then like equate it to 5 or something
#help-36
blissful meadow
Is it possible that 2p-q and 2p+q are integer factors of 25 if the only integer factor of 25 is 5?
mint forum
nooo
ohhhh
and p and q has to be the same
so like straight after that i'd say "therefore there are no positive integers"
or wtva
blissful meadow
and p and q has to be the same
Hmm not exactly. You would need 2p-q =5 and 2p+q = 5. You can check that this can't work unless q is zero.
And yeah you would explain why it's a contradiction and therefore conclude that no such positive integers exist.
mint forum
Hmm not exactly. You would need 2p-q =5 and 2p+q = 5. You can check that this ca...
trueee
Thank you so much azyr
Ok nicee im feeling more confident :) thank u
.close
final saddleBOT
final saddleBOT
Available help channel!
left trail
Since $T_n(\frac{1}{1+x})= \sum_{k=0}^n (-1)^k x^k$ then $T_{n+1}(\log(1+x)) = \sum_{k=0}^n (-1)^k\frac{x^{k+1}}{k+1}$ I see that if we say $j=k+1$ the sum is $\sum_{j=1}^{n+1} (-1)^{j-1} \frac{x^j}{j}$ which partial matches what we have but our upper obund and (-1) match. I don't see how we can alter the indicies of the sum anymore without changing $x^j/j$
final saddleBOT
Since $T_n(\frac{1}{1+x})= \sum_{k=0}^n (-1)^k x^k$ then $T_{n+1}(\log(1+x)) = \...
vital crag
Do you know how to simplify (-1)^2
left trail
in what sense? that is just 1
vital crag
And do you know what j - 1 + 2 equals?
left trail
j+1
vital crag
You didn't change your lower index in the sum to j
soft zealotBOT
BigBen
vital crag
What two things are you trying to match exactly
left trail
so in the first image we have a sum that is equal to the taylor polynomial of log(1+x)
vital crag
Of nth degree yea
left trail
so I applied the integration property to the nth degree taylor polynomial of 1/1+x
I integrated the sum but after changing bounds I don't see how some of the terms can match
vital crag
What terms exactly do you think not matches
vital crag
I integrated the sum but after changing bounds I don't see how some of the terms...
What order is the resulting Taylor series polynomial from the theorem
left trail
our sums are identical besides these two
left trail
What order is the resulting Taylor series polynomial from the theorem
n+1. but he somehow changes it to n
while keeping x^k/k the same
what am I missing?
vital crag
And do you know what j - 1 + 2 equals?
use this
vital crag
Do you know how to simplify (-1)^2
and this
vital crag
n+1. but he somehow changes it to n
then you shouldn't be comparing two things that shouldn't be the same and expect them to be the same
T_(n+1) should no be T_(n) in general
final saddleBOT
@left trail Has your question been resolved?
final saddleBOT
final saddleBOT
Available help channel!
final saddleBOT
robust palm
can someone explain the last part
final saddleBOT
Available help channel!
copper storm
“Rectangle with diagonals of length d. The diagonals form four angles, two of which are 30 degrees and the other two are 150 degrees. What is the area of the triangle?” The answer is d^2/4 . I get d^2/2. i dont know where i did my mistake. Can someone point it out or do it themselves but solve it by doing my method? Please and thank u!
final saddleBOT
“Rectangle with diagonals of length d. The diagonals form four angles, two of wh...
stable dew
final saddleBOT
@copper storm Has your question been resolved?
edgy mauve
“Rectangle with diagonals of length d. The diagonals form four angles, two of wh...
Here in case of area of the first triangle you determined ,
U forgot the 1/2 before xy
edgy mauve
So for the next one,same implies
final saddleBOT
final saddleBOT
Available help channel!
crimson zenith
can someone explain why we are saying |x|<|w| in the inductive hypothesis
crimson zenith
(definition of reversal)
for those who don't know ε is the empty string and • refers to concatenation
blissful meadow
can someone explain why we are saying |x|<|w| in the inductive hypothesis
They're doing induction on the length of w, so they're assuming that (xz)^r = z^r x^r whenever |x| < |w| and they're showing that (wz)^r = z^r w^r from that.
crimson zenith
They're doing induction on the length of w, so they're assuming that (xz)^r = z^...
what do you mean by 'on the length of w'?
blissful meadow
They kind of went out of order and proved the base case after but it doesn't matter.
blissful meadow
what do you mean by 'on the length of w'?
