lol
#help-17
vast shale
cedar kernel
one sec imma recheck if that was correct relation
damn you making me realise i need to make a short note copy for physical
godamn
vast shale
nope
short notes are meaningless
theyre not complete
you dont have explanations in them
cedar kernel
yeah , but when you have like 25 chapters and you are outside the exam hall , and cant recall a certain formula from a certain chapter , You can just quickly go through them
its alright different people have different opinions , thats completely normal and i dont want to argue on the same
vast shale
oh i personally wouldn't think about any chapter or syllabus outside the exam hell
midnight radish
i want to memorize table of 13 and i cant what to do
vocal sleetBOT
i want to memorize table of 13 and i cant what to do
frail jungle
instead
i do have a smart trick
but it requires some effort at the begining
basically, learn every table till multiple of 5 and then add or subtract the number from it
so if someone asks you what 13X8 is
just do 13X10 - 13X2
which is 130 - 26 = 104
vocal sleetBOT
@midnight radish Has your question been resolved?
tulip sleet
I have to perfect/complete this square
$-4x^2+4x+3$
vocal sleetBOT
I have to perfect/complete this square
$-4x^2+4x+3$
twin meteorBOT
rainy
tulip sleet
the answer is $4-(2x-1)^2$
twin meteorBOT
rainy
tulip sleet
for some reason i dont see it
usually my first step is to factor out a -4 from the a and b term
Like this
tribal moss
i made an example for you, look at it:
$-9x^{2}+12x+1=-9\left( x^{2}-\frac{4}{3}x \right)+1=\\=-9\left[ \left( x-\frac{1}{2}\cdot \frac{4}{3} \right)^{2}-\left( \frac{1}{2}\cdot \frac{4}{3} \right)^{2} \right]+1=\\=-9\left( x-\frac{2}{3} \right)^{2}+9\cdot \frac{4}{9}+1=\\=-\left[ 3\left( x-\frac{2}{3} \right) \right]^{2}+5=\\=5-\left( 3x-2 \right)^{2}$
twin meteorBOT
Joanna Angel
tribal moss
please analyse my example, and try to solve your one, in same way
tulip sleet
please analyse my example, and try to solve your one, in same way
I get line 1
What happens on line 2
tribal moss
i transformed parentthesis from first line into square of parenthesis in second life
line*
tulip sleet
huh
tribal moss
there is also faster way, and if you like i can give you such quick formula
to avoid losng time durign exams
tulip sleet
there is also faster way, and if you like i can give you such quick formula
im used to doing normal ones, just take half of the b term, ...
you know what i mean 🙂
tribal moss
depends on you 🙂
tribal moss
i wrote it very veyr slow and precisely for you
tribal moss
lok and lsite to me
tulip sleet
tribal moss
every and each trinomial written in geenral form
can be written in vertex form
called also as canonical form of quadratic function
that si very known formujla and it for thos estuents
who pass exams online
and they need onyl an aswer quick
unless oyu pas eam face to face
let me write it fo royu too:
$ax^{2}+bx+c=a\left( x+\frac{b}{2a} \right)^{2}-\frac{\Delta }{4a}\text{ }\text{ where:}\\\Delta=b^{2}-4ac$
twin meteorBOT
Joanna Angel
tribal moss
then you only need toknow
a
b
c
and that is all
🙂
but that is the same waht i did in longer way
tribal moss
no ? lol
tribal moss
if you like sure )
vocal sleetBOT
@tulip sleet Has your question been resolved?
tribal moss
thanks
$-4x^{2}+4x+3 = -4\left( x+\frac{4}{-8} \right)^{2}-\frac{64}{-16}=-4\left( x-\frac{1}{2} \right)^{2}+4=\\=-\left[ 2\left( x-\frac{1}{2} \right) \right]^{2}+4=4-\left( 2x-1 \right)^{2}$
twin meteorBOT
Joanna Angel
tulip sleet
Same, but this is way too complicated imo
tribal moss
it i squestion of practices
beleive me
solve such 10 examples and you see, you 'll be fine
tulip sleet
im saying this because my teacher did it like this
But i dont understand her method
tribal moss
he or she noticed that here, but it is not always possible
so his/ he rmethod depends on exercise
tulip sleet
ok
vocal sleetBOT
vocal sleetBOT
Available help channel!
vocal sleetBOT
vocal sleetBOT
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
peak shard
if you don't know how to solve a problem, try making it easier, maybe a natural question would be what the chance of the first pull being green would be
if you only pull once
well there is a formula for this, but it would be overkill, its probably easier to go threw this step by step and find your solution because remembering how to solve these kinds of problems is probably more useful and you understand why the formula works.
yeah so you have a 50% chance or 1/2 that your first pull is either red or green, now you can make a case distinction for both cases
if you want I can talk about the formula after we have solved this one if you dont mind
so if you dont know what to do or need help just ask
take your time
2/5 is kinda right, but not totally
lets go this threw
we have 2 cases
1: you get red in the first pull
2: you get green in the first pull
right?
why 3/10 or 2/10 ?
okay so I think you are mixing things up here now you are calculating the chance of
the first pull being green and the second pull being green
and
the first pull being red and the second pull being green
is that correct?
you asked this
"How can I calculate the chance of the second pull being green?"
