help-4

[ivory valley]

Du bestimmst a indem du einen weiteren Punkt einsetzt.

[livid gull]

Oh

[ivory valley]

Welche anderen Punkte stehen dir zur Verfügung?

[livid gull]

Keine ahnung

[ivory valley]

Schau aufs Bild.

Du hast drei Punkte.

Nur einen benutzt.

Es bleiben die Nullstellen.

[livid gull]

Uh die nullstellen

[livid gull]

ah sorry jetzt erst gelesen

[ivory valley]

Ja, du kannst zum Beispiel (96,0) wählen, einsetzen und dann für a lösen.

[livid gull]

Aber das verwirrt mich, ich verstehe nicht warum ich das so machen muss und dann ergibt irgendwie garnix sinn

[ivory valley]

Die Scheitelpunktform hat 3 Parameter.

a, d und e.

d und e kennst du schon.

a musst du noch durch nen anderen Punkt bestimmen, weil es gibt unendlich viele parabeln mit Scheitelpunkt (0,192)

[livid gull]

Ich glaub ich zieh mir doch kurz lieber youtube erklärungsviedeos von den basics rein bevor ich zu so aufgaben gehe weil ich versteh garnix irgwie

[ivory valley]

All diese Parabeln haben den gleichen grünen Scheitelpunkt (0,192) aber sind trzdm unterschiedlich, wegen des Streckungsfaktors a.

Dass du die richtige aber modellierst, dafür brauchst eine zweite Information, hier die Nullstellen, um das richtige a zu finden.

[vale dock]

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[ivory valley]

Schade.

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[plucky raven]

helloo, im revising maths at the moment & i was just wondering - i know that i have to multiply by 100x to get 15.5r, because 1.5r doesnt work. but why? 🙏

[hot tide]

,rccw

[plucky raven]

ah my bad

[hot tide]

    x =  0.151515
100*x = 15.151515

notice how the trailing digits align

and you get 100x = 15+x

[plucky raven]

ohhh yes now i see

[hot tide]

also 100x is 15.151515... not 15.5555...

[plucky raven]

i messed up when writing 100x

yeah

lmao

thank you

[hot tide]

in general if you have n digits that repeat you need to multiply by 10^n

but you don't need to remember that

[hot tide]

if you don't have any more doubts, type ".close"

[plucky raven]

alrighty, thank you so much

.close

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[worldly stirrup]

i got the answer as 26x26x25x24 but the correct answer is 26x25x24x23

why is my answer wrong

[odd jackal]

You cannot repeat any letter

[worldly stirrup]

first im counting the 26 alphabets of the first letter

then im counting the 26 alphabets of the second letters (cuz the letters can be repeated twiice)

then 25x24

[odd jackal]

without repeating the same letter twice means each letter can be used atmolst one time

[worldly stirrup]

oh i thought that repeating each letter twice meant it was repeated once again after being previously repeated

mb

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[left kernel]

can anyone help me

[odd jackal]

is this a test

[left kernel]

no

exercice on math academy

i forogot how to do this

[odd jackal]

set x to 0

and y to 0

to find the x and y intercepts

[left kernel]

?

btw im in 8th grade

[odd jackal]

so you have y=x+1

right

[left kernel]

yh

[odd jackal]

so what is x if y is 0

[left kernel]

um idk

wait is it 1

[fickle rose]

is this an introduction to linear equations?

if so, they might be looking for the OP to find the slope and using the +b for finding the y-intercept

[left kernel]

ok

i really dont get it

[fickle rose]

you are familiar with the equation (y=mx+b), yes?

[rocky lotus]

PajamaMamaLlama

[left kernel]

yes

[fickle rose]

we are given the line y=x+1, in this case what is m and what is b?

[left kernel]

is m the slope

whats b again

[fickle rose]

y-intercept, that is where the line intersects the y-axis

[left kernel]

idk

[fickle rose]

in y=x+1, what are the values of m and b

[y=mx+b]
[y=x+1]

[left kernel]

did u just react monkey

[rocky lotus]

PajamaMamaLlama

[left kernel]

yea i get that

[fickle rose]

compare the two side-by-side, can you now see what the values of m and b are?

[left kernel]

1 and 1 ?

[fickle rose]

Yep! So from this info we know that the slope and y-intercept are 1

do you know what the definition of slope is?

[left kernel]

no

[fickle rose]

so a slope if 1 means that if I move 1 step right in the x-direction I must also move 1 step up in the y-direction

[left kernel]

ok

[fickle rose]

like so (if you don't mind my amazing art skills)

[left kernel]

is it the black line

[fickle rose]

the black line was to demonstrate that going one unit right correlates to going one unit going up

i.e. a slope of 1

[left kernel]

alr

can we talk tmr i gotta go sleep

bye

[fickle rose]

I mean I probably won't be around but sure

bye lmao

.close

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[rigid stirrup]

Hello community, reaching out for help since ive been struggling in math, im now going into my sophmore year and recently failed my final exam and earned a D- i reach out because i know if something doesnt change and i dont get the help that i need then i will probably continue to fail, any tips will help

[fickle rose]

I was in the same position as you, and honestly what helped me most: practice... there are SO many resources online... we live in the information age

whatever course you're going into Geometry/Algebra II (typical of Sophomores) there's KhanAcademy (classic) but also I love the OrganicChemTutor on YouTube

[midnight pier]

Practice pratice till your eyes pop out of their sockets I mean it no matter how bad you are the more you practice the higher the grade, yeah maybe your friends don't have to practice as much as you but it doesn't matter you are different people. Practice till the previously hard questions look so trivial that's when you know you're gonna get a good grade in the exam. and don't be ashamed to ask for help and yeah btw try to have a notebook where you put all the questions that you feel are hard (your made a mistake in) and review them from time to time and start reviewing the questions less as you feel they are becoming easier.

that's it that's how you get an A

sorry for the long message.

