help-27

[serene swallow]

i just want know how to be more mindful during exams

[sand pumice]

do you go over your answers once you're done writing

[mild anvil]

draw a diagram

a tree diagram

diagrams are really useful

are you pressed on time?

[serene swallow]

answering to both of u my Yes i would love to recheck or draw diagrams but i run short on time

[mild anvil]

do past papers and give yourself less time maybe

[serene swallow]

i have done that

[mild anvil]

funny story, i was doing a pythag paper a few weeks back i think it was 5 weeks ago

and i was pressed on time

i rounded to 2 decimal places instead of 1

😭

[sand pumice]

when rechecking see if you can approach a problem in a different manner and if that gives you the same result

[serene swallow]

i have been there😭😭

[sand pumice]

my condolences

[mild anvil]

i only ended up getting a 93

[serene swallow]

the thing is that i have done that too but i dont know what happens to my mindfulness whilst exams that i mess up the eziest shit

[mild anvil]

when doing homework don’t rush

take it easy

[mystic hound]

tbh icl this feels like it might be a bigger thing 😭

[serene swallow]

i do that too

[serene swallow]

😭😭😭

[mystic hound]

do you have a teacher or someone you can talk to about it?

[sand pumice]

some ppl would be envious unless like everybody gets 95-100

[sand pumice]

do you feel stressed during the exam

[mystic hound]

i mean i obviously don’t know anything about how your school works but idk maybe you could get better help from people close to you or extra time or smth

[serene swallow]

i tried but they end up givign suggestions i have tried

[serene swallow]

nope

[mild anvil]

i know someone who got 3/43 marks and didnt give a shit 😭

[mystic hound]

this is such an interesting issue i’ve never heard of someone who’s felt like this

sounds rough tho

[serene swallow]

tbh it hurts more to score in the bracket of 90-93 in an ez paper

[mild anvil]

true

[serene swallow]

ikr

[mild anvil]

it was hella easy

it was just pressed on time

[sand pumice]

oh god 😭

[mystic hound]

i mean tbf it also might be a bigger problem if you get an exam at some point that’s relatively hard for you

it would suck to then have to worry about that too

[sand pumice]

so when you try a problem in a different way you still get the same incorrect answer?

[mild anvil]

do a bunch of practice problems and if you get them wrong get your mum to hit you

💀

[serene swallow]

from my understanding of this message, i am still scoring 90-92 but im stuck in this bullshit bracket

[sand pumice]

he has important exams next month 🥲

[mystic hound]

oh yeah shit

[serene swallow]

dont even remind me

im hella scared if this shit continues

[mystic hound]

honestly man if i were you id press on it more to teachers

parents

whoever

[mild anvil]

maybe its the exam anxiety

[serene swallow]

trust me bro they all give suggestions i have tried

[serene swallow]

im the chillest before each exam

[mild anvil]

and you feel nervous from your previous mistakes

[mild anvil]

everyone can lose their cool somehow

[mystic hound]

im assuming that you can read otherwise completely fine

and understand text completely normally

first time you read it

[mild anvil]

do a broad read of the question then close read

[serene swallow]

personally and respectfully speaking i dont think im that sorta guy and i have tried this anxiety management thingy too i think my lack of manfulness is far deeper than that

[serene swallow]

yes😭

[mild anvil]

hmm

[mystic hound]

weird

[mild anvil]

practice is key

[mystic hound]

this is so interesting

[serene swallow]

ahhh what i feel is i dont do that cuz i always finsih my paper cut to cut so 1 reading is fine for me
and by cut to cut i mean i only have 10 minutes left to revise and tie my answer sheet

[mystic hound]

but yeah idk what else to say other than ig practice a lot

give yourself a set amount of time when you practice and see if it still happens

[serene swallow]

trust me dude i do that😭
2 reference books

[serene swallow]

done that😭

[mystic hound]

does it still happen then?

[mild anvil]

clearly not enough

[serene swallow]

yes

[serene swallow]

ahhhh

[mystic hound]

i mean honestly idk much about psychology but maybe it’s like some kinda deeper anxiety thing or smth

[serene swallow]

chill out 😭😭

[mystic hound]

idk man

[mild anvil]

i think its just practixe

practice

[mystic hound]

i’ve never heard of it before

[serene swallow]

same here

[mystic hound]

for context how is a reference book?

like im guessing it’s just a bunch of math questions

[serene swallow]

uhm have u haerd of rd sharma and the ncert exemplar

like they are the best MATH books u can manage to get your hands on in india atleast

[sand pumice]

these two are the books everybody recommends for some reason

[serene swallow]

icl they are the best

[mystic hound]

nope

no idea

[serene swallow]

yea like they are the best

[mystic hound]

cool

[serene swallow]

here and pretty tuff too

[sand pumice]

pretty much

[serene swallow]

help me tho

[mystic hound]

is like a lot of just equations tho?

cause i’d guess the practice would just be doing a lot of practice questions

under time

[sand pumice]

one of them contains every type of problem that you would see the other i dont know

[serene swallow]

i am messing u

[serene swallow]

exemplar?

[sand pumice]

rds

[serene swallow]

should i read the question like he said

[mild anvil]

yes

you should

[serene swallow]

that is peak icl are u indian?

[sand pumice]

yes

[serene swallow]

but then my time is messed up

[serene swallow]

havent heard of rd😭👉

[sand pumice]

i was saying i didn't know much about exemplar

[mystic hound]

how much time do y’all have?

and like how many questions is it

[sand pumice]

try practicing with a restricted time limit more

[serene swallow]

oh yea that is peak too
high level for ez chapters i do exemplar
and tuff level for hard chpsters i do rd

[serene swallow]

rd is 600+ pages im sure
exemplar is 300ish

[mystic hound]

oh no i mean the average exam you have

[sand pumice]

usually it's 3 hours

[mystic hound]

how long are those relative to the amount of questions?

