help-27

[woven radish]

zgadnij

[deep relic]

I looked at the solution but the method was way different

(and longer)

[wicked turtle]

what does this mean, if x is a vector?

[deep relic]

just matrix multiplication

[wicked turtle]

but what is x^-1?

vectors don't have inverses

[deep relic]

fatal mistake

yeah

.close

[devout snow]

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[deep relic]

thanks

[wicked turtle]

yw

[devout snow]

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[real whale]

Hey, can someone explain me how to choose alpha and bêta please?

[real whale]

Or maybe how to integrate this integral please

[buoyant veldt]

U don't need to find the exact value, only if it is finite or not

[real whale]

Okay

How can I do that?

I mean, I have the feeling that the value of alpha and bêta doesn't really matter

@buoyant veldt

[buoyant veldt]

What techniques for testing convergence have u learned?

[real whale]

Cauchy's test?

Or Maclaurin–Cauchy test

I mean it's the same but in France we only say Cauchy

[pulsar sand]

You should think about this in terms of p-integrals

[real whale]

What do you mean by that?

[buoyant veldt]

As in $\int x^{-p}$

[woven radish]

Bob the Builder

[sand dove]

intégrales de riemann in french I believe

[buoyant veldt]

So compare with these integrals, whose convergence u alr know

[real whale]

Okay I will try

[wicked rover]

uhh i think that just means riemann integral? jessica specifally means integral of x^-p to infty

[sand dove]

sorry got confused with série de Riemann

[devout snow]

@real whale Has your question been resolved?

[real whale]

I have a question

If I have a sum of two integrals (with same bounds) where one converge and the other diverge, is the sum diverge too?

@wicked rover

[wicked rover]

can u recall basic limit fact to answer this?

[real whale]

lim(f(x)+g(x))=lim(f(x))+lim(g(x))

okay sorry

But thank you!

[wicked rover]

no need for sorry, its just a hint 🙂 and no prob

[real whale]

.close

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[wraith pumice]

factor 6x^2 + 11x + 3. can someone explain how it works solving via the box method or whatever

[wraith pumice]

smth like this

or any method...

[near stone]

ive never used this method before but you can try a different method

find the sum and the product

its just called x-box way?

this might help https://www.slideserve.com/miach/factoring-the-x-box-method

[rain token]

I can show another way

Its weird but works

[near stone]

i wonder how weird is

[rain token]

You take 6 and you mutliply it by 3

[wraith pumice]

xbox doesn't work for this

bc a is greater than one

[near stone]

thats what i just said

[rain token]

So you have x^2+11x+18

[wraith pumice]

typically for xbox you have to have 1x^2...etc

[rain token]

If you factor that you would have (x+9)(x+2)

Now you divide each term by 6

[near stone]

what happened to 6?

[rain token]

(x+9/6)(x+2/6)

[rain token]

Trust the proccess

[near stone]

!nosols also

[devout snow]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[rain token]

Now you simplify

[near stone]

who is the original poster

[rain token]

(x+3/2)(x+1/3)

Now you slide the denominator to your x term

[near stone]

depends on what "doesnt work"

can you show work

[wraith pumice]

💀

[near stone]

read the bot message

[rain token]

Hey its stupid but its the right answer

[rain token]

a

^ this should be the right answer

[near stone]

seriously you are not allowed to give them the answer

[near stone]

.

[rain token]

Oh wait what

Uhhhhhhhhh

[wraith pumice]

it's not that deep really

[rain token]

Whoops

[wraith pumice]

like ik how to use xbox and how to factor it's just on an exam I took I couldn't use xbox bc the ax^2 and bx weren't easiyl dividable

if that makes sense

[wraith pumice]

and someone after class told me to use this

I think if u use that method u can get the same answer. cxbra method was kinda fire though ngl

[rain token]

Go into google and search slide divide bottoms up

[wraith pumice]

bet

[rain token]

yeah

Thats what I learned

[near stone]

if you know how to find the sum and product then you should be fine

[rain token]

That too

[wraith pumice]

cool, thanks

[near stone]

the box thingy is just how you put numbers

[wraith pumice]

yeah

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[quaint cradle]

i need to find m

[quaint cradle]

i tried to write e^x as t and go from there

but im not sure what to do from here

[lusty sapphire]

+m-1>0?

m-n?

[quaint cradle]

m - 1 yeah

sorry for the confusion))

[lusty sapphire]

Find $m$ such that $\forall(x\in\bR)(e^{2x}+me^x+m-1>0)$?

[woven radish]

SWR

[lusty sapphire]

Are you trying to find the determinant here?

[quaint cradle]

yes

i thought maybe it would help

especially because its (m-2)^2

and i guess i could try to find the solutions for t if it helps?

is this relevant?

[lusty sapphire]

I think it should. The determinant will tell you the lowest point on the graph.

[lusty sapphire]

Find where t would be zero

[quaint cradle]

mmm ok ill try this

[devout snow]

@quaint cradle Has your question been resolved?

[modest spruce]

Think about e^x and how it is positive for all real x

[quaint cradle]

yep

thats true

[modest spruce]

so what does m need to be for a + ma + m - 1 > 0 when a is positive

[quaint cradle]

it would be a^2 + ma + m - 1 > 0 right?

