help-49

[runic hamlet]

or wdym

[keen harbor]

Or if it will be like: "well, that vector v is some tuple that can also represent some other vector w.

v would be the true vector there.

But, [v]_B is also a vector, right.

So, basically, you have a vector [v]_B, whose components are coefficients to some linear combination with some other vector from the basis, which results in a vector.

At that point, I was thinking: It's possible that this resulting vector v represents the coefficients to a linear combination with the standard basis. In this case, that would be an infinitely recursive thing that would actually still make sense.

[runic hamlet]

you can always take a tuple and interpret it as the coordinates with respect to the standard basis, then the relevant linear combination will give the tuple back

but I am still gonna say that this is not the "correct" way to think about vectors

vectors do not have to be tuples of numbers

and tuples of numbers do not have to be coordinates

[keen harbor]

Hmm, but isn't whatever representation chosen for a given vector at some moment, can be distilled into a tuple?

This is a legit question, not a retorical.

[runic hamlet]

v could be a polynomial like 1+x-18x^2

thats not a tuple

[keen harbor]

what type of struture is that?

[runic hamlet]

of course the whole point of doing this 'coordinates with respect to basis' stuff is that after picking a basis you can have [v]_B which is a tuple, yes

a polynomial

[keen harbor]

I mean, when you solve the polynomial, what are you left with?

I'm not trying to be snarky. I'm trying to understand what you mean.

[runic hamlet]

wdym solve

this cant be the first time youve seen polynomials

[keen harbor]

no

[runic hamlet]

polynomials are their own thing

[keen harbor]

but, it's like: 1 + 2 is a polynomial, no?

[runic hamlet]

1+2=3 is a polynomial, sure

[keen harbor]

But that's just 3, which can be a 1 element tuple.

That's what I'm confused about.

[runic hamlet]

yes but you cant combine 1+x like that

its just 1+x

[keen harbor]

the more I learn about vectors, the more confused I get.

Well, thank you for the conversation. I appreciate you taking the time to help. It has helped.

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[spark nebula]

$\sum_{n=3}^{\infty}x^{n\ \ }+\ \sum_{n=3}^{\infty}\frac{x^{n\ \ }}{n-1}+\sum_{n=3}^{\infty}\frac{x^{n\ \ }}{n-2}$

[grand pond]

rulzer.

[spark nebula]

i only know how to calculate the first sum

the others no

but i found a solution but i can't understand it

and that's the solution

[midnight plank]

@spark nebula Has your question been resolved?

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[keen harbor]

Are these 2 sentences equivalent?

[dawn dagger]

the latter seems a bit weird constructed

[keen harbor]

I'm trying to say: for each vectors in W, that implies the vector is in V.

It doesn't say that to you?

[last slate]

yesnt

the full sentence would be ∀x(x∈w -> x∈v)

an abbreviation is ∀x∈w: x∈v

[dawn dagger]

fast bozo

[keen harbor]

Like this?

Was the original one wrong, or just weird?

[dawn dagger]

id say both

like try to read

for all elements v in W implies v in V reads weird

[keen harbor]

because you read it like this?
(for all elements v in W) implies v in V

[dawn dagger]

yes

unless you intended it differently, then it came off wrong

[dawn dagger]

thats the way

[keen harbor]

Is this one equivalent to bozo's?

[last slate]

i think its easier to just write w ⊆ v

[keen harbor]

it is

I have that too

[dawn dagger]

yea i think its explaining what W \subseteq V really means

[last slate]

ok this question is more about notation than vectors

A⊆B is defined as ∀x(x∈A-> x∈B)

and there are shorthands for this notation

∀x∈A: x∈B

∀x∈A, x∈B

etc

i mean you could use ∀x∈A -> x∈B if you know you mean ∀x(x∈A-> x∈B) but i dont think anyone uses that shorthand

[keen harbor]

I agree the your parenthesized one is the better one. The one I originally had is no good. I asked about it because it didn't seem right.

I was also thinking of writing it like this:
#help-49 message

[last slate]

its ok i think

[keen harbor]

So, so that's why I also asked about that one

[last slate]

as long as the notation is not confusing so just a different punctuation symbol

[keen harbor]

thank you

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[mystic marsh]

How can I solve this using sin cos tan? I know I have to use arcsin and in the denominator we get (x+3)^2 + 4^2 but what else?

