help-33

[gilded smelt]

thought it was +4 😭

[tacit fjord]

well i think the top term

is a square

if u believe hard enough

[gilded smelt]

....

[tacit fjord]

im sure theres some crazy substitution that gets this no?

use complex

ez

[full flume]

aight i'm out, but if somehow doing this by hand get's us the book answer then hmu lol

[gilded smelt]

cya

[tacit fjord]

$$2\int\sqrt{\frac{5x^2-4}{x^2-4}},dx$$

[near rose]

💀

[gilded smelt]

$$\int\sqrt{\frac{5x^2-4}{x^2-4}},dx=7.2596$$

[elfin berry]

Scythe

[gilded smelt]

ta da, substitution done

[elfin berry]

Shuri2060

[near rose]

.close

[marsh citrus]

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[shrewd pawn]

how do we figure this out?

can someone give me some tips on how to think of it please

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[shrewd pawn]

its not working

im getting to this

.close

[marsh citrus]

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[wooden path]

I think it is: 5 times 4 times 3 times 2

i just wanna know if im doing it right

[sleek lake]

i think so

[wooden path]

cool thx

and for this it would be 5 times 5 times 5 times 2?

[sleek lake]

yeah

totally

[wooden path]

cool

for this one, i got (14 times 12)/2

[crystal sage]

it's 5P4

im pretty sure

idk im rusty on my combinations and premutations

[sleek lake]

yes checks out

[wooden path]

thx

[sleek lake]

(14 choose 2) − 7

[wooden path]

cool

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[novel fjord]

how did you get 12:40

[shrewd pawn]

but its wrong

[novel fjord]

im assuming you had a typo in your calculator somewhere

74/67 is not 1.15625

[tacit fjord]

,w 74/67

[elfin berry]

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[shrewd pawn]

12:36

[novel fjord]

there ya go

[shrewd pawn]

i guess i put it wrong in calculator

thanks

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[shrewd pawn]

i dont understand how they did this

can someone explain to me please

625 from where did it come?

[novel fjord]

completing the square after factoring out a 5

[shrewd pawn]

i dont understand it yet

[novel fjord]

do you know how to complete the square?

[shrewd pawn]

im getting to this point

isnt it divide by 2 and square it

[novel fjord]

yep

that's how they got 625

(50/2)^2

[shrewd pawn]

but its not working out with me

[novel fjord]

you're using different numbers lol

[shrewd pawn]

ahahah

[novel fjord]

ah lol

incorrectly factored out -5

[shrewd pawn]

whats wrong ?

[novel fjord]

sign errors

you have a +20t

[shrewd pawn]

so 10 meters?

[novel fjord]

now you have a different sign error lol

the +t^2 was correct

that's another easy fix though

[shrewd pawn]

pufff

my brain

[novel fjord]

the +100 should be within the parenthesis you have there

[steady sparrow]

Messi do you have derivatives able to use

[novel fjord]

that makes it a perfect square

[novel fjord]

his original problem shows completing the square as the method for getting an answer

[shrewd pawn]

i wil use derivatives later topics

so i got to this point

whats next?

[novel fjord]

+100 should be in parenthesis

t^2-20t+100 is a perfect square trinomial

[novel fjord]

right, now you can factor the parenthesis there

(what's inside of them, that is)

[shrewd pawn]

factor T?

[novel fjord]

factor t^2-20t+100

[novel fjord]

don't forget the -100+10 you have on the outside

[shrewd pawn]

ok

[novel fjord]

now you have an expression in vertex form

[shrewd pawn]

so how do i know how far did it travel?

[novel fjord]

you're looking for the maximum height

[shrewd pawn]

90m?

[novel fjord]

now that you have your equation in vertex form, you know the vertex

yep

[shrewd pawn]

but the correct asnwer says 510

[novel fjord]

huh

[shrewd pawn]

idk

[novel fjord]

gimme a sec...

aaah i'm a dumbass

been a while since i completed the square lol

even though it shows how in your original picture sent

[novel fjord]

in here, the -100 should be within the parenthesis as well

[novel fjord]

now you can factor the perfect square again

but you also have the -5*-100

[shrewd pawn]

it shouldnt make a difference?

[novel fjord]

yeah it will

you'll have $-5((t-10)^2-100)+10$

[elfin berry]

a disappointing son

[novel fjord]

distribute the -5

[novel fjord]

there ya go

[shrewd pawn]

this is crazy

so much pain to go through

thanks for the help

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[marsh citrus]

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[marsh citrus]

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[still temple]

Is it possible to find,
$\int_0^1 (x^x)^{(x^x)^{(x^x)^{...}}} dx$?

Or just it's indefinite integral?

[elfin berry]

!!!

[marsh citrus]

@still temple Has your question been resolved?