[sand pumice]

average as in periodic tests and stuff?

[mystic hound]

uh yeah sure

unless exams are like way longer ig

[sand pumice]

that varies everywhere

[deep abyss]

-# wut is tht pfp

[mystic hound]

well ig let’s say a final exam then

[mystic hound]

it’s tyler

[sand pumice]

3 hours

[mystic hound]

and how many questions

roughly

i’m trying to relate it to my exams

[deep abyss]

I think there r 30/40 ques

Ofc of different weightage

[mystic hound]

yeah

uhhh

[deep abyss]

@serene swallow maybe sample papers might help

[mystic hound]

oh wait that would like question 1 has task a, b, c so on or is each of those a separate question

cause we have 5 hours to do 10-15 tasks with anywhere from 1-4 questions per task

[deep abyss]

Like proof based ques or ques with mulitple sub parts

And there r mcqs

[mystic hound]

yeah we don’t have those

not at my level at least

high school last year (kinda)

[deep abyss]

Oh this is highschl last year then 11 12 is pre uni

[mystic hound]

you guys have stuff that’s pre uni but after high school?

well tbf our high school is later but

[deep abyss]

Where is OP

[mystic hound]

yeah

[deep abyss]

I thought everyone had tht?

[mystic hound]

uh i mean we don’t but like our school system is pretty different

i started in the school i’m in now at 18

but most start at 16-17

but no we just go directly to uni after this

[deep abyss]

Wyf

[mystic hound]

huh?

oh

denmark

[deep abyss]

Ahh is it like 6 25am?

[mystic hound]

yeah

i did not sleep

at all

[deep abyss]

Im such a genius

[mystic hound]

math and shit

this help channel is dead now

we killed it

teamwork

[deep abyss]

We who

[mystic hound]

dont try to pretend you didn’t take part in this

this is on both of us

[devout snow]

@serene swallow Has your question been resolved?

[deep abyss]

[devout snow]

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[devout snow]

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[main tendon]

Can someone guide me in solving this

[main tendon]

I found that the RHS is (n-1)³+n³

but how do I do for the LHS

[stone stump]

add up all the LHSs

and check where each of them start and end

[main tendon]

it is ending at n² and starting with (n-1)²+1?

[glossy dew]

yup

[olive snow]

Indeed

[tender wharf]

-# yes

[main tendon]

Now, how am I supposed to solve the (n-1)² terms

[tender wharf]

they are AP

[main tendon]

Yea but LHS became
[(n-1)²+1] + [(n-1)²+2] + ..... +n² right? So, how do I know the last term of that (n-1)²'s AP

[tender wharf]

last term is n^2?

[main tendon]

huh

am I getting something wrong here

[tender wharf]

for n> 0 every row starts with (n-1)²+1] and ends with n^2

so you have first and last term

you can figure the number of terms in every row

[main tendon]

oh mb when replacing the n with actual value, we do get the AP. Just the representation is different right?

[tender wharf]

yesh

[main tendon]

2n-2?

[tender wharf]

2n-1

[devout snow]

Channel closed

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[main tendon]

ty

[devout snow]

Available help channel!

Send your question here to claim the channel.

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• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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[broken narwhal]

should my first step be times 2 by 27/4

[broken narwhal]

if i didnt cancel

[mild anvil]

just cancel cirst

first

[broken narwhal]

but if it cant be cancelled

can that be my first step

[mild anvil]

but yes multiply 2 to remove the fraction

[pseudo basin]

can you do this step exactly as you intend it, ignoring whether it's legal, and show the result?

[devout snow]

@broken narwhal Has your question been resolved?

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lean lintel]

Four points A, B, C, D lie on a circle with diameter 6radical2 such that ABCD is a quadrilateral and BD is a diameter of the circle. If AC bisects angle BAD and the length of AD is 1, what is the perimeter of ABCD ?

Can anyone helo me solve this? I don't know where to start

[modern imp]

whats the diameter

[lean lintel]

6 radical 2

6 × radical2

Basically rad72

[modern imp]

ohk

i call it square root

so if i use that please bear with me

[lean lintel]

Same thing

[lean lintel]

👍

[modern imp]

alright, so since AC bisects angle BAD , we can locate C

wrt to the circle and BD

||its the arc midpoint||

[lean lintel]

Lemme draw a circle myself

[fluid token]

also, as BD is diameter, angle BAD is|| 90 degrees||

[lean lintel]

I'm trynna send a pic

[lean lintel]

I kinda get it until now

[modern imp]

nice

[lean lintel]

A is basically half the BD arc

So 90 degrees

[modern imp]

hold up

how is A midpoint of arc

[lean lintel]

Connect A to B and D

[modern imp]

yes

but that doesnt imply A is midpoint of arc

infact it isnt

[lean lintel]

Wait

We never said it is

[young spade]

Just as a side note, people usually write it as 6 sqrt2
mostly cause in this server we are used to writing in TeX markdown

[modern imp]

ideally i think C should be opposit A

cuz it bisects BAD

[lean lintel]

The BD arc on the other half of the circle should be 180 degrees

Wait

[modern imp]

yes but A and C should be on opposite sides

of BD

[lean lintel]

Ah

[modern imp]

now use pythagorus

[lean lintel]

U are right lemme check

Oki we are getting somewhere

Can u draw the circle and send me a pic

[modern imp]

i cant but i can explain

i dont have pencils and stuff

[lean lintel]

So here

What should I replace?

