Again
#help-28
neon basin
Exactly
It's also achieved by grouping if you don't see it
The top left and the one to the right of it can be grouped
And then their pair are the two right ones
Just differing in S_2
neon basin
And then the terms are disjunctively linked
Not conjunctively like you did in the beginning
plucky totem
Not conjunctively like you did in the beginning
where did I do that?
oh this one? Im meant to link with or?
neon basin
Left, Y = $\neg S_2 \land (\neg S_1 + F)$
glossy valveBOT
USS-Enterprise
neon basin
Yes
plucky totem
makes sense
neon basin
You are looking at terms where the output is 1. It can be either the top left input or the bottom left (or all the others with a 1)
That's why we have an or
plucky totem
👍
neon basin
I am guessing you have learned about
Disjunctive normal forms
and conjunctive normal forms?
plucky totem
I need to look into my notes everytime I hear those terms
but yes I have used it
learned about *
neon basin
Well this gives you the disjunctive normal form
plucky totem
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hallow walrus
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hallow walrus
So far I've proved that $u_n$ is a decreasing sequence of upper bounds of $S$, as $u_1$ is an upper bound of $S$, and for all $n\ge 1$, either $u_{n+1} = u_n$ or $u_{n+1}=\frac1{2}(u_n+s_n)\ge u_n$ is an upper bound of $S$. Either way $u_{n+1}\le u_n$.
Also, $s_n$ is an increasing sequence of elements of $S$, as either $s_{n+1}=s_n$ or $s_{n+1} = \frac1{2}(s_n+u_n) \ge s_n$. Either way $s_{n+1} \ge s_n$.
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kheer257
grave elm
So now u need that un is bounded to be able to claim that it converges (since by assumption monotone bounded sequences converge)
that should be pretty trivial
can u prove that its bounded?
hallow walrus
well yes
it's bounded by any bound of S
the second part is what I'm having trouble with
grave elm
showing that the limit alpha is an upper bound?
hallow walrus
It's pretty easy to get $s_n\le s_{n+1}\le u_{n+1}\le u_n$ and $\lim s_n=\alpha\le\beta=\lim u_n$
hallow walrus
showing that the limit alpha is an upper bound?
yes
glossy valveBOT
kheer257
grave elm
I'd go by contradiction probably. Suppose that there is some s in S > alpha.
grave elm
Right, it might be worth it to establish that un converges to alpha as well
which isnt too hard, since we know exactly how |sn - un| looks like (and can easily prove it converges to 0)
once we have that un converges to alpha, ii is pretty easy
yeah, that's probably a good way to do it
grave elm
If we look at the construction, un and sn are designed such that the gap between them shrinks by a half at every step, so the gap tends to 0 making un -> alpha
would u be able to write this out in full formality?
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hallow walrus
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✅ Original question: #help-28 message
hallow walrus
once we have that un converges to alpha, ii is pretty easy
it also makes iii really easy
grave elm
yep
torn jolt
@grave elm
grave elm
probably the best way to do this, knowing that both un and sn converge to the same thing makes it much easier
grave elm
<@951190447185874994>
ye?
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cold sluice
Can I get help with this? I'm not sure what the next step should be, this is my work so far.
buoyant ravine
first, that negative leading coefficient looks pretty gnarly, dont you think?
cold sluice
Well it's just a quadratic that points down
could i factor a GCF of negative 1?
Like this?
buoyant ravine
just multiply both sides by -1 
cold sluice
just multiply both sides by -1 <:happy:908558103992225822>
How can i get a perfect square though, if b and c are the same and a is different from them
isn't a perfect square either positive negative positive or all positive?
for coefficients of a b and c?
buoyant ravine
well, a and c must be the same sign for this to work
wicked crypt
I would factor out -1 so it becomes
-(x^2-7x-120)
Then simple trinomial factoring
= -(x-15)(x+8)
Roots are 15 and -8?
buoyant ravine
(also, congrats on blue!)
cold sluice
I would factor out -1 so it becomes
-(x^2-7x-120)
Then simple trinomial factori...
Omg i didn't know that it could be factored!
cold sluice
(also, congrats on blue!)
Ty ❤️
night lake
(also, congrats on blue!)
Nice Job bro
wicked crypt
Omg i didn't know that it could be factored!
Yea I was just trying combination on my calculator
cold sluice
i just assumed that there was no combination of multiplication that was the product of c and added to be 7
i thought it was too big
my bad
wicked crypt
What is blue
cold sluice
What is blue
active role
wicked crypt
Oh nice
cold sluice
Yea I was just trying combination on my calculator
someone told me that's not how you're supposed to do it (as in no brute forcing) but instead writing the factors of 120 and seeing which one if any get added to be 7 and stuff
and that's way better imo
but just looking at simple ones you can do it in your head
buoyant ravine
not exactly, i would get the leading coeff to 1 (not -1) first before you try whatever factoring method you were taught
in your solution, the constant term would be -(-15)(-8) which evals to -120, not +120
cold sluice
in your solution, the constant term would be -(-15)(-8) which evals to -120, not...