[twilit jetty]

you can rewrite $(x+3)^2+4^2$ as $4^2\qty(\textstyle(\frac{x+3}4)^2+1)$

[grand pond]

mtt

[mystic marsh]

Sure, but how does that help?

[twilit jetty]

now your integral looks really close to $\int\frac1{\sqrt{\ell^2+1}},d\ell$ with that $\textstyle\frac{x+3}4$ being the exception

[grand pond]

mtt

[twilit jetty]

now you can u-sub u = (x+3)/4

in general you can u-sub any mx + b without consequences since u' will be a constant that you can create when required

so after $\int\frac14\frac1{\sqrt{(\frac{x+3}4)^2+1}},dx$, youre one free $u$-sub away from making it look like a function you know

[grand pond]

mtt

[twilit jetty]

now are you allowed to use arsinh(x)

[mystic marsh]

Oh it makes sense, we couldn't do that b4 because of the 4^2 but now we can, cool

[twilit jetty]

yep

now keep in mind $\int\frac1{\sqrt{1+x^2}},dx=\operatorname{arsinh}(x)+C$ instead of $\arcsin(x)$

[grand pond]

mtt

[mystic marsh]

What does the h tell us

[twilit jetty]

so it looks like you dont get to use the hyperbolic trig functions

that means we have to do this the school way

[high gull]

burp

[mystic marsh]

Is it hard to learn hyperbolics? I just wanna know how to do these types of questions 😔

[twilit jetty]

you couldve gotten to use one, but we gotta take a harder way around which Im figuring out

do you remember the integral ∫ sec(x) dx?

[mystic marsh]

ln |Sec(x)+tan(X)|?

[twilit jetty]

yep

[mystic marsh]

This is the answer btw if you wanna know (I still wanna know how to get to it though 😔 )

[twilit jetty]

oh that might be using a different reason entirely

let me check how they did it then

[mystic marsh]

It is trig substitution

[fallow scarab]

You can always just differentiate the answer to work backwards

[mystic marsh]

I mean it works but IDK if it's faster to differenciate aprox 2 answers or just do the answer

[fallow scarab]

Why do you care if it's fast

Are you on a time limit right now

[mystic marsh]

Nah, I am cramming

Got an exam in 2 weeks so I wanna know which are the strats

[twilit jetty]

well the strats for this one dont seem to be promisingly faster than just d/dxing two answers

[lavish venture]

(x + 3)/4 = tan u

[twilit jetty]

maybe you should remember that arsinh(x) = ln(x + √(1 + x^2)) then use that as said earlier, but the chances youll see ∫ 1/√(1 + x^2) dx more than once I think are slim

if youre looking for strats though you might as well learn arsinh

[lavish venture]

i wouldn’t recommend learning hyperbolic stuff unless you’re doing an integration bee

[mystic marsh]

Does it reach ln at the end? If it's quicker than trig substitution I would say why not

[twilit jetty]

trick's a trick

you got to ∫ 1/√(1+u^2) du, right?

[lavish venture]

it’s niche

don’t try to memorize these things

youll forget it

[twilit jetty]

youll have to weigh how likely youll see a ∫ 1/√(1+u^2) du with remembering the formula it has

thats why I dont recommend this unless you have like 3 questions that all need the same integral

[lavish venture]

just do trig sub

trig sub is so simple

[twilit jetty]

we're talking time, not effort

[lavish venture]

unless its an integration bee

you won’t have to worry about time

people learn that hyperbolic stuff for integration bees because it’s quick

no need here

[mystic marsh]

Not much if none at all tho, so I don't think its worth it

[lavish venture]

like do you even know what the hyperbolic trig functions are?

your teacher probably wouldn’t even accept them as answers

[mystic marsh]

Not right now

[lavish venture]

then don’t waste your time

[twilit jetty]

I didnt say to just leave arsinh in there

[mystic marsh]

It's a computer exam DW

[lavish venture]

computer won’t accept it either if it’s coded to accept it in a particular form

[mystic marsh]

Ok, but I didn't really get how to do trig sub, I know we have to do some sort of triangle and I get that but how do we do the x, u and theta stuff?

[lavish venture]

trig sub is so much easier brother

[mystic marsh]

OK, what if we do trig sub ig

[lavish venture]

don’t look for shortcuts

[mystic marsh]

Understood, can you teach me trig sub for dummies?

