help-28

[boreal adder]

why

u dont need areas for anything

[idle jolt]

because?

well the triangle ABC is where I got it from

it's just writing its area in 2 ways so I can get the sine

[boreal adder]

well i mean

if it works it works but

[idle jolt]

what should I do instead?

[boreal adder]

my solution didnt need any areas

u should check my solution i think its neater

but ur solution worked too so

goodj ob

[idle jolt]

also please explain your solution just in case lmao maybe it's better than mine

omg I got it cosa=1/8

[boreal adder]

in triangle ADC, the angle bisector theorem says PC/PD = AC/AD

[idle jolt]

yeah

[boreal adder]

we know PC/PD = 3/2. it says it in the given that AP splits CD into 2:3

so AC/AD is also 3/2

[idle jolt]

oh yeah

[boreal adder]

now we draw the other diagonal

this makes AOD = 90 degrees
so AOD is a right triangle so cos(a/2) = AO/AD

[idle jolt]

so that's still the same as what I got with my theorem just in different order I think

[boreal adder]

AO is just half of AC so if AC/AD = 3/2, AO/AD = 1.5/2

which is 3/4

which is cos(a/2)

[idle jolt]

ohhh right I see now

[boreal adder]

now we use the double angle theorem
cos(2x) = 2 cos^2(x) - 1

cos(a) = 2 cos^2(a/2) - 1 = 2 * (3/4)^2 - 1 = 1/8

[idle jolt]

yup I see everything now

thank you so much!!!

[boreal adder]

np!

[idle jolt]

I hate geometry bc I struggle to think of stuff like "just draw a bisector"

[hushed barn]

I think you both got cos(a/2) but just in different ways lol

[idle jolt]

yup lmao but his idea was what let me solve the task

I really need to get better at this shit

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[worn lintel]

im confused

if it either wants me to take one every time i can or not

[full forum]

@worn lintel Has your question been resolved?

[worn lintel]

<@&286206848099549185>

[restive geyser]

"it" isn't clear, but I'm assuming you can only check on 0:20, 0:40 and 1:00

[worn lintel]

yes

[restive geyser]

(idk whether 0:00 should be included)

[worn lintel]

i think it is

since it says you need to wait 20 seconds

after draw

that makes it weird

[restive geyser]

So what is this about?

Oh that the 20s is just a cooldown?

[worn lintel]

im confused wheter i have to draw everytime i can

yea

or its jsut optional

[restive geyser]

as in, could you wait for 25 seconds

Er

[worn lintel]

but being optional is just weird

[restive geyser]

I'd go with 20s

[worn lintel]

so its every 20s cd?

[restive geyser]

yh

[worn lintel]

0:00, 0:20, 0:40, 0:60

does 60 even count

[restive geyser]

0:60 lol?

1:00

[worn lintel]

yeah

does it count

as the first minute

[restive geyser]

Honestly not sure whether to take that one

[worn lintel]

or no

[restive geyser]

I would personally, but still

Write down an extra sentence saying that's what you're doing

[worn lintel]

hmmmm

okay so from that i actually thought of a way

[restive geyser]

"I'm assuming this to mean that I check for a white ball at the start, at 20s in, at 40s in and at a minute in"

[worn lintel]

to solve it

[restive geyser]

If this is a high-school question I wouldn't think to overcomplicate it much more than that

[worn lintel]

its uh

olympiad

for 9th graders

does it work when i do this

1 - (probability of not getting white ball from the 4 draws)

[restive geyser]

That's a workaround that works yh

Because the inverse of that event is "a white ball was found at some point"

[worn lintel]

yes

okay hold on

0:00 = 4/5

0:20 = 8/9

0:40 = 26/29

0:60 = 12/13

2311/5655

[worn lintel]

yeah id say the main idea is to do this

thanks

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[signal beacon]

Jacinda plays a game with her friend
She can win lose or draw
THe probability that she wins the game is 0.28

Jacinda is twice as likely to draw the game as to lose the game

Work out the probabiliy that she loses the game

[signal beacon]

lose = x
draw = 2x
win = 0.28

sum of all probability is always 1

so x + 2x + 0.28 = 1
x = 0.24 ans

Jacinda plays the game 150 times
Find he expected number of times that she wins

[thorn marten]

28%

[signal beacon]

how'd u calculate it

pls explain

i dont want ans

i need to know how to solve it myself

[keen oriole]

is it not just cuz the win rate is 0.28 which means 28% of the time you win

[faint idol]

so, 42 timesss

[keen oriole]

yes

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@signal beacon Has your question been resolved?

[signal beacon]

rthans

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[solemn garden]

Hello need some guidance

[solemn garden]

Just need to translate it

[steel solar]

Ok.

[solemn garden]

Determine the best fitting polynomial of degree 3 for the view y = y(x) that contains the point (0,1)

And my function looks like this

Oh wait

[steel solar]

?

[solemn garden]

I forgot 1 crucial thing

[steel solar]

Okay.

[solemn garden]

That was really stupid of me

It's correct now

[solemn garden]

It contains P(0,1) now

[steel solar]

Nice!

