help-19

[sharp bridge]

does that look right

[limpid adder]

yes

[odd edge]

@sharp bridge Has your question been resolved?

[sharp bridge]

.close

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[solemn iris]

How does multiplying a number to the power of a number less than one work? I've tried it a few times in a calculator and it just give me whatever number I put in it.
For example, if I do 1^0.7, or 93^0.7, it just gives me 1 for the former, and 93 for the latter.

[low locust]

you definitely messed up when putting it in

[solemn iris]

Ok, but that doesn't answer how it works.

[tribal ingot]

its basically like taking some root of that number

because for example, $a^{1/2} = \sqrt{2}$

but 1/2 can also be written as 0.5

[tribal ingot]

so if you did $a^{0.5}$ that would be the same as $a^{1/2}$ which is $\sqrt{2}$

[clever fjord]

satvik

satvik

[solemn iris]

Apparently the calculator just doesn't like doing 1^0.7, because it seems to be working with 93 ._.

[tribal ingot]

1 raised to any power is equal to 1

[solemn iris]

Then why is this game saying that my debuff is total output to the power of 0.7 and it is working T_T

Isn't my total output 100%?

Which is equal to 1

[low locust]

in a sense, sure. but thats not like its meant

instead of doing 93 damage, you do 93^0.7 damage

or something along those lines

will depend on how its implemented

[solemn iris]

I see, so I just overcomplicated it

.close

[odd edge]

Channel closed

Closed by @solemn iris

Use .reopen if this was a mistake.

[solemn iris]

.reopen

[odd edge]

✅

[solemn iris]

Wait, I got another question.

[tribal ingot]

go ahead

[solemn iris]

What would be the reversal to do a number by the power by if I want to figure out what the original number was? 93^1.3?

[tribal ingot]

take that root of the number

if 93^1.3 = n

then n^(1/1.3) = 93

[solemn iris]

93 in this case being the reduced amount that I'm getting? Say debuffed earnings is x^0.38 and the outcome of that is 39.8 units per second.

What do I do to figure out what the original number was before the debuff

[tribal ingot]

take the 0.38th root of 39.8

or 39.8^(1/0.38)

they are both the same thing

[solemn iris]

So, this?

[tribal ingot]

the inverse operation of raising a number to the nth power is taking the nth root of it

yeah

[solemn iris]

I graduated hs years ago, I forgot all this stuff T_T

[dapper canyon]

$\sqrt[n]{x} = x^{1/n}$

[clever fjord]

69-category theorist

[solemn iris]

Apparently these are my actual earnings rn (I reduced 39.8 trillion to just 39.8)
[1.62262667 times 10 to the 16th]

[tidal matrix]

you can also verify this with exponent laws, if it helps: $$(a^n)^\frac{1}{n} = a^{n\frac{1}{n}} = a^1 = a$$

[clever fjord]

haseeb

[solemn iris]

Alright, thanks for the help y'all. I think I understand now.

.close

[odd edge]

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[normal sentinel]

Hi, I'm currently taking a calculus 1 course. I'm going over some integration by substitution problems.

there are two that are solved differently and I'm unsure of exactly why. I'll post the screenshots in a second

[normal sentinel]

so the integration on the left picture shows that we can rearrange u whereas the integration on the right picture does not represent u being rearranged much like on the left

my question is: what and why is the reasoning for this? my intuition tells me I could solve the left integration like the one on the right, but what is so different that I would treat u differently?

thank you

I do understand the process for solving both of these, I am just unsure of why these differences arise

or is this just another way of u integrating?

oh I just realized I didn't post the full process of the one of the left: one second

updated

[late sinew]

In the first one there's an extra x so to bring the integral into one variable u you need to write x in terms of u. In the other integral, there is no x term left over so you dont need to do this

Was that your question or do you mean smtg else

[normal sentinel]

where is the extra x variable on the left compared to the right? there is both an x outside the radical on both problems but the one to the right is x^3

[late sinew]

Welcome to mathcord, by the way

[normal sentinel]

thank you ❤️

[late sinew]

This extra x

That doesnt get cancelled

[normal sentinel]

OH

I see what you mean now

[late sinew]

These get cancelled

[normal sentinel]

so would it be correct to say then that we rearrange the u on the left problem so that we may cancel out the extra x variable?

[late sinew]

"Write x in terms of u" is a better phrase than "cancel out", but yes basically what we want to do is bring the integrand down to 1 variable

[normal sentinel]

thank you, I understand now. Appreciate your time

.close

[odd edge]

Channel closed

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[late sinew]

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[hidden escarp]

Hello I need help with this question please

It says Un is a convergent sequence

And I assume the rest is clear

[odd edge]

@hidden escarp Has your question been resolved?

[sturdy cape]

approche informelle : soient X et u les limites des suites respectives

[hidden escarp]

was just about to close the channel because I solved it lol

[sturdy cape]

ah lol

[hidden escarp]

but thank you

[vital holly]

Oui oui je parle drancais

[sturdy cape]

...drancais

[hidden escarp]

it's equal to 1

[sturdy cape]

ouais c'est ça hein

[hidden escarp]

merci oui la baguette

[sturdy cape]

j'y arrive là aussi ouais

[hidden escarp]

quelle methode tu as utiliser ?

[sturdy cape]

utilisé * (c'est le passé composé hein)

[sturdy cape]

mais celle-ci

[hidden escarp]

je suis pas francais lol

mais on etudie en francais

[sturdy cape]

une langue d'instruction, c'est ça ?