[rigid stirrup]

thank you this was very helpful ill keep this in mind for over the summer.

[midnight pier]

and yeah one more thing if your uni offers past papers

try to do those as much as possible

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[stark wedge]

du/dx isn't 12u

[dark epoch]

Ohh

[stark wedge]

and you didn't write down the chain rule in full even though you had one of the components for it correct

so do that

[dark epoch]

Okay

[dark epoch]

Oh chain rule applies to both dy/du and du/dx that's what you meant right

But i gotcha tysm<3.

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[dark epoch]

Oh you can countinue if you were telling something

Mb

[stark wedge]

no this time it's correct

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[twin warren]

Hi

[vale dock]

@twin warren Has your question been resolved?

[stiff lily]

if you have any questions,
please just ask it directly

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[midnight pier]

How do I write all the equivalence classes for R?

[ebon glade]

can you write down all numbers that are equivalent to 1?

[vale dock]

@midnight pier Has your question been resolved?

[midnight pier]

I can't understand.

[ebon glade]

can you write down all x with 1Rx ?

[stark wedge]

or in other words, all x such that (1, x) ∈ R

[midnight pier]

x= {3,5} ?

[stark wedge]

this is poor notation and also misses one number.

you would do better to write ∈ instead of = there

but what number other than 3 and 5 is 1 R-equivalent to?

[midnight pier]

I couldn't find that sign on my keyboard

[stark wedge]

the word in is a fine substitute.

anyway my question stands and i would still like you to answer it

[midnight pier]

1?

[stark wedge]

indeed.

[midnight pier]

Yep

[stark wedge]

so the equivalence class of 1 is {1, 3, 5}.

what numbers in A did you not yet touch and what equivalence classes do they carve up into?

[midnight pier]

2 and 4?

[stark wedge]

you might wanna try putting more confidence into your answers

yes the remaining numbers are 2 and 4

are they R-equivalent or not?

[midnight pier]

[midnight pier]

Nope

[stark wedge]

you claim 2 and 4 are not R-equivalent, do i understand you correctly?

i.e. you claim (2, 4) ∉ R

[midnight pier]

Yes ma'am

[noble anvil]

[midnight pier]

[stark wedge]

why are they not R-equivalent?

[midnight pier]

Because they weren't in our equivalence class

[stark wedge]

you either misunderstand or overthink or both

just because 2 and 4 both don't belong to some particular equivalence class of R, does not have any bearing on whether they are equivalent to each other or not.

ok, would you like to:
(A) attempt to answer the question "Are 2 and 4 R-equivalent?" yourself, or
(B) have me direct you step by step, or
(C) some secret third thing?

[midnight pier]

WAIT. Can you tell me what's the difference between equivalence class and being equivalent?

https://cdn.discordapp.com/emojis/903967475702591588.gif?v=1&size=48&quality=lossless

[midnight pier]

I'll go with B

[stark wedge]

an equivalence class is a subset of A in which all elements within it are equivalent to each other and not equivalent to anything outside the class

being equivalent is... being equivalent, a property of two elements. the kind of thing the equivalence relation directly supplies info for

[stark wedge]

ok in this case: can you read the problem and copy the definition of relation R as given by the problem

[midnight pier]

R= {(a,b) : |a-b| is divisible by 2}

[stark wedge]

ok

errr

the R= should be outside the curly braces.

R = {(a,b) : |a-b| is divisible by 2}

anyway, alright.

now write out, according to this definition, the meaning of the statement (2,4) ∈ R.

notably:

i want you to ACTIVELY DISMISS, for the time being, ANY AND ALL CONCERNS about the truth and falsehood of the statement.

and ONLY unpack the definition of R. nothing else. no less and NO MORE.

[midnight pier]

|2-4| is divisible by 2 so it belongs to R

Basically (2,4) belongs to R

[stark wedge]

well, you jumped ahead even though i tried to ask you not to.

[midnight pier]

😭

[stark wedge]

but yes you're correct.

(2,4) does belong to R and so 2 and 4 are R-equivalent and so they are in the same equivalence class

and so we have exhausted the entirety of A with two equivalence classes:

• {1, 3, 5}
• {2, 4}

[midnight pier]

Ohhh! Ohhh! okayyy

I think I am getting it

Yeah

Got it

Thanksssss

.close

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[olive grove]

got no idea how to starttt

[olive grove]

i swear the past papers are harder than everything i learn in class vro 💔

[manic cairn]

What does intersection mean?

[stark wedge]

do you know how to find when two curves intersect generally

like if you had y=f(x) and y=g(x) and wanted to see where they intersect, how could you do this algebraically

@olive grove

[olive grove]

oh

[olive grove]

nope

i got no idea

[olive grove]

when two of them got a point which is the same

[manic cairn]

Correct!

[olive grove]

[manic cairn]

A point is basically x and y coordinates

[stark wedge]

hmmmm are you sure about that

hang on @manic cairn lemme cook here a bit

[manic cairn]

Sure haha

[olive grove]

💔

ok hol up then

lemme think abotu this deeply

oh

the point would be one where both equations are equal?

i think*

[stark wedge]

yes

you set f(x)=g(x) and solve

in your case however you need to take a small step back

[olive grove]

o

[stark wedge]

and you want to first write the equation down

but then do nothing else until i instruct you otherwise

[olive grove]

ok

one sec

so Px^2 + Qx + 2P = 6x^2 + Qx + 4

[stark wedge]

yes

now what type of equation is this?

you should know the word.

