help-13

[long shale]

rt(x^2+4x)

and then in the answer they divided the whole f(x) by x when they found the limit at infinity

[dull oxide]

rt(x^2+4x)? What's rt?

[long shale]

root

square root

i dont know how to type that sorry

[dull oxide]

ah

gotcha

[dull oxide]

yeah this would work

[long shale]

is there a reason for that or is that just the method?

[dull oxide]

It has to do with leading terms, which may be a better way to analyize the problem.

In $\sqrt{x^2+4x}$, as $x$ gets very large, the $x^2$ term will grow much faster than the $x$ term. As this occurs, the $x$ term becomes neglible to the $x^2$ term. So for very large $x$, you can approximate $x\approx 0$ and leave $x^2$ as the only term. So $\sqrt{x^2+4x}\approx\sqrt{x^2}=x$.

[wraith dagger]

SWR

[long shale]

hmm okay i get it thank you

also

i have another question

is there a way to determine if a function has a slant asymptote by just looking at it because

okay so my test is tomorrow and we are graphing the complicated functions using limits derivatives and all that

and it takes a lot of time so i wanna know if theres a way to see that there is a slant asymptote by just looking at a function

to save time yk

[cedar kiln]

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[formal pasture]

how do i do #10

im confused

,rotate 270

[molten compass]

What are your thoughts so far

[crimson sedge]

just to make the question clear

you are considering the pendulum at the extreme position in the beginning?

[molten compass]

It says to consider it from the end of the swing

So I would say so

This is a prime case of shm

[crimson sedge]

|| so, my guess is that whatever wave you draw, it wont cross the time axis to change signs (considering a disp vs time graph) ? ||

[molten compass]

If this is truly a case of shm (probably is based on part b) we can assume that the centre of motion is 0 and thus there will changing signs for the displacement

[crimson sedge]

I assumed the centre of motion to be the extreme position, ig you could be correct in this situation though

[molten compass]

I agree with your assessment

Hopefully the op won’t just copy of you now

[crimson sedge]

hopefully

should I delete it?

[molten compass]

Prob better too

[cedar kiln]

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[fresh loom]

I'm somehow doing this wrong and I'm not sure. Can someone tell me.
I'm trying to find magnitude of the force on the 1nC charge
So what I did was first rotate this figure so the 1nC will be on the bottom right
Then i drew the forces, so for Force of 1 and -2, it will go to the right on the x axis because its attracted towards the positive
for force of 1 and 2, it will be repelled towards the southeast
So then i did my calculations, for force of 1 and -2, I got 1.8*10^-4 towards the right
for foce of 1 and 2, i got 1.8x10^-4 towards south east
Do components, 1.8x10^-4 x cos60 is 9x10^-5
1.8x10^-4 x sin60 is 1.56x10^-4
Since both the x components go to the right, i simply add them and get 2.7x10^-4
I do pythagorean theorem, so square the compoennts, and find the root, and I get 3.12x10^-4
but this is wrong apparently?

[cedar kiln]

@fresh loom Has your question been resolved?

[amber ridge]

@fresh loom did you draw a diagram? if so, can you show it?

@fresh loom

you still need help

because imma go sleep if not lol

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[crimson sedge]

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[crimson sedge]

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[cedar kiln]

✅

[crimson sedge]

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[spice kraken]

I think you can just substitute

[crimson sedge]

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[dry dirge]

can someone prove that circle which is going through points A B and C is inscribed in O_1 O_2 O_3 triange ?

in another words: is O1O2 and O2O3 and O3O1 tangent lines of that dotted circle ?

i dont thinc that radii of these big circles matters here but i wrote them just in case

i can see that big triangle is right triangle . But i dont think that it matters here

and the question original in this task was to find radius of that dotted circle . if itbis inscribed in that big triangle then i know how to calculate its radius .

[slate lintel]

the center of that circle goes through the angle bisectors of the triangle

other way around, the bisectors go through the center

which I think makes it the incircle but i would have to check that

[dry dirge]

ok the theorem is known to me but where do we need this theorem here?

[slate lintel]

just to show that this is in fact the incircle

[dry dirge]

can we do it this way. to say that O1O2 is not tangent but crossing the circle and discover some contradiction in that assumpsion ?

[slate lintel]

we could perhaps yes

depends on how those big circles are defined, i suppose

[dry dirge]

they are touching each other

and also the triangle is right if thet matters

sum of angles CBA and CDA must be 180 deg if circle goes though these four points

but idk if that helps

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[dreamy tree]

What is wrong with my answer ?

[surreal cave]

its doesn't look wrong...

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[fluid elk]

how do i solve for x

[raw gulch]

the hypotenuse should be 13 in length in your drawing

[crimson sedge]

^ ye what?

[fluid elk]

ye sorry

[crimson sedge]

@raw gulch are you able to assume that the two smaller triangles formed have side lengths of 6, 7.5, and x?

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[hollow storm]

Sorry guys, stupid question

[hollow storm]

If we had to calculate the limit of a point in which a function is not defined

[hollow minnow]

[modern compass]

that's wildly dependent on what else you know.

[hollow storm]

For example lim x-->-5 ln(x)

What is this?

Impossible?