I mean that we're considering the "previous cases" of induction to be strings shorter than w
crimson zenith
I mean that we're considering the "previous cases" of induction to be strings sh...
but arent the 'next cases' strictly strings that are 1 longer than w?
blissful meadow
This is more akin to strong induction if you're familiar with that.
crimson zenith
yeah i am
blissful meadow
So this is what they're doing. They're assuming what they want to show holds for any string shorter than w and then concluding that is holds for w as well
crimson zenith
oh wait, the basis step is w itself?
blissful meadow
The base case is |w| = 0, i.e. w is the empty string.
final saddleBOT
final saddleBOT
Available help channel!
worldly spruce
final saddleBOT
Available help channel!
gritty flume
final saddleBOT
deep mica
yep thats right
gritty flume
Wha
wary juniper
Wha
cos(360+x)=cos(x)
That's periodic property
Cos(180-x)=-cos(x)
That's supplementary property
brazen breach
Canu explain it to me
imagine a semi circle, it has an angle of 180 degrees
suppose an angle x inside the semi circle, 180-x would just be x mirrored along the center of the semi-circle hence the cos value is negative
might lowk be more confusing
final saddleBOT
@gritty flume Has your question been resolved?
late gazelle
cosine repeats its value every 2pi radians or 360degrees, so $\cos(x)=\cos(360+x)=\cos(2(360)+x)=\cos(n(360)+x)$ for positive integer values for n. \ So in our case of $\cos(540-x)$ we can split up 540 into 360+180 so $\cos(540-x) = \cos(360+180-x)$ and we know the 360 part adds nothing since it will repeat the value of $\cos(180-x)$ we can work with $\cos(180-x)$. \ And if you wanted to you could expand this using $\cos(A+B) = \cos(A)\cos(B)-\sin(A)\sin(B)$ which gives us $\cos(180-x)=\cos(180)\cos(-x)-\sin(180)\sin(-x)$. \ this simplifies to $-\cos(x)-0(-\sin(x)) \implies \cos(180-x) = -\cos(x) \implies \cos(540-x)=-\cos(x)$
soft zealotBOT
KB
late gazelle
hopefully that helps
gritty flume
Its a lot
gritty flume
.close
final saddleBOT
final saddleBOT
Available help channel!
outer hemlock
answer is 0 , but im getting a weird ans
final saddleBOT
answer is 0 , but im getting a weird ans
outer hemlock
rare girder
look closely at the brackets, the original expression is cos(a huge argument)
first two lines in your work are missing closing brackets as well
outer hemlock
fine lol i got it
sturdy cypress
@rare girder fr chill
final saddleBOT
final saddleBOT
Available help channel!
coarse loom
hi i've been trying to find the intersection of two subspaces for two hours and i cant get the way to do it
any helps?
final saddleBOT
hi i've been trying to find the intersection of two subspaces for two hours and ...
tropic crest
hi i've been trying to find the intersection of two subspaces for two hours and ...
please post the full question and what have you done here
coarse loom
like these two subspaces how can i find the intersection between them
south dirge
like these two subspaces how can i find the intersection between them
.pin
final saddleBOT
like these two subspaces how can i find the intersection between them
@coarse loom Has your question been resolved?
earnest shuttle
i actually dont ujnderstand direction ratios and cosines at all like how does he get these answers
final saddleBOT
i actually dont ujnderstand direction ratios and cosines at all like how does he...
final saddleBOT
final saddleBOT
Available help channel!
grizzled socket
Demonstrate each trigonometric identity
b) (1 - sin x)^2 + cos^2 x = 2(1 - sin x)
icy current
Demonstrate each trigonometric identity
b) (1 - sin x)^2 + cos^2 x = 2(1 - sin...
!status
final saddleBOT
Demonstrate each trigonometric identity
b) (1 - sin x)^2 + cos^2 x = 2(1 - sin...
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
<@&268886789983436800>
grizzled socket
4
icy current
!showyourwork
final saddleBOT
Show your work, and if possible, explain where you are stuck.
grizzled socket
thats what im trying to do
icy current
Kk
icy current
!done
final saddleBOT
If you are done with this channel, please mark your problem as solved by typing .close
grizzled socket
.close
final saddleBOT
edgy mauve
<@&268886789983436800>
lyric obsidian
<@&268886789983436800>
final saddleBOT
Available help channel!
zenith timber
im trying to solve 10! + 8! and so far ive done 10*9*8!+8! = 10²*8! .
but for some reason its false??? can i understand why
final saddleBOT
im trying to solve 10! + 8! and so far ive done 10\*9*8!+8! = 10²\*8! .
but for ...
zenith timber
to add up isnt the formula supposed to be n + (n-1)! + (n-1)! = (n+1) + (n-1)!
trail mango
what does ‘solve’ mean here
zenith timber
what does ‘solve’ mean here
like the simplest anwser
trail mango
im trying to solve 10! + 8! and so far ive done 10\*9*8!+8! = 10²\*8! .
