help-0

[ocean seal]

Enemagneto

[near apex]

You know that $(5+nx)^2 = (5)^{2} + (nx)^{2} + 2\cdot (5)\cdot (nx)$.

[ocean seal]

Enemagneto

[dense dove]

And they've written $\left(1 + \frac{3x}{5}\right)^{n}$ in sigma notation?

[ocean seal]

windowsncheese

[dense dove]

Well sigma + binomial

[near apex]

Yeah. In the answer that is given? yes

[dense dove]

Ah okay that is what was confusing me a bit I didn't realize it was just an expression of $\left(1 + \frac{3x}{5}\right)^{n}$

[ocean seal]

windowsncheese

[near apex]

No. It's the same thing written in compact summation form.

[dense dove]

How do they get from this

To this

[near apex]

Alright.

[near apex]

Firstly, you must know this:
(a+b+c) (m+n+ ... z)
When we open the brackets, a will be multiplied with each term inside the second bracket + b will be multiplied with each term inside the second bracket + c will be multiplied with each term inside the second bracket.

Brackets open like that. Do you know that?

@dense dove

[dense dove]

Yes I know that

[near apex]

I have to leave soon. Please, be present.

[near apex]

Cool

Now, let's consider 25 from first bracket.

When we multiply it with the terms in second bracket, we'll only get one term of x.

[dense dove]

From the 3x/5?

[near apex]

Term of x means that the term has x only with a coefficient.

[near apex]

Yes, but will the term be that?

[dense dove]

No because you have to multiply it I think

[near apex]

Wait. Do you understand what that summation notation represent ?

$\sum_{k=0}^{n} (n\choose k) \left(\frac{3x}{5}\right)^{k}$

[dense dove]

I think so, it means for every value, starting at k= 0, until n it will iterate through the next two sets of brackets with each increasing value of n until n is reached

i wrote that really weirdly

but it's basically just a sum of all the terms

[near apex]

Can you write the expanded expression on your notebook and show?

[dense dove]

How can I do that without knowing what n is?

[near apex]

In terms of n.

$\sum_{k=0}^{n} {{n\choose k} \left(\frac{3x}{5}\right)^{k}}$

[ocean seal]

Enemagneto

[near apex]

That when expanded looks like this:

$${n\choose 0} \left(\frac{3x}{5}\right)^{0}} + {n\choose 1} \left(\frac{3x}{5}\right)^{1}} + \cdots + {n\choose n} \left(\frac{3x}{5}\right)^{n}}$$

[ocean seal]

Enemagneto
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[near apex]

Get it?

[dense dove]

Ah I see, yeah I'm normally fine with binomial and know that. This is just the first time i've had to use n and seen summation notation in a question like this

[near apex]

Okay. Now, tell me the constant term in this

[dense dove]

The constant term would be the first one because you're putting 3x/5^0 making it just 1

[near apex]

Yes

[dense dove]

Then multiplying it by whatever (n,0) equates to on pascals triangle

[near apex]

Now, tell me the term of x.

[dense dove]

The one just after it

[near apex]

Good.

[dense dove]

Well (n,0) would always be 1 right?

So the constant terms of the expansion of $\left(1 + \frac{3x}{5}\right)^{n}$ is 1

[ocean seal]

windowsncheese

[dense dove]

term*

[near apex]

Now, when we multiply this expanded form with 25 from the first bracket: when we multiply with the first term, we'll only get a constant since 25 * 1 is just 25.
when we multiply with the second term, we'll only get a term of x, since 25 is constant so product will only have x from the second term of expanded form of second bracket.
When we multiply with all the other terms after this, we'll get a higher power of x like x^2, x^3, and so on until x^n. We can ignore these higher order terms since we are only looking for terms of x.

[dense dove]

Ok that makes sense

[near apex]

So, we get the term of x when multiplying with 25 as -
$$ 25 {n\choose 1} \left(\frac{3x}{5}\right)$$

[ocean seal]

Enemagneto

[near apex]

Get it?

[dense dove]

Yeah

And we know that = 100x

[near apex]

No.

Wait.

[near apex]

It's not.

[dense dove]

Oh

why not

[near apex]

We have more terms remaining in the first bracket. you have to multiply with them too. You might get terms of x there as well.

[dense dove]

Oh right of course

like "10nx"

[near apex]

Now, we see with 10nx.

Which term from right bracket expanded form will give term of x when multiplied with 10nx ?

[dense dove]

The constant term

first one

[near apex]

Yes.

So, we get:
$$ (10nx)\cdot {n \choose 0} \left(\frac{3x}{5}\right)^{0}}$$

[ocean seal]

Enemagneto
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[near apex]

Which is just:
$$ (10nx)\cdot {n \choose 0}$$

[ocean seal]

Enemagneto

[dense dove]

Which is just 10nx no?

[near apex]

Which is:
$$ (10nx)\cdot 1$$

[ocean seal]

Enemagneto

[near apex]

Which is just 10nx.

[near apex]

So far now, you have had two terms of x.

[near apex]

This

[near apex]

And this.

Now, let's see for $n^2x^2$.

[ocean seal]

Enemagneto

[near apex]

Which term from right bracket expanded form will give term of x when multiplied with $n^2x^2$ ?

[ocean seal]

Enemagneto

[dense dove]

It doesn't exist does it?

it'd have to be a negative exponent

but we know k=0

[near apex]

yes

[dense dove]

so can we ignore them?

[near apex]

So, we don't get any terms of x from that.

[near apex]

What do you think? Do they contribute to coefficient of x?

[dense dove]

They don't, and we are looking for the coefficent of x, so we can ignore them.