[olive grove]

polynomial

is this the right word or am i cooked

[stark wedge]

ok that's correct but be more specific

[olive grove]

quadratic

[stark wedge]

yes

now, do we know how to analyze a quadratic equation for how many solutions it has, without finding those solutions outright?

[olive grove]

discriminant

[stark wedge]

you might have also heard it phrased as "nature of roots"

[stark wedge]

yes correct

so put your equation into a form from which you can calculate the discriminant

or perhaps have a realization for an easier shortcut you could take

[olive grove]

lemme look at it

i infact can not think of the easier way

💔

ill write it down in the discriminant way

where i can get b^2-4ac

but wait a minute

[stark wedge]

well first write your equation into the form ax^2+bx+c=0

[olive grove]

yeah i was gonna do that but

dont i have both Px^2 and 6x^2

[stark wedge]

yes you do, and?

[olive grove]

how will i get the a value then

wait ill do the thing

[stark wedge]

after subtracting 6x^2 off both sides, these combine as (P-6)x^2

[olive grove]

huh

i didnt know it worked like that

hol up

Px^2 - 6x^2 + 0x + 2P - 4 = 0

i put 0x just to be able to think about b value easier

[olive grove]

is this right or nah

k cool

since ann said i can do that (P-6)x^2 thing

imma do it for the c value in the quadratic too

[stark wedge]

your equals sign went missing...

[olive grove]

oop

im not meant to do it like that?

= 0?

Px^2 - 6x^2 + 0x + 2P - 4 = 0

[stark wedge]

you need to include the =0 mate

(P-6)x^2 + (2P-4) = 0

[olive grove]

MY BAD

big my bad

[stark wedge]

now either you see the shortcut or you don't

if you don't, i will need you to first state the condition on D that guarantees at least one root

[olive grove]

i do not see the shortcut so

D = 0 guarantees 1 root

oh

D > 0

am i cooked then

wait i guess

i can write it like

D>= 0

im tryna get the greater than or equal to sign

💔

that will mean it got atleast 1 root

[manic cairn]

Continue the monologue buddy

[olive grove]

💔

bro idk what to do

im just yapping to myself

ill try i guess!

[manic cairn]

I am checking your steps

[olive grove]

oh ok

D≥0

so then

since a = (P-6) and c = (2P-4)

we can use the formula b^2 - 4ac

so D = 0^2 - 4(P-6)(2P-4)

expanding this

D = -8P^2 + 64P - 96

[manic cairn]

Do you need to expand?

[olive grove]

oh

wait a minute

i had a similar question earlier uh

wait we can turn this into inequality

i believe*

4(P-6)(2P-4) ≥ 0

[manic cairn]

• sign

[olive grove]

negative

-4(P-6)(2P-4) ≥ 0

[manic cairn]

-ve sign

Yes you missed it

[olive grove]

sorry

i can divide both sides by -4

(P-6)(2P-4) <= 0

lemme get the new sign since inequality flip

(P-6)(2P-4) ≤ 0

[manic cairn]

Continue

[olive grove]

P=6 P = 2

this means

wait

WAIT AHH

i got this

i think

either P > 6 or P < 2

AM I RIGHT

[manic cairn]

Try again buddy

[olive grove]

darn

[manic cairn]

Take an example

And try maybe

[olive grove]

💔

hol up some guy drew sum for me earlier

to try help me with a similar question

but this time the roots are 2 and 6

[manic cairn]

Yes

[olive grove]

he said if < 0 it has to be in the shaded area

if > 0 outside so

[manic cairn]

Yes

[olive grove]

infinity to 2 and 6 to infinity?

idk how to write this down

[manic cairn]

Are we in < 0 case or >0 case?

[olive grove]

OH WAIT

WE FLIPPED THE INEQUALITY

OHHHHHHHH

2 ≤ P ≤ 6

[manic cairn]

Correct!

[olive grove]

YAYAYAYAYAYAYAYAYAYAY

yeah i dont understand it fully

but i sorta get it now atleast

but this is only the condition on P

how do i work out condition on Q

[manic cairn]

Well the Q term got cancelled out

So what do you think it should be?

[olive grove]

erm

gimme a sec

gotta read back

q can be anything?

as long as 2 ≤ P ≤ 6

[manic cairn]

Yep

Any real

[olive grove]

cooooooool

aw hell naw

AW HELL NAW

i looked at the next question and i dont understand it either 💀

but whatever im done for tonight

@manic cairn tysm i hope you have a great day

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[hushed zealot]

Hello, I would appreciate help with this problem.

I have first attempted this problem by first approximating \frac{x^n}{2^n - 3^n + 5^n} as \frac{x^n}{5^n}

and then used the ratio test to confirm that for -5 < x < 5 the series converges (both endpoints leads to divergence)

However, I'm not sure whether approximating \frac{x^n}{2^n - 3^n + 5^n} as \frac{x^n}{5^n} is the appropriate step for solving this question because I feel like I should be using the comparison test..

Nonetheless even the comparison test doesnt work because \frac{x^n}{5^n} is is smaller than \frac{x^n}{2^n - 3^n + 5^n} so proving that \frac{x^n}{5^n} converges doesn't allow us to conclude \frac{x^n}{2^n - 3^n + 5^n} is convergent.