[hidden escarp]

j'ai pas bien compris

[hidden escarp]

oui je suis marocain

[sturdy cape]

ahhhh

ça a du sens

[vital holly]

Wist did u just take the like of both side

And solce

[sturdy cape]

"informal approach - let X and u be the limits of the respective sequences"

[sturdy cape]

yeah

[vital holly]

X = x+u / 1+xu

[sturdy cape]

There's enough in there that's assumed, that you can do this "legally"

And then the other key thing to note is that both sequences are positive

[hidden escarp]

oh ok so you assumed Xn is convegent ?

[sturdy cape]

X_n is convergent iff the limit exists

Thusly, if, when calculating the limit I actually do find one, I can make the claim that it is convergent*

-# *there are again some more requirements to this, but here it's safe enough

[hidden escarp]

I first showed that Xn between 1/2 and 1 by using recurrence puis j'ai etudie la monotonie de Xn pour dire qu'il convergente et finalement j'ai supposer la limite est egale a l et j'ai remplacer dans l'expression

[sturdy cape]

pour dire qu'elle (la suite) est convergente*

supposé que la limite est/soit égale à "l"

[hidden escarp]

oui elle la suite

[sturdy cape]

je l'ai remplacée dans...

[hidden escarp]

remplace*

[sturdy cape]

bah French grammar is a pain

bah enfin

[hidden escarp]

my friend is beyond cooked

[sturdy cape]

any grammar is a pain

[sturdy cape]

ton ami est plus du cuit*

[hidden escarp]

I can only use french in maths with "montrer que" "donc, alors, d'autre part" ...

but no grammar at all 😔

[sturdy cape]

Nah igu

[hidden escarp]

wait I meant french 😂

I even typed friend instead of french

my english is not any better lol

[sturdy cape]

I've also had to read a lot of French maths to figure out what the conventions are in French (because in English the conventions are different)

[hidden escarp]

wait so ur not french ?

[sturdy cape]

Non mdr

je l'ai appris à l'école secondaire

[hidden escarp]

damn ur french is good tho

[vital holly]

Wist this isn't an entirely valid prood

[hidden escarp]

ohh ok nice

[sturdy cape]

pas du tout hein 🤣

[vital holly]

U can say thst if X convergence

Then x is 1

But u still have to prove convergence

When u can show that x is bounded above by 2

[sturdy cape]

Yeah; I did say there's that caveat

[vital holly]

By just maximizing that fractuon

[hidden escarp]

better than mine who has been studying french for his whole life 😔

[vital holly]

Since u is bounded between 1/2 and 1

[hidden escarp]

we can show that its not hard here

but we have to show first that Xn is between 1/2 and 1

[sturdy cape]

du coup, tu venais de dire être marocain; ton français serait toujours dehors cuit
-# in any case, you'd just said you were Moroccan; your French would already be cooked outside

[hidden escarp]

What abt question 7 or 8 ?

Either one of them just the main idea im too sleeping rn to think too much

[sturdy cape]

T'as réussi à trouver f'(x), pour commencer ?

[hidden escarp]

Yep Ik

Bien sur

[sturdy cape]

Donc f''(x) ?

et ensuite...

Si tu peux trouver une...

fok

"pattèrne" (je sais qu'il n'existe aucun mot lol)

il ne faudra donc que le démontrer

Pour q (8)

Soit f une telle fonctionne

Pour commencer, soit x = 0, t'aurais quoi comme résultat ?
-# To start, if we let x = 0, what would you have as a result?

[hidden escarp]

Is there any "patterne" 💀

[sturdy cape]

ptn mais enseigner en français, c'est drôle
-# fok, teaching in French is wack

[hidden escarp]

You can just speak english lol

Btw can't we use taylor here ? Ik there is one usuelle for ln(x+1) but im not sure abt ln(x)

Maclaurin

[sturdy cape]

Try using the product rule, by writing $f(x) = x^{-1}ln(x)$

[clever fjord]

Waes (Wires)

[sturdy cape]

Désolé, moi aussi faut que je dorme, il est déjà 2:21 du matin
-# sorry, I also gotta go to bed, it's already 2:21 in the morning

Mais si t'as toujours besoin d'aide, fait ping @helpers
-# But if you still need help, ping @helpers

[hidden escarp]

Okk mrc beaucoup je vais la completer by myself

Completé*

[sturdy cape]

non ça c'est le futur proche

faut utiliser l'infinitif; donc "completer" c'est correct

[hidden escarp]

https://tenor.com/view/what-surprised-gif-25739412

I will complete it by myself ☺️

[vital holly]

@hidden escarp j'ai fait Q7

c'etait tres cool

[hidden escarp]

Comment ?

[vital holly]

sois g(x) = ln(x)

ok im going back to anglais

f(x) = g(x) * g'(x)

then u get this (f should be f^(n) )

since u get a recurrence relation

where

the coefficient of

g^(a) * g^(b) in f^(n) is the coeffiecient of g^(a-1)g^(b) plus the coefficient of g^(a)g^(b-1) in f^(n-1),
as a function this is written as c(a,b) = c(a-1, b) + c(a, b-1)

which is pascals triangle

then u can show g^(n) = (-1)^(n+1) * (n-1)! / x^n

[vital holly]

also i just realized u can simplify, which is so cool

[hidden escarp]

What is pascals triangle ?

