#help-0
1 messages ยท Page 500 of 1
nope
the
I suck at math
no, but i want to let new people help others aswell
yes!
oki
take the square root of both sides
bro which grade are you in??
stop doign this
break it down the 18 divide it by 2 until its no more divisible
Bonk
hey, listen, we've already had this conversation, and i'll keep doing it. If you're not happy just mute me
it just doesnt add anything to the conversation and its spamming
yay
? spamming ? You for sure know there is a notion of time in the the concept of spamming yes ?
you can go further if you want
now we have the sidelength of the triangle
now to get the area, we can break it up into two parts
how..?
do you notice how the area we want is equal to the area of the quarter circle minus the area of the triangle?
no
well just focus on bonk he is explainng well and patiently
well you are in 10th right .. from which country??
australia
OH
YEA
I SEE IT NOW
i think thats what my friend was trying to explain to me ngl
keep it up buddy you are working very hard
yay!
ok
so i work out the quartercircle
yes?
eys
yes
pi r^2 is the area of the full circle
indeed
that is for a whole circle you need to divide that by 4
and what is our radius?
square root 18
yup
so then what is the area of the quarter circle?
yes โค๏ธ
can you show how you got that answer?
1/4 x 18
hm?
you forgate the pi
area of quarter circle: $\frac{\pi r^2}4$
Bonk
you dont pull the 4 into the sqrt
square of what???
the radius is root of 18 right
looks good
so when we square that term the square root and the square will get canceled each other which will only lrft 18
ok...
you know the value of pi right??
leave in exact..?
3.14159
this is pi followed by
2653589
circumference over diameter
i forgto the rest
trust
yup just take upto 2nd digit after decimal
im better at chem
do not put it in decimals
ik
keep it exact
yes good
so, what did we calculate again?
4.5ฯ???
yes, but what is this for
of...?
indeed
and now we need the area of the triangle, right?
do you know how to calculate that
its 1/2 * base * height, yes
yup
so 9
that is indeed the area of the triangle
so we have the area of the quarter circle
and the area of the triangle
what do we do now?
do i have to leave it in exact form now
it jsut says find the shaded area
then exact form is fine
i dont like exact form
you can also do $\frac92\pi - 9\approx 5.14$ if you like
Bonk
youre going have to get used to it
you have to find the shraded part??
yes
what do you think you can do by seeing this image
minus little sector from big sector
yes indeed
so how we are going to minus?
$\tikz{\fill[blue!40!yellow] (0,1) --++(1,0)arc[start angle=0,end angle=90,radius=1];} - \tikz{\fill[blue!40!yellow] (1,0) -- (0,1) -- (0,0) -- cycle;} = \tikz{\fill[blue!40!yellow] (1,0) arc (0:90:1);}$
Bonk
yes!
i got 6.28cm^2
thats cool
alggg
i think that is not correct?? from what is asked in this question
ye it was the previous question, but i took too long haha
onh my bad haha
$\tikz{\fill[blue!40!yellow] (0,1) --++(1,0)arc[start angle=0,end angle=90,radius=1];} - \tikz{\fill[blue!40!blue] (1,0) -- (0,1) -- (0,0) -- cycle;} = \tikz{\fill[blue!40!yellow] (1,0) arc (0:90:1);\fill[blue!40!blue] (1,0) -- (0,1) -- (0,0) -- cycle;}$
Bonk
continue with your calculation
for which
is that solved??
ah, you did it already
i recommend going as far as possible writing it in exact form
then answering it in decimal
like here you should do $\\begin{aligned}&\frac{30}{360}7^2\pi-\frac{30}{360}5^2\pi\=&\frac1{12}49\pi-\frac1{12}25\pi\=&\frac{24}{12}\pi\=&2\pi\end{aligned}$
can you show the question?
i have a headache
look at each step
Bonk
hm
perhaps showing it like this is more obvious
where is the rectangular prism?
a..?
yes
this is correct
yay
youre doing b?
yes
ok
thats the diagonal
this is when you look at it from the front
can you see that?
unfortunately
unfortunately?
this is a bit more work than a, yes
lets goo
can you show your calculations?
then make sure its clean next time ๐
๐ฅฒ๐ฅฒ๐ฅฒ
yes
whats A5?
uhhh
you just wrote 604.8
i tryyy
the area looks correct
volume aswell
seems like youre already quite comfortable with this topic
youre doing very well
yay
this is just
this weeks hw
i still have
like
2 and a half pages
and its due on sunday
do as many as you can
and ask a question if you need it
or need your answer needs checking
!done
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use dm^3 = 1 L
..?
