh i can

[vale wigeon]

okay, then please do so

i'll be right back, but i'll compare my venn diagram against yours once i'm done.

[alpine sable]

e=18

b=30

d=28

f=23

a is coming out to be 24

[vale wigeon]

ok, yeah, your b is wrong

the count of 30 people who took math and phys includes both those who took only those two classes and those who took chem too!

it's b+e=30, not b=30

same goes for d and f

[alpine sable]

got it

[lone heart]

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• Show your work, and if possible, explain where you are stuck.
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[hallow ocean]

pls help me:
prove that this expression >0 with every x

[glass lichen]

What have you tried...?

[hallow ocean]

i tried to make x^2 - 4x + 2 become x^2 - 4x + 4 -2 so that i can do x^2 - 4x + 4 = (x-2)^2 > 0 but y^2 get me in trouble

y^2 - 2 that left can be <0 so i dont know what to do next

@glass lichen ?

also i tried:
x^2 - 4x + 2
= x^2 - 2x - 2x + 2
= x(x-2) - 2(x+1)

but it didnt work

what?

let me try

uhh correct it is

i tried x=1 and y=0 but well

<0

i think the the question has a problem

.close

[lone heart]

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[hallow ocean]

.close

[lone heart]

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• Show your work, and if possible, explain where you are stuck.
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[quick echo]

how do you simplify the answer?

[quick echo]

I should be getting this right but it's non in the choices

[alpine sable]

how u got this?

[vale wigeon]

you have $(-1)^{n-1}$, which is equivalent to $(-1)^{n+1}$

[ocean seal]

Ann

[vale wigeon]

(because (-1)^2 = 1 after all)

but also

(2 Points)

this really looks like a test

[quick echo]

this "was" a test earlier but didn't get to answer it

dunno how to simplify it

[vale wigeon]

ok, so it's a past test.

i already said what i intended to say, however.

[quick echo]

.close

[lone heart]

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[lone heart]

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• Show your work, and if possible, explain where you are stuck.
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• Be polite and have a nice day!

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[torn zenith]

@vale sapphire tu es la ?

[vale sapphire]

oui

[torn zenith]

C'était pour faire la q3 de l'exercice 2

Je remet

Déjà la q2 de lex 2 il faut faire la réciproque de Pythagore donc aucun problème

[vale sapphire]

c'est pour la construction au compas le problème?

[delicate cargo]

Oui comment on construit le projeter ?

[vale sapphire]

un petit dessin trouvé sur internet

On utilise le compas pour tracer deux points sur la droite qui sont à la même distance de A

[torn zenith]

Ça c'est un projeté ducoup

[vale sapphire]

Ce que ça veut dire c'est que la médiatrice de ces deux points elle va passer par A

Et donc leur milieu c'est le projeté

ouais

[torn zenith]

Attends 2 sec

[delicate cargo]

J’ai fait ça mais je suis pas sur de moi

J’me rapelle plus ou il faut piquer avk le compas

[torn zenith]

Fallait pas le faire dans l'autre sens ?

[vale sapphire]

c'est surtout que là tu viens de projeter B

pas E

Tu projettes E sur la droite AC

[torn zenith]

Faut partir de e

Pas de b

[vale sapphire]

Donc d'abord tu piques E, et tu traces un cercle qui coupe AC en deux points

Ensuite tu trouve le milieu de ces deux points et t'as fini

[torn zenith]

En Gros on trace la perpendiculaire de ac par rapport à e

ah ok

[vale sapphire]

Oui c'est exactement ça

Le projeté orthogonal c'est juste le point sur AC qui est le plus proche de E

et le chemin le plus court possible d'une droite à un point c'est une droite perpendiculaire

[delicate cargo]

Comme ça ?

[vale sapphire]

Ouais sans doute

Ensuite tu traces le milieu

[torn zenith]

Mais faut pas tracer la perpendiculaire en partant de E par rapport à ac ?

[delicate cargo]

Faut pas plutôt repiquer sur les deux point que je vient de faire ?