but now you where calculating the probability of
the first pull being green and the second pull being green
and
the first pull being red and the second pull being green
but you only want the chance of the second pull being green
first pull being green and the second pull
this and here tells you that these probabilities are depended on one another
but you are asked for the probability of the second pull being green which is for each case independent of the first pull
actually you dont know what the first pull was, but it was either red or green
(case 1) so lets say our first pull was red
what then is your probability that the 2nd pull is green
so what would be if your first pull was green, what would be your chance for the 2nd pull being green?
yeah and those are your answers because you are only asked what the probability of the 2nd pull being green is
Which is different if the first pull was red or green
yeah sorry I didn't know where your problem was, I think your problem is that you are not sure when to multiply and when to add probabilities is that a problem you have?
do you know what a tree diagram is, I could explain it with that
let me explain it with that
this is our representing tree diagram
do you understand why the answer is different for both starting cases?
cosmic heron
Write a linear equation with the given information:
Passing through (-2, 8) and PERPENDICULAR to y = 2x + 5
Idk what to do or how to start please help if you can thx:)
lament linden
Write a linear equation with the given information:
Passing through (-2, 8) and...
If two lines are perpendicular to each other, multiplication of their slopes equals to -1
cosmic heron
If two lines are perpendicular to each other, multiplication of their slopes equ...
wdym
lament linden
Then use the fact a point satisfies equation of a line that passes through it
cosmic heron
so first would i divide 1 by 2x+5?
lament linden
No
Lines can be presented in following 2 ways:
y = mx + c
ax +by + d = 0
(Where m,c,a,b and d are constants)
cosmic heron
yea
lament linden
Pick the format you like, say y=mx+c
cosmic heron
Pick the format you like, say y=mx+c
yea
lament linden
What is slope of y=mx+c ?
cosmic heron
m
lament linden
Also since we are trying to find the line that is perpenducular to y=2x+5, assume this y=mx+c is the lime we are searching
cosmic heron
and c is the intercept
lavish river
the slope is m. C is the y intercept
lament linden
So slope of these two lines are 2 and m, in order, multiply them, since they are perpendicular, we have
2m= -1
Find value of m, substitue that back into the equation y=mx+c ( since this is the line we are observint)
Then we are given that this line passes the point (-2,8)
Put that point in equation of line y= mx+c , then you'll find the c value
lament linden
You can use trigonometry to illustrate the fact multiplication of slopes of two lines that are perpendicular to each other is indeed -1
You will obtain something like $$\tan \alpha * \tan ( 90^ \circ+ \alpha) $$
twin meteorBOT
Cyrenux
vocal sleetBOT
@cosmic heron Has your question been resolved?
fleet frost
Is it possible to use the intersecting chord theorem to say that two chords are of equal lengths and give proof for it?
vocal sleetBOT
Is it possible to use the intersecting chord theorem to say that two chords are ...
thin vale
What's the intersecting chord theorem
vocal sleetBOT
@fleet frost Has your question been resolved?
vocal sleetBOT
vocal sleetBOT
Available help channel!
lavish river
we have (f(x), f(y), f(z)) ∈ F(B)
I don't understand how we get this part.
empty frigate
f is from A to B, so if x,y,z are in A, f(x),f(y),f(z) are in B
the reason that that triple still satisfies the equations is because f is a homomorphism
f(2x^2 + y^2 - 3z^2) = 2f(x)^2 + f(y)^2 - 3f(z)^2 by just repeated applications of f preserving the ring operations
lavish river
f is from A to B, so if x,y,z are in A, f(x),f(y),f(z) are in B
I think the fact that f takes in one variable but also satisfies the 1.1 and 1.2 given is the part that's kind of throwing me off? just feels weird. :(
empty frigate
well it takes one variable at a time, but we can just apply it three times, and map the triple (x,y,z) to (f(x),f(y),f(z))
and it's the entire triple (f(x),f(y),f(z)) that satisfies 1.1 and 1.2
lavish river
but how do we know that (f(x),f(y),f(z)) lies satisfies the stuff
what kind of ring homomorphisms even are there..
I got it. thanks.
.close
halcyon sluice
vocal sleetBOT
halcyon sluice
I have a basic question...more a confirmation, if you will. Let $c>0$. Is it true that $$\limsup\limits_{n\to\infty}cx_n=\infty\iff\limsup\limits_{n\to\infty}x_n =\infty?$$ I know the limsup of a sequence only diverges to positive infinity if the sequence $\sup_{n\geq k}x_n$ equals positive infinity for every $k$. I think this observation is enough to "prove" the above equivalence, right?
twin meteorBOT
Philip
halcyon sluice
the claim I was making about sup_{n>=k} x_n being equal to infinity for every k follows from the image I posted here
wraith mist
This seems obvious in terms of subsequences
limsup is just a regular lim for some subsequence ${ x_{n_k} }$ of ${ x_n }$:
$\limsup_{n \rightarrow \infty} x_n = \lim_{k \rightarrow \infty} x_{n_k}$
twin meteorBOT
EQUENOS
wraith mist
Multiplying the entire sequence by a positive constant c will multiply limits of subsequences by c
Due to the linearity of limits
vocal sleetBOT
@halcyon sluice Has your question been resolved?
halcyon sluice
.close
vocal sleetBOT
vocal sleetBOT
Available help channel!