[solemn garden]

Sorry for this, I figured out what was wrong

Have a nice day

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[lunar veldt]

Let q be a definite quadratic form, and I want to show that it is either positive definite or negative definite.
I then assume that q is neither positive nor negative definite, which leads to a contradiction. So $exists x \neq 0, q(x)>0$ and $y \neq 0$ such that $q(y)<0$.
Let's posit $\varphi: t \mapsto q(tx+(1-t)y)$ with $x$ and $y$ well chosen.
Thus $\varphi(t)=t^2 q(x)+(1-t)^2 q(y)+2t(1-t)b(x,y)$ where b is the polar form of q.
We have $\varphi(0)=q(y)<0$ and $\varphi(1)=q(x)>0$, $\varphi$ being continuous we deduce by the intermediate value theorem that $\exists t_0 \in (0,1) , \varphi(t_0)=0$. q being defined, it is necessary that $t_0 x+(1-t_0)y=0$, and here I'm stuck...

[glossy valve]

tm

[rough tundra]

you deleted the original msg the channell will close, open a new one

[lunar veldt]

okie thx

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[lunar veldt]

c’est quoi que tu comprends pas

[light creek]

i have a test tomorrow this is another class's test and i'm relizing that i'm cooked

[light creek]

got stuck on question 1 (you can talk in english if you want the test is just in french that's whhy i asked earlier

[lunar veldt]

tkt j’suis français

tu préfères l’anglais ?

[light creek]

oui s'il vous plait

[lunar veldt]

oki

what u tried for first question

[light creek]

honestly forgot the method since i was absent for half of the trigonometric functions lesson

[lunar veldt]

u need to show $\forall x, f(x+\frac{\pi}{2})=f(x-\frac{\pi}{2})$

[glossy valve]

tm

[lunar veldt]

in ur case

[light creek]

oh i see

so like i proove f (pi)= f(0).?

[lunar veldt]

no it’s for all x

[light creek]

oh i see

[lunar veldt]

calculates f(x+pi/2) and f(x-pi/2)

and see if it’s the same expression

[light creek]

thank you so much, i'll probably need this type of help through out the entire thing i'm kind of missing a lot of the cours😭

[lunar veldt]

lol np I’ll give u some hints

[light creek]

what is sin(x-pi/2)

[lunar veldt]

-cos(x)

[light creek]

uh huh...

there's a minus that i dont know where it came from

[lunar veldt]

where ?

[light creek]

one sec

very sloppy hand writing ik

oh btw you mentioned that you're french did you already finish the bac

[lunar veldt]

yes

[light creek]

how was it

[lunar veldt]

but wait i think i’m wrong

[lunar veldt]

facile lol

[light creek]

oh la literature aussi ?

[lunar veldt]

it was f(pi/2+x)=f(pi/2-x)

[lunar veldt]

ça non 😭

[lunar veldt]

sry

[light creek]

its okay fair mistake lol

[lunar veldt]

that’s why it didn’t work 🥲

[light creek]

its okay lol

[light creek]

what was your specialty

[lunar veldt]

maths and physics & chemistry

with expert maths option

[light creek]

oh cool

[lunar veldt]

and u

[light creek]

currently nsi next year maths and physics probably

[lunar veldt]

ahh ok nice

[light creek]

i'm tunisian if you're wondering ill have 2bacs next year 😭 😭

[lunar veldt]

lol u live in france ?

[light creek]

nope doing this thing where i do the french bac on top of the tunisian bac so that i'll pretty much be guaranteed that i'll go to europe

[lunar veldt]

t’es en terminale la ?

[light creek]

1er

[lunar veldt]

ahh ça va c’est dans longtemps le bac lol

[light creek]

it worked

[lunar veldt]

nice

[light creek]

how is f'(x) supposed to be like that i got 2sin(x) * cos(x) + 2cos(x)

[lunar veldt]

u have f’(x)=-2sin(2x)+2cos(x)

and then use $sin(2x)=2sin(x)cos(x)$

[glossy valve]

tm

[light creek]

yeah?

[lunar veldt]

so -4sin(x)cos(x)+2cos(x)

[light creek]

where did the - 4 come from

[lunar veldt]

because u have -2*2

[lunar veldt]

look

$(cos(2x))’=-2sin(2x)$

[glossy valve]

tm

[light creek]

ohh you derive the stuff on te inside too

[lunar veldt]

yes

and never forget the minus when u derive cos

[light creek]

got it

finished ex1 yayyaya

[lunar veldt]

gg 🎉

[light creek]

2 more to go ig

[lunar veldt]

ez geometry

[light creek]

hopefully

it feels like something got lost here

[lunar veldt]

no

[light creek]

oh i thought that was a minus 😭 😭

[lunar veldt]

lol no

[light creek]

how do you prove that something is othgonal on something else

[lunar veldt]

find a director vector of $\Delta$

[glossy valve]

tm

[light creek]

how is that done

[lunar veldt]

takes coefficients in front of variable (alpha)

so it’s (-1,2,-1)

[light creek]

ohhh

[lunar veldt]

if u have a director vector (a,b,c), the equation will be a(x-x_c)+b(y-y_c)+c(z-z_c)=0

[light creek]

i see

i missed a question what does this symbol mean

[lunar veldt]

included

[light creek]

got it

[lunar veldt]

$\forall x \in \Delta, x \in P$

[glossy valve]

tm

[light creek]

uh huh.. yeah i apperantly don't understand anyhting in this lesson \T-T we've studied it a while back and the teacher didn't like mz coming late

[lunar veldt]

ah

what was the lesson

[light creek]

espace

[lunar veldt]

ah

u know yvan monka in yt ?

