help-39

[uneven radish]

just do the same for those

[scarlet trellis]

i could write a program to do this more effiecently just checking each combo of powers like a=[1,10] b=[1,7] and c=[1,5] and just making sure the final number is less than 1000

and also running that for both (2,3,5) and (2,3,7)

i guess i should've looked further into the <13

that was quite simple

thanks all

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[scarlet trellis]

My method of solving is like this: $\2^(x^2+2x)=4^4=2^2^4=2^6\x^2+2x=6$

[jolly parrot]

UCYT5040

My method of solving is like this: $\\2^(x^2+2x)=4^4=2^2^4=2^6\\x^2+2x=6$
```Compilation error:```! Double superscript.
l.49 ...lving is like this: $\\2^(x^2+2x)=4^4=2^2^
                                                  4=2^6\\x^2+2x=6$
I treat `x^1^2' essentially like `x^1{}^2'.

Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2023/texmf-var/fonts/map/pdftex/updmap/pdftex.map}{/usr/l
ocal/texlive/2023/texmf-dist/fonts/enc/dvips/lm/lm-ec.enc}{/usr/local/texlive/2
023/texmf-dist/fonts/enc/dvips/lm/lm-rm.enc}{/usr/local/texlive/2023/texmf-dist
/fonts/enc/dvips/lm/lm-mathit.enc}] (./539123083751981077.aux)
 ***********```

[shadow stag]

$(a^b)^c=a^{bc},$ not $a^{b+c}$

[jolly parrot]

lpieleanu

[scarlet trellis]

ahh true

sorry i cant figure out my tex

thanks for your help

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[signal peak]

guys i dont understand this

[spare lark]

In what basis

I suppose it is canonical

Just do A + u

Cuz it represent translation wrt u vector

Let A be the bakery

You want to go to algebra pizza

But for this ubermath is telling you have to make a way straight of u = 1 step from to right and 2 step to the top

To get your pizza standing at (A+u) coordinates

Got it ?

[signal peak]

yes

appreciate it!

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[spare lark]

Yw !

[pearl pond]

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[signal peak]

only the zero vector is parallel right?

[signal peak]

bro

instead of just reacting

how about you help out

[toxic lichen]

please do not call me "bro".

and please edit that word out or delete it.

anyway look at vector b

[signal peak]

how is vector b parallel

[toxic lichen]

b = -2u

[toxic lichen]

please delete this message btw

[signal peak]

ooooooh

appreciate it so much

.close

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[toxic lichen]

bruh

was it that hard to delete one (1) message

[signal peak]

sorry

what

which message

[toxic lichen]

the one where you call me "bro"

[signal peak]

oh...

[toxic lichen]

which i pointed to, twice

[signal peak]

why if i may ask politely?

[toxic lichen]

i don't like being called "bro" or variations thereof.

[signal peak]

reason?

[toxic lichen]

i am a girl.

[pearl pond]

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[flint wyvern]

am i right with D here?

[flint wyvern]

i used the sine rule

[nimble socket]

what is this

hw, quiz?

[flint wyvern]

some practice thing

<@&286206848099549185>

[limpid lily]

Yes, that's correct.

[flint wyvern]

THANK YOU SM

[limpid lily]

No problem.

[nimble socket]

!15m

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[flint wyvern]

.close

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[maiden aurora]

I’m relearning polar coordinates right now, would you use the second derivative the same way to determine for which values its positive? I’m not sure what notation to deliver it in (specifically needing help with part D)

[pearl pond]

@maiden aurora Has your question been resolved?

[pearl pond]

@maiden aurora Has your question been resolved?

[pearl pond]

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[boreal cedar]

hello, what's the technique of balancing a chemical equation?

[mystic bison]

stoiceometry?

[plain bough]

iirc you js mark one with a number, calculate others

[boreal cedar]

yep

[plain bough]

nvm then i dunno waht stoiceometry is

[boreal cedar]

how do I balance
2 reactant and 3 product

[plain bough]

is that chemistry orrr

[boreal cedar]

it is chemistry

[plain bough]

oh

nvm i think i can help

i havent covered in school but read in books

[boreal cedar]

I'm kind of having a hard time thinking of a number to multiply

[plain bough]

do you have a pic or the quesiton or are you js looking for

[mystic bison]

send the problem

i havent done it in like years

but i could prob figure it out

[boreal cedar]

I don't have any but can you provide using commands?

[mystic bison]

im not a command junkie i got no idea how to use latex

[boreal cedar]

.

[plain bough]

imma try

so set any to one

so lets js use photosynthesis reaction

for simplicty

H2O + CO2 => C6H12O6 + O2

[boreal cedar]

C-2
H-4
O-2

[plain bough]

oh ok then

[compact ridge]

step 1 is to look at which elements only appear once in both sides

so that would be... ?

[plain bough]

ngl id js set one of the reactants to 1 and solve for others

then get rid of denominators

[compact ridge]

there's a mathy way using simultaneous equations but I feel that would be confusing

[boreal cedar]

then product:
C-1
H-2
O-3? cs 2+1

[plain bough]

could i try?

[compact ridge]

what's your thought process

[compact ridge]

I guess so

[compact ridge]

there's a clear method that works with pretty much any chemical equation and it's this

[mystic bison]

from what i remember this is morel ike solving a puzzle

than actually doing a problem

maybe thats a horrible take

[compact ridge]

oh absolutely

the way chemistry students do this is definitely nowhere similar to maths

it's more like follow this process and iterate until you get the correct answer

[boreal cedar]

My chat is sending

[compact ridge]

which is what optimisation in maths is and so on (gradient descent etc)

[boreal cedar]

look

[mystic bison]

i thought this was the best part of chemistry like actual work wise besides the experiments

atleast once ur good at it

[compact ridge]

you still haven't told me your thought process

that's correct but

you're going to struggle with other problems like this I feel

[compact ridge]

cause you don't know the technique

[boreal cedar]

Yeah so what's the technique

[plain bough]

is this a technique 😭

[compact ridge]

this

[boreal cedar]

how do I put the answers on the blank?

