help-10

[obtuse pebble]

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[fathom flicker]

Working on this proof. So far I've reduced it to, if a0=0 or a0=1 or a0 prime then we need to show an eventaully composite. Because otherwise a0 itself is composite and we are done

if a0=0 or a0=1 we eventually get 15=5*3 so those are also done

so essentially the problem is just

if a0 prime

show eventually a_n for some n is not prime

and while that is simpler I feel

I am also not sure how to deduce a statement about a numbers primeness

like that

[kind hawk]

feels like it should be possible to maybe give a closed form which you can maybe study

[fathom flicker]

wouldn't the closed form depend on the initial value

[kind hawk]

yes

[fathom flicker]

so how can we get something like that

I have done examples through a0=6

so for all 0,1,2...

but nothing stuck out really

a0 and a1 are the same just shifted

[kind hawk]

so lets do it abstractly

[fathom flicker]

sure

[kind hawk]

I am not sure if this leads somewhere but its my first idea

so a1=2a0+1

a2=2a1+1=2(2a0+1)+1=4a0+3

[fathom flicker]

a3=8a0+7

even

done

surely not

[kind hawk]

no

+7

[fathom flicker]

ah

but this looks useful

[kind hawk]

a4=16a0+15

what might we conjecture is the closed form?

[fathom flicker]

a5=32a0+31

always 1 less on the thing not a0

hm

[kind hawk]

what

[fathom flicker]

like

32 and 31

16 and 15

[pseudo swift]

And what's the coeff in front of a0 looking like

[fathom flicker]

8 and 7

a0 is always even

well

attached to an even thing

2^n look

[kind hawk]

yes 2^n

[fathom flicker]

2^n+2^(n)-1

[kind hawk]

so we conjecture that an = 2^n a0 + 2^n -1

[fathom flicker]

yes

[kind hawk]

and by induction we can prove that it is true

[fathom flicker]

an=2^(n)a0+2^(n)-1

and then we need to prove that an contains a composite number for all a0?

[kind hawk]

that for some n it will be composite, yes

[fathom flicker]

seems like the same question

haha

[kind hawk]

well we learned something more about our sequence. which could be useful

hmm

[fathom flicker]

if a0 odd

[kind hawk]

well we already saw that a0 is a prime

[fathom flicker]

2^(n) a0 is even and 2^(n)-1 is odd and their sum is odd

yeh a0 must be prime for us to care

their sum is always odd anyways

[kind hawk]

from the start clearly the sequence only consists of odd numbers cause its always 2*something +1

[fathom flicker]

yeah lol

it does generate primes nicely though

suprisingly enough

but that is off topic

an=2^(n)a0+2^(n)-1
the 2^(n)'s are starting to smell familiar to our last problem but I doubt they are actually related

[kind hawk]

2^n just appears everywhere

[fathom flicker]

a much stronger result holds

but that is expected

otherwise we have a nice prime generating function basically

[kind hawk]

yeah

[fathom flicker]

so not actually that interesting

[kind hawk]

hmm

I am not yet seeing where we want to go

I suppose an obvious observation is that the last digits either cycle 1->3->7->5->1 so eventually it will be divisible or they cycle 9->9

so we can assume that the prime we start with has last digit 9

[fathom flicker]

what'?

because if last digit of 2,4,6,8 or 9 they are composite?

not the case for the integer 2

or is 2*1 composite

[kind hawk]

an is odd. so it cant have 2,4,6,8 as last digit

[fathom flicker]

I wasn't saying no to that

hm

Denascite, it is 2:57

I must be going

sorry for the brain teaser

[kind hawk]

what

you and going to bed?

what an odd pair

[fathom flicker]

I promised you I'd go to bed by 3

[kind hawk]

yes

[fathom flicker]

so I have to erase my whiteboards and be on my way home

otherwise

i wont make it

[kind hawk]

you arent even home?

[fathom flicker]

just in a study room

close enough

very short walk

but now I might be 1 minute late because of u

3:01 it is

[kind hawk]

yes definitely because of me

all my fault

[fathom flicker]

indeed

glad we agree

[kind hawk]

good night

[fathom flicker]

goodnight!

thanks for your help

ill work on this tmrw

I'd say in the morning but I wont be up in the morning

.close

[obtuse pebble]

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[latent frost]

Derivatives!

[latent frost]

So, I can't learn the same way as others and would therefore ask if one of you amazing people kindly could help me get this concept

So there's a lot of different rules, and I know you can do (x+h) - (x) / h to get a proper result

but I'm not entirely sure how to do that when it gets "complicated" and would therefore prefer to learn the shortcut rules

so is my understanding correct here:

if I have (sin^2(2x))^3

I would have to use the "chain rule"

meaning I start from the outside and then work my way inside

So

3(sin(2x)^2)^2 * 2cos(2x) * 2

correct?

now, if we have (x^2)*sin(2x)

then we have to take the derivative of x^2 * sin(2x), then + the derivative of sin(2x) * x^2, also known as f'*g + g' * x

is this also correct?

then we have the quotient rule (?), meaning that if we have f(x) / g(x) we get f'x*g - g'*f / g^2

i believe im correct thus far (and hope so!), but I'm not entirely sure what were to happen if we combine these things..