[muted bear]

DNE

[modern compass]

oooooh, that kind of not defined.
yeah. the limit doesn't exist because the function isn't defined for x<0

[hollow storm]

Ok also

lim x -->0- of loga(x) with a>0

For 0+ it's plus infinite

[hollow minnow]

-inf

[hollow storm]

For 0- does not exist?

[hollow minnow]

This

[hollow minnow]

Yes

[hollow storm]

Ok so if there is no graph it doesn't exist

[hollow minnow]

You can’t approach from left side of 0 because f is not defined for x<0

[hollow storm]

Ya right

And last case:

If like lim to 0- and lim 0+ were equal

That's undefined?

If for example we had lim x-->5+ of some function and lim x--->5- of that same function and these are equal

What would happen?

[hollow minnow]

Then the limit exists at x=5

[hollow storm]

Oh yeah I'm an idiot

Thank you very much all!!

[hollow minnow]

[hollow storm]

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[hollow storm]

.reopen

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✅

[hollow storm]

@hollow minnow Sorry i didnt word this correctly

I meant if they are different

[hollow minnow]

Then the limit doesn’t exist at x=5

[hollow storm]

Like this

lim to 0- and 0+ are different

It doesn't exist?

[hollow minnow]

Yes

[hollow storm]

Ok thank you very much again

And sorry to bother :/

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[crimson sedge]

How do I solve $x^2-64=63x$

[wraith dagger]

Matt

[tropic oxide]

do you know how to solve quadratic equations in general?

[crimson sedge]

yes

the thing thats bothering me is the 63x

ion know if i can just subtract that & use the quadratic formula

[tropic oxide]

you can subtract it from both sides yes

there is nothing blocking you

it is not illegal

[crimson sedge]

cause if its $x^2-63X-64=0$ thats not a problem i solve that

[wraith dagger]

Matt

[tropic oxide]

lowercase x

but yes

why would it be illegal to subtract 63x from both sides?

[crimson sedge]

i was thinking it might not be possible because 63 is being multiplied by x

yk

but ig not

[tropic oxide]

you are not subtracting 63

you are subtracting 63x

also you could still subtract just 63, that's not illegal either

it just would not help you at all

[crimson sedge]

ok then, ty!

i get it now'

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[analog lichen]

Where do I go from here if I did the step right. Prompt: State if the given functions are inverses.

[dire geode]

Simplify and see if it satisfies the definition of inverse function

[analog lichen]

Can I see what’s the next step visually? Or

[dire geode]

The next step is algebra

Multiplication, distribution, addition

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[uneven pagoda]

an, bn are convergent sequences. an<=bn for all n
Prove that lim(an) <= lim(bn)
My proof:
bn - an>=0
define cn = bn - an; cn >= 0 for all n
lim(cn) = lim(bn) - lim(an) >= 0
==> lim(bn) >= lim(an)

[uneven pagoda]

Is it sufficient?

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@uneven pagoda Has your question been resolved?

[torn oriole]

looks good

altrough

[torn oriole]

what if cn were to be periodic in nature?

[uneven pagoda]

is it possible if an and bn converge?

[torn oriole]

hm

good argument

[uneven pagoda]

👍

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[echo atlas]

i need help figuring out a simple equation, i know how it works but i cannot seem to find the precise answer
the equation is sin(x) = x and i cant find x, desmos shows its between 0.9998 and 0.9999
ive tried using programming but that didnt work
any help would be greatly appreciated!

[sly abyss]

What

[tropic oxide]

... are you sure it isn't 0

[inland kayak]

yeah 0 works

[sly abyss]

Ye

[echo atlas]

no

[sly abyss]

It’s just 0 lol

[tropic oxide]

show what you've got on desmos

[sly abyss]

(0,0)

Is where they intersect

[inland kayak]

thats cos

[echo atlas]

damnit i mistyped

sorry

[inland kayak]

np

[tropic oxide]

,w cos(x) = x

[wraith dagger]

Results provided by WolframAlpha

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

Requested by ann_dec

[tropic oxide]

not quite as near to 1 as you claimed

[echo atlas]

then why does desmos show it between 0.9998 and 0.9999
im not contesting you im just interested

[tropic oxide]

it doesn't...?

shift your viewing window so you actually get a good look at it

[echo atlas]

i cant read i zoomed in and it shows what the bot showed

i must have tampered witht he eqation

[inland kayak]

@echo atlas

[echo atlas]

aah thanks

yea i must have edited the equation before posting here

[inland kayak]

yea

[echo atlas]

sorry for the watse of time 😭

[echo atlas]

is this the precise result or is it irrational

[livid hound]

irrational, that's only a decimal approximation

[echo atlas]

how could i find the exact result

[livid hound]

probs can't

[echo atlas]

ah damn

but thanks for the help!

i might be a frequent questioner because i always find ways to get myself confused

ima stop the channel now, thank you everyone!

.close

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[sonic fractal]

Hey

[sonic fractal]

I'm so lost

<@&286206848099549185>

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[austere hull]

did you just send your whole homework to us

lost on what

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[long fiber]

help

can anyone just explain the question

[idle cedar]

is this a geometry question ?