[compact ridge]

which elements only appear once on both sides?

[compact ridge]

you should not think of these problems as just getting the answer

[plain bough]

ngl i dont understand what this is supposed to mean

[compact ridge]

or any problems in general

[compact ridge]

there's O2 on the left and O2, O on the right

so O would not be one of those

[boreal cedar]

same

[plain bough]

oh you mean the symbol

[compact ridge]

yeah

[boreal cedar]

I dont get the technique

[compact ridge]

so C appears only once on both sides (C2, C)
and O appears only once on both sides (O2, O)

so we balance those first!

[mystic bison]

id bet organic tutor has a really good vid explaining how to do this step by step

[compact ridge]

C2H4 and 2 CO2
O2 and 2 H2O

[compact ridge]

oh absolutely damnit I should have just sent the video

[plain bough]

he does

[compact ridge]

https://www.youtube.com/watch?v=e_C-V5vJv80

[mystic bison]

shi chemistry is his fortay

id hope itd be good

[plain bough]

but this is the simplest way i can think of

maybe itll help for smaller equation

[compact ridge]

then you see what's up with the oxygens, so you have $C_2 H_4 + O_2$ and $2 CO_2 + 2H_2 O$ so far

[jolly parrot]

south

[compact ridge]

only the oxygens are unbalanced

so then yeah there are 2 Os on the left and 2 * 2 + 2 * 1 = 6 on the right

so change $O_2 \mapsto 3O_2$, and that's now complete

[jolly parrot]

south

[boreal cedar]

ooohhhhh

okay okay

I get it xDD

thankuuuuu

.close

[pearl pond]

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[compact ridge]

yeah org chem tutor says the same thing

no worries!

[boreal cedar]

.reopen

[pearl pond]

✅

[boreal cedar]

guys

can you give me examples for practice

for

L>M

[compact ridge]

what's that?

[boreal cedar]

here

mole ratio came from the balanced equation

[compact ridge]

you want mole ratio questions where you need to balance the equation first?

[bright lichen]

I NEED HELP

[boreal cedar]

yes, to solve problem like this

[bright lichen]

i was in iss cause of 5 tartdys and i was out so like

[plain bough]

wrong channel

[bright lichen]

oh my fault

thx

[boreal cedar]

I need to practice so that I can memorize the patterns cs formulas are not allowed

[plain bough]

np

[bright lichen]

wheres geometry 😭

[plain bough]

#❓how-to-get-help

[compact ridge]

#geometry-and-trigonometry

[plain bough]

wdym formulas arent allowed

[compact ridge]

doesn't your textbook have problems?

or go search up GCSE chemistry textbooks

[boreal cedar]

We're having quiz

And I need to practice more

so I'm asking you to provide me a practice problems, so I can master it :DD

[compact ridge]

and I'm saying to go look in your textbook

[plain bough]

i havent gone to chem class yt, thats next year 😔

[plain bough]

but this looks kinda wrong for some reason

5.86 hcl??

[boreal cedar]

noo, our teacher solved it

its a given

[plain bough]

oh

[boreal cedar]

from the problem

[plain bough]

whats the problem

[bright lichen]

no ones helping 😭

[plain bough]

#❓how-to-get-help

[boreal cedar]

I forgot cs its on the presentation

[plain bough]

ok lemme try and see what i can do

[boreal cedar]

oki

[plain bough]

never have i learned this topic before soo

[boreal cedar]

js give me random probs xD

[plain bough]

cant

cuz never studied before

so how can i 😭

[boreal cedar]

oh its okayyyy

ill close it now, thank uu

[plain bough]

is M grams per mol?

js gotta ask a few questions rq

[boreal cedar]

mol only

[plain bough]

cuz i understand it a bit

ok

[boreal cedar]

thats all?

[plain bough]

what is

mol a divided by

in the first equation

[boreal cedar]

its always mol/1L (L solution)

[plain bough]

oh ok

[bright lichen]

@plain bough i already have that but the instructions are dif then what i see

[boreal cedar]

L Solution x mol A/1L x mol B/mol A

then once u get the answer

[plain bough]

m with ? is mols for zn solution?

[boreal cedar]

you'll proceed to M= nsolute/L soln=

[boreal cedar]

from the given i sent

mol A is 2.6 M

then over 1 L

[plain bough]

i dont think i can help with this 😭

im so sorry

[boreal cedar]

next fraction is mol B over mol A
so mol B is the mole ratio of ZnCl2, mol A is mole ratio of HCL which is 2

[plain bough]

the third eqation is ticking me off

[boreal cedar]

its kay xD

[plain bough]

why are they cacelling 1L

[boreal cedar]

cs theu both have L

[plain bough]

ooh

ok

thats what that is

[boreal cedar]

yess

can i close now? xD

[plain bough]

yh

srry to waster your time :(

[boreal cedar]

oke thank u

nooo

itw fine rlly

dw abt it

.close

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[buoyant garnet]

One message removed from a suspended account.

[buoyant garnet]

One message removed from a suspended account.

[buoyant garnet]

One message removed from a suspended account.

[pearl pond]

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[spark lodge]

The answer sheet says it's 9

But how?