[light creek]

nope

[lunar veldt]

he is a french teacher

he explain very well geometry

[light creek]

got it i'll check him out after let's skip this for now

[lunar veldt]

oki

[lunar veldt]

proba 🤯

[light creek]

yeah denombrement apperantly it'll be proba next year

[lunar veldt]

ye

[light creek]

this is the lesson i understand best i think

[lunar veldt]

it’s the only one where I got the worst grade of my life 😭

[light creek]

oh sorr

[lunar veldt]

np, i’ll take my vengeance 🤓

[light creek]

sorry took a little brake (i got distracted)

[lunar veldt]

np np

[light creek]

the fist one is 5

right?

[lunar veldt]

no

[light creek]

what

[lunar veldt]

wait

ok no its 5

[light creek]

oh okay

the second one should be 4 right ?

[lunar veldt]

yes

[light creek]

what's D

[lunar veldt]

what u got for C

[light creek]

7

[lunar veldt]

no

[light creek]

oh

[lunar veldt]

remember the formula for $| A \cup B|$

[glossy valve]

tm

[light creek]

i

dpnt you just add them app

up

[lunar veldt]

if its disjoint yes

in our case no

it isn’t

[light creek]

oh

why

[lunar veldt]

because u can have 3 balls with the same color and with the same number

how much case tell me ?

[light creek]

1; 3 red ones

[lunar veldt]

yes exactly

so what is A U B

[light creek]

you add A and B the remove the common possibilities

[lunar veldt]

yes

[light creek]

so 6 ?

[lunar veldt]

||no||

[light creek]

5 final answer

[lunar veldt]

no 😭😭😭

[light creek]

bro how

[lunar veldt]

.

[lunar veldt]

.

[lunar veldt]

.

[light creek]

omg i sound s stupid right now

[lunar veldt]

we all makes mistakes lol

[light creek]

so it

it's 8

[lunar veldt]

yesss

🎉

[light creek]

how to do D

[lunar veldt]

idk how to explain 💀

list all the cases

[light creek]

you can't have only one white ball andone ball with the number 1 on it

because all the reds have the number 1 on them

[lunar veldt]

why

there is many cases 💀💀

if u take the white ball with the 1, u have many cases

[light creek]

what 😭

[lunar veldt]

ahhhh no

it’s only one white ball

idk how to read omg

[light creek]

and only 1 with the number one

[lunar veldt]

there is no cases ?💀

[light creek]

it seems like it if you take the white with 1 theb you cant have any red

but if yiu get another white and a red you cant have anything else

[lunar veldt]

if you take the white ball with the number 1 first, you have no choice, as all the red balls have the number 1

if you pick up a red ball with the number 1, you can’t pick up another red ball, and you can’t pick up two white balls because you can only pick up one white ball

[light creek]

let's say there's no possibilities and move on

[lunar veldt]

ye i think we have 0 possibilities

i think there is only 1 case for E

[light creek]

2 i think

[lunar veldt]

ah yes

im so dumb tonight

there is more than 2

[light creek]

like24

[lunar veldt]

i would say 4

because we have 4 choose 3 possibilities

[light creek]

you will use the arrangement thing

[lunar veldt]

i used combinations

$4 \choose 3$

[glossy valve]

tm

[light creek]

you can't question 2 changed the condition it's "succivement sans remise" now

[lime ether]

$\binom{4}{3}$

[glossy valve]

knief

[lunar veldt]

ah fck

[light creek]

oh hi

[lime ether]

hello

[lunar veldt]

i introduce u knief, master of fourier series

[light creek]

how are you

[lime ether]

wonderful

[light creek]

honor to meet you

[lunar veldt]

help me knief im so bad at combinatorics 😭

[lime ether]

combinatorics is fake math

[lunar veldt]

probability shouldn’t be in maths

[light creek]

dosen't change that i have a test tomorrow

[lime ether]

pray

🙏🏻🙏🏻

[light creek]

the only solution atp

[lunar veldt]

next questions are even worse 😥

j’espère y’aura pas ça au test 🙏🏼😥😥

[light creek]

the last exercise has a light bulb for some reason

[lunar veldt]

lol

[light creek]

bro reverted to french 😭 😭
my teacher always has the worst tests for some reason

[lunar veldt]

💀💀💀

oh integration

[light creek]

off limits

[lunar veldt]

ah

[light creek]

the guy that made education system here decided that itll be best if i studied derivatives this year and integrals next year

so cant use them

[lunar veldt]

how we do it without integration lol

[light creek]

also that isnt even a function isnt a function supposed to have one image per x

[lunar veldt]

image per f or x ?

[light creek]

image de x

also we're still in the other just wanted to show you the graph...

[lunar veldt]

nn ducoup, les fonctions qui n’ont qu’une seule image par x sont appelés injectives

f(x)=f(y) -> x=y

[lunar veldt]

lol so what u did ?

[light creek]

i mean ypu can not have f(x) equal 2 things at the same time

[light creek]

how is F done

[lunar veldt]

list all the cases

i think there is 4

[light creek]

i found more unless i

didn't understand the question

(my french isn't good)

[lunar veldt]

0 0 1 2 white balls
1’ 1’’ 1’’’ red balls
we have
2 1 1’ 1’’
2 1 1’ 1’´´
2 1’ 1’’ 1’’’
2 1 1’’ 1’’’

j’ai noté avec ‘ sinon on confond

[light creek]

troisieme means at the tird spot right

[lunar veldt]

ah mais c’est E ça nan

aaaaaa

I confused E and F

[light creek]

nope F

[lunar veldt]

omg

[light creek]

what d you think

[lunar veldt]

t’as combien pour F

[light creek]

i din't know how to do it

[lunar veldt]

i would say $A_3^5$

[glossy valve]

tm

[lunar veldt]

but we have 2 balls with 0

[light creek]

yeah?

