help-0

[clever pewter]

a>b aka a=bx

where x>1

[dark granite]

How bout a=b+L for some natural number L

[clever pewter]

ah that's simpler

[dark granite]

We don’t know a is a multiple of b so your way doesn’t work there

We only know a>b

So now show the inequality

[clever pewter]

I meant for x belongs to reals but sure

your way is just infinite times simpler

[dark granite]

Oh we only want to deal with integers here

[clever pewter]

right so (b+l)^2>b^2+1 as l>=1

[dark granite]

Good.

Very nice

So a^2>b^2+1

Divide a from both sides

[clever pewter]

yes after that the proof just follows

[dark granite]

Yeah so dividing a and multiplying by b gives ab>cb

This contradicts the minimality of ab so k must be 3

[clever pewter]

right

[dark granite]

Here’s my proof for it

[clever pewter]

so my approach was pretty much correct

[dark granite]

Yeah you’re were over half way there

[clever pewter]

oh this is one of the problems in the remark section in the book "Number theory: concepts and problems" by titu andresscu

(if you're wondering)

[dark granite]

Ah cool! I’ve seen this problem in some IMO books

[clever pewter]

yea the approach is same as of the legend of #6

1988 #6

[dark granite]

Not quite

With that one you can just go straight into assuming a>=b for some smallest pair

You don’t need to prove the equality case explicitly with that one

[clever pewter]

I mean the approach

not the complete proof

but yes

[dark granite]

Yeah the approach is similar I think

Though I did it purely by contradiction

If you like that problem I’d challenge you to prove that the fraction in problem 6 is the square of gcd(a,b)

[clever pewter]

wait what

[dark granite]

Yeah

[clever pewter]

btw osman nal proved something similar to that in the later half of his solution to 1988 #6 video

like not that

[dark granite]

Nice

Kk well nice working with you on this problem

[clever pewter]

same here

thanks

[dark granite]

You’re welcome. And this ain’t no elementary number theory problem either

[clever pewter]

https://youtu.be/osT2FEp7u8o this one

[dark granite]

I’d call this advanced

[clever pewter]

oops

[dark granite]

Nice thanks

[clever pewter]

I should have asked in some advanced channel like #advanced-number-theory or something

anyway

[dark granite]

Yeah

Oh well

This Channel is open for questions

[versed hazel]

Can I get some help pls

[vale wigeon]

this is a system of linear equations. have you ever solved systems of linear equations before?

[versed hazel]

Huh???

oh ok sorry

[vale wigeon]

i'm trying to understand how much you know, i.e. whether this is your first time doing any problems of this kind or just this system in particular that has you stuck

[versed hazel]

ya bro i am in y7 and im just shit at maths and thought i could get some help

[vale wigeon]

don't call me bro.

[versed hazel]

sorry

[vale wigeon]

so once again, let me ask: have you done anything like this before?

[versed hazel]

no not rly

[vale wigeon]

okay, first time then

[versed hazel]

ye

[vale wigeon]

have you been taught anything on this topic? perhaps some words like "substitution" or "elimination" might've been said in your class at some point

i'll have to explain at least one of them if not

[versed hazel]

yes i understand a bit

[vale wigeon]

okay, so let's look back at your system

2x + 3y = 1200
3x + 2y = 1300

[versed hazel]

ya

[vale wigeon]

so one of the strategies for solving systems of equations is called elimination

[versed hazel]

ok

[vale wigeon]

in general, elimination involves combining two or more equations such that the resulting equation has less variables than what you start with.

with linear systems, the "combining" usually happens through addition or subtraction.

[versed hazel]

ok i get it a bit now

[vale wigeon]

the equations may need to be modified somewhat before combining

such modifications typically involve multiplying both sides by a number

which number to multiply by is informed by the goal: make the resulting equation have less variables (typically one less variable; in 2-variable systems, like you have here, you'll be eliminating one variable and left only with the other)

[versed hazel]

ahhh

[vale wigeon]

alright, so does that mean you'd like to attempt this on your own? or should i keep going

[versed hazel]

hmmm

[versed hazel]

ya can u please keep going i dont rly understand still hmmm

[vale wigeon]

okay

i was fine either way, btw

anyway, ok

so what i'm going to do is multiply both sides by 2 in the first equation

4x + 6y = 2400

and multiply both sides by 3 in the second

9x + 6y = 3900

do you understand what i just did? (if your only issue is the why, it'll become clear in just a moment)

[versed hazel]

ahhh ok

oh i get it now

so now what do i do with the equation i remember smthin like needing to add the 2 together???

or no??

wait i think i did it

[vale wigeon]

with what i'm doing, you would be subtracting them instead

[versed hazel]

x is 300 and y is 200

[vale wigeon]

that way, the 6y's cancel out, and you're left with 5x = 1500

oh, ok, so you've done it already

great

[versed hazel]

thank you so much

oh wow ur good at maths

[hollow swan]

ann, can u help me?
can I confirm that the answer is 26^4 - 25^4?

[versed hazel]

lmao u explained it better than my teacher

[vale wigeon]

if you meant to ask me "can you confirm that the answer is 26^4 - 25^4?" then yes, i can and do confirm that the answer is indeed 26^4 - 25^4.

[hollow swan]

yeah sorry lol can u confirm

@vale wigeon can i also ask why i should use permutation or combination?

[vale wigeon]

you shouldn't

[hollow swan]

i shouldnt

sorry

im rlly tired lol

[vale wigeon]

you shouldn't "use" nPr and nCr.

[hollow swan]

yeah why?

[vale wigeon]

nPr and nCr are shortcuts for counts that arise naturally in certain scenarios

[dire pilot]

can I get help

[carmine lion]

no

occupied channel

[vale wigeon]

@dire pilot channel busy please move

[vale wigeon]

...and which can be calculated just by using the multiplication principle

and some intentional organized overcounting

i can present some examples if need be

[hollow swan]

so how do i know

when to use

x^something

or permutation/combination

[vale wigeon]

again

it all boils down to the multiplication principle

again, if you want, i can present some examples of multiplicative reasoning that i'm hardcore advocating for here

[hollow swan]

sure

[vale wigeon]

let's say that we have a deck of cards

(yknow, the standard, 52 cards, no jokers)

and suppose we shuffle the deck and deal out 5 cards on the table and arrange them in a row

how many different arrangements are possible?