so if one of you kind angels out there could show me the work for the derivative of these made up functions i'd appreciate it a ton

((sin^2(2x))^3)*((sin^4(5x))^6)

would look like this if it's hard to interpret

so in this case - is the rule still that you do f' * g + g' * f?

just that f' would be 3(sin^2(2x))^2 * 2sin(2x) * cos(2x) * 2?

also looking back at what i wrote i think it's good habit to change sin^2(2x) to sin(2x)^2 for beginners like me :D

and then my next question is - if I were to use the definition of derivatives

which i think is something like lim h-> 0 for f(x) = f(x+h) - f(x) / h

would I have to do that for each "part" and then add the mall together or should I just do it on one line?

like in this case - would I do

lim h->0 for f(x) = ((sin^2(2x+h))^3 * (sin^4(5(x+h))^6) - (sin^2(2x))^3 * (sin^4(5(x))^6)) / h

aka - put in the entire function, and whenever there's an X just change it to (x+h) ?

a lot of questions and I understand if it would take a lot of time to explain it and if no one wants to - but if one could I would be eternally grateful 🥺

if one of you end up responding please ping and ill check back

[obtuse pebble]

@latent frost Has your question been resolved?

[latent frost]

That’s not the ping i was hoping for

[latent frost]

<@&286206848099549185>

i read that i was allowed to use this ping once so I’ll try that 😶

And i'll also like to clarify that this is not a homework question, if you'd like to swap out the numbers I wouldn't mind, I'd just like some clarification as to what happens when we're looking at chain rule together with other rules as quotient rule etc

[brazen gorge]

oh man, how come no one replied

i'll help you out once i finish reading everything here lol

[latent frost]

im not sure because i dont think the question is that hard.. i just dont want to learn the wrong thing because unlearning it will be very hard :((

yeah sorry for the wall of text 😅

[brazen gorge]

one step missed

in order to grasp chain rule you really have to understand how composition of functions works

[brazen gorge]

our teacher taught us to go from inside to outside and i feel like that's easier

[latent frost]

3(sin(2x)^2)^2 * 2sin(2x) * cos(2x) * 2

is that it?

[brazen gorge]

yes

,w derivative of (sin^2(2x))^3

[latent frost]

ah yeah, just did it wrong but i think i would've done that correct

[warm shale]

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[brazen gorge]

huh

oh

yes, same thing as you have

[latent frost]

i found a calculator that I really can recommend though, https://www.derivative-calculator.net/

shows the actual steps

without having to pay for it as a student, very helpful

but there are still some things i don't really understand.. possibly because I haven't encountered it before

[brazen gorge]

okay, send one expression to differentiate so i can explain

because you wrote many

[brazen gorge]

okay

here, chain rule is not directly applicable

[latent frost]

so what they did was multiply the constants and take them out to make it easier to see the derivating

so you get 10 (x^2ln(x^4x))

[brazen gorge]

mhm

here you have two things involving x

and they are multiplied

so first you start with product rule

[latent frost]

and then it's f'*g + g'*f

right?

[brazen gorge]

yep

your f(x) = x^2 and g(x) = ln(x^(4x))

oh wait

[latent frost]

f' = 2x

[brazen gorge]

no

first, simplify

[latent frost]

but g' is where i struggle

[brazen gorge]

ln(x^4x) = 4xln(x)

[latent frost]

i know ln(x) is 1/x

[brazen gorge]

by log rules

[latent frost]

ok yeah that makes sense

[brazen gorge]

i was confused at first because x^x is far from elementary

[latent frost]

it probably is but I'm just trying to understand what happens when you have a chain rule inside of product rule or quotient rule

[brazen gorge]

so you have $x^2\cdot4x\cdot \ln(x)$

[warm shale]

artemetra

[latent frost]

and this was the "easiest" way I could come up with it

ok

[brazen gorge]

you can take out 4 again to have $x^3 \cdot \ln(x)$

[warm shale]

artemetra

[latent frost]

sorry how do I "take out"?

write 40 outside?

instead of 10?

[brazen gorge]

this

[brazen gorge]

mhm

[latent frost]

ok im with you

[brazen gorge]

now you can use product rule on that

f(x) = x^3, g(x)=ln(x)

[latent frost]

and then get 3x^2 *ln(x), + x^3 * 1/x

[brazen gorge]

mhm

simplify a little

and your are done

[latent frost]

so then you get like 4x^2*ln(x)

if i didnt do it wrong in my head

ok that makes sense

[brazen gorge]

uh

[latent frost]

if u do x^3*1/x

you get x^2 no?

[brazen gorge]

yes

[latent frost]

and then +

[brazen gorge]

!occupied

[obtuse pebble]

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).

[latent frost]

or wait you probably cant do that

nvm

i take it back

just realized it's multiplication there

[brazen gorge]

mhm

[latent frost]

so it's 3x^2 ln(x) + x^2

ok ok that feels promising

but is there an example where you have

chain rule inside of product rule?

if you understand what I mean?