[long fiber]

that’s not geometry.

that’s a theorem

[idle cedar]

lol yeah but in the context of what field of mathematics

also its not really a theorem , is it?

its rather an exercise

[long fiber]

it’s a circle theorem

[long fiber]

probably geometry

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[austere plume]

can some explain the 4 like to me please

[hollow minnow]

the 4 like?

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[austere plume]

Sorry I meant line 3

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[marble ermine]

hello again

[marble ermine]

$f(x, y) = \frac{x^2 + y^2}{\sqrt{x^4 + y^4}}$

[wraith dagger]

lilisworld

[marble ermine]

and f(x, y) = a when (x, y) = (0, 0)

is f continuous on R² when a = 1?

[final crag]

any guess?

[marble ermine]

i think so

uh

wait, is it true that the limit of f(x, y) when (x, y) approaches (0, 0) is 1?

[final crag]

well if there is a limit, it has to be the same along every path of approachright?

so let's try some simple paths

try along y = 0 and have x approach 0

what happens?

i.e. set y = 0 and then solve the single variable limit as x goes to 0

[marble ermine]

it's 1

[final crag]

same thing happens if you try setting x = 0 and have y approach 0

I have one more test case for you

what about the line y = x?

[marble ermine]

yes

[final crag]

so just set y = x and take the limit as x goes to 0

[marble ermine]

ok wait

[marble ermine]

it's 2/sqrt(2)

[final crag]

yes!

also known as just sqrt(2)

but there you go, you have different limits along different approaches

so it cant be continuous

My favourite approaches to test this are always x,y = 0, y = x and y = kx

[marble ermine]

ohh ok thank you very much 🙂

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[warm vector]

why is pi/3 not a solution?

[hollow trail]

your first step isn't quite correct

recall the angle addition identity for cosine is \ $\cos(A\pm B) = \cos(A)\cos(B)\mp \sin(A)\sin(B)$

[wraith dagger]

cloud

[hollow trail]

where the "$\mp$" indicates the sign is flipped compared to the $\pm$

[wraith dagger]

cloud

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[chilly valve]

I’m having trouble figuring out how to integrate this

[chilly valve]

I use substitution, right?

[hollow trail]

yes

[subtle harbor]

u=sin(t) works out most nicely :)

[chilly valve]

Ah yes, the derivative of sin in du is what cancels out the cosine

Thanks

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[crimson sedge]

determine the form of a particular solution
for the differential equation

so i got the auxilliary roots

its m= 3,3

how to find particular sol

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[crimson sedge]

.close

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[crimson sedge]

decide whether or not the method of undetermined coefficients can be applied to find a particular solution of the given equation

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[crimson sedge]

.close

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[subtle raven]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

@thorn sluice Has your question been resolved?

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[glossy oyster]

Thirty-nine percent of U.S. adults have very little confidence in newspapers. You randomly select ten U.S. adults. Find the probability that the number who have very little confidence in newspapers is (a) exactly six and (b) exactly four.

[glossy oyster]

yes

pls delete

[cedar kiln]

@glossy oyster Has your question been resolved?

[tropic oxide]

!status @glossy oyster

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[glossy oyster]

3

[tropic oxide]

show work & answer

[cedar kiln]

@glossy oyster Has your question been resolved?

[cedar kiln]

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[stray otter]

There is something wrong here, these 2 don’t add up, but idk where

[cedar kiln]

@stray otter Has your question been resolved?

[cedar kiln]

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[vestal hedge]

Need help with integral. Sorry gotta type it and idk how to use that bot.

Integral (2x^2+4x-1)/(x^3+2x^2++3x-6) dx

[vestal hedge]

So what I tried was (3x^2+4x+3-4-x^2)/(x^3+2x^2+3x-6)

And then we get two fractions first one having its derivative on top so it turns into ln(x^3+2x^2+3x-6) - integral (4+x^2)/(x^3+2x^2+3x-6)

And then I got stuck

pls ping if u respond

[slate lintel]

i would try factoring the denominator and use partial fractions

[vestal hedge]

No luck there

That -6 is bothering me too much

Cant factor it

[slate lintel]

it has at least one real root, maybe you can factor it a little bit?

[cedar kiln]

@vestal hedge Has your question been resolved?

[vestal hedge]

Tried my best with factoring, I always have -6 or something else in the end

[slate lintel]

(x+1)

[true violet]

$\frac{2x^{2}+4x-1}{x^{3}+2x^{2}+3x-6}\to\frac{2x^{2}+4x-1}{x^{3}+\left(-x^{2}+x^{2}\right)+2x^{2}+\left(-3x+3x\right)+3x-6}$

[wraith dagger]

CH3

[true violet]

then factor out (x-1)

[vestal hedge]

How do I factor (x-1) from x^3+2x^2?

