help-36

[supple mantle]

Add parentheses for (5+7*5^...)

Wait

Okay so

Yes your wrote it correctly

[sterile violet]

yeahhhhh

I think it goes like that and

add all of them

and get 0

above

[supple mantle]

Yep

[sterile violet]

0 over n-1

[supple mantle]

Mhm

[sterile violet]

😄

[supple mantle]

[sterile violet]

then I finally got it uu

thank you so much

I couldn't have done it without your help

I was stuck for a lot of time

uu

you're the biggest bird

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[shell monolith]

Im trying to find out how to get the formula (If that’s what it’s called in English), years on the left and value on the right (x,y). It could be an exponential or second degree (hopefully once again that’s how it’s called)

[shell monolith]

Im quite lost !

I mean I could easily do it if it was like 0, 1, 2, 3 but it’s 0, 3, 5??

[final saddle]

@shell monolith Has your question been resolved?

[shell monolith]

Im trying to find the amount of $ in 30 years

[final saddle]

@shell monolith Has your question been resolved?

[shell monolith]

Well well

[final saddle]

@shell monolith Has your question been resolved?

[shell monolith]

no

[final saddle]

@shell monolith Has your question been resolved?

[tranquil pine]

Et bah t'as de la chance que je sois là

Hmmm

,calc (30626/25000)^(1/3)

[soft zealot]

Result:

1.0699991265606

[tranquil pine]

,calc (35064/30626)^(1/2)

[soft zealot]

Result:

1.0700044644644

[tranquil pine]

Ça a l'air très proche d'une exponentielle

Ok, je vais expliquer ce que j'ai fait:

On sait que, si c'est une exponentielle, la valeur monte d'un facteur r (on pourrait utiliser n'importe quelle lettre mais je vais choisir r) chaque année

Donc de l'année 0 à l'année 1 c'est monté d'un facteur de r, puis encore de 1 à 2, puis encore de 2 à 3, ce qui veut dire que de 0 à 3 ça a été multiplié par r*r*r=r^3

Donc f(3)=f(0)*r^3

Si on rentre les valeurs, on obtient 30626=25000*r^3

Et du coup on peut résoudre r et on trouve r=(30626/25000)^(1/3)

Et on peut faire pareil de l'année 3 à 5 et on obtient f(5)=f(3)*r^2. Si on rentre les valeurs pour l'année 5 et 3, on obtient 35064=30626*r^2, et on peut encore une fois résoudre r et trouver r=(35064/30626)^(1/2)

Et au cas où tu ne sais pas ^(1/2) c'est la même chose que la racine carrée et ^(1/3) c'est la même chose que la racine cubique

[tranquil pine]

Du coup on a 2 valeurs de r

Et si elles correspondent alors on sait que ça peut être une exponentielle

[tranquil pine]

Et ça c'est le premier résultat

[tranquil pine]

Et ça le deuxième

Et les deux sont quasiment 1.07

En théorie ça pourrait aussi être un polynomial du second degré mais je pense que c'est pas une coïcidence que les deux résultats correspondent autant

Donc en supposant que c'est bien une exponentielle, on sait que d'une année à l'autre, la valeur augmente de 1.07. Donc la fonction est de la forme f(n)=c*1.07^n

$f(n)=c{1.07^n}$

[soft zealot]

Labyrinth

[tranquil pine]

Maintenant on a plus qu'à trouver c. On sait que à l'année 0 (n=0), la valeur est de 25000, ce qui est pratique car ça nous donne l'équation (n'oublie pas que n'importe quoi autre que 0 puissance 0 ça fait 1) f(0)=c*1.07^0=c=25000

$f(0)=c{1.07^0}=c=25000$

[soft zealot]

Labyrinth

[tranquil pine]

Enfin c=25000 c'est même pas vraiment une équation

C'est tellement pratique ça nous donne la réponse directement

Et donc résultat final: La formule c'est 25000*1.07^n

[tranquil pine]

<@&268886789983436800> Sorry to bother, but it seems this channel is broken

[sturdy ocean]

it should auto-time out

if youre confused by the lack of a rename, its because "boner" tripped our channel name filter

if it doesnt auto-time out later then let us know

[slender locust]

help

[plucky wind]

This defintion isn't correct, is it? Period is the distance between two consecutive maximum points, or two consecutive minimum points (these distances must be equal).

[silk saddle]

Periodstart color #aa87ff, start text, P, e, r, i, o, d, end text, end color #aa87ff
woah what lmao

[plucky wind]

That was HTML code that I didn't intend to bring over, it's fixed now.

[silk saddle]

ah

[plucky wind]

The definition given is not wrong but it's not complete either. Because you could also use the midline as a guide to determine the period of a sinusodial graph. If you were to use the midline to define the period it would be the distance between 3 consecutive midpoints.

[modest birch]

neither of those definitions generalize to anything other than sines/cosines

a period of a function, h is a number such that f(x) = f(x+h) forall x

and then the period is just the least of all of these numbers

[plucky wind]

But thats okay because the definition is in the context of sinusoidal graphics.

