#help-23
1 messages · Page 362 of 1
i doubt you will ever get complex analysis unless youre majoring in some strictly math-related major
complex algebra maybe as an introduction
Yeah I had that idea about hyperbolic functions, don't know much about them though
instead of the unit circle, you have the unit hyperbola
Depends on the country, usually taught in the first years of uni
and then take the x- and y- coordinates yep
Or, well, depending on how profound your complex analysis course is, it may be taught later, around third or fourth year
In what major?
A course literally called complex analysis
Mathematics, doubt you'll be taught that in any other major
complex analysis is usually offered in 3rd year if you're not in Europe/the UK
tbh complex algebra is just basically
$$a+bi, i^2 = -1$$
$$Re^{i\theta} = R(\cos \theta + i\sin \theta)$$
Maybe some bits in Physics, but not too much
but you can take it earlier if you have the right prerequisites
So just BSc in Mathematics right?
you don't need to take real analysis first, for example (depends on uni!)
Yup
but of course it's a proof-based course
Yeah I knew that, we're taught that in 11th here
Haven't seen any university offer complex analysis before real analysis
let me ask wai what year they take it, iirc it's similar?
to my knowledge most of the formulas and facts about complex algebra can be strictly derived from this two facts and knowledge from real algebra.
or for India
Here, for example, real analysis is a first year course and complex analysis an early third year one
my uni lets you do this
Is there such a thing as a non-proof based course in mathematics though?
Yeah I'm from India, we're taught a decent level of complex numbers in 11th itself, I'm not sure what they teach in college though
yeah, so computational linear algebra for example
im in engineering we basically dont do complex analyisis
Guess that's the thing, I'm in Europe
Fair enough, you don't really need it for anything in engineering
How about Astrophysics, what fields of mathematics do they cover there?
Astrophysics is a branch of physics, really, but it's just a lotta calculus and differential equations
They'll skim over the proofs in physics as well, be warned
for most people, lin algebra, (Diff, integral, vectorial and dynamical) calc and just a bit of complex algebra covers any realistic use-case of math in their career
Yeah, that's as far as most people go
Damn gotta look into linear algebra, heard it's tough
its a really extense topic, and it has a lot of uses for computers
Really glad I chose mathematics instead, though, would've hated it else as I really don't like calculus/analysis
Physics should be fine with me I hope, unless they bring it way too high way too fast
i havent had much of analysis, just calc, but i swear by it, lmao
Physics is a lotta calculus built really fast so you can apply it
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i need to solve by disk method
i have an integral, but its really weird to solve
is what my work looks like
i im trying expanding nr
How did you get the integral, by slicing out the part of the triangle and rotating it around y-axis?
seems like x-axis
I don't think that'd work
why not
The width above is not the same as the width below, if you rotate it along x-axis, then it won't work because it's not symmetrical
im pretty sure i got it now
no like
ill send a pic
what are you on about homie
,rotate ccw
Couldn't find an attached image in the last 10 messages.
,rotate ccw
and just draw an x axis thru it
and use that slant line as ur main thing
to solve with
Alright then, I was trying it with this figure unaltered
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i tried w(m-n) = 2mn but i dont understand what to do next
and i tried dividing
but that doesnt work
right because we still not n on both sides of the equation, is there a way to get all the n's on one side?
After this first open the bracket
in exercises like these, always start by pulling every term that contains your new subject to one side
How is this relevant here☠️
yes i understand that but i just dont know how to get all the ns to one side
Cross multiply first
Then say what you get
if a=b then a-b=0, apply this fact here
and w(m-n)=2mn is indeed the correct first step if that wasn't clear
wm−wn=2mn
wm=2mn+wn
um, you're not OP...?