Yea i kinda skipped a step
I suppose that here the factor becomes -(x+15)(x+8)?
buoyant ravine
easy way to check: get the sign of your constant term
cold sluice
positive
negative times negative equals positive
but b is positive, so that isn't possible
cold sluice
what about in your factored form
Also positive
but the - is because a is -
right?
what if i use the grouping method? can i?
buoyant ravine
Also positive
well, not quite, you can kinda think of it as -((x+15)(x+8))
cold sluice
well, not quite, you can kinda think of it as -((x+15)(x+8))
yea that's what i see it as
buoyant ravine
so 15x8 is positive, take the negation, it becomes negative
which doesnt match up with your standard form
cold sluice
so 15x8 is positive, take the negation, it becomes negative
that's when we wanna solve for x
buoyant ravine
wdym?
cold sluice
wdym?
that's when you distribute the negative right?
otherwise it's just gonna be like this -((x+15)(x+8))
buoyant ravine
I suppose that here the factor becomes -(x+15)(x+8)?
how bout this: you multiply the negative over
(the rhs is still 0, remember)
and factor like you normally do
buoyant ravine
avoids mistakes
cold sluice
15-8=7
cold sluice
how bout this: you multiply the negative over
Wait what about -((x+15)(x-8))?
buoyant ravine
lemme check
cold sluice
Or how about (-(x-15)(x-8))?
cold sluice
hint: the sign of your x term is flipped
that makes sense
buoyant ravine
mmh.. a bit farther 
cold sluice
if one if the x's is negative, then when multiplied by a non-negative it will yield a negative
cold sluice
mmh.. a bit farther <:sadcatthumbsup:932362879624163338>
Hmm
buoyant ravine
Wait what about -((x+15)(x-8))?
try switching the signs of your 15 and 8
,w expand -((x-15)(x+8))
glossy valveBOT
buoyant ravine
you should get this c:
buoyant ravine
mhm! np
cold sluice
i suppose my school just expected the quadratic formula NGL
this doesn't seem super easy
buoyant ravine
you'll get used to it
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prime harness
Idk how to start this...
snow moon
Is it redundant to write like this
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snow moon
<@&286206848099549185>
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elder sandal
at what values of n does the sum $\sum_{i=1}^{n}i$ become a perfect square?
glossy valveBOT
4-aminopyrimidin-2(1H)-one
frigid carbon
Have you tried?
old cave
Do you know the closed form of $\sum_{i=1}^{n}i$
glossy valveBOT
flynger
old cave
How to simplify it basically
elder sandal
yeh (n)(n+1)/2
still epoch
you need to find the rule that defines the outcome of the sum
elder sandal
mostly stuck on finding values other than 1 and 8
for n
those two are probably the only numbers
still epoch
you have to write a rule for n, because the number of values for n is infinite
I'm pretty sure
it's not confined to a specific domain is it?
in the question
elder sandal
problem is just finding values of n(n+1)/2 which are square numbers
elder sandal
https://en.wikipedia.org/wiki/Square_triangular_number
oh
frigid carbon
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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
old cave
To get rid of the pesky 2 I would try going by cases
frigid carbon
and (n+1) also has to be square
Right
old cave
since we know n or n+1 is even
old cave
m(2m+1)=k^2
or
m(2m-1)=k^2
frigid carbon
How is it same?
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vernal ice
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old cave
$(x-3)^2=|x-3|^2$
glossy valveBOT
flynger
old cave
since squaring removes the sign
old cave
are sure its not
vernal ice
old cave
Are you sure its not x-4?
vernal ice
Cancels
old cave
so d^2=
old cave
Yup
vernal ice
then apply the first derivative test?
still epoch
cant u use vectors
old cave
They dont teach vectors in calc 1
old cave
Vectors should be taught in like 8-9th grade but American education system
old cave
$d^2=x^2-4x+19$
glossy valveBOT
flynger
vernal ice
yeah thats what I got
vernal ice
old cave
Do you remember
how finding extrema works
check the boundaries of the domain and the points where f' is 0 or undefined
old cave
yup your first candidate is 2
f' gives you that
the other thing you have to check is if your domain abruptly ends
so the boundary points
whats the domain of y
of your original function
whats the domain of $\sqrt{4x+3}$
glossy valveBOT
flynger
vernal ice
(-3/4, inf)
old cave
wait im being dumb
f has a wider domain than the original function
the minimum of f over (-inf, inf) is the minimum of [-3/4, inf) anyway
so actually you're good
vernal ice
and if it wasnt in the original functions rsnge then i messed up smth with the Distsnce formula?