I guess I could have listed the category of question.

[final saddle]

@shell monolith Has your question been resolved?

[modest birch]

yeah but that's largely useless, because it doesn't even work well for things like sin x + sin 2x, and we actually probably do want to say something about that too

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[oak flame]

How do i solve this using the quadratic formula

[tranquil pine]

Well, do you know the quadratic equation

[oak flame]

I some what understand it but somehow i keep getting a different answer

[tranquil pine]

Show what you did

[tranquil pine]

,rcw

[tranquil pine]

Your simplification seems weird

[soft zealot]

starlight

[tranquil pine]

Yeah that

[oak flame]

what about the -10 at the start tho

[tranquil pine]

Seems legit

Leave it there lol

[oak flame]

what about the 2.a at the bottom

[tranquil pine]

Seems legit

[oak flame]

?

[tranquil pine]

Like leave it there

It is correct to have the 2×1 in the denominator

[oak flame]

Thanks i got it right

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[normal hill]

$(x+1)y' = y + x^2 + 2x$ solve this ODE

[soft zealot]

Mortta

[normal hill]

I have to use integrating factor

but like im confused on what that is

[mellow cedar]

wdym?

[normal hill]

Like i divide through by x+1, so i get y' = y+x^2+2x / (x+1)

split the fraction and so i get y/(x+1) as my integrating factor

i integrate 1/(x+1) get ln(x+1)

e^ln(x+1) = x+1

so times the equation by that i get the exact same place i started with

[mellow cedar]

and where does y disappears ?

[normal hill]

no clue i just followed my teacher on another question

https://gyazo.com/5fde81fd90589a9758cc68d9399d7813

he did the equation at the top

He just took the y out when he integrated so i did that too

[tulip coyote]

Think your integrating factor should be 1/(x+1) no?

[normal hill]

mhm thats what i used

[normal hill]

.

[tulip coyote]

As then you should have y’ - y/(x+1) = …

[normal hill]

Yea

[tulip coyote]

Then integrate -1/(x+1), get -ln(x+1) but then when you e that you get 1/(x+1) which is what you multiply everything by

[tulip coyote]

Giving you on the LHS here y’/(x+1) - y/(x+1)^2 = …

[normal hill]

?

[tulip coyote]

And the RHS should be (x^2+2x)/(x+1)^2

[mellow cedar]

$\qty(\frac{y}{x+1})' = \frac{y'}{x+1} - \frac{y}{(x+1)²}$

[soft zealot]

Mehdi_Moulati

[normal hill]

AHH i see my mistake thank you guys

legends

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[oak kraken]

Why am I allowed to remove these square brackets at the last step?

[paper zodiac]

where?

[oak kraken]

Light blue square brackets

[paper zodiac]

Aaah

[paper zodiac]

U don't need to but just to look nice. U can put the last one like this

• 6x(2x-3)(bla bla bla)

You can put 6x before (2x-3)

[oak kraken]

Is it the commutative property that says this does not need to be treated as a single factor grouped together?

[tranquil pine]

That thing ultimately only constitutes one term unless you expand the parentheses. Hence the non-necessity of the square brackets

[paper zodiac]

Yeah just to look neater

[paper zodiac]

And they're not really grouped together, they are their own separate factors now

6x and the other are separate factors

[oak kraken]

It looks like I forgot to add a trailing ] on the third step. These square brackets were not even necessary whatsoever to begin with?

[paper zodiac]

yeah, but doesn't mean they're wrong either

Y u ask? Did your teacher graded this?

[oak kraken]

Because multiplication is commutative

[oak kraken]

No I just forget sometimes when brackets are necessary

[paper zodiac]

Nah u can still use this

[oak kraken]

I see the term + term and think OK brackets

But distribution is different

[paper zodiac]

As long as they're gucci

[oak kraken]

Distribution works out to the same regardless of which factor distributes first

[paper zodiac]

nah just the same thing

[oak kraken]

As long as it’s not binomials * binomials I can remove brackets for single term multiplication at any time I think

[paper zodiac]

No I mean even if u add brackets to ur answer or not

[paper zodiac]

yes yes yes

[oak kraken]

OK great

Thank you!

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[fallen reef]

,w derivative 4\sqrt{x}

[soft zealot]

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Requested by dopediscorduser#4906

[fallen reef]

[4 \sqrt{x} = 4(1/2) x ^{1/2 - 1} = 2x^{-3/2} = \frac{2}{x^{3/2}}]

[soft zealot]

dopediscorduser

[fallen reef]

What am I dong wrong?

[amber holly]

1/2 - 1 is -1/2

[fallen reef]

Whoops

Thanks

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[meager mirage]

Hey I have a question about vectors and linear independence

[meager mirage]

Do 3 vectors need to be perpendicular to be considered independent?

[tranquil pine]

no

[amber holly]

No

[meager mirage]

Im struggling with what linear independence means in a 3d space

[tranquil pine]

its not on a plane

then its independent

on 3d space

[meager mirage]

How do you mean?