[modern imp]

yah, so basically since AC bisects BAD, AC must pass through BD, so A and C should be on opposite sides

[lean lintel]

Oh

Lemme test it

How can AC bisect BAD

[tender wharf]

he means like this

[deep abyss]

-# This is quad ABDC not ABCD

[modern imp]

i meant the angle mb

[lean lintel]

Smart

Jeez this Internet

Whatever I get it now

[tender wharf]

paint

[lean lintel]

Oki let's go to the next step

[modern imp]

alright

so we have AD

and we know angle BAD is 90

[lean lintel]

Yes

[modern imp]

what can we use

[lean lintel]

Same for BCD

[modern imp]

something famous about lengths in a right angled triangle?

[lean lintel]

Phytagrious guy

Spelled wrong tho

[modern imp]

pythagorus

iirc

anyways

[lean lintel]

Oki lemme see

We can find AD

[modern imp]

and since we know all the other sides....

[lean lintel]

We can find AB I meant

We already know AD

AB is obviously rad72 squared - 1

AB² I meant

[modern imp]

good fruit

[lean lintel]

So AB it self is Rad(rad72-1)

[deep abyss]

$\sqrt{72}^2-1$

[woven radish]

ch3rry

[lean lintel]

It is lol

It isn't?

[modern imp]

no you have to square the rad72

as mr cherry is doing

wait sorry ms cherry

[deep abyss]

-# mr cherry naurway

[lean lintel]

Square of rad72 is 72 which is AB²

[modern imp]

72-1

[modern imp]

im so sorry

[lean lintel]

Oh after that we take root again

[deep abyss]

-# how dare u assume my gender

[modern imp]

im sorry

[lean lintel]

Oki so AB is rad71

Now let's find other sides

[deep abyss]

-# its ok

[lean lintel]

Can we somehow find CD?

[modern imp]

its the midpoint of arc

[lean lintel]

The length of CD

[modern imp]

yes

we can use that

[lean lintel]

How

[modern imp]

OD=OC and angle is 90

[lean lintel]

Who is O

[modern imp]

centre lol

[lean lintel]

Oh

How come is OD = OC

It is

[modern imp]

its a circle

[lean lintel]

Radius

[lean lintel]

Yea

[modern imp]

so both are equal to radius

[lean lintel]

Ye correct

Now how we find length of CD

[modern imp]

pythagorus again

[lean lintel]

Smart

18 + DC² = OC?

[modern imp]

nope

[lean lintel]

What's wrong

OC squared ops

[modern imp]

OC^2+OD^2=CD^2

and we know OC and OD

[lean lintel]

U sure ?

CD isn't next to the 90 degree angle

[modern imp]

YES

CD is the hypotenuse

[lean lintel]

Hmh

But

OC should be hypotenuse

[fluid token]

angle COD is 90 degrees
-# (assuming O is the center)

[modern imp]

i gtg now, valar can you help pineapple

[lean lintel]

It also can not be 90 degrees tho

Cuz C can also not be in the middle of the arc

[lean lintel]

👍

[lean lintel]

Am I wrong?

[fluid token]

angleCAD = angle CAB (AC is angle Bisector)

so arc BC and arc CD are equal as they subtend equal angles on circumference

[lean lintel]

I think I'm starting to get it

Can u draw it for me?

[fluid token]

ye sure

[lean lintel]

Thanks ❤️

So

[deep abyss]

I also draw for you

[lean lintel]

Here we wanna figure out CD right?

[lean lintel]

❤️❤️

[deep abyss]

CD=BC

Equal arcs subtend equal angles

[lean lintel]

Yup angle C is equal to angle A

[deep abyss]

Or well chords

Huh

[lean lintel]

Language barrier makes this worse

[deep abyss]

I was referring to BAC=CAD=45°

These are subtended by CD and BC

[lean lintel]

BAD is 90 degrees

[deep abyss]

So tht means those r equal

BAC mb

[tender wharf]

what's your native?

[lean lintel]

Persian

[lean lintel]

Yup correct

[tender wharf]

ah

[deep abyss]

Okay so

[lean lintel]

Oki both 45

[deep abyss]

Now i suppose you can use right triangle BCD to ur advantage

[lean lintel]

We can

We can figure out BC if we had CD

Problem is we don't have CD but we can find it

[deep abyss]

...

[lean lintel]

Am I being dumb?

[deep abyss]

Say $(6\sqrt{2})^2=BC^2+CD^2$

[woven radish]

ch3rry

[deep abyss]

And $BC=CD$

[woven radish]

ch3rry

[deep abyss]

2 eqns 2 variables

[lean lintel]

Wait a minute

[lean lintel]

I get it BC squared + CD squared is diameter squared

But how do we know BC and CD are the same?

[deep abyss]

Do tk this theorem?

[lean lintel]

I do know it but the arcs aren't the same

[deep abyss]

.

[deep abyss]

.

[lean lintel]

Ah fuck

Jeez

They are both 45 BCD

And CDA

[deep abyss]

Yes pineapple

No CDA is 90

[lean lintel]

So yea then CD = BC

[deep abyss]

Right pineapple

[lean lintel]

CDB I meant

Noe

Their squares is 72

So they are each radical 32

Or 4 radical 2

[deep abyss]

Wut

[deep abyss]

Try thiss

[lean lintel]

CD² + BC² = 72

[deep abyss]

Turu then

[lean lintel]

They are equal So each of them is radical 32

Radical 32 is 4 × radical 2

Total 8 radical 2s is total for CD + CB

[deep abyss]

You mean rt36?