old cave
No, then that would mean you would have to check the boundary
Because your domain cuts off
vernal ice
im a lil confused
old cave
yes since its a quadratic
and its defined everywhere
even when ur original function was not
So if your minimum over f is still in the original domain you're chilling because no other point can have a lower f anyway
vernal ice
wait so like this
vernal ice
right
old cave
So the simple answer is just do what u did above
vernal ice
Okay I got (2, sqrt11)
old cave
the complicated answer is since theres only one extrema for f you dont even need to check the boundary technically
every other point has a higher f because like you said its cts, diff
vernal ice
ohhhh so if the boundary of f was more constrained than the original function I would have to check it?
old cave
it would be weird
for that to happen
but technically
yes
also
lets say the minimum of f was at -100
then you need to use where f cuts off
as the extrema
that wouldve been -4/3
vernal ice
f'(-100)=0
yeah thats what i meant
vernal ice
if that was your only extrema
I see
old cave
or if you had multiple
I dont want to confuse you so the surefire method is
- Check the boundary points of your original domain
- Check all critical points given by the f you're minimizing/maximizing
vernal ice
Nahh im not confused I get it
old cave
It can be simpler but you would need to make sure you don't need to check the other points
in order to do less than that
vernal ice
mhm
Basically all the x values that get me f'(x)=0 i would need to plug back into f(x) (if I had multiple) to see which one would give me the lower value if multiple minimums were present
old cave
yup
vernal ice
but if I was represented with one x balue it would be fine
old cave
and the same for the boundary points
vernal ice
because its the boundary
old cave
you're right but just make sure that if you see a different problem the steps could be a little different
so checking all critical points is safer
vernal ice
Yeah im gonna do an ellipse one
for max
Thank you tho i appreciate it
Math is so firee
still epoch
u need help?
tepid zenith
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tepid zenith
i dont understand the first problem
burnt cedar
@tepid zenith don't add or ask new problems here, just create another help tab and ask your doubt there
forest trail
i dont understand the first problem
Hi! The first subspace just represents the subspace of all the vectors where the first two components are the same
So for example, (1, 1, 1) or (2, 2, 0)
So, what do you think could be a basis for this subspace?
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static verge
whys it at 2i the line
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whys it at 2i the line
fast peak
|z| is the distance from z to 0 and |z-4i| is the distance from z to 4i
so |z| >= |z-4i| means "z is further from 0 than from 4i"
the line through 2i is exactly the line with all z that have the same distance to 0 and 4i
so all points above the line are further from 0 than they are from 4i
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fallen ivy
1104
In a corner bounded by two brick walls, a rectangular chicken yard is enclosed with a 10-meter fence, and its area is y square meters.
(diagram labels: the vertical side is x, the top horizontal side is 10 - x)
What values can
a) x take?
b) y take?
devout valley
Well, what's y in terms of x?
fallen ivy
x(10-x)
devout valley
And do you know how you'd find the maximum value that would take on?
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twin wolf
And do you know how you'd find the maximum value that would take on?
catbittttttttttttttttt
meager cypress
And do you know how you'd find the maximum value that would take on?
5?
the maximum y could be is 25 ig
meager cypress
OP?
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warped frost
hey OP, you have crossed the bot two times without responding to the question asked of you.
please at least voice your confusion instead of silently crossing the bot hoping for someone to read your mind.
brittle sun
@fallen ivy either respond to the people trying to help you, or close the channel
fallen ivy
Wait
delicate torrent
Wait
@fallen ivy it's been almost an hour since this was sent
Please actually respond to the people who are helping you
Don't just ghost people 😭😭😭
This would be closed eventually because of timeout actually, so if you want to get help, actually respond to the helpers
warped frost
OP's been crossing the bot, so not that fast. but still, courtesy
fallen ivy
I’m busy.
warped frost
you can close and reopen later
fallen ivy
It’s fine
sacred yarrow
.solved OP's not here
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grave elm
||It's literally neither||
tender axle
bro💀
zenith pilot
I think it is a parabola.
grave elm
I think it is a parabola.
thats probably the best answer u can give
tender axle
I think it is a parabola.
yeah that’s what it should be
chrome canyon
yes
grave elm
though the path starts kinda flat and the shape on graph isnt a parabola
zenith pilot
Could it possibly be a parabola?
grave elm
Could it possibly be a parabola?
It's closest to parabola
tender axle
why does this question even exist
knotty grail
wait is it ellipse
yeah
zenith pilot
I think it is not a parabola since at x = 2, y = x^2 = 2^2 = 4 but at x = 1, y is not equal to 1^2 = 1.
knotty grail
baited
tender axle
there could be infinite functions which look similar to that
grave elm
it doesnt look very perfect to me
knotty grail
hmm
grave elm
but its prolly closer to it than to the parabola
and that path on the left is completely misleading
i wonder what genius created this question
and having "oval" in the options is also incredibly stupid
frigid carbon
Parabola
fervent agate
and having "oval" in the options is also incredibly stupid
Ovals are ellipses tho no?
zenith pilot
An oval is a closed figure.
frigid carbon
Ovals are ellipses tho no?