[brazen gorge]

yes

[latent frost]

because that's where I think I'm not entirely sure what happens

I tried to come up with one here, but clearly you're way too good to treat it as such ;D

[brazen gorge]

let's do a simple one

$(x^2+1)^3$

[warm shale]

artemetra

[brazen gorge]

how would you go about this?

[latent frost]

3(x^2+1)^2 * 2x ?

i think?

[brazen gorge]

yes

good job

mmm let me think of another example

$\sin^2(\ln(x))$

[warm shale]

artemetra

[latent frost]

so here it's

i would rewrite it as

sin(ln(x)^2

and then we get

2sin(ln(x) * cos(ln)x * 1/x

[brazen gorge]

yes

you seem to be good at it lol

[latent frost]

XD

[brazen gorge]

idk where your confusion comes from

[latent frost]

i mean i'm good at this - but I just wanna make sure I understand what happens when you combine different rules

so what would happen if we do (x^2 + 1)^3 * sin^2(lnx)

would it be

[brazen gorge]

okay, here we have product rule first

[latent frost]

3(x^2+1)^2 * 2x * sin^2(ln(x)) + 2sin(ln(x) * cos(ln)x * 1/x * (x^2+1)^3?

[brazen gorge]

when you have an f'(x) you need to get on the rhs of some formula, you basically think of it as it's own thing and consider all of the rules all over again

[brazen gorge]

yes!

[latent frost]

in that case I think I do understand it

that was the only thing I wanted to make sure of

thank you so so so so so much

[brazen gorge]

you recursively differentiate everything until you get a form that doesn't involve any derivatives

[latent frost]

you're honestly a lifesaver

[brazen gorge]

no problem :D

[latent frost]

hope you have the best life you ever could have <33

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[timid silo]

Given a universal set 𝑈 and its three subsets 𝐴, 𝐵 and 𝐶.
Is required:

1. write the characteristic functions of the sets 𝐴, 𝐵 and 𝐶 in the form
   of binary vectors;
2. number each area of the Euler-Venn diagram with a binary
   code and indicate on each of the areas of the diagram the elements
   of the universal set that fall into this area;
3. make characteristic functions of sets from Problem No. 1
   (of its variant), write down lists of elements of these sets,
   if the set 𝑈 = {1,2,3, ... ,10}, and the sets 𝐴, 𝐵 and 𝐶 are given
   by lists:

𝐴 = {𝑥: 𝑥 −even and 𝑥 > 5}, 𝐵 = {2,3,6,7}, 𝐶 = {𝑥: 𝑥 > 6}
help pls ..

[timid silo]

Draw in the coordinate system 𝑥𝑂𝑦 the set 𝐴:
𝐴 = ((1,4) × (0,3))([2,3] × [1,2]);

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

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[graceful marten]

Hello

[graceful marten]

If I may ask why do they not do (3n-1)

[timber island]

#❓how-to-get-help

[bronze mica]

that case is covered by 3n+2

because n is somewhat arbitrary, if we pick n-1 instead we have 3(n-1) + 2 = 3n-1

or if you prefer the other way if we want to do 3n-1, then picking n+1 instead gives us 3(n+1)-1 = 3n+2

[graceful marten]

Ah ok, i think im following

If I may ask tho where did you get this +2

and -1 from

from 3(n+-1)

[bronze mica]

3(n-1) + 2 = 3n - 3 + 2 = 3n-1

and similarly for the other one

[graceful marten]

Oh nvm yes haha

got it

Thank you good sir

🗿

.close

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[timid silo]

When solving a separable differential equation

Like this one, is it recommended to keep the "P" with the "-K" or bring it with the (P-M) to the other side?

[pseudo swift]

it's not seperated if you leave the P on the other side

[timid silo]

Fair enough

[timber fox]

Have you tried doing it both ways and checking if there are any differences at all?

[timid silo]

Thanks for the help

[timid silo]

No, but I can already see the difficulties with keeping p(p-m)

That's going to be a more complicated integral

Given what the other guy says it makes sense to bring the p

[timber fox]

Oh, though you were asking about where to move the k around, if youre asking about the p then yes perry the platypus is right

[timid silo]

Thanks

.close

[obtuse pebble]

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[obtuse pebble]

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[blissful bane]

Describe the function f(z)= z^3, in the semi disk |z|<=2, and ImZ>=0

[obtuse pebble]

@blissful bane Has your question been resolved?

[blissful bane]

<@&286206848099549185>

[twin sapphire]

wdym describe?

[blissful bane]

plot in Z plane and W plane

like a conformal map ig

[obtuse pebble]

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[blissful bane]

.reopen

[obtuse pebble]

✅

[obtuse pebble]

@blissful bane Has your question been resolved?

[blissful bane]

.close

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[timid silo]

So basically I’m stuck in second question I have to prove this inequality

[timid silo]

And also in the first question why the domaine of definition is R+ ? I can’t find it

I found R

[twin sapphire]

,w plot x-arctan(x)

[warm shale]

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[timid silo]

Here is the function the domaine of definition is R?