Plane of the basevectors?

[tranquil pine]

it does not generate a plane

but R^3

[meager mirage]

Oh, So what plane then?

[tranquil pine]

for example this is linearly dependent

because it only generates a plane

[meager mirage]

Oook

And if theres a volume its independent

[tranquil pine]

because the vector can be a linear combination of the other 2

[meager mirage]

If we apply the same to 2 vectors does them generating a plane always mean linear independence

[tranquil pine]

yeah

in general if n vectors generate the R^n then they are a basis and linearly independent

[meager mirage]

So technically 4 vectors cant be linear independent in R3?

[tranquil pine]

no

[meager mirage]

They can?

[tranquil pine]

if V is a n dimensional vector space and ${ v_i }_m$ are vectors in V then they are linearly dependent

[soft zealot]

Jester

[meager mirage]

What does (vi)m mean?

[tranquil pine]

n and m are swapped here but the same thing

[tranquil pine]

its just a notation for ${ v_1, v_2, ..., v_m }$

[soft zealot]

Jester

[meager mirage]

Oooookkkk so they are dependent, I got it now, thank you very much 🙏

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[shut forge]

i wanna find all the solutions to the functional equation f(x^2)=x^4 my first guess is that since f(x^2)=(x^2)^2 f(x)=x^2 but is this the only solution? are there are discontinuous ones or something?

[amber holly]

No, I believe that is it

Something is forcing me to think that f should be defined for nonnegative x, although I clearly see that it can be defined for all reals

[rancid idol]

state the domain

[shut forge]

if the domain is the non negative reals then i can see that since it x^2 is a surjective map i.e. it is an epimorphism then it follows that f(x)=x^2 but i am interested in the solutions where domain is all of R

[final saddle]

@shut forge Has your question been resolved?

[final saddle]

@shut forge Has your question been resolved?

[final saddle]

@shut forge Has your question been resolved?

[desert mantle]

I mean does this give any restriction at all for f(x) with x negative?

[final saddle]

@shut forge Has your question been resolved?

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[mellow wharf]

would anyone mind checking my work? these are practice problems in my book, and the answers are not in the back, so i can't verify if i'd done them correctly, or if i've written them in the proper way

[final saddle]

@mellow wharf Has your question been resolved?

[final saddle]

@mellow wharf Has your question been resolved?

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[paper fractal]

Sorry but (x²-5) simplified

[tranquil pine]

simplified how?

well difference of squares?

i guess?

[paper fractal]

Yes maybe

[tranquil pine]

is this the original question ?

[tranquil flume]

(x+√5)(x-√5) is not simpler

[paper fractal]

That's what I meant ty

What is it called?

[tranquil pine]

difference of

squares

[paper fractal]

Ok ty guys i'm not english

Another question

Can I reshape an algebraically unsolvable equation to be solvable algebraically?

[tranquil pine]

well then wouldnt it not be algebraically unsolvable in the first place ?

[paper fractal]

Maybe in that form yes

But in another form solvable, so you call it algebraically solvable or unsolvable?

Take a look at what I mean

I posted it here earlier but they said it only has a numerical solution

[tranquil pine]

,r

[tranquil pine]

,rotate

[paper fractal]

I'm trying to reshape it to be able to solve it algebraically

[tranquil pine]

that is solvable im not sure what you mean

[paper fractal]

Can you solve it?

[tranquil pine]

hmmm you could graph it

[paper fractal]

without doing so

[tranquil pine]

lemme see

[paper fractal]

I think you can somehow reshape it and use logs

But I don't know exactly how

[tranquil pine]

$e^{2x} = (2x^2-5)$

hm

[soft zealot]

hibyehibye

[tranquil pine]

idk logs arent gonna help

maybe lambert w function somehow?

[paper fractal]

Idk

It's quite tough

Doesn't e follow the rules of logs?

[tranquil pine]

yes

[paper fractal]

So it can be e^2ln(x÷1)?

[tranquil pine]

what

no ?

not sure how u got that, could u show ur steps

[paper fractal]

I took ln for every x

I'm not sure how you say it in english

But if you understand me then alright

[tranquil pine]

then you get

2x = ln(2x^2-5)

which doesn’t help

[paper fractal]

Can't I say 2x+5 = 2 ln(x²)?

[tranquil pine]

no

[paper fractal]

Oh

Why not

[tranquil pine]

no clue what log rule you are using

[paper fractal]

If I take ln for ever x doesn't that mean I isolated it?

[tranquil pine]

you cant just ln for every x

you ln both sides of the equation

[paper fractal]

Oh

So not for every variable?

I got that wrong

[tranquil pine]

what you do to one side you have to do to the other

[paper fractal]

So 2x = 2ln(2x-5) would be correct?