[vital holly]

les binomiales

not rly pascal triangle

just the common recurrence relation

[hidden escarp]

Btw im still in high school so using any method beyond what im studying would be very confusing for me

[vital holly]

associated with the binomials

[hidden escarp]

Oh okk

[vital holly]

this problem is rly hard for high school tbh

like none of the concepts are above hs, but combing all that together

is

pretty tough

[hidden escarp]

Its an entrance exam for a university but I dont get to use much more than what I already studied in school

There are exceptions ofc

If its possible can you explain what you did with simpler terms ? If that is possible

[vital holly]

ok so

let g^n denote the nth derivative of g

then with g(x) = ln x

u have f = g * g'

f' = g' * g' + g * g''

and etc

the pattern here

is that

when u differentiate a term of the form g^n * g^m

u get g^(n+1) * g^m + g^n * g^(m+1)

due to the product rule

so lets say you have f^n written as a sum of such terms g^m * g^p

then when we differentiate f^n

how can we find the coefficient of the g^m * g^p term in f^(n+1), for any (m,p)

thats precisely the sum of the coefficients of g^(m-1) * g^p and g^m * g^(p-1) in f^n

i k this is rly complcated, but does everyhtng make sense

[hidden escarp]

Okk thank you so much I guess I understand a little

What abt the other question did you try to solve it ?

Anyways thank you Ill close this channel

.close

[odd edge]

Channel closed

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[hidden needle]

I'm factoring polynomials and getting confused when it involves pulling out factors from negatives

[wooden cypress]

First of all is that meant to be multiplication? because 6 * 2 isn't 6

[hidden needle]

factoring

it started without the numbers outside the square

finding the GCF

[wooden cypress]

Okay

[hidden needle]

betwen the two numers in each row and column

[wooden cypress]

oh okay icwym

[hidden needle]

🙂

the original problem was 2x^2-8x+6

[wooden cypress]

So if you have a negative term, then its factors are exactly the same as its positive version

Just with an extra negative sign

You just have to not overthink it

[hidden needle]

soo for -6x and 6 the factor is 6 or -6

GCF i mean

[wooden cypress]

Yes

Either works, neither is "greater" than the other in terms of gcf

But it's probably easier to go with the positive one, so 6

[hidden needle]

so the output i wrote down (the binomial resulting from this area chart), is (2x+2)(2x+6) which produces different results than the original problem when setting x to 0

which makes me think i got something wrong

like one of the GCFs outside the square should be negative

[wooden cypress]

I can't lie I've never come across this before so idk

[hidden needle]

i will share my beef stirfry and we will ponder

[wooden cypress]

But if you want to end up with the same polynomial you started with, then that's gonna have to become a multiplication table

As in the top terms are going to have to multiply with the side terms to get the terms in the square

[hidden needle]

WAIT

i understand now

i can just do what i did in the reverse.

if 2x and 2 don't multiply to -2x (which the obvoiusly dont) i can literally just correct one of them (but not both of course) to negative

wow

obvious

[wooden cypress]

I mean yeah you can do that

But then you get -4x, not -2x

Not the same

[hidden needle]

maybe

[amber veldt]

when you factor polynomials, you have to accept that the answers won't be unique up to sign

[hidden needle]

context

even so all the squares charts i'm producing are wrong.

ill send a ic

[amber veldt]

why not just factor it

[wooden cypress]

If that's the question then you'll find it a lot easier to just multiply out the options

[amber veldt]

(6)(2)=12, you want numbers that multiply to 12 and add to -8, easy

[hidden needle]

exactly

[amber veldt]

lol me and depression are saying the exact opposite suggestions

[wooden cypress]

Sorry it's probably not very helpful is it lol

[hidden needle]

lol

[wooden cypress]

You can do either

[hidden needle]

no ur good

[amber veldt]

oh i just do it , i think it's called factoring by grouping? never understood the name

you write

$$2x^2-6x-2x+6$$

[clever fjord]

gfauxpas

[amber veldt]

i dont use a punnett square or whatever you're doing

[tidal matrix]

i should start a counter for every new method i see of factoring a quadratic

[amber veldt]

2x(x-6)-2(x-6)

=(2x-2)(x-6)

=2(x-1)(x-6)

err

3

forgot to divide 6 by 2

don't do mistakes like that

2x(x-3)-2(x-3)

(2x-2)(x-3)=2(x-1)(x-3)

[wooden cypress]

Thing is they're not actually new methods, it's all just the same two methods but with extra steps / lines / drawings on top to disguise it

It's completely pointless

[hidden needle]

My output is still wrong. Im reading the response above now

[amber veldt]

if you're not getting the right answers with your square thing then don't do the square thing

[wooden cypress]

The problem with what you're doing, is you end up with 2x or -2x on both sides of your square on the top left

They're gonna multiply to get either -4x^2 or 4x^2

[hidden needle]

i'm ok not doing the square thing i found it silly

[wooden cypress]

Which isn't what you want

[hidden needle]

i just do'nt know another way

[wooden cypress]

I reckon you're not fully understanding what you've been taught

[hidden needle]

i agree

thats why i don't like this method

[wooden cypress]

Okay here's the easiest way

Firstly, you can divide all the terms by a number, so that the first term is just x^2

So in this case, you would divide by 2

That make sense?

[hidden needle]

dvide wha by 2, each part of the polynomial?

[amber veldt]

yes

[wooden cypress]

Yeah exactly

The whole thing

[hidden needle]

reverse distrubutive propert

[amber veldt]

2(x^2-4x+3)

[hidden needle]

ok

[wooden cypress]

Don't forget about the 2, that just becomes one of the factors

[hidden needle]

oh then just do normally

wow

[amber veldt]

but

[wooden cypress]

So yes you end up with this

[wooden cypress]

Oh okay, if you already know how to factorise those then that makes it a lot easier

[amber veldt]

factoring by grouping is what i recommend for when not all coefficients share a common factor nicely

want an example?