[tulip idol]

$y= (x^x)^{(x^x)^{(x^x)^{...}}} \implies y=(x^x)^y = (x^{xy})$

[elfin berry]

Ansh

[tulip idol]

hmm, can you establish a relation between dx and dy?

[elfin berry]

Ansh

[tulip idol]

which, doesn't have an indefinite integral 💀

but that's just my understanding and I've got no counter as to if there's some other y substitution that leads to a different expression which might be calculated!?

[still temple]

I thought I need to find the expression in terms of y. Right?

[tulip idol]

Yes ! the expression would just be y^y right?

I just wrote y = x^xy so that you can establish a relation between dx and dy

by taking log both sides

[still temple]

How can we solve $y = x^{xy}$?

[elfin berry]

!!!

[still temple]

For y

Taking log?

[tulip idol]

we aren't solving for it but rather looking out for dy/dx

[still temple]

🤔

[tulip idol]

,w \int_0^1 (x^x)^{(x^x)^{(x^x)^{...}}} dx

[elfin berry]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
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Requested by Ansh#3502

[tulip idol]

smh

,w integrate from 0 to 1 (x^x)^{(x^x)^{(x^x)^{...}}} dx

[still temple]

Lol

[elfin berry]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by Ansh#3502

[still temple]

I got $\frac{dy}{dx} = \frac{-y(\log(x) + 1)}{(x \log(x) - \frac1y)}$

[elfin berry]

!!!

[tulip idol]

,w integrate from 0 to 1 y^y/y^2 dy

[elfin berry]

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Requested by Ansh#3502

[tulip idol]

doesn't converge

easy

UwU

[still temple]

How did you get this?

[tulip idol]

magikkk

[still temple]

Bruh

[tulip idol]

anyways, magic integrals don't happen! Even I don't know yet how to check which integrals exist and which don't smh

[still temple]

I'm asking how did you get this?

[still temple]

And this!

[tulip idol]

that's not the point, the point is: you're wasting time on an integral that does not exist and neither of us know how to determine whether an integral exists or not 🤦‍♂️

hence, this discussion is fruitless

[still temple]

Ohh

But I didn't understand why does it diverge 🥲

Anyways thanks.

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[still temple]

$$\int_0^\infty \frac{x^{49}}{(1+x)^{51}} dx$$

I tried the following method:
$$I = \int_0^\infty \frac{x^{49}}{(1+x)^{51}}dx$$
$$I = \int_0^\infty \frac{x^{51}}{x^2(1+x)^{51}}dx$$
$$I = \int_0^\infty \frac{1}{x^2(1+\frac{1}{x})^{51}}dx$$

Now let (1 + 1/x) = u
-dx/x² = du

$$I = \int_0^\infty \frac{-du}{u^{51}}$$

Now how to deal with this? Upper limit is annoying me.

[elfin berry]

!!!

[still temple]

Hmm

Alright

1/50 [1/u⁵⁰] with limits ∞ and 0.

At ∞, it will be 0

And what about 0?

Ok solved it

Lemme write it down

GG well played

wtf 😭😭

what grade r u in

What

[still temple]

I forgot to change the limits after substitution 🥲

Ooooof

[still temple]

11th

I’m 12th

Me too

I mean I got out of 12th in 2021 august

Im now preparing for uni entrance exam

U applying to US unis?

Or IIT?

No india

Ooooo

ISI/CMI

Very hard

I don’t like IIT

Engineering and me have a love hate relationship

Especially chemistry

So I decided to move on

Ah good luck man

Im preparing for pure math exams

[still temple]

Oh are you Indian? Me too 🙂

Yes

But I'm commerce stream 🥲

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[junior robin]

I might actually keep a tab open as I encounter questions I'm stuck on, right now this is my problem

[junior robin]

what I did was sub in (r+2) in the equation
V(r)=4/3 pi r^3

[late grove]

you sub it into r

[junior robin]

was it expressing it in terms of pi and r that made it incorrect?

[late grove]

so it becomes:

$\frac{4}{3}\pi(r+2)^3$

[junior robin]

r+2*, but I think I did that correctly

[elfin berry]

alias

[late grove]

did you use any special methods to expand it?

oh, by the way

[junior robin]

For long and tedious expanding I do this lol

[late grove]

you need to multiply $(r+2)^3$, the WHOLE THING, by $\frac{4}{3}\pi$

[elfin berry]

alias

[late grove]

that was probably your mistake haha

[junior robin]

ahhh ok

[late grove]

btw the actual expansion is correct

that was your only issue

it's like if you had $4(x+2)^2$, you'd do $(x+2)^2$ and multiply that whole thing by 4

[elfin berry]

alias

[junior robin]

this is kind of annoying, I'm just adding 4/3pi before every variable 😭

I can simplify 2 of them I think tho

[late grove]

yeah, 6 and 12 can be divided by the 3

[junior robin]

I don't think so actually

gotcha. thanks

[late grove]

np!