[spiral pivot]

So let me see if I understand this correctly. Each card has 4 symbols. The 4 symbols are different from each other. And selecting any two different cards will give either 7 or 8 different symbols between them. However, it is possible for more than 2 cards to have any particular symbol. Yes?

[placid geyser]

i guess so

[spiral pivot]

Consider any 4 cards, if they all share a single symbol, then there are a total of 3*4 + 1 total symbols used. But this is 13 symbols. So only 3 cards can share a single symbol at most.

If we have 3 cards sharing a single symbol at most, then we have at most 3 * 12 = 36 total symbols on cards. There are 4 symbols per card. Therefore, the best we can do is 36/4 = 9 cards.

[pearl pond]

@spark lodge Has your question been resolved?

[spiral pivot]

@spark lodge ^

[spark lodge]

Hmm, yeah that makes sense

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[mellow idol]

Find all ordered pairs of primes (p,q) such that p²=3q²+3q+1

[mellow idol]

What I've got: q=3mod4

And ome ordered pair which is p=13 q=7

@cinder thistle what is bro cooking

[cinder thistle]

try bounding

||(p-1)(p+1) = 3q(q+1)
q divides p-1 or p+1
q<p<2p
=>
q<=p-1<2p-1
q|p-1
=>q = p-1
which is impossible
q+1<p+1<=2q
q|p+1
=>
p+1 = 2q
p = 2q-1
4q^2 - 4q +1 = 3q^2 + 3q + 1
q^2 - 7q = 0
q = 7 (or 0, but 0 isnt prime)||
do not open

[cinder thistle]

so, try proving that this is the only solution

[mellow idol]

Womt open

[mellow idol]

Bounding what

Gimme a hint G

[cinder thistle]

ok, so first, do you have any divisibilty relations?

[mellow idol]

Shat

Hat*

What*

[cinder thistle]

well
we are given
p^2-1 = 3q^2+3q

[mellow idol]

Mhm

[cinder thistle]

so
(p-1)(p+1) = 3q(q+1)

q is prime, so we know that q|p-1 or q|p+1

[mellow idol]

Yessir

[cinder thistle]

can you give some bounds for p interms of q?

[mellow idol]

First is not possible I think

[cinder thistle]

well, one bound is obvious
p>q

[mellow idol]

My bad I forgot to mention but the original question was find all ordered pairs of prime p,q st p²+q³ is a perfect cube, the equation i got was in a case of this question

[cinder thistle]

ah

oh, then you can open what i sent, i just solved that particular case

[mellow idol]

p²+q³=k³
p²=(k-q)(k²+kq+q²)
k=q+1 ( case im working with )
Not possible when k<p

q|p-1
q≤p-1
k≤p not possible

[cinder thistle]

how do you get q|p-1?

that is false

[mellow idol]

Thus

Thus

This

[cinder thistle]

so its possible q|p+1

[mellow idol]

Ye im checking that

Got it thanks

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[buoyant pasture]

( Z_1 ) and ( Z_2 ) are two complex numbers such that ( |Z_1| = |Z_2| + |Z_1 - Z_2| ) then

( Z )

(a) ( \text{Im}(z_1/Z_2) = 0 )

(b) ( \text{Re}(z_1/z_2) = 0 )

( \text{Re}(z_1/z_2) = \text{im}(z_1/z_2) )

(d) None of

[jolly parrot]

Andy

[buoyant pasture]

Actually I don't know how to start@frozen lantern

Where i start

[cinder thistle]

that means the distance between 0 and Z1 is the distance between O and Z2 + distance between Z2 and Z1 (O is the origin (0))

right

so what does that mean?

[pearl pond]

@buoyant pasture Has your question been resolved?

[quartz rampart]

Basically they both must have the same direction

[quartz rampart]

What happens when you divide to complex numbers having the same argument

[compact ridge]

yes more formally, $|z_2| + |z_1 - z_2| \ge |z_2 + (z_1 - z_2)| = |z_1|$

the equality case holds when the two vectors on the RHS are collinear

[jolly parrot]

south

[compact ridge]

look up a picture of the triangle ineq and you'll see

[pearl pond]

@buoyant pasture Has your question been resolved?

[quartz rampart]

@buoyant pasture what part isn’t clear to you?

[pearl pond]

@buoyant pasture Has your question been resolved?

[pearl pond]

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[buoyant pasture]

.reopen

[pearl pond]

✅

[buoyant pasture]

So what does it represent with respect to our question

[pearl pond]

@buoyant pasture Has your question been resolved?

[buoyant pasture]

@quartz rampart @compact ridge

[fair creek]

complex magnitudes are the same thing or at least analogous to side lengths

[buoyant pasture]

And our options are?

Then I don't understand the options

Real part of Z1 upon Z2

[fair creek]

what it means is that 1. imaginary part of Z_1 is zero, 2. real part of Z_1 is zero 3. Z_1 and Z_2 are parallel

the options are kinda misleading

hope that kinda helped

[buoyant pasture]

But z2 is also given

[buoyant pasture]

.close

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[viral wigeon]

i dont know how to do this i need this for math its basic proportions and im kinda slow please help

[frail temple]

do cross multiplication!

[viral wigeon]

can you elaborate i fell asleep in class today

[fringe robin]

multiply both sides by 2x+5, multiply both sides by 4

[viral wigeon]

.close

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[finite sigil]

you define x as different from zero that is the first step

[pearl pond]

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[pulsar stump]

How did they invert the matrix in LHS?

[nimble socket]

i will show you

first you calculate determinant

so 3(2)-(1)(-2)=6+2=8

[sharp vigil]

there is a simple formula for 2x2 inverses

[nimble socket]

also notice that $\textbf{M}^{-1}=\begin{pmatrix}a&b\c&d\end{pmatrix}^{-1}=\frac{1}{\det(\textbf{M})}\begin{pmatrix}d&-b\-c&a\end{pmatrix}$

formula's are not fun though

you might as well derive it

[pulsar stump]

Great, but was I supposed to know this? I'd there one for n>2?