[vale sapphire]

c'est ce qu'on fait

oui, pour tracer le milieu on pique sur chacun des points

on trace deux arcs qui se coupent, et le segment qui passe par leur intersection c'est la médiatrice

(= la droite perpendiculaire à AC passant par E)

[torn zenith]

Mais il a pas fait la perpendiculaire à partir de e

[delicate cargo]

C’est ça ?

[vale sapphire]

ah c'est pas ce qui est demandé

tu comprends mal ce qu'est un projeté orthogonal

[torn zenith]

c'est bon ce qu'il a fait ?

[vale sapphire]

ça doit ressembler à ça

[torn zenith]

mais enfaîte c parce que c a faire au compa

sans le compa oui d'accord

[vale sapphire]

mais oui il l'a fait correctement

[torn zenith]

pour le 3b on utilise la trigo ?

[vale sapphire]

ouais

tiens pour bien se convaincre que la construction elle marche

on va dire que X et Y sont à la même distance de H

Est-ce qu'ils seraient pas à la même distance de E?

[torn zenith]

Oui

Si

Vu que c'est perpendiculaire

[vale sapphire]

et donc t'es d'accord que t'as un cercle de centre E qui passe par X et Y?

[torn zenith]

C'est juste la longueur qui change

Ouais

calcule pas la qualité

C'est bon ?

[vale sapphire]

ouais trkl

tu t'en sors bien

[torn zenith]

Pour le b

[vale sapphire]

c'est juste une histoire de comprendre ce qu'on fait

trigo mec, pose la formule et réfléchis pas trop

[torn zenith]

Oui mais je sais plus si c'est à arc

ou sans ça

ATt

[vale sapphire]

la formule du cosinus c'est adjacent/hypothénuse

y'a que ça dont il faut se souvenir (et les formules correspondantes pour sin et tan)

Après si veux l'angle tu applique arccos

[torn zenith]

je sais ca tqt

soh cah toa

Sinon

sinus : opposé/hyp

Etc

Pour trouver l'angle de c on fait le sinus ducoup ?

[vale sapphire]

vu que tu connais toutes les longueurs tu choisis

[torn zenith]

sinus : longueur du côté opposé de à/longueur de lhypotenuse

[vale sapphire]

perso je préfère cos vu que tout est juste à côté

mais c'est toi qui vois

[torn zenith]

oui c'est vrai, on va prendre cos

Donc cos Â: 6/7,2

cos(6/7,2)

Ça fais ≈0,99

°

[vale sapphire]

 = arccos(6/7,2) attention

si tu veux inverser le truc c'est arccos, pas cos

[torn zenith]

Oui voilà

[torn zenith]

Je me disais bien

avec*

arc

[vale sapphire]

la formule c'est cos A = adj/hyp

mais si c'est A que tu veux, il faut inverser, d'où A = arccos(adj/hyp)

[torn zenith]

Oui

Arccos(6/7,2)

≈33,55

et pour le 3c

Faut déduire la distance du pont e à la droite ac

Trouver la valeur approchée en 10-²

[vale sapphire]

Regarde le triangle EFA

Tu connais l'angle A, tu connais AE

Et ce que tu cherches c'est EF

T'as tout ce qu'il faut

[torn zenith]

On utilise la trifo

Trigo

[vale sapphire]

ouaip

[torn zenith]

Sans arc cette fois ci

[vale sapphire]

de toute façon pour les triangles rectangles c'est simple, à partir du moment où t'as une longueur et une autre donnée on peut trouver tout le reste

j'te dis, essaye pas de te dire "ah est-ce qu'il y a un arc ou pas Aaaa"

[torn zenith]

Ya pas darc

Vu que on calcule une longueur

[vale sapphire]

parce que un jour tu vas te gourrer si tu comprends pas ce que tu écris

souviens-toi juste de par exemple cosA=adj/hyp

[torn zenith]

Ea c lhypotenuse on est d'accord ?