[lean lintel]

Yea I meant 36 sry

So total is 13 + rad71

The perimeter

Wow u solved it

6 + 6 + 1 + rad71 ( we found 1 + rad71 earlier )

[deep abyss]

Yayy

Go pineappleeeee

[lean lintel]

Hurray

I felt so dumb

[deep abyss]

No pinappleee

[lean lintel]

[deep abyss]

Thank me

[lean lintel]

Thanks alot for your help and see you!

.close

[devout snow]

Channel closed

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[devout snow]

Available help channel!

Send your question here to claim the channel.

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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!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[royal jay]

hi

[royal jay]

Hi I have this logic (r→(s∨p))∧(r∧q)∧(p→¬q) and I am supposed to prove that it's true when using both the deduction and the reduction method for logic. I have made steps for both but im getting stuck on the later parts so I would like some help

[royal jay]

1: r∧¬¬q assumption
2: r ,1 conjuctive simplification
3: q ,1 conjuctive simplification + double negative
4: p→¬q assumption
5: ¬p ,3,4 modus tollens

as for reduction method:

(r→(s∨p))∧(r∧q)∧(p→¬q) jag har gjort följande med reduktions metoden
1: assume that the conclusion is false but the assumptions are correct but S is false
2:r∧¬¬q is true so r and q must both be true
3:q true implicates that ¬q is false
4: ¬q false means that p must also be false since p→¬q is true

but from there im kinda stuck

[devout snow]

@royal jay Has your question been resolved?

[devout snow]

@royal jay Has your question been resolved?

[royal jay]

<@&286206848099549185>

[vale bolt]

hi

[royal jay]

hello sorry for pinging I posted my problem above, im new to this server and this service in particular so im not sure if I have done this correctly

[devout snow]

@royal jay Has your question been resolved?

[covert shale]

you're alright

where does this assumption ( r∧¬¬q ) come from?

from (r∧q) ?

[royal jay]

ahh my bad I

yes

its originally r∧--q

[covert shale]

got it

[royal jay]

so a double negative

but I dont know why he did that

[covert shale]

actually, can you post the whole original problem?

[royal jay]

sure gimme a sec

[covert shale]

idk why question makers write double negatives either it's pretty redundant

[covert shale]

,rotate

[royal jay]

its in swedish but nevermind that

[covert shale]

ah we want to show this implies s

[royal jay]

yes

[covert shale]

you can start by rewriting the implications to get a more normal form perhaps

[royal jay]

I might in a normal case but my teacher is very obtuse and im not risking it

[covert shale]

ah i see

[royal jay]

its the last steps that I need help with from both the reduction and deduction method

1: r∧¬¬q assumption
2: r ,1 conjuctive simplification
3: q ,1 conjuctive simplification + double negative
4: p→¬q assumption
5: ¬p ,3,4 modus tollens
as for reduction method:

(r→(s∨p))∧(r∧q)∧(p→¬q) jag har gjort följande med reduktions metoden
1: assume that the conclusion is false but the assumptions are correct but S is false
2:r∧¬¬q is true so r and q must both be true
3:q true implicates that ¬q is false
4: ¬q false means that p must also be false since p→¬q is true
but from there im kinda stuck

I did this for both

[covert shale]

got it

i'll take another look

perhaps it is best to consider the whole expression including the final implication

[royal jay]

what do you mean?

[covert shale]

no i thought i was onto something, wasn't lol

i'm stuck on where the assumption that (r and q) is true comes from, it doesn't seem immediate

unless we're just asserting that it is

[royal jay]

deduction or reduction?

[covert shale]

both

[royal jay]

ah I c

my teacher has these truth diagrams

and if we look at the top left one

we se that r and q can only be true if they are both true

thats for reduction

as for deduction I used subjunctive simplification

[covert shale]

i see, but the truth values for p, q, r, and s are not assigned yet, so we cannot make this assumption

[royal jay]

i think thats what its called

well we since we assume that the expressions are true and that the conclusion S is false

according to my teacher anyway

[covert shale]

hm, interesting

so he's trying to prove a contradiction then

[royal jay]

?

ah

yes

[covert shale]

so we want to reduce to (not s) basically

[royal jay]

we prove that the assumption of S being false is wrong

yes

sorry if im not conveying the problem

[covert shale]

no worries, just trying to see what your teacher is trying to convey to you

this is all pretty simple algebra when you rewrite the implications and distribute a little, but you said you don't want to do that

[royal jay]

well its not that I dont, its more like Im not sure if he will accept it

one of the students from his class said one wrong word for this particular problem and he sent the problem back and said redo

everything else was correct

but not that one word

so he is very on the nose with everything

[covert shale]

augh that's strange

[royal jay]

yeee

he is special

[covert shale]

lol

[royal jay]

but I need to pass this or I cant go for my masters degree next year

[covert shale]

oh god 😭

[royal jay]

I have done the exam but not these problems that we send in

and the deadline is 12 of january

[covert shale]

i'd first try using the rewrite as a back-up solution and then consult him to see if he'd accept that when the time comes

i need to go for now, maybe someone else can take a better look

[royal jay]

ahh i c

thanks

should I ping or just wait?

[covert shale]

hi, i'm back

you might want to consult #proofs-and-logic or something for another opinion

[devout snow]

@royal jay Has your question been resolved?

[devout snow]

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[neon wagon]

On another wheel of fortune with 100 sectors, 20 sectors are labeled “Win”, 50 are labeled “Lose”, and the rest are labeled “0”.

c) Calculate how many times the player must spin the wheel at least in order to have a probability of at least 94.88% of spinning “Win” at least once.

can someone help

[neon wagon]

like this formula is correct but idk why its >=

[full trellis]

Do you have the rest of your work or is this just copied from the solution?