[twin sapphire]

oh this is a 3rd root then yeah

[timid silo]

But the question say that it’s should be R+

[twin sapphire]

i think they did the same mistake as me

assuming root = needs then inside to be +

[timid silo]

no I think I get it since in our maths program we only study the restriction of Racine n ieme

N ieme square root

[twin sapphire]

nth*

[timid silo]

Idk how to say in English

[pseudo swift]

t'es au lycée ou pas ? c'est pour savoir pour la q2

[twin sapphire]

Nth square root

[timid silo]

Oui je suis au lycée

[pseudo swift]

ouais donc je vais pas te sortir des développements limités de la mort, t'as pas vu ça

alors ce que tu peux faire c'est

[timid silo]

Soustraction ?

Et encadre?

[pseudo swift]

yup

[timid silo]

Bah peut être je peux utilise le développement limite ? Car le prof a déjà parlé de ça mais vite fait

mais du coup comment on fait ?

Car la soustraction ça vas être trop long

[pseudo swift]

à gauche c'est le DL à l'ordre 3 de arctan(x), à droite c'est le DL d'ordre 5

[timid silo]

DL?

[pseudo swift]

mais bon je sais pas si c'est super utile pour l'inégalité

DL, développement limité

[timid silo]

ah

[pseudo swift]

le truc classique de lycée, c'est soustraction, puis regarde la dérivée pour minimiser/maximiser la différence en effet

[timid silo]

Comment ça j’ai compris le bail de soustraction mais dérive ?

[pseudo swift]

tu dérives, et avec la dérivée t'essaies de montrer que la différence est toujours <= 0 ou >= 0 selon le cas

[timid silo]

okay je vois

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

Bon jsp mais ca m’arrache pas

Est ce que je vais poser une fonction et la dérive ?

[pseudo swift]

oui

$$H(x) = \arctan(x) - x + \frac{x^3}{3}$$
$$L(x) = \arctan(x) - x + \frac{x^3}{3} - \frac{x^5}{5}$$

[warm shale]

aPlatypus

[pseudo swift]

H et L c'est juste pour High et Low, tu veux H >= 0 et L <= 0 pour résoudre la question

@timid silo

[timid silo]

Je vais dérive les deux fonctions ?

[pseudo swift]

yes

[timid silo]

Mais si H’(x) est positivé ça veut pas dire que la fonction est positive ?

Ça veut dire qu’elle est croissante

[pseudo swift]

oui

avec la dérivée tu peux tenter de trouver le minimum de H sur R

si ce min est >= 0, la fonction H est positive

[timid silo]

Ahh okayy j’avais compris autre chose tout à l’heure

[pseudo swift]

j'ai jamais dit que ça serait immédiat avec la dérivée lol

[timid silo]

Oui maintenant it’s makes sense

[timid silo]

Ouais voilà j’étais un peu confuse

[obtuse pebble]

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[pseudo swift]

.reopen

[obtuse pebble]

✅

[timid silo]

C’est bon j’ai trouvé merci beaucoup

[pseudo swift]

alright je ferme le channel alors

.close

[obtuse pebble]

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[west sierra]

Hello, I have a question from error analysis

[west sierra]

A=(20+-0.2)cm
Then is 2A = (40+-0.4)cm?

[brazen gorge]

yes

[west sierra]

Or is it 40 +- 0.2

[brazen gorge]

no, 0.4

[west sierra]

Because 2 is a constant?

Ohh

[brazen gorge]

there's a simple way to understand the intution behind this

[west sierra]

Can you explain that way please

Even though I can simply take that as answer

[brazen gorge]

20+-0.2 essentially means that the value lies within the interval of (20-0.2, 20+0.2)

right?

[west sierra]

Ohh

Minimum and maximum, yes

[brazen gorge]

so multiplying by 2 is like stretching that interval on a number line

[west sierra]

Yes I understand now

[brazen gorge]

np

[west sierra]

I see

Thank you

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[obtuse pebble]

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[raw plover]

Hay everyone.

Hope youre all well

Im struggling with this question (e)

[raw plover]

I got -70 as the answer

But apparently its 70 and not -70

[tranquil badger]

nah its -70

wait nvm

its 70

[raw plover]

Hi Foxfort

[tranquil badger]

hellooo

[raw plover]

Yeah I amm so confused how its 70 lool

Because you do - 2 x 20 = -40

then -3 x 10 = -30

so it = -70

[tranquil badger]

its (220)+ (410)-(420)+(710)

ugh

its

40+40-80+70

[raw plover]

oh yeahh

you genius foxfort

thank you

[tranquil badger]

no there are two negative signs being applied on the 7w so it becomes positive

[raw plover]

I was stuck on that question for hours lol

[tranquil badger]

lmao

silly mistakes

its ok

good luck

byeeeee

[raw plover]

thanks!

[obtuse pebble]

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[timid silo]

?????????? i don't even know where to begin

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

nah g

anyone 😭

[obtuse pebble]

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[hexed jewel]

help

[brazen gorge]

is it $(a,b) \in \mathbb{N}^2$ ?

[warm shale]

artemetra

[obtuse pebble]

@hexed jewel Has your question been resolved?