[lavish venture]

yea but he then has to memorize more shit

or know how to derive it on the spot

[twilit jetty]

I mean the guy has ∫ sec(x) dx memorized

I cant even do that

[lavish venture]

that’s a simple one

everyone knows that

[mystic marsh]

It's a MCQ, I'll probably derivate it if it seems hard enough but if trig sub is easy why not learn it too

[lavish venture]

he doesn’t even know what the hyperbolic trig functions are

[fallow scarab]

https://tutorial.math.lamar.edu/Classes/CalcII/TrigSubstitutions.aspx

Lots of examples with solutions

[twilit jetty]

if just to memorize a trick you dont need to know what the functions are for to know that ∫ 1/√(1+x^2) dx = ln(x + √(1 + x^2)) + C

its just another integral to memorize which you can if you ever see that integral a lot

[fallow scarab]

The blue table at the bottom tells you the sub knief mentioned

[twilit jetty]

in general any sort of √(1 + x^2) you generally go for replacing x with tan, since √(1 + tan^2) simplifies to sec

if you want to memorize which trig sub to use, you can go for that as a general rule

since you already have ∫ sec(x) dx memorized then you can go from there

[lavish venture]

memorizing which ones to use are simple if you remember the three pythagorean identities

1 - sin^2 x = cos^2 x
1 + tan^2 x = sec^2 x
sec^2 x - 1 = tan^2 x

the left sides mirror the situations where you use the appropriate substitution

[twilit jetty]

that handles √(1-x^2), √(1+x^2), √(x^2-1)

[lavish venture]

1 - u^2
1 + u^2
u^2 - 1

[twilit jetty]

after you use the correct trig sub then simplify, theres one last hurdle

do you know how to simplify sec(arctan(x))?

(...its got something to do with 📐)

@mystic marsh where are you at on the integral rn

[mystic marsh]

Uh I was tryna use my brain on the sec(arctan) question which I could only get to sec(1-x^2/1+x^2)

[twilit jetty]

well theres a geometry way to do it

[mystic marsh]

I guess I'll just learn how to derivate Ln with fractions chat 😔

Seems like the fastest and easiest way to do MCQ

[lavish venture]

how many days until the exam

[mystic marsh]

2 weeks

[lavish venture]

you’re chilling

[twilit jetty]

you still have time to just learn the geometry

[lavish venture]

this doesn’t take long to learn

[twilit jetty]

I was stuck on figuring out stuff like sec(arctan(x)) too until I heard what the trick is

do you want to hear it

[mystic marsh]

What is it

[lavish venture]

y = arctanx

x = tan y

[lavish venture]

draw a triangle satisfying this

[mystic marsh]

45 angle?

[lavish venture]

🤔

acute angles are y and pi/2 - y

but you only care about y

[mystic marsh]

So y=pi/2?

[lavish venture]

🤔

y = y

= arctanx

[mystic marsh]

Uh y=0=x?

[lavish venture]

draw a right triangle with an acute angle of y

such that tan(y) = x/1

use pythagorean theorem to find the hypotenuse

then write sec(y) in terms of x

and you’ve just solved for sec(arctanx)

[mystic marsh]

x/1? Did you meant 1/x?

[lavish venture]

no

i put /1 to emphasize opp/adj

[mystic marsh]

sec (y)= x /sen (y)?

[lavish venture]

in terms of x

[mystic marsh]

I suck at this challenge 😔 I don't even know what to put on the other sides of the triangle

[lavish venture]

@mystic marsh

[mystic marsh]

sec y = square root of x^2+1?

[lavish venture]

yes

[mystic marsh]

Oh you meant tan (x) = co/ca 😭 I am so pro at reading comprehension

[lavish venture]

whatever co/ca means

opposite/adjacent

[mystic marsh]

Opposite side/Closer side

Oh yeah that

Now with this new found power of sec y = (x^2+1)^1/2 what do we do?

[fallow scarab]

This explains the sub and even pictures of the triangles

[fallow scarab]

.

Should spend 10 minutes reading a worked example and be done

[lavish venture]

people aren’t that resourceful riemann

even if you feed it to then on a silver plate

😭

[mystic marsh]

I am tryna process this chat it's not that I don't read it 😔

Chat imma go to sleep I'll check the link before asking again dw, thanks for the help and have a great day

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[blazing moat]

idk what to do? do I js plug in (0, 0) ?

[gaunt nimbus]

ye

but why is ur slope

-sinx

not -sink

[blazing moat]

huh?

i js differentiated y=cosx

[gaunt nimbus]

ye

[blazing moat]

is that not right

cuz it says tangent

why k?