[hollow swan]

52p5?

[vale wigeon]

yes

but if we didn't know of the P function, we might instead reason as follows:

there are 52 options for what the leftmost card on the table may be. it can be anything we like, as we aren't imposing any restrictions on the layout or anyhting

[hollow swan]

yep

[vale wigeon]

once the leftmost card is placed, there are 51 options for the card next to it (the second card). why 51? because the first card's been placed already, and the deck is one card smaller

for the third card there are 50 options, then 49 for the fourth, and 48 for the last.

thus the number of arrangements is 52 * 51 * 50 * 49 * 48.

does that make sense?

[hollow swan]

yep

what would the question look like if my answer were to be 52nCr5 or 52^5?

[vale wigeon]

you mean 52C5

[hollow swan]

yep mb

52C5

[vale wigeon]

for 52C5, instead of laying them on the table, we deal them into a hand

so that the order of the cards among themselves no longer matters

(if you've played any card games, you probably know that you can arrange the cards in your hand in any way you like - normally people do stuff like sorting them by suit, but nothing's physically stopping you from putting them in whatever order makes you happiest)

[hollow swan]

yep

wht about 52^5

[vale wigeon]

for 52^5, instead of putting them all out on the table, we pull out one card at a time, record it (write it down on a piece of paper somewhere) and shuffle it back into the deck, and repeat that 5 times

[hollow swan]

oooo i see now

thanks this helped

[vale wigeon]

multiplicative counting can also explain why the number of ways to arrange n objects in a row is n!

(that's a factorial, not an exclamation mark)

[hollow swan]

yup i got that HAHA

thanks again

[dark granite]

And notice that @vale wigeon is implicitly talking about n DISTINCT objects being arranged in a row when she brought up n!. @hollow swan

[hollow swan]

so for 5a, what I got 15! / 4! x 11!
i cant seem to figure out 5b since its asking for at least
could u help me @vale wigeon ? ty

[vale wigeon]

missing parentheses, and the letter x used as multiplication

but yes, the answer for (a) is 15!/(4! 11!), or 15C4 for short

for b, try counting the outcomes which contain less than three heads (meaning either zero, one or two heads), and subtract that total from the count of all possible outcomes.

[hollow swan]

so all possible outcomes would be 15^2 right?

[hollow swan]

would the outcomes that would contian less than 3 heads be 3!(12!)?

[vale wigeon]

no to both

[vale wigeon]

2^15.

[vale wigeon]

15C0 + 15C1 + 15C2.

you seem to have blissfully ignored the (meaning either zero, one or two heads) bit

[hollow swan]

oops

yeah i forgot abt that

give me a sec, j calculating

@vale wigeon would this be correct?

[vale wigeon]

yes congratulations you understood and followed what i said

good job

[hollow swan]

yay ty

[hexed rose]

Based on this triangle how do I get the length of both the X and Y, how can I tell which function to use ? (sin cos tan) sorry if this is a dumb question noob here

[plucky crow]

sohcahtoa

sin = opposite/hypotenuse

cos = adjacent/hypotenuse

tan=opposite/adjacent

[prime stream]

use SAA

[hexed rose]

right but how do I which one to pick? out of sin cos tan

1sec need to lookup side angle angle again

nope lmao

[hexed rose]

okay so i think Ive got it, if I want to get X i have to use x and 9 (OH) so SOH / sin, if I want to get y II have to use y and 9 (AO) so TOA / tan

[nova anvil]

how to do this integral

[karmic juniper]

opposite side = O = 9
adjacent side = A = y
hypotneuse = H = x

s=o/h
c=a/h
t=o/a

so sin(40°) = 9/x
and tan(40°) = 9/y

then just do some algebra to get X and Y their own

not too difficult

[scarlet heath]

how do I solve for this, substitution?

[scarlet heath]

<@&286206848099549185>

[steep briar]

Yep

[scarlet heath]

what should I end up getting?

[steep briar]

What's your attempt?

[scarlet heath]

Im kinda stuck rn with uh x-8=y^2-6y

not sure what I should do with my uh-6y

[steep briar]

Might need to do the quadratic equation. Let me try that with what I got so far.

wait

[scarlet heath]

?

[steep briar]

so x=-(3-y)²-1, agreed?

[scarlet heath]

yeah

[steep briar]

Then isolate for y from there

(don't do (3-y)(3-y) )

[scarlet heath]

so move 1 to right so x+1 then square

squareroot*

then move 3

[steep briar]

yeah (don't forget the negative infront of -(3-y)

[scarlet heath]

kk

[steep briar]

So so far y=∓√[-(x+1)] if I did things right

[scarlet heath]

yeah I gjust got that I was p stupid

what would the domain n range be

[steep briar]

When is ±√[-(x+1)] not a real number?

(for what values of x is the inside of the squareroot not positive)

[rancid plinth]

yo can someone help me

[steep briar]

Channel's taken atm.

I guess not, carry on. I'll be off, remember to ask specific questions.

[torpid lagoon]

hi

i take this channel

[elfin snow]

damn

he really took the channel

[torpid lagoon]

the degree is 4
and x intercept is just when y=0
so, x=0 or x=3/2 or x=-5
End behaviour: The end behaviour will be as x approaches + infinity and as x will approach - infinity

[lilac vessel]

Is a vertical linear line a function or relation?

