u need to add
#help-39
blissful salmon
That’s the answer. And I really recommend you write exact answers until the end rather than rounding in your intermediate steps. It just prevents those tiny errors
midnight haven
Yea i used c²=a²+b²-2ab cos C
midnight haven
c?
oak ivy
the perimeter
we need to add c also
diagram
You just found the curved perimeter
What about that line on top
midnight haven
Do u mean thus
oak ivy
yes
midnight haven
blissful salmon
You forgot to square root it at the end
blissful salmon
OK I GOT IT
Great!
midnight haven
8.66+20.94=29.6
midnight haven
Thank u very much u two
midnight haven
Or 5sqrt3. I like to use exact values
Oo oki
blissful salmon
8.66+20.94=29.6
Or (20/3)pi + 5sqrt(3). It’s just good that you use exact values and then input this into your calculator. But if you arrive at the answer at the end then that’s fine too
blissful salmon
Thank u very much u two
No problem
midnight haven
Oo oki ill remember that for next time
blissful salmon
Oo oki ill remember that for next time
Great. Just in case you miss the answer just by a tiny error
dapper kraken
prove that there exists a constant $c$ such that for every positive integer $a,b,n$ where $a!b!\mid n!$, $a+b<n+ c\ln(n)$
jolly parrotBOT
skissue.in.a.teacup
dapper kraken
ln(n)<0 is only when n is either 1 or 2
if $n=1$, then $a=b=1$, which means $1<c\cdot 0$ which is impossible?
jolly parrotBOT
skissue.in.a.teacup
dapper kraken
am i dumb
woeful torrent
the right side is n + cln(n), so < 1 + cln(1) but your point still holds
because the left side needs to be 2
dapper kraken
the right side is n + cln(n), so < 1 + cln(1) but your point still holds
yeah but i moved it to the left side cause its 1+1<1+cln(1)
dapper kraken
ok fair
tropic saddle
the point is that if a! b! | n! then a+b is barely bigger than n
dapper kraken
wait cant you just pick an arbitrarily large c
dapper kraken
yeah but its bigger
if you pick some stupidly large close to infinity c then cln(n) is also stupidly large close to infinity
or is it like "you cant pick c to be infinity" argument
tropic saddle
well c has to still be a number
if you pick c as like 17 million, then thats huge, sure
but if n is 20 billion then c is small
dapper kraken
yeah ok fair
blissful salmon
ln(n)<0 is only when n is either 1 or 2
Uh I don’t think that’s true. Or I’m just being dumb
dapper kraken
yeah ignore that
woeful torrent
ln(n) <= 0 only when n <= 1
but like has been stated, you can care only for large values of n 😄
this is false as stated for n=1, as you have pointed
near sluice
blissful salmon
yeah but i moved it to the left side cause its 1+1<1+cln(1)
Yeah, only for the case n=1 it’s not true. They should’ve specified in the problem like for n>1 or something
cold pollen
Yeah, only for the case n=1 it’s not true. They should’ve specified in the probl...
can u help this guy in #help-27
blissful salmon
can u help this guy in <#903486480075354182>
Unfortunately not, it’s not in my field of expertise
pearl pondBOT
@dapper kraken Has your question been resolved?
dapper kraken
is this ai?
swift spruce
oops
pearl pondBOT
@dapper kraken Has your question been resolved?
vital crescent
**skissue.in.a.teacup**
hint: think about how many times 2 divides a!b! and how many times 2 divides n!
you should be able to construct an inequality from here
pearl pondBOT
@dapper kraken Has your question been resolved?
dapper kraken
hint: think about how many times 2 divides a!b! and how many times 2 divides n!
v_2(a!b!)=v_2(a!)+v_2(b!)<=v_2(n!)
a+b-S_2(a)-S_2(b)<=n-S_2(n) where S_2(r) is the sum of digits of r base 2
or maybe legrende's formula?
a+b<n-S_2(n)+S_2(a)+S_2(b)+1
blissful salmon
v_2(a!b!)=v_2(a!)+v_2(b!)<=v_2(n!)
a+b-S_2(a)-S_2(b)<=n-S_2(n) where S_2(r) is t...
I think you did it wrong. It should be v_2(n!) = n - S_2(n) not n + S_2(n) by alternate form of Legrendes formula
Otherwise what you have should have the right idea
dapper kraken
whoops
blissful salmon
v_2(a!b!)=v_2(a!)+v_2(b!)<=v_2(n!)
a+b-S_2(a)-S_2(b)<=n-S_2(n) where S_2(r) is t...
Ok this looks good now
blissful salmon
v_2(a!b!)=v_2(a!)+v_2(b!)<=v_2(n!)
a+b-S_2(a)-S_2(b)<=n-S_2(n) where S_2(r) is t...
Now, what is the right hand side of the last inequality less than?
dapper kraken
do you want to minimize rhs to get the thightest bound then prove that you can find c such that cln(n) is between the two or sum?
dapper kraken
Now, what is the right hand side of the last inequality less than?
uhhh like n+1+log_2(a)+log_2(b) or smth?
blissful salmon
do you want to minimize rhs to get the thightest bound then prove that you can f...
Probably, but it should be easy to see already
blissful salmon
uhhh like n+1+log_2(a)+log_2(b) or smth?
Think simple
blissful salmon
v_2(a!b!)=v_2(a!)+v_2(b!)<=v_2(n!)
a+b-S_2(a)-S_2(b)<=n-S_2(n) where S_2(r) is t...
Use only the n-S_2. What is this less than?
dapper kraken
a+b<n-S_2(n)+S_2(a)+S_2(b)+1
rhs of this
dapper kraken
v_2(a!b!)=v_2(a!)+v_2(b!)<=v_2(n!)
a+b-S_2(a)-S_2(b)<=n-S_2(n) where S_2(r) is t...
or this?
blissful salmon
should be less than n+1-log_2(n) or smth?
There’s a better bound
blissful salmon
even simpler?