[vale sapphire]

et après pour passer cos de l'autre côté ça devient arccos

c'est tout

[torn zenith]

Je connais soh cah toa

C'est plus simple

[vale sapphire]

ça marche aussi

mais y'a pas d'arc dans sohcahtoa

si tu passes une fonction de l'autre côté par contre y'en aura un

[torn zenith]

Une fonction ?

[vale sapphire]

cos, sin et tan c'est des fonctions

elles prennent une valeur et elles t'en donnent une autre

[torn zenith]

ah dac

[vale sapphire]

donc voilà, essaye pas de retenir 20000 cas dans ta tête, pars des formules simples et comprends comment tu les manipules

[torn zenith]

mais je sais pas quel formule utiliser

je connais l'angle a

[vale sapphire]

tu connais l'hypothénuse, et l'angle A

[torn zenith]

Je connais la longueur ea

[vale sapphire]

quel longueur tu veux calculer?

[torn zenith]

Ah bah non

Je connais pas

[vale sapphire]

si

E c'est le milieu

[torn zenith]

Oui donc 3

je veux calculer ef

[vale sapphire]

dans le triangle rectangle c'est quoi EF?

adjacent ou opposé?

[torn zenith]

Adjacent

[vale sapphire]

à A?

[torn zenith]

Oui

[vale sapphire]

adjacent ça veut dire "qui touche"

[torn zenith]

Il touche a

[vale sapphire]

est-ce que EF il touche A

on a le même dessin sous les yeux ou?

[torn zenith]

Ah non j'ai cru que tu parlais de ea

[vale sapphire]

ah ok

[torn zenith]

Non ef c'est l'opposé

[vale sapphire]

EA c'est l'hypothénuse fais attention

en fin de compte, t'as besoin de relier A à l'hypothénuse et l'angle opposé

[torn zenith]

ouais

[vale sapphire]

c'est quelle formule qui fait ça?

(1 chance sur 3)

[torn zenith]

Sinus

[vale sapphire]

correct

maintenant tu la poses, et tu réarranges pour obtenir EF

Y'a une chose qu'il faut retenir de tout ça

Et c'est vraiment quelque chose qui est fondamental en maths

Dans d'autres matières aussi, mais c'est surtout en maths qu'il y a des blocages

Il faut comprendre ce que tu fais.

[torn zenith]

Ouais

[vale sapphire]

L'essence des maths c'est de comprendre et de poser de manière un peu plus rigoureuse les choses qui nous entourent.

[torn zenith]

Mais faut s'entraîner aussi

c'est pas comme les autres matières

[vale sapphire]

Les formules à apprendre par coeur sans les comprendre c'est pas des maths

Après il y en a certaines qu'on apprend parce qu'on va pas les redémontrer à chaque fois

[torn zenith]

on connais af ?

[vale sapphire]

af?

[torn zenith]

La longueur af

Non on ne connais pas

[vale sapphire]

nope

[torn zenith]

J'ai une idée attend

[vale sapphire]

Les maths se simplfient grandement quand on commence à faire des choses parce qu'on les comprends plutôt que parce qu'on les a vaguement mémorisées et que "c'est le cours".

et toujours se souvenir de qu'est-ce qu'on fait

Tu vois par exemple l'histoire de AE. Faut garder en tête c'est E le milieu, pas un point quelconque.

[torn zenith]

oui

j'ai fais arcsin(3÷33,55)

≈5,13 cm

= ef

c'est ça ?

[vale sapphire]

ok mec explique moi

montre moi toutes tes étapes

la formule que t'as utilisée en toutes lettres

[torn zenith]

ok c'est pas ca mdr

bah je fais le sinus

J'ai lhypotenuse

L'angle a

[vale sapphire]

écris-la sur discord

[torn zenith]

Sinus : opposé/hypotenude

[vale sapphire]

ok

[torn zenith]

nuse

[vale sapphire]

maintenant écris ça avec les angles et les noms des longueurs

l'expression mathématique quoi

[torn zenith]

opposé c'est ef

attends jsuis perdu

[vale sapphire]

ça c'est bon

l'hypothénuse c'est quoi

[torn zenith]

3

[vale sapphire]

le nom

[torn zenith]

ea

[vale sapphire]

l'angle il s'appelle comment?