[true violet]

you don't

[vestal hedge]

What do you get after factoring

[true violet]

$\frac{2x^{2}+4x-1}{x^{3}+\left(-x^{2}+x^{2}\right)+2x^{2}+\left(-3x+3x\right)+3x-6}\to\frac{2x^{2}+4x-1}{\left(x^{3}-x^{2}\right)+\left(3x^{2}-3x\right)+\left(6x-6\right)}$

[wraith dagger]

CH3

[vestal hedge]

Ohh I see

[true violet]

now you can see how we can factor out (x-1)

[vestal hedge]

But I guess we need to factor out (x+1)

[true violet]

i tried that

won't get a good result

[vestal hedge]

Cause I have 4+x^2 on top

[true violet]

🤔

you'll get this after factoring

$\frac{2x^{2}+4x-1}{\left(x-1\right)\left(x^{2}+3x+6\right)}$

[wraith dagger]

CH3

[vestal hedge]

But thats not what I have in the nomerator

Its 4+x^2

[true violet]

it isn't 2x^2+4x-1?

[vestal hedge]

Im pretty sure its 4+x^2

[true violet]

??

[vestal hedge]

Oh thats where you wanna do it, I just had some steps done after that message

[vestal hedge]

.

Okay that works too

Yeah thats way easier

And one more thing. Was it possible to solve this with trigonometric substituion?

[true violet]

no idea lmao

[vestal hedge]

I heard people talking about it but I have no idea how to apply it there

[vestal hedge]

Okay thanks a lot

[vestal hedge]

I will leave the channel open for a while in case someone knows the answer for this

[true violet]

you can use trig sub after partial fractions

on one fraction though not all of them

[cedar kiln]

@vestal hedge Has your question been resolved?

[vestal hedge]

Well thanks ya all

[cedar kiln]

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[idle tusk]

trying to express y in terms of x in y^2 (y^2 + x^2) = 4x^2, looking at the solution by wolframalpha. why is the step substitute x=y^2 legal?

[idle tusk]

maybe i'm stupid but i really don't understand how this substitution makes sense

it then does the opposite

is wolfram alpha just assinging a new variable and coincidentally naming it 'x'?

[green thorn]

yeah seems like it, especially since it isnt simplifying (x^2)(x)

[idle tusk]

understood

[livid hound]

doesn't sound like something wolfram would do

[idle tusk]

let me name it u or something and try and solve it

[green thorn]

i mean look at the work though

it seems like that's exactly what it did

[idle tusk]

yeah it checks out

idk why it did that

thanks guys

.close

[cedar kiln]

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[fringe fractal]

suppose x,y and satisfy 7x^2-9xy+7y^2 = 9 , if a is the maximum value and b be the mininum value of x^2+y^2 then find the unit digits of 2^(10a+23b)

[fringe fractal]

plotting that we get an elipse , how to proceed further ?

[wraith dagger]

srant ganderson

[urban haven]

don't know anything about finding the unit digits
but the optimization problem should be easy using lagrange multipliers

[pastel vault]

try the change of coordinates $(x, y) \mapsto (x + y, x - y)$

[wraith dagger]

south

[pastel vault]

which will scale the ellipse by a factor of sqrt(2)

[fringe fractal]

technically we shoulint use that , cuz its a problem from early 11th grade

[pastel vault]

sorry, 1/sqrt(2)

[urban haven]

ah okay

[fringe fractal]

and then ?

[pastel vault]

you will have a rectangular ellipse

in the form x^2/a^2 + y^2/b^2 = 1

[fringe fractal]

oo wooww

[pastel vault]

yeah

[fringe fractal]

thenn ??? what can we do with that

how can we relate that to a circle

[pastel vault]

so you just need to find the x- and y- intercepts of the transformed ellipse

the major and minor axes lengths

the numbers are nice trust me

for 10a + 23b

[pastel vault]

x^2 + y^2 is the squared distance to the origin

[fringe fractal]

a is the length of semi major axis and b is the length of semi minor axis ?

[pastel vault]

yes

[fringe fractal]

and thats it ??

bruhh

[pastel vault]

yeahhh

[fringe fractal]

leeme tryyy

we get a^2 = 9/5 and b^2 = 9/23

so a and b will be roots of them and the calc get messed up

😭

sorry to tag you @pastel vault am i making some mistake here ?

[pastel vault]

yeah, that's for the transformed ellipse

so to transform back everything gets scaled by sqrt(2)

which means you have to multiply by sqrt(2)^2 = 2

[fringe fractal]

still we will have that ugly sqrt(23) and sqrt ( 5)

[pastel vault]

nah, look at 10a + 23b

oh wait

[fringe fractal]

yeah

[pastel vault]

dude, I swear it's 10a^2 + 23b^2

the heck

[fringe fractal]

i wish ITS JUST TOO PERFECT

[pastel vault]

I swear to god

[fringe fractal]

it was in our test😭

[pastel vault]

the question is wrong

RIP

[fringe fractal]

I SPENT 15 MIN

😭

the ans is given as 6

[pastel vault]

bro

dude ask for the solution then

[fringe fractal]

no they wont give

[pastel vault]

smh

[fringe fractal]

if we take 10a^2 + 23b^2 , we get 108

and 2^108 does end with 6 i think

[pastel vault]

yes

108 is a multiple of 4 and 2^4 = 16

[fringe fractal]

yeah

but anyways

they will delete the qn

thank you soo muchh 😊

.close

[cedar kiln]

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[pastel vault]

npnp

[cedar kiln]

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[verbal kestrel]

2x+3y+6z=0
3x+2y+7z=0

[verbal kestrel]

This system of equations will have infinity solution?