but for ...
why would that be true?
final saddleBOT
final saddleBOT
Available help channel!
final saddleBOT
fast kettle
<@&268886789983436800>
grizzled socket
i dont know how to demonstrate this trigonometric identity
final saddleBOT
i dont know how to demonstrate this trigonometric identity
grizzled socket
cos^2 x + tan^2 x - 1 = tan^2 x sin^2 x
1 - sin^2 x + sin^2 x/cos^2 x - 1
-sin ^2 x + sin^2 x/cos^2 x = tan^2 x sin^2 x
stuck here
blissful meadow
You ultimately want sin^2(x) and both terms on the left have sin^2(x) in common, so perhaps factor it out first.
blissful meadow
Not quite
You should still have two terms in the second factor after you factor sin^2(x) out.
grizzled socket
like that
blissful meadow
Yes,
blissful meadow
Can you write 1/cos^2(x) as something else?
grizzled socket
sec^2(x)
blissful meadow
Does rewriting -1 + 1/cos^2(x) with that yield some identity you know of?
grizzled socket
which identity
blissful meadow
Something like $\tan^2(x) + 1 = \sec^2(x)$.
soft zealotBOT
Azyrashacorki
blissful meadow
No. What do you have left in the second factor if you write 1/cos^2(x) as sec^2(x)?
loud sundial
cos^2 x + tan^2 x - 1 = tan^2 x sin^2 x
grizzled socket
-1 and sin^2(x)
blissful meadow
i replace 1/cos^2 x by 1 + tan^2(x)
I mean you could do that
blissful meadow
No. You had $\sin^2(x)\left(-1 + \frac{1}{\cos^2(x)}\right)$
soft zealotBOT
Azyrashacorki
blissful meadow
Yes
blissful meadow
Why is there an addition
grizzled socket
wasnt it -sin?
blissful meadow
Yes but you factored sin^2(x) out
grizzled socket
Why is there an addition
oh we factored it
blissful meadow
Leaving -1
grizzled socket
true yeah
grizzled socket
Why is there an addition
wait yeah we gotta remove it cause its multiplying the sin(x) with the parantheses right
blissful meadow
Yeah well you're just left with $\sin^2(x)(\tan^2(x) + 1 - 1) = \sin^2(x)(\tan^2(x))$
soft zealotBOT
Azyrashacorki
final saddleBOT
west moth
<@&268886789983436800> make a honey pot channel
final tangle
We can't afford honey
formal trail
if you have suggestions about server moderation dm @dusk parcel
i will say we have had this proposed many times before though
faint edge
Who says you aren't in the honeypot channel rn?
final saddleBOT
Available help channel!
west moth
would it not just have a mesage you can react to to get acces or do the bots get past that
final saddleBOT
would it not just have a mesage you can react to to get acces or do the bots get...
west moth
no way im pinned yay
lyric obsidian
uhh this isn't necessarily the purpose of a help channel lmao
but if you do have any concerns as they said before, bring it up to modmail
imma close this channel now...
.close
final saddleBOT
next thorn
would it not just have a mesage you can react to to get acces or do the bots get...
you claim the channel, if available, by simply sending a message
west moth
no i ment when you join the server
trail mango
<@&268886789983436800> make a honey pot channel
i am the honey pot channel
next thorn
no i ment when you join the server
you choose your roles when you first join ig?
west moth
thank you slayla
final saddleBOT
Available help channel!
minor rain
Anyone know the formula for this? I don’t want the answer cuz I wanna try it on my own 🙂
final saddleBOT
Anyone know the formula for this? I don’t want the answer cuz I wanna try it on ...
next thorn
Anyone know the formula for this? I don’t want the answer cuz I wanna try it on ...
I'm not familiar with this field but, if it will be helpful, you can think of the given information as (population grows / population shrinks)
for example, immigration is like the population shrinking.
marsh mountain
Anyone know the formula for this? I don’t want the answer cuz I wanna try it on ...
iirc, apes defines the growth rate as
$$\frac{\text{(births + imm)} - \text{(deaths + emi)}}{\text{total pop.}} \times 100$$
marsh mountain
np!
i havent tutored the class in a while
oh those should be in parenthesis as well lol
soft zealotBOT
blanketism
winter fox
Hey guys
do u know how is D right?
I spent way to long and somehow got the right answer
final tangle
do u know how is D right?
!occupied
final saddleBOT
do u know how is D right?