[torn zenith]

L'angle de ?

[vale sapphire]

que t'utilises dans la formule

[torn zenith]

À

Â

[vale sapphire]

ok

Donc la formule au final c'est sin(A)=EF/EA

On est d'accord?

[torn zenith]

Oui

sauf que j'ai mis langle

Ahhh

Faut faire un produit en croix

[vale sapphire]

c'est important de réfléchir avec les noms, pcq si t'as que les valeurs, tu sais plus qui est quoi, et tu peux rapidement te perdre

oui tu peux le voir comme ça, mais multiplier les deux côtés par EA ça marche aussi

[torn zenith]

Sin(33,55/1 ef/3

J'ai un problème

Jsuis perdu

[vale sapphire]

recommence par repasser avec les noms

j'te dis, c'est un coup à se perdre de garder les chiffres

les variables c'est pas une lubie des matheux, le but c'est quand même de donner un nom aux choses desquelles on parle

[torn zenith]

Sin: opp/hyp

[vale sapphire]

ahem

sin(A)=EF/EA

je parle de ça quand je dis les noms

[torn zenith]

J'allais y venir

[vale sapphire]

on y est déjà venu de toute façon

maintenant, tu veux trouver EF

[torn zenith]

c'est pour les détails

oui

[vale sapphire]

comment tu fais pour avoir EF seul d'un côté du signe égal?

[torn zenith]

on le déplace

[vale sapphire]

(si tu écris l'équation en remplaçant les lettres par les valeurs je vais faire une syncope)

[torn zenith]

mais ça dépend

si ef est pas tout seul

On soustrait l'autre valeur

Ou addition à gauche du égal

additionne

[vale sapphire]

Si tu as x/5=3, c'est quoi x?

[torn zenith]

X=15 ?

[vale sapphire]

oui

3*5

[torn zenith]

oui

[vale sapphire]

quand on passe un truc divisé de l'autre côté on le multiplie

[torn zenith]

oui

c'est ça

[vale sapphire]

tu peux le voir comme x=5*x/5=5*3

maintenant revenons à nos moutons

si tu as EF/EA=sin(A), c'est quoi EF?

y'a rien qui a changé, juste les noms

[torn zenith]

genre on donne les valeurs ?

Ah

L'opposé ducoup

[vale sapphire]

non pas les valeurs

EF = quoi

c'est strictement la même chose que ce qu'on vient de faire

si tu vois quelque chose de différent ici à part les noms, je t'en prie dis-le moi

[torn zenith]

bah

C'est une distance je sais pas

Ah ok

[vale sapphire]

mais une distance c'est un nombre comme un autre, les règles des maths n'ont pas été changée derrières ton dos par des matheux maléfiques au moment où on a commencé à écrire les choses avec des lettres

[torn zenith]

Ef on a pas encore sa valeur

[vale sapphire]

mais on peut l'exprimer en fonction d'autre chose

je répète la question

[torn zenith]

l'opposé

mais c un nom

[vale sapphire]

et alors, ça représente quand même un nombre

le but des variables c'est de pouvoir manipuler des nombres sans savoir forcément ce qu'ils valent

mais ça reste des nombres

dont on connaît pas forcément la valeur

[torn zenith]

on sait pas le nombre justement

On cherche la distance

[vale sapphire]

pour EF oui

[torn zenith]

X ?

[vale sapphire]

je réitère EF/EA=sin(A)

comment tu fais pour passer EA de l'autre côté

[torn zenith]

On le multiplie

[vale sapphire]

OUI

[sweet gulch]

Guys

[vale sapphire]

EF = EA×sin(A)

[torn zenith]

oui

[sweet gulch]

What's the answer on

[vale sapphire]

go somewhere else

this channel is already occupied

[sweet gulch]

Oh alr

[torn zenith]

yes

[vale sapphire]

je disais donc

EF = EA×sin(A)

on est arrivé à ça

[torn zenith]

ef = ea x sinA

[vale sapphire]

Mais tu connais EA

Tu connais A

Il te reste plus qu'à sortir ta calculette

[torn zenith]

A la 33,55

c'est *

aaaah ok

[vale sapphire]

https://cdn.discordapp.com/attachments/490557019623915520/905798212382760990/20211104_112458.jpg je remets ça ici

tu trouves combien au final?