[tranquil pine]

no

ln(x^n) = nln(x)

[paper fractal]

Fk

[paper fractal]

Yes

[tranquil pine]

that isnt x^n

its x^n + c

[paper fractal]

It has to be x^n purely

I see

[tranquil pine]

you cant solve this with just logarithms

[paper fractal]

I can see

But it should be solvable somehow other than graphing it

[tranquil pine]

maybe lambert w function somehow by multiplying by e^-2x

[paper fractal]

Yeah

But I would do that to the other side too?

[paper fractal]

Sorry but what is lambert w function

Nvm i looked it up

It's not within my studies

[tranquil pine]

then just graph it I gues

[paper fractal]

If I manage to solve it can I dm you the solution to make sure it's correct?

Or should I post here

[tranquil pine]

sure

[paper fractal]

Ok then

Ty for your time

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[tranquil pine]

Can you help me with This question?

[tranquil pine]

It’s like a Competition Math Problem That my Friend sent me (Since I do a lot of Competition Math), and I can’t figure it out

hmm

Can you Help?

im thinking

probably to do w/ the chords

Okay

[tranquil pine]

Yeah, So, I think You have to find the Areas then use the Chord Theorem

But, I don’t know where to go after That…

okay I think I got it

whats ur work so far

u calculate the horizontal chord?

I tried do it in my Mind (since I didn’t have paper on me)

[fresh star]

@tranquil pine nice name

[tranquil pine]

But, Um, I think you see that the Lengths of the Squares are obviously 10, 5, and 4

yeah

and then theres a unknown

Then, You use the chords theorem

yes then u have area of red square

Yeah, What after that?

okay so

wait lemme try to draw it

okay it didnt turn out well

i basically just found an equation for the circle

using those lengths

bec u can find perpendicular bisector

Can you try and draw it?

cuz the lines are perpendicular in the first place

I am not following on what you are saying

[tranquil pine]

of the chords

Oh, I See

Okay, Then?

just solve algebraically for r

Wait, So what did you get?

actually no point

you have r^2

so just multiply by pi

What did you get?

[robust mulch]

have you heard of power of a point

[tranquil pine]

well you have to do it i cant just give the answer

[robust mulch]

oh wait you need the radius

[tranquil pine]

Okay, Following what you said, Can I at least ask you if my answer is correct at the end?

ok sure

Ok, Give me a Sec

Okay, Is it 51.25 Pi?

yea

1.5^2 + 7^2

,w 1.5^2 + 7^2

[soft zealot]

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[tranquil pine]

Okay, Thanks for All the Help

np

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[icy plaza]

Having trouble figuring out what I did wrong

[dry light]

$(2 - \sqrt 5)(2 + \sqrt 5) ≠ 4 - \sqrt 5$

[soft zealot]

Umbraleviathan

[fresh star]

are you allowed to use lhopital

[icy plaza]

Whats that?

[fresh star]

wait what

its like the cheatcode

for limits

in that form

[dry light]

They prob don't know derivatives

[dry light]

Anyways this

[fresh star]

oh right

[icy plaza]

We dont

[fresh star]

yh

[dry light]

You cannot cancel factors from terms

That's just not how addition works

[icy plaza]

Im getting pretty confused trying to foil this

Also the 5s are s I'm just bad at drawing larger Ss

[dry light]

(a-b)(a+b) = a^2 - b^2

[final saddle]

@icy plaza Has your question been resolved?

[dry light]

Simplify the denominator

But even then

Wait you don't even need to rationalize

Lmao just factor 4-s using difference of squares

[icy plaza]

I dont know what you mean by that?

[dry light]

Read from here

I realized you don't need to simplify the denominator or even rationalize

[scarlet badger]

rationalize the denominator, you get (4-s)(2+sqrt(s)) on the top, then you can cancel 4-s because (2 - sqrt(s))(2 + sqrt(2)) = 4-s
then replace s with 4, you get 2 + 2, the function approaches 4 when s approaches 4

[dry light]

Or be a genius and realize that 4-s = (2-sqrt(s)) • (2+sqrt(s)) and simplify

You don't even need to rationalize

[scarlet badger]

yeah works either ways

[icy plaza]

Ok I got it

[scarlet badger]

what umbraleviathan says is also correct

[icy plaza]

I got to stuck in sqrt(s)*sqrt(s) = (sqrt(s))^2

Thx

[final saddle]

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[flat lion]

I have three polynomials that form a subspace.
p1(x)=x^3+x
p2(x)=x^2+1
p3(x)=x^2+x

using p1 p2 and p3 i need to get another polynom a(x)=-x^3+5x^2-x+5

how to generate polynom a(x) using p1,p2 and p3?