10cm * 10cm * 10cm = 1 Liter
hm
=> 0.1m * 0.1m * 0.1m = 1 Liter
oh.
i just times by 1000
im doing this question now
but with this you can see why thats the case ๐
hmmm
okay 
this step is wrong
cm^3=L
anyway
you pretty much know how to answer these questions
i recommend you close this channel
ok
and open a new one if you have a new question
ima go bed anyways
(also i gotta go for lunch lol)
oof
bed? where are u from
h, +10
europe..?
ofc
ye
england?
nope
smth close..?
i seem british ig lol
across the channel
france..?
i doubt youre gonna guess it lol
netherlands
iceland..?
ohhhh
๐
DONT BEAT ME UP
we barely learn about the geography of other countrues
you only learn about australian geography?
netherlands is dutch..?
yes
not realy either
oh wow, this side of the country is alloutback
this other side
is also basically all outback
but some cities on the coast
๐
geography of australia done
cause im moving to europe this year
fr?
oh? why
exchange student
and doing exchange??
yes
for how long
6 months
yes
lol
so youre just oging to school in europe for 6 months just cuz you can
i mean
its a good change
ive been going to basically the same schhool my whole life
how old are you bonk???
with most of the same people
im 22
yeah thats usually how it works lol
what are you doing these days
not really
wdym not rly
gotcha
here youre in the same class for primary school
then highschool basically same ppl in the class aswell
in a big city like mine
youd move around
alot
but
ive been stuck in the same place
i mean
i travel alot
with my family
only time when you have different classmates is when you go from primary school -> high school and high school -> uni
but never had a change in culture or environment
no?
even in a big city youll still just go to the same school
i mean
there are only like 4 in australia
its โdifferentโ highschool
its just
its right next to eachother
so most people
1/3 of our grade
is from our primary
ye okay sure
i have family that lives in sydney ๐
lmao
too many murders now
???
so many
i read that NSW is the cocaine and gambling capital of the world lol
give me permissions to nuke lmao xd
they just busted the big weed place near me
lemme take my dogs too
dont kill the dogs
lmao sure
what time it is??
thanks for the help
np !
11:30 pm
wowowowow its 6 pm here
ooo
wya??
,ti
The current time for .bonk__ is 01:25 PM (CET) on Fri, 14/02/2025.
,ti
You haven't set your timezone! Set it using the interactive timezone picker with ,ti --set.
hmmm
hmm intresting
i thought you were going to bed
can i send friend request to u???
!done
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was
turn off the phone
or PC
whatever youre on
close your eyes and go to sleep
(also close the channel)
thats hard
,ti
The current time for xyruseditz is 05:57 PM (IST) on Fri, 14/02/2025.
now it is good
whats ist
Indian standard time
go to bed
right..
the other night
i was like
ima go bed at 9 pm
i went to bed at 2 an
okay
but now
you can change it
and go to bed now
if you dont close the channel
and my crush is texting me
i will
ATLEAST ACCEPT MY FRIEND RQUEST
I ALREADY DID
go to bed
you did..?.????
no stay up
CAN I FRIEND REQUEST U
why not
๐
i have friends
๐ฆ
i swear
!done
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please
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Why is it that instead of subtracting the intervals at the end, we add them instead?
that exercise looks so familiar...
It does? lol
i think i just did it this morning
Oh actually
Well im confused here and the teacher didnt explain why but dont we subtract at the end cause FTC or
This is for distance
Lemme write it
no no lemme write
why at the end is it not 22/3 - 17/6
shouldnt it be that cause of FTC?
@fierce cipher
Im just confused why it switches to addition at the end
,w -32/3 + 27/2
right but then we subtract the results cause of ftc right
rip
Wait so with FTC
We disregard the negative?
Here?