[neon wagon]

no this is mine

like

i just kinda memorized how to solve these type of questions

but i dont really udnertsnd

[full trellis]

Alright let's just try and solve it from the beginning then

Whats the probability to win in one spin?

[neon wagon]

0,2

to get the label win

yeah

[full trellis]

Yeah, then we use the complement event 'no win in n spins '(= every spin is not a win)

[neon wagon]

yeah so 0,2 ^n for the at least thing right

[neon wagon]

why no win tho

were looking for atleast one win

[full trellis]

Because 'at least one win' is difficult to count directly

But the complement is a lot easier

What are some possible cases of getting a win at least once in n spins?

[neon wagon]

uhh

wdym

i dont understand

[full trellis]

Like say you do 10 spins. What different ways are there to get at least one win?

You could get 1 win, 2 wins, 3 wins...

[neon wagon]

one at the begning

oh

so uhh

like every possible way i mean

10 wins or 9 or 8 or 6

etc

[full trellis]

Yeah any one of those can mean 'at least one win'

[neon wagon]

oh yeah ngl

ur right

[full trellis]

So when we take the complement, there is only one case to look at

So whats the probability of no win in n spins?

[neon wagon]

what event would we be looking at in the complement

[full trellis]

Complement is 'no win in n spins' in this case

[neon wagon]

ohhh

were seperating the wins from the other things

so we are just left with the wins

[neon wagon]

like 0%?

[full trellis]

How do you take the complement? The chance to win in one spin was 0.2

Whats the chance a spin is not a win?

[neon wagon]

0.8

%

[full trellis]

Right

[neon wagon]

oh yeah

[full trellis]

No just 0.8

That would be 80%

[neon wagon]

the n confuses me

yeah

[full trellis]

Okay so we have 0.8 probability in one spin. What is the chance for n spins?

[neon wagon]

n spins? wdym

[full trellis]

What is the chance to keep getting 'no win' when we do an n amount of spins

[neon wagon]

also 0.8?

[full trellis]

Let's think about just doing two spins

[neon wagon]

okay

[full trellis]

So n = 2 right?

[neon wagon]

yeah

[full trellis]

So you have (not win spin 1) AND (not win spin 2)

[neon wagon]

yeah

[full trellis]

How do you calculate the chance here?

[neon wagon]

0,8^2

?

[full trellis]

Yep

Notice the exponent is 2

[neon wagon]

yeah

[full trellis]

So what is it for n amount of spins?

[neon wagon]

n

infinite

idk

like

oh

0.8^n

[full trellis]

Mhm

[neon wagon]

oh

yeah

[full trellis]

So how do we properly write P(at least one win) knowing this?

[neon wagon]

1-0.8^n

i understand

[full trellis]

Exactly

And that's how you get the expression in your photo

[neon wagon]

we seperate teh wins from the other

oh but

heres the question

why the >=

whats up with that

the >

[full trellis]

Because it is at least one win. You can also get 2, 3, 4, 5, ... , n amount of wins

That would also satisfy 'at least 1'

[neon wagon]

ohhhh

like

so "<" would be then

at most?

right

[full trellis]

Yes

[neon wagon]

oh okay

man i got alot of subjects left to study

[full trellis]

As you work out this expression, the inequality sign will switch though if you haven't worked it out already

[neon wagon]

oh yeah ik about that

[full trellis]

Good luck!!

[devout snow]

@neon wagon Has your question been resolved?

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[devout snow]

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[grand edge]

Have you tried Change of base?

[bleak stag]

i was tryin to use ln

[inner summit]

what have you tried

[inner summit]

do you know what the change of base formula is

[grand edge]

pretty sure they don't and that's why they need help

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

[woven radish]

Roy

[grand edge]

Change of base is how you're going to get any natural logarithms and get the x term out of the base. Once you do that, you have a simple equation in terms of ln(x)

[bleak stag]

ok i will try

now idk how to go from this

[pearl relic]

wrong order

flip each fraction

the base goes on the bottom

[torpid cloak]

She fixes it in second img

[stoic lotus]

Why would you convert to natural log instead of base 4?

You'd get a nice quadratic equation

[pearl relic]

second img uses 6 instead of 16

[stoic lotus]

I guess it also works

[bleak stag]

ok i finished

[stoic lotus]

You just factor out ln4, pretty much the same thing

[bleak stag]

but i lowk dunno the restrictions

[stoic lotus]

Wdym

[bleak stag]

like the restrictions on the domain

[stoic lotus]

Logarithm bases cannot be negative and cannot equal to 1

[bleak stag]

hopefully this is right

[stoic lotus]

Yep

Nice

[bleak stag]

thanks guys

.close

[devout snow]

Channel closed

Closed by @bleak stag

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[devout snow]

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[forest flower]

hi can anyone help me with this problem i dont get how im supposed to do it

[devout snow]

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

[forest flower]

1

[sand quarry]

Would you like a hint

[forest flower]

yeah please

[uncut crow]

i have a hint. what would be the answer if you surveyed 1 person?

go away

[sand quarry]

Could you translate "48 out of every 120 people" in mathematical terms

[violet wind]

!nosols

!noai

[devout snow]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[sand quarry]

<@&268886789983436800>

[forest flower]

wuh

[smoky hull]

what an encounter

[forest flower]

48/120

[sand quarry]

Simplify that

[arctic field]

using AI to respond to help channels is not helpful

[forest flower]

12/30

[sand quarry]

More

[forest flower]

2/5

[sand quarry]

Ok awesome

[forest flower]

😲😲

wait hows that possible

[sand quarry]

So out of 500 people, how many like baseball given that ratio

[sand quarry]

Hmm?