Yes, it’s easy to see
dapper kraken
nvm i did it wrong
blissful salmon
There’s a better bound
I don’t really mean better, but much simple bound. What is n-S_2(n) less than?
prisma elk
sehingga kekw
dapper kraken
I don’t really mean better, but much simple bound. What is n-S_2(n) less than?
ohh wait im dumb i read that as "thats" lmao
blissful salmon
ohh wait im dumb i read that as "thats" lmao
lol
blissful salmon
idfk just n or smth?
Yes
dapper kraken
blissful salmon
v_2(a!b!)=v_2(a!)+v_2(b!)<=v_2(n!)
a+b-S_2(a)-S_2(b)<=n-S_2(n) where S_2(r) is t...
Ok now from here what does this also tell us about a+b?
blissful salmon
<:sully:651816820122189834>
I was kinda waiting for you to say n. That’s why I said simple haha
dapper kraken
i was skeptical as it was too simple
blissful salmon
i was skeptical as it was too simple
Sometimes simple is best
dapper kraken
Ok now from here what does this also tell us about a+b?
its less than n+S_2(a)+S_2(b)?
blissful salmon
its less than n+S_2(a)+S_2(b)?
Yes! Great job! We’re nearly there
blissful salmon
its less than n+S_2(a)+S_2(b)?
What is S_2(k) less than now?
blissful salmon
log_2(k)?
Kind of. It’s less than the number of digits of k in base 2. What’s a formula for that?
blissful salmon
or simpler like k
Don’t think too simple now
dapper kraken
Kind of. It’s less than the number of digits of k in base 2. What’s a formula fo...
is the number of digits not log_2(k)
or do you want to floor it
blissful salmon
or do you want to floor it
Floor it
dapper kraken
floor(log_2(k)) then
blissful salmon
floor(log_2(k)) then
Very close but not it
dapper kraken
honestly i dont get how you can even prove c exists
like what do you want to prove? how can you conclude c exists?
blissful salmon
like what do you want to prove? how can you conclude c exists?
Uh we will get to that later
dapper kraken
Very close but not it
oh do you add 1 or smth
cause S_2(k)<=floor(log_2(k)) if im not mistaken
blissful salmon
oh do you add 1 or smth
Yes
dapper kraken
How about k=1?
huh
oh
S_2(k)<=floor(log_2(k))+1 right
or am i doing a dum again
blissful salmon
S_2(k)<=floor(log_2(k))+1 right
Right
blissful salmon
or am i doing a dum again
No you aren’t dumb
dapper kraken
so a+b<n+S_2(a)+S_2(b)<=n+log_2(a)+log_2(b)+2?
blissful salmon
so a+b<n+S_2(a)+S_2(b)<=n+log_2(a)+log_2(b)+2?
Yes, now what can you do further?
dapper kraken
=n+log_2(4ab)?
blissful salmon
=n+log_2(4ab)?
Yes and even further. What is ab less than?
dapper kraken
uhh
dapper kraken
no way its n right
blissful salmon
no way its n right
That’s what a is less than. How about ab?
dapper kraken
so n+log_2(4ab)<n+log_2(4n^2)=n+2log_2(2n)
oh wait wtf this is so close to the original
blissful salmon
so n+log_2(4ab)<n+log_2(4n^2)=n+2log_2(2n)
Yes
dapper kraken
so whats left is to get rid of the 2 coefficient infront of n and changing it into ln instead of log_2
blissful salmon
oh wait wtf this is so close to the original
It is indeed very close
dapper kraken
well you can just split the 2 coefficient into 2log_2(n)+2 and since were dealing with massive n that +2 wont contribure to anything
blissful salmon
well you can just split the 2 coefficient into 2log_2(n)+2 and since were dealin...
Yeah do this
dapper kraken
so n+log_2(4ab)<n+log_2(4n^2)=n+2log_2(2n)≈n+2log_2(n)
wait
what if you just let c=2/ln(2)
dapper kraken
so n+log_2(4ab)<n+log_2(4n^2)=n+2log_2(2n)≈n+2log_2(n)
cause n+cln(n)=n+2log_2(n) by change of base and its exactly this
then idk just say multiply c by like 1 million to completely ignore that 2
blissful salmon
what if you just let c=2/ln(2)
Uh that doesn’t work
blissful salmon
then idk just say multiply c by like 1 million to completely ignore that 2
You can do this instead
dapper kraken
Take n=2 for example
ok to be fair i shouldve wrote like c=2k/ln(2) for some massive k
blissful salmon
ok to be fair i shouldve wrote like c=2k/ln(2) for some massive k
True, then it would work. I think even a fairly small k works like k=3
dapper kraken
why think small when you can think big
blissful salmon
why think small when you can think big
Big is too much lol. And small is actually impressive
pearl pondBOT
blissful salmon
ok, tysmm <:pandahugg:513069003120574464>
No problem :)))
pearl pondBOT
Available help channel!
hollow vector
Find out the last three digits of 2025^2025
pearl pondBOT
Find out the last three digits of 2025^2025
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
toxic lichen
last three digits = remainder mod 1000
hollow vector
How to do that mod ? what do I do after writing 2025^2025 mod 1000?
hollow vector
No actually. I heard about it from chatgpt , still don't understand . Isn't there any other intuition to do it as we do for finding the first digit( recognizing repeating patterns of powers)
toxic lichen
!nogpt
pearl pondBOT
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).
toxic lichen
No actually. I heard about it from chatgpt , still don't understand . Isn't ther...
you can go for that sure
im a bit skeptical that a pattern will emerge quickly enough tho
hollow vector
As it has to do with 3 digits (not 1 ) I haven't found any patters
Which method should I learn?(You mentioned one before)
toxic lichen
well first off you can say that the last 3 digits will be the same as for 25^2025
which is 5^4050
hollow vector
Yes
toxic lichen
and then you can look at the last three digits of 5^n for n=1,2,3,...
hollow vector
Umm, okay let me have a try
Wow I got a pattern . 5^1 and 5^2 it doesn't work but for later values of n it looks like the odd powers have 125 as last three digits and even ones have 625. 4050 is an even number hence the answer is 625?
Is there any general way of doing these types of maths? What about 2023^2023?
grim fractal
Another good way is the binomial theorem
grim fractal
Is there any general way of doing these types of maths? What about 2023^2023?