[modest spruce]

it doesn't matter what a is just that its positive, as e^x is always > 0 and e^2x > 0 we can think of any e^bx as being > 0

[quaint cradle]

hmmm

makes sense

[modest spruce]

you get my logic ?

so lets assume a=0.001 plug that in to the inequality

[quaint cradle]

yeah, since e^x is always positive obviously e^2x is also positive so we can just use "a" for both

right?

[modest spruce]

yeah

[quaint cradle]

ok cool

so

a ( 1 + m) + m - 1 > 0

[modest spruce]

0.0001 + m*0.0001 + m - 1 > 0

[quaint cradle]

oh

basically m > 1

[modest spruce]

almost

if e^x equaled zero at some point that would hold

but e^x never equals zero so 1 also works

[quaint cradle]

right, 1 also works

meaning m is bounded from [1, infinity)

[modest spruce]

yeah m greater than or equal to 1

[quaint cradle]

now just one more question

why did we have to assume a approaches 0?

because we wanted to find the lower limit

ok duh

logical

[modest spruce]

when faced with questions like this it's a good idea in my experience to try some values, i used a = 0.0001 to show that for small values of a m being 1 still keeps the expression positive, it's always handy to draw the graph of e^x on the side

[quaint cradle]

i ll keep it in mind, it also helped me here

thanks for the idea!

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[wind root]

help ☹️ how th is this a vertical asymptote?
4. Suppose f increases without bound as x approaches some number c from the left. How does the line x=c related to the graph of f?
A. x=c is the tangent line to f.
B. x=c intersects f at (c,0).
C. x=c is a vertical asymptote to f.
D. x=c is a horizontal asymptote to f.

[wind root]

like how ☹️ I dont understand 😭

[mystic scarab]

And recall definitions/meaning of horizontal and vertical asymptotes

You'll have them for sure in your textbook or notes

[wind root]

there arent any definitions thjat were provided to us.. as far as I know we just solve and do activities

also how do I graph them in the first place

nvm let me check sum

yep, not in our notes... I think that wasn't taught since its my first time encountering the word "bound"

pls help I just need an explanation <@&286206848099549185>

[mystic scarab]

Vertical asymptotes occur when the function goes to +∞ or -∞ if x approaches some finite number
For example, y = 1/x has a vertical asymptote, i.e. x = 0

[wind root]

Ohhhh

what abt horizontal asymptotes?

[mystic scarab]

It's somehow the "opposite"

y = k is a horizontal asym if f(x) → k when x → +∞ or -∞

Note that you could have two different horizontal asymptotes, one for x → -∞ and the other one for x → +∞

I believe you understand why they're called vertical and horizontal

[wind root]

yeea

this one confuses me 😭

[forest tree]

It’s not that complicated, the asymptote is a x or y valie where there is no answer, like Alberto said, y=k and x=h are the asymptote

[misty crest]

what’s confusing

[wind root]

nvm I figured it out lol the word polynomial got me overthinking 😭 math rlly isnt my thing fr

[wind root]

ohh

[forest tree]

But for the limits you can do it with Lim X—> 0 or Lim x —> infinity, for the asymptote

[wind root]

oh, then whats a left handed limit is it the one where e.g. c^-?

[mystic scarab]

Exactly

[wind root]

then the answer here is when x approaches to c^-?

srry im really cooked, I only remember how to solve these, but not the concept

WTH IS THIS ☹️😔

[mystic scarab]

Well, you HAVE to know that thing is the tangent

You first need to revise the theory

[wind root]

HOW 😭

[wind root]

what theory exactly... cuz like the only theory we discussed is those constant, radical/root, division..etc

[mystic scarab]

Theory about derivatives

[wind root]

ye that was when I was absent

its this right?

[mystic scarab]

Yep that's it

[wind root]

how am I supposed to understand when it's all solving...

like these are all solving

[mystic scarab]

https://youtu.be/v1OkAmd_AD8?feature=shared

https://youtu.be/x3iEEDxrhyE?feature=shared

[wind root]

THANK YOU 😭😭😭🙏

my head is hurting trynna understand these 😭

what is that

what topic

[foggy ermine]

It should be A. x

I think

I just used 1/x as my sequence of x sub-base n.

[devout snow]

@wind root Has your question been resolved?

[wind root]

thank you

[foggy ermine]

Am I correct?

[wind root]

ye

how abt this one?

[frozen aurora]

do you understand what that statement means?

``$\exists m$ and $M$ such that $\forall x \in [-1, 1]$, $m\leq f(x)\leq M$''

[woven radish]

artemetra

[frozen aurora]

if yes, try describing in plain english what it's supposed to mean

[wind root]

no...

😭

[frozen aurora]

if i am to translate it literally it says that there exist two numbers m and M such that f(x) is between then for all x between -1, 1

those are bounds on f(x) on that interval

[wind root]

ohhhh

[frozen aurora]

so to find a function for which this is not true, you need to find a function that is unbounded (= grows to + or - infinity) somewhere on [-1, 1]

[wind root]

then B?

[frozen aurora]

yep

[wind root]

O

what abt this?

[frozen aurora]

ok

the limit is to 0+

right?