[junior robin]

alright back to square one

the question says to express the answer in terms of pi and r. does that have something to do with it?

[marsh citrus]

@junior robin Has your question been resolved?

[oak bay]

I think you have to take the difference b/w the 2 volumes

[junior robin]

tbh I just skipped the question, I'll ask about it during class

I do have a question about how to move on from here though

I don't remember the fraction rules at all

damn I guess the tutors just took a break lol

[surreal basin]

Is that the full question?

[junior robin]

yeah

I posted the question earlier

yes this is a new question because I decided to skip the volume question

[junior robin]

the last 1/x-1/a thing is the first step to solving the problem but idk what to do next

[surreal basin]

What’s the question asking for?

[junior robin]

it's giving f(x)
and then asking you to solve f(x)-f(a)/x-a

[oak bay]

Is some limit given?

Like sat lim x-->a or something

[junior robin]

nah, it just asks to simplify

I feel like there's a rule here but I just don't know it

this is purely algebra review for calc 1

so no limits yet

[oak bay]

Hmm

Like this maybe

Ignore the triangle

@junior robin

[junior robin]

oh wow whats the thought process behind that

I just put it into symbolab and that's the right answer

symbolab said to use something that made
1/x-1/a equal to a-x/ax

[oak bay]

Yeah that's what i did..

I just multipled by taking LCM in the numerator

[junior robin]

the triangle threw me off

lol

I thought you used something trig related

[oak bay]

Nah just ignore it lol

[junior robin]

but thanks regardless

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[still temple]

guys im stuck on trig substitution

if i draw the reference triangle, it should have a /a in the denominator, but the answer's incorrect

?

Same with this one

[tacit fjord]

This the original one?

[still temple]

yes

[tacit fjord]

I'm sure that's the standard form for one of the inverse trig

Look them up or try differentiating them all

If its not inverse trig itll be inverse hyperbolic ig

[still temple]

umm but hte instruction was specifically to use the trig sub

[vestal stratus]

arcsinx I believe

d/dx arcsinx = 1/sqrt(1-x^2)

[tacit fjord]

Yeah sure trig sub

I mean you're gonna end up with the same thing

[tacit fjord]

If you notice this, then this strongly suggests which sub to use.

Huh its minus the inside though 🤔

@still temple im a bit confused, can you state what the original integral you're stuck on is btw

[still temple]

this one

[tacit fjord]

What's wrong with this answer

oh you're saying its wrong

[still temple]

see the x/a + sqrt(x^2 - a^2)/a at the bottom?

that was what I got

but the answer is apparently the stuff in the ln =

so my question is where did the 'a' go?

[tacit fjord]

are you sure this is how it works

[still temple]

yeah

we build a reference triangle

[tacit fjord]

never done this kinda thing before to check

[still temple]

and shove it back in

[tacit fjord]

So sub is sec theta = x/a

Why did you write sqrt(x^2 - a^2) for sec theta

instead of x/a

?

wut

[still temple]

oh that's not in the particular order

the stuff inside the ln()

[tacit fjord]

ok sure sure

[still temple]

but where did the /a go

[tacit fjord]

Where does this come from

[still temple]

that's apparently the answer

you substitute sec theta from the reference triangle back in

tan theta too

[tacit fjord]

well clearly not though.

im gonna suspect theres something wrong with the d(theta)

[tacit fjord]

ok ok

a sec theta...

[jagged kestrel]

should be a factor of a^2 on the bottom

[still temple]

where?

[tacit fjord]

yh I sus your substitution is not quite right

[jagged kestrel]

in the very first substitution

[tacit fjord]

I would say redo this step

[jagged kestrel]

before cancellation

[tacit fjord]

bit by bit.

[still temple]

uhhhh

you're right 1 sec

[jagged kestrel]

actually wait

nvm

[tacit fjord]

what

are we still hunting for it

[jagged kestrel]

break up the logarithm. you will have your answer -log(a)+C which can be absorbed into the constant

[tacit fjord]

ahahaha

yh stuff like that happens ofc.

@still temple

[jagged kestrel]

it's just a fancier way of simplifying lol

[still temple]

wait no

im back at square 1

[jagged kestrel]

no you did it right

the first time

[still temple]

How does the ln become this?

[jagged kestrel]

plug in your values for sec and tan

you will have the form of ln(M/a)+c yes?