[near apex]

Yes

So, we have:
$$ 25 {n\choose 1} \left(\frac{3x}{5}\right) + (10nx) = 100x$$

[ocean seal]

Enemagneto

[near apex]

Solve this. Easy peasy.

${n\choose 1} = n$

[ocean seal]

Enemagneto

[dense dove]

yeah i was about to say so

[near apex]

Now, you can solve. Right?

[dense dove]

15nx + 10nx = 100x

[near apex]

Yes

[dense dove]

25n = 100

n = 4

[near apex]

Cool. Really gotta go now.

Good day.

[dense dove]

Thanks so much, you were really helpful and patient with me 🙏🙏🙏

[lone heart]

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[warped topaz]

https://i.imgur.com/NiZvRsg.png

[warped topaz]

Question b

For this its just the same method as a solid of revolution but I dont square it, right?

And dont multiply by pi at the end

[lone heart]

@warped topaz Has your question been resolved?

[warped topaz]

<@&286206848099549185>

[crimson hearth]

what?

[warped topaz]

Like to get the solid of revolution you do this right

https://i.imgur.com/ptqQJlR.png

Except you square the x first

So for this question you do the same, but you dont square the x

[crimson hearth]

my brain is too small for that

[warped topaz]

You just do this, right?

[crimson hearth]

but i can try

[warped topaz]

No like im asking

Is this the method

[crimson hearth]

oh

[crimson hearth]

i think its right

i will double check

[warped topaz]

I also think its right but im not 100% haha, thats why I created this thread to ask

can maybe another helper confirm?

[crimson hearth]

<@&286206848099549185>

[frigid mirage]

isnt it just asking for this

[warped topaz]

yeye

[warped topaz]

Is this the method?

Same as a solid of revolution except you dont square it

[frigid mirage]

I don't think the pi should be there, this doesn't feel any different from a normal definite integral from 0 to 1

[warped topaz]

Ah yeye sry I mentioned that before just forgot to reapeat it lol

But other than that its correct right?

[frigid mirage]

$ \int_{0}^{1} (1-x)e^{-x} dx $

I mean

why isnt it the integral of f(x) dx from 0 to 1

[warped topaz]

It is right?

[frigid mirage]

if im misunderstanding something im sorry lol

this has a x here

[warped topaz]

Just a random image i pulled from google

I just want to know if my method is correct

[frigid mirage]

$\int_{0}^{1} (1-x)e^{-x} dx$

[ocean seal]

nebula40

[warped topaz]

Yes

[frigid mirage]

yeah I think this makes sense

[warped topaz]

okok good

Thank you!

❤️

.close

[lone heart]

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[wheat isle]

how can I make a table for part c?

[solemn juniper]

Why do you need a table?

[wheat isle]

to solve it?

[solemn juniper]

How exactly have you been taught to solve expected value problems?

Don't say "with a table"

[wheat isle]

well i know how to find expected values when im provided the data distribution in a way thats easy for me to understand like an expression or a piecewise looking thing but word problems makes it confusing for me

[solemn juniper]

Well, you're given that there are three children, and we tend to assume that males and females are equally likely to be born (so 50% chance)

Does that sound like a distribution you've heard of?

[wheat isle]

yeah the heads and tails thing is similar right

[solemn juniper]

Pretty much

[wheat isle]

so the outcome x would be either

0,1,2,3 girls

?

because those are the possibilities of getting however many number of girls in a 3 child family right

[solemn juniper]

Yes

[wheat isle]

so 0 and 1 girls would have the same probability of 1/2

what about 2 and 3? is that 1/2 times 1/2 and 1/2 times 1/2 times 1/2

[solemn juniper]

Careful

0 and 1 don't both have probability 1/2

Have you heard of the binomial distribution?

[wheat isle]

if you bring up what it looks like it might ring a bell

[solemn juniper]

Well you have n independent trials, each with probability p of landing on a certain outcome (girl)

So for there to be 0 girls, we would need all three children to be boys. Given that the probability of getting a single boy is 1/2, the probability of getting 3 must be...

[wheat isle]

1/8?

[solemn juniper]

Quite so

[wheat isle]

okay

and for 1 girl that would mean

getting 2 boys so the probability of getting one girl is 1/4

[solemn juniper]

No

Getting one girl means that you get one girl and two boys

[wheat isle]

so what would that probability look like

[solemn juniper]

Well, that situation describes getting either BBG, BGB, or GBB

Each such case carries what probability? (Two boys and one girl)

[wheat isle]

so isnt the probability of getting two boys 1/2 * 1/2

[solemn juniper]

It is

But the third child can be either a boy or girl

[wheat isle]

So if the probability of getting 0 girls is 1/8 how do we account for that

[solemn juniper]

Well in that case, the only arrangement of the three children is BBB

So indeed it's as simple as just 1/2*1/2*1/2

[wheat isle]

Yeah

[solemn juniper]

This is similar, though it requires one more step

[wheat isle]

Hm

not sure

(Also I gotta eat dinner BRB if you’re still on later that would be great)

[solemn juniper]

,ti

[ocean seal]

The current time for steakanator is 02:07 AM (PDT) on Wed, 23/08/2023.

[solemn juniper]

Probably not

[stone sphinx]

,ti

[ocean seal]

You haven't set your timezone! Set it using the interactive timezone picker with ,ti --set.

[solemn juniper]

I suggest looking up the binomial distribution to help you understand this scenario

[lone heart]

@wheat isle Has your question been resolved?

[wheat isle]

Alright

[lone heart]

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[bleak citrus]

I got x = - 7/4 but im not sure if its correct

I made 2 the same denominator as 4x/7

then I subtracted 14 from both sides

then divided both sides by 4

[alpine sable]

thats correct

[bleak citrus]

alr

thanks

i was js checking

.close

[lone heart]

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[modest wraith]

I dont know how to do that

[paper mango]

The table?