[royal gust]

We can write the polynomials as coordinates from a basis. Obvious basis, let's use (1, x, x², x³)

You have the vectors
(0,1,0,1), (1,0,1,0), (0,1,1,0)

And you want to find some linear combo of them to write:
(5,-1,5,-1)

[flat lion]

yes i know that. I am lost with how to create the linar combo

[royal gust]

That's the same as solving this block matrix:
[0 1 0 | 5 ]
[1 0 1 | -1]
[0 1 1 | 5 ]
[1 0 0 | -1]

[flat lion]

i know that q* p1(x)+w* p2(x) + r * p3(x) = x(x)

solved that, but like what does that solution mean

[royal gust]

Where your q, w, r are the solutions to it

You can imagine multiplying the left by (q,w,r), and getting the right. Whatever (q,w,r) can satisfy that satisfies the equation you just wrote

[flat lion]

omg thank you

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[sick peak]

when differentiating something let's say there is a function f(x) and i want to find it's derivative; now we can use the derivative definition which is as follows: f(x+a)-f(x)/a (where a tends to 0) now my question was that for any arbitrary function how do we know that "a" always cancels out(might be a dumb question since I haven't formally studied derivatives but was confused about this)

[sick peak]

like since the derivative is always in terms of x; how do we know a is always cancelled out some way or the other

[vital crag]

you don't know. that's when the function is non-differentiable at x

[sick peak]

so when we can't do that it means the function is non-differentiable there?

[vital crag]

yes

you should formalize what "can't" means in that sentence

https://tutorial.math.lamar.edu/classes/calci/onesidedlimits.aspx

[sick peak]

i mean how do I do that?

[sick peak]

okay I'll check it out

[vital crag]

first calculate each one-sided limit, then if they both exist and are equal, then that's by definition differentiable at one point

[sick peak]

oh okay

so if it doesn't follow that definition there is a term of "a" in when using the limit definition there?

[vital crag]

write out using formulas what you mean

and be more specific

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@sick peak Has your question been resolved?

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[kind inlet]

how to integrate x^5(2x^3-3)^2/3?

[kind inlet]

using u substitution

[dusky violet]

well that's the first thing you can do

[kind inlet]

i can substitute the x^3

but im not sure how to get rid of x^2

[fluid sky]

X² will get canceled

[kind inlet]

i mean this

i feel im supposed to express x^3 in terms of x and x^2?

[fluid sky]

So x⁵ will turn to x³

[kind inlet]

yes

[fluid sky]

U can actually take the substitution for the entire 2x³-3

[fluid sky]

And u have a substitution for x³ so put thatthere

[kind inlet]

i got x^3(u)^2/3 / 6

OH u-3/2 ?

[fluid sky]

X³ is (u+3)/2 ?

[kind inlet]

OH

[fluid sky]

Yes

[kind inlet]

ok thanks

that was impossible to spot

[fluid sky]

😄

[kind inlet]

.close

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[rapid chasm]

Can I have help doing $3y^2+40+117=0$ by completing the square

[soft zealot]

MathematicsPractice

[final saddle]

Show your work, and if possible, explain where you are stuck.

[rapid chasm]

ok

[tired walrus]

did you perhaps mean 3y^2 + 40y + 117 = 0?

but yes, show your work regardless.

[rapid chasm]

yes sorry

it becomes a really weird yucky decimal answer and its not meant to

[magic sparrow]

Are you trying to solve for y-values that make the equation true? or are you trying to simplify it by completing the square

[rapid chasm]

trying to solver for y values

its part of a larger question

[teal kayak]

why not just divide by 3 everywhere

[tired walrus]

i think i see the first mistake

[rapid chasm]

the first mistake noooo

[tired walrus]

from $3\paren{y + \frac{40}{6}}^2 = \frac{49}{3}$, you attempted to take the square root of both sides, but incorrectly wrote the right-hand side as $3 \paren{y + \frac{40}{6}}$ instead of $\sqrt{3} \absv{y + \frac{40}{6}}$ as it should have been (or without the absolute value bars -- but then you would want a $\pm$ on the other side.)

[soft zealot]

Ann

[tired walrus]

also, yeah, you're making your life difficult in some places by not simplifying / dividing out

it would've been better to divide out by 3 and then you would not even have to deal with radicals

you would have (y + 20/3)^2 = 49/9 from which you get y + 20/3 = ±7/3, nice and rational

[rapid chasm]

so i needed to root both threes....

[tired walrus]

"needed" in a context-dependent sense.

[rapid chasm]

^3, hwere do i get that

[tired walrus]

ah, sorry. typo

[rapid chasm]

oh algfsd

[rapid chasm]

but what happens to the 3 in front of (y^2+40/6

[tired walrus]

there is no (y^2 + 40/6

not only did you put the ^2 in the wrong place but you also forgot a closing parenthesis

[rapid chasm]

sorry

[tired walrus]

and

it would've been better to divide out by 3
which is what i did

[rapid chasm]

Tthen i would end up getting y/3?

[tired walrus]

??

[rapid chasm]

Im so confused rn

what you did

[tired walrus]

okay then maybe i should write it out on paper, give me a minute

[rapid chasm]

its okay

Dont hassle yourself

I will do it eventually

[tired walrus]

i will go an alternative route which i think is less messy

[rapid chasm]

i think i got the right answers in the end

thanks so much

.close

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[tawny flame]

I'm a little unfamiliar with how to answer this
(ping me btw)

[tranquil pine]

what is the standard equation of the line that you know of

[tawny flame]

As in y = mx + c?