[stark wedge]

well you can use the limit comparison test

if that's authorized

if not, you can still compare against Cx^n/5^n for some suitable constant C

[hushed zealot]

oh wait yeah

so if you use the limit comp test itll give = 1 which is greater than 0 so both are divergent or convergent but since x^n/5^n is convergent for -5 < x < 5 the original series also converges for the same interval?

[edgy atlas]

or you can use root test. the bottom part might look monstrous but it really is just a little puppy inside. factor 5 out from bottom and you can see the rest just goes to 1 as the limit goes to infty, leaving you with x/5 i guess

[hushed zealot]

right
i was confused whether the interval of convergence for (x/5)^n would be the same for the original series

and im not sure how to justify this in my answer

[edgy atlas]

also you'll have to manually check for x = 5 and -5

[hushed zealot]

if the interval of convergence is indeed the same

[hushed zealot]

yeah i did

the divergence test was enough for that

[edgy atlas]

the limit? oke so you have 5((2/5)^x - (3/5)^x + 1)^1/x . you should know that when 0 <= a < 1, a^x goes to 0 as x goes to infinity and 1/x just goes to 0. thus you only have 1 left inside.

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@hushed zealot Has your question been resolved?

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[echo hearth]

Hey guys, what’s the formula for a cup?

[echo hearth]

So yeah, a cup could debate be a cylinder

But that’s not truly a cup

A cup comes in many different shapes and sizes

And even the accounting for that, we have to remember the width

[sleek nebula]

what

[echo hearth]

So you know how a cup has width

[sleek nebula]

what do you mean cup

like x^2 + y^2 = z?

[echo hearth]

No, like a 3-D cup

A physical cup

How do I calculate that?

[marsh forge]

depends on the type of cup

[stark wedge]

calculate what exactly

the volume?

[echo hearth]

Yes, the volume

[stark wedge]

you don't calculate, you measure

get your cup, get a kitchen scale, put the cup on it and then pour water into it until it's full

and take the difference between the starting and final readings in grams

that'll be equal to the volume of the cup in mL

(or if you have a scale with a TARE button, put the cup on that, then press TARE so that the reading resets to 0, and then pour water and you'll just have the right number when the cup is full)

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[lean dawn]

This is true for graph laplacians, but is it also true for adjacency matrices that, if you apply A to a vector (representing node values) k times, that the node values will be propagated to its k-hop neighborhood?

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[tough maple]

Hello, 👋 on question 37) in the first picture that’s the answer for it but the next picture is how I answered it by myself is my reasoning valid? Or is it wrong?

[tough maple]

@ me when someone answers thanks !

[frozen ledge]

how do you know the limit exists for g(x)?

when you split the limit into the product of limits you assumed that both limits existed

which is not necessarily true

we only know that g is bounded

doesn’t mean the lim as x —> c g(x) exists

[tough maple]

true

but I don’t get the answer like I can memorize it but what does it mean by |g(x)| <= M like it could be any number?

[frozen ledge]

it means it’s bounded

so it doesn’t shoot off to infinity

it avoids any infinity times 0 nonsense

[tough maple]

When we say that they match at everything but one point does that mean at that one specific point like does that mean it’s the squeeze theorem?

[frozen ledge]

to see why we want g to be bounded consider [\lim_{x \to 1} (x - 1) \cdot \frac{1}{x^2 - 1}]

[rocky lotus]

knief

[frozen ledge]

as an example

taking g(x) = 1/(x^2 - 1)

g isn’t bounded

so we can’t just say look it’s zero because of the x - 1

this limit is actually 1/2

[frozen ledge]

wdym match?

[tough maple]

But we could say the same about lim x->0 x(cos1/x)

The limit approaches 0 I thought even if cos1/X didn’t exist since X becomes 0 then it gets multiplied and boom

[frozen ledge]

cos is bounded

even if the argument acts funky

[tough maple]

“The functions f & g for all X except for X=C”

[wraith heart]

,w plot x cos(1/x) for -1 < x < 1

[frozen ledge]

it oscillating like a mad man as x -> 0 doesn’t matter when x is sufficiently small

[rocky lotus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by rie.mann

[frozen ledge]

limits dont care about what happens at x = c

they care about neighborhoods around c

[frozen ledge]

they don’t "match"

it first says lim f = 0

and then says |g| <= M

they don’t both satisfy this

not sure what you mean

[tough maple]

Look in the squeeze theorem here it says something like that

[frozen ledge]

i’m not sure what your question is

elaborate

[tough maple]

Oki wait

there’s something like that in question 40)

If everything matches up except for when x=0 then how are they the same limits like how’d they get the same value

That’s my question

sorry though it’s kindaaa different from 37 but it has the same point kinda 😭

[frozen ledge]

it reads lim x -> c f(x) = 0. also, |g(x)| <= M for all x except for x = c

the all x thing is for g being bounded

for 40 it’s for f and g being equal

[tough maple]

Ohh oki I get that now

[tough maple]

But for 40 when it says except for X=0 does it mean f(0)≠g(0) ?

[frozen ledge]

yea

[tough maple]

Ohhh oki

Thank you I get it now

[frozen ledge]

you’re welcome

[tough maple]

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[echo hearth]

So guys, you see the 10f

And you see all the zeros

Well, basically, I wanna fill all those zeros with

[wraith heart]

did you have a question after this ?

[echo hearth]

It’s OK I found out how to do it

[echo hearth]

Nope, I’m OK. I think I’ll come back if I need something.