2023^2023 is equivalent to 23^2023 mod 1000
= (20 + 3)^2023
If you expand this with the binomial theorem, most of the terms are divisible by 20^3 (and hence have 000 as the final 3 digits, and can be ignored)
or ig now you have to find 3^2023 last 3 digits
hollow vector
Isnt there any general approach?
grim fractal
I'm pretty sure 3^2023 = 3^23 mod 1000
or generally a^(m + 1000) = a^m mod 1000
but I don't remember the proof
Oh yes this is Euler's Theorem
hollow vector
How to solve then
grim fractal
If $\mathrm{gcd}(a,m) = 1$, then $a^{\phi(m)} \equiv 1 \pmod{m}$
jolly parrotBOT
Ari
grim fractal
where $\phi(m) = # {n \leq m : \mathrm{gcd}(m,n) = 1}$
jolly parrotBOT
Ari
grim fractal
And you know $\phi(1000) | 1000$
jolly parrotBOT
Ari
grim fractal
So $23^{1000} \equiv 1 \pmod{1000}$
jolly parrotBOT
Ari
grim fractal
And so $23^{23} \equiv 23^{23} \cdot (23^{1000})^2 \equiv 23^{2023} \pmod{1000}$
jolly parrotBOT
Ari
hollow vector
Ah, I don't understand a single thing. where can I know more about this?
grim fractal
And now the binomial theorem method is easier
grim fractal
Ah, I don't understand a single thing. where can I know more about this?
The theorem's a little easier to understand if you know abstract algebra but it's not necessary
pearl pondBOT
@hollow vector Has your question been resolved?
plush bramble
<@&268886789983436800>
wet osprey
<@&268886789983436800>
sharp smelt
<@&268886789983436800>
pearl pondBOT
wet osprey
Love
light helm
np
prime bramble
pearl pondBOT
Available help channel!
hushed sky
How many arrangements of the letters in MISCELLANEOUS have no pair of consecutive identical letters?
This problem has to deal with the Principle of Inclusion-Exclusion
jolly parrotBOT
joseph
pale holly
I need help regarding simplifying algebra
pearl pondBOT
I need help regarding simplifying algebra
pale holly
I can get K on its own. But it still has L on the other side of the equation. How can I get an answer for K & L. without having them in the answer
plush bramble
what are production and cost functions and lagrange equations
pale holly
They're basically a concept in economics. The actual process is just multi calculus variable and then solving simultaneously
plush bramble
what are the definitions how they relate to each other
pale holly
Tbh u don't need to know the detail of what they mean. You just need to know objective function and constraints. In this case objective function is the cost and constraint is Production. The constraint should equal to P
pale holly
Langrange equation is basically an equation which combines objective function, constraint and constant. It's as follows objective function + (constraint- constant)
Tbh I just need help simplifying algebraic equation
plush bramble
what is there to simplify then
pale holly
I have 3 equations and I need to find L, K and lambda.
plush bramble
where is it here
plush bramble
what are production and cost functions and lagrange equations
could have just showed those when i asked here
plush bramble
are the unknowns alpha, beta and lambda?
pale holly
No they are L,K and lambda.
plush bramble
By no means is this the most optimal. Solve eqn3 for L^alpha and plug into eqn2. This will eliminate L as a variable from eqn2 and only have lambda and K. Solve for lambda in eqn2 and plug that into eqn 1. This will give you a solution for K. Then plug that K into eqn2 to get an eqn for lambda. Repeat for 3 to get an eqn for L
jolly parrotBOT
riemann
plush bramble
you'll use those a lot
pearl pondBOT
@pale holly Has your question been resolved?
pale holly
Could you help me on the part where I sub lambda into eqn1. I don't know how can I get rid of L
That looks horrible to simplify. Have I done something wrong
Also wouldn't it be better to divide eqn1 by eqn2. Because they have quite a lot of similar variables
I got this before. Could this way work?
plush bramble
Also wouldn't it be better to divide eqn1 by eqn2. Because they have quite a lot...
no because that's not how dividing sums by sums works
,tex .wrong cancel
jolly parrotBOT
riemann
plush bramble
you can solve for K^beta from eqn 2 and plug that into lambda to make it easier
Plug this into eqn 1
that'll give you an eqn for K
pale holly
Then how would I get rid of the L^alpha -1
plush bramble
**riemann**
use this
pale holly
Oh so separate L by have 2 L's
plush bramble
L^alpha = [L^(alpha-1)] ^ (some power)
plush bramble
Plug this into eqn 1
i'd just solve for L directly by taking the 1/alpha power to both sides of here
plush bramble
and plug that into the L here
plush bramble
Is this P th same P as in the production function. Why is that an equation at all?
pale holly
No that P is the constant
plush bramble
oh lower vs. upper case. got it
pale holly
The first part is the constraint and the P is the constant
What should I do from here
plush bramble
okay scratch what i said before. multiply eqn1 by L and you'll get A * L^alpha K^beta somewhere. Use eqn 3 to substitute that for P. Similarly, multiply eqn2 by K and you'll get P again. eqn1 will become
wL + P * (stuff) = 0
eqn2 will become rK + P * (other stuff) = 0
plush bramble
,rotate
jolly parrotBOT
plush bramble
solve eqn1 for L and solve eqn2 for K. Then plug into 3 and solve for lambda
pale holly
I got this for lambda
Could you check my working out
Actually I forgot to divide the right side by A
I should've done that on the second last step
plush bramble
,rotate
jolly parrotBOT
pale holly
Which part
plush bramble
**riemann**
use distributivity and a lot of other rules here
pale holly
Could this way not work
plush bramble
,rotate
jolly parrotBOT
jovial kiln
<@&268886789983436800>
woeful torrent
<@&268886789983436800> scam, similar to #help-49 message.
also same scam has been sent in:
#help-42 message
#help-21｜아리스킨충1 message
...
all help channels
pearl pondBOT
pearl pondBOT
Available help channel!
broken chasm
how did they get the integral...
pearl pondBOT
how did they get the integral...
boreal raptor
how did they get the integral...
sum is not smooth
but lim n-> inf
means infinite blocks
becomes smooth
becomes integral
broken chasm
wdym smooth
urban ivy
how did they get the integral...
riemann sum with delta x = \frac{T_f-T_i}{N}
the denominator of the sum requires T_i to be added due to the bounds of integration from T_i to T_f.
grim phoenix
Would this be adequate proof?
grim phoenix
<@&286206848099549185>
grim phoenix
<@&286206848099549185>
pearl pondBOT
@grim phoenix Has your question been resolved?
hazy bear
Idk about that ~ symbol but here's my proof:
Join OB.