[pallid scarab]

It's frenching time

Tu parles de l'exercice de remplir le tableau ?

[modest wraith]

ooeeee mon garssssssss

its french timee mdrrrrr

yeah bro its the table

[weary wyvern]

(f o g)(x) = f(g(x))

[pallid scarab]

Ok donc tu vois comment appliquer la composition ?

Quand t'as une composition de plusieurs fonctions à la suite, tu commences par la fonction la plus à droite, puis tu fais de droite à gauche

Je te fais l'exemple de la première, h○g○f(x)

[weary wyvern]

\circ

[pallid scarab]

Too bad latex is not compatible with this

(The ○)

[weary wyvern]

$f \circ g$

[ocean seal]

giannis_money

[pallid scarab]

Yes I know

I was talking about my symbol

I'm lazy 😂

Anyways

Donc quand t'as une suite de compositions, t'écris h○g○f(x) = h[g(f(x))]

Pour la question 1 donc, on commence par f qui est le plus à droite

Donc on remplace f(x) par son expression, cos x

h○g○f(x) = h[g(cos x)]

On continue donc avec g, mais il faut faire gaffe un peu

Comme on cherche g(cos x) et qu'on connait g(x) = x² +1

Chaque fois que "x" apparaît il faut le remplacer par cos x comme c'est notre entrée

Donc g(cos x) = (cos x)² + 1

Tu vois comment ça marche ?

[lone heart]

@modest wraith Has your question been resolved?

[modest wraith]

ca va donner racine de (cos x)² + 1 ?

@pallid scarab

[pallid scarab]

Oui c'est ça pardon @modest wraith

[modest wraith]

mhh kk lets me try the other oneeeeee

[pallid scarab]

$h\circ g\circ f(x) = \sqrt{(cos(x))² + 1}$

[ocean seal]

rafilou2003

[modest wraith]

ca donne (racine carre cos x)^2 +1 ?

@pallid scarab

et le dernier cos racine de x^2 + 1

et tu peux faire le 2 du tbl stp

cooach tes passé ou?

[pallid scarab]

Pardon je suis à l'ouest

J'ai mis ce serveur en mute, essaye plutôt de me ping en MP

[pallid scarab]

Le deuxième c'est bon

Tu peux même simplifier en cos(x) + 1 mais on s'en fiche

[pallid scarab]

J'arrive pas à comprendre celui là

C'est peut-être bon mais la façon dont tu l'as écrit me perturbe

Mets des parenthèses

[modest wraith]

att je te send picture

[pallid scarab]

C'est pas ça :(

Recommence avec juste f○h(x)

[modest wraith]

cos racine de x?

[pallid scarab]

Oui

$f\circ h(x) = cos(\sqrt{x})$

[ocean seal]

rafilou2003

[pallid scarab]

Et donc quand t'appliques g à ça, qu'est-ce que tu obtiens ?

[modest wraith]

japplique ca au x de h?

ou alors ca donne (cos (racine de x))^2 +1

@pallid scarab

[pallid scarab]

C'est mieux !

Alors le 2)

[modest wraith]

fais le toi

att jai une question

genre

si on commence par h

ensuite et que ya encore 2etapes

on fait ca ? g(h) puis f(g ((h))

genre f sapplique pour g et h ?

[pallid scarab]

f s'applique pour g(h(x))

Je te fais un exemple de comment ça aurait pu se passer autrement

pour $g\circ f \circ h(x)$

[ocean seal]

rafilou2003

[pallid scarab]

j'aurais pu écrire que c'est g(f(h(x)))

et ensuite j'applique g à f(h(x))

ça me donne $f(h(x))^2 + 1$

[ocean seal]

rafilou2003

[pallid scarab]

ensuite j'applique f à h(x)

ça me donne $(cos(h(x)))^2 + 1$

[ocean seal]

rafilou2003

[pallid scarab]

et enfin je remplace h(x) par son expression

$(cos(\sqrt{x}))^2+1$

[ocean seal]

rafilou2003

[pallid scarab]

@modest wraith ça dépend de la méthode que tu préfères

[modest wraith]

daccord

[pallid scarab]

Si tu veux je te montre f rond f pour la 2)

$f\circ f(x) = f(f(x))$

[ocean seal]

rafilou2003

[pallid scarab]

on remplace f(x) par son expression :

$f(f(x)) = f\Big(\frac{1}{1-x}\Big)$

[ocean seal]

rafilou2003

[pallid scarab]

et maintenant faut appliquer f à ce machin entre parenthèses

donc $f(f(x)) = \frac{1}{1-\Big(\frac{1}{1-x}\Big)}$

[ocean seal]

rafilou2003

[pallid scarab]

Là il va falloir simplifier si on veut continuer

[pallid scarab]

@modest wraith tu penses pouvoir simplifier ça ? C'est assez facile mais il ne faut pas se tromper avec les signes -

[modest wraith]

@pallid scarab non j'ai aucune idée

[pallid scarab]

multiplie le numérateur et le dénominateur par (1-x)

[modest wraith]

Ahh bah oui

Enfin nn

Tu sais l'écrire pour que je puisse visualiser

[pallid scarab]

$\frac{1}{1-\frac{1}{1-x}} = \frac{1-x}{(1-x)\Big(1-\frac{1}{1-x}\Big)}$

[ocean seal]

rafilou2003

[modest wraith]

Et du coup ça ns donné 1- 1/1-x ?