cyan crest
Hi, can I please check my $\epsilon$-$\delta$ proof?
vocal sleetBOT
Hi, can I please check my $\epsilon$-$\delta$ proof?
twin meteorBOT
_Kookie
cyan crest
I'm aware of that - it's mostly to check my understanding
brisk moss
some of the things are like, how you came up with the proof and don't need to (shouldn't) be there in the final written up proof
but yes you're understanding looks good
cyan crest
if I were to write this up in some sort of assignment I would cut out most of what's in the image already
brisk moss
just one small thing
cyan crest
should that second < be a <=?
since it could be possible for them to be equal
the deltas
brisk moss
i think it should just say |x-c| < delta_1 and |x-c| < delta_2 or something
cyan crest
but didn't I make the assumption that delta_1 is smaller than delta_2 in that very moment only?
brisk moss
you've already declared delta_1 to be the delta associated with f and delta_2 to be the delta associated with g
thin vale
🧙♂️
brisk moss
like, they're fixed at this point
cyan crest
yes they are, but they are still arbitrary as they are dependent on f and g
because f and g are arbitrary
you can't be sure if delta_1 is smaller or larger than delta_2
brisk moss
isn't that my point
thin vale
you can do something better
let delta=min(delta1,delta2)
and just make that work
instead of dealing with 2
delta
thin vale
idk i didn't read ur work
brisk moss
come on austin
cyan crest
I think it would help if you did?
thin vale
just one small thing
I just saw this part
brisk moss
@cyan crest shoot sorry i missed this if
cyan crest
bruh
brisk moss
but i still think it's awkward to write
cyan crest
yeah I did say I would cut out a lot off stuff in the image
this would be included
but still
brisk moss
ok, fine then
cyan crest
if I have an understanding to the point I can talk about this to others
then I'm a happy person
:>
brisk moss
yes, your understanding looks good, sorry i missed that
vocal sleetBOT
thin vale
no problem
brisk moss
ty austin
brisk moss
yea
vocal sleetBOT
Available help channel!
naive heron
Suppose that a parabola has vertex $\left(\frac{1}{4},-\frac{9}{8}\right)$ and equation $y = ax^2 + bx + c$, where $a > 0$ and $a + b + c$ is an integer. Find the minimum possible value of $a$.
vocal sleetBOT
Suppose that a parabola has vertex $\left(\frac{1}{4},-\frac{9}{8}\right)$ and e...
twin meteorBOT
Aurora
glad python
aime
glad python
I don’t think this is the best way to approach it
naive heron
how would f(1) help because theres no way ill be able to find that
glad python
So you can remove b and c from the question entirely
glad python
how would f(1) help because theres no way ill be able to find that
Write it in vertex form
glad python
Now what’s f(1) in terms of a
naive heron
and we want to find min value of a
naive heron
Why do we want min(a)?
its what question asked
naive heron
idk how f(1) would help though
glad python
In terms of a
glad python
idk how f(1) would help though
We want f(1) to be an integer
naive heron
f(1) = 9/16a-9/8
glad python
And we want the smallest a such that this is an integer
naive heron
oh so we could just minimize a for it to be an integer
glad python
Yeah
glad python
There might be smaller, I didn’t check
twin meteorBOT
naive heron
oh what hmm
glad python
That’s the answer
glad python
Np
naive heron
.close
cyan crest
I'm back, and this time I'd like to check my limit of product of functions proof is correct
vocal sleetBOT
I'm back, and this time I'd like to check my limit of product of functions proof...
lucid bane
median crane
why is (\eps_f = \4\eps{\abs{\map fc}})
twin meteorBOT
Pure
cyan crest
why is \(\eps_f = \4\eps{\abs{\map fc}}\)
lmao that must be a typo again
it's meant to be epsilon/|g(c)| isn't it
median crane
should be epsilon/2|g(c)| i think
cyan crest
yeah you're right
median crane
Ignore the reactions.
cyan crest
why are they even there in the first place?
median crane
No clue.
brisk moss
yea who is doing that
median crane
Yeah, seems okay. I'm not a big fan of how you structure your proof, but it's alright.
Actually, maybe double check your value for (\eps_g).
twin meteorBOT
Pure
cyan crest
fair enough
done
so basically epsilon_g is the same
I had the working done on paper
I'm just bad at copying stuff from paper to latex
skill issue ikr
vocal sleetBOT
@cyan crest Has your question been resolved?
vocal sleetBOT
vocal sleetBOT
Available help channel!
vestal comet
Is this correct?
vocal sleetBOT
@vestal comet Has your question been resolved?
wraith bear
What happens at the third step?
vocal sleetBOT
What happens at the third step?
twin meteorBOT
wraith bear
.close
vocal sleetBOT
vocal sleetBOT
Available help channel!
loud mango
vocal sleetBOT
loud mango
here is my awnser
and this is the teachers:
they said do not simplify
so would my awnser be wrong..?
or not
loud mango
huh
vast shale
send it on a mistake
loud mango
k
vast shale
i got the link on some web
strange crater
you'll have to ask your teacher if yours is wrong
loud mango
but technically its the same
strange crater
so what's your question?
loud mango
well 2x2 = 4
if i write 2x2
is it wrong
both gives 4
thats what im asking
.close
unborn wadi
vocal sleetBOT
strange crater
looks right
hard atlas
well you expressed the results in terms of the standard basis of R^4
vocal sleetBOT
@unborn wadi Has your question been resolved?
vocal sleetBOT
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Available help channel!
waxen flower
"Tie N balloons in a row. Each balloon is red (R), green (G) or blue (B). Petriukas wants to have exactly one balloon of each color. Calculate the minimum number of times you will need to use the scissors?"
if line is RRGRGRRB answer is 3 and if RGRGBB 2
I am asking for a formula / solution so I could find answer with any line
vocal sleetBOT
"Tie N balloons in a row. Each balloon is red (R), green (G) or blue (B). Petriu...