[rigid smelt]

Relation

There is one x-value that correspond to many y-values

[lilac vessel]

Thanks so much @rigid smelt

[wheat pewter]

Hey I need help

What is the smallest value of a, in terms of e, for the equation

a mod e = e -1

I got this far but got stuck at this point

[vale ridge]

can someone help me
. James looks in his TV cabinet and finds
some old Beta and VHS tapes. He has
17 tapes in all. He finds that he has three
more Beta tapes than VHS tapes. How
many of each type does he have?

using linear systems

[wheat pewter]

or in particular - I need to find a in terms of e so that d is an integer for d= (a+1)/e

Sure lol @vale ridge

17 = x + (x+3)

17 = 2x+3

14 = 2x

[alpine sable]

@vale ridge https://www.youtube.com/watch?v=xKH1Evwu150

[wheat pewter]

7 = x

7 VHS tapes and 10 Beta

[alpine sable]

@wheat pewter $a \equiv e - 1 \pmod{e}$?

[wheat pewter]

huh?

[ocean seal]

n/c

[vale ridge]

thank you man

[vale ridge]

thnx too

[alpine sable]

You should watch the video so you understand how to solve them in the future

[wheat pewter]

I'm soryy to bother you lol thanks for the help ... can you explain?

[alpine sable]

Is that the congruence you are referring to?

[wheat pewter]

That's what I'm looking for yeah ... smallest a in terms of e so that d is an integer

[alpine sable]

And I'm guessing all 'smallest' a are equivalent in their respective equivalence classes?

[wheat pewter]

Yep

See it's $d = \frac{k\left(\Phi\left(n\right)\right)+1}{e}$

[ocean seal]

Dasho

[wheat pewter]

And $\Phi\left(n\right)$ is known

[ocean seal]

Dasho

[wheat pewter]

so I'm trying to find a value of k and d so that d is whole and k is as small as possible

Think the unit circle... after the revolution of the angles, they'll start and finish at the same place @warped phoenix

[alpine sable]

@wheat pewter I don't remember, but why isn't it just e - 1? All equivalence classes are unique

So how could it be smaller than e - 1?

[wheat pewter]

My god I'm an idiot

Okay yeah ofc that makes sense lol thanks

[alpine sable]

[wheat pewter]

Man how did I not see that like wow

okay thanks @alpine sable

[obtuse warren]

hey i dont wanna interrupt, did you get helped?

[wheat pewter]

Yes I did ... go ahead

[obtuse warren]

okay thank you

This is part of my homework for my college math class. I am confused on the process really. This is not an exam question or test question so I believe by asking this I am following the rules.

[alpine sable]

So evaluate g(0)

[wheat pewter]

3?

[obtuse warren]

yeah but i dont know what to do with the inequality

[wheat pewter]

@obtuse warren 0 is greater than 3 so you use the top option in this case

likewise for 2

and then the opposite for -7 because -7 is < -3

[obtuse warren]

ohhhhh

okay okay i get that now

[wheat pewter]

You just use them as like conditional rules

Think like scratch program

If x >= 3, then return x+3; elseif x < 3, then return -(x+3);

[wheat pewter]

@obtuse warren

np

[obtuse warren]

Okay thank you for the help!

One more question, how would I do range for a piecewise function??

[hollow swan]

@vale wigeon i sent you a msg in dms

[obtuse warren]

This is the home work question for more specific help.

[shell widget]

@obtuse warren Just do the unions of both the ranges

[obtuse warren]

so (-inf, inf) u (-inf, inf) ???

[shell widget]

You have (-infinity, 8) U (-2, +infinity) = (-infinity, +infinity)

[obtuse warren]

where did you get the 8 and -2?

[shell widget]

Look at the graph

8 is where the first line stops, -2 is where the second starts

[obtuse warren]

ohhhh

I see

[ionic jewel]

I'll give you a lpt

don't dm people

50/50 you get ignored/you get flamed lol

[river glade]

How do I solve this? Also what is this topic called? I'm asking so I can learn about it by myself. I literally don't even know how I would even start googling this

don't mind the 1-3 part, that's just an indication of where to write the answer

[alpine sable]

You translate the point (2,1) to the origin

Then you rotate by 45 degrees. That can be done using a rotation matrix or complex numbers

Then you translate it back

[drowsy fable]

so basically
0 = 0/1 = 1/0
how.

( "/" = Fraction sign)

out of everywhere you had to write a question you chose this place bruh.

[river glade]

what does that mean

[alpine sable]

Does anyone know how to simplify this equation? I want all X'es on the left, but i dont know how to handle parenthesis:

x = a(b * c * x)

[vale wigeon]

$x = abcx$?

[ocean seal]

Ann

[vale wigeon]

is that your equation?

[alpine sable]

Not sure what u mean with $.
My equation is x = a * (b * c * x) . I want the X only on the left side

[vale wigeon]

look at what the bot wrote

okay, so what you have is x = abcx

this can only be true if x = 0 or abc = 1.

did this equation arise from solving a problem, or did you just come up with it randomly?

[alpine sable]

I was working with a different formula, which involved the solution (X) in the right hand side of the equation. This is a simpler one, but same thing: X is both on left hand and right hand side.

[vale wigeon]

can i see this other formula?

[alpine sable]

If it was like this x = a + x I would know how to get X out, i would just subtract on both sides

[vale wigeon]

x = a+x has no solution for x either, just so you know

subtracting x from both sides leaves you with a = 0

anyway, again, can i see that other formula?

i'm llike 95% certain you're messing something up

[alpine sable]

I had this:

        eX + fY = ai + bj
        eX = ai + bj - fY
        X = ai + bj - fY / e

Is this place in use

[vale wigeon]

yes it is in use

[alpine sable]

Here I managed to get X on left hand side, alone. And then:

gX + hY = ci + dj
gX + hY - gX = ci + dj - gX
hY = ci + dj - gX
Y = (ci + dj - gX) / h

[vale wigeon]

okay so you're missing some parentheses.

X = (ai + bi - fY)/e

[alpine sable]

Yeah, i noticed on last line of X

[vale wigeon]

...so what you're actually saying is

you were working with a system of equations?

[alpine sable]

anyway, you see these 2 formulas - they are for solving X and Y

[ocean seal]

Ann

[alpine sable]

Now, if we take the first formula - it depends on the variable Y, which is defined as Y = (ci + dj - gX) / h

[vale wigeon]

you started with these two equations. yes or no?