[pastel vault]

yes

it's underdetermined, as you have 3 variables but only 2 equations

[verbal kestrel]

.close

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[queen token]

hey, I need help proving this:
let I = [a,b] with a < b and f: I -> I be a continuous function. using the intermediate value theorem show that f has at least one point in I where f(x) = x. Hint: look at g(x) = x - f(x) and create an argument with a root

[queen token]

what ive done so far is:
let
g(a) = a - g(a)
g(b) = b - g(b)
with case distinction
case 1: g(a) > 0 and g(b) < 0
then after the ivt there must be an x in I with g(x) = 0 => f(x) = x

but that only works for the g(a) > 0 and g(b) < 0 case and not g(a), g(b) > 0 or g(a), g(b) < 0

[finite raven]

What about f(x) = x+1

[queen token]

then f: I -> I wouldnt hold

[finite raven]

Oh

ok

[upper abyss]

What's I here?

[finite raven]

[a,b]

[upper abyss]

Oh lol it's right there oof

[queen token]

im not even sure if my approach is right so

it might be better to start somewhere else

[slate lintel]

what you might do is draw a square which is your [a,b]²

[hollow minnow]

I think youre on the right track

hi hayley

[slate lintel]

and draw a continuous function in that square, try to make it not intersect f(x) = x

hi pure

[queen token]

im not sure i follow

[slate lintel]

if the function f(x) is from [a,b] -> [a,b] then it fits within a square

[finite raven]

f(x) <= b. So g(b) = b - f(b) >=0. (Because f(x) in I = [a,b])

[finite raven]

f(x) >= a. So g(a) = a-f(a) <= 0

If you assume, f(x) is never x, we get g(b) > 0 and g(a) < 0

g is continuous (sum of continuous functions)

[queen token]

but why would we assume f(x) is never x?

[finite raven]

By contradiction

[queen token]

ah

[finite raven]

Or you could say, if g(a) = 0 or g(b) = 0, we are finished

Assuming g(a) < 0 and g(b) > 0

[queen token]

how did you get g(b) = b - f(b) >=0 from f(x) <= b ?

[finite raven]

f(x) <= b = >
f(x) -b <=0 =>
b - f(x) >= 0 (for all x)

[queen token]

ah

[finite raven]

We have that because f(x) in I = [a,b]

[queen token]

makes sense

[finite raven]

We didn't use surjectivity tho

There must be a hole in the proof

[queen token]

so we assume f(x) is never x and our contradiction is that g is continuous?

[finite raven]

We have g continuous, g(a) < 0 and g(b) > 0

by intermediate theorem we have .... ?

[queen token]

there must exist an x in I with g(a) < x < g(b) ?

wait

no

x in I such that g(x) = 0

[finite raven]

yes

[queen token]

okay huge i got it

tysm

appreciate it

.close

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[cedar kiln]

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[soft jay]

To find an inflection point I need to differentiate twice and set it =0 but what do I do if I end up with something like this?

[soft jay]

I'm trying to find the inflection point of x^3+y^2+xy+y=0

@slow jewel any ideas bro?

[slow jewel]

Hmm let's see

Ok I'm back, sorry

[slow jewel]

So 6x +2(dy/dx)^2 +2(dy/dx)= 0

You could solve the quadratic now

In dy/dx

[soft jay]

I can?

But what about the 6x

[slow jewel]

I think so

Ig that would be the "constant" term in this case

[soft jay]

So let dy/dx =y for instance?

And then we get 6x+2y^2+2y

But we can solve this because there are 2 variables?

[slow jewel]

Yeah

,align 2y^2 + 2y + 6x&=0 \
ax^2 + bx + c &=0

[wraith dagger]

Lorentz

[slow jewel]

Looks similar

[soft jay]

Oh OK I'll give it a try

[slow jewel]

Sure

[soft jay]

I wad thinking to equate to the dy/dx from before but there would still be the issue of 2 variables

[slow jewel]

Hmm

I thought we would get smth

Wait but doesn't the point of inflection also satisfy dy/dx = 0?

[soft jay]

If it does it would certainly make things another easier

[slow jewel]

I think it does

What we could do is

Find the points at which dy/dx= 0

And then plug in those values in d²y/dx²

[soft jay]

OK but doesn't d2y/dx2 have to be =0 too

Or perhaps I misunderstood

[slow jewel]

Yes it does

[slow jewel]

If the value after substitution gives 0 then it's a point of inflection

Else not

[soft jay]

And then dy/dx=0

So 6x=0

X=0

[slow jewel]

Ye

[soft jay]

And then to find y I sub back into the original equation?

[slow jewel]

Yes

[soft jay]

Pls be -1 🤣

[slow jewel]

It is, then?

If you put x=0, y=0,-1

Both would be points of inflection then

[soft jay]

Awesome 👌

Thanks again Lorentz absolute genius

.close

[cedar kiln]

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[slow jewel]

Nah mate me no genius

.And np

[cedar kiln]

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[wraith ice]

How to find the commutative matrix with this matrix?

[cedar kiln]

@wraith ice Has your question been resolved?

[wraith ice]

<@&286206848099549185>

[granite knoll]

what do you mean the commutative matrix? a matrix B such that AB = BA?