[still temple]

yes

/a won't disappear

[jagged kestrel]

but you can use the properties of logs to break apart the logarithm

[still temple]

OH MY LORD

[jagged kestrel]

ln(M/a)=ln(M) -ln(a)

[still temple]

and since

ln(a) is a constant

we can yeet it into C

[jagged kestrel]

and then absorb yes

[still temple]

I love you guys so much

[jagged kestrel]

tricky tricky

lol

[still temple]

See this is why calc should be banned

the amount of dopamine kick I get when something gets solved is

great

too great

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[still temple]

@jagged kestrel@tacit fjordthank you

[marsh citrus]

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[ripe hull]

Can someone help guide me through this? I'm having a hard time writing a proof

[ripe hull]

I know there's supposed to be two cases as to where C can lie

I was given a hint by the professor to make sure on when I reach the contradiction with one of my axioms (two distinct points determine a line)

[marsh citrus]

@ripe hull Has your question been resolved?

[tacit fjord]

explain the notation.

[ripe hull]

When we assume C lies on the opposite side of AD, then is it its own point? Like out of the plane and not on the line?

[tacit fjord]

Can you translate the symbols to english

In particular, I need the H(...)

[ripe hull]

Ah

Probe that if B is not in line AD and B is between A and C, then C is in the half plane H (B, (line) AD)

Prove*

[tacit fjord]

I am not familiar with the geometrical axioms you are using, sorry.

So I wouldn't be able to help

[ripe hull]

Its okay, no worries

Thank you for trying tho, I appreciate it

[marsh citrus]

@ripe hull Has your question been resolved?

[still temple]

Could i ask for help here please?

[marsh citrus]

@ripe hull Has your question been resolved?

[ripe hull]

.close

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[clear kettle]

Can I have help please?

[clear kettle]

<@&286206848099549185> How does this help chat work? Do I wait for someone to come?

[bright jay]

Yes

You just wait

[jade iron]

Yes, when you take the derivative of the variable you are working with. The other variables are held constant. Good job

For future reference:

f_x: 2xyz
f_y: x^2z
f_z:x^2

[clear kettle]

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[still temple]

If f={(2,-4)} what is f^-1(f inverse)?

[warped sedge]

im not sure but wont it just be f?

[novel fjord]

...no

[warped sedge]

ah sorry then

[novel fjord]

what happens to domain and range when you take the inverse of a function?

[warped sedge]

they get swapped

[bright jay]

Not you

[novel fjord]

you are not the person asking the question

[still temple]

lol i am very bad at function 😓

[novel fjord]

well luckily for you, somebody gave you the answer

[still temple]

hehe yea

thank you lol

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[serene oar]

im not sure how to sketch this, i cant find the x and y intercept

[mystic minnow]

For the y-intercept set x=0.

[serene oar]

yes but the thing is if you solve for x and y both intercepts = 0

and if i check the answers the intercept is (1,4)

x=1, y=4

[mystic minnow]

that's not an intercept. that's just a point on the graph

Are you sure you're trying to find the right thing?

[mystic minnow]

And just as a point, there are no intercepts. They don't equal 0 because they don't exist.

[clear acorn]

can anyone help me?

[wind lion]

#❓how-to-get-help

[mystic minnow]

claim an open room. See #❓how-to-get-help

[clear acorn]

@wind lion@mystic minnow thanku so much

[serene oar]

oh i see

so how to i find the point on the graph

?

[mystic minnow]

Any point? plug in an x value and get the y value. If x = 2, then y = 2/2 = 1. If x = 10, then y = 4/10 = 2/5

I don't know what the question here is. You just showed me an equality.

[serene oar]

im sorry i just got pretty confused

it kinda makes sense now ig

thanks

.close

[marsh citrus]

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[mystic minnow]

np

[marsh citrus]

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[sturdy forge]

1, 3, 7 I got true for all

[wind lion]

What is your reasoning for 7

[sturdy forge]

If it's discontinuous then wouldn't the point not exist?

[wind lion]

That's true, but a limit can exist in a situation like

At x = a

The limit still exists, even though the point doesn't

[sturdy forge]

true good point

[wind lion]

So this function is discontinuous at x = a

Yet the limit is 4 or whatever

[sturdy forge]

not sure how I didn't think about that 😂 makes perfect sense though lol

thanks for the help!

[wind lion]

Yep, no problem

[sturdy forge]

.close

[marsh citrus]

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[rapid island]

open

Suppose that $\sin(\theta) = \frac{2x}{1+x^2} ~~~ -\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2}$

[elfin berry]

azeem321

[rapid island]

What value do we give for $\cos(\theta)?$ So far, I have $c =\pm \sqrt{1-s^2} \implies \cos(\theta) = \pm \frac{|1-x^2|}{1+x^2}$

[elfin berry]

azeem321

[rapid island]

Do we drop the $\pm$? if so, why?

[elfin berry]

azeem321

[static quarry]

cosine is nonnegative for angles in that range.