[alpine sable]

Yes

[vale wigeon]

and you want to solve these for X and Y

[alpine sable]

Yes.

[vale wigeon]

okay

[alpine sable]

And so far I have this:

X = (ai + bj - f*((cj + dj - gX) / h)) / e

[vale wigeon]

you're going about doing that in a way that is technically valid but full of pitfalls

[alpine sable]

This is where i take the value Y and plug it into the first forumla

[vale wigeon]

X = (ai + bj - f*((cj + dj - gX) / h)) / e
now THIS is a very different equation from the x = abcx you tried to show earlier

you do realize there's more than just multiplication of x by constants going on here, do you not???

[alpine sable]

its the same problem im facing however - you see, both equations have X on the right hand side of the equation. And i need to get it out

[vale wigeon]

"both equations have X on the right hand side" is too vague\

the similarities between the equation you showed at the start and the equation that you now reveal you have end there

anyway, do you want to consider the equation $X = \frac{ai + bj - f \cdot \frac{ci + dj - gX}{h}}{e}$ in isolation, making us both wade through a forest of algebra that nobody needs nor benefits from... or do you want an actual solution method for your original problem, i.e. solving the system $$\begin{cases} eX + fY = ai + bj \ gX + hY = ci + dj \end{cases}$$ for $X$ and $Y$ in a way that's not nearly as painful?

[ocean seal]

Ann

Ann

[alpine sable]

i just wanted to know how to isolate x on the left side of the parenthesis

[vale wigeon]

.-.

you're not answering my question

you're being really vague now and it's getting somewhat annoying

[alpine sable]

well, sorry but i dont know what you mean. i dont know know what 'solving the system' means.

[vale wigeon]

you have a system of two equations in X and Y and you want to know the values of X and Y that, when plugged into the equations, make them both true at the same time.

finding such values of X and Y is called solving the system.

just like finding the value of an unknown that makes one equation true is called solving that equation

[alpine sable]

this is what i want yea. and it's going pretty well, except, right now im facing the issue where i'm trying to solve x and its both on the left side, and right side of the equation

You see in this 1 you linked here? x is both on right side, and left side of equation

[vale wigeon]

yes, i'm quoting the equation you wrote

"trying to solve x and its both on the left side, and right side of the equation" is a way less informative description than you think it is

[alpine sable]

Not solve, just simplify it. Rearrange the equation

So that X is only on left side

If what i'm asking is impossible, pls let me know

[vale wigeon]

it is possible

[alpine sable]

and what i should read

[vale wigeon]

you're just insistent on wording things in a way that just... is bordering on nonsensical

i suppose you're now going to tell me that you're completely unfamiliar with algebra?

[alpine sable]

im not

[vale wigeon]

does that mean 'i'm not familiar with algebra' or 'i'm not unfamiliar with algebra'

[alpine sable]

i just don't do math that often, so im unfamiliar with solving equations

[vale wigeon]

solving equations basically is algebra.

[alpine sable]

right

owell, thanks for your time

[vale wigeon]

...

i was about to upload a picture of the work to isolate X as i would've done it

but if you don't want that, it's fine

i'll just screw off

[alpine sable]

that would be sweet. i figured u didnt bother with me anymore, since u said, it's possible but didnt say how 😄

[vale wigeon]

hope my handwriting's legible

[alpine sable]

thats nice, thank you. i think this is what i wanted, but i have to study it for abit

[lapis imp]

Help pls

[vale wigeon]

@lapis imp we aren't quite done talking here yet. use one of the other 9 questions channels.

@alpine sable from the looks of it, you need to go through a course of (high)school level algebar

algebra*

[alpine sable]

Damn 🙂

on third line its a 'c' not 'e' right?

right after f

[vale wigeon]

yes

[alpine sable]

to clear up some confusion, you are 'expanding' on line 3 - this is why i showed x = a(b * c * x) example. this is the step i couldnt do

but i see u just put f infront of everything in the parenthesis, then u can remove them

just making sure, this is true right? f(cj + dj - gx) = (fcj + fdj - fgx)

Yes, $x(a + b) = xa + xb$

[ocean seal]

n/c

[alpine sable]

In this case, $x = f$ and $a + b$ is denoted as $cj + dj - gx$

[ocean seal]

n/c

[alpine sable]

And putting parentheses around a particular value does not change that value

[novel crypt]

I'm not sure if I actually need to prove this or not, but it may be a helpful lemma for a homework im working on: Does anyone know how to approach:

Prove that for any $A = {x | x > 0}$, $B = {y | y > 0}$, $\exists x. x \in A \land x \not\in B$, that the sum of all of the elements of a is not equal to the sum of all of the elements in b

oops thats not quite right hold on

[ocean seal]

jcob

[novel crypt]

ok I fixed it. Any idea?

forgot also $|A| = |B|$

[ocean seal]

jcob

[vale sapphire]

@novel crypt Why is it not the case that A = B ?

They're defined the same way

[novel crypt]

oops because I'm an idiot

I meant to say that $\forall x \in A . x > 0$, $\forall x \in B . x > 0$

[ocean seal]

jcob

[novel crypt]

I defined them incorrectly

I just meant to say that they're greater than 0 because obviously, if they could be pos or neg, then sum {-2, 2} = sum {-1, 1}

but because of the structure of the problem (its about minimum spanning trees) the costs cannot be negative, so I'm just trying to establish that if all of the costs are different sets of the same length then it has to be the case that the sum of the costs are not equal

[vale sapphire]

You'll have trouble proving this because it's not true

A = {1,4}, B = {2,3}

They satisfy your conditions (afaict), but have the same sum

@novel crypt

[novel crypt]

Ah ok

Thanks! Hmm.

brain malfunction 😂

[vale sapphire]

Yeah, two things having the same sum isn't a strong condition at all

[novel crypt]

ok well in the problem, A and B index the same set, and A and B never use the same index twice

Thanks btw

I'm thinking maybe bcoz of that, you can show sum A =/= sum B?