Now in Right side take n common
What's OP?
wm=n(2m+w)
I would think that OP's supposed to be the one working through the problem here
original poster
Now leaving n, divide both side by (2m+n)
aka the person who asked the question
the person you're responding to is not the person asking the question
I did not even see that LOL
no no
just my thought
try not to spoil it for OP, I suppose
i tried the distributive property
See the process and you will get it most likely
i got wm-wn = 2mn
correct
Yeah That's correct
and that's it
just be careful to put parentheses
wm/2m+w is correct but would be better to write it as wm/(2m+w) so as to be interpreted as one fraction
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im working on my undergrad final year project mainly about CVRP, i mainly done with like anything but the result. Anyone have a website or code that can turn matrix into graph
yea i tried it, and it doesnt give the one that i wanted
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nvm
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how do i do this
can you write down what it means that
matrix A has eigenvalue λ and corresponding eigenvector e
though actually it feels very fishy that it's apparently claimed AB and BA have all eigenvalues the same just bc they share one eigenvector...
you didn't write out what i asked for here
@rich dune Has your question been resolved?
what shld i write down
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for part d i dont understand how 4+3 gives 9
what makes sense is they add A = 2 to 4 and 3 to give 9
but idk why A is needed if im not expanding it
@fringe dock where is part d in the image, also please ellaborate your present working
4+3=7 btw
ive gotten to the point where i got
4 -8x + 16x^2
and
3 +3x +3x^2
when i add them up its fine, but the first value is meant to be 9 but i get 7
also answer the first part, where is d?
ooh ok
entire ms
why do you add
should it not be the product
aah ok
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Two vertical poles, 80 m and 20 m high, are 13 m apart on level ground. Straight wires
are attached from the top of each pole to the bottom of the other pole. Find the vertical
distance, in metres, from the ground to the intersection of the wires.
Please don't occupy multiple help channels.
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How to solve que c
@lilac epoch Has your question been resolved?
do you have your drawing?
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in answer sheet answer is given A but mine answer is coming D is aswersheet is wrong T-T ??
could you show your steps?
local minimum is -0.343 and endpoint min is -1.57
have you not done derivatives?
true
so how would you solve cos(2theta) = 1/2
when is cosine equal to 1/2?
think of the unit circle
in interval of -pie and pie
forget that for now
there would be two solution
when is cosine equal to 1/2
-pie/6 and pie/6
firstly solve this: cos(x) = 1/2
yeah x = pi/6 and - pi/6
true
now we have that 2theta = pi/6 or -pi/6
so theta = pi/12 or -pi/12 right?
alright first lets solve cos(x) = 1/2 without any constraint
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this is the solution
but I want to solve this using desmos
also why does the solution make the slope of f(x) as -2
isn't the slope -4?
I got -16 with desmos
and I get -8 (the solution given) if I take -2 instead of -4 as slope while regressing n(x)
I am taking n(x) as equivalent to h(g(x))
also I got the answer but I still don't properly understand the logic behind using x1 values to regress n(x)
I was using another seperate variable x2 before and didn't get the answer
but why is it neccessary to use x1 here, logically?
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stfu ripico
the gradient of f(x) is definitely -4. Maybe there was a typo in the question or the solution of the question
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How to factor something like this more quickly i usually do guess and check but this is my complex
How to factor something like this more quickly i usually do guess and check but this is my complex
The second equation isn't the same as the first. If that was your attempt to factor it, that's wrong.
I kinda forgot how to do this, but you got the right idea with the -15 and +1. The only issue is the 8
Edit: that's how you figure out when a=1. It's different when a≠1
Imma look this up real quick
Its the ac method
First, simplify the equation. So in your case, it would be 0=4x²-7x-15
Multiply a and c to get -60. What adds to -7 but multiplies to -60?
I immediately think 12 and 5 multiply to 60, which works when it's -12 and 5
So split it into 4x²-12x+5x-15=0
Group and simplify to 0= 4x(x-3)+5(x-3)
Then distribute so (4x+5)(x-3)=0
Check your work. This multiplies out to 4x²-12x+5x-15=0. Multiply that by 2 and you get the original equation, so this factoring was correct.
Does my explanation make sense?
Basically, first you simplify. Then, you multiply a and c to find a number. You then figure out numbers that multiplies to ac and adds to b. Use those numbers to split up b, and then group and simplify.
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Hello, ive tried to solve this question , ive done this g v (c\wedge b)=(g v c)\wedge(g v b)=c\wedge a=c
and g\wedge(c v b)=(g\wedge c) v (g\wedge b)=g v h = g , this means that the lattice isnt distributive right?