[lunar veldt]

ahhh but we can have a 0 ball at the fourth draw

then $A_3^6$ and since we have 2 balls with 0, it is $2 \cdot A_3^6$

[glossy valve]

tm

[light creek]

so 240

[lunar veldt]

oui

[light creek]

merci

really thank you for your time, i'm going to sleep now

so sorry for taking so much time

[lunar veldt]

np u welcome

at least it made me revise proba lol

[light creek]

how do i close this

[lunar veldt]

.close lol

[light creek]

.close

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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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[carmine lichen]

heyyy

[carmine lichen]

how do i value the integral from 0 to infinity of 1/(1+x^4)dx

using the beta integral

[glossy valve]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[lethal cobalt]

the integral just SCREAMS to use cauchy residue

but if you insist on the Beta function...

[carmine lichen]

my exercise says to use the beta integral

and also i do not know what is the cauchy residue

[lethal cobalt]

yeah, get that... Just seems like a bad way in general

understandable... I'm trying to figure out how to do this w. the beta function

[slender bane]

since this function is defined for all x values, why not just take the integral?

[carmine lichen]

i have to use the beta function

[slender bane]

oh ok

[lethal cobalt]

You can sole it from 0 to 1 by using an identity w the beta funvtion

that's B(z_1, z_2) for all positive real n

[carmine lichen]

ohhh

first time seeing this

[lethal cobalt]

with that you can just plug in z_1, z_2 and n

[lethal cobalt]

yeah... Don't know if your teacher allowes it (they might want you to solve it another way)

and extending this to 0 -> infinity might not be easy

and it's -t^4

so this is more a happy observation, rather then a way to solve the problem

[carmine lichen]

yes i get it

can we solve this using gama instead

[remote kite]

I think we can?

[carmine lichen]

i will go and ask my proffesors for this

the man himself

[lethal cobalt]

wait... You should be able to solve this w. this

and a t = u^4

and that you should also be able to derive so your teacher can't complain on you using it

[carmine lichen]

how did you get this?

[lethal cobalt]

that's a (semi) well known identity

[carmine lichen]

can you help me just solve it using the gama?

i am sure that i am missing something and i will go and ask about it

[lethal cobalt]

you sure that you can get a good solution from just the gamma function?

[carmine lichen]

the answer in the end should be pi root(2)/4

[remote kite]

One should start with the Beta Function and moves on to the Gamma function naturally.

[lethal cobalt]

got that w. the beta function

[carmine lichen]

no idea what to do

i did find this identity

[lethal cobalt]

okay... Are you okay with this identity?

start of with that identity, and substitute u^4 = t

[carmine lichen]

that is b(x,y)

[lethal cobalt]

I don’t know how that would help you…

[carmine lichen]

me too

this is the best i could do

here y=t^(1/4)

i do not know if you still want to help or not

[lethal cobalt]

I got it

[carmine lichen]

really?

[lethal cobalt]

ok

[carmine lichen]

tell me

[lethal cobalt]

yeah. 1 sec

[lethal cobalt]

First you can prove this is the same as the Beta function by the substitution t = u/(1-u)

there you have that

[carmine lichen]

ohhh

[lethal cobalt]

then you use the sub u^4=t

and try to figure out z_1 and z_2 s.t. the rest works out

It's a pain

[carmine lichen]

yeah i can see that

i am trying to wrap my mind around it

what was the u sub/

[lethal cobalt]

here?

[carmine lichen]

yes

[lethal cobalt]

t=u/(1-u)

I'll try to rewrite it in latex, might be easier

[carmine lichen]

no idea what latex is but ok

[lethal cobalt]

program for writing math equations/symbols

[remote kite]

Did you need help using the Gamma function to solve the integral?

[carmine lichen]

sure

[carmine lichen]

hmm you actually solved it

now what next?

[lethal cobalt]

there you go

[remote kite]

That's using the Beta Function. Yes?

[lethal cobalt]

yea

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[junior pond]

This is secondary school GCSE level, so it shouldnt be too complicated
The question said to factorise a quadratic, but I got a different answer to the mark scheme.
I don't notice any mistakes in my workings, so if a helper could please lmk about a mistake I made, you'd be a lifesaver thanks

[narrow ermine]

They are the same. $$(2x-4)=2(x-2)$$

[glossy valve]

Azyrashacorki

[narrow ermine]

If anything I’d say your answer is more factorized

[junior pond]

wouldnt the 2 outside the brackets make turn the (2x-3) into 4x-6 as well?

making mine 2x-4)(4x-6)

[narrow ermine]

It just distributes into one of brackets

If it was both you’d have a factor of 4

[junior pond]

ahh that makes sense

tysm

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[weak wadi]

Can I solve part b using workdone = gpe

[sturdy valve]

yes

what will u take as the workdone

[weak wadi]

One moment let me see

[weak wadi]

Friction + weight component perpendicular to the slope?

Parallel*

[sturdy valve]

yeah parallel

perpendicular would be 0

[weak wadi]

Yeah

Nb

[sturdy valve]

and what will u take as friction?