[hexed jewel]

N-{0}^2

@brazen gorge

[brazen gorge]

ah

[sage geode]

The author really couldn't write $a, b \in \bN^*$ could they

[warm shale]

A Lonely Bean

[brazen gorge]

so non-zero natural numebrs

[fervent cradle]

interesting question

[hexed jewel]

i think it's the same thing

[sage geode]

That's almost the point, but nvm

[brazen gorge]

it's 1/sqrt(2) right?

[hexed jewel]

yea

[brazen gorge]

in the question

okay

[hexed jewel]

also this question is freshman level

so i think you cant use advanced concepts

[fervent cradle]

maybe i have something

if i haven't messed this calc up

lemme see

[hexed jewel]

k

[obtuse pebble]

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[fervent cradle]

wait

[hexed jewel]

.reopen

[obtuse pebble]

✅

[fervent cradle]

nvm

ok this seems somewhat hard and interesting

[hexed jewel]

yea i thought at first it would be easy but there i am staring at the paper for an hour

[fervent cradle]

what subject is this

just like, random proofs?

[hexed jewel]

yea

[fervent cradle]

ouch

[hexed jewel]

it's precisely called "inequalities in R"

idk if jensen theorem would be useful

the absolute value x function is convex afterall

[tardy epoch]

Here's a recent related problem
#help-19 message

[fervent cradle]

oh bruhhhh

[hexed jewel]

tbh i still didnt get it

[obtuse pebble]

@hexed jewel Has your question been resolved?

[obtuse pebble]

@hexed jewel Has your question been resolved?

[hexed jewel]

i got it

thanks

.close

[obtuse pebble]

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[summer robin]

hi

[summer robin]

sorry man i typed first

[ornate barn]

ohh mb

[summer robin]

sorry

[ornate barn]

I'll wait 😦

[pulsar quarry]

just open a new channel?

[summer robin]

is it best to absorb the C1/2 into like a C2 or just continue with C1?

[pulsar quarry]

lol

im in a good mood

how is that rude

sorry if you think thats rude

[alpine bison]

i accept your sorry

[summer robin]

question

[obtuse pebble]

@summer robin Has your question been resolved?

[severe depot]

i don't think it really matters

[summer robin]

ok sure, i just carried on anyway i just wanted to know if there was like a standard way ppl usually do it

[severe depot]

if C1 = 4 (for example idk) then it's either (x+4/2)² if you keep C1/2 or (x+2)² if you write C1/2 as C2

[summer robin]

ye

[severe depot]

yea usually just the simplest forl to work with in the next step

form

[summer robin]

okay

thanks

.close

[obtuse pebble]

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[timid silo]

hi

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[real prism]

sending over pics, one quick second.

[echo gazelle]

what do u need help with

[real prism]

sorry, had to take a snapshot.

i just need help with problems 2 and 3, and if i can get those, then 4-6.

this teacher is a nightmare

@languid sage

.... but nobody came

[obtuse pebble]

@real prism Has your question been resolved?

[timid silo]

which problem do you need help with

[obtuse pebble]

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[real prism]

.reopen

[obtuse pebble]

✅

[real prism]

problem 2 and 3

mainly jsut those two

if i can get those ill be fine

[timid silo]

ok

for two

what does <N = <B

tell you about the triangle as a whole

(specifically QN and QB)

[real prism]

uh

[real prism]

im confused what

[timid silo]

it tells you the triangle is isosceles

[real prism]

oh

yeah

[timid silo]

so if we have a isosceles triangle

that looks like this

it tells you something about two of the sides

[real prism]

theyre congruent

[timid silo]

yea

so going back to this

[timid silo]

we know that QN = QB

correct?

[real prism]

we can induce that ebcause its isoceles yeah

its not given thoguh

here do you want a higher quality image?

[timid silo]

there is enough

sure

[timid silo]

ok so we know this

correct?

my point is that since you know N = B

[real prism]

correct.

[timid silo]

we can also say their corrosponding sides are equal

[real prism]

im following

[real prism]

using which explanation

just definition of iscoceles?

the def of isoceles says 2 sides are congruent, but we cant use that because we dont know that

[timid silo]

if two angles are the same

their corrosponding sides are congruent

since N = B

then QB is congruent with QN

since QB corrosponds with N

and QN corrosponds with B

do you see?

[real prism]

....no

im so slow

my brain is liquidating this topic why dont i get it

[timid silo]

ok

its ok

[real prism]

...its not though

i have midterms in like two weeks

a test in TWO DAYS

and i barely even get whats going on anymore

this is the only class im struggling in and i hate it

[timid silo]

have you learned this

at all?

if a = c

then their two corrosponding sides are equal

[real prism]

no.. our teacher is really weird; the way he teaches geometry is almost in a "control freak" way

he only teaches us about certain proofs and certain ways to solve

and then he'll leave something random out and we have to use those specific proofs to get there even though theres something like

[real prism]

this

right in front of us

[timid silo]

ok

well its ok

just know this is a fact

[real prism]

ok ok

[timid silo]

if two angles are =

in a triangle

then their corrosponding sides are congruent

[real prism]

ok

[timid silo]

so then would you agree based on that fact that QN = QB

[real prism]

yes

[timid silo]

ok does this make sense

the purple lines just mean the entire QN and QB

sorry if its a little confusing

[real prism]

yes

[timid silo]

ok

so now we know QN = QB

and we know QR = QU

can we say anything about RN and UB?