[rapid island]

nvm, got it

thanks

.close

[marsh citrus]

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[cyan loom]

Could someone compute this for me

[cunning citrus]

no

use a calculator

[cyan loom]

I can't seem to get the right answer

[sturdy forge]

I'll try to enter it sec

[cyan loom]

I just wanna see if people are getting thr answer or I'm doing it all wrong

I'm getting about 200

But it should be 70

[cunning citrus]

,w ln(2.510^(-3))/(2.9710^(-2))-ln[210^(-2)]/(2.9710^(-2))

[static quarry]

Would be very surprising if the answer is actually an integer.

[cyan loom]

Sorry it's ln of the top expression only

[sturdy forge]

I got -3.168 for mine

[cyan loom]

Not the whole thing oops

[sturdy forge]

gotcha

[coral hazel]

welcome to chemistry where everyone gets a different answer

[cyan loom]

Frick idk how to solve this then

Unless you guys wanna give it a go 🤣🤣

[cunning citrus]

#old-network for the Chem server.

[coral hazel]

are you very sure your equation manipulation is right

[cyan loom]

I'm using the wrong formula then or something

[sturdy forge]

is this ln [200*10^-2]?

[elfin berry]

Results provided by WolframAlpha

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

Requested by Omegabet_#9771

[sturdy forge]

^ that formula is wrong

[cyan loom]

Wow

[sturdy forge]

in the wolfram alpha

oh nvm

now it's right

[cunning citrus]

messed up signs and messed up dividing by k

[sturdy forge]

@cunning citrus beat me to it xD

[cyan loom]

K I'd actually psoitivebi just didn't show full equation

My b there

But why did that computer 70 now?

[cunning citrus]

$\ln(A)=-kt+\ln(A_0) \ kt=\ln(\frac{A_0}{A}) \ t=\frac{1}{k}\ln(\frac{A_0}{A})$

[elfin berry]

Mosh

[cyan loom]

Oh when subtracting them you can rewrite it as a fraction

I forgot that rule

[cunning citrus]

log laws.

[cyan loom]

Yeah okay that makes more sense

Thank you for the help guys !

.close

[marsh citrus]

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[marsh citrus]

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[devout compass]

I did tan-1(15.1/31.9)

25.33073234

<@&286206848099549185>

<@&286206848099549185>

[iron portal]

Hello

@devout compass

[devout compass]

hi

[iron portal]

do you stilll need help?

[devout compass]

:d

you don't know how long I've been waiting X_X

[iron portal]

Sorry!

[devout compass]

atleast an hour and 30 min

😦

it's like everyone skiped me

😦

[iron portal]

ok let me see

what math are you in?

[devout compass]

uhh

[iron portal]

well ike name of class?

precalc?

[devout compass]

oh uhh

it's

trig

trig right angles

right triangle

lol

[iron portal]

myguy

legit

use inverse tan

[devout compass]

I did

I got 25.33 something

[iron portal]

then youre chilling

[devout compass]

maybe I got decimal placement wrong?

I put 25 and it was wrong

[iron portal]

why does it matter, wdym

Oh

you in degree mode?

[devout compass]

ye

[iron portal]

ok lemme see

25.3307?

[devout compass]

ye

[iron portal]

well...

type the wholething in?

idk what's wrong

do 25.3307

in the input thing

[devout compass]

oh

I got it!

it was 25,3

weird

25.3*

[iron portal]

HA

OMG

I FEEL SO BAD

[devout compass]

i don't understand the rounding thing with that dumb chart :/

[iron portal]

you spent a whole hour

[devout compass]

ikr wtf

[iron portal]

dw not ur fault

[devout compass]

ugh so stupid bro

[iron portal]

do you know sig figs?

it's liek how many digit sare there that are "important"

[devout compass]

sig and figs wdym by that

oh

sig digits

[iron portal]

ye

[devout compass]

naw

can you explain better than the chart?

lol

[iron portal]

hm

This man can teach better than anyone i know

https://www.google.com/url?sa=i&url=https%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3Dl2yuDvwYq5g&psig=AOvVaw2XC-xgJYzjJUfptTBU7H34&ust=1643945901340000&source=images&cd=vfe&ved=0CAkQjhxqFwoTCKDD7e3N4vUCFQAAAAAdAAAAABAD

[devout compass]

bet

I'll watch rn

[iron portal]

👍

lmk if u need help

[devout compass]

dam man

15min long

lol

[iron portal]

i mean if it helps it helps

you alr wasted an hour

[devout compass]

this 1

I got c wrong

how come

I did a2+b2=c2

@iron portal

[iron portal]

nonono

[devout compass]

no?