[torn zenith]

Mais je trouve 100,65 a chaque fois

tout à l'heure j'ai trouvé ça et ca peut pas être ça

[vale sapphire]

t'as bien fait 3*sin(33,55) ?

[torn zenith]

J'avais fais sin(33,55x3)

Je trouve 1,65

[vale sapphire]

oui

sin(ab) ≠ a*sin(b)

[torn zenith]

oh d'accord

[vale sapphire]

ça marche pas du tout fais attention

par exemple

sin(2×45°)=sin(90°)=1

mais 2*sin(45°)=√2

[torn zenith]

faut mettre le chiffre avant en gros

[vale sapphire]

surtout, t'as pas le droit de passer quelque chose à l'intérieur de sinus

[torn zenith]

d'accord

[vale sapphire]

c'est comme si j'écrivais 5*1/2 = 1/(5*2) en passant le 5 à l'intérieur

[torn zenith]

ouais

d'accord

[vale sapphire]

ça marche pas trop 😅

[torn zenith]

ouais j'ai vu

Du coup la faut une valeur approximative à 10-²

1,65 cm ducoup

[vale sapphire]

au juste pour la valeur approximative faut arrondir ou pas?

[torn zenith]

Valeur approchée à 10-² près

[vale sapphire]

1,129 faut arrondir à 1,12 ou à 1,13?

faut prendre la valeur la plus proche

[torn zenith]

10-² donc c'est les centièmes ?

[vale sapphire]

oui

[torn zenith]

1,13 je pense

Mais j'arrondis à 1,66 ducoup

[vale sapphire]

oui

on prend ce qui est le plus proche

0,9 on l'arrondirait à 1, pas à 0

[torn zenith]

1,657

[vale sapphire]

ouaip

[torn zenith]

donc 1,66

[vale sapphire]

👍

[torn zenith]

dans ce cas ef = 1,66 cm ?

[vale sapphire]

oui

En conclusion, faudrait que tu t'entraînes un peu à manipuler les variables

[torn zenith]

ouais mais j'avais mal compris

[vale sapphire]

je te pose un mini-exo

mais honnêtement c'est une règle de calcul littéral (et mental) qui sert absolument tout le temps

donc vraiment, c'est un truc à retenir

[torn zenith]

d'accord

[vale sapphire]

Si a,b,c sont des nombres et que a=b/c. b est égal à quoi? c est égal à quoi?

je veux une expression littérale (de toute façon je te donne pas de valeurs)

[torn zenith]

b=ac

B =à x c

[vale sapphire]

oui

c est un peu plus dur, mais une fois que t'as compris c'est un truc qui est très utile

parce que bon, les divisions c'est pas non plus l'opération la plus rare

surtout en physique

[torn zenith]

ouais

bon du coup j'ai finis mes exercise

exercices

merci beaucoup vraiment

merci d'avoir pris du temps pour moi

[vale sapphire]

tssk, t'as oublié l'expression de c

elle est tout aussi importante

mais pas de problème

[torn zenith]

c'est quoi

L'expression de c

[vale sapphire]

tu sais que ac=b

[torn zenith]

oui

[vale sapphire]

est-ce que tu peux faire une opération pour ne te retrouver qu'avec c

[torn zenith]

diviser

non

[vale sapphire]

par quoi?

fais-toi confiance

(je te rappelle que si tu divises n'importe quel nombre par lui-même ça fait 1)

[torn zenith]

bah

a

[vale sapphire]

Oui

[torn zenith]

car cx1 = c

[vale sapphire]

et tu te retrouves avec c=b/a

remarque qu'on est parti de a=b/c

[torn zenith]

oui

[vale sapphire]

t'as vu comment on a pu juste échanger les deux?