[real prism]

congruent

[timid silo]

exactly

perfect

heres the updated picture

[real prism]

got it

[timid silo]

ok so we want to show JV = VF

do you have any guess what to do from here

[real prism]

yes

no, haha
this is actually where i got stuck when i tried doing this on my own

[timid silo]

no problem

lemme think and make sure my steps are logical

ok i think this is how the proof is suppossed to go

look at NF

and BJ

can we say anything about those?

[real prism]

i know theyre congruent but i have no way to prove it

[timid silo]

well you know JF is equal to itself

correct?

and NF = JF + NJ
and BJ = JF + FB

we know JF is congrurent to itself
andwe know NJ is congruent to FB

so we know NF and BJ are congruent

[real prism]

ok

that makes sense

ok

[timid silo]

ok to update

So to prove VF = VJ

We either need to show that <VJF = <VFJ

or show that the three other sides in RNJV = UBFV

basically the easiest way is to do this

since we know NF = JB and RN = UB

then we know the last remaining side is equal

that is

RF = UJ

therefore all the sides are equal

so <VJF = <VFJ which are corrosponding angles to UB/RN

so VJ = VF

[obtuse pebble]

@real prism Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[red mirage]

Is the answer D

[vivid blaze]

wait one sec

[red mirage]

When I plug in -2 it equals 10/0, when I plug in 3 it's 0/0

When it approaches infinity (10/0) there's a V.A. right?

[latent walrus]

if the limit is infinity or -infinity then yeah theres a VA

ugh

[vivid blaze]

its 12

-12

,w factorise (x^3+x^2-8x-12)

[warm shale]

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Requested by reyioa

[latent walrus]

merci

[vivid blaze]

np

[latent walrus]

yeah D seems valid

[vivid blaze]

,w factorise (x^3+9x)

[warm shale]

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Requested by reyioa

[latent walrus]

its -9x

[vivid blaze]

oh

then it would be x(x^2-9x)

[red mirage]

Thank you, I have another one

[latent walrus]

x(x^2-9)

[red mirage]

Guessing its C

[latent walrus]

i believe so

[vivid blaze]

we know that A cant be the anwer

bc x can be infinity exactly

i meant g(x)

[red mirage]

I think it's C because v.a. is infinity and it says lim g(x) as x > 4 = infinity

[vivid blaze]

yes i'm thinking that too

[red mirage]

Thank you guys

My last one is

[vivid blaze]

what did you get

i got d but I dont know if that is right

[red mirage]

I don't know how to solve, I don't know if g(x) in the problem is relevant or not and I should only look at I, II and III equations

[vivid blaze]

no its C

probalay not

[red mirage]

Either C or D

[obtuse pebble]

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[delicate estuary]

Are vertical asymptotes basically the same thing as infinite discontinuity?

[obtuse pebble]

@delicate estuary Has your question been resolved?

[delicate estuary]

<@&286206848099549185>

[obtuse pebble]

@delicate estuary Has your question been resolved?

[obtuse pebble]

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[thorn mountain]

i am having trouble finding wat the angles are

[heady walrus]

Is this geometry?

[timid silo]

lol

is the guy peeing

[heady walrus]

I am not even sure what the coordinates mean?

[thorn mountain]

wait, i think i got it

[real prism]

what subset of math is this? i might need it for a project

[thorn mountain]

this is from angular kinematics in biomechanics

but i think this is basic geometry

i had to just go back a few yrs

i got 84.66 but the closest answer is d

anyone else got an answer?

[obtuse pebble]

@thorn mountain Has your question been resolved?

[thorn mountain]

.close

[obtuse pebble]

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[dark idol]

What is the y intercept

[dark idol]

It wants round 2 decimal places

[obtuse pebble]

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).

@dark idol Has your question been resolved?

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[strong barn]

how do you solve this :[

[obtuse pebble]

@strong barn Has your question been resolved?

[strong barn]

<@&286206848099549185>

[obtuse pebble]

@strong barn Has your question been resolved?

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[red mirage]

Would -3 on both sides be 5?

[timid silo]

why 5?

[red mirage]

There's no -3 that approaches the right side, only left. So I suppose it would be 5.

[timid silo]

ah

yes it would be 5

[ornate goblet]

Why do you think this?

[red mirage]

I don't see any line that approaches the right where it lands on -3, only -2.

[ornate goblet]

Both parts of those are in the function f(x)

Lemme mark it

[split dew]

i don't think any of those should be DNE

[ornate goblet]

'

[ornate goblet]

^

Both the left hand and right hand of this part of f(x) approach a number

and it looks like that number in both cases would be 5

sorry for the shoddy drawing

[ornate goblet]

Im not sure if mentally you are separating the two bisected parts of f(x) for this to happen 🤔

[red mirage]

On -3 - (left side) I see it lands on 5 but I don't see -3+

[ornate goblet]

look at that top part of the slope

between -3 and -2

as it goes from -2 to negative -3

[red mirage]

over here?