[wind root]

ye

right handed

[frozen aurora]

this means that we are only interested in stuff that happens around zero, and more conceretely to the right of 0

[wind root]

so 1?

🥹

[frozen aurora]

why 1?

i don't want you to just guess lol

[wind root]

um cuz its at the right of 0 and both points point at it

this

[frozen aurora]

yeah but this happens around 1

[wind root]

oh

[frozen aurora]

you are interested in what is going on around 0

so this is your only area of interest

[wind root]

Ohhh

[frozen aurora]

the fact that there are jumps at -1 and 1 does not influence the limit at 0

[wind root]

mb

[frozen aurora]

so what's the answer now

[wind root]

idk 😭 0?

[frozen aurora]

yep

[wind root]

How

[frozen aurora]

it goes to zero in this area

[wind root]

wth

[frozen aurora]

are you familiar with the formal definition of a limit or nah

like epsilon delta

[wind root]

nah we literally just delve straight to solving

[frozen aurora]

alright

[frozen aurora]

what is confusing you?

[wind root]

why is it zero and why is it not focused on this:

is it cuz its not abt the asymptote or whatever

[wind root]

nvm u answered that alrdy

[frozen aurora]

you are looking at where x = 0

imagine zooming in on the function around 0 as arbitrarily far as needed

at some point stuff that happens around -1 and 1 doesn't matter

you start getting interested in region [-0.1, 0.1], then [-0.001, 0.001], then [-0.000000000001, 0.000000000001] and so on

[wind root]

oooooh

Ok

[frozen aurora]

so again you only really care about what's happening super super close to zero

[wind root]

alright I understand it a bit now

what about this? is the answer b?

[frozen aurora]

yes

[wind root]

Lol cuz its literally pointing there

[frozen aurora]

lmao

↘️ asymptote

[wind root]

A?

[wind root]

actually ignore that I need help in this instead

whats up with the arrow pointing to two

[frozen aurora]

alright you want the limit as it goes to -infinity

so you want to look far into the left side

someone continued the curve for you

[wind root]

oh then it is two?

[frozen aurora]

yep

[wind root]

what abt this? i dont get it

[frozen aurora]

what do you not get

[wind root]

why its pointing up

nvm

[frozen aurora]

because the graph is going up?

[wind root]

LOL

😭

ye

is the answer perhaps infinity?

[frozen aurora]

yes

[wind root]

how will it be -infinity then?

or dne

[frozen aurora]

huh?

the answer is infinity

[wind root]

ye but what does a -infinity and dne graph looks like?

[frozen aurora]

-infinity

[wind root]

how about a dne?

[frozen aurora]

dne: this at 1 or -1

(if we talk about a two-sided limit)

or for example 1/x

[wind root]

ohhh

[frozen aurora]

limit dne at 0

[wind root]

what about this? is this infinity?

[frozen aurora]

no

[wind root]

dne?

[frozen aurora]

yes, but why?

[wind root]

Ohhh

[wind root]

cuz it looks like the graph u showed LOL

😭😭😭😭

[frozen aurora]

that's a fair point

[wind root]

Ye imma need to study all of it by myself

[frozen aurora]

well stuff left of 2 is going to -infinity

but stuff right of 2 is going to +infinity

you cannot go to both + and - infinity at the same time

thus the limit does not exist

[wind root]

ohhh

[frozen aurora]

(nb: this is not the only way a limit can be nonexistent, but it's the most common one)

[wind root]

how did number 36 go both + and - infinity?

[wind root]

oh

ok

[frozen aurora]

green goes to +infinity, blue goes to -infinity

[wind root]

IM SO DUMB

😭😭😭😭😭😭

[frozen aurora]

lol

[wind root]

this is -infinity right?

[frozen aurora]

yeah no this is just visual stuff

yep

[wind root]

what abt this?

is it infinity since its 2 on the left

wait u alrdy said the answer

nvm

[wind root]

lol

LOL

thank you 🥹🥹🥹🫡🫡🫡

Whoever you are

I hope u have a blessed day fr

🙏🙏😔

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[frozen aurora]

wish you good luck with studying

[devout snow]

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[restive river]

How would I find the range of this function? I don't know the proper mathematical way of finding ranges yet, I usually just look at the denomitator and exclude the value of x when the denomiator = 0

[wheat pawn]

well, one way is representing the function

if you dont want to do the full study, you can just go for asymptotes, which will give the most information

[devout snow]

@restive river Has your question been resolved?

[granite zenith]

I though you just do domain of the inverse

Switch the places of x and y, solve for y, find the domain

But also in this scenario it has no way of reaching positive infinity because the bottom is negative after you pass 1

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[rain token]

Woah

[rain token]

Is that an aops problem

.close

[devout snow]

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[lapis mirage]

Complete the unit circle reference, then use the circle to find the exact value of each trigonometry function

[wintry flare]

which question

#15?

[lapis mirage]

Yes

[wintry flare]

what is troubling you on this part

[lapis mirage]

It’s like trigonometric values is 0 over -1

I just don’t know what to do

[wintry flare]

correct

so this means your answer is 0

thats all

well actually

its not 0

that would be for 180° sorry

at 90° and 270°, out of sine and cosine, which value is 0

?