[iron portal]

in this case c is actually b

[devout compass]

hmm

[iron portal]

8.68 squared + c squared = 10.85 squared

[devout compass]

ok so hmm

me so confused lol

[iron portal]

c squared = 117.7225-75.3424

c squared = 42.3801

c = 6.51

[devout compass]

ohh ok ok

alr I get it now

thanks

this 1

why I get wrong?

B..... cos-1(9.17/14.66)

C..... sin-1(9.17/14.66)

b=14.66-9.17

b=214.9156-84.0889

b=130.8267

b=11.43794999

Nvm I figured it out

I round wrong

[marsh citrus]

@devout compass Has your question been resolved?

[devout compass]

this 1

finding a

I do uhh

a=7.12-6.04

a=50.6944-36.4816

a=14.2128

a=3.769986737

@iron portal

cuz that wouldn't make sense cuz the hypotinuse is smaller :/

.close

[marsh citrus]

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[rapid sapphire]

How would you go about proving $a^n +b^n \neq c^n, n>2$ and $a,b,c \in \mathbb{Z} $

[elfin berry]

suck2015

[rapid sapphire]

I know if it's $a^n + b^n = c^n$, then it's pretty much Fermat's Last Theorem?

[elfin berry]

suck2015

[knotty frost]

http://www.lbatalha.com/blog/feynman-on-fermats-last-theorem

yeah it would be fermat's last theoirem

[marsh citrus]

@rapid sapphire Has your question been resolved?

[iron portal]

?

[rapid sapphire]

Oh i'm just curious as to how you would prove it

[main idol]

http://scienzamedia.uniroma2.it/~eal/Wiles-Fermat.pdf

slightly more accessible proof
https://people.math.wisc.edu/~boston/869.pdf

[rapid sapphire]

ah ok , so i got this problem assigned for a homework, where a,b,c are equal to integers and the sides of a right triangle

I'm not sure how to go about the problem ( I guess the sides of the right triangle part makes it easier to approach than the actual FLT)

[tulip idol]

would've helped better if you shared the actual question along with ur query

[rapid sapphire]

Sure, prove for any $n, a^n +b^n \neq c^n, n>2$ and $a,b,c \in \mathbb{Z} $ and $a,b,c$ are the sides of a right triangle

<@&286206848099549185>

[main idol]

the proof is here, please stop pinging

[rapid sapphire]

sorry , that wasn't my question

By induction, how would one prove that if $a,b,c \in \mathbb{Z} $ are the sides of a right triangle, then for any $n, a^n +b^n \neq c^n, n>2$

[rapid sapphire]

lmao this is a hw problem not a research problem

[main idol]

i'm gonna block you if you keep pinging me

[rapid sapphire]

i literally just responded to you

you linked a research paper for a hw problem

[tulip idol]

You have $a^2 + b^2 = c^2$ yes?

[elfin berry]

Ansh

[rapid sapphire]

yeah

pythagoras

[tulip idol]

Now you only need to show that: \ $a^2 \cdot a^{n-2} + b^2 \cdot b^{n-2} \neq (a^2 + b^2) c^{n-2}$

And that's a hint

[elfin berry]

Ansh

[tulip idol]

May I ask which chapter or lesson are you taking right now? for your instructor to give this question

Is it a geometry lesson? surely not

[rapid sapphire]

ah abstract algebra

[tulip idol]

abstract algebra, I see

[tulip idol]

this the way then

[rapid sapphire]

ah ok thanks. I'll try that then

oh wait, that's pretty neat

I'm almost ashamed that I didn't see that from the get-go

[rapid sapphire]

@tulip idol sorry, one last ping: so in induction, usually you assume some c and show c+1 holds. I'm not too sure how to do that here. I think my approach works via strong induction because i end up satisfying c-1 though

because (c+1) = (2)+(c-1)

[tulip idol]

induction you show your proposition holds for k = 1

assume true for k = n

show for k = n + 1

Check out the article on strong induction on brilliant.org

[rapid sapphire]

oh yeah i know what they are but i don't think i

i'm allowed to use strong induction for this. I guess my induction is weak rn but I think I'm being thrown off by the "/="

[tulip idol]

what if you showed the inequality?

[rapid sapphire]

Oh wait

[tulip idol]

the RHS far succeeds the LHS for bigger and bigger ns

[rapid sapphire]

lol just as you said that, i realized how to do it, since showing the RHS>LHS is simple by multiplication

[tulip idol]

proving the last theorem in general is a pain, although for the triangle question, it's just a sweat

[rapid sapphire]

wait what do you mean by this? literally multiplying both sides by c and showing that they aren't equal?

i'm sorry for all of these q's, i've never really encountered an induction problem using "/=" before

[tulip idol]

[tulip idol]

show that for n > 2, the left hand side > right hand side

[marsh citrus]

@rapid sapphire Has your question been resolved?