Not all
grave elm
"oval" is generally something that kinda looks like ellipse imo
knotty grail
and having "oval" in the options is also incredibly stupid
ok yeah some points seem a bit off
knotty grail
is this even a legitimate question
random totem
What grade are you in
tender axle
this looks like a prabola right guys?
random totem
@zenith pilot
knotty grail
maybe its just spiral man idk
zenith pilot
<@1426225845126561793>
10th
frigid carbon
Bro is there answer key
grave elm
ok yeah some points seem a bit off
and they couldve at least made it "part of an ellipse"
tender axle
wrong question. just ignore it
random totem
Not to actually scale the figure
tender axle
Then they just want you to tick parabola i guess
yeah exactly
grave elm
closest is ellipse imo
zenith pilot
wrong, its e^x+e^-x-2
Well, at x = 2, y is not, here, 4.
grave elm
they prolly want u to tick ellipse
zenith pilot
It is clearly 4 in the original graph.
tender axle
Well, at x = 2, y is not, here, 4.
yeah the grpah can always be adjusted, to match those values
fervent agate
There’s no numbers on the graph tho are there 
tender axle
my point was that you can make any function with a similar graph
it cant exactly always be y^2=4ax
knotty grail
it cant exactly always be y^2=4ax
three points are enough
tender axle
wrong, its e^x+e^-x-2
an equation using e^x somehow having a prabola type of shape proves it
frigid carbon
Conclusion its a curve
tender axle
three points are enough
yeah but where was it given in his original q
fervent agate
No way it’s coshx or sinhx tho right 
knotty grail
not an accurate parabola
tender axle
it just showed a graph with vertix at 0,0 and with arms up
grave elm
this is ellipse fit
knotty grail
might be it
zenith pilot
yeah the grpah can always be adjusted, to match those values
We have four options.
grave elm
much better than this
random totem
They said they're in 10th grade, idt a 10th grader is supposed to know about all this. The creator of the question wanted students to tick parabola
tender axle
they prolly want u to tick ellipse
ijust hate having to tick options based on what the examiner thinks instead of them just making a definitive question with a fixed answer
grave elm
||It's literally neither||
yeah, that was my first thought
tender axle
Conclusion its a curve
yeah thats all u can say atp
tender axle
They said they're in 10th grade, idt a 10th grader is supposed to know about all...
yeah agreed
zenith pilot
Assume that it is a parabola. Then we have the vertex at (0,0) with the parabola passing through (2,4). This implies that the equation is y = x^2. But at x = 1, the value of y is clearly not 1 = 1^2, so it cannot be a parabola, can it?
frigid carbon
Why does it matter
zenith pilot
What is the best answer disregarding the context of the question?
tender axle
so theres no use of wasting time on a question thats very clearly vague bro
frigid carbon
If answer is wrong it's the questions fault
tender axle
What is the best answer disregarding the context of the question?
parabola
tender axle
If answer is wrong it's the questions fault
yes
questions like these are just the fault of the question maker. any remotely important exam wouldnt keep a question like this
frigid carbon
Also the path looks different than graph
tender axle
and even if they do itd be bonus by the time marks are given
zenith pilot
ellipse.
I just justified why it cannot be a parabola and the graph of this ellipse looks much similar.
frigid carbon
I just justified why it cannot be a parabola and the graph of this ellipse looks...
Ellipse it is then
zenith pilot
Okay. It is obvious anyway.
knotty grail
I just justified why it cannot be a parabola and the graph of this ellipse looks...
but but
oval
anyways im not sure what theyre asking for
does this come under a specific chapter
zenith pilot
I think an oval is always closed.
knotty grail
so is an ellipse
zenith pilot
does this come under a specific chapter
Yeah, quadratic equations.
tender axle
Yeah, quadratic equations.
💀
knotty grail
i think examiner just whipped up a random curve and just hoped the students would put parabola
tender axle
bro just mark parabola and move on gng
zenith pilot
I know they want us to tick parabola.
tender axle
I know _they_ want us to tick parabola.
in that aspect parabola is def correct, but if u forget about that there are infinite curves that look exactly like this
which could chnage after the point the image in the graph cuts off
frigid carbon
End it
tender axle
yes
zenith pilot
Yes. Two options, oval and ellipse, are both definitely correct.
tender axle
🥀
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plucky totem
I dont realy understand why theres a Parity bit on the left side, isnt the one on the right side in charge of checking if the parity condition is met?