[lapis mirage]

?

So it’s 0 right

[wintry flare]

no

tan=sin/cos

at 90° and 270° sine is equal to 1 and -1, respectively

and cosine is equal to 0

so youd have -1/0

does this make sense?

this statement is not equal to 0

[lapis mirage]

Ok

[wintry flare]

so your answer would be undefined

does this make sense ??

[lapis mirage]

Yes

[wintry flare]

whats the question bog

[lapis mirage]

?

[wintry flare]

which one

[lapis mirage]

3

My bad got confused with bog

[wintry flare]

do you know how to start it

[lapis mirage]

Yeah

[wintry flare]

alright looks pretty good to me

sine is negative though

which part are you stuck on?

[lapis mirage]

How did they get the quad

[wintry flare]

it tells you that sinθ < 0

meaning its negative

sinθ < 0 implies it lies in either Q3 or Q4

do you know which quad its in based off tanθ=-21/20 ?

[lapis mirage]

Idk ngl I thought it would be quad 3

[wintry flare]

what value is cosine in q3

sign wise

[lapis mirage]

Idk ngl, am using the unit circle reference

Both neg

[wintry flare]

that is correct yes

however the tangent value would be positive

a negative over a negative = a positive fraction

[lapis mirage]

Make sense

[wintry flare]

so itll be in q4

[lapis mirage]

Ight thank you

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[hybrid veldt]

Help

[hybrid veldt]

Can someone check my work

[junior flax]

where's your answer 🗿 . How am I supposed to chack when there's no answer

[hybrid veldt]

I got

13.1

[wind mason]

,w 16*cos(53 degrees)

[wind mason]

Nope

Do it again, full work shown

[hybrid veldt]

Is that right

[junior flax]

should be

[hybrid veldt]

Ok thanks

[wind mason]

Much better.

[hybrid veldt]

For this problem I got 74

Degrees

Can you guys check it ?

[devout snow]

@hybrid veldt Has your question been resolved?

[devout snow]

@hybrid veldt Has your question been resolved?

[devout snow]

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[restive river]

no

[restive river]

i had to go

ive been trying for hours

just to fail again and again

[summer summit]

this was a while ago, restate your question again

[restive river]

i used quadratic formula

How would I find the range of this function? I tried using quadratic formula but i dont think im doing it properly

[devout snow]

@restive river Has your question been resolved?

[wintry flare]

did you solve for y ?

[devout snow]

@restive river Has your question been resolved?

[devout snow]

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[vocal mural]

,,$a_n = 5^{n+1} - 4^{n+1}$

[vocal mural]

How can I find a recursive formula for this?

$$ a_n = 5^{n+1} - 4^{n+1} $$

I've plugged in the values:
$a_0 = 5 - 4 = 1$
$a_1 = 25 - 16 = 9$
$a_2 = 125 - 64 = 61$
$a_3 = 625 - 256 = 369$

[woven radish]

eugene

[devout snow]

@vocal mural Has your question been resolved?

[vocal mural]

<@&286206848099549185>

[thin willow]

Sure

[vocal mural]

thank you

for context, this formula finds a coefficient for some z

[vocal mural]

hello...

🙁

[teal charm]

dont worry

im just as confused as you

we cant write it as arithmetic oviously

nor geometric

[vocal mural]

Thanks i am glad i am not alone

[vocal mural]

why not geometric

[teal charm]

it doesnt have a common ratio

9/1=9

61/9 is definitely not 9

so it has to be a combine

which is hard

in case you dont know

combines are either add then multiply, or vise verse

[vocal mural]

Maybe using the 5 and 4 could do something

[teal charm]

wait we are looking for a recursive formula

[vocal mural]

Yes

[teal charm]

oh that changes everything

so it has to be written as a_1= 9

then a_n+1=a_n, but a_n has to go through a function

brb gonna math experiment

[vocal mural]

yes

I have a harder problem after this 😭

[teal charm]

and there you have it

a_n1 is just a_n+1

what could possibly be harder

[vocal mural]

this only works when n = 0

[teal charm]

no it does not

does?

hold on

ohhhh

[vocal mural]

youre saying 5^10 - 4^10 = a_2?

[teal charm]

hmm

yeah now i see it

i fu up

hmm

now its hard

[vocal mural]

i got more in stock due tomorrow morning

this is supposed to be the easiest one

[teal charm]

oaky

if you simplify the og formula, you get $$ 5^{n} - 4^{n} + 1 $$

[woven radish]

FlamingChez

[teal charm]

i wonder...

[vocal mural]

Its a closed formula for a sequence

[teal charm]

no you cant

the best way to do it

is to simplify 5^n - 4^n

but you cant

according to mathway its not simplifyable

[vocal mural]

check out help-45 as well if you're able to help with that

[teal charm]

yeah no.

but the simplified doesnt match with the original

idk

i think i did something wrong

imma grab someone

[vocal mural]

doesnt help that i have a massive headache rn

[visual hazel]

dont really know if this counts as recursive

but i got an = 5^n + 4a_{n-1}

[vocal mural]

need to find for a_n+1

[visual hazel]

i see

[visual hazel]

it should all be in terms of a_n only?