[pallid scarab]

ça va pas nous donner ça non

je te le fais :

[solemn axle]

Infintely many solutions

[pallid scarab]

bruv doin sarcasm and I replied with sarcasm that would not be considered as sarcasm

$\frac{1-x}{(1-x)\Big(1-\frac{1}{1-x}\Big)} = \frac{1-x}{(1-x)-\frac{1-x}{1-x}} = \frac{1-x}{1-x-1}$

[ocean seal]

rafilou2003

[pallid scarab]

tu vois comment finir la simplification @modest wraith ?

[modest wraith]

Nn

[pallid scarab]

1-x-1 =?

[modest wraith]

2

[pallid scarab]

euh

1-1 = ?

[modest wraith]

0

[pallid scarab]

oui

[modest wraith]

Mais j'ai repris -

Le x

Oe lstmb mdrr

[pallid scarab]

ok xd

1-x-1 = -x

donc on peut simplifier en $\frac{1-x}{-x} = \frac{x-1}{x}$

[ocean seal]

rafilou2003

[pallid scarab]

et là c'est bon

[modest wraith]

D'accord je vois

[pallid scarab]

en fait c'est même mieux si on l'écrit comme $1-\frac{1}{x}$

[ocean seal]

rafilou2003

[modest wraith]

Ah d'accord

[pallid scarab]

mais là y a plus rien à faire

prêt pour $f\circ f\circ f(x)$?

[ocean seal]

rafilou2003

[modest wraith]

Vsy att

[pallid scarab]

alors indice : on a pas calculé $f\circ f(x)$ avant pour rien

[ocean seal]

rafilou2003

[pallid scarab]

aie bonne chance

ce que tu as écrit est correct

mais pour le simplifier bonne chance mec

[modest wraith]

Ah

[pallid scarab]

je remets ça

[modest wraith]

Kk

[pallid scarab]

oui !

et maintenant essaie de simplifier ça

[pallid scarab]

euh

1- (1-(1/x)) = 1/x

je vois pas comment tu trouves 0/x au dénominateur

[modest wraith]

Le -1

Il est pas au numérateur?

Ou alors je dois le mettre au même dénominateur

Du coup ça donne -1x

[pallid scarab]

attention tu fais -(-1x)

[modest wraith]

J'ai déjà changé les signes

[pallid scarab]

$\frac{1}{1-(1-\frac{1}{x})} = \frac{1}{1-1+\frac{1}{x}} = \frac{1}{\frac{1}{x}}$

[ocean seal]

rafilou2003

[modest wraith]

T'as vue le 1 devant

Je l'ai zippé

Voilà pk j me suis trompé

[pallid scarab]

ok

[modest wraith]

Du coup ça va j'ai capte

Merci bcp

[pallid scarab]

ok

attends on a pas fini

[modest wraith]

Ah bon

[pallid scarab]

C'est quoi 1/(1/x) ?

[modest wraith]

Bah je multiplie nn?

[pallid scarab]

ouais et ça te donne quoi?

[modest wraith]

Par ×/+

×/1

Ça donne x?

[pallid scarab]

oui x

[modest wraith]

Yupiii

Mais att

1×1 ça donne

1

Nn j'ai tripe tkt

[pallid scarab]

dac

[modest wraith]

Merci bcp lamie

[lone heart]

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[lone heart]

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[lone heart]

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[weary quartz]

hi

[weary quartz]

can anyone tell me if i’ve done this right

[lone heart]

@weary quartz Has your question been resolved?

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[wide gyro]

hi i was doing a question in contradiction, and i got to a point where b^2 = 2(something)

[wide gyro]

my teacher said if b^2 is even, even b must be even

i remember her proving it but i forgot how

can someone explain why b must be even if b^2 is even

[pallid scarab]

use contraposition

what's the contraposite of (b^2 even => b even) ?

[wide gyro]

idk what contraposition is

but i guess its contradiction

so b is odd?

[pallid scarab]

the contraposition of A => B is (not(B) => not(A))

and those are equivalent statements

[wide gyro]

ive only learnt 3 proofs till now

[pallid scarab]

So the contrapositive of a statement is true, the original statement is true as well

[wide gyro]

lhs=rhs, induction and contradiction

[pallid scarab]

well then contradiction is the closest

suppose b^2 is even and b is not even, ie b is odd

[wide gyro]

yes b = 2a+1

oh wait

i can square the binomial

[pallid scarab]

yes

[wide gyro]

and i guess its even then

which contradicts

alr ty

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[lone heart]

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[fickle musk]

i need some help understanding combination and binomial relation

[fickle musk]

are there two formulas for combinations?

Isn't

 nCr= n!/r!

what is up with this:

nCr = n!/(r!(n-r)!)

[median oar]

This is nPr

[fickle musk]

umm 5c3 i wrote out 5 * 4 * 3/3*2=10

how is this valid?

if that is npr

[vale wigeon]

$\frac{5 \cdot 4 \cdot 3}{3 \cdot 2} \neq \frac{5!}{3!}$

[ocean seal]

Ann

[vale wigeon]

also parentheses around denom

[fickle musk]

umm so how is this equal to the second formula

[median oar]

They aren’t

[fickle musk]

then why is the answer same

[median oar]

nPr and nCr are different

[fickle musk]

not my question

[vale wigeon]

do you see the ≠ sign in my tex thing

[fickle musk]

this

[fickle musk]

yeah

but if that is the accurate ncr formula

then how is that answer valid since it is not following the formula

[vale wigeon]

$\frac{5!}{3! \cdot 2!} = \frac{5 \cdot 4 \cdot 3 \cdot \cancel{2 \cdot 1}}{(3 \cdot 2 \cdot 1) \cdot \cancel{(2 \cdot 1)}}$

[ocean seal]

Ann

[vale wigeon]

does this answer your question?

[fickle musk]

oh so if i do till r factor in nr writing (n-r)! is not necessary

got it

as it cancels out

thanks

.close

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[limpid spade]

?