F(b) - F(a) but ohhh
Is it cause distance
not displacement
yea
right so essentially the absolute value of the entire function right so that turns to always positive
ye
So in every case regarding distance would I always use positive using the FTC
think simple
- negative areas turn into positive
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@little bough just wanna ask something
Yeah
what country are u from
i knew it
Yep ๐
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need help if my proof is correct
24
ax โ by (mod 5)
(ax) - (by) = 5k
a*x * (-1) * b * y
(-1) * a * b * x * y
where a * b = 5n, x * y = 5m
(-1) * 5n * 5m = 5k
-25nm = 5k
5(-5nm) = 5k
let k = (-5nm)
5k = 5k
.close
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need help on simplifying this
was told this can be simplified to a cos(b) where a and b are constants and a is an integer
@vivid heart Has your question been resolved?
I've tried rewriting it with cos (3x) = 4cos^3 x- 3 cos x and the arctan into arccos but it didn't help
hm well
hm
let theta = arctan(3sqrt(3))/3
then we know tan(3 theta) = 3 sqrt(3) and are interested in cos(theta)
does that help us at all
@vivid heart Has your question been resolved?
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!status
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
1
we'll start by breaking it into two simpler terms
[
S = \frac{1}{(1 + \log_e 10^n)^r} \sum_{r=0}^{n} (-1)^r \binom{n}{r} + \frac{\log_e 10}{(1 + \log_e 10^n)^r} \sum_{r=0}^{n} (-1)^r \binom{n}{r} r.
]
oh right
yajat
do you recall this property? [
\sum_{r=0}^{n} (-1)^r \binom{n}{r} = (1 - 1)^n = 0, \quad \text{for } n \geq 1.
]
yajat
Yes
It will be inside the summation now
The denominator
Its not independent of r anymore
[
S = \sum_{r=0}^{n} (-1)^r \binom{n}{r} \frac{1}{(1 + \log_e 10^n)^r}
- \sum_{r=0}^{n} (-1)^r \binom{n}{r} \frac{r \log_e 10}{(1 + \log_e 10^n)^r}
]
yajat
Do you recall this binomial summation property? [
\sum_{r=0}^{n} (-1)^r \binom{n}{r} x^r = (1 - x)^n
]
yajat
Yes
i think it was something like
,tc white
I don't recognise the option white. Use ,texconfig to see the list of options.
,tc dark
I don't recognise the option dark. Use ,texconfig to see the list of options.
,tc color dar
Unknown colourscheme dar. Valid colourschemes:
โ โ โ โ โ โ white: Pure white background, with black text.
โ โ โ โ โ โ light: Very light grey bckground, with black text.
โ โ โ โ โ โ โ grey: Discord-grey background, with white text. (Recommended)
โ โ โ darkgrey: Dark grey background, with white text.
โ โ โ โ โ โ โ dark: Dark background, with white text.
โ โ โ โ โ โ black: Pure black background, with white text.
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trans_black: Transparent background, with black text. (May cause issues)
,tc color dark
You have switched to the dark colourscheme.
,tex yes this is how you do it
yajat
Thank you
try to use this on the first sum
(-ln10)^n
[
\sum_{r=0}^{n} (-1)^r \binom{n}{r} \frac{1}{(1 + \log_e 10^n)^r}
= \left( 1 - \frac{1}{1 + \log_e 10^n} \right)^n
]
[
1 - \frac{1}{1 + \log_e 10^n} \quad = \quad \frac{\log_e 10^n}{1 + \log_e 10^n}
]
[
\left( \frac{\log_e 10^n}{1 + \log_e 10^n} \right)^n
]
yajat
this is what you're supposed to get, and not (-ln10) raised to the power n
I think I got how to do the second summation
for the second summation, we differentiate the binomial property
$\binom{n}{r} = \frac{n}{r} * \binom{n - 1}{r - 1}$
once you're done simplifying both the summations, you will just need to add them and it should be equal to 1
Dhruv
Then cancel out the r
bro is this JEE
Yes
oh alr
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Is writing sum of sigmas like this correct?