SR divides triangle CO and CB of triangle OBC in equal ratios (1/2) so SR is parallel to OB by Thales Theorem.
So angle CSR=COB, CRS=CBO, C=C.
Hence CRS is similar to CBO.
CS/CO=RS/BO
1/2 = RS/BO
BO=2RS
Similarly, you can find that BO=2PQ
Hence RS=PQ
But really what is that symbol? '~'
weary ledge
!nosols
pearl pondBOT
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
hazy bear
Would this be adequate proof?
Yes correct!
I understand you used ~ symbol for vector
Lol I use → for vector so I was confused
hazy bear
!nosols
Sorry, but he already solved it and his solution is correct, I just provided an alternate solution 
hazy bear
Would this be adequate proof?
Just 1 wrong thing, you should write CR→, SR→, PQ→, CA→ instead of plain CR, SR, PQ, CA or it will make your equations wrong
timber tartan
Just 1 wrong thing, you should write CR→, SR→, PQ→, CA→ instead of plain CR, SR,...
eh let's not be pedantic
hazy bear
.close
buoyant pasture
Integration ( (0 ) to ( \pi/4 ) ) ( \sec^7\theta )
pearl pondBOT
Integration \( (0 \) to \( \pi/4 ) \) \( \sec^7\theta \)
jolly parrotBOT
Andy
buoyant pasture
( \int_0^{\pi/4} \sec^7\theta , d\theta )
jolly parrotBOT
Andy
bitter herald
Use the reduction formula
toxic lichen
well you can also substitute u := sin(θ) and do an 8 term partial fraction decomposition 
thoroughly unpleasant but it exists
toxic lichen
ehhh
complex how?
i did think about making like cos(x) = (e^ix + e^-ix)/2 but in your case all that mess will be in denom
stoic harness
That’s false right?
pearl pondBOT
That’s false right?
crystal hill
Are endpoints considered as local extrema
huh
no, when the domain is restricted, we find absolute extremas
pearl pondBOT
@stoic harness Has your question been resolved?
dusty flame
?
i mean
consider this
by using second derivative test
only x =2 is a rel maximum
however x=4 is the actual maximum
crystal hill
however x=4 is the actual maximum
we sub in the end points in the first order derivative
x = 2 is the local maxima is the constraint {1 < x < 4} is not present
pearl pondBOT
@stoic harness Has your question been resolved?
plucky python
That’s false right?
yh that's false
pearl pondBOT
pearl pondBOT
Available help channel!
sharp smelt
I'd like to show if dim(null(S)) = dim(null(T)) ... then $S=E_2TE_1$
jolly parrotBOT
wai
sharp smelt
.close
pearl pondBOT
sharp smelt
.reopen
boreal raptor
lowk idk what ur asking for
pearl pondBOT
crystal hill
wow
boreal raptor
uh
crystal hill
start again mister
boreal raptor
yea
crystal hill
transition as in the point of intersection?
boreal raptor
🥀
boreal raptor
transition as in the point of intersection?
oh i was thinking like
crystal hill
whatever it is claim a new channel, this one is going to die
crystal hill
im not sure either
boreal raptor
like how to convert a graph into the other
crystal hill
no it will close on its own
lone nimbus
when do i make the new post
boreal raptor
now
lone nimbus
ok
boreal raptor
another channel
crystal hill
pearl pondBOT
Available help channel!
dense ledge
find p prime so that 7^p+9*(p^p) is a square number
pearl pondBOT
find p prime so that 7^p+9*(p^p) is a square number
dense ledge
.close
pearl pondBOT
spiral pivot
@dense ledge did you figure it out?
I was looking into the problem and I'm curious to know if it has a solution
Seems pretty difficult if so, but I'm skeptical about it.
pearl pondBOT
Available help channel!
dapper kraken
how many triples of natural numbers $(a,b,c)$ exist such that
$$a+2b+3c=455$$
$$5a+3b+c=420$$
jolly parrotBOT
skissue.in.a.teacup
dapper kraken
im p sure you can make it into 3(5a+3b+c)-(a+2b+3c)=805
14a-7b=805
2a-b=115
pure rapids
yep i think
dapper kraken
**skissue.in.a.teacup**
im pretty sure if you pick (a,b) such that 2a-b=115 then there always exist c that holds for the original
how do i write that
toxic lichen
im pretty sure if you pick (a,b) such that 2a-b=115 then there always exist c th...
will c ≥ 0 tho
pure rapids
and reduce it to b and c
jolly parrotBOT
Result:
1855
Result:
265
hm yeah
spice condor
which now has a finite number of _natural_ solutions...
the original also had finite solutions
so
what is your point
dapper kraken
the original also had finite solutions
2a-b=115 didnt
spice condor
2a-b=115 didnt
i am referring to the original problem good sir
toxic lichen
i was highlighting why my route was better than skissue's.
dapper kraken
i am referring to the original problem good sir
ann was talking about how when i changed it to 2a-b=115 it has infinite solutions, while changing it to b+2c=265 had finite solutions good sir
spice condor
i was highlighting why my route was better than skissue's.
alright
spice condor
ann was talking about how when i changed it to 2a-b=115 it has infinite solution...