[jolly parrot]

ooaa

[nimble socket]

yes yes

which i was going to show

[nimble socket]

so you set this aside

[sharp vigil]

it's very common, so you do just end up learning it

[nimble socket]

now what you want to do is calculate adj(M), or the adjoint of M

[sharp vigil]

the same formula involving the determinant and adjoint apply for larger matrices, but they become more annoying to calculate so usually you use an algorithm instead

[nimble socket]

you do this by multiplying by a cofactor matrix, which just has 2 rules:

1. take the determinant minor of each entry in the matrix and replace it with the entry
2. multiply by (-1)^{i+j} for an entry a_{ij}
   then transpose the resulting matrix

[pulsar stump]

Oh so it's just one of adjA's propeties? Ty

[nimble socket]

yes

[pulsar stump]

Great

.close

[pearl pond]

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[pearl pond]

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[scarlet glade]

I don't understand why this is

[robust kraken]

because $a \cdot b = ||a|| \cdot ||b|| cos(\theta)$

[jolly parrot]

tm

[warm current]

How are you defining dot product?

[fringe robin]

Definition of dot product on R^n

(Or derived from)

[scarlet glade]

I don't even remember

Hmmm

[warm current]

That would help

[scarlet glade]

It seems like it quantifies the difference in direction

But weird because the smallest difference is the largest output

[jovial kiln]

aka the angle between vectors, yes

these are two vectors

[jovial kiln]

the relation of the vectors is s.t it satisfies this equation

this is known as the dot product

[jovial kiln]

when you rearrange for the angle theta, you get this formula

[scarlet glade]

Why does it use cos?

Dies 2 vectors multiplied together give a vector that's an equal difference of direction to each vector?

[warm current]

"Multiplying vectors" is an odd thing to define. It's not really defined explicitly. You have dot product, which returns a scalar, and cross product (in 3d), which returns a vector

[scarlet glade]

I yeah...

The dot product isn't the magnitude of that... Huh

[scarlet glade]

I just can't wrap my head around that

I can't figure this part out:|a|*|b|*cos(x)

[pearl pond]

@scarlet glade Has your question been resolved?

[warm current]

Again, you should find the definition of dot product you were given. It's hard to explain without a starting point

[warm current]

Because this is often the definition of dot product

[scarlet glade]

I think i was looking for a definition like this

is that right? i think so

[scarlet glade]

That is not right

IMA try to figure this out on my own for a while

.close

[pearl pond]

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[scarlet glade]

ohhh

its the projection of a vector multiplied by the vector being projected onto

[pearl pond]

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[robust nexus]

the continuity part was fine
but for differentiability
we have to take limit h to 0+ and h to 0-

[robust nexus]

i dont get what we write tho

like f(0+h) - f(0)

would it be h^2sin(1/h) - 0 ?

should be right?

[merry carbon]

Don’t forget being over h

[robust nexus]

ye ye

i just dont get what we write in the numerator

or should it be 0 - 0 in the numerator?

[merry carbon]

What you wrote as is

[robust nexus]

h^2sin(1/h) - 0

[merry carbon]

You’re doing e.g. (f(0 + h) - f(0))/h and the limit as h -> 0 (which you needn’t separate out)

[robust nexus]

wait so we got f(h) - f(0) at the top

but h is approaching 0, so shouldnt we do f(0)-f(0)

[merry carbon]

You wouldn’t, no for one, that would get you a 0/0 if you wanna do it that way, which is not helpful

[merry carbon]

And the aim of writing this out is that you realise you’re taking the limit of h * sin(1/h) as h approaches 0, which should be recognisable

[merry carbon]

(That’s upon dividing by h, of course, as per here)

[robust nexus]

true ok thank you

.close

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[robust nexus]

.reopen

[pearl pond]

✅

[robust nexus]

wait also

what does it mean by f

f' not being continous everywhere

if something is differentiable, doesnt that mean f' is continious

@merry carbon

[merry carbon]

Not quite no: if f is differentiable, the original function, f is continuous, but that doesn’t mean the derivative, f’, is continuous itself

The aim is that you notice what the derivative, f’, is, and notice that at 0, there’s a discontinuity there

(At all points other than 0, you can differentiate the “x != 0” part “normally”, but the derivative at 0 is what needs careful consideration each time)

[robust nexus]

hmmm ok ill have a look

thank you

[pearl pond]

@robust nexus Has your question been resolved?

[pearl pond]

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[brisk garden]

Hi all, I'm working through my undergrad discrete math homework but I'm really stuck on this proof. Prove that there are infinitely many odd integers n ∈ N for which n, n + 2, n + 4, n + 6, . . . , n + 1000 are all composite

[brisk garden]

I'm not really sure how to start this proof but earlier in the hw there was a similar question proving there's infinite values of n that make n and n+100 composite so I'm thinking it's something similar. But I'm genuinely not sure how to do this proof

[pearl pond]

@brisk garden Has your question been resolved?

[brisk garden]

<@&286206848099549185>

[molten widget]

Sorry for the wait. Without spoiling the entire argument, I'll sketch the main idea.

The goal here is to show there are infinitely many odd integers ( n ) with the property that every number in the sequence ( n, n+2, n+4, \ldots, n+1000 ) is composite. Notice that's 501 numbers altogether, and so we'll index them with some (k = 0, 1, \ldots 500 ).

What you're going to do is select 501 distinct odd primes—call them ( p_{0}, p_{1}, \dots, p_{500} ) (for each ( k ). Obviously, infinitely many odd primes exist, so grabbing 501 of them isn't a problem.