[ornate goblet]

Okay I think i was right In my assesment

Not that part

We are still looking over a little more to the left

let me drawy

This part up here

mentally you wanna be dividing both sides of the graph into two parts from the point -3

the left hand side and the right hand side

I think you are getting distracted by the fact that the graph does a big jump at -2

Even thoguh that happens, there is a small part between -3 and -2 for the slope to be continuous

Look at how that little blue segment looks as if it is approaching 5

[red mirage]

So the value on the right is 5?

[ornate goblet]

Yes

Right hand limit should be 5

[red mirage]

I put the value in the square

[ornate goblet]

Yep

See how the blue is touching that part

Now, if we were finding the right hand limit of -2, it would be 2 and the left hand limit would be 6

[red mirage]

Oh now I see

[ornate goblet]

because of that huge jump

[red mirage]

So for this all the values are equal to 5?

[ornate goblet]

hope the graph helped, i think it is easier to visualize

[ornate goblet]

If the left hand and right hand limit are the same, the limit is that number

Oh I apologiaze

[ornate goblet]

f(-3) should be 6

[red mirage]

For f(-3) the black hole lands on 6

[ornate goblet]

Indeed

The answer in order should be 5, 5, 5, 6

The left hand limit is trending towards 5

same with the right hand limit

since they are the same the limit should also be 5

but the actual value at f(-3) is where the point actually lands

And that is on 6

Does this make sense?

[red mirage]

Yes thank you, I got it right

[ornate goblet]

Nice

[red mirage]

Can you tell me my mistake on this one

[ornate goblet]

Why did you put DNE for f(-3)?

[red mirage]

I think it's undefined instead of DNE

[ornate goblet]

Why do you say that

[red mirage]

Might be undefined

[ornate goblet]

I just want to know the thought process

[red mirage]

The black hole isn't present at f(-3)

[ornate goblet]

I see where the confusion is

For f(-3) to be undefined

There has to not be a point there at all

A black hole is one example of there being a point

But also

Just anywhere that there is a line is valid

So all numbers between -6 and -2 should have valid real number solutions

[red mirage]

Are all values 4 again?

[ornate goblet]

Yes

I made the point bigger for emphasis

but if you see black at that point on a graph, there is a value there

Whereas if you did -2 for example, where it is white, that would be DNE

[red mirage]

Thank you again

[ornate goblet]

All good

Are you going to try one more?

[red mirage]

I could, but it's for horizontal/vertical asymptotes.

[ornate goblet]

Well if you have any more questions for the next little while feel free to ping 👍

[silk hill]

i believe they're all 4

[ornate goblet]

Indeed

[red mirage]

This is the problem

[red mirage]

Yeah it was

[ornate goblet]

I have tutored plenty of people on this stuff so I should be able to help with anything

[ornate goblet]

It looks like to me that there is only one asymptote at 3

And you got that part correct

[ornate goblet]

Why do you think there is an asymptote at -6 and -4?

[red mirage]

That was right. I don't know how to find horizontal asymptotes though, even off this example

I guess they have to have a straight line

[ornate goblet]

Horizontal asymptotes are going to be denoted by that red line

They will continue into infinity

You want to check for two things

What value the red line is on

So that would be -4 and -5

And also what direction the graph is trending towards that asymptote

Notice how the left part of the graph is the one thats jumping on the -4 line

That means it is trending towards -infinity

[red mirage]

It only gives a red line in a example

I looked at this problem and saw a straight line which is what the red line linked to

[ornate goblet]

Thats unfortunate

Silly by whoever made this program

[ornate goblet]

In that case you might just have to assume

Look at the leftmost and rightmost parts of the graph

and if they are trending in a straight line

Look at what part of the Y axis that is on

And that will be the asymptote

In most cases I am familiar with, asymptotes are denoted with dotted lines, because it is not easy to interpret just by looking at a small section of the graph

[red mirage]

Yeah we know that at 6 it approaches - infinity

But the horizontal asymptotes are confusing since they don't have the red lines

[ornate goblet]

I agree, and I dont blame you for that

[ornate goblet]

In this one it looks like the asymptote is at y=4 and not 6

With a vertical asymptote at x=6

maybe that is what you intended

[red mirage]

What about this line over here

[ornate goblet]

You got that correct

It approaches y = -2 as it goes to negative infinity

[ornate goblet]

Make sure you put here though that its -infinity

Because +infinity means its moving to the right side, which in this case the arrow isnt doing that

[red mirage]

So this is the answer?

[ornate goblet]

Yep

So in total 3 asymptotes

y = -2 , y = 4, x = 6

[red mirage]

Oh I only put 2 asymptotes, I'll fix it

[red mirage]

Only have one attemept left

[ornate goblet]

Alright lemme check over

The only one incorrect is that last one

Should be +4, since if you draw the line its at that point

And the limit is approaching +infinity, as that part is going to the right

So horizontal asyptote at y = 4, where limit as x -> +infinity of f(x) equals 4

[red mirage]

Oh I didn't see the 4 above, so this is the answer?