[weak wadi]

Mb*

The friction I calculated in part a, u can see there on the picture

[sturdy valve]

cant really understand but ill assume u got it correct

[weak wadi]

Fmax=( 1/4 )cos(alpha)x(g)x(mass of A)

[sturdy valve]

yh

[weak wadi]

Okay thx

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[sullen wave]

Help, back again

[sullen wave]

How am i supposed to find D

[narrow path]

!qa

[sullen wave]

I got the same thing

I found

B

[narrow path]

can you show the full question

[sullen wave]

Okay wait

Actually it's A

Here Ex11

[narrow path]

,rccw

[sullen wave]

WIAT

WEONG ONE

EX13

[narrow path]

,rccw

[sullen wave]

C

[sullen wave]

Answer for a b

[narrow path]

if you draw a diagram its really easy

[sullen wave]

Only example they gave

A | | BC
A
y= 3/2x + 7
BC | | AD

AD
y= 3/2x + 7
AD || BC

Uh

AB || C
y= 2/-3x + 2⅔

AB || DC

y= 2/-3x + 2⅔

Isn't it's
CD || AD

I FUCKING GOT IT

FINALLLY

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[spring vigil]

can someone explain

[spring vigil]

why its 1/x^4 on the bottom instead of 1/x^2

for

5/x^2 + 49 / x ^2

[onyx glen]

1/x^2 = sqrt(1/x^4)

this is what happens when you push the 1/x^2 under the root

[spring vigil]

wait im having trouble seeing why

oh ic

did u just sqrt the 1/x^4?

[woeful pasture]

${\frac{1}{x^2} \sqrt{5+ 49x^4} = \sqrt{\frac{1}{x^4}}\sqrt{5+ 49x^4} = \sqrt{\frac{1}{x^4}({5+ 49x^4)}$

[glossy valve]

k
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

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[golden shore]

Hello

[minor crater]

send your thing

[glacial stream]

Please only use help- channels to get help when you're stuck on a math problem

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[scenic brook]

for a function like this

how can we determine if as x->+-oo, whether the function approaches 2 from above or from below

[onyx glen]

consider the expression f(x) - 2

see if you can determine whether it's positive or negative for large x

after simplification obv

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[lyric tusk]

It’s my mock for my national MO and no solution yet

[lyric tusk]

Here’s my work

I given p is prime such that $gcd(a_i,p) = 1$ for $i=1,2,\dots,m$ and given f(n) for " "by FLT I got
$$ f(p-1) \equiv 1 ,(mod ,p)$$
$$ f(k(p-1)) \equiv 1 ,(mod ,p) $$
$$ f(n+k(p-1)) \equiv f(n) ,(mod ,p) $$
then if $f(n)$ is divisible by some $p$...
but I don't know how to show here
or if I considered $f(k) = p$ for some $k\in\mbb{N}$
Is this valid?

[glossy valve]

Sea

[lyric tusk]

I’m stuck to show that f(n) con 0 (mod p)

or else I was wrong

I tried for case p=2,3 and in case p=3 I found something like
$$f(n+2) \equiv f(n) ,(mod ,3)$$
So if I scope for $$ f(n+p-1) \equiv f(n) , (mod ,p) $$
It would be easier right?

[glossy valve]

Sea

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@lyric tusk Has your question been resolved?

[lyric tusk]

I think it's time to <@&286206848099549185>

[rocky delta]

hi sea

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@lyric tusk Has your question been resolved?

[charred raft]

consider $S(n) = \sum_{k=1}^m k^k a_k ^n$ you'll consider the case where $S(1)$ is a prime otherwise its a trivial case, so let $p = S(1)$ by FLT , for $n \equiv 1 [p-1]: \$ if $$a_k \equiv 0 [p] \implies a_k ^n \equiv 0 [p]$$ -another trivial case- $\ \$ if $$a_k \not \equiv 0 [p] \implies a_k ^{p-1} \equiv 1 [p] \implies a_k ^n \equiv a_k [p] \ \$$ Thus you can see that $S(n) \equiv \sum_{k=1}^m k^k a_k = S(1) \equiv 0 [p]$

[glossy valve]

Goëtia

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[wary condor]

a math teacher i watch on yt started teaching Permutations by giving us the notation, explained the steps on how to solve stuff like 5P2 and then said "n has to be a positive integer" and then i immeditaly paused. I want someone to actually explain wtf is happening

[wary condor]

what are Permutations

not the whole thing i mean- an introduction

to make me UNDERSTAND bc this is not math.

[torn jolt]

lmaoo

i really hated stat

but i remember

permutation was when the arrangement matters

and combination when it didnt

[wary condor]

elaborate?

[weak bolt]

Well, you're not going to pick one thing out of the negative things.

[wild sleet]

it's talking about like, if you have 5 letters, you can make 20 words out of it, if they are 2 letters long, because 5×4

[weak bolt]

Imagine you have a box with -1 ball inside.

[wary condor]

im not talking about why n has to be positive. im asking- what is n what is r what is P what is it used for

[weak bolt]

$A_n^p$ ?

[glossy valve]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[weak bolt]

Like that?