0k
dapper kraken
thus b + 2c = 265
soo im guessing you want to bound one of the variables (b is prolly the easiest one) such that a is still natural?
toxic lichen
sure but better to say 1 ≤ c ≤ floor(265/2) and then b = 265 - 2c and then what'll a turn out to be?
dapper kraken
$$a+2(265-2c)+3c=455\iff a-c=-75$$
$$5a+3(265-2c)+c=420 \iff 5a-5c=-375 \iff a-c=-75$$
jolly parrotBOT
skissue.in.a.teacup
dapper kraken
so $a=-75+c$?
jolly parrotBOT
skissue.in.a.teacup
dapper kraken
oo so 76<=c<=132?
wait we have the 3 variables dependent solely on c and we know the bounds of c
so isnt the amount possibilities of (a,b,c) just the amount of possibilities of natural 76<=c<=132
132-76+1=57?
iron basin
or you can just assume a to be known and then derive b and c like ||b = 115 - 2a and c = a + 75||
and find the tightest inequality with all of them together
only b here fixes the upper limit
oh nvm that's what just happened lol
dapper kraken
.solved thanks yaal 
pearl pondBOT
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Available help channel!
broken chasm
i am not sure if i am completely hallucinating but this seems wayyy too easy especially since what was done in class seems to take much longer than this
leaden nebula
i thought the final result would be
$$(sin^-1 (x^2))^2 * 1/4$$
jolly parrotBOT
Snek
spice condor
i thought the final result would be
$$(sin^-1 (x^2))^2 * 1/4$$
rip
broken chasm
i am not sure if i am completely hallucinating but this seems wayyy too easy esp...
question is here
west vault
i am not sure if i am completely hallucinating but this seems wayyy too easy esp...
What's your question?
You need help with understanding the proof?
What is 15.10a)?
broken chasm
it's just the sequence a_n=(-1)^n\cdot n
I'm just not sure if this proof is right because it seems way too simple
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Available help channel!
buoyant pasture
pearl pondBOT
buoyant pasture
By integration i got xy=constant
so if constant is 0 then xy=0 which means A and B
And if not equals to 0 we got symmetric y=x,-x so C
I guess answer should be D
brittle tinsel
if you get xy = C, then y = C/x
as long as x is nonzero
and that's symmetric about what?
spiral pivot
So if the function is symmetric about y = 0, this implies you can replace x with -x and get the same equation.
Is this true? What about the other two?
(no need to actually solve the DE in this case)
pearl pondBOT
@buoyant pasture Has your question been resolved?
rigid geyser
just confused on what to do first?
pearl pondBOT
just confused on what to do first?
light helm
you should draw the line for the tower slightly titled to distinguish it from north
empty rover
for box b, why the take 200-T=20a?
pearl pondBOT
for box b, why the take 200-T=20a?
crystal hill
What are the forces acting on the box B?
empty rover
tension force and the weight
crystal hill
What about the gravitational force?
empty rover
What about the gravitational force?
that would be the acceleration=a right?
crystal hill
No, a is the acceleration of the box after calculating all the forces
On the box, one force acts upwards and one force acts downwards
The sum of those forces will give the net force
empty rover
No, a is the acceleration of the box after calculating all the forces
I dont get what you mean for this
empty rover
On the box, one force acts upwards and one force acts downwards
upwards is tension, downwards is the weight right?
verbal whale
Yes
crystal hill
upwards is tension, downwards is the weight right?
What is the weight of the box?
crystal hill
Both boxes will experience same acceleration
Same magnitude not direction
empty rover
What is the weight of the box?
A is 300N and B is 200N
empty rover
Now what is the net force acting on B?
200N
crystal hill
Tension and weight is the force that is acting on the box
empty rover
Nah 200 - t
wait so net force = weight - tension?
crimson birch
wait so net force = weight - tension?
Yes
crystal hill
wait so net force = weight - tension?
Do you know why tension is negative?
crystal hill
It's because it acts opposite to the direction of weight
crystal hill
is it because tension is acting downwards?
Tension acts upwards
empty rover
It's because it acts opposite to the direction of weight
so it acts upwards
crystal hill
so it acts upwards
Yes
empty rover
so there will not be a case where tension acts downwards right
since weight always acts downwards
crystal hill
so there will not be a case where tension acts downwards right
No
crystal hill
since weight always acts downwards
Yes correct
If weight is the only force acting on a box, tension will always act upwards
empty rover
If weight is the only force acting on a box, tension will always act upwards
oh then what are the other forces that will act on the box that causes tensions act downwards instead?
crystal hill
There will be cases, but you may never come across them at your level
Let's say a balloon tied to the ground filled with helium
It will have buoyant force acting up, which will have tension acting down
empty rover
There will be cases, but you may never come across them at your level
then for a why they took T=30a?
crystal hill
Yes
empty rover
Yes
actually, you said that tension acts opposite direction as weight right?
then how come for box A, the tension is not acting opposite the weight of box A
crystal hill
Net forces
For box A a force acts towards the right
The pulling force
So tension acts opposite to it, to the left
empty rover
So tension acts opposite to it, to the left
oh that makes sense
but if there's no pulling force, then tension goes up right?
crystal hill
If there is no force acting on a box, tension is 0
empty rover
If there is no force acting on a box, tension is 0
but there's weight tho right? so it'll act upwards?
crystal hill
the normal force acting on box A would cancel it
empty rover
the normal force acting on box A would cancel it
oh so normal reaction force would also be 300N right?
same as the weight?
crystal hill
yes and they would cancel each other out
empty rover
yes and they would cancel each other out
ok thanks for helping me out bro, really appreciate it
crystal hill
yw, good luck
empty rover
can I add you by any chance cus I might need help in the future
crystal hill
sure
empty rover
.close
midnight flicker
The figure shows the graph of the function y = f(x). Which of the functions CAN be an antiderivative of f(x)?
autumn fossil
The figure shows the graph of the function y = f(x). Which of the functions CAN ...