[torn zenith]

ouais mais ca retourne le cerveau mdr

[vale sapphire]

on s'y habitue

je pense qui si tu cherches sur internet tu peux trouver des exos interactifs si t'es assez déter

mais ça vient avec l'habitude

le maitre mot c'est que x/x=1 et x-x=0

[vale sapphire]

d'ailleurs ça marche aussi avec la soustraction

[torn zenith]

ouais

[vale sapphire]

si a = b-c, c = b-a

[waxen oxide]

someone english?

[vale sapphire]

in another channel

[torn zenith]

ouais

d'accord

[waxen oxide]

where?

[vale sapphire]

read #❓how-to-get-help

[torn zenith]

go to help 12

[vale sapphire]

voilà, ça c'est des trucs dont tu peux te convaincre, et vraiment à partir du moment où tu peux ne plus réfléchir à ces manipulations pendant 1 minutes

ça dégage la voie et tu peux te concentrer sur la compréhension

c'est nécessaire de prendre le temps

[torn zenith]

d'accord

[vale sapphire]

j'ai vu personnellement que l'enseignement des maths en France c'est vraiment pas ça

moi ça m'a pas arrêté, mais moi j'ai un grain aussi

[torn zenith]

oui niveau enseignement en France on est dernier

quasiment

[vale sapphire]

donc voilà, je les connais les lacunes et les problèmes courants

En somme, bien comprendre ce qu'on fait avec les variables

[torn zenith]

et sinon tu as quel age ?

oui

[vale sapphire]

18

[torn zenith]

ah d'accord

[vale sapphire]

et pas essayer d'appliquer les formules, mieux vaut les retrouver en manipulant

[torn zenith]

ouais

d'accord merci bcp

je ferme le salon ?

[vale sapphire]

ouaip tu peux

[torn zenith]

et à bientôt peut-être

[vale sapphire]

qui sait

[torn zenith]

.close

[lone heart]

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[neon latch]

Hii, can someone help me with a graphic? It's about underdamped harmonic oscillator. I need to do a graphic of |Ts| vs w(transfer function module vs angular acceleration)

[neon latch]

And I think I'm doing it wrong

I also need to do a phase vs w graphic

I've already found the Ts and the roots

I'm getting this as a |Ts| vs w

(I gave r, k and m initial values in order to be underdamped )

And this as a phase vs w:

Can someone tell me if I'm doing it wrong ?

I can send the deduction if needed

[lone heart]

@neon latch Has your question been resolved?

[neon latch]

And s = w initial

<@&286206848099549185>

[lone heart]

@neon latch Has your question been resolved?

[bronze heron]

@neon latch I'm not really sure what this is, what is the context ?

have you identified the cutoff freq ?

what is s ?

[neon latch]

In Ts? Is the variable of the transform function

In the graphic s is w0, sorry it was confusing

[bronze heron]

so you're using a laplace transform ?

[neon latch]

Yes, in order to obtain the transfer function

[bronze heron]

I am a bit confused by the way you treat your variables, how do you get from your laplace transform to frequency domain where your variable is w

[neon latch]

I did the lapace transform of the derivatives

[bronze heron]

alright

the next step might be to find the break frequency

[neon latch]

Ok, how can I do that? Sorry, my main language is not English, I don't know if I 've heard about that

[bronze heron]

this https://en.wikipedia.org/wiki/Harmonic_oscillator#Damped_harmonic_oscillator

here they call it undamped frequency

ok wait

s = iw

[neon latch]

Ok

[bronze heron]

so you get that Ts is a complex number

[neon latch]

Yes

[bronze heron]

so replace s with jw in your transfer function

and take the module

[neon latch]

Yes, i did it

I took the module of the two roots

Separately

And got to this

[bronze heron]

yeah

[bronze heron]

but when you computed the roots

you considered the variable to be s

wich is jw

however your variable really is w

[neon latch]

Okay

I have to take the module of the (S-So) with s = jw

Then take the complex part as the complex part of the root + w right?

so I can do sqrt(a^2 + b^2)

with b = w + complex part of root

?