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
winter fox
ok thankyou
tranquil pine
final saddleBOT
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
cerulean epoch
<@&268886789983436800>
oak shore
How can i do this question easily and otheother queations like this?
oak shore
it is gcse standard question
timber leaf
It just itself + 0.2 of itself
oak shore
so i just do 350 x 0.2?
timber leaf
350+350*0.2
oak shore
it says 420
timber leaf
yea
oak shore
whewhen i see thrse sort of questions how do i remember ?
plucky rover
70% of the number is 350, how are you saying 120% is 420
plucky rover
Please forget everything min has said in this channel
plucky rover
Are you familiar with any algebra
oak shore
yes
timber leaf
Ah wait I forgot about that one 😂
plucky rover
You're given 70% of x is 350
oak shore
yes
oak shore
70x = 350
plucky rover
Then find 120% of x
oak shore
70/100 * 350
severe canyon
No...
timber leaf
you divide by 70/100
severe canyon
You're given 70% of x is 350
@oak shore
plucky rover
Brother write an equation for x first
Let's not skip steps
Especially when you don't understand what you're doing
tidal drum
How can i do this question easily and otheother queations like this?
if u kno 70% is 350 then divide to find 1% 350/70 = 5 then multiply by the precent u want 5 x 120 = 600
plucky rover
!nosols
final saddleBOT
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.
oak shore
if u kno 70% is 350 then divide to find 1% 350/70 = 5 then multiply by the precent u want 5 x 120 = 600
plucky rover
For the love of christ, there are three people helping. We don't need you to drop in with the full solution while we are trying to help OP understand
oak shore
i have the answer
if u kno 70% is 350 then divide to find 1% 350/70 = 5 then multiply by the precent u want 5 x 120 = 600
plucky rover
<@&268886789983436800> not sure what in the troll is happening here, one of you should have eyes on it regardless
tidal drum
For the love of christ, there are three people helping. We don't need you to dro...
he wanted a universal solution for a peroblem like this i jsut gave it to him
srry
tidal drum
he gave it to me
yes he is ok with it see so the next time he runs into a problem like tht he will just use the method i had given him
oak shore
can u see it
tidal drum
can u see it
see wht exactly
trail mango
tidal drum
russell newcombe
oak shore
im doing gcse maths
foundation tier
foundation tier
in england
as adult learner
whqwhatcawhqwhatcan i do to pass this
?
stray hill
@oak shore whats your question ?
oak shore
In recent mock exam i got 87/240 marks in a recent mock exam past paper
two papera
papers
severe canyon
@oak shore focus on your question, please
oak shore
it was paper one
anananand paper three
my teacher said i probably would of needed another 10 marks
what do u recommend i do?
this is my question
to pass
i have the real exmexaexmexam
exam papr one
paper
in 20 days
what do i do to pass?
severe canyon
Do
Not
Write
Like
This
?!?!
oak shore
Upcoming GCSE Exams
GCSE Maths
Paper 1: Thursday 14 May 2026 (AM)
17d 23h 48m 10s
Paper 2: Wednesday 3 June 2026 (AM)
Paper 3: Wednesday 10 June 2026 (AM)
severe canyon
And???
steep hatch
.close
final saddleBOT
final saddleBOT
Available help channel!
late gazelle
How do I go from here?
final saddleBOT
How do I go from here?
late gazelle
forgot to add the - between the two logs in final line of working
eager shore
Oh this integral
I'd not start with this approach in general
,w inverse of f(x) = (1 - x)/(1 + x)
soft zealotBOT
eager shore
right so if you multiply the inside of the log with e^(-x), and sub e^(-x), the resulting function is its own inverse
late gazelle
so like $\frac{e^x-1}{e^x+1} \frac{e^{-x}}{e^{-x}} = \frac{e^x-1}{e^x+1}$ ?
soft zealotBOT
KB
late gazelle
if multiplying by e^-x/e^-x does nothing then whats the point in doing it?
eager shore
It's just a manipulation that helps us get the integrand into a more workable form
late gazelle
It's just a manipulation that helps us get the integrand into a more workable fo...
mhm ok, so just multiply it through and dont simplify then
then sub what we get as u and go from there?
eager shore
Yeah
late gazelle
ok, lemme see
eager shore
I meant rewrite the argument as $\displaystyle \frac{1 - e^{-x}}{1 + e^{-x}}$ and then substitute it for $u$
soft zealotBOT
wilson
eager shore
Else you could first substitute e^-x for u and then (1 - u)/(1 + u) for v
late gazelle
**wilson**
ok yea i got that
so do we get $\int_{\frac{1}{2}}^{0} \ln{u} \frac{(1+e^{-x})^2}{2e^{-x}}du$
soft zealotBOT
KB
eager shore
late gazelle
oh i missed up didnt i
eager shore
I made a YouTube video about this integral let me go have a quick look
[ -\int e^x \ln \frac{1 - e^{-x}}{1 + e^{-x}} e^{-x} \dd{x} ]
soft zealotBOT
wilson
eager shore
excuse parentheses
late gazelle
?