[steep perch]

.close

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[agile kernel]

Let $f:\mathbb{R}\to\mathbb{R}$ be a function satisfying $f(x)=f(x+1)$ and $f(x)=f(x+\sqrt2)$ for any real number $x$. Is it necessary that the function $f$ is a constant function?

[ocean seal]

Puntre

[ivory fern]

No, but if you add the hypothesis that f is continuous then yes

[agile kernel]

Could you give an example of the function please.

[mortal trellis]

how many x are there which need to satisfy f(x)=f(0)?

[ivory fern]

f(x) = 1 if there exist integers a and b such that x = a + b sqrt(2); f(x) = 0 otherwise

I think that works

[oak perch]

No, if you add some other conditions like f being continuous then yes

[agile kernel]

I see. Thank you all.

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[golden glade]

can someone help me find the local maximum and minimum for this function?

[golden glade]

ive tried everything and i have no idea

[alpine sable]

Cos(0x^2)?

[golden glade]

umm the function as a whole

[alpine sable]

No I'm making note on how that is probably written wrong

[golden glade]

this is roughly the function with its asymptote (x is roughly estimated)

[alpine sable]

Because in that case it's is just simply cos(0) which is 1. So redundant completely to write it as cos(0x)

[golden glade]

ohh okay

[alpine sable]

Your question is still unclear! Why are you considering asymptotes when looking for extrema?

Open up your own help channel. This channel is already occupied

And please delete your message after you do

Thanks

[burnt rivet]

oh i thought i did do that wrffff

[golden glade]

im not sure i just put them there for reference

here, one second

oh wait

wrong one

i'll change it to cos(0)

[alpine sable]

Okay

Sooooo

Let's be clear

We are trying to find the local extrema of [
\map f x = \f3{x-6.2}+1
]
?

[golden glade]

well wouldn’t it be -1 because its cos(0) -2

so 1-2

[alpine sable]

It's -(1-2)

[golden glade]

OH

yes okay

my bad

yes that’s right

[alpine sable]

Oki doki so do you know first derivative test

[golden glade]

no i don’t think so

[alpine sable]

What do you know about finding extrema?

[golden glade]

nothing really we barely covered it which is why im struggling so much 💀

actually now that i look at it i dont even think this graph has a max and min

[alpine sable]

You're correct but you have no way to assert that this is the case

[golden glade]

ok true

so i have to find derivatives to find this out

[lone heart]

@golden glade Has your question been resolved?

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[vague dust]

Been a long time since ive done math and this question is simple is but bothers me i forget how to do it can someone help me out real quick

[vale wigeon]

given you have these answer options, the least headachey way of doing this would be to expand each one until you get a match.

[native cloud]

!status

[lone heart]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[native cloud]

!show

[lone heart]

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

[frigid mirage]

the usual way to do this when you dont have options is to split the middle term and factorize

[native cloud]

!try

[vague dust]

alright thanks

.done

[native cloud]

It's .close

[vague dust]

.close

[lone heart]

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[lone heart]

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[dull prawn]

"A" walks 10 hours daily and covers 480 km in 12 days. In how many days will he walk 20 hours daily to cover 360 km?

Can anyone solve it and explain the solve?

[vale wigeon]

we won't solve this problem for you @dull prawn

[native cloud]

!status

[lone heart]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[dull prawn]

2

[wary stream]

Who walks for 20 hours tbh?

[dull prawn]

Idk

This math's is on my syllabus

[native cloud]

"Application" lmao

!show

[lone heart]

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

[dull prawn]

I was stuck in the 20 hour to day transformation

[vale wigeon]

"A" walks 10 hours daily and covers [REDACTED] in 12 days.

how many hours does A spend walking this way?

[lone heart]

@dull prawn Has your question been resolved?

[brave grove]

(x-2)(3x-3)

[lone heart]

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[simple saddle]

how to solve this?

[lone heart]

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

[simple saddle]

i tried finding individual sum say sum a= 24*7.8 like that and 700 =rel total and ~618= promtness total

[vale wigeon]

that looks correct to me

the ranking stuff seems irrelevant to the problem (unless it comes up in later problems with the same data...)

,calc (6.5 + 6.9 + 4.1) * 100

[ocean seal]

Result:

1750

[vale wigeon]

,calc (7.8 + 7.5 + 4.9) * 100

[ocean seal]

Result:

2020

[simple saddle]

ohh

[vale wigeon]

700 =rel total and ~618= promtness total
these totals look strange though

[simple saddle]

i didn't sum those ratings

i was doing people who gave A first*7.8

[simple saddle]

thanks

.close

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[lusty thunder]

hi

[lusty thunder]

We have defined cosine and sine using the unit circle. Recall how. Now consider an arbitrary right triangle. Show how, using that definition, we can show:
The cosine of an acute angle in the triangle is the adjacent leg divided by the hypotenuse.
The sine of an acute angle in the triangle is the opposite leg divided by the hypotenuse.

Im not very sharp at geometry or trigonometry so for this i might need some help

[cursive badger]

you know how the right triangle in the unit circle works yes?

[lusty thunder]

no, im not sure

[cursive badger]

oh good place to start then lol

so you have a circle of radius 1, your unit circle, yes?

[lusty thunder]

yes

[cursive badger]

and you draw a line from the center point to the circle, length 1, yes?

[lusty thunder]

yes

[cursive badger]

you rotate it versus the +x axis, so angle of 0 is pointing +x, angle of π/2 is pointing +y, etc...

angle of 3π/4 is pointing diagonally top left, etc...

still with me?

[lusty thunder]

right

like this?