@waxen flower Has your question been resolved?
civic pendant
@waxen flower are you looking for an algorithm?
waxen flower
somewhat, yes
something that I could put into code to solve random lines of such thing
civic pendant
I think there are 5 possible outputs
waxen flower
writting code is the easy part, knowing how to solve this is the problem
civic pendant
either it is impossible, or you need 0,1,2,3,4 cuts
you will never need more than 4 cuts
waxen flower
"We guarantee that every symbol is used at least once."
civic pendant
Oh, so it's only 0,1,2,3 or 4
you just need to find out which case is it...
It's 0 if the string is of length 3
It's 1 if it starts or ends with the three colors
it's 2 if it contains the string "RGB" (or other order of all colors) or if it has two consequitive colors on the one end and it the remaining color on the other end
Such as RGGGGGGGGGGGGGGB
for example
Now we can prove that it's at most 4
Do you know how?
waxen flower
Thanks a lot for all of this but if you can write when it is 3 and 4 it would be perfect 😄
just the "if's"
civic pendant
ok
waxen flower
thanks
civic pendant
But I might be wrong, so you should verify that
civic pendant
So for three
you look at the colors on both ends
if you have, say "R" on one end, you need to look for "GB" or "BG"
and so on...
so something like if string[0] == "R" or string[-1] =="R" and string contains "GB" or "BG"
and for all 3 colors...
then you can do it in 3 cuts
Is that clear?
4 otherwise...
waxen flower
so something like if string[0] == "R" or string[-1] =="R" and string contains "...
this part not too sure, will know after I try it in code
civic pendant
I mean, if it starts with R or ends with R and it contains "GB" or "BG"
or if it starts with "B" or ends with "B" and it contains "RG" or "GR"
or if it starts with "G" or ends with "G" and it contains "BR" or "RB"
that's it..
Goodluck, let me know how it goes
waxen flower
I mean, if it starts with R or ends with R and it contains "GB" or "BG"
I understand that, more so thinking how I will write it in code, but again, will let you know soon if I succedd or hit a wall
civic pendant
👍 +
uneven finch
number 16
vocal sleetBOT
@uneven finch Has your question been resolved?
uneven finch
<@&286206848099549185>
twin meteorBOT
vocal sleetBOT
@uneven finch Has your question been resolved?
viral quarry
Hello , i have an exam soon and its about quadratic equations and inequations
viral quarry
and i have 1 thing that i do not understand
So i get quadratic equation , and i have condition like
That both solutions are the same sign
What condition i should use?
i know for
x1*x2>0
But i think there is one more, so can someone explain it to me
Also there is condition like both solutions are different signs
viral quarry
Yes, we learned that
viral quarry
Okay
tribal moss
$\text{Let}\\ax^{2}+bx+c=0\text{, }\text{ where}\text{ }a\neq 0$
twin meteorBOT
Joanna Angel
tribal moss
it is a general form of the quadratic equation
then if yo talk about zeros = roots
you have to show condtion
that makes existsnce of roots posisble means:
$\Delta=b^{2}-4ac\ge 0$
twin meteorBOT
Joanna Angel
lyric relic
sorry wrong pic
twin meteorBOT
Joanna Angel
viral quarry
Yes
tribal moss
$a\neq 0\\Delta\ge 0\x_{1}\cdot x_{2}=\frac{c}{a}>0$
twin meteorBOT
Joanna Angel
tribal moss
if roots should be of different sign
viral quarry
That is if they are the different sign?
tribal moss
then you write third case diffeernt means
same sign +
difeernt sign minus
$a\neq 0\\Delta\ge 0\x_{1}\cdot x_{2}=\frac{c}{a}<0$
twin meteorBOT
Joanna Angel
viral quarry
So wait in both D should be >=0 ?
tribal moss
yes
viral quarry
I thought i should not include that
viral quarry
If i saw same sign, x1*x2>0
tribal moss
there is not a hing inequalioty between compelx numbers
tribal moss
$\text{ Same sign of roots:}\\a\neq 0\\Delta\ge 0\x_{1}\cdot x_{2}=\frac{c}{a}>0$
viral quarry
both solutions are positive
twin meteorBOT
Joanna Angel
tribal moss
$\text{ Different sign of roots:}\\a\neq 0\\Delta\ge 0\x_{1}\cdot x_{2}=\frac{c}{a}<0$
twin meteorBOT
Joanna Angel
viral quarry
okay thank u ill wait
tribal moss
$\text{ Positive roots:}\\a\neq 0\\Delta\ge 0\x_{1}\cdot x_{2}=\frac{c}{a}>0\x_{1}+x_{2}=-\frac{b}{a}>0$
twin meteorBOT
Joanna Angel
tribal moss
and
$\text{ Negative roots:}\\a\neq 0\\Delta\ge 0\x_{1}\cdot x_{2}=\frac{c}{a}>0\x_{1}+x_{2}=-\frac{b}{a}<0$
twin meteorBOT
Joanna Angel
tribal moss
okay thank u ill wait
can you see it ?