What purpose does the left P serve?
plucky totem
also I dont know what is meant by "odd parity"
tepid plover
I am not well versed here. But afaik "odd parity" means left P ensures that the number of 1s in input (including left P itself) is odd. Since there are 4 actual input bits, if P is odd parity, P should be 1 if there are even number of 1s in x's.
vestal ridge
CAN SOME 1 HELP ME AT MY CHANNEL PLS HAVE BEEN WAITING FOR 16 MIN NO ONE KNOWS THIS QUESTION?
frigid carbon
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@plucky totem Has your question been resolved?
queen crater
Can you show the original?
plucky totem
I am not well versed here. But afaik "odd parity" means left P ensures that the ...
I thought Parity on the right side checks if the amount of incoming 1s is even, wouldnt it be 5 if left P was 1 too?
or does P not get parsed through "Parity"?
plucky totem
Can you show the original?
its in german
queen crater
So?
queen crater
I'm asking for the original because you posted a poorly translated version
queen crater
Is there a question associated with that? Why does the box have a question mark?
tepid plover
or does P not get parsed through "Parity"?
I don't know what the "Parity" means there. What I know is that the "parity bit" is usually attached with the data bits for checking errors in transmission, so I thought the "parity bit" mentioned is the left "P".
plucky totem
Is there a question associated with that? Why does the box have a question mark?
here og and translated, I wanted to understand the text before going to the tasks
queen crater
here og and translated, I wanted to understand the text before going to the task...
Ok but that's important context
Namely, (b) kinda answers your question, doesn't it?
P is supposed to be the parity bit for X (X_0 through X_3), and your task is to (a) design a circuit to generate it (again), which goes into Parity on the right, and to (b) design a circuit to check whether P is valid (so, whether P and your answer for Parity match)
The check goes into OK
plucky totem
so in a) the parity bit is the Output
Im not sure what you mean by this [...] to check whether P is valid (so, whether P and your answer for Parity match)
queen crater
In (a), ignore P and OK, make Parity
In (b), take P, check if it's correct, make OK
plucky totem
how do I check if P is ok if it has no input?
queen crater
Wdym
queen crater
Whether P and what you get for Parity match
Let me rephrase
In (a), you take X and build a circuit that generates the corresponding parity bit and outputs to Parity
In (b), you take X and P and build a circuit that checks whether P is the parity bit for X, and output to OK, 1 if it is, 0 if not
plucky totem
ok I had to "ELI5 with an analogy" chatGPT ur message, but its clear now thx
can you read circuits and the logic gate symbols? I would show you what I come up with if so
queen crater
...sure
plucky totem
...sure
ok I replaced x_0 to x_3 with a-d
is that right for b)
oh wait
Im not supposed to check if P and parity are both 1
im just meant to check if they match
wait...
queen crater
(a) is fine but the output is Parity, not P
plucky totem
(a) is fine but the output is Parity, not P
oh... the task says the output should be the Parity bit which is P
this is my a)
I think my b) was wrong, I shoulda used XNOR to generate P not XOR
are either of these right you think?
in a) it should be a XNOR not XOR right before P*
queen crater
oh... the task says the output should be the Parity bit which is P
I don't get how you still don't understand that "Parity" and "OK" are the outputs and "P" is an input
plucky totem
the term "Parity bit" does it refer to P or to "Parity"?
queen crater
Both...
P is the parity bit given to you
Parity is the parity bit you generate
Anyway, if you take your diagram for (b) and link that middle XNOR with Parity instead of P, it looks correct
Though you probably want to draw Parity on the very right and P on the very left, and link both XNORs instead of using Parity as input
plucky totem
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verbal sky
hey
onyx glen
|5x-1| = -(5x-1) when 5x-1<0
verbal sky
|5x-1| = -(5x-1) when 5x-1<0
ah okay
so basically the two tails can be thought of as kinda different graphs
like this
onyx glen
in a way yes
verbal sky
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long helm
hi, i need help with this one.
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hi, i need help with this one.
long helm
Thanks in advance!
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long helm
or 3d in general
yess
i have
but the function for the half circle would be f(x) = sqrt(r^2-x^2)
right?
oak ruin
wdym half circle
long helm
but the function for the half circle would be f(x) = sqrt(r^2-x^2)
and this is going to rotate around the x-axis so that we get the volume
oak ruin
oh I see what you mean
long helm
mhm, but the half circle is greater under the x-axis so i dont think it would work
oak ruin
Yeah it'd be easiest to only consider the top half
long helm
or maybe it could actually
long helm
Yeah it'd be easiest to only consider the top half
oh okai
oak ruin
and we assume 80% of the upper hemisphere is in water
oak ruin
Well we're only in the upper half
long helm
what do you mean?