Then we set up a system of congruences as follows. You pick an integer ( n ) that’s odd, so you have your first condition: ( n \equiv 1 \pmod{2} ). Then you arrange things neatly so that:

[
\begin{aligned}
n &\equiv 0 \pmod{p_{0}} \
n &\equiv -2 \pmod{p_{1}} \
n &\equiv -4 \pmod{p_{2}} \
&;;\vdots \
n &\equiv -1000 \pmod{p_{500}}
\end{aligned}
]

(This just means each prime divides exactly one of your sequence terms.)

Since all these moduli (2, ( p_{0}, p_{1}, \dots, p_{500} )) are pairwise coprime (no common factors, by design), CRT guarantees there's a solution ( n_{0} ) to this big set of congruences, and not just that, but it guarantees infinitely-many indexed by some positive integer parameter, which means there are infinitely many solutions (you shoulduse the explicit CRT construction to show this, likely).

I'll leave the rest for you to piece together.

[jolly parrot]

binglesnatch

[brisk garden]

Oooo I see. Thank you very much :)

.close

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[pearl pond]

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[proper flare]

Prove that (a+b, a-b) >= (a,b)

[proper flare]

Thanks.

Solved.

.close

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[toxic lichen]

wow ok doorslam

[pearl pond]

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[sharp smelt]

Trying to find the local extrema of this function

[frozen lantern]

[cinder thistle]

first hold y constabt and find possible extrema for x

[sharp smelt]

I know I can use calculus

[compact ridge]

AM-GM

[sharp smelt]

was wondering if AM_GM can help here

[frozen lantern]

it screams am gm

[toxic lichen]

if x and y are both positive then yes

[frozen lantern]

oh thats fair

[haughty lynx]

yess

[sharp smelt]

Yea, so the functions would be $xy$ and $(1/x) and 1/y)$

[jolly parrot]

What a wonderful world !

[sharp smelt]

$\frac{xy + 1/x+1/y}{3} \geq (xy \cdot 1/x \cdot 1/y)^{1/3}$

[jolly parrot]

What a wonderful world !

[sharp smelt]

so the minima when x,y \geq 0 is $3$

[jolly parrot]

What a wonderful world !
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[cinder thistle]

AMGM will only give you global extrema, (in positive reals) you have to prove that all extrema are global

[frozen lantern]

well

if we're being pedantic

oh nvm

sorry

i almost thought they said local extremum and i was like wellllll

[compact ridge]

they did tho

[frozen lantern]

no they said extrema rip

[sharp smelt]

I mean all global extremas are local as well

this just enabled me to find one of them

[compact ridge]

this function has no global extrema

[toxic lichen]

you... do need to just find the gradient and set it to zero

[cinder thistle]

when domain restricted to positive reals, it does

[toxic lichen]

yeah so it is still true that all global extrema are local

[cinder thistle]

(AMGM anyway restricts to positive relas)

[inland ivy]

Vacuous truth

😼

[toxic lichen]

yes hence the sotrue emoji

[sharp smelt]

okie

[autumn trellis]

Average helper help channel

[compact ridge]

all pigs that fly are smarter than Asteroid

[autumn trellis]

True....

[sharp smelt]

souh

[frozen lantern]

im smarter than all pigs that fly

[sharp smelt]

I start by finding points where $\grad{f}=0$

[jolly parrot]

What a wonderful world !

[sharp smelt]

$\left(y- \frac {1}{x^2}, x - \frac{1}{y^2} \right)$

[toxic lichen]

sqrt?

[sharp smelt]

my bad

[jolly parrot]

What a wonderful world !

[sharp smelt]

so we need y=1/x^2 and x=1/y^2

so x=y=1

[autumn trellis]

Your assessment seems accurate.

[sharp smelt]

I then have to evaluvate $f_{xx} f_{yy} - f^2_{xy}$ at this point

[jolly parrot]

What a wonderful world !

[sharp smelt]

wait, Ann. Why is that wrong

[toxic lichen]

x=y=±1

[sharp smelt]

ah, right

[inland ivy]

No, x and y are positive

[sharp smelt]

wait, no

[autumn trellis]

One must note, (-1)^4 =/= -1

[sharp smelt]

I also have $f_{xy} = 1 = f_{yx}$

[jolly parrot]

What a wonderful world !

[sharp smelt]

thus at (1,1)

$d<0$

[jolly parrot]

What a wonderful world !

[sharp smelt]

(1,1) is thus a saddle point

[cinder thistle]

not+-1 just 1

[autumn trellis]

Two people have already alerted the person in question. Thank you for your valuable third contribution

[toxic lichen]

errr

wait

oh yes my bad they do both have to be positive

wait what even happens to our function in the x<0, y<0 quadrant then lol

[cinder thistle]

its a minima? not a saddle?

[autumn trellis]

x=y=-1 isnt even a solution to the system?

[sharp smelt]

idts

[cinder thistle]

i am sure it is a minima, you can prove minima by amgm

[sharp smelt]

That's a minima in the first quadrant

,w xy+1/x+1/y minima

[jolly parrot]

Results provided by WolframAlpha

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Requested by math_rocks

[sharp smelt]

wait

what

oops

got it

thanks

.close

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[fading raft]

The cardboard in the shape of an isosceles triangle, green on the front side and blue on the back side, is placed on the x-axis from its base with corner B at the origin as shown in Figure 1.
The triangle in Figure 1 is first folded along the y=4 line as in Figure 2 and then folded again along the y=n line as in Figure 3.
Since the area of the triangle CGE' in Figure 3 is 4 square units, what is the distance between points D and G?