[vocal mural]

actually, i think you can have the recursion formula = a_n but you would have to use n-1 or n-2 etc to calculate it

if youre using a_n+1, you can use a_n in the expression which is better since you can plug in 0 for n

it shouldnt really matter though

as long as we can use the existing term and previous term to find the next term

[devout snow]

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[daring hollow]

help

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[haughty monolith]

DO NOT EVALUATE

[winged bluff]

To be honest im not 100% sure this is correct but if ive understood the question correctly then maybe you could:
first set up the integral that youd use to evaluate the area normal ly(cosx-ln(x²+1) yada yada),
then since youre summing up infinitesimal bars, when you use those bars as a base for the cross region of a right isosceles triangle they instead of having length f(x)dx theyll have area 0.5f(x)^2 dx (since the base and height are equal cuz its iso and right angled, so you just got the area of the cross region to be 0.5 x the base (f(x) dx bar) x the height (same as the base cuz iso right angled)
so itll the integrand will be 0.5(f(x)-g(x))² maybe

correct me if ive misunderstood anything/lmk if any part doesnt make sense cuz the explanation is kinda shaky tbh

so you dont get the solution?

o h

well idk much about calc ab but pretty sure you need do be able to do that for it yeah

[devout snow]

@restive river Has your question been resolved?

[shadow dirge]

should be this

these are how the triangles are laid out

[winged bluff]

isn't that what i said just without giving out the solution directly...

[shadow dirge]

the bottom left expression is the volume of an infinitesimal "element" of the solid

[shadow dirge]

i didn't read your text

[winged bluff]

oh ok

[devout snow]

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[sterile jungle]

I wish to prove or disprove that there are infinitely many n such that tau(n) = 2025 and n divides sigma(n) + phi(n)

[sterile jungle]

my instincts are to consider all possible forms of n (p^2024, p^674*q^2, etc)

[mild comet]

is sigma the sum-of-divisors function

[sterile jungle]

im not too sure if this is the intended idea or is there another elegant way?

[sterile jungle]

yep

[mild comet]

and phi the totient function

[sterile jungle]

yep

[mild comet]

by the way, how do you know that there are infinitely many n such that tau(n) = 2025 in the first place?

[sterile jungle]

the prime divisors could be any selection of primes

2^2024, 3^2024, 5^2024, 2^674 * 3^2

[mild comet]

tau of those numbers isn’t 2025 though?

[sterile jungle]

tau = number of divisors

[mild comet]

oh, I thought it was the ramanujan tau function

then that makes sense

[restive river]

hi soup and harry

[mild comet]

sorry, I’m not good at number theory, and I don’t know how to solve it. Maybe you could try bounding sigma(n) + phi(n) to restrict the number of divisors. Is this part of a problem set?

[sterile jungle]

yea it's part of an olympiad training module

[devout snow]

@sterile jungle Has your question been resolved?

[devout snow]

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[inner ibex]

Hi how would I graph this

I have plotted the asymptotes of +-3

[gray plover]

Hey I was looking for some help in this question. just not getting the right approach on how to solve it

[wind mason]

There are also other asymptotes you are missing (horizontal!)

[wind mason]

!occupied

[devout snow]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[inner ibex]

how do i fidn that one?

im not familiar with this form

normally the horizsontal asymptote is on the side like +k

[inner ibex]

wat

3/1?

[wind mason]

And one more thing, note the end behavior as the function is approaching these asymptotes

[wind mason]

Yep

[inner ibex]

wow

is there any other way of finding it?

[wind mason]

Uhh

Not that I know

You could apply limits but it's the same thing

[inner ibex]

how did they find the one on the top right

[wind mason]

The factored form?

[inner ibex]

ophh mb

this one

oh..

thats the top left..

[zenith star]

The fourth second quadrant?

[inner ibex]

wth i thought quadrant was unit circle

[zenith star]

Nah, it's relative to the axes.

[wind mason]

Well.

[wind mason]

^

[inner ibex]

wat

[wind mason]

A quadrant refers to a space within the Cartesian plane

[inner ibex]

oh ok

how did they get that

[inner ibex]

reghjerjhgre

[wind mason]

This!

the function gets larger as x->-3 from the left

[zenith star]

You could isolate this observation by looking at a graph of 1/x?

[inner ibex]

i dont understand dis

[zenith star]

As it is approaching -3.

[wind mason]

See how the asymptote is x=-3, where you're referencing?

[inner ibex]

yes

[wind mason]

Well, a general rule of thumb is that unless there is a root of multiplicity 2 in the denominator, that the direction of the graph will alternate near the asymptotes

[wind mason]

So how did you get that graph from (3, infinity)?

[inner ibex]

the x intercept of 3

i mean 43

[wind mason]

Yeah, you can just start from there and apply this.

[inner ibex]

4

[wind mason]

Well, I was just looking for "x-intercept".

Anyways.

[inner ibex]

the x intecept is 4 soz

[wind mason]

You can also just plug in a really small value, like -3.1, to see if it stays positive or not (or even plug in a whole number value!).

[inner ibex]

how do i graph the line with no point

[wind mason]

What do you mean?