eager shore
so after this you should have
soft zealotBOT
wilson
eager shore
,w differentiate (1 - x)/(1 + x)
late gazelle
thats a great question
eager shore
no worries
late gazelle
<@&268886789983436800>
eager shore
<@&268886789983436800>
eager shore
**wilson**
Right well do you get how we got to here?
late gazelle
**wilson**
although there is one thing, where does the - sign come from here?
eager shore
Oh there's meant to be a second - there to cancel it out mb
I multiplied and divided by -e^(-x) to set the differential up
It makes carrying substitutions out mentally very fast
eager shore
late gazelle
oh i c
eager shore
So over here -e^-x dx becomes du
late gazelle
yep
i did 2 e^-x
rather than 1 e^-x and 1 e^x
okay that looks better
then we set v = 1-u/1+u
eager shore
Did you do it properly?
late gazelle
i hope i did
late gazelle
so like b4 i did the v-sub
late gazelle
ah how do i rewrite u in terms of v
late gazelle
so when i get $\int_{0}^1 -\frac{1}{2} \frac{1}{u} \ln{v} (1+u)^2 dv$ is that correct (ish)
soft zealotBOT
KB
late gazelle
**wilson**
its just getting to this part i dont understand
eager shore
so u is interchangeable with v yea?
Yep, in the definition
[ u = \frac{1 - v}{1 + v} \Leftrightarrow v = \frac{1 - u}{1 + u} ]
late gazelle
so $-\frac{1}{2} \int_{0}^1 \frac{1}{v} \ln(v) (1+v)^2 dv$
eager shore
**wilson**
This is what I meant
late gazelle
v=1/u
eager shore
But this was your sub?
late gazelle
yes
late gazelle
**wilson**
mm, i still dont really understand but its just the one step I can go over later, so uh to go from here im also lost 😭
i dont think another sub would help would it
oh well we can simplify the 1+v and the 1/(1+v)^2
soft zealotBOT
wilson
late gazelle
yes
eager shore
So we can substitute 1/u = (1 + v)/(1 - v), ln (1 - u)(1 + u) = ln v, and du as well
eager shore
**wilson**
And you end up with this
late gazelle
mhm that makes more sense
you do sub a little different than me so i think that's why mine looks weird
late gazelle
im working through it
so just to clarify, we should simplify it to: $2\int_{0}^1 \frac{\ln v}{(1-v)(1+v)} dv$
soft zealotBOT
KB
late gazelle
since we can cancel a factor of 1+v from the numerator and denominator
eager shore
Yep
late gazelle
then expand it out to get 1-v^2 on the denominator?
eager shore
Yep that works
late gazelle
if only the ln(v) was like k, so we had k/1-k^2 and this would be easy
eager shore
Have you done competition/advanced integration before?
late gazelle
nope
eager shore
nope
Okay, so here the most straightforward thing you could do is use the Taylor series
eager shore
Are you familiar with $\displaystyle \frac 1{1 - z} = \sum_{k=0}^{\infty} z^k$ for $\abs{z} < 1$
soft zealotBOT
wilson
eager shore
bc lnv isnt defined at v=0
That's fine
We can write that with a limit if you want to be formal
late gazelle
**wilson**
z being a complex number a presume
late gazelle
mhm ok
soft zealotBOT
wilson
late gazelle
we could have something like $\int_{0}^1 \ln(v) \frac{1}{1-z}dv$ where $z=v^2$ thus we get $\int_{0}^1 \ln(v) \sum_{k=0}^{infty} z^k dv$
idk how to do sums in latex
eager shore
Yes
late gazelle
but hopefully that is readable
eager shore
\[ \int_0^1 \ln v \sum_{k=0}^{\infty} (v^2)^k \dd{v} \]
soft zealotBOT
wilson
eager shore
(hopefully that TeX is readable to you)
eager shore
**wilson**
Do you understand the math?
late gazelle
in what way
eager shore
**wilson**
Do you get why this is the same integral
late gazelle
yea
eager shore
Right
So now we can take the summation outside the integral
I'll also add the 2 back that we've been dropping
[ 2\sum_{k=0}^{\infty} \int_0^1 v^{2k} \ln v \dd{v} ]
soft zealotBOT
wilson
eager shore
All good?
late gazelle
so uh, why can we take the sum out the integral - ive seen this done many of times so it feels as if you can always do it but isnt there conditions that have to be satisfied for it?
i pretty much take it as if i have an integral and a sum i can always interchange them bc ive never had a case where i cant
eager shore
Well you can't always do it, you need absolute convergence
But for most integrals, you can sort of cheese it by knowing that you can do it
Because if you couldn't you wouldn't be able to solve the problem
late gazelle
that's funny
eager shore
**wilson**
Do you have an idea with how to proceed
late gazelle
ibp?