oak ruin
20% of the upper half is equivalent to 10% of the whole thing
oak ruin
(0.20)(0.50) = (0.10)
long helm
ok so 10% of the whole circle is above the water
oak ruin
yeah
is the same thing as saying 20% of the upper hemisphere is above the water
bc we know the entire bottom hemisphere is under water
ancient solar
can you use the formula for the volume of a spherical cap directly, or are you supposed to derive it using calculus?
long helm
can you use the formula for the volume of a spherical cap directly, or are you s...
hii
i think i am supposed to integrate it..
i can use integral in order to find the volume
however, i am not sure which function to integrate
i wanted to use this fomrula: y = sqrt(r^2-x^2)
but then from the picture i see that origo is not in the centre of the circle
ancient solar
There's no coordinate system in the figure above
ancient solar
I think a good start is to find a formula for a spherical cap, ie the top part of the sphere
Formula for the volume of a spherical cap I mean
ancient solar
Are you familiar with finding the volume of a shape using integration?
long helm
yess
the function of a half circle is y=sqrt(r^2-x^2)
V=pi*Integral of f^2 from -H to h
and then divide it by two
long helm
I think a good start is to find a formula for a spherical cap, ie the top part o...
to get this
ancient solar
Did you mean f(x) instead of y^2?
long helm
the function of a half circle is y=sqrt(r^2-x^2)
f(x) = sqrt(r^2-x^2)
long helm
Did you mean f(x) instead of y^2?
yes
ancient solar
Hmm, why from -H to h?
I'm thinking just derive the volume formula in full generality first, then apply it to the ice sphere later
Brb 5 min
ancient solar
So I would just let the top of the sphere be 0, and let the other end of the integral be z or whatever. Then you get a formula for the volume in terms of the radius r and the "height" z
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earnest nexus
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wary condor
m is a wire and on the left it was put in the center of c circular wire (lets call it a) once and and on the right it has a coil wire (lets call it k) wrapped around it
two questions, first, where is the north pole for k and c located in the two situations?
second, is the angle between the current direction of m and the B of k/c in both situations 180?
wary condor
like Fb = 0 in both situations
wary condor
well another helper did tell me that
but i got confused bc i never saw a north pole thats
outwards
brittle sun
I mean why not
brittle sun
The current and the magnetic field are always perpendicular to each other
Atleast I think so
wary condor
The current and the magnetic field are always perpendicular to each other
but theres a sin(theta) in the law 😔
Fb= B i l sin(theta)
brittle sun
Right that means I'm stupid
torn sand
m is a wire and on the left it was put in the center of c circular wire (lets ca...
what is c?
wary condor
what is c?
c is the name of the circular wire
k is the name of hte coil-shaped wire
oops!
ok i just changed names
its been a while since i've slept! excuse me
torn sand
m is a wire and on the left it was put in the center of c circular wire (lets ca...
Is that a current arrow in the left diagram? Or is it a label for c?
wary condor
no 😭 thats the current arrow
I got answers from the teacher! ty for reading
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noble citrus
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noble citrus
i got (sqrt(25/2),sqrt(25/2)) as the maximum but idk how to find the length and width of the rectangle
green kite
what is (sqrt(25/2),sqrt(25/2))
ivory pilot
i got (sqrt(25/2),sqrt(25/2)) as the maximum but idk how to find the length and ...
what are you referring to by those coordinates?
please do tell how far you've gotten
noble citrus
the area is a=xy
y=sqrt(25-x^2)
i substituted
i did the derivitave
then fdt
found the maximum
and thats the maximu m
ivory pilot
alright
@noble citrus so the point you've found is essentially the intersection between the semicircle, the puple line and the blue line
can you define what the y coordinate of a point actually represents on a graph
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dawn owl
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dawn owl
hey so ive found out that for c=2a there will be no sol at all as you can see:
but they want me a t1,t2,t3 and im not sure what it is
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pallid urchin
pallid urchin
<@&286206848099549185>
sudden chasm
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pallid urchin
without derivatives
manic sand
You factorise and you will get the answer.
sudden chasm
use identity \
$a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$ \
glossy valveBOT
T&C
pallid urchin
use identity \\
$a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})$ \\
thats the easy part
sudden chasm
also, $\sin x+\cos x =\sqrt{2}\sin \qty(x+\frac{\pi}{4})$
glossy valveBOT
T&C
manic sand
so i js replace?
sudden chasm
and, $\sin (x) \cos (x)=\frac{1}{2} \sin (2x)$
manic sand
What about this method?
glossy valveBOT
T&C
pallid urchin
yeah but what after
pallid urchin
What about this method?
ive already done that
manic sand
ive already done that
It’s solving quadratic.
frigid carbon
what if they are reciprocal
slate violet
so i js replace?