Will have to think on that more haha

[lavish girder]

<@&286206848099549185>

is this correct

[vale sapphire]

please ping helpers after asking, if your question hasn't been answered for a while

[lavish girder]

okay

[vale sapphire]

regardless, if 4 is a root of this, then x - 4 divides it

that's almost a definition

[lavish girder]

wdym

if 4 is a root of this

[vale sapphire]

if it makes 0 when you plug in 4

[lavish girder]

ok let me try

it is

so what does that mean

it means its right?

[vale sapphire]

Yeah. A way you can see this is that if your polynomial equals 0 when x=4. Then you can express it as x-4 (which is 0 when x=4, and simple) times something else.

That's what you're saying here

[sick oyster]

how to get the range for [ sqrt(1+x^2) ]/(x-1)

[simple hamlet]

is that even a question ?

[lavish girder]

yeah y

[simple hamlet]

it's soo easy bro

[lavish girder]

wdym

aight

sure

whats the answer then

[alpine sable]

someone help me plz

[simple hamlet]

(x-4)(x squared+4x+16)

[lavish girder]

so its x-4

[simple hamlet]

yeah

[lavish girder]

so i got it right

[simple hamlet]

u r right

[lavish girder]

okay

[simple hamlet]

another question ?

[sick oyster]

.

[alpine sable]

@simple hamlet

[wind bane]

what have u tried?

[sick oyster]

=y then squaring both sides, some solving and got y^2 > 1/2

[simple hamlet]

one minute

Adjacent angles

[alpine sable]

adjacent and

linear?

[wind bane]

yes both

[simple hamlet]

how please ?

[sick oyster]

but according to the graph y is not > 1/sqrt2 intead y>1

[lavish girder]

is this correct

[sick oyster]

no

[lavish girder]

wait im dumb

[wind bane]

so i would break this into cases, one where x is larger than one and the other car where x is less than one

[lavish girder]

i thought it said remainder

[simple hamlet]

i think the second

one minute

[lavish girder]

are u talking abuot mine

[simple hamlet]

the problem that im bad at english

[lavish girder]

@simple hamlet

[simple hamlet]

no no just one minute give me one minute

[sick oyster]

so basically the squaring both sides thingy is causing a problem

[wind bane]

yea

so if we start with x > 1, then we know x - 1 > 0 so that’s good

we also know that x^2 + 1 > 2 from that info

[simple hamlet]

idk tbh

[sick oyster]

so when do we know that we'll have to break it into cases?

[simple hamlet]

@lavish girder

i hope this pic will help

[wind bane]

there is an asymptote at x = 1

you may run into issues if you try to do it all in one piece

[lavish girder]

i dont get it

[sick oyster]

ohk got it

thanks a lot 🙂

[wind bane]

yup. so just to continue, you can get that sqrt(x^2 + 1) > sqrt(2). then divide by (x - 1) and look at what happens as x approaches 1 and as x approaches infinity

[simple hamlet]

i think the third one im not sure

hey guys

<@&286206848099549185>

[alpine sable]

what about this?

[wind bane]

what are ur thoughts, kirk?

[alpine sable]

i think

vertical

and adjacent?

[simple hamlet]

put the third one depends on this pic

[wind bane]

two angles cannot be both adjacent and vertical at the same time

[lavish girder]

okay ill put the third one

[alpine sable]

adjacent only then

[wind bane]

no they are not adjacent either. adjacent means that the angles are right next to each other; one literally touches the other

[simple hamlet]

one minute

[alpine sable]

linear

[wind bane]

i would suggest going over your definitions once more, as it seems like you’re just guessing here

but no. they are not linear either

[alpine sable]

bruh

so its just vertical

[simple hamlet]

there is two answers then

[alpine sable]

which

[alpine sable]

.

[simple hamlet]

the first one pic

[alpine sable]

adjacent and linear?

[simple hamlet]

yes

for me i will but linear and adjacent

i think these two photos can explain

[simple hamlet]

none of these

because

srr

it is not linear first because both of them are not 180 according to this pic

it is not vertical

and it is not adjacent

[simple hamlet]

hello

[alpine sable]

hi

[simple hamlet]

i have just answered u

[alpine sable]

so for this

its none?

[simple hamlet]

the second pic none of these

[simple hamlet]

yeah none

[red spear]

Do

Do I differentiate this using the product rule ?

I assume these are 2 functions being multiplied by each other :DD

[wind bane]

yes the product rule is correct

[red spear]

@wind bane

Thanks

[wind bane]

no

prob

[red spear]

can you help me with one more thing @wind bane

[wind bane]

sure

[coral pagoda]

Enthusiast, can you please take it easy on the pings?

[wind bane]

lol

[ionic jewel]

[red spear]

Wait wut

I haven’t tagged anyone

[wind bane]

don’t worry about it lol just ask ur question

[red spear]

oh yeah sorry

So if I differentiate it using the product rule I’ll have (3x)(e^x) + (3)(e^x)
but how do I simplify (3x)(e^x)

[ionic jewel]

you cant

[vale sapphire]

you don't

[red spear]

oh

do I just leave it as it is?

[ionic jewel]

best you can do is $3e^x(x+1)$

[ocean seal]

bunny

[ionic jewel]

or what you have now

[red spear]

Oo

Thanks mates

[alpine sable]

Can someone help me with this?

Dont know what to do

I first thought I had to solve for roots but i cant really solve without s...

[wind bane]

you can too solve. just use the quadratic formula

[alpine sable]

but what would b be?

oh wait

[rigid ocean]

@alpine sable solve with complete sqvare

[alpine sable]

🤦‍♂️

[rigid ocean]

help bhai

how do you put in that form?

[final breach]

the quadratic formula

[alpine sable]

lool

[rigid ocean]

Ohk bhai I am apologize

[upper pebble]

"We know that the probability of getting 'Honey Bunny' in one try is p.

Therefore, the probability of getting a candy other than 'Honey Bunny' in one try is (1-p)."

why is it (1-p)?