[wraith heart]

!done

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[wraith heart]

<@&268886789983436800> scam

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[ornate dragon]

Find a linear homogeneous second order ODE such that the indicated set forms a solution basis for the ODE

solving i)

i'm not too sure what to do. I think I can simplify the first one since e^x is never 0

then i have a system of equations?

[wraith heart]

yea looks like it. should have many solutions for P and Q

[ornate dragon]

alright. i’ll work on it and let you know if anything comes up. thanks

@wraith heart would it still be valid even though 1 is not in the domain?

I could multiply it all by (-x+1)

(x−1)y′′+xy′-y=0

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[nimble sage]

guys i gave an extremely difficult questions of Sequence and Series. Q) If (a+1),(b+1),(c+1),(d+1) is in GP then find the value of (b+c+2)^2 = _____ options:- (a) (a+b+2)(b+c+2) (b)(a+b+1)(c+d+1) (c) (b+c+2)(c+d+2)

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Please don't occupy multiple help channels.

[quasi valve]

.close

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[river snow]

hello! i did this question earlier in my exam and couldnt figure it out! i’m told ABCD is a parallelogram. it asks me to plot point D.

[river snow]

i put down like -3, -2 and now reflecting on that im pretty sure it has to be 0 on x axis 🤦

[stable parcel]

There are several possibilities for where to place D. You can place it further up beyond C and B. You can place it further right beyond B and A. You can place it further left beyond A and C. They probably mean the last one.

So, what you can do is to see how far up and how far to the right of C that B is.

How far up from C is B?

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[glass kelp]

i believe the naming hints at a specific order?

a and c as Chai. T Rex has suggested

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[sacred pasture]

Hello, next year I'm going to enroll in university for computer science, I've worked my way up through the education system without really having any math. One of the classes is Calculus. I never really had any math experiences besides basic arithmetic and some geometry. I've looked at some of the weekly assignments and this is one of the questions.

Obviously I have no idea what is going on but I was wondering what math subjects I would need to study to be able to solve one of these questions.

The question translates to: determine the whole numbers p and q

[frozen ledge]

algebra

[viscid sedge]

algebra

oh and trigonometry

[dusky pulsar]

does logarithm come under algebra?

lowkey i have no clue on what algebra is defination wise

[viscid sedge]

no

[frozen ledge]

yes

[fathom quartz]

p=24
q=-28

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[storm ferry]

how do i do this

find the equation of a plane

i though

we just

do

[x,y,z] = [-2,3,12] + s[-2,1,5]+t[-6,3,15]

[glass kelp]

S and t are two lines on the plane, right?

[glass kelp]

If this is the case

Consider cross product to find the normal vector that describes the plane

[storm ferry]

ya

why

canr i just use dot product

[glass kelp]

Do u know why we use normal vector for equations for planes?

[storm ferry]

np/.

no..

[glass kelp]

Ok so

Assume x_0, y_0, z_0 to be a known point on the plane

And let x,y,z be another point on the plane

[storm ferry]

ye'

[glass kelp]

Then the vector (x-x_0, y-y_0, z-z_0) lies on that plane

Now

For a point to be on that plane as well, it must be on the same level of gradient

So all of vectors on that plane must be perpendicular to whatever is perpendicular to that vector all

To do that

[glass kelp]

Take this vector and dot prod it the normal vector denoted as (a,b,c)

And set result = 0

That is the equation of the plane

a(x-x0) + b(y-y0) + c(z-z0) = 0

[glass kelp]

U can use this info to solve this

Find a,b,c

[glass kelp]

Plug into this

[storm ferry]

my head hurts from trying to understand this 😭

why is it so complicated

[storm ferry]

i dont get this bruh

[glass kelp]

Is this intro for Mvc?

[storm ferry]

whats mvc?

[glass kelp]

Dot product between normal vector and vector that lies on the plane is zero right?

Cuz they are perpendicular

[storm ferry]

yes

tha makes sese

[glass kelp]

Multivariable calculus

[glass kelp]

Ye. We use that eq to describes the plane

[storm ferry]

oh no

this course is calculus and vectors

[storm ferry]

wahtt

so

the 2 vectors

in the vector wequation

are perpendicular..?

[glass kelp]

No

The two vectors that are given

Are on the plane, right?

[glass kelp]

They are perpendicular to the normal vector we’re finding tho, yes

[storm ferry]

oh..

think

i think

my concept and understanding is very bare minimum

i lowkey dont even understand what the vector equation means 😭

[glass kelp]

The thing after the s and t

Are the vectors describing the line

The two things in front of it just fixes the lines to their originating points

https://youtu.be/rL9UXzZYYo4?si=_KFZgc-plwSkZENL

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[valid epoch]

can anyone help me out here, i have multiple other questions but i just cant understand

[valid epoch]

its saying my answer has a wrong variable letter, i cant find anything wrong though..

[steady charm]

you should not include the "g(x) =" part in your answer since it's already there outside the answer box

[valid epoch]

could you help me out with the full assignment?

[valid epoch]

it worked, thanks

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@valid epoch Has your question been resolved?

[valid epoch]

@scenic kettle

[still talon]

you can post ur problem

[noble anvil]

? Hasnt it been solved

[valid epoch]

sure

[valid epoch]

more questions..

so it should be -5.5, 4 and -6,7 but it keeps telling me its wrong ( i know i messed up the graphic its random numbers)

[still talon]

can u show ur work

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[high tide]

Why do we define y as -x ?