[tranquil pine]

yes

so you want to find what m and c are

do you know what m and c are?

[tawny flame]

m is the gradient and c is... the constant?

[tranquil pine]

the constant of what

also they are both constants

[tawny flame]

y intercept??????????

[tranquil pine]

yeah

[tawny flame]

oh

[tranquil pine]

yeah so from the picture, the y intercept should be clear right?

[tawny flame]

just guessed 😅

[tawny flame]

3

[tranquil pine]

yes

so you know that y = 3

aka c = 3

thats one out of the way

one more to go

the gradient, do you know the formula to calculating the gradient?

[tawny flame]

rise/run :D

[tranquil pine]

yep

[tawny flame]

wait brb

k back
uh wait is the gradient 8/4

no wait o

[tranquil pine]

no

[tawny flame]

therefore gradient is 2

[tranquil pine]

yes

[tawny flame]

forgot to subtract 3 sorry

so y = 2x + 3

[tranquil pine]

yes

[tawny flame]

wait is it alright if i ask another question with a similar concept

[tranquil pine]

yeah

although it would be best if someone else took over, i have to go for ab it

[tawny flame]

oops

wait no i got the other q alg

thx for your help :)

.close

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[tawny flame]

K i thought I had a lead but I have no idea how to find c

[tranquil pine]

substitute in some value of y and x that you know of

like (2,5)

and solve for c

[tawny flame]

wha-

[tranquil pine]

you know that y = -3/2 x + c

substitute in some values for y and x such as (2,5)

and solve for c

[tawny flame]

like 5 = (-3/2)*2 + c or is that not what you meant

[tranquil pine]

yeah

exactly

[tawny flame]

so 5 = -6/2 + c is it

[tranquil pine]

wot

[tawny flame]

confused

[tranquil pine]

that sentence did not make sense

[tawny flame]

im confusing myself

💀

[tranquil pine]

but yes it is 5 = -6/2 +c

[tawny flame]

OH I JUST REALISED HOW I PHRASED THAT SENTENCE LMAO SORRY

[tawny flame]

where am i supposed to go from there-?

[tranquil pine]

basically solve for c

i.e. make it be the only thing on one side

[tawny flame]

uh.. 😅

make c the subject/transpose?

[tranquil pine]

i...sure

although i never heard someone say it as "make it be the transpose"

[tawny flame]

💀

5 - (-6/2) = c?

[tranquil pine]

yes

[tawny flame]

5 - (-3) = c
5 + 3 = c
c = 8

[tranquil pine]

yes

woo

good job

[tawny flame]

ty 😭

y = -3/2x + 8 😭

omds

ok i got it thanksss

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[tranquil pine]

yepp

[final saddle]

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[soft zealot]

Snowen

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[white rock]

how do i do this

[fresh star]

you litreally cropped out

the question

[white rock]

mb

[fresh star]

it says the direction b, so i assume its the projection on the unit vector that is in the direction b.

so starters

write those A B C D options

as the unit vectors

[white rock]

2i+3j/sqrt13

3i+2j/sqrt13

3i+4j/5

4i+3j/5

[fresh star]

now

do the dot product

with each of those vectors

to the i + 4j

given

and see which one gives you 11/sqrt13

you can already tell it wont be C or D

[white rock]

14,11,19,13

B?

[fresh star]

C and D are 19/5 and 13/5

[fresh star]

yh

[white rock]

ok thanks

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[white rock]

.close

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[white rock]

idk how to do this

[fresh star]

have you learnt chain rule

well this isnt specifically chain rule itself

but the application of it

other way round

not the application

but the chain rule itself

dy/dx = dy/dt * dt/dx

[tranquil pine]

remember that
[
\dv[y]{x} = \f{\ff{dy}{dt}}{\ff{dx}{dt}}
]

[white rock]

ok

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[distant spindle]

I'm really confused with no.7

[distant spindle]

I have no idea how to do ut

[fluid sky]

If a line is tangent to a circle what must be the distance from the center of the circle to the line

[distant spindle]

Wdym? The radius?

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[weary lantern]

Does a derived function when sketched look like the flipped version of its non-derived self?

[weary lantern]

(I'm doing these questions on graph sketching)

or rather, like this?

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[hushed ingot]

.open

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[tranquil pine]

FUCK

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[frosty musk]

if N a normal subgroup of G, for any a in G, show that (Na)^(gcd(ord(a)),ord(Na))=N
any hints?
#groups-rings-fields seems busy so reposting here

[final saddle]

@frosty musk Has your question been resolved?

[tranquil pine]

hint: NaNa=Na^2

[hearty zephyr]

doesn't ord(Na) divide ord(a)?

[desert mantle]

I mean that's essentially what you have to show

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[fossil zodiac]

okey wai

[fossil zodiac]

t

yu have this shjt

wait

i have done this

im tired ok

maybe $ln(2) + 7 * ln(z) + 3 *ln(y) - ln(5) + ln (x)$

[soft zealot]

Lupin

[vital crag]

what's the question

simplify?