[hidden needle]

yes please

[amber veldt]

$$6t^2-11t-10$$

[clever fjord]

gfauxpas

[amber veldt]

(6)(-10)=-60, we want numbers that multiply to -60 and add to -11

[hidden needle]

ok i'm following

[amber veldt]

any suggestions?

[hidden needle]

4 and 15

[amber veldt]

great

[hidden needle]

-4 and 15

err

-15 and 4

[amber veldt]

$$6t^2-15t + 4t -10$$

[clever fjord]

gfauxpas

[amber veldt]

and if you switch -15t and +4t it should work anyway

so

look at the first part

6t^2-15t

factor it?

[hidden needle]

wait

so thats it

just splitting the middle part

until those two numbers and adding the variable

and its the same

[amber veldt]

no square nonsense required

[hidden needle]

i mean i guess it makes sense

huh

ok

so now where are we

[amber veldt]

I'm an old man, I was taught this before they invented the square m ethod. we didn't even have squares back then

[hidden needle]

lol

[amber veldt]

factor 6t^2-15t what do you get?

[hidden needle]

no squares back then only rhombuses. building was alot harder

[amber veldt]

indeed

perpendicular slope both ways

3t factors out

[hidden needle]

both 6 and 16 divide by 3

yeah

3t

[amber veldt]

$$3t(2t-5)$$

[clever fjord]

gfauxpas

[amber veldt]

okay

next chunk

$$4t-10$$

[clever fjord]

gfauxpas

[amber veldt]

has a common fctor of 2

[hidden needle]

2

[amber veldt]

$$2(t-5)$$

[hidden needle]

yeah

[amber veldt]

$$6t^2-15t+4t-10=3t(2t-5)+2(2t-5)$$

[clever fjord]

gfauxpas

[amber veldt]

oh i made a typo above

this is what i meant

hey look (2t-5) is a common factor

[hidden needle]

ah

ok

[amber veldt]

so the quadratic = (3t+2)(2t-5)

[hidden needle]

wut

[amber veldt]

i factored out (2t-5)

[hidden needle]

i don't understand

[amber veldt]

which part

[hidden needle]

how we simplified from the right side of hte texit to that

[amber veldt]

okay sure

let's define A=2t-5

just to make the pattern clear

so we have

[hidden needle]

got it

[amber veldt]

3tA+2A

factor out A

A(3t+2)

[hidden needle]

ok. i think i understand. but then how (3t+2)(2t-5)

[amber veldt]

because A=2t-5

[hidden needle]

oh

huh

[amber veldt]

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratics-multiplying-factoring/x2f8bb11595b61c86:factor-quadratics-grouping/v/factoring-trinomials-by-grouping-4 more examples, with randomly-generated exercises if youd like

[hidden needle]

i will take this knowledge and spread it around the earth

[amber veldt]

hold on

[hidden needle]

thank you for the help. i will try to master grouping

[amber veldt]

we need a logo

dont go

one moment

[hidden needle]

lol

[amber veldt]

okay

[hidden needle]

lol

i can share my webcam to show the notebookim using now for clarity

doing a problem

wait were not in a channel no i cant

nevermind lol nbd

[odd edge]

@hidden needle Has your question been resolved?

Channel closed

Closed by @hidden needle

Use .reopen if this was a mistake.

[odd edge]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[pulsar kelp]

what equation would I set up to find NM?

[pulsar kelp]

wouldnt i need LM first?

[brittle plinth]

your answer for the second statement is wrong

[pulsar kelp]

i wasnt worried about that

[brittle plinth]

what trig ratio should you use instead

[pulsar kelp]

cos

i think

[brittle plinth]

take a look

LN is the opposite side with respect to angle M

and NM is the adjacent side

so which function should you use

[pulsar kelp]

oh wait tan stands for tangent

tan

[brittle plinth]

alr

what ratio is tan of angle M?

[pulsar kelp]

tangent is opposite over adjacent right?

[brittle plinth]

yeah

[pulsar kelp]

k then it would be 21/NM

[brittle plinth]

and we know angle M is 20°

so tan(20°) = 21/NM

[pulsar kelp]

then NM(tan(20°)) = 21

and then NM=21(tan(20°))

ok thanks

.close

[odd edge]

Channel closed

Closed by @pulsar kelp

Use .reopen if this was a mistake.

[brittle plinth]

wrong

[pulsar kelp]

i meant 21/(tan(20°))

[brittle plinth]

ok

[odd edge]

Available help channel!

Send your question here to claim the channel.

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[twin vigil]

is there any mistake in my solution or mathematics

[twin vigil]

Options do not match

Well question 3

[mystic saffron]

i dont think you integrate position to get velocity

wat

im dumb

eh no that confused me, yeah its derivative instaed

[odd edge]

@twin vigil Has your question been resolved?

[odd edge]

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[dreamy totem]

can soneone give me a hint as to how I can do this quesiton

[dreamy totem]

Ping me pls

[narrow crypt]

do u know the limt definition of a derivative

@dreamy totem

bruh

[dreamy totem]

Wait is it just 1

Or am I tripping

[narrow crypt]

no

):

[narrow crypt]

js use this

hint: answer will be in terms of x and some constant

[lime sapphire]

I attched a hint...not the completed work

[narrow crypt]

oh dear

its not that deep

[wooden python]

is this handwritten AI output

[brittle plinth]

account just made today

very sus

[narrow crypt]

@dreamy totem have u found how to solve this

[regal dirge]

essentially, you need to use the definition of the derivative and apply the top property on f(x+h)

then everything simplifies nicely

[narrow crypt]

bruh

!noans

[odd edge]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[regal dirge]

alright fine

is it fine if i truncate it

[narrow crypt]

ye

[regal dirge]

cool

[odd edge]

@dreamy totem Has your question been resolved?