[wary condor]

like we started calculus by an introduction saying "oh yeah we're gonna solve some slope problems that are otheriwise impossible"

and getting some area under curves too

so what are permutations

[wary condor]

perhaps? our notation might be different bc i study in arabic

[weak bolt]

Owh

[weak bolt]

Just write it

[wary condor]

so whats happening over there

what does this mean

[torn jolt]

nPr

[wary condor]

yeah

what is nPr

i dont understand what its used for

im sorry if im asking for too much

[weak bolt]

n: عدد الكرات الكلي
r: عدد الكرات المسحوبة
P: طريقة السحب

[wary condor]

im just- confused and idk what we're doing with permus

[wary condor]

okay so its about arrangments! can you give an example like a question that we use permus to solve

[weak bolt]

$\frac{n!}{(n - p)!}$

[glossy valve]

╰ 𝕃 𝕌 ℂ 𝕀 𝔽 𝔼 ℝ ╮

[weak bolt]

Do u know multinomial coefficient?

[wary condor]

nope

this is my first time learning arrangments

or

combiantaitosoon-ics idk how you spell that

nPr = (n-1)(n-2)(n-3)...(n-(r+1)) YEAAAAAH IT MAKES SOO MUCH SENSE OH MY GOD I UNDERSTAND

please help me.

idk what the fuck this guy is talking about

[weak bolt]

عندما نأخذ ثلاث كرات معا، لا يهمنا ترتيبها
أ ب ن = ب أ ن = ن أ ب = احتمال 1
لكن في حالة الترتيبات، يعمنا الترتين،
لأن: أ ب ن = احتمال 1
ب أ ن = احتمال 2...

لذلك نستعمل المعامل متعدد الحدود

[wary condor]

okay so

you're saying

we use permus

to find the possible arrangments

for a certain event?

when arranging actually matters.

[torn jolt]

yea

[weak bolt]

لنفترض أنه لدينا 5 كرات.
3 حمراء
2 زرقاء
لترتيب كرتين حمراء وكرة زرقاء
نقوم بــ:
5!/( 1! × 2!) × 3P2 × 2P1

[torn jolt]

exactly

[weak bolt]

بحيث 5! أتى من عدد الكرات الخمسة
و 1! أتى من عدد الكرات الزرقاء التي نريدها
و2! أتى من عدد الكرات الحمراء التي نريدها

[weak bolt]

بدونها، سيكون الأمر وكأنك تقوم برفع 3 كرات مرة واحدة، دوناً عن الاهتمام بترتيبها
بحيث أن ترتيب abc يختلف عن acb على الرغم من وجود نفس الأحرف

[wary condor]

you can speak english with examples like these. i will understand. just for better communication

[wary condor]

so when we want the amount of ways to arrange something we factorial it

like the possible ways to arrange two red balls is 2!

[weak bolt]

If u have 1 color, if would be 2!/2! = 1

[wary condor]

yeah okay

ty for the introduction

have a good day.

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[idle bridge]

I need help

[idle bridge]

For your information, this is my work so far so do check it. i haven't completed (c) because i need to find the best approach - so far I think the pattern the red extends 1 only in direction and blue extends in both directions? i dont think my reasoning is correct

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[nimble crane]

.,ui

test

[nimble crane]

,ci

[glossy valve]

Channel information for help-28｜higher.

​ ​ ​ ​ ​ ​ Name: help-28｜higher [#help-28]
​ ​ ​ ​ ​ ​ Type: Text channel
​ ​ ​ ​ ​ ​ ​ ​ ID: 903486480985489478
​ ​ ​ ​ ​ ​ NSFW: No
​ ​ Category: ⏳Math Help (Occupied) (360643799707549696)
Created at: <t:1635478363:F>

Topic

Math Help. Please read #❓how-to-get-help and #rules.

[nimble crane]

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[nimble crane]

okay that’s all I wanted

[twilit leaf]

integration relaxes the-

never mind

[gritty rose]

BOWELS

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[glad heart]

quick question on how to connect piecewise

am I just connecting the first line to the second?

[minor crater]

why are there two graphs?

[glad heart]

the top is my attempt

the bottom is the computers correct answer

[minor crater]

first off the bat, you have two lines for x ≤ 2

this can't be a function

fails vertical line test

[glad heart]

okay I see how do I pick because its saying either greater -2 or less than 2

[minor crater]

huh?

what made you put the line at y= -5

[glad heart]

the -2<x

-3 for x

gives me -5 when I plug

open dot

should be x greater than -2

[minor crater]

you sure?

[minor crater]

it's -2x+1, not 2x+1

[minor crater]

i am not sure what you mean by this

[glad heart]

what do you mean

[minor crater]

for x in between (-2, 2], f(x) is -2x+1

[glad heart]

-2x+1 if -2<x<2

I messed up thinking x needed to be less than -2

what I am asking is how to pick

because its giving me two ponts on problem 2

so which do I pick and how do I connect

[minor crater]

pick between what?

if x ≤ -2, it's 5
if it's above -2 but below 2 it's -2x+1

for each x there is exactly one y

[glad heart]

I have two options right here right

-2x+1 if -2<x<2

I have two possilibilties to plug in for x? ?

I think it would be better

just to ask what I do on this step>>>>>>>>>-2x+1 if -2<x<2

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[slow sleet]

am i right with C here?