[ornate goblet]

Other way around for that last part

limit as x -> infinity of f(x) equals 4

For horizontal asymptotes, x will always -> +/- infinity, and for vertical it will always -> a real number

[red mirage]

It was correct, thank you again bro

[ornate goblet]

You got it

Is this calculus or precalculus?

[red mirage]

Calculus

[ornate goblet]

I see I see

[red mirage]

I'm a junior but take calc ab

[ornate goblet]

I took calc ab back in highschool as well

Taking it as a junior is impressive

This part is going to be really important to understand what its doing and why, so getting a good grasp of this stuff is going to pay off for the more complicated work

[red mirage]

I got this right based off what you taught me, thank you

[ornate goblet]

Awesome, looks like you got it

[red mirage]

I'll close to not take up your time, have a good night and thank you again

[ornate goblet]

Imma be up late so if you still have questions feel free to reopen and ping me

Good luck on your work

[red mirage]

If you insist I could show a differential equation one, after I finish the asymptotes

[ornate goblet]

Sure 👍

Are you sure though that its differential equations though, thats a big leap in difficulty

Im pretty sure those are taught in higher calcs

[red mirage]

I did horizontal asymptotes and graphs before but it's confusing on this website

I'm on unit 2 at the moment

I can find horizontal/vertical asymptotes off equations and piecewise functions but haven't seen one on graphs, so I needed help

[ornate goblet]

Sounds good

It is much easier on graphs

[red mirage]

I need help with c and d. For a it was

y-f(0)=6(x-0) or y-2=6x which equaled to 5. b is -4 [ -2*2] times 6 = -24.

[red mirage]

Need help with c mostly

[ornate goblet]

Gimme a second to read

Uhhhhh

I wont hold you I am confused why you are doing this in AB

[red mirage]

I think I can find a question on brainly, but I don't understand it. I was told to separate the variables and integrate. I never learned that before.

[ornate goblet]

So you know how to integrate?

[red mirage]

Nah

That's Calc BC

Maybe I can skip this question

[ornate goblet]

Yes

I may be confused about something

But you should not be able to do this without integration

[red mirage]

There's a brainly question on it I can send a link to, the explanation is confusing

[ornate goblet]

Yes

[red mirage]

https://brainly.com/question/31432819

I know how to solve a and b based off x-x1 = m(y-y1)

[ornate goblet]

interesting

Is this something you need to know how to do for your class?

[red mirage]

No we never went over this, its just a ap classroom question

[ornate goblet]

Alright that makes sense

I do not believe there is a way to do this without integration then

I would skip and talk to whoever is teaching you about it

Since I might be having a lapse in my knowledge

[red mirage]

Yeah I'll skip it. My last question is if this is a jump discontinuity

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

$\frac{2^x}{3^x - 1}$

[warm shale]

disciple

[ornate goblet]

Nice latex

[red mirage]

I'll get the full problem

[red mirage]

Thx

[ornate goblet]

Are you asking to satisfy continuity?

[red mirage]

It's asking what type at x = 0

[ornate goblet]

Alright so you are trying to satisfy for k

[red mirage]

For k I think it's 1/6

[ornate goblet]

Let me check

[red mirage]

For the first equation if you plug in -3 it equals 0

[ornate goblet]

Nice

[red mirage]

So I did -3k+1/2=0

[ornate goblet]

That is how you would solve this type of problem

[red mirage]

Yeah but I don't know about (b)

[ornate goblet]

I can only see (a)

[red mirage]

Sorry the question is b) What type of discontinuity does f have at x=0? Give a reason for your answer.

[ornate goblet]

Oh yeah lemme check

Is this assuming k?

from (a)?

[red mirage]

I would assume yeah

[ornate goblet]

Ok

Do you know the different types of discontinuity>

?

[red mirage]

Hole jump vertical

If something like 0 on both sides equals -1, but on the piecewise it says "0= -2" there would be a hole

[ornate goblet]

Yep

[red mirage]

I think if you plug in 0 in the last equation it would be 1/0, it might be a V.A. since it's infinity

[ornate goblet]

Yes

So

You plug in 0 for the k equation

(no matter what k is it will be 1/2)

And you are correct that it would be 1/0 which is invalid

And if you look at the right hand limit of that equation it approaches infinity

so this would be vertical as you said

[red mirage]

So b would be a V.A. How do I know on a piecewise function if there's a jump? If they have different values for both sides?

So for example if 0 on the left was 2 and the right was 3 there would be a jump?

[ornate goblet]

Mhm

[red mirage]

Thank you it makes sense now, I finished my entire packet now

Thank you so much bro, hope you have a goodnight

[ornate goblet]

You too, glad I could be of help

[obtuse pebble]

@red mirage Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

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[fiery sandal]

Hi!I have a question,y’all.Could i show that a set is a vector space by proving it’s a subspace every time?To me,it doesn’t make much sense,can anyone explain this to me if it’s indeed true.If it is why would we ever want to go through those 10 axioms?

[obtuse pebble]

@fiery sandal Has your question been resolved?

[dark stirrup]

How would you prove it's a subspace without first showing it was a regular vector space?

[brave bramble]

In a beginner LA course, you will always take a larger vector space (usually Rⁿ or Cⁿ) then just prove certain subsets are actually subspaces