[dreamy totem]

I was lagging

[dreamy totem]

Degree 1 x?

[dreamy totem]

I don’t get how yiu got rid of the x

[narrow crypt]

What x

[narrow crypt]

Yes

[narrow crypt]

they got this from

[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}]
Consder ${f(x+h) = f(x) + f(h) + x^2h + hx^2}$. We have
[ f'(x) = \lim_{h \to 0}\frac{\cancel{f(x)} + f(h) + x^2h + hx^2 \cancel{- f(x)}}{h} = \lim_{h \to 0}\frac{f(h) + x^2h + hx^2}{h}]

[regal dirge]

^ yep

the f(x) cancels out

[clever fjord]

k

[regal dirge]

oh wow i've never seen \cancel in latex

noted

[silk fox]

$= \lim_{h \to 0 } \frac{f(h)}h + 2x^2 = 1 + 2x^2$

[clever fjord]

creeperdoesredstone

[silk fox]

wait hang on a minute

[regal dirge]

it's $xh^2 + hx^2$, not $x^2h + hx^2$

[clever fjord]

M1n3c4rt

[silk fox]

$= \lim_{h \to 0 } \frac{f(h)}h + x^2 + xh$

[clever fjord]

creeperdoesredstone

[odd edge]

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[silk fox]

which is equal to 1 + x^2

[odd edge]

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[toxic rose]

for $x \in \left( 0, \pi \right)$, how many solutions does the equation $\sin x + 2 \sin 2x - \sin 3x = 3$ have?

[clever fjord]

rak³en

[toxic rose]

now the solution I found in my modules solution book pdf is manipulating this into $4 \cos x - 4 \cos^2 x = 3 \csc x - 2$ and arguing that the $\text{LHS} \leq 1$ and $\text{RHS} \geq 1$

problem is

how tf was i supposed to see this

[wooden python]

csc

[toxic rose]

😅

[wooden python]

uhh hold on tho

let me do my magic

[zinc glacier]

surely a major part of the solution is doing the expansion in the first place

[toxic rose]

even if I had remembered this way of solving problems (using inequalities on both sides) i wouldnt have been able to find the right manipulation without spending like 5-10 minutes on this trying out stuff

[toxic rose]

well yes

[toxic rose]

[wooden python]

ok so i also ended up with f(cos(x)) = k/sin(x)

where f is a quadratic and k is a constant

cbf to figure out which ones

probably same ones as you

orrrrr wait hold on

not sure how you got cos^3(x) cause i didnt

[toxic rose]

yeah should be ^2 mb

[clever fjord]

rak³en

[toxic rose]

what i need is intuition behind getting my expression into this form

hopefully other than just 'seeing' it

[wooden python]

no fucking clue tbh

[odd edge]

@toxic rose Has your question been resolved?

[prime basalt]

I guess the "idea" was to express everything in terms of sin(x) and cos(x), and doing so forces us into the expression you mentioned

[odd edge]

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[quasi shard]

I have questions regarding 4a,b and c. For a can I say that if x is not an integer [x] = x-r . Where x-r=t. t≤x<t+1. And if it is an integer [x]=x

[hasty vortex]

I’m sorry I’m not used to the notations but is this the floor function

[2.1] = 2?

[silk fox]

seems like it

[quasi shard]

Yes

[agile sinew]

what you've claimed is right

[quasi shard]

So can I say that about the function because I don't know if it is a given property

[agile sinew]

usually we denote r by {x}

[stoic cloud]

Isn’t that the nereast integer function you’re showing? The floor function is ⌊𝑥⌋

[agile sinew]

that just means the fractional part of x

[quasi shard]

Or do I have to prove it

[agile sinew]

Floor function can also be written like that

we'll accept the notation in this context

[hasty vortex]

I know I don’t have that on my keyboard

[stoic cloud]

ahh okay

sorry

[hasty vortex]

Nah I’m confused as well

[quasi shard]

Because I feel like I can keep doing this type of thing and use it prove all the questions but feels like cheating

[hasty vortex]

[x] = x -r , 0<= r < 1

This is given

[quasi shard]

Like for [-x]= -x+r-1

[agile sinew]

no that's a definition!

[agile sinew]

Mhm

this is the definition you're allowed

[quasi shard]

And this

Also part of the definition?

[hasty vortex]

Take x = 2

That doesn’t work

If 0<=r<1

[quasi shard]

For negative x

If it isn't an integer

[hasty vortex]

My bad I didn’t see

[quasi shard]

If it is -x+r

[hasty vortex]

It should work

As you just used 2 definitions

[quasi shard]

So this is all part of the definition then

Now for 4c

[hasty vortex]

Oh wait ur suppose to prove those properties

[quasi shard]

That's what I'm confused about

[hasty vortex]

So what did you do to prove [-x] = -[x]-1

[quasi shard]

Well if I say that [-x] = -x+r-1. When -x is not an integer

But I don't know if I can do that

[hasty vortex]

Nah

[quasi shard]

@hasty vortex ?

So is [x] = x-r then allowed or not basic enough

[hasty vortex]

Do you know how to compare [x]

Idk what it’s called in English

like . < . <

[quasi shard]

Ye

[hasty vortex]

a<b<c

[quasi shard]

So t≤x<t+1

[hasty vortex]

Where t = [x]

Right?