In the solution it says it shoudlnt be, but my teacher said during the lecture that it should, so im lost. can anyone help me figuring it out and what i do wrong?
Why would those statements make the lattice not distributive?
@timid token Has your question been resolved?
because g\neq c?
im unsure how to solve it
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✅ Original question: #help-23 message
@timid token Has your question been resolved?
That is correct the lattice is not distributive.
is my solution correct? and also do anyone know how to check if its complemented? i just know a v a'=1 and a\wedge a'=0
It should be right
do you know how to check if its complemented?
yeah the soultion to the exam says its not distributive but during a lecture he could show it
Well you could identify the bounds then define the complement afterwards
Yep
thats the bounds i suppose but idk with the complements
Well, you could "test" the elements. You've already proven that e fails it, no?
Pick an element to test then it must meet the conditions. It Must join the biggest element and meet the smallest
Just take note that as long as one element fails, then it isn't distributive
i think i prove it not being distributive, but when looking at the lattice i cant think of two elements that doesnt join the biggest and the smallest
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✅ Original question: #help-23 message
Well, one element is just enough to prove it's not distributive, if that's what you mean
SLR
One element failing is enough proof it's not
Wait hold up
*complemented rather
It's not about two elements that don't join the smallest/biggest, it's more about a specific pair of elements for example x and x' that must simultaneously satisfy the conditions
i dont really understand, any element b,d ,f with c,e,g in my opinion share a and h . can u give a example that shows its not complemented?
You're right that some pairs share the greatest element which in this problem is a like take for instance pair b and c. However, to be complement they should also be share h as their meet
x v x' = a
x ^ x' = h
Just assume that ^ is the "meet"
If we look at element e in the same problem and it's join with a, b, and c however the meet of e of those elements are...
e and b -> d
e and c -> g
Neither of d and g, doesn't the least element of h, no?
So, we could conclude that there's no element x' that satisfies both conditions for e
And because one element has not complement, the entire lattice is not complemented
Also the lattice non distributive in a way, it violates the distributive law
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✅ Original question: #help-23 message
how does e and b -> d?, i would understand if it was d and e -> b , like (d v e)=b
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x(=\frac{1}{4}+\frac{2}{16}+\frac{3}{64}+\frac{4}{256}+...)
Boogie
aight yeah
yeah i did
we are dividing by 4 because if u look at the denominators they are all powers of 4
i got sumn like x/4= 1/16+2/64+3/256+4/1024+...
yes correct
now do x-(x/4)
would it be legit to quadruple it instead of dividing by 4
ig u can do that as well
oh word lemme see
It would js give a negative gp upon subtraction then
didnt u have helper role
Huh?no why
so i kinda got why u did dat
but like can u explain why so like i dont have any doubts or nun
ohh mb i think it was @stoic saddle
To convert it into a gp.
When you're subtracting u subtract the ones with common denominators
i think i fucked sumn up
And as u can see the subtraction of numerators would give one
(I explained tht so bad ngl)
Yh
ig redo it
Are u finding 4x or x/4
4x was easier to calculate cuz less fractions
Okay
so i got 4x-3x=(1+8/16+12/64+16/256+...)-(3/4+6/16+9/64+12/256+...)
4x-3x?
yea
Why
But we already have x?
act good point
What did u get for 4x?
i got the left side of ts equation
1+8/16+...
(is "word" some slang?🫠)
Yh
aight so i got 1+2/4+3/16+1/64+...
How about u write 1/2 as 2/4
I said don't budge the numerators at all
You'll understand why ,when we get the answer
No tht would change the original numerator
Now just do 4x-x
Only subtract terms with like denominators
word
omg wdym by word?
nah
u want $x(=\frac{1}{4}+\frac{2}{16}+\frac{3}{64}+\frac{4}{256}+...)$ right?
Viͥђaͣnͫ
Here, dont budge denominators
yeah
u see the general form here?
word mb its takin a lil long i do it in my head
its smth like $x = \sum_{n=1}^{\infty} \frac{n}{4^n}.$
Viͥђaͣnͫ
holy fuck what is dat
u didnt read summation yet?