tribal moss
so as you have noticed, all is based on Viete'a formulas
viral quarry
Because pos + pos will for sure be + number
and neg + neg will be - neg so we use x1+x2
viral quarry
yes i wrote them rn
tribal moss
great
viral quarry
so i will always get more than 1 range right?
for values
then i find intersection and write them?
viral quarry
Okay, thank u so much this has helped me alot because i didnt know for the D>=0 condition
Thank u for your time]
tribal moss
yvw 🙂
left cape
can anybody help me with my math project:In the coordinate system, draw the graph of the function f(x)=3sin(2x+6π)
. By proper translation in the same coordinate system, also show the function g(x)=3sin(2x+6π−2)
.
vocal sleetBOT
can anybody help me with my math project:In the coordinate system, draw the grap...
left cape
and please help me with this problem also i am stuck:raw the graph of the function f(x)=cos(2x−6π)
. Don’t forget that the minus sign indicates direction.
vocal sleetBOT
@left cape Has your question been resolved?
slow tundra
vocal sleetBOT
slow tundra
Help please
blazing robin
!show
vocal sleetBOT
Show your work, and if possible, explain where you are stuck.
slow tundra
Tried breaking the terms
Number of terms increases
Couldn't figure out a reduction
vocal sleetBOT
@slow tundra Has your question been resolved?
slow tundra
.close
vocal sleetBOT
unborn sentinel
sa
shadow raft
how else can you write the equation in the form of the roots?
glad python
You know beta and gamma in terms of alpha so you can find alpha since their product is 8
And solve for everything
covert granite
x= what
its an eqn
oh is that a 0
look at numerator
if a =0 then a/b = 0
thats all ill give u
wooden stump
vocal sleetBOT
wooden stump
i have a question about jordan block form
in this video by sheldon axler, he defines the basis for nilpotent operators
he also says that N^(m_n +1) v_n = 0
However, in our basis, we had Nv_1 and N^2 v_1. Wouldn’t the fact that Nv_1 being in our basis indicate that N^(1+1) v_1 = 0?
vocal sleetBOT
@wooden stump Has your question been resolved?
wraith mist
Substitute v1 to the formula with N(z1, ..., z6)
vocal sleetBOT
vocal sleetBOT
Available help channel!
eager fulcrum
Let r,s be real numbers. Find the maximum t such that if $a_1, a_2...$ is a sequence of positive real numbers satisfying $$a_1^r+a_2^r+...+a_n^r\le 2023n^t$$ for all $n\ge 2023$ then the sum $$b_n=\frac1{a_1^s}+\frac1{a_2^s}+...+\frac1{a_n^s}$$ is unbounded i.e for all positive reals M there exists n such that $b_n > M$
vocal sleetBOT
Let r,s be real numbers. Find the maximum t such that if $a_1, a_2...$ is a sequ...
twin meteorBOT
Rhema Singh
eager fulcrum
I don't really understand where to start at all
wicked violet
which chapter u r studing
eager fulcrum
this is an olympiad problem
eager fulcrum
yes, but how does this problem relate to sets?
wicked violet
i am having problem in few question of sets
wicked violet
yes, but how does this problem relate to sets?
no i am asking for my problem
wicked violet
open a different channel
can u tell me how to open i am new on this platform
eager fulcrum
read #❓how-to-get-help
wicked violet
okk
i created it but no one is coming in that
how much time it will take to solve the question?
eager fulcrum
.close
wide sundial
vocal sleetBOT
glossy gull
Hi guys, could someone help me do this :V
If p = (a^3 + b^3 + c^3 + d^3)^(1/3)
Then when p is selected again it will be cubed.
So let's say the next 4 numbers selected are p, e, f, g then we will get (p^3 + e^3 + f^3 + g^3)^(1/3) = (a^3 + b^3 + c^3 + d^3 + e^3 + f^3 + g^3)^(1/3)
Also, there will always be 1 number remaining on the board
wide sundial
If p = (a^3 + b^3 + c^3 + d^3)^(1/3)
Then when p is selected again it will be c...
i don't get you, wdym when p is selected again it will be cubed
won't it be preserved
since it's cubed then cube rooted
glossy gull
Yes its cube rooted and then when it's selected again it is cubed again
So the cube root is basically irrelevant until the last cube root
wide sundial
yeah in that case nothing changes
glossy gull
So unless I'm missing something, the end result will simply be (1^3 + 2^3 + 3^3 + ... + 100^3)^(1/3) and the order in which the elements are selected is irrelevant
glossy gull
Oh yeah mb
tidal dock
sum of cubes from 1 to 100
glossy gull
(3^100 - 1)/2
barren falcon
use that invariant in this task is sum of cubes of all the numbers
wide sundial
somebody else said the same thing about that being the answer
glossy gull
(3^100 - 1)/2
This?
wide sundial
So unless I'm missing something, the end result will simply be (1^3 + 2^3 + 3^3 ...
this
wide sundial
If p = (a^3 + b^3 + c^3 + d^3)^(1/3)
Then when p is selected again it will be c...