@quasi shard

[quasi shard]

Ye

[hasty vortex]

What ur looking is to find [-x]

[hasty vortex]

What do u think u can do here

[quasi shard]

Well if x is an integer we know that t=x

I don't think you need to back it up with anything

And then we have t<x

And for this case I'm saying we can say that then t=x-r

[hasty vortex]

No no forget about x-r

[quasi shard]

Ok well then t<x

[hasty vortex]

What do u think happens to this if u multiply everything by -1

[quasi shard]

+t≥-x>-(t+1)

[hasty vortex]

-t => -x > -t-1

[quasi shard]

Ye

[hasty vortex]

So : -t-1 < -x <= -t

[quasi shard]

Ye

[hasty vortex]

So by definition [-x] = -[x]-1

Do u get it ?

The greatest integer under -x is -t-1

And t = [x]

[quasi shard]

One sec

Let me process this

[hasty vortex]

There is an unique integer t such as t < x < t+1 if x isn’t an integer

[quasi shard]

But isn't this only true if x is an integer x?

[hasty vortex]

No [x] is always an integer

It’s the greatest integer under x

[5.6] = 5

[quasi shard]

Yes but how do we know that it integer for a general x is x is not an integer

We just know that there is an integer t which x is greater than

[hasty vortex]

t is not any integer

It’s the greatest one

There is only one t such as t < x < t+1

Because there can’t be any other integers between two consecutive jntegers

2 < 2.3 < 2+1

Get what I’m saying ?

[quasi shard]

Ye

Let say for question 4a then we have [x+n] = t and [x] =t1

[quasi shard]

Also for this we say that if -x is an integer than it must be -t because of the equality sign in the inequality?

[hasty vortex]

What?

Oh yes

[hasty vortex]

But here we’re talking about x non integers so there won’t be any equality signs here

[quasi shard]

@hasty vortex

[hasty vortex]

Alright so x+n = [x] + n + r

Where 0<=r<1

We’ll start from here

[quasi shard]

How did we get this?

[x] ≠ x if x is not an intger

[hasty vortex]

the definition we used earlier

x = [x] + r

0< r< 1

[amber veldt]

aka x mod 1

[hasty vortex]

alright so if x+n = [x] + n + r

then

[x+n] = [ [x] + n + r]

[quasi shard]

Then [[x]+n+r is bound by t+n≤ [x]+n+r < x+n+1 right?

[hasty vortex]

yes

but basically since r is between 0 and 1

[ [x] + n + r] = [x] + n

since [x] and n are both integers

[quasi shard]

So your saying instead of t we have [x] because it always must be an intger

[hasty vortex]

t = [x] we can call it wtv we want

[hasty vortex]

do u get this part?

[quasi shard]

Ye because it will always equal the lower bound since it can never be an integer

[hasty vortex]

So in conclusion we have [x+n] = [[x]+n+r] = [x]+n

So [x+n] = [x]+n

Where n is an integer

[quasi shard]

Yes

[hasty vortex]

which is what we wanted to prove on 4a

all good?

[quasi shard]

Ye

[quasi shard]

Quick question how can we just assume this to be true?

[hasty vortex]

x = [x ]+ r

add n both sides

[quasi shard]

Ok

Thx

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[simple vortex]

I'm at the point where I discovered the angles but idk what to do to find the measurements of each one

[simple vortex]

do I add the 90 and 60 to a and the 45 +90 to B or what exactly

[lean yew]

consider triangle ABD first

what kind of a triangle is it?

[simple vortex]

45-45-90

[lean yew]

what are the relationships of the lengths of its sides

[lean yew]

i mean, yeah

but what kind of triangle?

[simple vortex]

right triangle

[lean yew]

other than that?

hint: the two 45s

[simple vortex]

isoceles right triangle

if that's what you're referring to

[lean yew]

correct

so what are the relationships of the lengths of the sides of an isosceles triangle?

[simple vortex]

2 equal sides

on the adjacent and opposite

[lean yew]

could you label the two equal sides?

by letters

[simple vortex]

AD and AB

[lean yew]

good

next, notice that AB is one of the sides of the triangle on the right

[simple vortex]

yep

[lean yew]

now i'm gonna use a bit of a trigonometry "hack"

[simple vortex]

Ohh its equal

[lean yew]

see the 30 and 60 there?

[simple vortex]

yep

[lean yew]

now, which is bigger: sin 30, or sin 60?

[simple vortex]

60

[lean yew]

ok good

[simple vortex]

So it's bc

[lean yew]

and the sine is the ratio of which two sides of a right triangle?

[simple vortex]

opposite to hypotenuse

[lean yew]

excellent

so now let's compare the two sines

[simple vortex]

sorry for the early conclusions

but go on

[lean yew]

since they are sines of two angles in the same triangle, the hypotenuse is the same, and we are really comparing opposite side to opposite side

the side opposite 30 is AB, and the side opposite 60 is BC

you concluded that sin 60 > sin 30, so it must then follow that BC > AB

but you also noted that AB = AD

so final conclusion?

[simple vortex]

BC>AD

[lean yew]

there we go

[simple vortex]

well thanks

[lean yew]

nps

[simple vortex]

I'll be closing this chat now but you were a great help. Hope it works for everyone else too the same way it worked for me

have a great day

.close

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[vapid estuary]

where do i start..