[fading raft]

I cant even move pencil💀

Help

[safe prairie]

first off n=2

[placid geyser]

Then height = 8

[safe prairie]

height of abc = 6 what

[placid geyser]

Abc

Triangle

[safe prairie]

why 8

[placid geyser]

Nvm, it's 6

[fading raft]

This?

[safe prairie]

yes

[fading raft]

Bro im still donkt know what to do💀

[placid geyser]

If it's 2, then the height of GCE' is 2, then it's area is 2 × ½ × E'C = 4

And I think all the triangles are similar

Are they?

[fading raft]

Hmmmm🤔

Not sure

[placid geyser]

Yep, they are

[safe prairie]

since the cuts are parallel and equally spaced, the bases of all 9 small triangles are the same

[placid geyser]

Then E'C = D'E' = DE = 4?

But it's not an option

[fading raft]

Guys

All the small ones are same

Right?

[safe prairie]

yes

[placid geyser]

Yes

[fading raft]

.

[safe prairie]

yes we said yes

[fading raft]

But what to do now hmmmmm🤔🤔🤔

[placid geyser]

Height = 2, area = 4

[fading raft]

Slow internet bro..

[placid geyser]

We get base = 4

Then answer should be 4

[safe prairie]

base of all small triangles are 4

[placid geyser]

But it's not an option

[safe prairie]

no

[placid geyser]

Oh, wait

Nvm

I thought they asked D and E

Not D and G

[placid geyser]

2√10?

[fading raft]

Guys?

[placid geyser]

Yes

You'll get 2√10

Option B

[fading raft]

How

Oh

H.h=4+36

Yep

Gg

.close

[pearl pond]

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[pearl pond]

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[bitter flame]

A two-digit number a in base 11 is one-third of the number formed by reversing its digits when considered in base 19. How many such numbers are possible?

[bitter flame]

I got the equation 2x=y

after simplifying

so im getting the answer 4

but it's 5 and u count x=5, y=10 pair too

but why? y = 10 cant be considered a single digit number where it normally has to be right?

is this something regarding the bases? I don't quite understand

[fringe robin]

consider the first digit x and second digit y

the first number is 3(11x+y)=19y+x

16y=32x

ok

[fringe robin]

because y is a less than 11

[quartz rampart]

10 is a single digit in base 11

[fringe robin]

want a pancake?

[quartz rampart]

No I’m busy chasing butterflies

[fringe robin]

I'll send a skarmy your way

[quartz rampart]

Basically you can use a symbol like A to represent 10 in base 11

[pearl pond]

@bitter flame Has your question been resolved?

[bitter flame]

oh, ok ty

.close

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[fading raft]

A polynomial P(x) of degree 4 with real coefficients P(x) >=x for every real number x
It ensures inequality.
Accordingly, P(4)=?

[pure rapids]

Intresting question

[fading raft]

Guys do i need to draw graph for solving this?

[pure rapids]

Perhaps

Idk if it helps

[fading raft]

Hmmmm🤔🤔

[midnight haven]

And we know that p(1) = 1

p(2) = 4

p(3) = 4

Using this we can probably find the coefficients of P(x)

[pure rapids]

It inflects at 0 and does not go to negatives

[fading raft]

Hmm🤔

But how

[vital crescent]

P(x)\geq x for all real number? are you sure its not for all positive real number?

[pure rapids]

3*

[midnight haven]

yup sorry my bad

[fading raft]

Yes bro it says for every

[pure rapids]

Make simultaneous equations

[vital crescent]

oh wait nvm

[midnight haven]

$$P(x) = ax^4 + bx^3 + cx^2 + dx + e$$

[pure rapids]

Let fx be ax^4 + bx^3 + cx^2 + dx + e

[jolly parrot]

Edmund Cloudsley

[midnight haven]

I noticed :)

[pure rapids]

Sry

[midnight haven]

no problem mate thanks for pointing it out nevertheless

[pure rapids]

Wait what does 'it ensures inequality' mean

[fading raft]

Should we draw a tangent because it could be equal?

[midnight haven]

That wouldn't make much sense

[pure rapids]

How would you proceed

[fading raft]

1 min

[midnight haven]

we can use the fact that $P(x) \geq x$ for all $x \in R$

[jolly parrot]

Edmund Cloudsley

[pure rapids]

Hmm how

[vital crescent]

yes, because P(3)=3, and P(x)\geq x, it must be the case that P'(3)=dx/dx=1

[fading raft]

Ah

True

Hmmmmmm🤔🤔🤔🤔

[vital crescent]

so you have 4 equations

[vital crescent]

and hopefully this gives you a valid a,b,c,d

[pure rapids]

Theres an e

The constant term as well

[vital crescent]

Oh, then do the same for P(1)=1

[midnight haven]

Perhaps P(1) = 1

yeah

[pure rapids]

@fading raft welcome to mathcord btw

[vital crescent]

5 equations for 5 variables

[pure rapids]

Ah mb yes

[fading raft]

Ty

[midnight haven]

solving for five unknowns gonna take some time :)

[pure rapids]

Yea

[fading raft]

Guys sory for badbdrawing

Can we go from this way?

[vital crescent]

hopefully they know some linear algebra stuff for this

[midnight haven]

exactly!

gonna be a nightmare without it

unless ofc there is a faster method out there

to do this question

[pure rapids]

...