[inner ibex]

the one on the top left

i still dont know how they got it

i got the one on the bottom right because there was a point i could use

x = 4

and 2 asymptotoes

but now i have no point

but 2 asymptotes

[wind mason]

I gave you two method

[wind mason]

This is the most logical one

[wind mason]

This is just a general rule

[inner ibex]

.close

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[bleak arrow]

why are there extranaeous solutions in radical equations?

[wind mason]

Because they aren’t in the square root’s principal domain/range.

[bleak arrow]

I know that's why solutions are extraneous, but I asked a different question, I asked why do we get extraneous solutions in the first place?

why isn't everything perfect like regular linear equations

why do we get extraneous solutions

[wind mason]

Because when we square both sides, we introduce a new domain that the square root is not used to.

Think about the inverse of x^2.

[bleak arrow]

ok

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[idle citrus]

i just need someone to check if this is correct

[mild comet]

You could check the answer by imagining it drawn out. Since 2.58 + 14 < 40, the triangle would be impossible, because of the triangle inequality.

(You took square roots instead of squaring, in the second line)

[idle citrus]

youre right lemme try and resolve i always get confused 😭

[idle citrus]

okay i got 37.47 this time

well approx 37.47

[mild comet]

good

[idle citrus]

alright thank you!

[mild comet]

this time, if you imagine it drawn out, the base is about as long as the hypotenuse

so you can confirm your answer that way too

[idle citrus]

thanks!

[mild comet]

and not incredibly short like 2 ft

[idle citrus]

.close

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[viral lynx]

I got the derivative of g(x) and h(x) and set them both equal to 0

for g'(x) = 0 I got x = +-1 and for for h'(x) = 0 I got x = 1 and x = 0

Just so that I understand, these are the x values of the maximums, right?

[bitter quarry]

for maxima g''(x) <0 and for minima g''(x)>0

[viral lynx]

Like I still need to sub these x values back into g(x) and h(x) to check if theyre valid, right?

[bitter quarry]

the values u got are just when the slope is zero

they could either be maximum or minimum

[bitter quarry]

so this

or yea u could just plug in the numbers and check the value, the bigger value is maximum and smaller value is the minimum

[viral lynx]

Ah ok got it

Thank you so much!

❤️

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[devout snow]

@light hazel Has your question been resolved?

[mild comet]

can you show your work?

I’m not familiar with the method you’re using (I only know of the method of lagrange multipliers), could you describe it in more detail?

[devout snow]

@light hazel Has your question been resolved?

[mild comet]

I tried graphing it in desmos 3d, and came up wth the idea of performing a linear change of coordinates: let s = x_1+x_3, and let t = x_1-x_3

Then the constraint simplifies to a quadratic in two variables

Let s and t be as above. Then calculate q and r in terms of s, t, and x_2. Then you will find that t can be set to 0

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[uncut tiger]

where do i start with this question

[pseudo basin]

if you didn't know that there was only one intersection with the x axis

could you find all such intersections?

[uncut tiger]

does that mean x-4=2x+K

or do u use simultanious equations

[ripe hollow]

When a curve intersects x axis

the y becomes 0

[uncut tiger]

yes

[ripe hollow]

so we may equate the equation to 0

(x-4)(2x+k) = 0

We can find the value of x and k by this

[uncut tiger]

2x^2+kx-8x-42=0

[ripe hollow]

It's rather a complex turn

Why don't we just divide both sides first with (2x+k)

What will the equation become?

[uncut tiger]

just x-4

x=4

[ripe hollow]

exactly

We have the value of x

[uncut tiger]

then sub it in?

[ripe hollow]

Exactly

[uncut tiger]

wait so how do uk to put y=0

i understand x int is when y=0

but like hm

[ripe hollow]

If a curve intersects x axis

y is 0

[uncut tiger]

what would happen if i expanded that equation and solved it from there

instead of dividing both sides with 2x+k

[ripe hollow]

Well you certainly can expand

But can you get the value of x and k by doing that?

not really

[uncut tiger]

uh

[ripe hollow]

It's similar to solving quadratic

[uncut tiger]

but theres 2 values that are unknown

[ripe hollow]

Yes that's why we have two parts of the equation

It's broken down for us

So we first find the value of x

Then substitute

[uncut tiger]

yes

[ripe hollow]

To find the value of k

However are you still unsure

How intersection of x axis

Results in y=0?

[uncut tiger]

no ik to find x into y must be 0 and opposite w y axis

thats when x=0

idk how i didnt see that

[ripe hollow]

It's all chill

Good job tho

Now you know

[uncut tiger]

alr ty

yep

ima do more question cya

[ripe hollow]

No problem have a great day

[uncut tiger]

u2

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[devout snow]

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[deep relic]

$$
Ax^{*}Ay - Ax^{*}\lambda_y y - \lambda_x x^{*}Ay + \lambda_x \lambda_y x^{*}y = 0
$$

why AxAy can be reshufled and simplified to ||A||^2? the same with the third term

[woven radish]

zgadnij

[deep relic]

b)

vector matrix multiplication is not commutiative

and how A*A gives norm or whater this is

A*A should be another matrix

.close

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[deep relic]

.reopen

[devout snow]

✅

[deep relic]

I substituted using the Ax

but I still don't understand how they get ||A|| there

[pseudo basin]

i think there's an easier way to prove the same problem tbh

consider $x^*Ay$ and show that it's equal to both $\lambda_x x^*y$ and to $\lambda_y x^*y$

[woven radish]

Ann

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[devout snow]

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[jagged lily]

How do I show that this series is probably divergent

S_2

I’m not too sure what test I use

I tried the ratio test it just gave me 1

Inconclusive

I’m not too sure how I show lim_n->inf a_n = inf

hold up

hospital rule

wth

im done

-close

,close

?close

!clpse

!close

bruh

.close

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[grand warren]

.close 😭

[jagged lily]

I gotta remember that

🙏

thank you

[devout snow]

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[lavish canopy]

Is 101.3 deg rlly a solution here?