oh okay i see, i just kept misreading what you typed
you just map the variables to new stuffs
glossy gull
Let's start with a simpler version
We select two elements instead
We get p = (a^2 + b^2)^(1/2)
Then we do it again and select p as one of the new numbers and get
k = (p^2 + c^2)^(1/2) = (([a^2 + b^2]^(1/2))^2 + c^2)^(1/2) = (a^2 + b^2 + c^2)^(1/2)
wide sundial
Let's start with a simpler version
We select two elements instead
We get p = (a...
mhm i get it now, idk i was misreading
but wait
what if i keep picking 1,2,3,4
glossy gull
You cant
tidal dock
you can't
glossy gull
Because its erased
tidal dock
after one pick they are gone
glossy gull
Then they will be combined with new numbers
wide sundial
1,2,3,4 -> replaced to the other thing with the (a^3 + b^3 + c^3 + d^3)^(1/3) -> keep picking that even after being replaced
wide sundial
where does it say that?
tidal dock
look at it like this
a,b,c,d,e,f,g,... (initially)
take a,b,c,d
result is p=(a^3 + b^3 + c^3 + d^3)^(1/3)
p,e,f,g,... (after one step)
take p,e,f,g
result is (p^3 + e^3 + f^3 + g^3)
the p^3 essentially "unwraps" what we got last time
so result is (a^3 + b^3 + c^3 + d^3 + e^3 + f^3 + g^3)^(1/3)
wide sundial
look at it like this
a,b,c,d,e,f,g,... (initially)
take a,b,c,d
...
wait i think i misunderstood again 😭
i thought (a,b,c,d) swithces to (p,p,p,p)
tidal dock
and by induction you'll have sum of cubes of all numbers
wide sundial
didn't realize it was (a,b,c,d) -> (p)
glossy gull
where does it say that?
I think its implicit that when a new number is written on the board it will be erased after being selected again
tidal dock
i thought (a,b,c,d) swithces to (p,p,p,p)
no, then this process will be infinite
glossy gull
You can only perform this operation 33 times in the case where the new numbers can also be erased
wide sundial
,w 4 + 3n = 100
twin meteorBOT
wide sundial
no, then this process will be infinite
right i get it
wide sundial
I think its implicit that when a new number is written on the board it will be e...
mhmmm
so wait
you were right about your answer right
wide sundial
okay one more question in this context
glossy gull
Smh
wide sundial
lol
tidal dock
wide sundial
okay that part is fine, i just 🔭
tidal dock
okay one more question in this context
go for it
wide sundial
no i mean the same problem but what does it mean when the "sum of cubes is invariant"
ofc related to this
glossy gull
What that person meant was that the sum is invariant under cube rooting
wide sundial
it doesn't change cuz you keep mapping the variables to new things?
wide sundial
What that person meant was that the sum is invariant under cube rooting
oh no, the other person who got your answer said something about
sum of cubes is invariant
glossy gull
That doesnt really make sense without further context
It has to be invariant with respect to something
wide sundial
What that person meant was that the sum is invariant under cube rooting
okay so what does this mean
sum in invariant under cube rooting 
glossy gull
It means that the cube root has no effect on the answer since it is just cubed again
wide sundial
i don't get it tbh:
(a,b,c,d) -> (a^3 + b^3 + c^3 + d^3)^(1/3)
p = that thing
(p,e,f,g) = (p^3 + e^3 + f^3 + g^3)^(1/3)
i get that (1/3) is evaluated at the very end
at least i don't see how it has no effect
does it not? it does at the end where you cube root
tidal dock
p has a cube root, but then you do p^3 so it doesn't matter
moreover your final answer wants N^3 from you, since N has a cube root there too
wide sundial
wait do you mean that the final tuple will be 4 numbers that all have (1/3) in them?
glossy gull
i don't get it tbh:
(a,b,c,d) -> (a^3 + b^3 + c^3 + d^3)^(1/3)
p = that thing
...
Instead
Let p = (a^3 + b^3 + c^3 + d^3)
Then the next sum including p will be
((p^(1/3))^3 + e^3 + f^3 + g^3)^(1/3)
wide sundial
like on the 33th iteration
tidal dock
wait do you mean that the final tuple will be 4 numbers that all have (1/3) in t...
after the final iteration you will have just one number left
wide sundial
Instead
Let p = (a^3 + b^3 + c^3 + d^3)
Then the next sum including p will be...
right i get that but p^(1/3)^3 is okay not for e,f, and g though
glossy gull
right i get that but p^(1/3)^3 is okay not for e,f, and g though
Assuming they are 3 of the original numbers on the board yeah
wide sundial
after the final iteration you will have just one number left
right
oh wait that number maybe is x = (sum of cubes)^(1/3)^3
glossy gull
The questions asks for N^3
So the last number is (sum of cubes up to 100)^(1/3) and then to answer the question you cube that
glossy gull
Yes
wide sundial
So the last number is (sum of cubes up to 100)^(1/3) and then to answer the ques...
yeah okay that was too hard lol
thanks @tidal dock @glossy gull
tidal dock
no problem
wide sundial
@glossy gull just curious but when you wrote out the wrong sum right
where did that come from
(3^100-1)/2
wide sundial
how
glossy gull
(3^100 - 1)/2 = 3^1 + 3^2 + 3^3 + ... + 3^99 + 3^100
glossy gull
Np
wide sundial
.close
vocal sleetBOT
vocal sleetBOT
Available help channel!
sick owl
How, just how?
a = w^2 r rcap
Did they write d (r cap)/dt = omega?
paper depot
what's rcap?
paper depot
sick owl
Polar coordinates?
mild flower
we'd usually call that r-hat
vocal sleetBOT
@sick owl Has your question been resolved?
haughty narwhal
How, just how?
a = w^2 r rcap
Did they write d (r cap)/dt = omega?