[rapid sapphire]

ah ok thanks! sorry it took so long

[marsh citrus]

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[delicate lagoon]

Just want to know whether my answers are right: (fyi, this ain't a test, so I'm not cheating lol)

[delicate lagoon]

1. Lower limit, l = 2x + 1
2. 0
3. a) Mean
4. 3+9+13 = 25
5. 55
6. b) Bar Diagram
7. Temporal classification
8. x = 22

[marsh citrus]

@delicate lagoon Has your question been resolved?

[delicate lagoon]

<@&286206848099549185>

[delicate lagoon]

Hello?

[delicate lagoon]

[marsh citrus]

@delicate lagoon Has your question been resolved?

[twin girder]

5 is wrong

[thick sapphire]

Anyone can help me with Differential Equation?

[marsh citrus]

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[delicate lagoon]

.reopen

[marsh citrus]

✅

[delicate lagoon]

Oh, I see. So, it should be 55-51= 4?

Are the rest correct? Thank you for responding

[marsh citrus]

@delicate lagoon Has your question been resolved?

[delicate lagoon]

[delicate lagoon]

Help

<@&286206848099549185>

[delicate lagoon]

...

[faint pumice]

Seems about right but I don't think we can really help. To be clear, it is very not-math and more of vocabulary

[marsh citrus]

@delicate lagoon Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[delicate lagoon]

Oh okay, that's understandable. But "seems right" helps me and thanks for responding!🍀

[delicate lagoon]

.close

[marsh citrus]

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[light pecan]

i kinda need help with the procedure

[bright jay]

Plug in 3 as n

[light pecan]

ok sure again

thats what i did

so (3- 2) + (3-1)

right?

[bright jay]

It's functions so g(3 - 2) and g(3 - 1)

[light pecan]

ok

So, g1 + g2

would be 3

so g(3) = 3

?

[bright jay]

Not exactly

What's g(1)?

And g(2)?

[light pecan]

g(1) = -5

ohhhhhhhhhhh

that makes sense

ok

g (2) = 3

[bright jay]

And so you get?

[light pecan]

-2

[bright jay]

Yes

[light pecan]

oh

thanks

.close

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[light pecan]

i cant, each time there is a variable of sequences i keep. getting. it. wrong.

[light pecan]

so here my reasoning was f(n-1) = f(2)

therefore f(2) x -0.2

=

f (-0.4)

which i know is wrong

[rapid island]

What is f(2)

[light pecan]

its doesnt say

[rapid island]

Use this formula

[light pecan]

well yeah, thats what i typed here

[fierce pond]

$f(2) = f(1)*(-0.2)$

[elfin berry]

Erzis エルジス

[fierce pond]

that's how you get f(2)

and based on that you calculate f(3)

[light pecan]

ok, but how do you get this formula

[light pecan]

.

[rapid island]

It's given in your formula

You sub in 2 for n

[light pecan]

oh okay

i was replacing with 3

[rapid island]

What we have here is a recurrence relationship. It gives you a way of finding the next term in the sequence provided you know the previous term

in order to find f(3)

we must know the previous term

i.e. f(2)

[light pecan]

so f(2) = 25/-0.2

right?

= -125

[fierce pond]

no

it's multiplication

not division

[light pecan]

right

yeah mb

f2 = -5

?

[fierce pond]

correct

[light pecan]

so what now...

subtract -30

?

[fierce pond]

what? why?

just calculate f(3) using the formula

[light pecan]

ok def not

[fierce pond]

you have f(2)

so $f(3)=f(2)*(-0.2)$

[elfin berry]

Erzis エルジス

[light pecan]

ohh

[fierce pond]

the same way we calculated f(2)

[light pecan]

-5 x (-0.2)

[fierce pond]

yes

[light pecan]

1

right?

[fierce pond]

https://tenor.com/view/one-number-one-numero-uno-numeral1-number1-gif-20152870

[light pecan]

finally hahaha

thanks

[marsh citrus]

@light pecan Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

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[subtle pawn]

So if anyone knows in math, i got question.
In the real numbers there isn't something called 0.0000…1
But in hyperreals there's, and it's equal to 1/∞= 0.000000… 1, but hyperreals follow the rules of real numbers, it just add to them,
So if 0.999…=1 in real numbers
0.99999… =1 in hyperreals but that's wrong cuz
0.999…=1-(1/∞)
And 1-(1/∞) ≠1 in hyperreals
So am I wrong cuz i know hyperreals follow the rules of real numbers

[marsh citrus]

@subtle pawn Has your question been resolved?