When trying to solve a local extremum

[plush crater]

y+x = 0

[copper stump]

coz one line earlier, it says y+x=0

[plush crater]

is one of the equations

[high tide]

I can chose one of the equations

But why -x ?

[plush crater]

u r aware of the substitution method right?

to solve systems of equations

[copper stump]

either is fine, you can substitute y = 2-x^2 in the second equation

still gives the same equation

[high tide]

how can i visualize it

i have no idea what operations are used

[stark wedge]

y+x=0, subtract x from both sides, y=-x

[high tide]

:((

1 operation that i have no idea how it works and the whole problem cannot be solved

why does this happened , why do we even do it

the algorithm in my literature doesnt explain anything

and y is not even there , tf are we replacing

[copper stump]

wdym y is not there?

we replaced it thats why it disappeared

[copper stump]

yea, that -x used to be a y

[high tide]

ohhhhhhhhhhhhhhhhhhhhhhh

[copper stump]

replacing the y is exactly what the substitution did

[high tide]

yeah got it

then why do we swap -2 and -x

[copper stump]

for lucidity

just so that you can recognize that this is a quadratic equation

and thus proceed to finding its roots

[stark wedge]

for neatness

[high tide]

i see

i still dont understand why y = - x

[glass kelp]

(y + x) - x = -x
y = -x

[high tide]

what

but why - x , where tf does it come from

[pallid smelt]

do yk how to rearrange equations?

[high tide]

no

i dont know when i can do it

and when i cant

[lyric sundial]

You can always do it (if you do it correctly of course)

[high tide]

how tf did i pass calc 1

[lyric sundial]

Exactly our thoughts

[high tide]

i knew how to solve derivatives

[lyric sundial]

Especially very basic algebra

[high tide]

i have so many holes

in my knowledge

that just fuck up my ability to solve these equations

[pallid smelt]

picture this x+y = 0 this is the same as a number + a number = 0 this implies that one number has to be negative because if we had two positive numbers then it ≠ 0

[high tide]

my fucking god

XDDDDDDDD

i think i got it

[pallid smelt]

now when we add these they = 0 so the like amount it increases by has to be decreased by the same amount

lets use 2 + x = 0

[high tide]

• x

*-x

[pallid smelt]

to find x we need to take the 2 to the other side by subtracting it

if x+2=0 then (x+2) - 2 = -2

[high tide]

another example

?

[lyric sundial]

x + 3 = 0

How would you solve it?

[high tide]

(x + 3) - 3 = -3 ?

[lyric sundial]

Awesome, so x = -3

[pallid smelt]

because +3 -3 = 0

[high tide]

i knew this , but i didnt knew how to apply it to variables with no given number

[high tide]

yeah i know it

[pallid smelt]

x+y=0

[lyric sundial]

It's exactly the same

[pallid smelt]

solve for y

[high tide]

(X + y) - x = -x ?

[lyric sundial]

In other words you want to write it in the form y = ... where no y appears in the right-hand side

[high tide]

corrected it

[lyric sundial]

Yep alright

[high tide]

y = -x

[pallid smelt]

yup

[high tide]

oops

[pallid smelt]

now u were given x^2 - 2 + y = 0

[high tide]

(x^2 - 2 + y) - y = -y ?

[high tide]

ill hold into that for later

[pallid smelt]

no we don’t need to rearrange that now

u were given that

and we know y = -x

[high tide]

wait

huh

[pallid smelt]

x^2 - 2 + y = 0 we don’t need to rearrange

what we can do is use substitution

because we know that y=-x

[high tide]

how do we know that y = - x , is it from the previous equation ?

[pallid smelt]

yup

[high tide]

ahhhhhh

u shoudve said so

[pallid smelt]

the one we rearranged

[pallid smelt]

this one

[high tide]

x^2 - 2 - x = 0

[pallid smelt]

yup

[high tide]

why do we rearange it tho ?

[pallid smelt]

then it had that => x^2 - x - 2 = 0 because the typical way you write out a quadratic is ax^2 + bx + c where a, b and c r constants

[high tide]

ahhhh

[pallid smelt]

so ppl do it for comfort ig

[high tide]

i can do it without rearaging in that case ?

[pallid smelt]

it looks weird with a constant in the middle and an x term on the end

[high tide]

in theory

[pallid smelt]

u can but after i see the questions asks u to use the quadratic formula

and u have to be careful to assign the correct a, b and c to the different parts of the formula

[high tide]

it says that we need to find the critical points

[high tide]

understood

[high tide]

in the algorithm

then why do we use the quadratic formula , in another example it doesnt use it

hold on

[trim dock]

It uses the quadratic formula there

At least the discriminant

[high tide]

in that case - (b / 2a) ?

if the discriminant is equal to 0

[trim dock]

Yes

That is the unique solution to the equation, with multiplicity 2

[lyric sundial]

Yes, because what happens to the quadratic formula $x_{1,2} = \frac{-b \pm \sqrt{\Delta}}{2a}$ when $\Delta = 0$ ?

[rocky lotus]

Alberto Z.

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[winged rose]

Why is y=x^2 removed ?

[leaden cargo]

you have
$$y + dy = x^2 + 2xdx$$
now substitute $y = x^2$ and simplify

[rocky lotus]

Luigi

[winged rose]

Whats the rule for substitution though?

[leaden cargo]

I mean literally write x² in place of y

y is equal to x², so we can write x² whenever we see y, they are the same thing

[winged rose]

Alright i get it.