[strange aspen]

just apply the properties of 30-60-90 and 45-45-90 triangles

[vivid wyvern]

Start by finding the value of BC

[vapid estuary]

right right so what does this mean

[strange aspen]

i'll provide an image just a moment

[vapid estuary]

okay

sub hour celeste run

[strange aspen]

never saw someone acknowledge it lol

[vapid estuary]

bring me that image

[strange aspen]

here

[vapid estuary]

im actually confused, what is this image supposed to be pointing out

[strange aspen]

in 30-60-90 triangle if the side length adjacent to the 30 is x then you can find the other sides by multiplying by sqrt(3) for the other leg and 2 for the hypotenuse

and in reverse if for example the side adjacent to the 60 divide by sqrt(3) to find the side adjacent to 30
and multiply by 2/sqrt(3) to find the hypotenuse
it's eaiser when you work it out rather than me saying it

[vapid estuary]

i see

thanks

.close

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[remote axle]

I found that its not continuous when x=2, and its partial derivative with respect to y is not continuous at only that point as well, therefore the interval must not include x=2, but must include x=3 due to the initial condition. However there are 2 answers that satisfy this, (2,inf) and (2,pi), so im not sure if i did something wrong or if i am supposed to pick (2,inf) just because its a bigger interval, or what

[meager juniper]

What happens if sin(x) = 0?

[remote axle]

so 0 is also banned, but that doesnt change my problem right?

[meager juniper]

Well, if 0 is also banned, does this limit the interval at all?

[remote axle]

it cant contain 0 or 2, but must contain 3, so none of the interval can be less than 2, less than 0, or between 0 and 2, so then anything over 2 is fair game no?

[meager juniper]

Notice that sin(x) = 0 is banned

Not x = 0

[remote axle]

sin(0)=0 no?

wait

[meager juniper]

Sure, but are there other values of x for which that is true?

[remote axle]

i see

thank you

my eyes have been opened

[remote axle]

the gears started turning as soon as i said this lol

[meager juniper]

😄

[remote axle]

.close

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[iron sky]

im supposed to know how to add radicals and im mixed the rules with multiplying so all my numbers are off

[wooden python]

show the question(s)?

[iron sky]

3 ab

[wooden python]

what's this gotta do with radicals

[iron sky]

uhh that’s what the unit is

😭

[wooden python]

you sure you're not looking at the wrong unit's name?

radicals are like, square roots and shit

[iron sky]

wait that was

last unit

i think is is

[wooden python]

this looks more like rational functions

or algebraic fractions

[iron sky]

YES

LOL

[wooden python]

whatever you wanna call them

[iron sky]

😭😭

[wooden python]

anyway ok what's your status for 3a

!status

[odd edge]

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

[iron sky]

okay i am stick on a small part

i checked the answer key

and i’m missing a negative

[wooden python]

well show your work then

[wooden python]

,rccw

[iron sky]

wait what do i do with that

[wooden python]

all i did was rotate your picture

[iron sky]

ohh

[wooden python]

anyway

you multiplied top and bottom of the second fraction by x-3, which is good

but i don't quite follow what you did afterward

[iron sky]

i rewrote it

like the first part

of the question

[wooden python]

i mean it sounds like you kinda just

got lost in the sauce a bit

cause i see you just trailing off and then writing down what appears to be the final answer with a "Why?" on top of it

you could've simply written your next step as $$\frac{1-4(x-3)}{(x-3)(x+1)}$$

[clever fjord]

Ann

[iron sky]

and then do i foil

[wooden python]

you expand

distribute

there's no FOIL to be had (nothing that looks like(__+__)(__+__) ) and i recommend against that mnemonic

[iron sky]

wait

is this on the wrong track

[wooden python]

yes, very!

this isn't (1-4)(x-3).

it's -4(x-3)+1.

your infatuation with the FOIL mnemonic makes you hallucinate it in places where it is not applicable

[iron sky]

is four negative because of the subtract sign

[wooden python]

yes

it was being subtracted and it still is being subtracted

[iron sky]

but what do i do after

[wooden python]

after what

[iron sky]

because the answer is supposed to be

sory is a bit blurry

[wooden python]

you shouldn't be peeking at the answer key all the time.

peeking at the answer key and trying to strategize from there is a bad idea

[iron sky]

yea i was checking

[wooden python]

anyway, after getting to $$\frac{-4(x-3)+1}{(x-3)(x+1)}$$ you distribute the $-4$ into the brackets.

[clever fjord]

Ann

[wooden python]

you should stop peeking at the answer key.

and also you should answer the questions i ask you. "the answer should look like this" does NOT answer the question "after what?"

[iron sky]

uh

[spring arrow]

Stop looking at the answer key how are you supposed to learn

[iron sky]

i distributed

the 4

to x-3

[grizzled gazelle]

The correct way of learning shouldnt be looking at the answer key :/

[wooden python]

@spring arrow @grizzled gazelle i said the very same thing btw

[wooden python]

ok, show what you get after doing that

[grizzled gazelle]

ikr, like literally

[iron sky]

-4x-13+1

[spring arrow]

Someone I taught used to do that and she failed most of her classes

[wooden python]

-4*(-3) is not -13

[spring arrow]

This is a lesson learn 100%

[wooden python]

you messed up the multiplication AND the sign

[iron sky]

-4x-12

[wooden python]

AND the sign

negative times negative = ?

[iron sky]

12

oops

sry

[wooden python]

did you fix this error now

[iron sky]

its at -4x+12

[wooden python]

your numerator should be -4x+12+1

that +1 never went anywhere

[iron sky]

-4x+13

did the one turn positive bc the munis is connected to the four

[wooden python]

it didn't "turn" positive...

it always has been

[iron sky]

ohh

thanks

for 3b i made them all the same denominator but cant figure where to from there

[wooden python]

,rccw

[wooden python]

$x^2 - 4 \neq (x-4)(x+4)$

[clever fjord]

Ann

[wooden python]

you messed up the difference of squares

remember it's $a^2 - b^{\color{red}2} = (a-b)(a+b)$

[clever fjord]

Ann

[wooden python]

can you tell me what the correct factorization is for x^2 - 4?