eager shore
Yep
late gazelle
we need so many variables and subs loll, let alpha = v^2k and dalpha = 2kv^2k-1 dv = ln(v), v = vlnv - v
nevermind that
thats terrible
eager shore
Choose to differentiate ln v and integrate the polynomial term
eager shore
Sure, whatever works, I don't use that terminology at all
late gazelle
or a and b
whatever works
ah so for integral of v^2k with respect to v, is it 1/2k+1 v^2k+1
eager shore
yep
eager shore
where did the v^2k+1 go
eager shore
Yep
late gazelle
this looks terrible
eager shore
That term outside the integral collapses to 0
late gazelle
so limits
late gazelle
so we get -2sum[integral thing]
soft zealotBOT
wilson
late gazelle
what happened to the v
eager shore
So now you're just left with a series
soft zealotBOT
wilson
late gazelle
oh so the integral just becomes 1/(2k+1)
eager shore
Do you know how to evaluate this?
late gazelle
we never learnt how to evaluate series...
but just add first few terms to get an idea for the answer
eager shore
Do you know the Basel problem
late gazelle
pi^2/6
eager shore
[ \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6} ]
soft zealotBOT
wilson
late gazelle
the 2k+1 doesnt really matter in this context does it
eager shore
wdym
late gazelle
as in, let W = 2k+1, so we get the form of -2sum[1/W^2] = -2 pi^2 / 6
so we get -pi^2/12
eager shore
Oh you can't do that
late gazelle
oh well
soft zealotBOT
wilson
eager shore
Any ideas?
late gazelle
subtract them
eager shore
Works
soft zealotBOT
wilson
eager shore
What next?
late gazelle
im still writing it 1sec
i should really try to figure this one out, bc you have pretty much walked me through each step of this
eager shore
Try writing it back in sigma notation on the left
eager shore
lol
late gazelle
well
eager shore
these are the odd numbers
soft zealotBOT
KB
soft zealotBOT
KB
late gazelle
and thats just basel problem multiplied by 1/4
eager shore
**wilson**
now sub it back
late gazelle
S=pi^2/6 - pi^2/24 = pi^2/8
late gazelle
so -pi^2/4
ehem
you saw nothing
i needa work on my integration holy
tysm for the help!
.close
final saddleBOT
final saddleBOT
Available help channel!
rotund stag
final saddleBOT
rotund stag
Again i know youre going to apply the perimeter of a sector formula, and get an equation probably that gives the two answers, but i cant figure out how to do it
rotund stag
Do you have the perimeter formula?
yeah
rotund stag
its going to be twice the radius, and the arc length
its gonna be like 2r + theta / 360 x 2πr
rotund stag
what is theta
60/360
rugged merlin
simplify that fraction
rotund stag
simplify that fraction
1/6
rotund stag
so what is your equation now
2r + 1/6 x 2πr
rugged merlin
what is 1/6 * 2
rotund stag
what is 1/6 * 2
1/3
rotund stag
2r + 1/3πr
rugged merlin
can you make this in the form of r(cpi + k)
rotund stag
c is 1/3 and k is 2
rugged merlin
and you are done!
final saddleBOT
@rotund stag Has your question been resolved?
rugged merlin
<@&268886789983436800>
harsh stratus
hey need help with this
final saddleBOT
hey need help with this
drowsy epoch
did you multiply x inside
harsh stratus
ye
onyx peak
can you show the full context?
drowsy epoch
Can you post the original
harsh stratus
okay one second
okay it's in swedish but
rotate y around the y-axis
determine the volume
drowsy epoch
Oh and so you are apply the shell method
harsh stratus
heck yeah
drowsy epoch
But one important note, you must not pull out x outside the integral
harsh stratus
oh okey
soft zealotBOT
harsh stratus
yes
drowsy epoch
One hint: add and subtract one in the numerator
harsh stratus
harsh stratus
oh right
drowsy epoch
x/(x²+1) and -1/(x²+1)
drowsy epoch
what is (x^2+1)'
drowsy epoch
I meant the derivative of x²+1 which is simply 2x
harsh stratus
yes
drowsy epoch
so notice you can make the numerator the derivative of the denominator
harsh stratus
ye
soft zealotBOT
drowsy epoch
Yeah you can also do substitution with u=x²+1
harsh stratus
drowsy epoch
yes
final saddleBOT
final saddleBOT
Available help channel!