I'd try differentiating all the options
or at least those you cant immidiately eliminate
lapis flicker
y = lgx
autumn fossil
The figure shows the graph of the function y = f(x). Which of the functions CAN ...
or if you could guess what function f(x) is, you can just integrate it
pearl pondBOT
boreal raptor
The figure shows the graph of the function y = f(x). Which of the functions CAN ...
what is tgx?
lapis flicker
tanx? ig
alpine surge
Must be
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Available help channel!
dense reef
is this doable?
pearl pondBOT
is this doable?
spare lark
sum of arctan(k3^k + (k-1)3^k)?
dense reef
sum of arctan(k3^k + (k-1)3^k)?
yeah
spare lark
you can simplify the inside
dense reef
you can simplify the inside
can you hint at what do i do?
spare lark
factorisation
dense reef
and do i need to occupy two channels when i have two seperate questions?
spare lark
one question at a time
dense reef
factorisation
but that doesnt get me anywhere
dense reef
one question at a time
okayy
dense reef
but that doesnt get me anywhere
or maybe im not able to proceed
spare lark
factor by 3^k
also i have no idea how to go for summing from 1 to 10
except computing each value
dense reef
factor by 3^k
yeah but doesnt make it any easier
spare lark
it does
dense reef
also i have no idea how to go for summing from 1 to 10
well in things like this its usually given by arctan(1/smtg smtg) then using formula it gets converted to a telescopic series
spare lark
you have (2k-1)3^k
dense reef
so is there any way to make it into that??
timber tartan
,w sum k = 1 to 10 arctan((2k-1)*3^k)
spare lark
see
timber tartan
it's a bs question
dense reef
it's a bs question
alright thank you
dense reef
.close
pearl pondBOT
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muted sierra
4sec^2 θ /(1-3sec^2 θ ), manipulate it such to bring 2 Sec2 θ i.e. 2(sec^A)/( 2-sec^A)
eager jewel
sec^A?
muted sierra
Actually the q was to prove sec(45deg+ theta) sec ( 45deg- tta) = 2sec2 tta, i have to prove this, so like first of all i converted the the given into 1/cos(45+tta)cos(45-tta), then by the help of formula , i got 1/cos^2 45deg - sin^2 tta, then by further trigonometric man i reached there, so first of all do tell if i am doing wrong or right by solving that way
i want u guys to like do something to that eq such to convert it into rhs like by dividing/multiplying...anything and bring the rhs without actually changing the ratios directly
do tell me if it is possible or not
?
plush bramble
,w sec(45deg+x) sec ( 45deg-x) = 2sec(2x)
jolly parrotBOT
plush bramble
yes it's possible
muted sierra
how?
plush bramble
try using addition angle formulas for 1/cos(a+b) on the left
muted sierra
4sec^2 θ /(1-3sec^2 θ ), manipulate it such to bring 2 Sec2 θ i.e. 2(sec^A)/( 2-...
We're talking about this right?
plush bramble
Actually the q was to prove sec(45deg+ theta) sec ( 45deg- tta) = 2sec2 tta, i h...
maybe. you said your actual question was here
plush bramble
since it's true, it should be able to prove with trig identities
muted sierra
We're talking about this right?
yeah but i want help from here, if it is plausible
plush bramble
4sec^2 θ /(1-3sec^2 θ ), manipulate it such to bring 2 Sec2 θ i.e. 2(sec^A)/( 2-...
how do you know this is right
muted sierra
well i brought that
muted sierra
Actually the q was to prove sec(45deg+ theta) sec ( 45deg- tta) = 2sec2 tta, i h...
explanation
i'm not, 100% sure may be i'm wrong
plush bramble
not enough details to verify your steps are correct
muted sierra
w8
plush bramble
• Show your work, and if possible, explain where you are stuck.
muted sierra
1/cos^2 45 deg sin^2 theta, this is right okay?
lol am wrong after that..damn
w8 lemme solve again
timber tartan
1/cos(45+x) = √2/cos(x) - sin(x)
1/cos(45-x) = √2/cos(x) + sin(x)
multiply to get
2/cos^2 (x) - sin^2 (x)
which is just 2/cos(2x) or 2sec(2x)
plush bramble
1/cos(45+x) = √2/cos(x) - sin(x)
1/cos(45-x) = √2/cos(x) + sin(x)
multiply to g...
!nosols
pearl pondBOT
1/cos(45+x) = √2/cos(x) - sin(x)
1/cos(45-x) = √2/cos(x) + sin(x)
multiply to g...
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
muted sierra
NVM i solved thanks btw, actually i took cos 45 1/2 by mistake , ultimately getting 1/4.. sry guys
Thanks again
.close
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Available help channel!
prisma vortex
pearl pondBOT
strange fox
,rot
strange fox
,rccw
jolly parrotBOT
strange fox
kinda weird how they dont even give you what corners are ABCD and which midpoints are EFGH
prisma vortex
kinda weird how they dont even give you what corners are ABCD and which midpoint...
ig we can just put in random ones for each
doesnt rlly matter
strange fox
yeah
strange fox
<@1183148179319959595> ?
no idea but i would probably guess 4
prisma vortex
well, we may say that S+5+7+8 is the area, yes?
yeah
prisma vortex
my teacher said theres a way to do it without assuming so i kinda wanna find it
frigid lark
my teacher said theres a way to do it without assuming so i kinda wanna find it
geometry
spam circles
frigid lark
i think its to do with the lines idk tho
thing is, 7 and 5
for example
the bigger the 7 is, the smaller the 5
now, if youll stretch it
very far
so one line cuts the square in half, to its width
so, then you form on the bottom left square a trianglw
and top left, becomes a hole, that exact triangle sizw.
so, two opposing parts add up to the sum of two
which means the sum of two is 12
so one is 6
and 8+S=12
so S=4
but thats some insane logic
strange fox
real
pearl pondBOT
@prisma vortex Has your question been resolved?
midnight haven
I have a question to ask here is the image:
pearl pondBOT
I have a question to ask here is the image:
midnight haven
I would just want to know what's the difference between the rate and cumulative rate
midnight haven
And maybe an explanation on why does getting 1 copy have a lower chance than getting 3 copies
shrewd mist
rate is the individual likelihood of any given constellation occurring (C1, C2, etc.), while the cumulative rate is the likelihood of any given constellation or any constellation "below" it occuring. So the cumulative rate of C4 would be the likelihood of C4 + C5 + C6, while the rate would just be the likelihood of C4 only
midnight haven
rate is the individual likelihood of any given constellation occurring (C1, C2, ...
But why is the individual likelihood of c1 0% even though c1 is like "first copy" or "first success"?
midnight haven
rate is the individual likelihood of any given constellation occurring (C1, C2, ...