[bronze heron]

oh ok don't take the roots

I was trying to understand how you ended up with w⁴

[neon latch]

me too lol, I feel it is wrong

[bronze heron]

ok so replace s with jw

[neon latch]

ok

[bronze heron]

in the first expression on Ts

and then take the module

[neon latch]

ok

[bronze heron]

because when you replace s, you get a clear understanding of what's in real and what's in the complex part

[neon latch]

ok, then I take the module of that?

[bronze heron]

yes

[neon latch]

ok

wait, I have to find the roots first

right?

[bronze heron]

no no

no roots 😉

[neon latch]

ohh hahaa okok

but I can't distribute the module in sums

🤯

[bronze heron]

$$|Ts| = \dfrac{1}{sqrt{(k-mw^2)^2+(rw)^2}}$$

[ocean seal]

Grubinski

[lone heart]

Channel closed

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[neon latch]

.reopen

[lone heart]

✅

[neon latch]

ohh thanks, I´ll try with this

[bronze heron]

hahh i hate tex, you get the idea

the denom is under the root

[neon latch]

ahhhh yes, I see

[bronze heron]

are you supposed to graph this by hand or on a computer ?

[neon latch]

in a computer

[bronze heron]

ok, because by hand it can be tricky, so they are techniques to get an idea of the curve, that's when having the break freq is useful

[neon latch]

thanks!

I have a classs now, but I'll try this

can I write you later?

👉 👈

[bronze heron]

yeah

[neon latch]

thank you very much!

.close

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[topaz flower]

The half-life of radium-226 is 1590 years.
(a) If a sample has a mass of 150 mg, find a function that models the mass that remains after t years.
(b) Find the mass that will remain after 1000 years.
(c) After how many years will only 50mg remain?

[topaz flower]

How would I go about solving this problem, not really sure how to start

[bleak ridge]

Well uhhh

You'd want an exponential decay function of the form

m(r)^t

Where m is starting mass

r is the rate at which it decreases per year

And t is the number of years

You need to solve for a value r such that it's halved at 1590 years

So m/2 = m(r)^1590

Or just 1/2 = r^1590

[topaz flower]

so 150/2=(150)r^1590

[bleak ridge]

Ye

Actually it looks like there's an easier way

[topaz flower]

so thats just the function that correctly models the mass after t years?

[bleak ridge]

Without having a rate of 0.00000001

Ye

Use a function of the form

m(1/2)^(t/half)

Which makes intuitive sense

[topaz flower]

Ok makes sense

[bleak ridge]

Because you want it to be 1/2 ^ 1 when the time equals the half

[topaz flower]

Alright

[topaz flower]

I figured the other thing out, thank you

.close

[lone heart]

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[harsh acorn]

can somebody explain me the question please? I don't know any idea to solve this

[sand geode]

hi @harsh acorn

[harsh acorn]

hello

[wanton jasper]

65% of students of languge, study not english
it's 13% of all students

65% * l=13% * x

100% * l = 20% *x

so 20% is answer

[harsh acorn]

I understood the first part but how did you get 100%=20%*x?

[wanton jasper]

there

l is number of language students

[harsh acorn]

oh

I get it now

tysm again

.close

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[alpine sable]

I need help showing uniqueness for a function f = E + O where E is even and O is odd. I already know there exist a E and O which E(x) = $\frac{f(x)+f(-x)}{2}$ and O(x) = $\frac{-f(-x)+f(x)}{2}$. How should I show uniqueness, I tried considered another even and odd function h and g then I went to show h = E, and g = O. Is this the correct way to show uniqueness?

[ocean seal]

Outlander

[fresh quail]

"$$ answer {r^{2} {\left (\cos \left (\theta \right ) - \sin \left (\theta \right )\right )}}\ dz\ dr\ d\theta $$"

[alpine sable]

.close

[lone heart]

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