PLEASE DO NOT ASK TWICE
frigid carbon
atp just differentiate
pallid urchin
It’s solving quadratic.
cant js assume both are 1s
and if we did it would be and
not or
knotty grail
cant js assume both are 1s
hi
im lonely...
its been three days...
please come back to #help-43 soon... plea..
or ill just paste the suggestion here
pallid urchin
alr
knotty grail
$\sin^3(x) + \cos^3(x) = 1 = \sin^2(x) + \cos^2(x)$\
$\implies \sin^2(x) \cdot (\sin(x) - 1) + \cos^2(x) \cdot (\cos(x) - 1) = 0$
glossy valveBOT
knotty grail
there
knotty grail
ts leading to same fate 🥀
wym
pallid urchin
differentiation is better
Differentiation?
knotty grail
final line\
$\implies (1 - \cos(x))(1 + \cos(x))(\sin(x) - 1) + (1 - \sin(x))(1 + \sin(x))(\cos(x) - 1) = 0$
thats it
glossy valveBOT
pallid urchin
how does any of that conclude the 2 solutions
slate violet
how does any of that conclude the 2 solutions
you can take 1 - cos(x) as a common factor
knotty grail
how does any of that conclude the 2 solutions
just factor it
pallid urchin
sec
pallid urchin
yeah i am still not getting those results @knotty grail
@slate violet
why am i getting either cos x=-1 OR (sinx=1 or (cosx=-1 and sinx=-1))
knotty grail
wait why am i getting sinx + cosx =-2
easily provable to be impossible
knotty grail
why am i getting either cos x=-1 OR (sinx=1 or (cosx=-1 and sinx=-1))
where did you get cos x = -1 from
pallid urchin
kept factoring
knotty grail
can you show what your next step from here was
knotty grail
okay.. show the entire expression
youll get cos(x) - 1 = 0 as one solution
which means cos(x) = 1
pallid urchin
yeah but to think about it
when we have alot of factors that equal 0
even if just 1 is a solution it would still be right
in our original equation
sinx would still be unknown
idk if u get my logic or not
knotty grail
even if just 1 is a solution it would still be right
yeah because all of them are solutions
frigid carbon
yeah because all of them are solutions
except the -1 ones
knotty grail
sinx would still be unknown
i can see what you mean
were not trying to find a unique value of sin x
knotty grail
finding roots means finding all values which satisfy
we conclude that these values work
we dont actually have a value for x
knotty grail
like the entire one?
pallid urchin
here
knotty grail
$\sin^3(x) + \cos^3(x) = 1 = \sin^2(x) + \cos^2(x)$\
$\implies \sin^2(x) \cdot (\sin(x) - 1) + \cos^2(x) \cdot (\cos(x) - 1) = 0$\
$\implies (1 - \cos(x))(1 + \cos(x))(\sin(x) - 1) + (1 - \sin(x))(1 + \sin(x))(\cos(x) - 1) = 0$\
$\implies (1 - \cos(x))(1 + \cos(x))(\sin(x) - 1) + (\sin(x) - 1)(1 + \sin(x))(1-\cos(x)) = 0$\
$\implies (1 - \cos(x))(\sin(x) - 1)(2 + \sin(x) + \cos(x)) = 0$\
$\implies \cos(x) = 1$ or $\sin(x) = 1$ or $\sin(x) + \cos(x) = 2$ (the third case is not possible)
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pallid urchin
.reopen
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✅ Original question: #help-28 message
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cinder thorn
$| \cos^3(x) | \leq \cos^2(x)$ with equality iff $|\cos(x)|=1$ or $\cos^2(x) =0$
glossy valveBOT
bloubbloub
cinder thorn
Same with sin
Then we have equality in $$1=|\cos^3(x) +\sin^3(x)| \leq |\cos^3(x)| + |\sin^3(x)| \leq \cos^2(x) +\sin^2(x) =1$$
From that you can easily finish
glossy valveBOT
bloubbloub
torpid perch
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Renato
torpid perch
(a:b) = gcd(a,b)
keen vector
which one do u wanna start with
torpid perch
3
keen vector
hmm
keen vector
3 must divide one of the number a-1, a, a+1
keen vector
And I have a question on pie
not here
limpid forum
Can sm1 explain why pie is
keen vector
open a new channel
keen vector
this channel is occupied
keen vector
!help
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keen vector
3 must divide one of the number a-1, a, a+1
@torpid perch 3 doesn't divide a^2-1 = (a-1)*(a+1)
so 3 doesn't divide a-1 or a+1
so 3 divides a
now we need to show that 3*7 divides (a^2 + 3)*(a^2 + 5)
torpid perch
torpid perch
so 3 divides a
no spoilers
keen vector
well yeah you basically showed a = 0 (mod 3) using this
torpid perch
torpid perch
well yeah you basically showed a = 0 (mod 3) using this
yes
look at this one now
i adapted your idea
torpid perch
now we need to show that 3*7 divides (a^2 + 3)\*(a^2 + 5)
if 3^4 x 7 divides X then 3 x 7 divides X aswell? that's the rationale?