U kinda have a direct formula for this. But i suppose udk tht
ive only seen sigma in physics for now
Wait a sec do yk GP?
gary payton
yk about series? geometric series
nah havent learnt it yet
What'd u get
Then who gave u this problem😭
bro ur teacher is cooked
so i got 3/4+6/16+9/64+12/256
Ehh u sure sure?
For this?
lowkey stealing is a more ethical way to get money than ts mf teaching
yeah
what did i fuck up😭
bro
How would u get tht
this mf didnt teach u geometric series it would take a lot of time to solve this
Are u doing it on a pen and paper?
yeah
Can u show me ur work?
icl i cant
im on pc rn and my phones camera dont work
what should it equal so i can figure out what i did wrong
i got my terms tmr yall am i cooked😭
(2/4 - 1/4)+ (3/16- 2/16).....
wait lemme teach u a formula
If u dont know a thing about series and sequences then yes
what shouldnt it be 4/4-1/4
word
didnt i time it by 4
This is 4x
yeah
This is x
the 1 is 4/4
bro how the hell he gonna solve this rn???
word mb gng
a ques, where is this question from?
my teachers exam from like 5 years ago
broooo
Okay , do yk which topic it is from?
fuck ur teacher question its old
uh i think telescopic fractions?
literally he sure taught geometric series to them
omm idk what dat is
look bro
notice that
lemme explain
A geometric series is a sum of numbers where each number is multiplied by the same ratio to get the next one
for example
1 + 2 + 4 + 8 + 16 + ...
Anyways did u get 4x-x?
Its not telescoping
Here, every number is multiplied by 2 to get the next one. The 2 is called the common ratio
lemme explain geometric to this guy and let this shit end
i redid it and it was the same shit i said i think
This simplifies to that again??????
I really would need ur work tbh
@untold marlin
Honestly its the same thing
But if u want to go ahead
And again DONT BUDGE THE NUMERATORS IN THIS STEP
word
i aint fucking with this shit anymore do anything u want
Woah thts ryttt
ok so what now
But wait thts 4x-x
yeah
Not x- (x/4)
Oh okay
i meant i added it up again
Now write the denominators in the form of 4ª
Leave out 1.
$ S_\infty = \frac{a}{1-r} + \frac{d r}{(1-r)^2} $
tf
$S_\infty = \frac{a}{1-r} + \frac{d r}{(1-r)^2}$
Viͥђaͣnͫ
ah
(=\frac{1}{4}+\frac{2}{16}+\frac{3}{64}+\frac{4}{256}+...)
Boogie
what no
mb
it didnt copy
i write my shi in a calc and paste it so i get the correct fraction symbols
Smart
$1, 2, 3, 4, ...$ is the AP term and $\frac{1}{4}, \frac{1}{4^2}, \frac{1}{4^3}, \frac{1}{4^4}, \dots$ is the GP term
Viͥђaͣnͫ
(\frac{1}{4^{0}}+\frac{1}{4^{1}}+\frac{1}{4^{2}}+\frac{1}{4^{3}}+)...
Boogie
Had to google tht person but okay
$S_\infty = \frac{a}{1-r} + \frac{d r}{(1-r)^2} = \frac{1}{1 - 1/4} + \frac{1 \cdot (1/4)}{(1 - 1/4)^2} = \frac{4}{3} + \frac{1/4}{(3/4)^2} = \frac{4}{3} + \frac{1/4}{9/16} = \frac{4}{3} + \frac{4}{9} = \frac{16}{12} + \frac{4}{9} = \frac{4}{9} \text{(after simplification)}$
the answer
Viͥђaͣnͫ
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Watch this first
after using AGP
(Mate, i got an exam tmrw i shouldn't be here)
fuck it im js memorizing the anwser😭
Watch this, sir.
dont fuck with the question at all if they didnt teach u
also would u tell me your syllabus for the tes?
idk lowkey whats a syllabus
As in what topics or units will be asked in the test
Okay and?