tacit shuttle
help question 3 part 2 please(idk if that correct cuz i used translate)
tropic crest
help question 3 part 2 please(idk if that correct cuz i used translate)
pretty sure it is
if you need to you can post the original here for me to verify
tacit shuttle
tropic crest
yup looks fine
thin cloud
help question 3 part 2 please(idk if that correct cuz i used translate)
what have you tried
tacit shuttle
add some midpoint to make triangle midsegment
tacit shuttle
i tried to prove its parallel and base on the isosceles triangle to prove perpendicular bisector
my grammar kindda bad srry if it made u confused
soft zealotBOT
final saddleBOT
@tacit shuttle Has your question been resolved?
wary juniper
Chứng minh BKG cân tại K là đc
supple jolt
final saddleBOT
Available help channel!
void ridge
hello
void ridge
im new
woven ledge
you have a question?
void ridge
no
modest sequoia
im new
if you're looking to ask a math question feel free to ask in a help channel. if you are looking to just talk go over to #discussion or #chill
short berry
i am trying to reach math above what we get in 8th grade , help , what do i start with
modest sequoia
i just wanna get to know yall so that when i do i can ask it
you can go over to #discussion or #chill to just talk to people. these help channels are purely for math questions
woven ledge
i am trying to reach math above what we get in 8th grade , help , what do i star...
starting 9th grade
void ridge
what grade math do yall know
modest sequoia
theres all levels of math experts here
woven ledge
can you train fast calculations
you can talk in #math-discussion or #study-discussion
short berry
i guess man
woven ledge
or is it smth you get at birth
yeah you can train it defo
modest sequoia
thats done by practice. we arent here to "train you" also this is not your help channel so either create a help channel if you have a math question or go to #study-discussion
if you get stuck on a problem definitely feel free to make a channel to get help from someone, #❓how-to-get-help
final saddleBOT
@void ridge Has your question been resolved?
onyx peak
<@&268886789983436800>
polar spruce
<@&268886789983436800>
how the FUCK
final saddleBOT
polar spruce
he sent it like 2 seconds ago
onyx peak
he sent it like 2 seconds ago
im the best anti-scam bot in here
final saddleBOT
Available help channel!
worn cypress
How do i determine if word problem wants combination or permutation. I know the differnce is that combination is when the order dosen't matter and permutation is when it matters.
final saddleBOT
How do i determine if word problem wants combination or permutation. I know the ...
whole halo
see if switching the order around changes it to a different thing or it doesnt matter
if you need to hire 3 people, the order you hire them in doesnt matter
if you need to hire 3 people to three different places though, the order you hire them in does matter because a different person being in a different job counts as being a different thing
"5C3 to choose 3 people out of 5 to hire"
"5P3 to hire, out of 5 people, into these 3 different job positions"
worn cypress
So shoudl the first thing i see is if I swap the roles around will the whole thing change?
worn cypress
So the question state, A team needs to select two seniors to be co-captains. There are five qualified seniors, how many different choices are there.
Would it be combination because even if I switch the orders there will still be 2 senior captains
whole halo
yea
whole halo
np
worn cypress
.close
final saddleBOT
final saddleBOT
Available help channel!
topaz kite
<@&268886789983436800>
warm python
final saddleBOT
warm python
I'm interested in showing $\mathbb{Q}(\sqrt{2},\sqrt{3})\subseteq \mathbb{Q}(\sqrt{2}+\sqrt{3})$
soft zealotBOT
Wai
warm python
$(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}$, then $5-2\sqrt{6}$ is in the field too. So we have $\sqrt{6}$ is in the field
soft zealotBOT
Wai
soft zealotBOT
Wai
warm python
so $\sqrt{2},\sqrt{3}$ are too
soft zealotBOT
Wai
warm python
I don't get why I need to show the degrees of the extensions are the same?
glossy zephyr
Can we assume that 1,sqrt2 and sqrt3 are linearly independent?
warm python
Can we assume that 1,sqrt2 and sqrt3 are linearly independent?
they are over Q yea
oh, those are seperate questions
mbmb
thought those were hints 😭
glossy zephyr
It should also be kind of a hint
Try to think of the elements in those two as vectors in R^3
glossy zephyr
well, in reality it would be Q^3. you get the idea anyways.
glossy zephyr
If R^3 is the set of all real valued vectors, Q^3 is the set of all rational valued vectors.
warm python
√2,√3 aren's spanned by rationals
glossy zephyr
Yea, but you can think of them as directions
glossy zephyr
Q1 = Q[√2,√3] = {a+b√2+c√3 : a,b,c in Q}
Q2 = Q[√2+√3] = {a+b√2+b√3 : a,b in Q}
warm python
Q1 = Q[√2,√3] = {a+b√2+c√3 : a,b,c in Q}
Q2 = Q[√2+√3] = {a+b√2+b√3 : a,b in Q}
sure, but √6 is part of the field extension too
final saddleBOT
@warm python Has your question been resolved?
warm python
Q1 = Q[√2,√3] = {a+b√2+c√3 : a,b,c in Q}
Q2 = Q[√2+√3] = {a+b√2+b√3 : a,b in Q}
Mind if I close this now?
glossy zephyr
yea, go ahead
warm python
.close
final saddleBOT
final saddleBOT
Available help channel!
chilly current
Here is my question:
The question below is just as it is. There is no other information given. I can simply just graph it on Desmos, however, I won't understand the math behind it. Any help?