[swift cypress]

0.999... is defined as a limit of partial sums, it's a real number, and it's equal to 1

introducing hyperreals cannot and does not alter that

[subtle pawn]

But hyperreals have infinisimal, and as i just said 0.9999… could be defined as 1-infinisimal which isn't equal to 1,so are U sure

So unless 1-infinisimal =1 which will mean infinisimal = 0
And they are not

[swift cypress]

there are actually infinitely many of them

however they have standard part 1, meaning that they are closer to 1 than they are to any other real number

[subtle pawn]

But they aren't 1,
1-infinisimal = the closest number to one Other than one.

[marsh citrus]

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[subtle pawn]

So am I wrong or something or anything Deals with infinisimal numbers is the exception not the rule

.reopen

[marsh citrus]

✅

[subtle pawn]

[marsh citrus]

@subtle pawn Has your question been resolved?

[subtle pawn]

<@&286206848099549185> 🦆 hi, i know it's simple question but it important to me

[marsh citrus]

@subtle pawn Has your question been resolved?

[eager sphinx]

when do you use hyperreals

[subtle pawn]

Tbh never, but i just wanna know

But in Calculus they do

[eager sphinx]

honestly hyperreals are just inconceivable to me

its hard for me to get my head around them

idk

[subtle pawn]

Yeah there aren't simple ideas

Like 0.0000…1
When the 1 will ever come

[subtle pawn]

$X=Y

so $f(3)=f(2)*(-0.2)$

Oh nice

so $f(x)=x²+1

Very nice thing, i actually like this

OK, i got out without answer but at least with nice bot, i will leave and accept that no one will answer me

.close

[marsh citrus]

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[knotty ember]

I need to find the angle between the adjacent lateral faces of a square based pyramid

but idk where to begin with this question

[robust flax]

love your pfp

[marsh citrus]

@knotty ember Has your question been resolved?

[knotty ember]

ty lol

[marsh citrus]

@knotty ember Has your question been resolved?

[knotty ember]

.close

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[silver hamlet]

how do i get started on this?

[silver hamlet]

I have to simplify.

[glacial hedge]

write it as one fraction. You should recognize a trig identity

[silver hamlet]

alright one sec

a fraction is x/y right?

didn’t get maths in english

[glacial hedge]

yes. Go from 2 fractions to 1 fraction

[silver hamlet]

alrighty lemme try

[glacial hedge]

how else are you going to work with this anyway. Think about what you're able to do. When there's so few options you should just try them

[silver hamlet]

that’s true, i rarely try all my options if i’m even a bit insecure about what i think it should be

[tacit fjord]

You should try them all, otherwise you're solving nothing.

[silver hamlet]

i will, before going in here i’ll try everything from now on:)

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[marsh citrus]

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Remember:
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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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[pliant quartz]

If I am finding the difference between two numbers on a number line, am I counting the number 0?

[runic crag]

yes

You cant skip from 1 to -1

[pliant quartz]

On thank you

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[marsh citrus]

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[wicked sparrow]

occupied

[wicked sparrow]

hello someone help with this english

thanks 🙂

[novel fjord]

this is a math server lol

[wicked sparrow]

can you pls 😦

[bright jay]

Not really because this is a math help server, not an English help one

[wicked sparrow]

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[opaque wolf]

@wicked sparrow There might be an english-learners server out there that might be able to help you

[marsh citrus]

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[main heart]

its C right

[main heart]

cause 3 is cubed a to the power of 2

and the others take away 2

[opaque wolf]

$125^{\frac{2}{3}} \cdot a^{\frac{2}{3}}$

@main heart

[elfin berry]

sills

[main heart]

oh

so yes C is right

thanks

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[spark canopy]

I know this may sound dumb for a bc student lol but how do you simplify sin(-55pi/6)? in fractions

<@&286206848099549185>

[rich coral]

Wait 15 min before pinging helpers @spark canopy give us a bit to respond first we already get pinged when you make it

[spark canopy]

my bad

[rich coral]

And to answer the question why not just plug it in to your calculator

Get an answer

Unless you’re not supposed to round

[spark canopy]

no its strictly in terms of fractions

[still temple]

what do you mean in terms of fractions?

[opaque wolf]

An exact answer

Not with decimals

[still temple]

ok

[opaque wolf]

e.g $\frac{\sqrt{3}}{2}$

[elfin berry]

sills

[still temple]

-54 = -9*6

so

[opaque wolf]

??

@spark canopy Sorry, I would help, but I haven't done differentiation with parametric functions yet

[true flume]

@still temple you are very much on the right track

[still temple]

-48/6 - 7/6

[true flume]

this is just doing the co coterminal angles

not calculus

[still temple]

we don't care about 2pi

2pi*k

so = sin(-7/6 pi )

[thorny jungle]

right, since it will just repeat that oscillating shape 2pi*k times

[still temple]

= -sin (pi + pi/6)

= sin(pi/6)

[thorny jungle]

right

[still temple]

now what was it