[alpine sable]

bc 1 try

[rigid ocean]

[noble sinew]

Because all possible outcomes must sum to 1 (100%)

So there are only two options either you get it or you don’t

[upper pebble]

oh i see

thanks for your help

[glass lichen]

You either get it or dont, so the prob's have to add to 1

[rigid ocean]

You are smarter than me.

[glass lichen]

Use the hint at the bottom

radial distance

it's also defined in the hint...

radius

it's just redefining the trig functions onto circles

Yes, though it's bad to still require right angle trig definition

sin(t) is opp/hyp in a right triangle.

Like if you have a right angle triangle then it's fine, but trig is much more than just triangles

it's not ideal to think of sine as a ratio of triangle sides for example, since then you cant graph sin(x) as an example

[vapid bridge]

1. Given the Exponential Growth Model f(t) =Aekt where A and k are real numbers. Explain what the parameters A and K mean for the exponential equation.

can someone help

[glass lichen]

sin(t) is positive in quadrant I and II

[final breach]

the first parameter is a

[vapid bridge]

?

[final breach]

the parameter a tells us where it touches the y-axis when t = 0

[vapid bridge]

oh

[final breach]

the parameter k tells us how fast the function grows with time

the higher the value of k, the faster the function grows

[vapid bridge]

oh i see

thanks!

[remote iron]

f is 6 and c is 10

Solve are of.both triangles

[wind bane]

u tried anything yet?

[woeful pulsar]

yeah the x is related to cos(t)

hold on

[nimble meteor]

here gimme a sec

[woeful pulsar]

hmm this is nonstandard

yeah you have the wrong ratios

sin is y/r

cos is x/r

[nimble meteor]

this is the unit circle

[woeful pulsar]

yeah

[nimble meteor]

as you can see x is positive to the right

and y is positive upwards

and negative downwards

exactly

because if you think about what the unit cirlce is

*circle

its basically just an extension of our soh cah toa defnition to include all angles

BUT

on the graph

and do you know what the radius of the unit circle is?

yes exactly

I'm guessing your in alg 2 rn learning about trig?

[icy trail]

sorry its sideways

using graph transformations i don't see where I've gone wrong here

[candid torrent]

,rotate 90

[candid torrent]

,rotate 90

,rotate 90

[icy trail]

ty!

[candid torrent]

what's your question here?

[icy trail]

i thought sin(90-x) = cos(x)

[nimble meteor]

I think it does equal cosine since they are cofunctions

[nimble meteor]

yes I'm fairly sure that is correct

[icy trail]

so I'm not sure why transformations are wrong

[nimble meteor]

what's your question again>

[icy trail]

where*

[nimble meteor]

they're not where did you get that idea?

ok give me one sec

[icy trail]

the resulting graph dosent look like cos(x)

[remote iron]

Nothing

[nimble meteor]

here let's go one at a time

@warped phoenix was here first

[remote iron]

Dk what to do

[icy trail]

yes np

[nimble meteor]

@icy trail is after

sorry

[icy trail]

np

[wind bane]

do you know what similar triangles are?

[remote iron]

Yes

And no

[wind bane]

try and use those to get…

wdym yes and no

[nimble meteor]

ok I think you are just confusing yourself

the easiest way to think about this is graphically

[wind bane]

hop over to trig and geometry channel

[final breach]

area is (1/2) *b *h

so 30

then use the property of similar triangles

[nimble meteor]

so you remember where they say pi/2 is less than theta which is less than pi?

pi is 180 so on a graph/unit circle it would be at the coordinates -1,0 right?

ok yea so that's the problem

check this out

[remote iron]

Lol ik that

[nimble meteor]

@warped phoenix do you get how I got this?

its basically just the unit circle without coordinates

well yeah technically its both though

cause that's where it starts

like its 0pi and 2 pi

if that makes sense

no no

ok so in a circle

how would you define this point

(sorry I know it looks like a lemon)

I'm using a mouse on paint lol

no lol just consider a normal circle

like let's just say that's a normal circle

how many degrees is that?

well what if you haven't taken the rotation yet?

then what

YES

exactly

so it could be either

depending on the situation right?

that's the point

anyway back to your actual question

yes

[nimble meteor]

you see this?

basically the sin is always positive in the first two quadrants because the y axis is above the x axis

think about it like that

and sin is always negative in the third and fourth because the y axis is always below the x

axis

make sense?

YES

exactly

so in what quadrants do you think cos is negative?

well not quite

think about it again

ah man no

dang this is so hard to explain

if it was in real life itd be much easier

ok dw I have an idea

pls excuse my horrible paint drawing skills

but basically do you see how to the right x is positive?

yes

so do you know that sin=y and cos=x?

ok you are missing some fundamental concepts

I suggest you watch a few youtube videos and khan vids to clear up your doubts

then come here and ping me if you still don't understand...sound good?

[icy trail]

with the unit circle hyp=1
so with your sin = opp / hyp you get sin = opp

and the opposite side is your height

[whole prairie]

Can someone help me with this please? I can’t figure it out from the explanation my prof gave me

[icy trail]

are you still free or do you need a break :)

picture also says 2pi and 0pi are the same place

learning!

i think they're offline ;-;

you're thinking about when cos is negative?

yeah i see your logic

[final breach]

complicated question, so you'll have to fill in some of the dots yourself, but
let y be the car traveling south, so y = 20 - 160t
let x be the truck traveling east, so x = 30 + 140t
By virtue of the Pythagorean Theorem, denoting z to be the distance between the police car and the truck,
l^2 = x^2 + y^2 = (30+140t)^2 + (20-160t)^2
Differentiate the equation to calculate the rate with respect to t. dl/dt would be the rate

[plucky crow]

But the angle is inbetween 90 and 180 degrees

Where sin is positive

[final breach]

basically d/dt(l^2)

[plucky crow]

Look at the cast diagram

[whole prairie]

Okay thank you

[snow cedar]

would the basis of a trivial null space just be the zero vector?