[winter patrol]

no, they messed up their identity

they should've added 180°, instead of subtracting 78.7 from 180

[devout snow]

@lavish canopy Has your question been resolved?

[lavish canopy]

Why add 180?

[winter patrol]

period of tan

[lavish canopy]

Right, but why doesn’t subtracting by 180 work?

Would u always just add 180 in this case?

[gritty juniper]

It does work

[lavish canopy]

So then 101.3’s also a solution?

[gritty juniper]

No

[lavish canopy]

Oh, wait

She subtracted 180 from theta

[wind mason]

$\tan(180^{\circ}-x^{\circ})\neq\tan(x^{\circ})$, since $\tan(-x)=-\tan(x)$.

[woven radish]

;(

[lavish canopy]

But even if you subtract by 180, it falls outside the domain

[lavish canopy]

Okay

[wind mason]

Elaborate.

[lavish canopy]

Look at the domain

Nvm, it’s not in the pic

But it must fall within unit circle values

[wind mason]

Which are?

[gritty juniper]

Ur teacher did a mistake...

[lavish canopy]

Okay, so the answers are 78.7, 45, 225, and 258.7?

Right?

[wind mason]

We already identified that, from this

[gritty juniper]

Yea

For 0 to 2Pi

[wind mason]

Yes. Also, you never clarified the domain.

[lavish canopy]

Thanks

Forgot to

[devout snow]

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[devout snow]

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[limpid epoch]

Does this look right?

Point D (2, 1)

Point E (3, 5)

Point F (6, 2)

PART A

Side DE = √((3 - 2)^2 + (5 - 1)^2) = √(12 + 42) = √(1 + 16) = √(17) ≈ 4.1231

Side EF = √((6 - 3)^2 + (2 - 5)^2) = √(-32 - 32) = √(9 + 9) = √18 ≈ 4.2426

Side FD = √((2 - 6)^2 + (1 - 2)^2) = √(-4)^2 + (-1)^2) = √(16 + 1) = √(17) ≈ 4.1231

PART B

Side DE = (5 - 1)/(3 - 2) = 4/1 = (y = 4x)

Side EF = (2 - 5)/(6 - 3) = -3/3= (y = -x)

Side FD = (1 - 2)/(2 - 6) = -1/-4 = ¼ = (y = x/4)

PART C

Equilateral - Not all sides are congruent in length, only DE and FD are.

Right - Sides DE and FD are opposite recipricals, therefore they create a 90° angle.

Isosceles - Side DE and Side FD are congruent in length, proving this correct.

Scalene - Side DE and Side FD are congruent in length, thus disproving this.

Triangle DEF is an isosceles, right triangle.

[dense jay]

your steps confuse me
where does the sqrt(12+42) come from, the stuff before and after it are fine

i havent checked beyond part A yet

oh i see

its your exponents

[limpid epoch]

OH I AM SO SORRY

Let me fix that

That should be better, I thought the exponents would copy as written, but I guess I was wrong, my bad.

[dense jay]

for part B, your slopes are fine but the lines themselves arent

[limpid epoch]

Do you mind explaining how?

[dense jay]

you have them all passing through the origin

[limpid epoch]

ohh, I see, I didn't add the intercepts?

[dense jay]

yeah

[limpid epoch]

ah, I see what you mean

[dense jay]

like for DE
D should be on there, if y=4x then y=4(2)=8, but it is actually 1

[limpid epoch]

oh, I see

So using mx+ b = y it would be 4x + 1 = y

Okay

[dense jay]

i dont think b would be 1

[limpid epoch]

Really?

[dense jay]

that would make point D into (2,9)

[limpid epoch]

Oh, true

Wait, nvm, I think I just need to find the slope so the equations don't matter

[dense jay]

ah alright

[limpid epoch]

mhm, thank you so much!

[dense jay]

DE and FD dont make a 90 degree angle

[limpid epoch]

I really appreciate it (geometry isn't my strong suit)

[limpid epoch]

I thought since they were opposite recipricals they would be?

[dense jay]

has to be the negative reciprocal

[limpid epoch]

Ohh, and since -1/-4 would simplify to 1/4, it doesn't work?

[dense jay]

yup

would have to be 4 and -1/4

[limpid epoch]

ah, okay

So it's just an isosceles?

[dense jay]

that seems to be the case

[limpid epoch]

Alright, I see what you mean

Thank you so much!