P.S. I prob won't reply for a long time, I'm going to bed.
final saddleBOT
Here is my question:
The question below is just as it is. There is no other info...
rocky tusk
Here is my question:
The question below is just as it is. There is no other info...
if f is increasing then f(x) > f(y) when x > y
so it suffices to show 3x + 1 > x - 5
sturdy cypress
yeah they mean square both sides
grave trout
Here is my question:
The question below is just as it is. There is no other info...
If you can't reply I think you shouldn't occupy a help channel. Maybe try a help forum instead. Also this channel probably will close due to timeout
rocky tusk
well
i mean i guess my proof just solves the inequality
but maybe they want you to do some algebra like frowny said
chilly current
wait is squaring both sides all i have to do..?
im pretty sure i have to find the domains i think...?
wait also sqrt(3x+1) always be greater than sqrt(x-5) since it rises faster and starts and (-1,0) and since sqrt(x-5) rises slower and starts and (5,0)?
oak bobcat
umm rethink that
sturdy cypress
yeah the domain thing is still important
chilly current
okay i wont be here for long so imma close this channel and use the help forum.
(sorry for the trouble im kinda new here)
.close
final saddleBOT
final saddleBOT
Available help channel!
topaz kite
<@&268886789983436800>
gritty flume
can someone help me, I’m doing trig functions and learning about the period but I haven’t learnt it in pi form before
final saddleBOT
can someone help me, I’m doing trig functions and learning about the period but ...
viral wave
can someone help me, I’m doing trig functions and learning about the period but ...
what is the issue mate
severe canyon
can someone help me, I’m doing trig functions and learning about the period but ...
π means 180°
Then you can find all the other angles, such as π/2 = 90°, 2π = 360° and so on and so forth
soft zealotBOT
drowsy epoch
For example if n=180° then θ=π
n is the angle in degree
If n=360° then θ=2π
It's just the formula for converting the units
gritty flume
Ic ic
Tysm everyone
but how do calculate the period
he’s doing this but idk wht this is
rugged merlin
but how do calculate the period
For Asin(Bx - C) your period is going to be 2pi/|B|
B = 1/2 in the pic u sent
gritty flume
do I change the pi to 180
rugged merlin
no
gritty flume
oh
rugged merlin
I mean thafs ok then
But u gotta learn
Radian is everywhere
You cant escape it
The formula becomes 360°/|B| in degrees
gritty flume
oki Ty
erm how do Ik one full cycle for sin
Is it when the graph touches the line again
The x line
rugged merlin
Ok
I feel like you are really better off watching some khan academy vids about trig functions
gritty flume
ok
final saddleBOT
@gritty flume Has your question been resolved?
orchid coral
Factorize 8a^6 + 5a^3 + 1
final saddleBOT
Factorize 8a^6 + 5a^3 + 1
orchid coral
Putting a^3 = x, the expression becomes 8x^2 + 5x + 1, which does not have a nice factorization.
radiant igloo
try making it into a^3+b^3+c^3-3abc
radiant igloo
Putting a^3 = x, the expression becomes 8x^2 + 5x + 1, which does not have a nic...
good first thought 👍 (even if it's wrong here)
gritty chasm
well if you observe 8a^6 and 1 are already perfect cubes; you can't tamper with them. try to rewrite the 5a^3 term.
orchid coral
E = (2a^2)^3 + (1)^3 + (-a)^3 - 3(2a^2)(1)(-a)
gritty chasm
$E = (2a^2)^3 + (1)^3 + (-a)^3 - 3(2a^2)(1)(-a)$
soft zealotBOT
Annie Maqionde
timber leaf
tiem
gritty chasm
Is there another way?
i dont feel so
gritty chasm
E = (2a^2)^3 + (1)^3 + (-a)^3 - 3(2a^2)(1)(-a)
well it looks right to me
<@&268886789983436800>
grave trout
Is there another way?
There is definitely a messy way taking a^3 as x as you mentioned but let's not go there
gritty chasm
@orchid coral has your question been resolved?
timber leaf
Why bro is messaging like a bot ;-;
radiant igloo
clankery
final saddleBOT
final saddleBOT
Available help channel!