Thanks dude :))

.close

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[slim wraith]

How Far Can You See? The conning tower of the U.S.S.
Silversides, a World War II submarine now permanently
stationed in Muskegon, Michigan, is approximately 20 feet
above sea level. How far can you see from the conning tower?

[slim wraith]

I do not know which geometry theorem or formula will apply here

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@slim wraith Has your question been resolved?

[slim wraith]

...

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[burnt crystal]

PLEASE HELP

[plush crater]

have u tried u sub and partial fractions

[burnt crystal]

yes

[plush crater]

what did u get

[burnt crystal]

tanx(
2
1
​
ln2−
2
1
​
ln(1+
e
2

1
​
))+cotx(−
2
1
​
ln2+1+
2
1
​
ln(1+
e
2

1
​
))

THIS

[brave grail]

how are we supposed to read that mess

[burnt crystal]

this

i dont get how to further simplify this

oh

thats fine

[fickle rose]

I think cosove is wrong, assuming that expression is indeed the correct antiderivative, this doesn't reduce to 1:

[\tan(x) \qty(\frac12\ln2-\frac12\ln\qty(1+\frac{1}{e^2}))-\cot(x) \qty(\frac12\ln2-\frac12\ln\qty(1+\frac{1}{e^2}))+\cot(x)]

Of course now the simplification should be rather trivial

[rocky lotus]

PajamaMamaLlama

[burnt crystal]

got it thanks!

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[winged kite]

In thousands of dollars, the cost of producing x cars per day is defined by the following function: C(x)=0.1x^2+1.2x+3.6

The question a is: Determine a U(x) function representing the cost per unit.

The answer according to the book is U(x)=0.1x+1.2+3.6/x

I assume they just divided C(x) by X, but I’d like to know why they did that? What’s the logic behind it?

[drifting hornet]

The cost of producing x cars in one day is 0.1x^2+1.2x+3.6

how much does 1 car cost?

x cars ...................................... 0.1x^2+1.2x+3.6 K$
1 car ......................................... ? K$

[winged kite]

I’m sorry, I don’t really understand what the K$ here means

[drifting hornet]

just thousands of dollars

[winged kite]

But I have a general Idea of the logic now, Thank you

[drifting hornet]

i didnt know how to write it

so i just wrote K$

it's the same thing as
If 5 apples cost 15$, then 1 apple costs 15/5$

[winged kite]

Oh my gosh I feel so stupid now, I was too hung up on the day part

Thank you very much! Sorry for wasting your time,

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[open niche]

I need help finding my mistake while using matrixes to solve system of equations idk where i messed up

[open niche]

Γ3 means like R3 like row 3

[drifting hornet]

seems like only R3 is incorrect

[open niche]

Wdym

Yea at the end the 3rd equation

But idk which step has a mistake

In solution it does steps with different order but as long as the steps are mathematically correct the matrices should be equivalent

[drifting hornet]

the last step seems wrong

R1 -> R1 - R3 - 2R2

[open niche]

Ok which part did you notice ?

[open niche]

What should the first row become after that ?

[drifting hornet]

oh nvm i just misread 1 as 2

[open niche]

-R3 is +0 +1 -1 +3 so first becomes 1 2 0 4 and then -2R2 is +0 -2 +0 +2

Oh wait so R1 at the end should be 6

But still not matching

Imma do it in 2 steps cause im rushing all at once

[drifting hornet]

ohh

you forgot -

3 - 3 -6

became 3 - 3 6

[open niche]

Wdym

[drifting hornet]

row 3

[open niche]

Oh

Damn 😭

I been looking at it for 20 mins couldnt find

[drifting hornet]

20 mins?

[open niche]

Thanks bro you the goat

[drifting hornet]

when that happens, take a blank sheet and try solving it again

[open niche]

Yea i didnt check that row even once

[drifting hornet]

you probably wont do the same mistake twice

[open niche]

Yea true i started wondering if the order matters

But its equivalent so i realised it shouldn't

[open niche]

Yea you are right helps to start from nothing

Thanks man

[drifting hornet]

np

[open niche]

You free rn by chance? If i can dm you i need some advice on next topic

Also idk how to close this help ticket/channel

[drifting hornet]

.close

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[drifting hornet]

maybe just open a new channel

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[sand jacinth]

hi there! working on this question (1st photo) and ive attached my work (2nd photo)! the pdf answer key im looking at says the answer should be $3788.00. can anyone help me figure out where im going wrong? thanks!!

[jade ivy]

You need to calculate 4 years worth of contributions and then treat that as the principal for the remaining 72 periods.

[sand jacinth]

i see! thank you so much :)

[jade ivy]

I misread the problem and it should be 120, but in either case what you wrote isn't that.

[sand jacinth]

alright, thanks!

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[worldly lotus]

So i got it right but with the wrong equations according to my teacher someone please explain to me what I did

The question is: the sum of two numbers is 12 the difference of their squares is 48 what are the two numbers

[marsh forge]

what did you do here..?