[glass lichen]

the 0 space has a basis {0} yes

[snow cedar]

okay thank you

[final breach]

the angles are given

pi/4 is 45 degrees, so the ratio of the sides are 1:1:sqrt(2)

i don't see where x is assumed to be 1 though

x just cancels out

[lime harness]

We're on horizontal shifts, and this question came up and my teacher doesn't know how to explain it, can someone explain this question?

"why isn't it 2(t-6)"

here is the original problem

ping me back please

[elfin snow]

channel is currently in use @alpine sable

[alpine sable]

@elfin snow sorry sorry

[alpine sable]

Go on a graphing calculator. Graph these three expressions:
(x - 6)^2
3(x - 6)^2
(3x - 6)^2

Once you have graphed them, analyze them.

Do you see how when you compare f(x) = (x - 6)^2 and f'(x) = 3(x - 6)^2, the transformed graph hasn't shifted its position, but simply got skinnier? This is a vertical compression.

But, when you compare f(x) = (x - 6)^2 and f''(x) = (3x-6)^2, the transformed graph has actually moved its position and gotten a little skinnier towards the y-axis. This is a horizontal compression.

You don't see it as easily with quadratics. You can see this happen more with equations such as f(x) = Asin(bx), where A is a number that results in a vertical stretch or compression, while b is a number that results in a horizontal stretch or compression.

[lime harness]

Thanks!

[alpine sable]

Np.

Also, if you want to horizontally stretch, you simply use the formula (x/b)^2, where b is any real number not equal to zero.

Which, when you want to compress, you multiply x by b instead of divide.

[rotund verge]

is this channel still in use?

[alpine sable]

Yep, it's free

[sinful monolith]

nvm

sorry

wait is anyone using it?

if not someone help me

pls

[thorn kindle]

The arithmetic mean of a set is the sum of the items in the set divided by its size

[sinful monolith]

i dont get it

[final breach]

solve the angles first

[thorn kindle]

No its easier if you take the average first

[final breach]

(3a+a+2a+3a+3a)/5

right

[gray isle]

there is absolutely no need to determine the sizes of a or the individual angles

[thorn kindle]

You need a itself

But not every angle

[final breach]

yeah mb

[thorn kindle]

Wait this is actually even simpler

It doesnt even matter what a is

You have a circle whose measure is 360 degrees split into 5 sections

[final breach]

right

[thorn kindle]

The average therefore must be 360/5

[alpine sable]

find a first a + 2a+3a+3a+3a = 360 , then find their average by dividing 6a/5

[thorn kindle]

You dont need to do that

Also its 12a/5

[sinful monolith]

then what do i do?

if i dont need to do that

[thorn kindle]

360/5

[sinful monolith]

can i ask how did we get 360?

get*

[thorn kindle]

How many degrees are in a circle

[gray isle]

(total angle sum around a point) divided by (the number of angles)

[alpine sable]

it should be 180/5

[final breach]

a circle has 360 degrees

[sinful monolith]

i see

[alpine sable]

sum of all angles is 360

[thorn kindle]

There are 360 degrees in a circle last time i checked

[sinful monolith]

"A circle is divided into 360 equal degrees"

i looked up how many degrees in a circle

[alpine sable]

yes its 360/5

[sinful monolith]

how about this question

how do i even find lowest grade

[thorn kindle]

(a+b+c+d+e+f)/6=75

a+b+c+d+e+f = 450

(b+c+d+e+f)/5 = 85

b+c+d+e+f = 425

[sinful monolith]

wait what

[thorn kindle]

What is unclear?

[sinful monolith]

what is this part

[thorn kindle]

Thats called multiplication

Look at the line above it

[sinful monolith]

how do you get the 450?

[thorn kindle]

(a+b+c+d+e+f)/6=75

Do you understand how i got this?

I'm given that the average of 6 numbers is 75

[sinful monolith]

yeah i get the (a+b+c+d+e+f) is the amount of math test right?

[thorn kindle]

That is the sum of all the test scores

[sinful monolith]

ok

[thorn kindle]

Ok then. So we have

(a+b+c+d+e+f)/6=75

What happens when you multiply both sides by 6?

[sinful monolith]

=450

[rotund verge]

if everythings solved with the problem above may i get some help with this exercise about differentials and linear approximation?
The exercise reads as follows:
A car is traveling down a steep hill. The number of feet s(t) traveled at "t" seconds is given by s(t) = 5t^2 +2
what's the velocity in t = 1 s?

[dense carbon]

Help?

[thorn kindle]

10

[thorn kindle]

Look up refraction rules for spherical mirror

[steel spire]

How do you do this

[thorn kindle]

Is that multiplication or composition

[steel spire]

this is a relations and functions topic so I'm guessing composition maybe?

[steel spire]

oh nvm

I got it now

[thorn kindle]

You're welcome

[left whale]

help

[thorn kindle]

No

[rotund verge]

how'd you get 10

waitwait so if i use s(t) = 5t^2 + 2 = 10
thus s'1 = 10(1) = 10

[alpine sable]

Why is d + v = c

[wind bane]

instantaneous rate of change of P(t) at t=-1 is just the first derivative of P evaluated at t=-1

right. so to differentiate $x^r$ with respect to $x$, then we have $$\frac{d}{dx}x^r=rx^{r-1}$$ for all real numbers $r$

[ocean seal]

coycoy

[wind bane]

so i would write this as $$\frac{3}{t+3}=3(t+3)^{-1}$$ and apply the power rule (the above rule) along with the chain rule

[ocean seal]

coycoy

[wind bane]

didn’t see this.

so for example, if i have $f(t)=1/t$, then $f(t)=t^{-1}$ so $$f’(t)=-1\cdot t^{-1-1}=-t^{-2}=\frac{-1}{t^2}$$

[ocean seal]

coycoy

[wind bane]

by the power rule

[ocean seal]

clyda

[wind bane]

$3(t+3)^{-1}$

[ocean seal]

coycoy

[wind bane]

this part isn’t the hard part, just for sake of notation. the rule is that $1/x=x^{-1}$