Thts all?
yea
Well
the book is like 8 lessons
This is a part of Sequence and series
Mb, yea that should've been e and d -> b
And will not be asked in ur test
So I'd recommend u wait till ur teacher teaches this topic
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If $a_1,...,a_7$ are not necessarily distinct real numbers such that $1<a_i<13$ for all i, then show that we can choose three of them such that they are the lengths of the sides of a triangle.
T&C
hi this is my first time solving questions such as these. I'm actually practicing from a problem compendium without the help of any mentors.
all that comes to mind is triangle inequality and some combinatorics when i think about tackling this. but i don't quite know how to begin.
vectors eh
hi this channel is already occupied. you can create your own help channel. go through #❓how-to-get-help
ok back to the original problem
I’m pretty sure there is a theorem that says the sum of two sides of a triangle is always greater than the third side
so there are 11^7 possible combinations
Oh wait it’s real numbers
yeah both of our approaches are the absolute same
hence i rejected the combinatorial angle. but if we restrict ourselves to the naturals maybe we can use some combinatorics
sure
it might be pretty dumb
take intervals between two umbers
yeah that’s also equally useless
wait
I think yiu can do proof by contradiction
oh can you tell me how
but the statement would be proven false
however
there is a bit of a catch here
yeah that doesn’t work as well
I was thinking you could take some particular values and prove that no single combination forms the sides of a triangle
im so confused rn. im thinking of several ways at the same time
I’m literally in the same state as you
ah okay, ty i understand!
say for now, we stick to naturals.
then we have the numbers {2,3,...,12}
can we figure out the number of ways to choose 3 numbers so that the sum of any two is always greater than the third. then can we use the IEP?
I have an idea of a proof by contradiction by sorting the seven values in non-decreasing order (b_1 lowest, b_7 highest).
then, suppose no triple forms a triangle. then, for all i < j < k, b_i + b_j <= b_k (where b_i <= b_j <= b_k)
consider all five consecutive triples (which gives the largest possible LHS), and bound all five of them below using the above inequality, then express all of those inequalities in terms of b_1 and b_2.
might come across something here
7?
yeah the question says to pick a set of 7 out which any 3 must form a tringle
oh yeahh right mb
okay im sort of getting this. ill try
This seems to be the method that works rn
I will have to note that the final contradiction requires you to remember the strict inequality of b_i < 13.
btw why are we considering 5 consecutive triples?
(b1, b2, b3), (b2, b3, b4), etc.
ah i see
as for why not something like (b1, b3, b4), because b2 >= b1, so even if we disprove b1 as a candidate element in a triple where b4 is largest, we cannot yet disprove that a triple with b4 largest cannot exist, because b2 is possibly larger than b1 and therefore could satisfy a triple with b4 largest
a similar argument works for any triple with a largest element that isn't b3.
so we are picking the triples with the largest possible LHS
(for each candidate largest element in the triple*)
ohhh i understand now. this is such a smart proof. thank you so much. I'll try to redo it myself now from the beginning
Ohhh
glad to have given you some insight
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hello. so i wanna convert z_2 to rectangular coords and i let z = z_2 ^(1/24) to convert that into polar and then evaluate z^24 to get z_2. Is my method wrong? I cant evaluate the angle and it should be staged for the angle to work so i mightve gone wrong somewhere
hint: convert your complex number inside the (...)^24 into re^it form
i mean thats what i tried in the pic no?
i found r and tried to find t
unless thats not the method to find re^it. i searched it up on google but didnt find what i wanted
it only showed me calculators and no methods
oh i can just write our wanted z as 2-√3 +i ? and we concern ourselves with 1/2 later?
its alright
But, basically, given x + yi, we have r^2 = x^2 + y^2 and t = arctan(y/x)
which you've done. ok I see what your problem is now.
sorry, I jumped the gun
hahah its okay
so if this way doesnt give convenient results is there a different one?
this would be the easiest way
hmm
otherwise, you'd need to do (x+yi)^24 which is a huge polynomial to evaluate
mayb our prof wrote smth wrong bc were not supposed to use inverse trig functions yet
nono she told us to use what weve learnt in complex analysis to solve it
anyway ill ask her tomorrow
thank you
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the angle is 5pi/12, but I'm not sure what trig you know to show this
ill work it out and see if its possible
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I might be romanticizinf math a bit but what went wrong
You can check your graph in desmos
It seems like u completely ignore negative r
and at pi/2, u forgot to multiply by 2
otherwise u just need to fix the sign (it doesnt change much, the circle will just be under the x-axis)
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can anyone help me to solve this puzzle? it's not necesarry but i will accept any kind of help.