.close

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[ivory salmon]

could someone help me with these? i got the right answers by just making tables but im not sure how to actually solve it algebraically or what steps i should take

[faint gorge]

the usual way is to observe the dominant term in the numerator and denominator and factor that

and you are using that 1/x -> 0 as x goes to infinity

[ivory salmon]

by dominant term do you mean the highest degree?

[wicked rover]

yes

[ivory salmon]

would it be something like this then?

[wicked rover]

smth like that but ur missing details

u must factor out the dominant on top and bottom

in this case x^2

[ivory salmon]

like this?

[wicked rover]

yes

when u take lim x->infty u can cancel x^2

[gleaming fog]

oh i see it now too

[ivory salmon]

okay! sorry what do i do next 😭

[gleaming fog]

you cancel out the x^2 from top and bottom

then plug in infinity to the denominator

because an infinite denominator is infinitely close to 0

we can assume the x^2 in the denom means that whole thing is 0

so it simplifes into -2/-1

[wicked rover]

for large $x$, that fraction becomes $\frac{\frac1{x^2}-2}{\frac4{x^2}-1}$

[woven radish]

ロケットジャンプ

[ivory salmon]

ohh so like this?

[gleaming fog]

(at least i think))

(im not a helper)

[wicked rover]

thats essentially what happens but a teacher would yell at u for "plugging" infty

[ivory salmon]

oh 😭 what would they want instead?

[wicked rover]

on the side write "we know $\lim_{x\to\infty}\frac1{x^2}=0$" etc

[woven radish]

ロケットジャンプ

[ivory salmon]

oo alright 🫡

[gleaming fog]

for this part we simplify by using the conjugate right?

[ivory salmon]

i was thinking about using the conjugate but i wasnt sure if that was the right way to go

[gleaming fog]

this is what i write on tests

[wicked rover]

and u should either write lim in each expression or put $\app$ when u finally take the limit

[woven radish]

ロケットジャンプ

[wicked rover]

smth like $\dots=\dots\app\dots=\dots$

[woven radish]

ロケットジャンプ

[ivory salmon]

its all multiple choice w no partial credit so i havent really been careful with the steps

[wicked rover]

in that case its fine

[ivory salmon]

but ill keep this in mind!! thank you

[wicked rover]

but u know to be careful if its short response 🙂

[ivory salmon]

should i open another help for the 2nd question i posted?

[wicked rover]

r u done with this one

[ivory salmon]

im done with the top one but not the bottom one

[wicked rover]

ok just continue here

[ivory salmon]

🫡

should i multiply by the conjugate? it doesnt seem like theres anything i can easily factor out

[wicked rover]

if u squint hard does the top dominant look like |x|?

[ivory salmon]

uhh if i desmos it, it looks like it

[faint gorge]

when it comes to dominance you can neglect smaller terms basically

[wicked rover]

so aim on doing algebra to look like it

basically u use sqrt(x^2)=|x|

can u show me the first step in algebra

[ivory salmon]

ohh okay, i dont think ive ever used sqrt(x^2)=|x| before lmao i didn't realize it was the same

[gleaming fog]

i thought u cant do a sqrt of x because it becomes imaginary or smth like that

https://cdn.discordapp.com/attachments/1274477651041456223/1281255374162493510/look-up-scared.gif

[faint gorge]

x^2 is always positive

non-negative to be precise

[gleaming fog]

ah

[wicked rover]

first just look inside the sqrt and factor, what do u get?

[ivory salmon]

do i factor out x^2?

[wicked rover]

yes

[ivory salmon]

so like this?

[wicked rover]

$\sqrt{1+16x^2}=\sqrt{x^2(1/x^2+16)}=\sqrt{x^2}\sqrt{1/x^2+16}=|x|\sqrt{1/x^2+16}$

now sqrt distributes over product

now use my rule

[woven radish]

ロケットジャンプ

[ivory salmon]

would it be like this or is this malalgebra

[wicked rover]

its exactly what i have

[ivory salmon]

🫡

[faint gorge]

(dont forget the = )

[ivory salmon]

then 1/x^2=0?

[wicked rover]

cant do that yet

we're not looking at the whole fraction

$\frac{|x|\sqrt{1/x^2+16}}{x+5}$

[woven radish]

ロケットジャンプ

[ivory salmon]

that's what i have so far, what next?

[wicked rover]

we can replace |x| by being smart

lim x->-infty

ie we only consider large negative x

then what does |x| become?

[ivory salmon]

inf? 😭 sorry im not sure

[wicked rover]

if x is large enough and negative we can safely say x<0

[ivory salmon]

so just -x?

[wicked rover]

yes

[ivory salmon]

ohhh alright

[wicked rover]

$\frac{-x\sqrt{1/x^2+16}}{x(1+5/x)}$

no more can be done on top

what about bottom?

[ivory salmon]

factor out x?

[wicked rover]

yes

[woven radish]

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[ivory salmon]

so like this?

[wicked rover]

ready to take limit or no?

[ivory salmon]

or am i skipping steps

[wicked rover]

its a bit bad

u had x on top and bottom

u can cancel

$\frac{-\sqrt{1/x^2+16}}{1+5/x}$

[woven radish]

ロケットジャンプ

[ivory salmon]

o okay so cancel first then sub 1/x=0?

[wicked rover]

yes

also why urs doesnt work is that

after u take limit u should have no x anywhere