[ocean seal]

coycoy

[wind bane]

two different ways of writing the same thing

one is just easier to use/more applicable to the power rule

$3/(t+3)$ is equivalent to $3(t+3)^{-1}$

[ocean seal]

coycoy

[wind bane]

u forgot to multiply by -1 out front

remember, derivative of $x^r$ is $r x^{r-1}$

[ocean seal]

coycoy

[wind bane]

the the $x$ in your case is $x=(t+3)$ and the $r$ is $-1$

[ocean seal]

coycoy

[wind bane]

lol

this is just notation. so if i want to write $1/3$, it is the same thing as writing $3^{-1}$

[ocean seal]

coycoy

[wind bane]

because $(t+3)^{-3}=\frac{1}{(t+3)^3}\neq\frac{1}{t+3}$

[ocean seal]

coycoy

[wind bane]

these are called rules of exponents, or exponent laws if your interested more

sure

yes but differentiation doesnt really care about scalars

you can just drag them out front

example: $\frac{d}{dx}(3x^2)=3\frac{d}{dx}(x^2)=3\cdot (2\cdot x^{2-1})=6x$

[ocean seal]

coycoy

[wind bane]

right. then simplify

remember to pull the scalar out. should still have 3 on the top

no nine anywhere

sry. didn’t parse this correctly. should be 3•-1(t-3)^{-1-1}

yes

you would technically use the chain rule as well but you don’t need to here

anyways. just plug in t=-1 and you’re done

awesome

[marsh cradle]

quick question

is it possible to evalute alpha and beta

based on the ratios of their sines

and cosines

[undone dock]

Do you mean like if you know the sine of alpha is n/m, can you find the value of alpha?

[marsh cradle]

no

sinalpha/sinbeta = m/n and cosalpha/cosbeta=a/b

where i know the values of a,b,m,n

[jagged imp]

yes

[elfin snow]

technically yeah

[jagged imp]

in this context it means the same thing

[elfin snow]

ye

k well I guess this channel is free so

question: I know you can switch the bounds of an integral (e.g. $-\int_a^b=\int_b^a$) but can you do the same for series in a way?

[ocean seal]

gmod

[ionic jewel]

no

series have to go from a lower number to a higher number

otherwise they equal 0

[elfin snow]

hold on let me show you what im trying to do here

this "anagram function" would for example take in 1789 and spit out 9871

I have the concatenation function already and I want to make this summation go from the last term to the first term

could I just do m=k to 1 for the summation?

but that wouldn't make sense so idk

@ionic jewel

or not

[ionic jewel]

you can't do this, but it's adding a bunch of things together

going the correct way (up) will yield the same answer

i don't feel like figuring out how your function works

[elfin snow]

not really because of the way the function works

order matters in the summation

but thanks for helping :)

[ionic jewel]

well a summation isn't the right tool then

[elfin snow]

no but it is the actual function for concatenation

[ionic jewel]

because order objectively doesn't matter when you add things together

associative property or something

[elfin snow]

no ik that but the $10^{\lfloor \log n \rfloor}$ term adds a certain number of zeroes to move the digits place over a certain amount

[ocean seal]

gmod

[elfin snow]

so it boils down to something like 9000+300+70+1 to make 9371

so what im trying to do is get it to do 1000+700+30+9 go make 1739

[ionic jewel]

well do it the other way to get 1+70+300+9000

just swap the upper and lower bounds on the sum

[elfin snow]

that would yield the same number as the input

[elfin snow]

im trying to get this

[ionic jewel]

no i get it

I'm just saying

[elfin snow]

but then the sum would go from k to 1 which doesn't seem to make sense

[ionic jewel]

make your sum

go from the lower value

to the higher value

[elfin snow]

k to 1?

[ionic jewel]

what are the possible values of k?

as long as they are less than 1, go for it

[elfin snow]

positive integers since they denote the k-th term of the sequence

[ionic jewel]

then

do

1 to k

lower to higher

[elfin snow]

but that's the same as before

[ionic jewel]

and why is it wrong again?

[elfin snow]

im trying to invert the number, hence anagram

it's not wrong, it's the input

[ionic jewel]

let me just tldr this

[elfin snow]

ok

[ionic jewel]

For any $\sum_{n=a}^b f(n)$ where $a < b$ would be equal to $\sum_{n=b}^a f(n)$ if you were allowed to go from higher to lower

[ocean seal]

bunny

[ionic jewel]

the second form isn't permitted

but (a+b+c) = (c+b+a)

[elfin snow]

alright so in my original image I state the fact that the concatenation of k terms is this certain summation. the entire summation represents a single number, and my goal is to invert the number so instead of 12345 it would be 54321. in 12345, 1 is the first term, 2 is the second term, etc etc, and 5 is the k-th term. my objective is for 5 to be the first term, etc etc, and 1 to be the k-th term so that the number can be inverted properly

reposted for reference

ok now that im thinking of it, anagram id the wrong word - I'm really just inverting the number (making it backwards)

[ionic jewel]

wait

i see your problem

[elfin snow]

yay

I want the sum to start at the last term and go up to the first term

[ionic jewel]

you don't actually need the sum to go backwards, you can simulate that

same sum of m=1 to k

but then whenever you would use (m), use (k-m) instead

gets the effect you want of "reversing"

[elfin snow]

ohhhhhh

so I replace all my m's with (k-m)'s?

[ionic jewel]

my bad, i didn't understand before

[elfin snow]

no worries

[ionic jewel]

yes, you see how that works right?

[elfin snow]

yeah I do see

[ionic jewel]

well

maybe (k-m+1)

[elfin snow]

ok tysm that makes sense

[ionic jewel]

since m starts at 1 and you want the first one to be k

[elfin snow]

yeah

[ionic jewel]

so k-1+1

yea

ok yeah mb

glad it helped

[elfin snow]

lol dw

:)

[ionic jewel]

you didn't even post the problem

do u know derivatives

uh wait sec

so you could use derivatives to find the velocity at each of the relevent points (2.1,2.01...) and then just average them

there's also an actual equation of you want exact average velocity

assuming you know integrals

add up all the sides