Yes they did
r cap is cos theta i + sin theta j
Differentiate it and then write dtheta/dt as w
vocal sleetBOT
@sick owl Has your question been resolved?
sweet ingot
F((3x+1)/(x-3))=x f(x)+7=0 whats x?
vocal sleetBOT
F((3x+1)/(x-3))=x f(x)+7=0 whats x?
vocal sleetBOT
sweet ingot
.reopen
vocal sleetBOT
✅
sweet ingot
xy+2(xy-x)+1=3x y=fx f(x-2)=16/9
Find x
How do i turn the f(x-2)=16/9 to f(x) there isnt any x’s on the right side
vocal sleetBOT
@sweet ingot Has your question been resolved?
sweet ingot
Also dont know if i can ask 2 questions in 1 but asking anyways will make another ticket if i cant, f(2x-5)=8x^3-12x^2-6x+3 What is f(x+2) equal to
vast shale
Also dont know if i can ask 2 questions in 1 but asking anyways will make anothe...
It’s fine
$xy+2(xy-x)+1=3x , y=f(x) , f(x-2)=16/9$
This it?
sweet ingot
Its
twin meteorBOT
Kenshin
sweet ingot
Yeah
vast shale
What have you done up till now?
sweet ingot
I think i need to turn f(x-2) into f(x) but i cant do that since the only way i know how to do that is if the right side has x
Then since y=f(x) i can replace the ys
vast shale
Then since y=f(x) i can replace the ys
yes
we can write the first equation as something like this
$3xy +1 = 5x$
twin meteorBOT
Kenshin
sweet ingot
Yeah found that as well
Dont know how to turn the f(x-2) to f(x) which is the next step i think
twin meteorBOT
Kenshin
sweet ingot
Yeah
vast shale
So we can say that $ f(x-2)-16/9 is a factor of f(x)$
$ f(x-2)-16/9 is a factor of f(x)$
Bot fail sob
vast shale
Wait a second, lemme solve it
vast shale
**Kenshin**
We were here
sweet ingot
Can u explain that i think the term is named differently
Ight
vast shale
**Kenshin**
$(5x-1)/3x =y$
twin meteorBOT
Kenshin
vast shale
Get what I’m saying?
sweet ingot
Also dont know if i can ask 2 questions in 1 but asking anyways will make anothe...
What abt this one
vast shale
$f(2x-5)=8x^3-12x^2-6x+3 What is f(x+2) equal to$
twin meteorBOT
Kenshin
vast shale
Can u finish this im doing something wrong but dont know what
Okay okay
vast shale
$5/3 -1/3x = f(x)$
twin meteorBOT
Kenshin
sweet ingot
Yeah i made a major mistake how did you get that
vast shale
**Kenshin**
$5x-11/(3x-6)=16/9$
twin meteorBOT
Kenshin
vast shale
$45x-99 = 48x - 96$
twin meteorBOT
Kenshin
sweet ingot
Found it ty
vast shale
cool then
sweet ingot
**Kenshin**
What abt this question
vast shale
I’m thinkin
sweet ingot
Yeh i have time solvinf other questions in the meanwhile
Is ax+b an line function
Dont know the translation for it
vast shale
Is ax+b an line function
yeah
vast shale
**Kenshin**
put x = 5/2
vast shale
abcde?
vast shale
Ah okay okay sure
sweet ingot
Ight
Also postinf the other question i got stuck on doğrusal means line
Keep finding 9
Btw should i make another ticket for each question? Or is posting it on one ticket and waiting fine
vast shale
Btw should i make another ticket for each question? Or is posting it on one tick...
Its fine
sweet ingot
$f(2x-5)=8x^3-12x^2-6x+3 What is f(x+2) equal to$
Checked there isnt a typo where are you stuck on?
vast shale
**Kenshin**
put in x+5/2 instead of x in the equation to get f(x)
sweet ingot
Ight
sweet ingot
put in x+5/2 instead of x in the equation to get f(x)
Already did that btw wasnt it x/2+5
Because its 2x?
Cant remove the x^2
Or find a way to
vast shale
I can’t understand
…
sweet ingot
**Kenshin**
This
vast shale
,translate
sweet ingot
put in x+5/2 instead of x in the equation to get f(x)
I think d is (x-1)^3+14 what did you find in that question
obtuse knoll
I want to prove that, if a surjective morphism from A to B exists, then there exists a injective morphism from B to A.
For a function to be a morphism, the neutral element (let it be 1) from on set has to map to the neutral element of the other set and the function has to be associative => g(m*n) = g(m)*g(n) for all m, n
I proved that, if e is a surjective morphism from A to B, then there exists a injective function g from B to A. Because e is surjective, for every b in B there exists at least one a in A. If multiple a's map to the same b, I just choose one via the axiom of choice. So g maps every b in B to exactly one a in A. Because e is a morphism, e(1_A) = 1_B (the neutral element of A maps to the neutral element of B) we can choose for g to map 1_B to 1_A.