[forest flower]

hmm

[forest flower]

mvm mb

200/500

[sand quarry]

Yeah

So how many

[forest flower]

300 play it

wait

how many what

[sand quarry]

How many people like it

[sand quarry]

From this

[forest flower]

200

[sand quarry]

Yeah cool

So 3/5 of those people play it

What'd that number be

[forest flower]

hmm

lemme think

how did i get 600/2500

do i simplify it

[sand quarry]

Idk either but thats correct

[sand quarry]

Yes

Maks the denominator 500

[forest flower]

120/500

[sand quarry]

Yeah

[forest flower]

😯😯

[sand quarry]

So how many people played baseball

[forest flower]

120

😲😲😲😲😲😲😲😲

[sand quarry]

Aight

[forest flower]

is that the answer

[sand quarry]

Yep

[forest flower]

THANK YOU

[devout snow]

@forest flower Has your question been resolved?

Channel closed

Closed by @forest flower

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[glossy obsidian]

FInding the natural domain and range of this function

[glossy obsidian]

I've been thought to find the factors when it comes to quadratics which means x values they cannot be..

but in this, they are unfactorisable due to having a negative discriminant

anyone mind telling me how to start in these type of questions?

[void fox]

for this expression to make sense you basically only require that the argument to the logarithm is positive

ie it is asking you when is x^2+2x+3 positive

[tall comet]

if D<0 then the range of the quadratic is always positive

[crystal dawn]

can be always negative too

but yea in this case

[tall comet]

oh yeah i forgor

[tall comet]

can be -ve too so just check that once

[glossy obsidian]

so if the domain is greater than 0, that would mean the range is always positive?

[crystal dawn]

range of ln is always all reals

[glossy obsidian]

ahh ok

what about normal log

[crystal dawn]

same

[tall comet]

if it is positive then there are no restrictions for the log

[glossy obsidian]

so any type of log will always have a all real y?

[crystal dawn]

$\log_a(x)$ produces all reals

[void fox]

Maybe the question meant to ask what is the image of this function

[woven radish]

Hanako(x, y); ∂(fox)/∂x

[void fox]

otherwise you can always just say the range is ℝ

[crystal dawn]

and this is regardless of its domain

[glossy obsidian]

its asking for the natural domain of the function

[glossy obsidian]

ok thanks

[crystal dawn]

in this case, the only thing that concerns you is that the argument to the log must always be positive

then check the argument of the log

[glossy obsidian]

what do u mean by arguement of the log?

[crystal dawn]

in here, x is the argument to the log function

[glossy obsidian]

oh yeah

[tall comet]

the thing inside the log function

[crystal dawn]

so check whatever inside the log function. ban all values of the variable that makes the argument negative or 0

[glossy obsidian]

so in this case the arguement of the log is positive, how does that effect the function?

would that mean the domain and range will always be all real?

[crystal dawn]

yes

wait actually no

the domain is all reals, yes

[glossy obsidian]

what about range

[crystal dawn]

but because you have a minimum on the quadratic, you'll need to find the minimum of the log separately

let the quadratic be at its min, then solve for the log

[glossy obsidian]

so for this question here, i know that if i sub in anynumber it will spit out a positive number, which means that it is all real x?

[crystal dawn]

all real x is the domain, yes

[glossy obsidian]

i dont get this sorry

[crystal dawn]

you know how quadratics either have a min or a max?

[glossy obsidian]

yeah

[crystal dawn]

since log is strictly increasing over its domain, and the argument is a quadratic that is never negative, the minimum of the quadratic argument is also the minimum of the log function in this case

put another way, the quadratic will never get under a certain value

the minimum of the log function is thus ln(whatever value the quadratic will never fall below)

[glossy obsidian]

so, whenever there is a quadratic as a arugement of log.. the means that the quadratic represents the minimum of the log(quadratic) graph?

[crystal dawn]

no

its minimum does, assuming its minimum is not negative

and it applies in this case because the minimum of this quadratic is positive (the whole quad is always positive anyway)

[glossy obsidian]

is what you're saying here is that the minimum of both the log(quadratic) and quadratic are the same? but only when the quadratic has a minimum?

[crystal dawn]

no

[glossy obsidian]

sorry i dont quiet get it 😓

[crystal dawn]

ok so let's find the minimum of this quad

[glossy obsidian]

-2/3

?

[crystal dawn]

the minimum of this quad is f(-2/2) = f(-1) = 1 - 2 + 3 = 2

but that is just the minimum of the quadratic

so now, ln(2) is the minimum of c(x)

[glossy obsidian]

ok

[crystal dawn]

and ln(2) is

,w ln(2)

[woven radish]

Results provided by WolframAlpha

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Requested by flowerychild

[crystal dawn]

that thing

so your range is all real y greater to or equal to that thing

[glossy obsidian]

ahhh ok icic

so id have to find the minimum of the y then put that into the log function and that will spit out the minimum of the function

[crystal dawn]

mm...

[glossy obsidian]

and in this case since its a minimum, its anything greater than the new minimum

[crystal dawn]

ok let's make it clearer

suppose you have a function $\ln{c}$ where c is some expression (linear, quadratic, etc.)