[cursive badger]

yes

so, you see the radius line

and you can draw lines down to make a right triangle

the hypotenuse is ALWAYS 1 in the unit circle yes?

and the sides range 0 to 1, can't even be bigger, right?

[lusty thunder]

hypotenuse is cos right?

[cursive badger]

don't go there yet and no

[lusty thunder]

im trying to translate english in my head so yea its a bit slow for me

[cursive badger]

hypotenuse is the long side of the right triangle

[lusty thunder]

yeah sorry, cos is the katete

[cursive badger]

which is the radius from the center to the edge

[lusty thunder]

or whatever the english word is

[cursive badger]

anyways

we have a right triangle, the longest edge (hypotenuse) is stuck at length 1

the sides range from 0-1, ok

so, the literaly definition of sin and cos is

at any given angle ø, sinø is the length of the vertical side of the triangle formed by the radius at angle ø

and at any given angle ø, cosø is the length of the horizontal side of the triangle formed by the radius at angle ø

still with me?

sin and cos are just the lengths of sides in a right triangle in a unit circle

[lusty thunder]

vertical in english means up or down, yes?

[cursive badger]

yes

horizontal is left right

horizontal is x axis, vertical is y axis

[lusty thunder]

At any given angle ø, sinø is the length of the marked in red

[cursive badger]

yes!

[lusty thunder]

okay:)

[cursive badger]

if the sinø is negative this means the triangle is going downwards instead of upwards

you understand that yes? just reflected down

[lusty thunder]

yes, its reflected in the x-axis

[cursive badger]

yup

so, we know what sin and cos are, and what ø is, and what the unit circle is

for any given ø there is only one pair of sin and cos, and vice versa

because of this, we can use a known angle to find the size of sides of a right triangle with hypotenuse 1

however, right triangles are rarely hypotenuse 1 right?

[lusty thunder]

right

[cursive badger]

in fact, sin and cos are the side length of a scaled right triangle, scaled to be hypotenuse length 1

let's say we have a triangle with sides 3, 4 and 5

we want to find the angles

we know the 90º is opposite of the 5 side, but how do we find the others?

we shrink the triangle by a factor of 5: the angles do not change

[lusty thunder]

can you elaborate on what you mean with 90 degrees is the opposite of the 5 side

[cursive badger]

the right triangle with sides 3 4 and 5 has short side 3, medium size 4 and long side 5

the longest side in a right triangle is opposite the right angle that gives the right triangle its name

[cursive badger]

so we shrink it all down and while the sides change, who cares? we know the sides. we want the angles, which do NOT change when shrinking.

we have a new triangle with sides 3/5, 4/5 and 1 with the SAME angles

here lemme desmos this

[lusty thunder]

Yeah, okay

[cursive badger]

(just a moment I'm bad at desmos)

[lusty thunder]

Thank you:)

[cursive badger]

we start with the big red triangle

bottom edge 4, vertical edge 3, long hypotenuse length 5

we scale it down by a factor of 5 to make its hypotenuse 1 (and therefore fit inside the circle)

[lusty thunder]

right ,yea

[cursive badger]

so we get a new, smaller, black triangle with side length 3/5, 4/5 and 1

but notice how the angles have not changed.

effectively, sin and cos are not just the measurements of the sides of a triangle in the unit circle

they are the ratios of the a side in a right triangle over the length of the hypotenuse (scaling it down by dividing by the length of the hypotenuse)

[lusty thunder]

so cos is 4/5

[cursive badger]

yes

[lusty thunder]

sin 3/5

[cursive badger]

(I can't read sorry) yes

[lusty thunder]

right okay, i get it now

So what they mean is that a triangle is formed for every ø

[cursive badger]

yes

[lusty thunder]

And since this triangle can be upsized, its properties are applicable to all other triangles of that sort

[cursive badger]

yup

all right triangles can be scaled like this

smaller ones too, like 5/100, 12/100, 13/100

[lusty thunder]

and so the purpose of the unit circle and why they mentioned is, is because we can scale the hypotenuse down to 1, making it fit inside the circle

That makes a lot of sense, thanks. Do you wanna do the follow up with me as well?

[cursive badger]

sure, I can help!

[lusty thunder]

but wait, what about cos(0)

that wouldnt form a triangle, right?

[cursive badger]

it's technically a triangle, for the sake of generalization

sides of 0, 1 and 1

[lusty thunder]

oh, okay

[cursive badger]

0^2+1^2=1^2

[lusty thunder]

A right triangle has sides a=1, b=√3 and c=2. What is the exact value of the angles expressed in radians?

so here i should do the same i guess

[cursive badger]

draw it out first, what is the hypotenuse?

[lusty thunder]

hypotenuse is 2

[cursive badger]

mhm

so now scale it down

[lusty thunder]

so to get hypotenuse of 1, we have a = 1/2, b = sqrt(3)/2

[cursive badger]

mhm

[lusty thunder]

and from this i can look in my table

so that would be

[cursive badger]

you drew this out yes? which of a and b is sin and which is cos (vertical and horizontal)?

[lusty thunder]

pi/3

[cursive badger]

π/3 is which angle though?

[lusty thunder]

60

[cursive badger]

I mean, opposite of a? b? c?

[lusty thunder]

opposite of b

[cursive badger]

mhm

they asked for it in radians not degrees fyi

just remember that

[lusty thunder]

yeah, so it would be pi/3 right?

if its radians

[cursive badger]

so what are the other two angles?

just checking since you said 60 earlier

[lusty thunder]

pi / 3 = 60
one is 90
last one must be 30

[cursive badger]

yup

convert to radians though

does that all make sense?

[lusty thunder]

pi / 2 = 90

pi / 6 = 30

[cursive badger]

yup

30/60/90 triangle 1/sqrt(3)/2 one of the most common ones you'll see

[lusty thunder]

nice, thanks

[cursive badger]

anything else you need help with?