As in I have a sum called S, and it can be expressed as
Or
Then we want to calculate S+S
Can I write it like I just did
yes
seems right
k and -k cancel out
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use linearity
what did you do on the test?
flipped the sign of the g(x) and then replaced the numbers in the function and got the answer 42
my friend said the answer will be 90 but were not sure whos correct yet
so i came here to see whos correct
how?
ill write it out
$\int_2^5 2f(x)+3g(x)+3\dd x\\2\int_2^5 f(x)\dd x-3\int_5^2 g(x)\dd x+\int_2^5 3\dd x\\2\cdot 30-3\cdot 7+3\cdot 3\\60-21+9 = 48$
Bonk
unless i made a mistake somewhere
i trust you stranger on the internet
we will see how the test will go from here
thank youuu
.close
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Your opponent is playing a virtual computer game where their character opens a mystery box on each turn. The mystery box either has a green pill (55% chance) or a red pill (45% chance) inside. If the pill is green, they get +1 point, and if the pill is red, they get -1 point. If at the end of the game, your opponent has a positive score, you must give them $1000, and if they have a negative score, they must give you $1000. You can choose for them to either have to play exactly 3 turns, exactly 33 turns, or exactly 333 turns. Which number of turns do you choose? What if the green pill gives your opponent +3 points instead of +1? Explain your answers
what is the probability of you winning the game when you play 1 turn?
@urban rain Has your question been resolved?
no
45%
if they get a red pill they lose 1 point
and if their score is negative you win
the more turn the lower the probability of you winning
because youre taking out the luck
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Can someone help explain why the trapezoidal rule divides f(X0) and f(Xn) by 2?
What is the formula for area of a trapezoid?
I looked it up, still trying to figure out why the formula looks like that
Ah, okay
Think about a width n, and two points x_0 and x_1. What would be the area under the curve on that interval 0<x<n?
According to the trapezoidal rule.
I would calculate h as n - 0 /1 so h = n? seems weird? Then its h/2 * (f(x_0) + f(x_1)) ? Idk I have programmed it for a couple functions successfully but this notation confuses the heck out of me
Precisely.
Yea but what confuses me so much then is we do basically the trapezoid area formula for the end points summed up but then we have a sum we dont divide by two in the middle?
Like how is that?
Or How come is that?
Oh that
I haven't dealt with that specific simplification method yet
๐
Nw! I appreciate the honesty!
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Hi guys
So I'm writing a combinatorial proof
for a summation identity
for integer n>= 1
How do I go abt this
Maybe I shouldn't have made this my first message bc now this is what's pinned
!xy
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
well, ${n \choose i}$ is indep of j, thus $\sum_{i=0}^{n-1}\sum_{j=0}^{n-1} {n\choose i}{n\choose j}=\sum_{i=0}^{n-1}\left({n\choose i}\sum_{j=0}^{n-1}{n\choose j}\right)$
Bonk
I don't quite understand
you can pull one outside of the sum
nah you need to show a bit more
Do we not consider two subsets?
wdym?
but wdym two subsets
also, are you sure this is correct?
Yeah that's the correct question for sure
,w sum sum (n choose i)(n choose j), j=0 to n-1, i=0 to n-1
cuz WA says (2^n-1)^2
!nogpt
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).
all AI is shit for maths
It's a custom GPT that I gave a bunch of videos and slides to
I mean
Can I show u anyways
Everyone says AI is bad for math but tbh it's done me pretty well in recent history but probably because the math I do is really simple
is this the AI answer?
Yeah
yeah, youre wasting my time
Bro
I'm just wondering if it's a good answer to reference and learn from
๐
I'm here to learn and I'm using AI as a tool
@vague jolt Has your question been resolved?
I mean can u help me out
unfortunately the bot wouldnt respond to you
Can u help me out
indeed
I can't
Why not
I'm dumb
combinatorics is not my area, so unfortunately not, hopefully someone who can help comes soon
This gotta be a troll
No way bro just @ all the moderators
Surely that doesn't actually work
you called me a bum
I called u a bum ironically
it does
yeah thats how it works here
though idk what they will do here
I mean
lol
indeed
If anything he should get in trouble
take it back
For @ ing unnecessarily
read the rules
He's also a new user
Me calling u a bum boy is obviously an ironic joke and ur the one trolling right now so yeah if I was a mod I'd mute you
I'm not trolling
I take jokes seriously
Bro ur literally a child
what's wrong with that?