[woven radish]

Hanako(x, y); ∂(fox)/∂x

[crystal dawn]

then, if c is always positive and has no minimum (or has a minimum of 0), the range of ln c is still all reals.
if c is always positive and has a minimum, the range of ln c is all real y greater than, or equal to, the minimum of ln c

[glossy obsidian]

right

in expressions like ln( sinx or e^x ) that just means its all real

[crystal dawn]

careful. the argument to logs cannot be 0

and sin(x) = 0 for all x = npi (n integer)

[glossy obsidian]

ahh ok

[crystal dawn]

also uh

ln(e^x) is just x

by definition

[uncut crow]

sin(x) is also negative in some places

[crystal dawn]

oops yea forgor

[glossy obsidian]

i just realised 😓

[glossy obsidian]

oh yeah so it just wouldnt work

[crystal dawn]

it would work where sine is positive

but anywhere outside and ln just dies

well any log just dies, not just ln

[uncut crow]

it wouldn’t parse when x is such that sin(x) is negative

[crystal dawn]

so the domain of ln(sin(x)) looks more like {x | 0 < x - 2n pi < pi} (integer n)

[glossy obsidian]

why wouldnt ln(sinx) not work out
if i rmbb sinx has positive and negative

[crystal dawn]

are you asking why would it work out or why would it not work out?

[glossy obsidian]

why it wouldnt

[crystal dawn]

it would! you'll just need to limit the domain

[crystal dawn]

^

[glossy obsidian]

as long as the log function is positive in all areas, the log will work out

but in the case of sinx since it has both positive and negative, we would have to chop off the domain to make sure the function is always positive?

[crystal dawn]

correct

otherwise log dies and we are sad

[glossy obsidian]

ahh ok makes more sense now

and to recall

log cannot be zero either is that right

[crystal dawn]

yup, hence why I used < instead of $\leq$

[woven radish]

Hanako(x, y); ∂(fox)/∂x

[glossy obsidian]

ahh ok

js wanted to clarify

if the log is greater than 1 does that mean, x1 <= 1?

[crystal dawn]

what's x1

[glossy obsidian]

a function

[crystal dawn]

and where is this function

[glossy obsidian]

wait sorry the base*

not function

[crystal dawn]

so you're asking, if $\log_{x_1}(c) > 1$, then ${x_1}$ is necesarily $\leq 1$

[woven radish]

Hanako(x, y); ∂(fox)/∂x

[glossy obsidian]

yes

[crystal dawn]

no, of course not.
ln(e^2) = 2 but e > 1

[glossy obsidian]

because i got some help from someone last time and thats what i was told, but i wasnt too sure about it

[crystal dawn]

was there some context to the question?

[glossy obsidian]

they were asking to solve the inequation

[crystal dawn]

but the base here is e

doesn't make sense, cuz e > 1

[glossy obsidian]

i dont quiet get it

[crystal dawn]

your question earlier does not make much sense with this context, because here you are given a fixed base that is quite obviously > 1

as a reminder, you asked whether the base of a log is necessarily leq 1 if the log itself is greater than 1

but here the base is already greater than 1

perhaps there is some other context I'm missing or I'm misreading this hard

[glossy obsidian]

the base that was given was > 1, and i was told that whenever it is larger than 1, then x1(function) <= 0

when the base is < 1 , then x1 >= 0

the inequality sign is switched up

i was just asking whether that is correct or not

sorry that i didnt clarify

[crystal dawn]

uh... let me try to understand

what is x1(function)?

[glossy obsidian]

i was saying that x1 represents a function

[crystal dawn]

ok so now x1 is a function and not the base. what kind of function is x1? what has it to do with the log in question?

[glossy obsidian]

ignore what i asked u in the very beginning because i got things messed up, but lets say the function is this

i was saying that since e is around 2.71, and is > 1

the function is <= 0

but lets say instead of e its 0.4, the function would be => 0

[crystal dawn]

but the value of the function depends on x!

you can't just say the function is less than 0 or greater than 0 without saying what value you're using for x

[glossy obsidian]

ohh sorry

ur right

[glossy obsidian]

im just wondering, how would u approach this question?

[crystal dawn]

the whole log is \leq 0 if its argument is \leq 1

so check where the rational function inside the log has the same sign on both num and denom, and abs(num) < abs(denom), and (num \neq 0 and denom \neq 0)

[glossy obsidian]

sorry what does leq and neq mean?

[crystal dawn]

to slow down and clarify:

• same sign on numerator and denominator to prevent negative argument to log (negative divided by a negative is positive, recall)
• abs(numerator) \leq abs(denominator) to ensure a fractional argument, fulfilling the \leq 1 criteria
• numerator \neq 0 to prevent 0 argument to log
• denominator \neq 0 for obvious reasons

\neq = not equal to, \leq = less than or equal to

the terminology comes from latex: $\neq$ and $\leq$

[woven radish]

Hanako(x, y); ∂(fox)/∂x

[glossy obsidian]

ahh okay i understand

i just tested, and whenver i put it any number a positive number will come

[crystal dawn]

might wanna double-check

[glossy obsidian]

oh hm

so far some fractions do not work

ok all good i think ive gotte it now

thank you so much for your time and help @crystal dawn

hope i can get help from u again sometime, cos im for sure asking again sometime again 😅

[crystal dawn]

nps

!done if you're done

[devout snow]

If you are done with this channel, please mark your problem as solved by typing .close

[glossy obsidian]

!done

[devout snow]

If you are done with this channel, please mark your problem as solved by typing .close

[glossy obsidian]

.close

[devout snow]

Channel closed

Closed by @glossy obsidian

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[short hare]

f(x) is a piecewise bitch of
3x^2 + 4 : x >= 1
6x+1 : x < 1
I wanna find the local maximum/minimum/(s)/ifthereis

I started by differentiating the two pieces and rewriting
6x : x > 1
6 : x < 1
first question; what do i write for x=1?

the critical points are supposedly the turning point (x=1) and f'(x) = 0 which from the 6x piece we get x=0

[short hare]

but is wrong 😔 pls help

[crystal dawn]

uh, what happened to x = 1 on the first definition in the derivative?