[fossil zodiac]

Decompose the following terms into a representation with "simplest numerals", i.e. without products,
quotients, powers and roots within the logarithm!

[fossil zodiac]

re u from germany?

[vital crag]

use this

first split the product

[fossil zodiac]

is that wrong at all?

[vital crag]

just hard to follow

do one step at a time

[fossil zodiac]

i didnt get the solution

wait

where the 11 comes from

lol

[vital crag]

no point in jumping ahead

just simplify one step at a time

[vital crag]

using this

[fossil zodiac]

okey lets do it together if u want: my first step would be this: 3/2 * ln(x)

then + 5 * ln(y)

[vital crag]

.

thing under square root is m and everything else is n

[vital crag]

using the product rule here

[fossil zodiac]

Okey. If we look at the root and the other stuffs we see a multiplication, so its m + n. Is it that, what you meant? Under the root we can do: x + y. at the other side, we see z+y...

[vital crag]

i don't know what you're saying

[fossil zodiac]

i am too

[vital crag]

log(mn) = log(m) + log(n)

[fossil zodiac]

yes i know

[vital crag]

find what m and n are

[fossil zodiac]

i know allt the four

[vital crag]

i explained it here

m includes the square root

[fossil zodiac]

Under the root you see x * y = ln(x) + ln(y)

bro we cant go further like that.. You have to say more idk

[vital crag]

no

[fossil zodiac]

i need that

[vital crag]

you're skipping too many steps

[fossil zodiac]

name one before

only one

[vital crag]

get the first step

[fossil zodiac]

tell me

My way how I did that on all the other task, which works except this one: $(3/2 * ln(x) + 5/2 ln(y) - ln(x) - ln(y)) + ( ln(2) + 7 * ln(z) + 3 * ln(y) - ln (5) + ln(x)$

[soft zealot]

Lupin

[fossil zodiac]

my head hurts, beats explods etc etc

i do it tomorrow again alone lol

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[vital crag]

but yea this is wrong on a lot of levels

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[coral ravine]

hello how do i solve this limit

[coral ravine]

x and y are vectors in a bounded domain in R^2 and R is a scalar

it is (in theory) log R

i cannot figure out how to put it into wolfram

x/|x| is unit vector so that shouldn't contribute to that term in the absolute value, but aside from that...

if it were parenthesis I suppose I could distribute the |x|/R and get R - |x|y/R which would go to R in the limit

is this legal

wait, it is if R is positive isnt it

im a genius

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[vital crag]

.coose

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[crude dust]

Any hint on how to solve this?

[warm ether]

for the cos^2 you can try rearranging the double angle formula for cos(2x). for the 6sinxcosx you can use the double angle formula for sin(2x)

the sinx doesnt really need any work to be done on it

[crude dust]

The cos^2 is multipled by 9 so should I subtract 9?

[warm ether]

wdym?

[crude dust]

Its 9cos^2-sin^2x

So should I subtract 9?

So I can make it cos2x

[warm ether]

no, temporarily ignore the 9

just focus on the cos^2(x)

[crude dust]

Okay

Should be like this right?

[warm ether]

no

cos^2(x) does not equal cos 2x

[crude dust]

Wait I made a mistake writing the question

[warm ether]

$cos(2x)=2cos^2x-1$

[soft zealot]

AℤØ

[crude dust]

This is the question im sorry

I forgot the power of 2 to the sinx

[warm ether]

ah

its not a big deal, use sin^2+cos^2=1 to make the sin^2 into 1-cos^2 then simplify

after that you just do what we were doing before

[crude dust]

Like this right?

And cos^2x =(1+cos2x) or =1/2(1+cos2x)?

[warm ether]

looks alright so far. And its the second one

[crude dust]

This should be the answer right?

I can simplify it more but ima leave it be

[warm ether]

looks alright

[turbid cypress]

hahya

what is 1 +1

[modest birch]

two

[final saddle]

@crude dust Has your question been resolved?

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[snow harbor]

hii

[snow harbor]

I need help with an issue

[calm junco]

go ahead

[tranquil pine]

,rccw

[versed crater]

Have you tried anything

[snow harbor]

I tried but I didn't understand anything

I only made it to the first fraction

1/16^-0,75

[calm junco]

so that is 16^3/4 right

[snow harbor]

yes

[calm junco]

right

then you need to find x^4 = 16

what is x in that case

[calm junco]

@snow harbor i'll explain do this first

[snow harbor]

this part i did

took the fourth root of 1/4096

[calm junco]

no no no

[snow harbor]

I'm using Google translator so if something is wrong you already know

[calm junco]

its alright

[calm junco]

so what is x in here

2 right

[snow harbor]

but why 2

1 minute I'll try to do it from the beginning

[calm junco]

so for example
when solving 16^3/4
you need to first find x^4 = 16 after that you'll do 2^3

what do you get afterwards

[snow harbor]

can you start from the beginning please? I don't understand

25^1/2

[calm junco]

okay

[snow harbor]

=

5 yes?