[wraith ice]

yes :Đ

[granite knoll]

tbh I'd probably try to brute force it at first

but it's symmetric so there might be a more clever approach

[wraith ice]

um

[wraith ice]

.close

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[molten schooner]

Hello

[molten schooner]

My problem:

[bright agate]

[molten schooner]

A person eats 1/3 of a jar with colas. After that another person eats 1/3 of what’s left in the jar after the first person ate. After that a third person eats 1/3 of what’s left after the two other people ate their parts. When he’s done only 16 colas remain. How many colas did each person get

P1, p2 and p3

I don’t remember who got who but I do remember it was divided into 8, 18 and 12. Does anybody else get a similar answer?

[slow jewel]

Let initial no be x

No left after first person= 2x/3

Coz x- x/3

[slow jewel]

And now the second eats 1/3rd of this

Just repeat the same thing

[cedar kiln]

@molten schooner Has your question been resolved?

[molten schooner]

So 1/3 x 2/3 which is 2/9

And 2/9 + 1/3 which is 10/18

That’s how much they ate together

The third person eats 1/3 out of 8/18

Which is 8/54

[granite knoll]

yes that looks right

[molten schooner]

So 8/54 + 10/18 which is 38/54

That’s what they are

Ate*

And 16/54 of the total amount of colas is 16 colas

Yeah this is somewhat right

Right?

[granite knoll]

looks right yes

[molten schooner]

Well I must have been stressed or something on my test because I remember also writing as a part of my answer that they ate 38 cookies in total

[granite knoll]

they did

[molten schooner]

Most of my friends got that part to 54, which is also a number that appeares in my calculation

I was right then?

that’s good

[granite knoll]

well if there were 54 to begin with, how many did P1 eat?

[molten schooner]

18

[granite knoll]

and P2? and P3?

[molten schooner]

And person two 12 p3 8

[granite knoll]

then add them

[molten schooner]

38

That’s right

Nice

Well how did my friends add it up two 54

But most of them did get those same individual numbers

I suppose when I asked what they got I formulated it in an odd way

Thanks for the help

[granite knoll]

ask them how they did it/

[molten schooner]

Ooooh yeah no of course I get it now

When I asked them they thought I meant how many colas existed in the first place

Which is 54

[cedar kiln]

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[lean sable]

i want to calculate particular integral for the following equation
$2-D^2-D=\cos{2x}$

[wraith dagger]

Hack With Techno Boy

[lean sable]

can anyone help

i currently solved using basic principle of replacing $D^2$ by $-a^2$ and got
$\frac{\cos{2x}}{6-D}$

[wraith dagger]

Hack With Techno Boy

[lean sable]

idk what to do next

[cedar kiln]

@lean sable Has your question been resolved?

[lean sable]

<@&286206848099549185>

[cedar kiln]

@lean sable Has your question been resolved?

[cedar kiln]

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[supple niche]

two integers a, b are relatively prime. if a|c and b|c (divisible) then ab|c. Since they are relatively prime gcd(a,b) = 1, ax + by = 1 and c = aj = bk. However i am not sure how to use that to show c = abz???

[candid coyote]

Hi

@supple niche

[candid coyote]

Oh sorry

ax+by=1 multply by c

[supple niche]

you would get cax + cby = c but that is not a * b

how would u get a * b?

[candid coyote]

Look you can say that ab|cax

Then ab|cby

[supple niche]

how did you get ab|cax?

[candid coyote]

You have b|c then ab|ca then ab|cax

[supple niche]

nvm ok that makes sense

[candid coyote]

Then ab| cax+cby

End

[supple niche]

ok i see

👍

thanks

.close

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[stable sundial]

to find x intercept

[stable sundial]

what if i accidentally using quadratic formula instead of factor?

[tropic oxide]

nothing

it doesn't say what method to use

so you can use any method you want

[stable sundial]

it still show same result?

[tropic oxide]

of course it will

(unless you mess up)

[primal socket]

It's math

[stable sundial]

after factor i got 5,1

but quadratic formula is 12?

[primal socket]

What do you mean, 12?

[stable sundial]

may be i did wrong?

[primal socket]

What did you do after $3 \pm \sqrt{16}$

[wraith dagger]

USS-Enterprise

[primal socket]

How'd you get sqrt(4)

[stable sundial]

4 and then time 3

get 12

[primal socket]

[stable sundial]

-6 divide by -2 i got 3

[jaunty plume]

no

[stable sundial]

factor of 16 is 4 and i take 4 time 3

[primal socket]

huh

First of all.

You are equating everything to each other

We have $x_{1, 2} = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

[wraith dagger]

USS-Enterprise

[primal socket]

This is an equation

To get x1 and x2, the solutions to our quadratic equation

You can't just equate everything to the previous

[stable sundial]

got it

[primal socket]

So anyway, after some work

We get $x_{1, 2} = \frac{-6 \pm \sqrt{16}}{-2}$

[wraith dagger]

USS-Enterprise

[primal socket]

This is correct

But what you do afterwards I don't understand

The square root of 16 is a nice number

[stable sundial]

i still confuse to pull out radical

but now i get it

[primal socket]

What's the sqrt(16)

[stable sundial]

4

[primal socket]

Good, so we get $x_{1, 2} = \frac{-6 \pm 4}{-2}$

[wraith dagger]

USS-Enterprise

[primal socket]

And we take x1 with plus and x2 with minus, or the other way round

[stable sundial]

yes

which is 5 and 1

[primal socket]

And we get our solutions

Yes.