[lusty thunder]

I have to explain why cos(v)^2+sin(v)^2 = 1 is known as the "simple rule"

but I guess it relates to what you explained earlier with the hypotenuse always being 1

[cursive badger]

yup exactly

they're sides of a right triangle with hypotenuse 1 by definition

that is literally what they are derived from P: that and the pythagorean theorem

[lusty thunder]

Thanks a lot

There is more if you have time

[cursive badger]

I do have time actually lol

ask away

[lusty thunder]

I have to get arccos(1/2), arcsin, etc

[cursive badger]

huh ok

[lusty thunder]

I don't actually know the difference between arccos and cos

[cursive badger]

let's start with the simple one

cosø is the horizontal length at the angle ø

arccosx is the angle at horizontal length x

[lusty thunder]

expressed in radians?

[cursive badger]

whatever unit you want

usually radians though yeah

degrees are hateful and want us to fail

[lusty thunder]

lol

okay, so arccos(1/2) would be pi/3 ?

and cos(pi/3) = 1/2

[cursive badger]

yes and

yes

[lusty thunder]

arcsin(1) must be 0 though

or well

[cursive badger]

nope

what is the only value at which sinø is 1? (not π that was a typo)

[lusty thunder]

pi/2

[cursive badger]

yes

[lusty thunder]

right okay

[cursive badger]

so what is arcsin(1)? what is the angle where sin is 1?

[lusty thunder]

90 degrees

[cursive badger]

also note that the answer here is ±π/3 sin cos(π/3)=cos(-π/3)=1/2

but arccos and arcsin have limited ranges by common consensus

[lusty thunder]

right, because it can be under and upper?

this is in relations to workign with complex numbers and usually they can tell you what quadrant you are in

[cursive badger]

yup yup

[lusty thunder]

arcsin(-sqrt(3)/2) = -pi/3 and pi/3 ?

[cursive badger]

no

sin(π/3) is sqrt(3)/2

uh wait -π/3 is right

sorry, used to seeing only positive angles P:

[lusty thunder]

alright

solve cos(x) for A, B and Real numbers

[cursive badger]

what are you confused about?

[lusty thunder]

how to get cos(x) = 1/2 when it comes to the interval [-pi, pi]

normal interval makes sense

[cursive badger]

wdym?

why would it not work with [-π, π]?

[lusty thunder]

if the interval is -pi to pi

i dont know how to calculate it

i mean i dont even know how to find -pi on a unit circle

[cursive badger]

the angle -π?

think of it as -180º

[0, 2π] covers the whole circle, as does [-π, π]

[lusty thunder]

right

so we go from -180 to 180

not sure how this affects radians though

[cursive badger]

hmmm

radians and degrees describe the same thing

it's just while degrees go from 0 to 360 radians go from 0 to 2π

they just have different scales

I called degrees awful earlier but really they're the same thing

[lusty thunder]

i wouldnt know where to place -pi on a unit circle though

[cursive badger]

same place as -180º

at first converting between the two helps a lot

since most people find degrees more intuitive

[lusty thunder]

i have a hard time visualing it hehe

oh

[cursive badger]

take your line facing directly right

rotate it 180º downwards

[lusty thunder]

right okay

so its like the opposite way

[cursive badger]

yup

coincidentally, 180º and -180º end up at the same place

as with π and -π

[lusty thunder]

so sin becomes cos and cos becomes sin

[cursive badger]

no

no no no do not get that kind of idea

[lusty thunder]

oh

okay

[cursive badger]

hmm how do I put this

you know how a full revolution changes nothing?

you go +360º and your line hasn't moved?

so the place shown by 27º is the same as (27+360)º

the angle 90º is the same as 450º (360+90)

so with -180º

we add 360º, full revolution it doesn't move

we find that -180º shows the same place as 180º

[lusty thunder]

Right

makes sense now

What I dont understsand is how to find the value for R @cursive badger

[cursive badger]

R is the set of all real numbers

you are not trying to find R xD

by saying x ∈ R you're being told the angle in sinx is a real number

like 2π/e or 3 or 378912*10^17

[lone heart]

@lusty thunder Has your question been resolved?

[lone heart]

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[vapid shuttle]

Working on part (a)

I have done the r_(k+1) part

but I am stuck on the r_(k+2) part now

I tried rewriting it using the r_(k+1) part but that didn't go anywhere

[echo socket]

You can just apply the first identity twice btw

[echo socket]

Wait wdym

Show your work

[vapid shuttle]

wdym by this

[echo socket]

First replace k with k+1 and you have [ r_{k+2} = \frac{r_{k+1}+1}{r_{k+1}} ]

[ocean seal]

A Lonely Bean

[vapid shuttle]

oh I tried doing the opposite

[echo socket]

Oh

Well yeah this is easier

[echo socket]

And just replace r_{k+1} with (r_k + 1)/r_k

[vapid shuttle]

ater that

after*

I just end up with

$\frac{\frac{1}{r_k}+2}{\frac{1}{r_k}+1}$

[ocean seal]

Austin

[echo socket]

Multiply top and bottom by r_k and you're good

[vapid shuttle]

I see

ty

[lone heart]

@vapid shuttle Has your question been resolved?