Ur acting like one too bro
Nothing wrong w it
Just making a statement
Bro anyways
Stop this nonsense
I need help
LOL
stop laughing at me
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Is it possible to construct 3 nondegenerate right angle triangles with integer side lengths and the same hypotenuse such that the area of two of the triangles sum to the area of the last?
I'm trying to either prove it isn't possible or find a counterexample and I haven't really been able to make progress
@tough minnow Has your question been resolved?
by "degenerate" do you mean "similar" or "congruent"?
I mean the areas are nonzero
a triangle with a side that's 0 length would have 0 area and be a degenerate triangle
just means the side lengths are all nonzero
@tough minnow Has your question been resolved?
So, in theory, you could use two similar or congruent triangles?
I suppose, yes
but even then I'm not sure it makes the problem that much easier
ive tried playing around with algebra and not really gotten anywhere but i might just be missing smth
my gut says there probably does exist an example but probably you need really big numbers
yeah I did think of and then see this general circle construction before
if you fix the hypotenuse at 1 then the legs just have to be rational
but I don't think that helps much
Hello Iโm currently studying for a big test on algebra, Iโm trying to find the square root of pie
Hello huzz @everyone
Do you guys have any suggestions to get goon of your paper
make your own help channel, read #โhow-to-get-help
I got a little to excited
yeah that's what I thought, account created today
can someone get these trolls outta here
ok made this so it now shows all of the points on the circle that give integer side lengths
ah yeah thats a good idea, could be a good way forward
desmos links don't update, they snapshot
you need to make a new link
didn't know you could to that elipsis syntax to get a list of integers, cool
yeah its super useful
ig shouldnt be too hard to throw together some crappy python code to do some manual searching
I already told chatGPT to
ages ago
up to 10k
and it didn't find it
which is promising
because I don't believe it's out there
fine I'll code it myself later
oh you mean it wrote some code
oh ok lol i thoguht you meant you asked it to check and it said "no" LOL
i mean so long as you checked the code over so that it was acc doing what you wanted then yeah thats fine
but i feel like the numbers may just be BIG
well I'm pretty sure if they do exist they're bigger than 10^14 but not certain
whered that number come from ?
I derived this problem from another problem
as a condition that must be met for it to be possible
computer searches of the original problem have gone to 10^14 and found nothing
but my version with the triangles is a condition that's needed for it to be possible which means if I prove it's impossible I prove the original problem impossible
however if I prove it possible it doesn't nessisarily mean the original problem is possible
whats the original problem?
the 3x3 magic square of perfect squares
for one to exist, there must be a solution to my triangle problem
ah
yeah ok i well now i dont feel so good about finding one of these sets of triangles ahha
that's quite a bit of math but I have a document of it
I'll get it
wrong one, that's older
please
yeah that's it
for each diameter going through the center you get this general form of (c^2-o,c^2,c^2+o) which can be converted into a Pythagorean triple in this weird way where o is 4 times the area of a right triangle of integer sides and c is the hypotenuse
for the two diagonals the value of o is x and y, and by considering the vertical column in the middle, it's o value can be called s for sum where s=x+y
now you get that x+y=s and those 3 values can be seen as 4 times the area of a triangle but the 4 cancels and you simply get the fact that the area of two of the triangles adds to the area of the third
and they all share the same hypotenuse because c is the central square they all share
and that forms it
oh i just finished reading that one ๐ญ
are the differences significant
yeah nice
I'm about to give up so I was hoping that maybe I could nerdsnipe someone
haha
and with the problem in a different form I thought that maybe someone could get a new perspective on it
im still of the opinion that finding 3 triangles will be possible
which as you say wont really help your cause
true
to fully encode the whole problem
you just add one more triangle
which must have an area being the difference of the first two
but I feel like you don't need it for it to already be impossible
starting to sound less possible but idk my intuition says thats out there
one thing that is known is that if it is out there, there are only finitely many relatively prime/coprime solutions
for the full thing
ignoring the difference triangle I can't say that for sure
I see you have been keeping up to date