[stable sundial]

thanks for help

[primal socket]

No problem

Just remember

Don't equate everything to the previous

You have equations

Not expressions

[stable sundial]

sure

[primal socket]

Go to a new line with every step

[stable sundial]

.close

[cedar kiln]

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[austere surge]

Good morning,

I am working on a personal project, and to complete it I need help understanding hidden Markov processes. I tried just looking it up but kind of got bogged down in notation. If someone could help me understand what the definition in this Wikipedia article is saying, I would very much appreciate it https://en.wikipedia.org/wiki/Hidden_Markov_model (or if you have a better one, even better!). In the meantime, I'll continue looking over it, and I'll close this channel if I end up working it out. Thanks a ton if you decide to help!

[cedar kiln]

@austere surge Has your question been resolved?

[cedar kiln]

@austere surge Has your question been resolved?

[dire geode]

Get a probability book instead

Or read lecture notes

[austere surge]

actually fair point

.close

[cedar kiln]

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[cedar kiln]

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[vapid harness]

I think im doing something wrong, idk.

The question is to derive this:

[vapid harness]

and then plug in for x = 1

NVM

.close

[cedar kiln]

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[devout matrix]

help

[calm elm]

insert a value for the numerator and the denominator. lets say u pick say 1 (ie change in time), u then go up to the graph line and go across to get a value for lawns mowed. lmk when u figure out what the matching value should be.

[devout matrix]

yeah I guessed randomly and got one correct 😭

thank you alot tho

[calm elm]

fair lol

gl with the rest

[devout matrix]

thank you

.close

[cedar kiln]

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[wintry ingot]

How would you guys begin solving this, I got -70 by subbing in values lol

[upper abyss]

You'll need h'(x). What is it?

[wintry ingot]

It's not given in the q?

Well that's what I'm supposed to calculate

Do I need to derive a formula or something?

[upper abyss]

You'll need to take the derivative of h(x), in order to get h'(x)

[wintry ingot]

Yeah that's the part Im stuck on

Not sure how to differentiate h(x)

[granite knoll]

product and chain rules

[wintry ingot]

.close

[cedar kiln]

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[brittle rock]

Hello! i was wondering if anyone around could help me with some vector rotation math. Im trying to find out how to convert a euler rotation into a quanternion, i have a 3d software open here that i can help visualize what im doing if you would want that to help. im relitively fresh with these concepts so just looking at the equation online didnt make much sense to me.

[upper abyss]

Go for it

[brittle rock]

do you mean the visualise thing?

my software automaticall has a conversion between these two that works, i want to find out the math behind it so that i can use it in my programing

[sleek condor]

did you read about quaternions?

i assume euler angles are pretty straightforward

[brittle rock]

i did a bit. it uses the 4th value with a normalized unit vector for the others

eulers are yea, i just need to find a way to convert a euler into a quanternion, and that im having a harder time finding an easy explanation on

[sleek condor]

you essentially need to find the axis of rotation from euler angles

[brittle rock]

yes i belive, the "w" right?

[sleek condor]

no, the w represents the amount by which you rotate

i think the most straightforward way would be to obtain a rotation matrix first from euler angles

and then convert that to quaternions

but regardless of maths, why do you need euller angles in your code?

they pretty much are bad especially nowadays

the only advantage they have is they use less memory

[brittle rock]

due to the way my software works, when referencing the rotation of a different object in the scene, it only reads its x y z values and has no option to read it as a quaternion. Since i am using another object to reference and influence the roation of another, with the method i currently have i am running into several gimbal lock issues where the rotation is not properly calculated. this would be solved if i could make the euler turn into a quanternion to then influence the rotation

[sleek condor]

oh and are you sure this is euler angles and not euler axis?

[brittle rock]

since its possible for me to make python code and equations i should be able to use whatever formula the software already uses to convert the two between each other,

[brittle rock]

euler angles i belive

[sleek condor]

with which convention?

because euler angles depend on which rotation you do in which order

[brittle rock]

x>y>Z currently

i could change it

but the issues would just change to a diffent angle

[sleek condor]

yeah

but thats important for the translation

what i would suggest if you dont care about performance for your code

use rotation matrices

transform the x axis rotation into a matrix (easy)

then transform the y' axis rotation into a matrix (a bit harder since the rotation is around a new y' not y)

then do the same for z

but its z''

after being transformed by the two rotations

and then multiply those three matrices

[brittle rock]

https://en.wikipedia.org/wiki/Rotation_formalisms_in_three_dimensions#Euler_angles_↔_quaternion

i found this

[sleek condor]

https://en.wikipedia.org/wiki/Conversion_between_quaternions_and_Euler_angles even this

[brittle rock]

yea i cant quite understand what that formula is saying

this is the matrix you were speaking about right?