i can tell more info if anyone helps
I suggest you just lay it all down here so that if somebody who knows see it, they can help
hmm right
I can't read your handwriting
normal sudoku rules aply+ the squares connected with the lines must difer min. by 5+ the squares connected with the dark dots must be x-2x + the squares connected by the empty dots must be X--X+1
its not even fully in english
Ah that's why I was confused 😭
anybody?
I don't know what the lines in the boxes mean. That's different from sodokus I do
The ones I do just have numbers
Still numbers 1-9?
yes and each printed square(not which i drew with my pencil)counts as a regular sudoku square
Ohhhh ok lemme grab a piece of paper. You're 100% sure it's solvable?
idk i got it from the math-discussion chanel yesterday
Ohhh ok. Well, we'll see
the only thing i know is that the squares connected by the line cant be 5 because then theres no number biggr or smaller by 5
chat gpt say that its solvable-didn't ask for sollution tho
Just to confirm, dark dots could be 2 and 4, and light dots could be 4 and 5?
That's what it means?
not excatly : the squares connected by the dark dots are like 1:2 :one of them is twice as much as the other, the white dot: one of the squares is one less(or more(doesn't mater))than the other
Ok. So my example works?
the dots are not numbers
Oh, yeah that makes sense
I meant the boxes could be that sorry
But I understood it right lol
Id start with the middle left column and the middle middle. Where they intersect
Ohhh ok
Seems like there are lots of dots and lines there
never use chatgpt
hmm yea it would be really good if i would have even one number writen in 🙂
also, the original puzzle is unreadable
I said that I didn't use it for sollution
bruuuuh
you used it to tell you that it's solvable. So the information of "it being solvable" is suspect, since it came from gpt which is terrible
even chat gpt said that he wasn't sure, can you tell me then?
not until you actually put the puzzle in readable form. As i said, your original one is unreadable
theres no need for reading
ah yes. You definitely dont need to read the instructions to solve a problem
the instructions are here man
and heres the nicer image for you
that's not a set of instructions. That's unreadable and incoherent notes
tell me what you don't understand
literally nothing is understandable there
"normal sudoku rules apply" on something that has no numbers already fails to be understandable
do you know sudoku?
yes
Oh good point sudoku do normally start with some numbers. This one is really difficult
then ? "normal sudoku rules aply"
yep thats the hard part
i think tipmix is the key to victory 🙂
so what you're saying is, this is a sudoku with extra restrictions?
and no starting numbers
okay, lemme try to translate your crappy notes
when you say "the squares connected with the lines" do you mean ONLY the endpoint squares of the lines, irrespective of any angles within the lines?
also on those, does that mean they need to be a pair 1-6, 1-7, 1-8, 1-9, 2-7, 2-8, 2-9, 3-8, 3-9 or 4-9?
with dark dots one of them needs to be twice the other, and with empty dots one number is the next of the other?
the ends and the difference is minimum 5
with exactly 5 being valid?
if you mean next to like 1,2,3,4,5,6,7,8,9 then yes (one more/+1)
yes
5 is the only number which cannot be connected with the line(5-5=0/5+5=10
I think i will need to ask a math expert to say if its solvable or not 🙁
where did you get this from?
also, a puzzle is not necessarily "math"
but it feels like if it has a solution, it is not unique
I give up on this puzzle lol. I don't have the time to solve it
they sent this to me in the math-discussion chanel yesterday
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We need a clearer picture
You're over 13... Right???
yup
,calc 3/100
Result:
0.03
calculators are your friend
Aside from that, it looks like you already converted all the numbers into decimals?
oh :3
Now you just have to order each of the numbers based on their decimal value :)
to least to greatest
So which one's the biggest? Is it:
0.3
2/5 (which is 0.4)
0.68
5/8 (which is 0.625)
you can't use calculators but you can ask people for help?