[grave elm]

,w cos(2xpi) = sin(5xpi)

[glossy valve]

Results provided by WolframAlpha

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

Requested by methisalwaysright

[grave elm]

29pi/14 isnt even in the interval [0, 2pi) btw

so I'd guess B assuming that the rest is correct

[slow sleet]

ohh your right i didnt realize

[grave elm]

you can verify it by cross-checking it with this (youll probably have to approximate your results, because WA aint cooperating today)

[slow sleet]

not D right

just wanna double check

[slow sleet]

got it okay

[grave elm]

Cant be D, cause x = 0 doesnt work

its actually doable with nearly no calculations just by eliminating all the answer choices lol

cant be A or D, cause x = 0 doesnt work

cant be C, cause 29pi/14 isnt in the interval

[slow sleet]

i understand okay

thank you so much for helping

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[azure inlet]

why is an wrong

[onyx glen]

put the (-1)^(n-1) back in

they didn't ask you to take absolute values at that juncture yet

[azure inlet]

oh okay

thanks

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[orchid minnow]

Please for the love of god help me finish this question step by step

[versed hamlet]

What have you tried so far?

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[covert goblet]

Curious as to how we were supposed to guess ln x as the test function for the limit comparison test

I get how the answer key works, but is there any easier reason we chose this apart from being able to see that if we L'Hopital it, we can get a limit with an x/arctan x term

or was i just supposed to see that

[fast peak]

well the idea is more or less that for x near 0 you have arctanx roughly equal to x, so then lnarctanx should roughly be lnx

or at least you hope it is

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@covert goblet Has your question been resolved?

[covert goblet]

is this generally true for all sin,cos,tan,arctan,arcsin,arccos etc near 0

[zealous lily]

yes, C is right.

C is right answer.

[covert goblet]

<@&286206848099549185>

[gritty rose]

no

cos(0) = 1

,tex .unit circle

[glossy valve]

riemann

[covert goblet]

oh

rightie

xd

but definitely for sin and tan

because it's just le olde taylor series

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[last knot]

I need help with mathematical analysis problems related to integrals, functions, the area of a region between a curve and the axis, calculating limits, and so on. Willing to pay too.

[green merlin]

If you have a particular problem, you can always post it here

[last knot]

Not particular, but that kind of math problems mentioned above. I will really need a helping hand

[green merlin]

hmm

okay

[last knot]

This for example

[fast peak]

these channels are for specific questions. and this server does not offer any kind of paid tutors or anything

[modern horizon]

You could dm me

[last knot]

you need to do it cuz you turned off your dm's or smth

[modern horizon]

Ah my bad ;o

Just turned on

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[torpid perch]

help with 2)

[thick hedge]

Determine the dimension of T, where k is any value in R?

[torpid perch]

,w det {{1,k,0},{0,-1,k},{2,k,1}}

[trail parrot]

Whats the max dimension a 3x3 matrix can have

[glossy valve]

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[torpid perch]

(k-1)(k+1) = 0

[torpid perch]

max is rank 3

[trail parrot]

And whats the min dimension, and when does it have that

[thick hedge]

Instead of using the determinant here, you';d probably want to use row reduction to determine the values of k for which the dimension is 3,2,1

[torpid perch]

why row reduction

[trail parrot]

You can do it by simple observation tbh

[torpid perch]

?

[trail parrot]

When would a 3x3 matrix have 0 dimension?

[torpid perch]

when we got 3 free variables

[thick hedge]

If you're able to get a row with only zeros, then the dimensin would be 2

[torpid perch]

minimum dimension of T is 1

[trail parrot]

Is dimension 1 possible?

[torpid perch]

determinant already tell us all the possible k values such that we have linear dependency

in fact is only two possible ks, agreed?

[trail parrot]

This isn't accurate. The only transformation which has 0 dimension is the one which squishes the entire space into the 0 vector

Which would be a matrix with all entries 0

Free variables just mean you can make it so that the matrix has all entries 0

[torpid perch]

,w rank {{1,k,0},{0,-1,k},{2,k,1}} where k = 1

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[torpid perch]

,w rank {{1,k,0},{0,-1,k},{2,k,1}} where k = -1

[glossy valve]

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[torpid perch]

it was a trick question what you asked, how am I supposed to kmow that

[trail parrot]

Whats the dimensions of a matrix?

dim(M)

What does it give yoy

[torpid perch]

the image

the dimension of the colspace/rowspace

idk, the definition of dimension is counting the number of vectors of the basis

[trail parrot]

Yes

[torpid perch]

the cardinality of the basis

idk if its defined for matricea

[trail parrot]

So the dimension of R³ would be 3

[torpid perch]

you cab argue that a linear transformation can be represented by a matrix

in that case it will be the image

[slender onyx]

we find dimT=3 for k!=1 or -1

how do we find dimT for k=1 or k=-1?

[torpid perch]

yes dim T = 3 for R - {1,-1}

[torpid perch]

rank

[slender onyx]

and how to find rank?

[torpid perch]

dim T = 2 for {-1+,1}

[torpid perch]

ref

[slender onyx]

cool done 🙂

[torpid perch]

i appreciate it

. close

.close

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[formal quiver]

The sum of all 4-digit numbers formed with the digit 1, 2, 3, 4 without repetition, is-

[urban jacinth]

What have your tried

[thorn marten]

What can you not do?

There are a total of 24 arrangements of 1, 2, 3 and 4.

[urban jacinth]

Hush hush let him try

[formal quiver]

I made several numbers and I try to add randomly but I don't think it will work it took lots of time

[formal quiver]

Repetition

[urban jacinth]

Hint: the number of times any digit appears in all arrangements is same

[formal quiver]

24×10

[urban jacinth]

Why 10?

[formal quiver]

24×100

?