[void fox]

f(x) is what??

[short hare]

idk what to write for x=1

[crystal dawn]

it should be the first definition as well, no?

[short hare]

it is

[crystal dawn]

your first piece is x \geq 1, but your derivative's first piece is x > 1

[short hare]

yeah arent we supposed to seperate the x= piece from the x> piece

bc

we're not sure if

its even differnetiable there

[crystal dawn]

I must be tweaking out then, ignore me

[winter torrent]

they're essentially the same. differentiating on a closed interval is kinda cringe

[crystal dawn]

ah
every day my lack of calc knowledge shows some more

[winter torrent]

no it's fine that's a reasonable doubt

[short hare]

what do i do hayley 🙁

[winter torrent]

well since the limits of your derivatives agree at x=1 you can say that the derivative at x=1 is also 6

[short hare]

great i thought so too

now, what are the critical points for hte fucntion

[winter torrent]

u tell me

[short hare]

why am i constnatly being hamster held

[crystal dawn]

because you are car
now give it a shot :D

[short hare]

i said, at the turning point which is x=1 and from f'(x) = 0 which is x=0

is that correct

[winter torrent]

i guess it depends on your definition of critical point but x=1 is definitely worth keeping in mind

f ' (0) does not equal 0 though

[short hare]

hm?

[winter torrent]

you have f ' (x) as this right?
6x : x > 1
6 : x < 1

[short hare]

yes

[hollow ice]

for x<1, you have a different definition of f'

you cant use 6x there

[short hare]

oh is 0 less than 1?

[winter torrent]

sure is

[short hare]

my bad

[crystal dawn]

learning calculus to forget arithmetic, respect

[hollow ice]

TIL - car, 2026

[short hare]

6=0 doesnt sound promising

[hollow ice]

no one said it is...

[short hare]

undefined...??

[hollow ice]

undefined what? where?

[short hare]

if we set f'(x) = 0, we get 6=0

what does that mean

[hollow ice]

doesnt mean anything

f' is never 0 anywhere

for your choice of f

[short hare]

so the only critical point is at x= 1

[crystal dawn]

um, car? f'(x) = 0 doesn't mean you get to auto-sub x = 0

[short hare]

elaborate?

I mean- yeah true but where are you getting that form

[crystal dawn]

you saw f'(x) = 0 and you somehow got 6=0
from your earlier confusion I deduced that you had to be using x = 0, which lies in the second definition of f'(x), giving you 6

[short hare]

the piece i was incorrectly usng was f'(x) = 6x whicwhen =0 means x=0

[crystal dawn]

oh lol ok

[short hare]

but yeah ty for pointing that out

I can be silly sometimes 🤩

[short hare]

so-

yeah

I usually dont put ? at the end of my questions

so yeah, that was a question

[hollow ice]

yes, thats the only point in the domain of f where the derivative is undefined

[short hare]

great

ty bacter, hana and hayley!

.close

[devout snow]

Channel closed

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Use .reopen if this was a mistake.

[devout snow]

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[thorny rune]

I'm trying to understand convergence radius.
this was on the website i was using. can anyone explain why this is 1?

[void fox]

What definition of limits are you comfortable with? In any case this is immediate because you can bound the difference |(n+1)/n|-1 by taking n as large as you like

[thorny rune]

but it wouldn't (n+1)/n result in a fraction?

[void fox]

What does being a fraction have anything to do with the question

[covert inlet]

Try to think it in this way imo: n is so much bigger than 1 so we can just remove +1. Try to group by n and you will see

[thorny rune]

ummmm im gonna sound really dumb, but im not sure what defintions you are referring to, if there is more than one.

i know that when you graph, its the thing you get close to, but don't touch.

[wind mason]

what about algebraically?

like do you know epsilon-delta or roughly how to calculate limits algebraically

[void fox]

I think they are probably just doing an intuitive (non-rigorous) definition of limitsa

[thorny rune]

if n were 10,
it would be 11/10, and the answer wouldn't be 1. there would be a remainder

[covert inlet]

But n is not 10, n is a much much bigger number. It goes to infinity. If you collect n you get n(1+1/n) /n

[thorny rune]

i don't quite understand epsilon-delta, tbh. i was trying it out but got lost.

[wind mason]

k but what happens as n gets arbitrarily large

like lets say 99999999

what’s the remainder on that

[covert inlet]

1/n is 0

[wispy river]

Remainder tends to 0

[thorny rune]

aah, so the remainder becomes so small, it's inconsequential.

[wind mason]

not you sir! Asking OP

[thorny rune]

like how 1/x could be 0

[wind mason]

what you have said is basically, (n+1)/n=1+1/n

[thorny rune]

ahh

[wind mason]

and the RHS will tend to 1 as n->infinity

[wispy river]

1/infinity tends to 0

[covert inlet]

Only if x goes to inf

[wind mason]

because the remainder becomes arbitrarily small

[wispy river]

Yeh

[thorny rune]

okay, that makes sense. thanks

[covert inlet]

Try with the calculator: 1/99999, 1/677567577 etc

[thorny rune]

on a side note, does anyone have anything to recommend for learning epsilon - delta?

[covert inlet]

You will see a number that is almost 0

[wind mason]

You should do more algebraic practice with similar limits, like $\lim_{x\to\infty}\frac{5x+2}{x+1}$

there’s a neat formula for these kinds of limits of rational functions

[wispy river]

Do different kind of questions

[woven radish]

;(

[thorny rune]

how do i practice that?

[wind mason]

Once you experiment with higher orders