I have problems proving that g is also associative, I've thought about function compositions because g(jk) = g(e(j) * e(k)) = g(e(jk)) which is a composition of a surjective function e with a injective function g. But I haven't found a property of this composition that helps me
vocal sleetBOT
I want to prove that, if a surjective morphism from A to B exists, then there ex...
obtuse knoll
I don't even know if there has to exist an injective morphism
vocal sleetBOT
@obtuse knoll Has your question been resolved?
obtuse knoll
I dont quite understand, I need to prove that g(j * k) = g(j) * g(k) for every j, k
wooden stump
if e is surjective, we know that it maps to every element in B. but e(j) doesn’t necessarily map to j. We know e will map some element in A to j
obtuse knoll
I dont understand your argument. My problem is that a * a' = c is in A but because g is injective, there doesnt need to be a element in B such that g maps to c. Therefore g(j)g (k) could produce such a c and no jk in B could map it with g to c
obtuse knoll
Maybe I have to change my construction of g. But I dont see an easy construction such that for every c = m*n in A there exists a j in B with g(j)=c
vocal sleetBOT
@obtuse knoll Has your question been resolved?
obtuse knoll
.close
vocal sleetBOT
vocal sleetBOT
Available help channel!
west coyote
Guys help me please. We got three digit numbers like this 3XY - ZX3=106 what is exact value of 3YZ-YZ3=? Teach says it is possible to find them without knowing exact value for x,y or z. I tried every equation nothing happens
vocal sleetBOT
Guys help me please. We got three digit numbers like this 3XY - ZX3=106 what is ...
west coyote
no ideas?
tidal dock
3XY is 300 + X*10 + Y
ZX3 is Z*100 + X*10 + 3
tidal dock
idk what your teacher meant but Z is 2 ans Y is 9
just sub those in
value of X doesn't matter
west coyote
how did you find z and y
west coyote
but equation ends up y-100z=-191
west coyote
i dont understand
how can you find them while we got two unknown
did you try it manually and find if its true or something
tidal dock
3XY is 300 + X*10 + Y
ZX3 is Z*100 + X*10 + 3
106 is 100 + 0*10 + 6
tidal dock
did you try it manually and find if its true or something
no, just logic
west coyote
can you write it step by step if its possible
west coyote
3XY is 300 + X\*10 + Y
ZX3 is Z*100 + X\*10 + 3
106 is 100 + 0\*10 + 6
i get 300+Y+Z*100+3
tidal dock
and when subtracted are zero
tidal dock
yes ik
but that means that after subtracting, we didn't have to do any carries or anything
west coyote
so you re basically tying because they re numbers from 1 to 9?
tidal dock
idk the term in English
tidal dock
so you re basically tying because they re numbers from 1 to 9?
well that's for sure
they are digits
and can't be 0
west coyote
3XY is 300 + X\*10 + Y
ZX3 is Z*100 + X\*10 + 3
106 is 100 + 0\*10 + 6
doesn't that make z 2
because we need 100 right? for 106
west coyote
and makes y 9
tidal dock
yep
west coyote
oh how stupid i am
tidal dock
precisely
west coyote
i was just trying to solve it with 2 unknown equations or something
vocal sleetBOT
@west coyote Has your question been resolved?
rain hollow
Need help with thees 3 equations need to prove that there correct
blazing robin
!show
vocal sleetBOT
Show your work, and if possible, explain where you are stuck.
rain hollow
Idk where to start
blazing robin
are you familiar with some identity?
gritty breach
how would I prove this to be true?
both are formulas for the sum of a sequence
and what I have to do is prove them to be the same
flat whale
how would I prove this to be true?
both are formulas for the sum of a sequence
...
!original
vocal sleetBOT
how would I prove this to be true?
both are formulas for the sum of a sequence
...
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.
gritty breach
the sequence is an=2n-19
for the 2nd task I have to prove that the sums of the first few digits can be calculated with the formula of n(n-18)
vocal sleetBOT
vocal sleetBOT
Available help channel!
sinful magnet
why -1 + sqrt(n+1)
vocal sleetBOT
why -1 + sqrt(n+1)
dull bear
The yellow paired terms cancel out
sinful magnet
The yellow paired terms cancel out
ok but why does it result in -1 + sqrt(n+1)
dull bear
The only terms that don't cancel out are literally the sqrt{n + 1} and the -sqrt{1}
dull bear
that's what a_{n-1} becomes
sinful magnet
that's what a_{n-1} becomes
ohhh each time you get a number u substract the other term by 1
like iff n =100
sqrt(100)-sqrt(99)
ok i get it
ty
dull bear
Yea, like that 
sinful magnet
.close
vocal sleetBOT
vocal sleetBOT
Available help channel!
clear sundial
vocal sleetBOT
clear sundial
i dont really understand what is happening in this solution
i agree with the n(n+1)/2
but what is the point of subtracting k?
regal bane
You're not subtracting k. You're subtracting 1 + 2 + 3 + ... + (k - 1)
sweet birch
basically you're removing all the terms before k
clear sundial
ahhh i see