I understand the second part except the part that the cumulative rate should exist at all
I understand the fact that the chances of getting 6 copies of the same certain thing in a drop pool in such few trials is low, as shown in the table where 6 copies have a 1.36% chance of occurring
And would make sense to get 1 or 2 copies considering there is a guarentee system at play
Which is why it is shown as 100%
But what confuses me is that it's individual rate does not make sense to me, unless the rates is telling me the chances of landing on a particular outcome in those amount of trials
midnight haven
Ok so here is what I think from studying the table and the graph
Cumulative rate is the chances of any of the copy levels or constellation levels to occur in N amount of trials depending on value N
The rate is the chances of the result to end up after N amount of trials
midnight haven
Let's say I'm a photon
And another photon is coming my direction
So frame of reference speed is c
Object speed is -c
c-(-c)=2c
world ends
There must be something preventing this right?
pearl pondBOT
@midnight haven Has your question been resolved?
pearl pondBOT
midnight haven
.reopen
pearl pondBOT
✅
plush bramble
c-(-c)=2c
Should try asking in #old-network message
midnight haven
The rate is the chances of the result to end up after N amount of trials
Is this right though?
pearl pondBOT
@midnight haven Has your question been resolved?
pearl pondBOT
@midnight haven Has your question been resolved?
pearl pondBOT
@midnight haven Has your question been resolved?
pearl pondBOT
@midnight haven Has your question been resolved?
tiny pawn
Test
pearl pondBOT
@midnight haven Has your question been resolved?
midnight haven
The rate is the chances of the result to end up after N amount of trials
So is this correct?
hazy bear
The rate is the chance of a specific result happening in a single trial.
The cumulative rate is the chance of that result happening at least once over N trials.
General definition, it maybe different in your context
pearl pondBOT
@midnight haven Has your question been resolved?
versed remnant
prob a dumb question but
pearl pondBOT
prob a dumb question but
versed remnant
so set b is basis for P2 and i rewrote them as
{1 0 1, 0 1 1, 1 2 1} and we know that p(t) should also span that set so
i combined both into the augmented set and that didnt get me the answer
and then i dropped the question into chatgpt (ik bad mistake but i was tired) and this gives the answer
why does this work and the method i tried fail?
near mauve
so set b is basis for P2 and i rewrote them as
{1 0 1, 0 1 1, 1 2 1} and we kno...
p(t) wouldn't span B, p(t) is just a vector
near mauve
and then i dropped the question into chatgpt (ik bad mistake but i was tired) an...
yep finding the coordinates means finding the coefficients
versed remnant
p(t) wouldn't span B, p(t) is just a vector
p(t) should span P^2 and b set is the basis for that , so i thought it would be in set b
near mauve
p(t) is a linear combination of the basis elements
the thing a single vector v spans in a vector space V is the set {xv, where x is field member}
toxic lichen
so set b is basis for P2 and i rewrote them as
{1 0 1, 0 1 1, 1 2 1} and we kno...
you know p(t) should belong to span(B) you mean
versed remnant
yeah that , like we can write p(t) as linear combination of vectors in B
toxic lichen
p(t) alone doesn't need to span shit
like you're not using the verb "to span" correctly
versed remnant
well , im still confused
why isnt my way not correct tho? (we skipped that part :c)
like im confused as to why the answer (from chat gpt) rewrites the matrix as
(a 0 c, 0 b 2c, a b c)(t^0
t^1
t^2)
i thought we were to find a b c , the scalars for which the linear combination is possible
but we are finding t?
toxic lichen
... obligatory !nogpt
near mauve
i thought we were to find a b c , the scalars for which the linear combination i...
nah we're comparing coefficients.
a + c = 1
b + 2c = 4
a + b + c = 7
toxic lichen
but also maybe you are overthinking the matrix thing???
versed remnant
... obligatory !nogpt
yes ik but ||the final answer from chat gpt is the correct one||
near mauve
to solve for a b c and find the coordinates
toxic lichen
yeah idk im confused wtf you did or why you did it and your wording is kinda not helping
i think im out
versed remnant
hmm
versed remnant
nah we're comparing coefficients.
a + c = 1
b + 2c = 4
a + b + c = 7
ik we are trying to find a , b and c
near mauve
hmm so what are you stuck at
versed remnant
like i understand ur compare coefficient method , but i need to like use matrix
and this example isnt helping either
wait nvm
i figured it out
ty both <3
.close
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Available help channel!
unkempt yacht
is my explanation correct?
pearl pondBOT
is my explanation correct?
woeful torrent
that is true 😄 matrices that are row equivalent to the identity are linear bijections, so there can only be one solution
sharp vigil
you left off a couple - signs in the matrix
woeful torrent
yes, because the corresponding system changes 😄
sharp vigil
well it doesn't matter in the sense that it doesn't matter what the entries in the coefficient matrix are called. but it does matter in the sense that if the system of linear equations is notated as it is, it doesn't necessarily tell you anything about the matrix you wrote down
woeful torrent
you could have, however, just left out the matrix and your solution would be just fine 😄
unkempt yacht
well it doesn't matter in the sense that it doesn't matter what the entries in t...
ahh i see
unkempt yacht
you could have, however, just left out the matrix and your solution would be jus...
yeah
alright thanks everyone
.close
pearl pondBOT
cerulean smelt
how do I calculate that
pearl pondBOT
cerulean smelt
dapper kraken
this channel is closed as the original message was deleted, make a new channel
pearl pondBOT
Available help channel!
sick moon
pearl pondBOT
sick moon
Of course, I personally believe that this statement is true; however, if there happens to be a counterexample, I would also be truly grateful if you could help identify it.
I have considered the problem by dividing it into two cases: when the graph G is a convex grid-graph and when it is a non-convex (or concave) grid-graph. As a result, I was able to successfully prove the uniqueness of the decomposition in the case where G is a convex grid-graph. However, when G is a non-convex grid-graph, I find myself unable to see a clear strategy or approach for proving uniqueness. I have not been able to identify a valid method of attack, and the structural irregularities in non-convex regions make the problem especially challenging.