Yep.
sounds wrong
Hi? I did give you a hint above :)
the biggest is 0.625
mmmm are you sure?
yeah
Remember that when we are counting decimal closest to the decimal point is larger.
5/8
ok
Let me ask you a question,
ok
Which number is bigger?
0.62 or 0.68?
0.68
Good, why?
beacuse theres is an 8 and the other one has a 2
Good!
But, if you wanted to be more specific,
The second decimal place of 0.68 is larger than the second decimal place of 0.62

I.e 0.68 is larger than 0.62.
Now, that being said,
Which one is bigger?
0.68 or 0.625?
Good! Why?
becuase 0.625 is it on there???
wdym
or it to big
Not sure what you mean
0.625 is to big
i know
The fact that the 8 is bigger than 2 means that anything after the 2 doesn't matter.
So for example,
0.68 is bigger than 0.625 because the 8 in 0.68 is bigger than 0.625.
ok
nice
So, would you agree 0.68 is the biggest? What goes next?
then 2/5 goes next
wait
Are you sure?
2/5 is only 0.4
What's the next number smaller than 0.68?
it 0.3\
fine
0.3,
2/5 (which is 0.4),
0.68,
5/8 (which is 0.625)
After 0.68, what is the next biggest number?
0.628
Perfect! There you go
What's the next biggest number after 0.628?
wait wait
Should be the other way around
Because it's
... < ... < ... < ...
Like
0.68 goes at the right
... < ... < ... < 0.68
Because the < means
Smaller < Bigger
So for example, 0.625 < 0.68
Other way
Swap the 0.68 and 0.625
Because 0.68 is bigger than 0.625
So it should go on the other side
so o.68 should take 0.628 side
Yes
done
Good, now fill in the rest according to their order :)
Yes
ok
Just to be consistent with the original question
Try it first. Again, convert all the numbers into decimals.
Do you know long division?
no :3
How did you convert these numbers into decimals then?
idk i just had someone to help me
😭
As mentioned earlier, the easiest way to find decimals is to use a calculator. However, if you are not allowed to use a calculator, the next best strategy is to use long division. Now, long division is a pretty long method, so it's best if you learn it via youtube or another source.
https://www.youtube.com/watch?v=usxTJbQbAIg
Welcome to How to Divide a Smaller Number by a Larger Number with Mr. J! Need help with dividing smaller numbers by larger numbers? You're in the right place!
Whether you're just starting out, or need a quick refresher, this is the video for you if you're looking for help with how to divide a smaller number by a larger number. Mr. J will go thr...

For example, in order to convert 4/5 into a decimal, you would do 4 divided by 5.
im just watching it on discord
0.3
Yeahhhh well
Technically
1/3 is 0.333333....
But for convenience,
I'm going to convert it to just 0.33
So 1/3 is close to 0.33
ok so0.48 is the largest
So nice, you have:
4,
4/5 (which is 0.8),
0.48 and
1/3 (which is around 0.33).
So, which one is the greatest?
nope
nope
...
Look carefully at each of the numbers
Very carefully
There should be a number that is much bigger than the others
4 is the largest
so put it at the end
ok
What is the next largest number?
Nope
0.8
There you go!
Remember
The number closest to the decimal places is the largest
So for instance
ok
ok
80 is bigger than 48, so 0.80 is bigger than 0.48
Yep
time for 6
This one's a little tricky, but let's see how you go :)
Good! Why?
beacus eit lower
oh whoops I didn't read the question
Well hold on
Remember
The first thing we need to do with these questions is..?
least to greatest
to figure out the demicals
Exactly
So figure out the decimals first, then tell me
Remember also that the signs flip again
ok
So the lowest will go first and the highest will go last.
0.6
0.66
Hint: ||Use addition||
Nope
idk
5.6
There's one more number you haven't converted into a decimal
wait where'd you get that?
counted :3
how many time 3 can go into 5
These are the numbers, so
You have 5.7,
Oh I mean
5 3/5 is equal to 5.6
What's the other number that we need to convert?
orz
hi pigeon
I NEED HELP I DONT KNOW
Using the method here