keen vector
now we need to show that 3*7 divides (a^2 + 3)\*(a^2 + 5)
actually this is sufficient but not necessary
im just thinking for any a and b is there an x such that x = a mod3 and x = b mod7
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torpid perch
**Renato**
can i get some help with the number 4
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thin flint
if you know gcd(a, b)=5 what do you know about a and b
thin flint
yes
torpid perch
thin flint
in an other form it would be?
thin flint
slightly different still?
torpid perch
in an other form it would be?
torpid perch
slightly different still?
a = 5k
b = 5q
thin flint
well if 5 is the greatest common divisor
then k and q have property that is important to the result
allows us to say something about the steps we perform
torpid perch
k is coprime to q
thin flint
exactly
torpid perch
a^4 + b^4 = (5k)^4 + (5q)^4
a^4 + b^4 = 5^4. k^4 + 5^4 . q^4
2ab^2 = 2(5k)(5q)^2
2ab^2 = 2(5k)(25q^2)
gcd(a^4+b^4, 2ab^2) = gcd(250kq^2, 5^4 . k^4 + 5^4 . q^4)
,calc 5^4/125
glossy valveBOT
Result:
5
gcd(a^4+b^4, 2ab^2) = gcd(250kq^2, 5^4 . k^4 + 5^4 . q^4) = gcd(2kq^2, k^4 + q^4)
@thin flint is best I can do, correct or no?
🥺 🤓
shit is getting nasty
thin flint
the equals isn't correct
torpid perch
last one?
thin flint
not quite
torpid perch
thin flint
gcd(na, nb) = n * gcd(a, b)
torpid perch
gcd(na, nb) = n * gcd(a, b)
that
mb
gcd(a^4+b^4, 2ab^2) = gcd(250kq^2, 5^4 . k^4 + 5^4 . q^4)
thin flint
now you have 2 options
you have the option that the two remaining terms are coprime
in which case the gcd is 5^3
torpid perch
can u help me simplify this
torpid perch
,w gcd(250, 5^4)
glossy valveBOT
torpid perch
or not coprime in which case it is 5^3 * m where m is the common factor
m = 125?
or what?
I might be misunderstanding
torpid perch
but ok, so if the gcd of m is made of comprimes then m = 1, if the things inside the gcd are not coprime what are thr possible outputs of m?
that's what u said
thin flint
i know one example for which m is not 1
if k = 5
and q is coprime
if that happens then you get
gcd(5^5 + 5q^4, 2×5×q^2)
which has a gcd of at least 5
gcd(5^5 + 5q^4, 2×5×q^2) = 5 * gcd(5^4 + q^4, 2×q^2)
same thing happens when q = 5
slightly different result
but gcd will still be 5 * gcd(...)
so you now know that gcd(a^4 + b^4, 2ab^2) = 5^3 gcd(5k^4+5q^4,2kq^2)
with at least 2 possible answers
one being that the result is 5^3 because the remaining gcd = 1
and the other is 5^4 because the remaining gcd = 5
uneven pike
Hi. Suppose that there is a prime p different than 5 and 2 that divides both a^4 + b^4 and 2ab².
So, we have : p divides 2ab², which means that p will have to divide either a or b, but not both because (a:b)=5. That means that p can't divide a^4 + b^4 which means that for a first step we can deduce that (a^4 + b^4 : 2ab²) if of the form 5^k
torpid perch
wait
uneven pike
Also, if we write a = 5l and b=5m we can see that 2ab² = 125 * 2 * lm² and a^4 + b^4 = 125 * 5 *( l^4 + m^4) meaning that 125 = 5^3 divides both quantities hence a further optimization will be that the gcd is of the form 5^(k+3).
torpid perch
I UNDERSTAND NOTHING AT ALL
uneven pike
Okay my english is frankly not good so bear with me lol
torpid perch
is not about the english is about the number bashing
like you guys are throwing equations at me and numbers and factorisation but
is overwhelming me
uneven pike
Okay I'm going to try to explain it to you slowly step by step
torpid perch
can you makle a drawing
torpid perch
please no AI
leaden ermine
torpid perch
the fuck is wrong with this people
hybrid yacht
please no AI
i know you guys have progressed well beyond the starting point
but start these with small test cases
i'll give a non-AI answer if need be
devout valley
(As others have mentioned, please don't use AI to answer people's questions, it can be unreliable and we can remove people from the server for doing so 
)
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delicate torrent
**Renato**
Which question are you at now?
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gleaming turtle
Hello, i have a test tomorrow and i need to remember if im getting no solutions/only 2 solutions/a range of solutions. I can remember the rest like how the parabola opens or if it touches the x axis once/twice or not at all