I would therefore like to ask if there are any known techniques, structural insights, or efficient strategies that might be helpful when trying to prove properties like uniqueness in the decomposition of non-convex grid-graphs. Any guidance or suggestions on how to approach this more effectively would be sincerely appreciated.
pearl pondBOT
@sick moon Has your question been resolved?
sick moon
<@&286206848099549185>
frigid lark
WHAAAAAAT
pearl pondBOT
@sick moon Has your question been resolved?
sick moon
I think it's tooo long to read. Sorry about that
Let $G$ be a finite grid graph (with edges connecting adjacent vertices in $\mathbb{Z}^2$), and define the \textbf{broadcast domination number} $\gamma_b(G)$ and the \textbf{radius} $\mathrm{rad}(G)$ in the usual way.
We define a \textbf{convex grid-graph} as one satisfying $\gamma_b(G) = \mathrm{rad}(G)$.
A \textbf{convex grid-graph decomposition} of $G$ is a disjoint partition $G = G_1 \cup \cdots \cup G_n$ such that each $G_i$ is convex, and
[
\min\left( \sum_{i=1}^n \gamma_b(G_i) \right) \le \gamma_b(G),
]
with $n$ minimized among all such decompositions.
I was able to prove that such a decomposition is \textbf{unique} when $G$ itself is convex.
However, when $G$ is \textbf{non-convex} (i.e., $\gamma_b(G) < \mathrm{rad}(G)$), I am unsure how to prove or disprove uniqueness.
So my question is:
\textbf{Given a non-convex grid graph $G$, is the minimal convex decomposition always unique?}
And if not, might there exist a counterexample?
jolly parrotBOT
{ :(
sterile jolt
hi
pearl pondBOT
@sick moon Has your question been resolved?
pearl pondBOT
@sick moon Has your question been resolved?
lapis elk
gee, i would like to help, but i am also super confused, confused as f 
dense lodge
I need to Solve A, B , C, X, Y
The only info that is given is that the centre of the circle is x40 y60
dense lodge
I know X and B and now im stuck
dusty flame
U found A?
dense lodge
no
dusty flame
It’s a right angle, ya?
dense lodge
What do you meanwith right angle
(im swedish)
dense lodge
yes
dusty flame
So A + B = 90 deg?
dusty flame
A is not the angle?
dusty flame
Oh there’re 2 triangles sharing it
dense lodge
ye
pearl pondBOT
@dense lodge Has your question been resolved?
dense lodge
<@&286206848099549185>
pearl pondBOT
@dense lodge Has your question been resolved?
pearl pondBOT
@dense lodge Has your question been resolved?
nocturne lily
The correct answer my professor gave us is at the top left of the page. I’m not getting anything close to that. So, can someone find my mistake? I tend to mask small mistakes that I don’t pick up on, so I’m afraid it’s that
nocturne lily
Ping me if you reply
dapper kraken
The correct answer my professor gave us is at the top left of the page. I’m not ...
heres the mistake
$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$
jolly parrotBOT
skissue.in.a.teacup
nocturne lily
thank you ! I'll fix that and see if I have any other issues
hmm i still didn't get what my professor got so i'm a bit confused
So close
why did i do 3x9=18
huh
pearl pondBOT
@nocturne lily Has your question been resolved?
nocturne lily
<@&286206848099549185>
alpine surge
<@&286206848099549185>
What do you want to know?
nocturne lily
What do you want to know?
If I’ve done anything wrong bc my answer is different from the one my professor gave
alpine surge
If I’ve done anything wrong bc my answer is different from the one my professor ...
There is no π in the value of h u found
It is 5/√3
nocturne lily
Ah the pi cancels out ?
alpine surge
Yes
nocturne lily
Tyyy
There’s another problem I need help with but I’ll just try to figure it out tmw in school
alpine surge
Mm
pearl pondBOT
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pearl pondBOT
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Available help channel!
steady quartz
pearl pondBOT
steady quartz
Hi. I am doing this past paper Q, it is GCSE level Question, and I think I've got the correct final answer, I did use a calculator but very rarely and is a practice paper at home past paper and this is non calculator Section A
pearl pondBOT
@steady quartz Has your question been resolved?
balmy crystal
steady quartz
How many marks have I likely got out of /16 marks for this section? it is Section A Non calculator paper.
pearl pondBOT
steady quartz
How many marks I likely got out of /16?
How can I improve?
what is a prediction of /16?
It says this
Total Marks: 15/16
Based on the above breakdown, you are likely to score 15 out of 16 for this non-calculator Section A paper.
The deductions primarily come from small mistakes in the Meal Deal and Division questions. You performed very well, with only minor errors. Well done!
Am I right ?
timber tartan
i've seen this paper like twice, what exam is this
pearl pondBOT
@steady quartz Has your question been resolved?
frosty finch
How many even numbers are there with 4 different digits arranged in ascending order from right to left (ones to thousands)?
hollow cobalt
How many even numbers are there with 4 different digits arranged in ascending or...
Any progress?
swift spindle
pearl pondBOT
fickle schooner
The volume of a sphere is not pi r^2
crimson birch
4/3pir^3
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robust kraken
Let's show that any totally ordered set $(E, \prec)$ is a lattice. Let $x$ and $y$ be in $E$, we construct $P \subset E$ such that $P=\left{ x , y \right}$. If $x \prec y$ then $x$ is a lower bound of $P$ and $y$ is an upper bound of $P$, we reason similarly for the case $y \prec x$. Thus, any two-element part of $E$ has an upper bound and a lower bound. is the reasoning right?
jolly parrotBOT
tm
tardy bay
if the task is to show that all two-element subsets have a lower and upper bound, then yes, this looks good
robust kraken
I want to show that every two-element part of a totally ordered set has an upper and lower bound so yeah
thx
.close
pearl pondBOT
pearl pondBOT
Available help channel!
proven kraken
Does someone know a website like https://artofproblemsolving.com/alcumus
where I could practice math online. specifically I want to focus on improving my ability to prove stuff.
wooden merlin
If you're just going to solve math problems, there are tons on the internet, and you could just borrow a book from the library.
proven kraken
If you're just going to solve math problems, there are tons on the internet, and...
but I want to practice on a gamified website
pearl pondBOT
@proven kraken Has your question been resolved?
stone imp