[lone heart]

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[spiral pumice]

doing limits in calc. this question kinda stumped me as i’ve never seen “int” before outside out python lol. “lim int(x) as x approaches 0 from the right”

[maiden glen]

it might represent the integer part of x

as in, the round down function

[spiral pumice]

so the graph would look like steps

with the closed circle on left, open on right

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[alpine sable]

Hi I had this question

[alpine sable]

I got c

But apparently the answer is a

Here’s my reasoning:

for the u vs x graph, we know, acceleration = v du/dx

this only works when acceleration is time independent

So acceleration is constant wrt time

So the acceleration is constant wrt x

But I can’t seem to understand what went wrong

[crisp pine]

I would have said you were correct with c

Does it give any explanation as to why it's a

[alpine sable]

Yes

But I don’t understand what’s wrong with this

The instructor explained casually

That since slope of u-x graph is negative and constant and velocity is positive but increasing

[crisp pine]

Ah wait sorry, my mistake

It's a velocity displacement graph, not a velocity time graph

So you can't just use the derivative

[alpine sable]

Why not?

I used (dx/dt)(dv/dx) = v(dv/dx)

[crisp pine]

Sure, so the acceleration at a point is the velocity multiplied by the gradient of the graph

[alpine sable]

Yes

I have a question

[crisp pine]

This is why it's a not c

[alpine sable]

when I have dv/dt = a

What are the conditions

[crisp pine]

What do you mean by the conditions sorry?

[alpine sable]

Ohhhj The velocity is changing

[alpine sable]

Like when does that fail?

[crisp pine]

I'd be inclined to say never

[crisp pine]

Yah exactly

[alpine sable]

I’d been stuck on that for so damn long

Thank you so so so much

I really appreciate your time

Have an awesome day man

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[crisp pine]

No worries :)

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[sly spindle]

How do you find the slant asymptote of the equation f(x) = x/lnx? I can see there's what appears to be a slant asymptote, but idk how to calculate it

[sly spindle]

in precalc the method i learned for finding slant asymptotes is if the degree of the numerator is 1 more than the degree of the denominator, then you divide the two to get the equation for the asymptote but that doesnt seem possible here

also what even is the degree of ln(x)?

[sly spindle]

i dont really understand why that method works either so it would be great if somebody could explain that to me 🥲

[naive valley]

do you know about derivatives?

[sly spindle]

yes

[naive valley]

what's the derivative of x/ln(x)?

[sly spindle]

(lnx - 1)/(lnx)^2

[naive valley]

ok

what does that do as x->infinity?

[native cloud]

You might have to use limits for this one

Why is my message being sent now..

[remote heron]

https://math.stackexchange.com/questions/1978602/how-to-find-the-oblique-asymptote-of-this-non-rational-fuction

[sly spindle]

uhhhhhhh

[naive valley]

well when x is large, the -1 in the num is negligible, so your function just looks like
ln(x) / ln^2(x)
which is
1/ln(x)

what does that do as x->infinity?

[sly spindle]

is that like 1/infinity???

[naive valley]

yes, in the limit

[sly spindle]

is it 0??

[naive valley]

yes

[sly spindle]

ok

[naive valley]

so if there is an asymptote, its slope is gonna be zero

now you need to check if there actually is an asymptote

[sly spindle]

how do you do that

[naive valley]

well an asymptote with slope zero is horizontal, yes?

[sly spindle]

oh wait so not a slant

[naive valley]

and if a function is going to have a horizontal asymptote, then it needs to have a limit as x->infinity

so check if your function does

[sly spindle]

is it just

infinity?

[naive valley]

yes

x grows way faster than ln(x)

x/ln(x) is unbounded

[sly spindle]

oh ok

that makes sense

[naive valley]

so those two facts combined:

• function goes to infinity as x->infinity
• function's slope goes to zero as x->infinity
  means that there can't be an asymptote

btw, f(x) = ln(x) is another function with the same two properties

[sly spindle]

ohhhhh

so like

it goes to infinity instead of approaching another line

as x goes to infinity

so no asymptote

and cus slope is 0

right

ok

that makes sense

[naive valley]

arguing in words, if there's an asymptote it has to be horizontal, but this function grows past any candidate horizontal line

[sly spindle]

ok

thanks

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[hot bluff]

number 19

no examples in the book on this type of problem

[vapid shuttle]

What have you tried

[hot bluff]

my first try with nothing to go on

its wrong af

book answer is h != 2, but i have no idea how to derive that.

[vapid shuttle]

If h = 2

Then column 2 is a scalar multiple of column 1

[hot bluff]

this is literally first section of first chapter in the book, what does it mean when a column is a scalar of another?

[vapid shuttle]

Scalar multiple

It means column 2 is column 1, but just multiplied by a constant

In this case, if h=2 then column 2 is column 1 times 2

[hot bluff]

okay, which would mean we only have one line/equation?

when h = 2, both rows become scalar multiples as well, right?

oh no, third column doesnt

alright, so we would have a scalar multiple of another column if h=2, but how does that help us determine possible values?

this would create an inconsistent system because the columns would be scalar multiples but the rows would not be scalar multiples?

so the approach here is to find what would invalidate the system. since 2 would create a scalar multiple of column 1 but not column 3 then it would be inconsistent. how can we prove that 2 is the only value that would produce an inconsistent system?

was able to find it by algebra

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[fresh nebula]

Here is my work for all 3 of them

I think I got 1 and 2 right but im not sure about the third one

can anyone oversee if my work is correct? or where i made a mistake

nevermind the first power set, I understand that it is R=1 (-1,1) now.

<@&286206848099549185>

only need help on the second and third power series (which are labeled 9 and 10 on the screenshots)

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@fresh nebula Has your question been resolved?

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@fresh nebula Has your question been resolved?

[fresh nebula]

pls

[nimble fern]

firstly, i found this site is helping me to check.
https://byjus.com/maths/radius-of-convergence/

secondly, you just have to consider R=radius of convergence,
so for 9), you can just state R=1

oh wait, haha i misread, you also need the interval

so, you're right for (9)

checking (10) now