I would be cooked rip

Well theres the wonderful ,w

,w general equation for 5x5 matrix determinant

[jolly parrot]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by craze.cv

[fading raft]

Looks like this is nonse cuz no one replied xd😓

[pure rapids]

Nvm not wonderful

[vital crescent]

the basics are not that hard, its just a systematic way of solving linear equations

[pure rapids]

Ah ok

[pure rapids]

Looks ok

[vital crescent]

it looks fine

[fading raft]

Hm hmm

[vital crescent]

combined it with this

and solve for

$$\begin{cases} P(1)=1\ P(2)=4\ P(3)=3\ P'(1)=1\ P'(3)=1\end{cases}$$

[jolly parrot]

qwertytrewq

[fading raft]

Guys can we say f(3)=3

[vital crescent]

what is f?

[fading raft]

And let's take these as tangents

[vital crescent]

take what as tangents?

[fading raft]

İ just said p(x)=f(x)

can we do this?

Or this nonse

[vital crescent]

f(3)=3 is given in the question so yeah

[fading raft]

İ think we made some progress

[vital crescent]

Suppose $f(x)=ax^4+bx^3+cx^2+dx^1+e$, and plug in $x=1,x=2,x=3$

[jolly parrot]

qwertytrewq

[fading raft]

Hmmm🤔

[vital crescent]

since f(x) is degree 4, it must be of the form $ax^4+bx^3+cx^2+dx^1+e$

[jolly parrot]

qwertytrewq

[vital crescent]

and since $f(1)=1$ we have $a+b+c+d+e=1$

[jolly parrot]

qwertytrewq

[fading raft]

Makes sense but🤔🤔🤔

[vital crescent]

yeah you essentially just solve this system of equations

[fading raft]

Hmmmmmm🤔

[vital crescent]

ngl its probably annoying but

[fading raft]

Bro this problems are annoying

No idea who wrote thia

[pure rapids]

Fr

Which contest

[fading raft]

And why

[fading raft]

You mean grade?

This 10th

[pure rapids]

Ah yes

I do too much mo

[pure rapids]

Damn

Which country (if ok to share)

[fading raft]

turkiye

[pure rapids]

Seems hard af for tenth

[iron basin]

turkey

[pure rapids]

Ah

[vital crescent]

yeah, they should make nicer problem

[fading raft]

İ think i solved

İdk if its luck or smth

[iron basin]

were you taught differentiation?

[quartz rampart]

Let px-x=Fx

[fading raft]

Guys is this true? Or nonsense

[iron basin]

smarttttt

[fading raft]

🤔makes sense

[quartz rampart]

If you do this you can find the function

[fading raft]

What u find answer?

İ find 22

[iron basin]

i think you did exactly what mathanos proposed

[quartz rampart]

1 min I just saw the Qs rn

[fading raft]

Oh

İn 1 mi n he solved

Smart dude

[iron basin]

although the method could've been a bit more thoughtful

it's correct

[fading raft]

There should be shortest way ig

[quartz rampart]

I got 28

[fading raft]

Bro u sure maybe some miss calculating its not in the answers

[quartz rampart]

Oh lemme check

[iron basin]

yes

[fading raft]

@pure rapids

[iron basin]

i think it's 22

[fading raft]

Same way or u did somethign different

[pure rapids]

Hm

[iron basin]

same way

[vital crescent]

looks correct

[pure rapids]

Yes

Should be ok

[fading raft]

İf yiu guya know any shortee waycan you please write

xd

[iron basin]

this is quite short

but i don't know

[fading raft]

İdk bro.. xd

[quartz rampart]

Oh Shi I read p(2)=2

Mb

[fading raft]

Yea answer is 22 ichecked

Gg

.close

[pearl pond]

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[sharp smelt]

Is there any simple way to differentiate this

'cause the gradient 'aint gong to be pretty

[toxic lichen]

it is not that bad? just chain rule it

the -(x^2-1)^2 will be killed when you find f_y

[sharp smelt]

hmm

okay

$\grad{f} = ((-2(x^2-1)(2x) -2(x^2y-x-1))(2x-1), -2(x^2y-x-1)(2x))$

[jolly parrot]

What a wonderful world !

[toxic lichen]

f_y is not that

f_y = 2x^2(x^2y - x - 1)

[sharp smelt]

wait

what

[toxic lichen]

also better to write it as f_x = ... and f_y = ...

in separate lines

[sharp smelt]

yeah

fair

[toxic lichen]

so as not to have to carry 500 kilograms on your back

[sharp smelt]

$f_x = -2x(x^2-1) - 2(2xy-1)(x^2y-x-1)$
\
$f_y = -2x^2(x^2y-x-1)$

[jolly parrot]

What a wonderful world !

[toxic lichen]

ok yeah that's more like it.

from f_y=0 you get that either x=0 or y = (x+1)/x^2

really just algebruh from there

[sharp smelt]

hmm, okay

showing they have 2 critical points isn't the boring part

having to find $f_{xx}, f_{yy}, f_{xy}$ is

[jolly parrot]

What a wonderful world !

[sharp smelt]

fair though

thanks

.close

[pearl pond]

Channel closed

Closed by @sharp smelt

Use .reopen if this was a mistake.

[pearl pond]

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[midnight haven]

Yoo

I got no idea what this is

This is for math

I guess something about Two Sample 1 Sample

[full agate]

I mean You gotta know what decision is

Idk what it is

[midnight haven]

Here man

I'm guessing it's a One tailed hypothesis

Because it says there "employees who have been employed for 5 years or less"

I guess There is comparison

[full agate]

Is this decision thing?