[calm junco]

yes

you take the square root

when its 1/2

i will send an image so you can see how to do these

[snow harbor]

ok

thanks

[calm junco]

hold on

[snow harbor]

ok

[calm junco]

does this explain a bit

hello?

[snow harbor]

like this?

ops

wrong picture

like this?

[calm junco]

yes but you forgot that 0,75 has a - before it

so you need to change 1/16 to 16

[calm junco]

yes

now what does that make up to

[snow harbor]

16x16x16?

[calm junco]

so you see the 4 in there?

[snow harbor]

yes

[calm junco]

that means there is a number, lets say a, a^4 equals 16

if it was 3 it would be a^3 = 16

so if a^4 = 16 what is a?

[snow harbor]

a=16 right?

[calm junco]

no but a x a x a x a = 16

then what is a

[snow harbor]

aaaaaaa

2

[calm junco]

yes!

then after you find that

[snow harbor]

Now I'm understanding

[calm junco]

you need to do 2^3

because theres a 3

what does that equal to

[snow harbor]

8

[calm junco]

yes

5-8

[snow harbor]

-3

[calm junco]

yes! now we just need to do the rest

can you do them as i explained or do you want me to go through

[snow harbor]

my problem is (0,001)^-1/3

[calm junco]

okay

now that is (1000)^1/3 right?

[snow harbor]

root 3?

3 _____
/1000

[calm junco]

i just changed 1/1000^-1/3 to 1000^1/3

could you follow until there?

[snow harbor]

ok

1 minute

[calm junco]

,rccw

[calm junco]

yes you are doing well

now can you tell me what this equals to

[snow harbor]

that

[calm junco]

yes but thats a^3 = 1000 right

[snow harbor]

yes

[calm junco]

then what is a

[snow harbor]

10

[calm junco]

yess

now do the last one

[snow harbor]

10^3 isn't it 10000?

[calm junco]

no its 1000

[snow harbor]

oh

ok

[calm junco]

you need to add three 0s

dont mind about the already existing 0 in 10

[snow harbor]

ok

[calm junco]

now the last one

[snow harbor]

16/1^1/2

[calm junco]

yes

can you solve that now

[snow harbor]

-2?

[calm junco]

nope

see you need to do this again

a^2 = 16

a x a = 16

[snow harbor]

4

[calm junco]

yeah

now you can solve the whole thing

[snow harbor]

-3+10-4

[calm junco]

just remember this

and youre good

[snow harbor]

ohhhh

THANKS

I spent 1 hour and a half trying to do this.

[calm junco]

no worries

[snow harbor]

but a question

[calm junco]

sure

[snow harbor]

I have another question in doubt

(125/343)^1/3

[calm junco]

okay so

you need to first find 125^1/3

and then

343^1/3

then divide the results

[snow harbor]

ok

1 minute

a x a x a = 125

right?

a=5

[calm junco]

yes

[snow harbor]

Is there any way to find out without guessing?

343

[calm junco]

not really but heres a strategy

5^3 is 125 then its higher than that just try from there
6^3 = 216
7^3 = 343

[snow harbor]

understood

[calm junco]

whats the answer now

[snow harbor]

5/7?

[calm junco]

yep

you got it

[snow harbor]

omg

Very thanks

Very Very Very thanks

[calm junco]

no problem

[snow harbor]

it was just that

thank you very much

[calm junco]

😌

[final saddle]

@snow harbor Has your question been resolved?

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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>.
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• Be polite and have a nice day!

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[fervent ferry]

Hey, I'm doing a physics problem.

[fervent ferry]

A baseball hit vertically towards the sky stays 6,25s in the air. Determine the distance that separates the contact point from its maximum height

I don't know what my three variables would be to find the distance

I know acceleration is -9.8m/s

and time is 6.25s

[modest birch]

I'm assuming the contact point is on the ground

then it reaches its maximum height at 6.25s/2

[final saddle]

@fervent ferry Has your question been resolved?

[fervent ferry]

contact point on the ground ;-;

but then what are you doing when you say 6.25s/2

what's the thought process?

(This is the first few weeks of 11th grade physics)

[final saddle]

@fervent ferry Has your question been resolved?

[final saddle]

@fervent ferry Has your question been resolved?

[final saddle]

Channel closed

Closed due to timeout

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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.

[languid adder]

hey i have a question

[languid adder]

so i have a candy bar that is 11 cm width and 9 cm length

i have a larger candy bar that has a scale factor of 4 which means the sides are 44 cm wide and 36 cm length

for the nutrition label

do i just multiply everything by 4

like if that small sized candy bar had 85 calories

do i mutiply by 4 to find the calories of the larger bar

anyone?

[ionic tulip]

is it a 2d candy bar?

[languid adder]

no drawing

why does that matter

[ionic tulip]

well if it was 3d then the calories would increase based on the volume of the candy bar

[languid adder]

nah its a drawing

so look

[ionic tulip]

if its just 2d then yea you would just multiply by 4