[gaunt nimbus]

tangent at point k

[blazing moat]

oh it should work like that

i got ksink + cosk = 0 for (0,0) do i put this into desmos to get the k?

[brisk widget]

yea that should work

[blazing moat]

i got the right answer

thanks!

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[blazing moat]

wait are you the same guy?

[brisk widget]

probably, from ap server?

[blazing moat]

yea

[blazing moat]

bruh are you in college or hs

[brisk widget]

hs

the undergrad role is for what math you do, not for actual status lol

[blazing moat]

i thought youre in college 😭

[brisk widget]

oh naw 💀

[blazing moat]

which grade?

[brisk widget]

11

[blazing moat]

same

[brisk widget]

🔥

[astral talon]

. (i would like to suggest that since it IS a multiple choice you COULD have just sketched a graph and estimated. it's clearly above 2.5 even on a sketch)

[brisk widget]

tru

[blazing moat]

yea couldve js done plug n chug

[midnight plank]

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[fringe quiver]

What is the difference between taking the first derivative and drawing the sign table AND doing the second derivative test?

[midnight plank]

@fringe quiver Has your question been resolved?

[fallen vigil]

The both allow to characterize a critical point. The first derivative requires you to take a 1st derivate, find roots, then evaluate multiple points. The second derivative test required you to take a 2nd derivative then evaluate a point

[fringe quiver]

Finding f'(x)=0 and then looking for sign changes gives the min and max, but finding f'(x)=0 and plugging into f''(x) also gives the min and max

Is there any reason why to choose one over the other for min and max?

[fallen vigil]

The 1st derivative test works when the 2nd derivative may not exist. The second derivative test is usually easier to use though since you don’t need to find the roots of the first derivative

Most of the time it will just be whichever you find more convenient

[fringe quiver]

Alright

[fringe quiver]

Don't we take the roots of the first derivative and plug into the second derivative?

[fallen vigil]

Yeah sort of

In order to find critical points you usually take the 1st derivate and find its roots. It is possible though that you may find a critical point by other means.

For the 2nd derivative test, you just need to know where the critical point you care about is

For the 1st derivative test you need to know about the critical points “near” the root you are trying to characterize

[astral talon]

(basically for a second derivative = 0, you usually have to take first derivatives around the critical point to determine nature)

there are also other methods

[fringe quiver]

I got it now, thanks a lot

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[glossy oak]

Question 28 please

[cursive swan]

Find angle TPR, and from it subtract angle SPR, which is half of angle QPR because PS bisects angle QPR.

[glossy oak]

82/2

[sage galleon]

41

[glossy oak]

so 41

76-41

[sage galleon]

idk tbh not good in shapes and geo

[glacial lodge]

@glossy oak i think I got it

TPS is 32

@glossy oak do you have the solution?

[glossy oak]

nope

[vocal raft]

16°

?

[glossy oak]

yes

[glacial lodge]

how did you?

make that

[vocal raft]

Bro it's not that hard

Just use angle sum property

[glossy oak]

My mind not working tho

I have been doing these sums

For like 3 hours now

[vocal raft]

On triangle PQR to get angle P.

Then use the fact that PS bisects angle P.

Then use angle sum property on PQT to get angle QPT.

So angle TPS = QPS-QPT

[vocal raft]

Which angle you marked 82?

[glossy oak]

Whole

[vocal raft]

Yea then good

[vocal raft]

You do this and tell if you still have problem

[glacial lodge]

give me a minute to remake

this

please

i show you

[glossy oak]

I got it

but I did it kinda differently

I did 25+x=57-x

[glacial lodge]

give me just a minute @glossy oak

@glossy oak @vocal raft
i think is this

TPS is 24º

[vocal raft]

Bro is upto nothing

[glacial lodge]

is this right

i think so

it all looks right

[vocal raft]

Given that PS bisects the angle QPR

Which means QPS = SPR

[glacial lodge]

and that happens

[vocal raft]

Your figure violates that

25+24≠33

[glacial lodge]

oh

with that logic its 8º

because 33-25=8

[vocal raft]

You didn't need to draw that extra perpendicular line

[glacial lodge]

but 90+66+8 is not equal to 180

[glossy oak]

Tf

8th grade question bro

Chill out

[glacial lodge]

to calculate PSR

[glacial lodge]

dont you have the solution from the book?