[midnight haven]

Well

That's a 2 tailed hypothesis

I sorta learned it when like there's only 1 reject null hypothesis

and not 2

and also there is 1 z table

or z number

if that's aight with you

[full agate]

You gotta use some formulas ig, play with number of samples and this salary

[midnight haven]

Well that's the thing

I don't know what kind of question is that

<@&286206848099549185>

<@&286206848099549185>

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

<@&286206848099549185>

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[high mesa]

if anyone can help out with trig identities/symmetry

[high mesa]

please advise

thanks

[toxic lichen]

!noclopen

[pearl pond]

Please don't repeatedly close and claim a new channel with the exact same question. This erases all previous progress made towards your problem and is confusing for helpers, making it more difficult to help you. Please be patient, even if your channel has not received much attention.

[high mesa]

yea cause there were people in the other chat that was clearly not trying to help

was just becoming a distraction

i really need to learn this shit

[hazy pilot]

what do you not understand?

[high mesa]

so once again, if you anyone is willing to help, please chime in

thanks

how these symmetries and identities work

[plush bramble]

i mean you made it harder than it needed to

[high mesa]

that's how the website asks you to login

i dont think people here even know what canva is

it's not some random website

a quick google search of canva will quickly show that this is not some obscure website

anyways

we're getting off track once again....as usual as it seems in here

people love to join these help rooms and discuss anything other than helping

this is server is probably 90% highschool students based on #discussion

[toxic lichen]

ok

for the record

this is what i was going to show you

and explain roughly as i typed just now

[shrewd mist]

Hard to figure out what you're specifically stick with from the lack of context but cosines are symmetric along the horizontal axis and sines are symmetric about the vertical axis. They are the x and y values of the input angle respectively.

[toxic lichen]

he is, as far as i can understand, stuck understanding 3 ideas stacked on top of each other in one rather messy diagram.

i tried to separate them out, and to my slight chagrin OP could not even decide which one he wanted to look at first despite me offering him the choice explicitly. (but in fairness, there was a buridan's-ass-proof solution to that conundrum.)

anyway yeah @high mesa if you do in fact still want to look at what i had to say, here it is.

[high mesa]

@toxic lichen yea im still interested

im still here

trying to work it out on my own until somebody chimes in

with some advice

[toxic lichen]

i mean, my advice was and still is to look at the 3 (or maybe 4) symmetries in the pic in separation from one another.

and i've separated them out for you.

[high mesa]

ok thanks

[pearl pond]

@high mesa Has your question been resolved?

[pearl pond]

@high mesa Has your question been resolved?

[high mesa]

.close

[pearl pond]

Channel closed

Closed by @high mesa

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

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Remember:
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[fast pilot]

is there anything wrong with this qu

[pearl pond]

@fast pilot Has your question been resolved?

[fast pilot]

<@&286206848099549185>

[placid geyser]

Which one are you stuck on?

[fast pilot]

a)

i think the question may be wrong

[placid geyser]

No, it is correct

[pine temple]

The component displacement is correct

[fast pilot]

wait so my DA line is right yeah?

@placid geyser

[placid geyser]

??

What does it mean

[fast pilot]

(3,3) + lambda(1,1)

@placid geyser

[placid geyser]

Oh, u mean that

Yes

[pine temple]

Remind me what that second form is

This

[fast pilot]

ok so when i substitute a value for lambda, i should get the E position vector right?

[placid geyser]

Yes

Wait, no

It's wrong

It shouldn't be (3,3)

It should be either the vector A or D

So, (1,11) + lambda (1,1)

[fast pilot]

ohhhhhhhhh

[placid geyser]

Then, lambda = -3 gives E

[fast pilot]

wait

[fast pilot]

whered u get the z coord from?

the 1

[placid geyser]

Which 1

The (1,1)?

[placid geyser]

It's for the equation of a line

It's the same as telling x = 3 + lambda, y = 3 + lambda (parametric form)

In other words, the graph y = x

[fast pilot]

the 1 from the position vector in the z coord

[pine temple]

Oh! Believe it or not, I’m supposed to be learning that next week

[placid geyser]

Oh, good luck

[pine temple]

I’m only in Calc II, so no wonder is there some elements that confuse me 😅

[placid geyser]

There's no z coord. (1, 11) x = 1, y = 11; (1, 1) x = 1, y = 1

[fast pilot]

oh sorry i misread the 11

[fast pilot]

wait what is this meant to be then?

[placid geyser]

The line AD

[fast pilot]

are u sure?

[placid geyser]

Yes

[fast pilot]

wait

[placid geyser]

Ig

[fast pilot]

oh ok

i see

[placid geyser]

Hey the line BC too and find them intersection point

And tell if you got it or not

[fast pilot]

ok

[fast pilot]

i got it

tysm for ur help

.close

[pearl pond]

Channel closed

Closed by @fast pilot

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[midnight haven]

Hey, can anyone help me with some basic physics? Electric circuit stuff
Or this not the place for this

[last summit]

Nah we dont do physics here

only real math

[midnight haven]

Ah, that's unfortunate

Was assuming people here would know basic physics and could maybe help

[plush bramble]

#old-network for physics help

[loud phoenix]

how basic?

[midnight haven]

Just tryna

Set up Kirchoff's equations

On a circuit

[loud phoenix]

ok

do you have a problem

[midnight haven]

Yes

I need to consider this circuit

The capactior has some charge Q, and is neither fully charged nor completely uncharged

The first part is asking how many distinct currents there are in this circuit

I said the number of distinct currents is 3, since it corresponds exactly to the number of branches

Is that correct?

[loud phoenix]

yoo sry my bad i can't do this

[midnight haven]

Ah, okay

All good

<@&286206848099549185>

[fair creek]

you can use Kirchhoff's laws

[midnight haven]

I'm trying to

Don't know how to

[fair creek]

my hands are kinda tied because they would call what i'm about to do as "spoonfeeding"

but do note that V1 and V2 cells produce I1 and I2 currents