[sleek condor]

they use tait bryan angles which are slightly different

but yes

but this is the rotation matrix claculated from the quaternion form

[brittle rock]

i see this one

i think i get it, the symbols are just xyz in that order right?

[sleek condor]

no they are the corresponding angles

[brittle rock]

yaw pitch and roll

[sleek condor]

you have to look out for the convention used

its z x z here

[brittle rock]

oh'

[sleek condor]

(euler angles reallly do suck)

you say euler angles

[brittle rock]

lol my b

[sleek condor]

it can mean 25 different things

[brittle rock]

yea it uses specifically XYZ Euler

so thats z at the top of the heirarchy

[sleek condor]

intrisic or extrinsic?

rotation around the original axis?

or modified axis?

[brittle rock]

original

[sleek condor]

so extrisic

[brittle rock]

ok

[sleek condor]

do rotation matrix then its even easier

this is it

each rotation about each axis

is just a matrixc

[brittle rock]

ill do a test euler

[sleek condor]

then you can use the quaternion matrix formula to convert into quaternions if you feel like it

[brittle rock]

x= 25 y=0 z=39

[sleek condor]

25 radians?

[brittle rock]

sorry i should convert it back to radians yea

[sleek condor]

no you dont need

since you put everything into cos

trig functions in numpy gotta have a degree option

[sleek condor]

what am i supposed to do with this?

lol

[brittle rock]

no im just slow and sending my progress

[sleek condor]

i thought you would try to code it

[brittle rock]

im testing the math

coding it would take too much time here with you

no it seems i got it wrong. hm

it wasnt the same when i tested it

[sleek condor]

i mean it isnt that hard, you have to write three functions (one line functions) for the three matrices and you are done

or one big function

still one line

but with a bigger matrix

[brittle rock]

hm

its just not working.

ill figure this out sometime ill let you go rn

i have someone i can chat with

thanks for helping tho!

.close

[cedar kiln]

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[mental quail]

why is the example in the text telling me to take the magnitude to find the acute angle?

in other words, why is it asking me to change the sign from negative to positive?
does it also mean taking only the positive value?

[sly abyss]

Right

So if we keep it negative

One thing first

We know that arctan is restricted to quadrants 1 and 4 correct?

arctan is the inverse of tan if you don’t know

[mental quail]

i guess so. but if i say no i wont get my answer for like another 30 minutes and ive wasted hours and hours on this alreawdy

[sly abyss]

So if you do arctan of the number

You get -1.4117 radians

but it’s asking fo the acute angle

So you just take the absolute value

you can do it either way

But remember

Love is not about finding the right person, it’s about becoming the right person

Do not go looking for the person you want to spend the rest of your life with, become the person you want to spend the rest of your life with.

[mental quail]

does this apply for sin and cos problems as well?

[sly abyss]

Ye most likely it’s a weird and useless question tho

Not for cosine

Actually

[mental quail]

it being weird and usless is subjective...i have to solve it the way they present their problems or i wont get marked correctly

[sly abyss]

I take that back

Cosine is restricted to 1 and 2

[mental quail]

ug

[sly abyss]

So ur calc can give an obshte angle

[mental quail]

nobody knows on first try

[sly abyss]

So you would absolute value of theta

Minus pi/2

[mental quail]

so to find the acute angle of a problem like cos(theta)=-1.234

i would need to..... ?

[sly abyss]

U wouldn’t

That doesn’t work

[mental quail]

imagine a negative number that DOES work

its an example please

[sly abyss]

Ok

Let’s say

Cos = -1/2

Put in ur calc it tells u 2pi/3

Ph wait u don’t need the absolute value oops

[mental quail]

can it be cos = -0.54321

[sly abyss]

Ye

[mental quail]

i need someone who knows how to solve these simple trig problems and not confuse me more and more 😭 please

[sly abyss]

I’m doing it right

It’s just .99654

That’s acute

[mental quail]

okkkk so cos = -0.54321 's acute angle is found how?

[sly abyss]

normal

[mental quail]

what is normal please show me the steps

[sly abyss]

Arccos of the value u gave me

…

[mental quail]

and how it relates to taking the magnitude or not taking the magnitude

so when and why would i use the magnitude of the number and when not to?

[sly abyss]

I don’t know tbh

I don’t know what it’s asking

For

[mental quail]

😦

the acute angle of cos = -0.3178 would be found how?

angle = arc cos (-0.3178) ?

or

angle = arc cos (0.3178) ?

[sly abyss]

Top one

And then do

Pi-that found angle

[mental quail]

doing the top way doesnt lead to the right answer in my problem set 😐

[mental quail]

angle = arc cos (-0.3178) ---> 1.894 radians * 180/pi = 108degrees for an acute angle. acute angles are not greater than 90 degrees

@sly abyss is my math right for that? or is it wrong? its okay to say youre wrong too im just trying to learn

@old ridge if anybody else can please help with understanding this trig acute angle problem please weigh in

[cedar kiln]

@mental quail Has your question been resolved?

[sly abyss]

It is 1.8942

When u plug in ur calc

But that’s onsite

Obtuse

Pi/2 is roughly 1.57

[mental quail]

the question was the acute angle of cos ____ would be found how?

you told me the top option which found an obtuse angle :/

i still dont understand this problem for anything looking