[worldly lotus]

I messed up so I scribbled it out and continued from there

[worldly lotus]

Also i think x was supposed to end up as the bigger number but y came out bigger

[toxic ridge]

Interesting

You've forgotten to square while substituting for y in x^2 - 48 = y^2

[frigid harbor]

indeed it is

[worldly lotus]

How tho

[toxic ridge]

@worldly lotus

[worldly lotus]

I see

[worldly lotus]

Uhm so does it count as correct? My teacher marked it wrong

But I got the right answer

[toxic ridge]

No

If it's a subjective q

Yes if it's an mcq

[worldly lotus]

I got it wrong then

And thats the eisiest question 😭

I won't know my score till monday

He just told me that one for some reason

[toxic ridge]

Probably cuz it's a unique error

Doing something wrong to get a right answer is interesting

[worldly lotus]

Lol

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[lost locust]

guys anyone in 11th preparing for JEE

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[lime notch]

So this came up trying to find a generating function for edge weighted trees with weights [1,2,3] and no child edges are equal to their parent edge.
The g.f. can be written G(x) = 1/(1 - ax - bx^2 - cx^3)
With:
a = 1/(1 - bx^2 - cx^3)
b = 1/(1 - ax - cx^3)
c = 1/(1 - ax - bx^2 )
Can you solve for a,b,c?
Trying substituon I can get say a quadratic equation relating a and b and solve but then plugging back in a you get a equation of degree higher than 2.

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[weak juniper]

so, i said fx = 2x-2y

fy=2y-2x

and constraint equation: x^2+y^2=4

so i set fx and fy to 0

0=2x-2y —> 2x = 2y —> x=y

same with the second one

then i plugged into constraint

2x^2 = 4

x = +/- sqrt(2), y = +/- sqrt(2)

but if i plug in (-sqrt2, sqrt2)

0=-2sqrt(2)-2sqrt(2) so that’s false

(also considering x=y)

so i said C was correct, but ig it was actually B

[rough talon]

notice that the extrema of f(x,y) is x=y as f(x,y)=(x-y)^2

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[tidal cedar]

In the attached step for the Integrating Factor Method (Differential Equations) why is there no + c after integrating the function in the exponent

[steady charm]

because you only need "an" antiderivative of p(x), not "the most general" antiderivative of p(x)

if we did add a $+C$ then we would get [ \mu(x) = e^{x^2+C} = Ce^{x^2} ] so now we're just multiplying both sides by some additional nonzero constant

[rocky lotus]

cloud

[steady charm]

at which point you could just divide both sides by C again and you're back where you started

[tidal cedar]

I see now you get the same thing as your final answer

Thank you for explaining

[steady charm]

another way to think about it is that adding the +C gives you all possible integrating factors, but you only need to multiply by both sides by one integrating factor for the algebra to work

and as we've seen it doesn't matter which one we choose

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[hazy dirge]

HELP PLEASE :{

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No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[hazy dirge]

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[prisma flint]

Hello! 👋
How do I get the Values in the tableu after the Ford-Fulkerson-Algorithm?
"Erhöhung" means increase and "Flussstärke" means value of the total way you go in the end.

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[clever sentinel]

I saw you post the same question yesterday too. Looks like you aren't getting any help here, maybe you should try sending this problem in anyone of #discrete-math , #optimization or #theoretical-cs for better help, whichever category fits your question best.

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[prisma flint]

<@&286206848099549185> 🙏

[astral depot]

hey brot, what is Knoten
Edit: sorry, i googled it and it translates to 'node'.

Under each iteration, we are given 2 numbers. The number in the form 'N+' denotes that the predecessor of the node in N and the other number denotes the residual capacity along the path of the predecessor.

we assume that the bottleneck capacity of node 1 is infinity, because it helps us find the Maximum flow

in each iteration, we find the increase in residual capacity to each node and label it with its predessor to denote the path along which it is maximum.

hope this answers your question

[prisma flint]

Thank you very much, but I dont understand why the 6 in iteration 1 is 10 and not 14 or 8 and i dont get the entire iteration 2-4

[astral depot]

the 6th node in iteration 1 is 10 because the 1 -3 -4 -6 path has the highest residual value , which is 10

[prisma flint]

Ah, got it thank you!

But what are the iterations 2 3 and 4?

[prisma flint]

Please dont leave me hanging

[astral depot]

oh sorry

just a min

From Iteration 1, we already found a path from node 1 -3 - 4 - 6 with a bottleneck (minimum residual capacity) of 10, and the total flow is now 10.

Now we fill the path 1-3-4-6 with 10 in each path

from here , we do the same thing but with the new information we gained from iteration 1 and improve on it.

[prisma flint]

sorry, i dont get it

[astral depot]

okay

did you understand how we got the values in iteration 1

[prisma flint]

yes

but after that im lost

[astral depot]

okay

the maximum flow we found in iteration 1 is 10 right?

along the 1-3-4-6 path

so before proceeding to iteration 2, we fill 1-3-4-6 path with flow of 10

then we repeat the same process, with the updated information

[prisma flint]

i dont understand that part

[astral depot]

oh okay

try thinking about it like this:

filling each path with flow of 10 is like subtracting 10 from its maximum capacity

so basically, for iteration 2, the 'new' 1-2 path has maximum flow of 12-10=2, the 'new' 2-4 path has flow of 10-10=0 and the 'new' 4-6 path has max flow of 14-10=4

[prisma flint]

sorry, i dont understand

[astral depot]

hmm i made that a bit confusing sorry

okay

after iteration 1 fill the maximum path with flow of 10

once you fill it, you can't use the capacity of 10 it lost (because you filled it) again right?

[prisma flint]

yes

[astral depot]

i mean if the flow from 4 to 6(max is 14) is 10, you can only use the remaining 4 for further iterations

[prisma flint]

ah ok

[astral depot]

so you basically subract 10 from each path on 1-3-4-6 and repeat the process

you do this until you can't improve on flow of system anymore

[prisma flint]

but why then 4 iterations and not 2?

[astral depot]

notice that they stopped this process at iteration 4 because iteration 4 and 3 have same max resiudal flow(which is 18)