@iron sky

[iron sky]

(x-2)(x+2)

[wooden python]

right

now redo 3b from scratch with this factorization in mind.

[iron sky]

okok thx

[wooden python]

(your other denominator was factorized correctly.)

[iron sky]

just making sure that when it’s add or subtract you can’t cancel out what’s on the bottom right

[wooden python]

when it's add or subtract
too vague.

show me exactly what you're trying to cancel here.

[iron sky]

instead of multiplying radicals

[wooden python]

there are no radicals here...

there were never any radicals in the questions we've been doing.

[wooden python]

,rccw

[wooden python]

this one is unreadable.

[wooden python]

and in this one your last step is not legal.

idk what you're trying to do but it looks like you got lost in the sauce.

so let me try to set you back on the right track:

you want to add these fractions.

in order to add these fractions, you want their denominators to be the same.

in order to make their denominators the same, you need to convert each fraction into an equivalent fraction with the common denom which you want.

the COMMON DENOMINATOR in your case is (x+2)(x-2)(x-10).

therefore, your next step ought to be as follows, with the newly introduced factors highlighted in red: $$\frac{{\color{red}(x-2)}(x-5)}{{\color{red}(x-2)}(x+2)(x-10)} + \frac{(2x+1){\color{red}(x-10)}}{(x-2)(x+2){\color{red}(x-10)}}$$

[clever fjord]

Ann

[wooden python]

you don't need to get a red pen or anything, but i'm writing it this way specifically to make it a bit clearer what is going on.

the words COMMON DENOMINATOR should ring a bell. do they? yes or no.

[iron sky]

yes

my current is (x-17)(2x+1)\the common denominator

[wooden python]

(x-17)(2x+1) ???

how did this happen?

[iron sky]

uhh

i put them tgt

[wooden python]

again show your work.

put WHO together.

[iron sky]

it might be hard to read

[wooden python]

oh dear.

i mean ok i have like

suspicions as to what might've happened.

[wooden python]

,rccw

[wooden python]

ok so you jumped directly from
(x-2)(x-5) + (2x+1)(x-10)
to
(x-17)(2x+1)
with nothing in between

is that what i am seeing?

[iron sky]

heh yes…

[wooden python]

once again

how did this happen?

[iron sky]

gulp…

[wooden python]

like it's pure BS but i want to see if there was any logic at all behind it.

or if you just made a number smoothie.

thinking you were gonna cook but you actually burned the pan

[iron sky]

i missed the lesssons heh

[wooden python]

thankfully not the entire kitchen, but still.

[wooden python]

ok so it was just something random that you hoped was right, but no reasoning behind it.

got it.

[iron sky]

yea

[wooden python]

in that case let me once again put you on the right track.

we're again working on just the numerator here.

[iron sky]

okok thx again

[wooden python]

which right now is (x-2)(x-5) + (2x+1)(x-10)

if any of these factors were matching (i.e. one in the first product with one in the second), we could pull that one out and save ourselves some effort.

unfortunately, we have no such luck here.

so we will have to do it the "hard" way.

namely, you expand everything (if you're FOIL-brained, that's what you do -- twice) and then collect like terms.

and in doing all this, DON'T TOUCH THE DENOMINATOR AT ALL.

do you understand what to do? yes or no.

[iron sky]

okok yes

[wooden python]

aight do it

[iron sky]

ill brb inna min

can i collect everything after expanding

currently here

[wooden python]

let me just double-check your expansions

[iron sky]

praying it’s right

[wooden python]

there is no need for prayer. i am not a goddess.

anyway yes your expansions are correct, and yes you should collect like terms just as i told you the first time around.

[iron sky]

okay u have a question that’s going to sound really dumb

i

[wooden python]

well, better to be a fool now than to be a fool 30 years later.

so ask away.

[iron sky]

is 2x^2+x^2

2x^2

[wooden python]

no

2x^2 + x^2 isn't equal to 2x^2 any more than two dollars plus a dollar would leave you with two dollars still

[iron sky]

is it 3x^2

[wooden python]

it is indeed

remember that when a term has an unwritten/"missing" coefficient, that means the coefficient is 1.

or -1 if the term has a minus sign in front.

[iron sky]

im at -3x^2-26

[wooden python]

wait what

how did the 3x^2 grow a minus sign

[lean yew]

huh

x^2 + 2x^2

[wooden python]

x^2 + 2x^2 = +3x^2 just as we went over now

[lean yew]

= -3x^2?

[iron sky]

NO NEG

[wooden python]

also how did a raw -26 happen

[iron sky]

IT WAS AN WRASER BIT

[lean yew]

how did an eraser get onto discord

[iron sky]

i thought i wrote it but it was eraser dust

[lean yew]

but anyway

[wooden python]

-7x - 19x is not a raw number. i think you forgor the x

[lean yew]

i think you would benefit from grouping the like terms first

x^2 with x^2, x with x

[lean yew]

looks ok to me

[tacit wasp]

By the way, try to make photos that aren't that blurred @iron sky
It's very difficult to read

[iron sky]

sorry yea

[wooden python]

3x^2 - 26x is the correct simplified form for the numerator.

so to bring us back to the original fraction, you now have: $\frac{3x^2-26x}{(x-2)(x+2)(x-10)}$

[clever fjord]

Ann

[wooden python]

i would call it done at this point, as the numerator could factor to x(3x-26) but that's not going to cancel with anything from the bottom.