[urban jacinth]

Oh yes

Youre on the right track

[formal quiver]

Thanks

[urban jacinth]

Keep adding till 1000

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[sly solstice]

any help on this please, ive tried looking for ways to utilise gassiuan elimaination to have the 2nd row, 2nd column be = 1 but i literaly cannot find a way
and ik its smth eventually to do with a quadratic, thx and plz

[full forum]

Please don't occupy multiple help channels.

[sly solstice]

tbf the other one was closed 8min ago

[full forum]

@sly solstice Has your question been resolved?

[slate violet]

probably make the 2nd row be $0, \lambda^2, \frac{3}{2} \lambda^2 + \frac{1}{2} \lambda^3 \ \ 2 \lambda^2$ first

[glossy valve]

south

[slate violet]

then subtracting gives (0, 0, 3/2 l^2 + 1/2 l^3 - 5l | 2l^2 + 4l)

so 2l^2 + 4l = 0 implies l = 0, -2

don't forget then that lambda = 0 also implies infinitely many solutions

[sly solstice]

dont we want the 2nd row 2nd column to be 1

afai have been told

[slate violet]

you don't need to to figure it out

[sly solstice]

per gaussian

[slate violet]

,w {{1, 2, x, 1}, {0, 1, (3+x)/2, 2}, {0, -x, -5, -4}} rref

I mean

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[slate violet]

okay fair enough you better find the RREF

[sly solstice]

idk how you got that

[slate violet]

multiply row 2 of the 4th line by lambda

[sly solstice]

oh the 4th line

aight

[slate violet]

no but you should do something different from this step

multiply the 2nd row here by lambda, then multiply the 3rd row here by 2

now subtract the two rows

,w x * (0, -2, -3 - x, -4) - 2 * (0, -x, -5, -4)

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[slate violet]

yes then $\frac{8 - 4x}{-x^2 - 3x + 10}$ simplifies further

[glossy valve]

south

[slate violet]

ping

[sly solstice]

aight jus took some time to write it down to this

thx for ping asw

[sly solstice]

would this be a valiud statement

i hear dsomewhere regarding this question that when you have the quadratic, only one of the values like work or smth

i.e not the -1 idk why though

and i may be wrong

[slate violet]

so when lambda = 2, 8 - 4 lambda = 0

and then if you check, -lambda^2 - 3 lambda + 10 is also 0

[sly solstice]

oh so when lambda is 2, then its 000 | 0 hence infintie

and when its -1 its 12 = 12 so unique?

[slate violet]

yeah so if you set -lambda^2 - 3 lambda + 10 = 0, then that gives you (0 0 0 | something)

[sly solstice]

or is that a no solution

[slate violet]

you should not do -l^2 - 3l + 10 = 8 - 4l

that makes no sense to do

[sly solstice]

oh

interesting

[slate violet]

so then since lambda isn't 2 here for 1 value

[sly solstice]

is it because its instead (-l^2 - 3l + 10)*z = 8 - 4l

[slate violet]

that other value is when the system is inconsistent

[slate violet]

yeah

[sly solstice]

also could you pelase clarify the wording of the question when it says inconsistent

like infinite?

or

[slate violet]

so this is z

[slate violet]

no solution

like 0x + 0y + 0z = 5

[sly solstice]

yea

[sly solstice]

ok so to locate 1. infinite number of oslut ions you want 0 0 0 | 0
hence you want -l^2 - 3l + 10 = 0 and 8 - 4l = 0
using the right question you find when l = 2, it equals 0, check with left equationm, it also equals 0, hence the only value its infintie is at l = 2
(skipping finding the quadratic that gives -5 and 2 for when that left side = 0)?

for one of the proofs?

[slate violet]

yes! now that's correct

[sly solstice]

does the left hand side must be 0 0 0 | x where x is any value to get a "no solution" i.e inconsistent?

[slate violet]

well you can't have x = 0 for inconsistent

but yes, x can be any other value apart from 0

[sly solstice]

yea

interesting

[sly solstice]

is there any better way, other than using the quadratic formula to solve -l^2 - 3l + 10 = 0 to get -5?

or is that the only way disregarding other methods ofc like compelteing the square

[slate violet]

no better way, yeah

as you said yourself it involves quadratics

[sly solstice]

yup

[slate violet]

you could factor the quadratic in the denominator and make it cancel also

using that method you'd get (0 0 1 | 4/(x + 5))

then that would be undefined when x + 5 = 0 or x = -5

[sly solstice]

what would be a valid proof that for all values but -5 and 2 there is a unqiue solution? determinent or something

[sly solstice]

oh interesting yea nice, thx

[slate violet]

well the RREF is the proof

[sly solstice]

oh right sorry i saw u type that somewhere above

how would you use the row echelon form to prove it sorry

usually tis 0 0 0 | x

and you set the last 0 to say alpha or lambda

but in this case you have both sides with lambda

[sly solstice]

pingidity ping

[slate violet]

what do you mean by prove

[sly solstice]

well what do you state to show that all other values but -5 and 2 would be a unqiue solution

[slate violet]

there's only three outcomes for any system of linear equations

[sly solstice]

yep

[slate violet]

infinitely many solutions, no solutions, or exactly one solution

if an x-value doesn't give infinitely many solutions or no solutions

it must automatically imply exactly one solution

[sly solstice]

ah alrigfht

thank you sm

[slate violet]

no worries!

[full forum]

@sly solstice Has your question been resolved?

[sly solstice]

just want to confirm does everythinghere look good