[glossy oak]

U can do that without perpendicular

[glossy oak]

I only have answer

Sorry

[glacial lodge]

but from the book or your

[last slate]

I am more failure

[vocal raft]

Idk how you doing but let me tell u
Angles P+Q+R=180
65+P+33=180
P=180-98=82

Given PS bisects angle P so,
QPS=SPR=P/2 = 82/2 = 41°

In triangle PRS,
P+R+S=180
41+33+S=180
S=180-74=106

So angle PSR=106°

What are you doing with that much wrong calculation you should start practicing

@glossy oak you can close the channel now?

[glacial lodge]

ok, I understand it now

I was kinda confused

[vocal raft]

Lol

You got caught in bro's diagram trap

🪤

[glacial lodge]

fr

your are in which grade?

[vocal raft]

12th

[glacial lodge]

damn, I am in 9th almost 10th, but my school is paid, it's a college, makes easier things

[vocal raft]

good luck bro

[glacial lodge]

but I would like it to be harder I get easily a 98 in the global test

[vocal raft]

We should leave chatting in help channel bye

[glacial lodge]

i really love math, and complex things

[glacial lodge]

yeah

@glossy oak please close the channel

TPS is 16º

[midnight plank]

@glossy oak Has your question been resolved?

[midnight plank]

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[scenic nest]

How do you solve this?

[scenic nest]

This is what im getting, where am i going wrong?

[meager wraith]

Are u required to find f(x)?

[scenic nest]

yeah

well like y(x) but same thing

equation in terms of y

huh

[meager wraith]

[scenic nest]

what do you mean take dx to the right

gng 🥀

[woeful turret]

solve what? what do u have to find?

[scenic nest]

y

everything in terms of x

gang 🙏

[meager wraith]

U can treat e^-2c as some constant A. so e^2(x-c) can be written like Ae^2x

[scenic nest]

even then that constant cant be equal to -1

right?

can it

well if you look at the answer i guess that is the only way you get the answer

cooked piece of math

[meager wraith]

Don't need to. At the end, its only some constant

[scenic nest]

fair enough

thanks

[meager wraith]

constants in integration can treated as flexible unlike variables. like -c can be another constant C and so on.

[scenic nest]

fake constant 🥀

makes sense

ty

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[twilit field]

Quick question, every element of an ordered field is vacously an upper bound of $\varnothing$ right

[grand pond]

What a wonderful world !

[lyric charm]

that is indeed so.

[twilit field]

Cool, that's it.

Thanks!

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[raw vector]

Given circle (O), diameter AB = 10 cm. C ∈ O (AC<AB). CH ⊥ AB (H ∈ AB). HI ⊥ BC (I ∈ BC). Calculate HI² + OI²

[raw vector]

How do I do this?

[tawdry laurel]

lemme see

[raw vector]

Seems basic but i feel like i'm missing something

Cuz i don't see the way

[tawdry laurel]

this the diagram right

[raw vector]

Yes

[tawdry laurel]

||coord bash||

thats barbaric tho

OI is the ugly length here

[raw vector]

There's gotta be an easier way

I don't wanna use coordinates though i know that would work

[tawdry laurel]

ya

I feel that use 直角相似

like similarity

[raw vector]

I don't know Chinese

[tawdry laurel]

mb

i think it is right angle similarity

[raw vector]

Where's OI from though?

[tawdry laurel]

like ACB similar to CHB similar to HIB

[raw vector]

HI² = IB . IC but that also doesn't help

[tawdry laurel]

hm

[tawdry laurel]

its HI^2 but okay

[raw vector]

Edited it

There's no similar triangles with OI

Also it's not easier to calculate IB . IC, so...

[tawdry laurel]

trig might work

wait

[midnight plank]

@raw vector Has your question been resolved?

[tawdry laurel]

ehh this question a bit weird tbh i would use coord

Sry i cant rlly find a nice sol

[raw vector]

Eh, i guess i'll have to overkill it with coordiantions

.close

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[viral dagger]

3^3 is from the 3 domain each having 3 possivilities in the range right?

[lyric charm]

yes

[viral dagger]

actually nvm im just dumb

i wanted to ask why doesent it also rule out a straight vertical line but i just realized its not a function and its not counted at the start anyways

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[exotic stratus]

!status

[midnight plank]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[cursive swan]

I don't understand the second sentence.

m_1 and m_2 aren't themselves in the equation p/q + q/p = 2/3

[exotic stratus]

Yea

[tough shale]

It means that p and q follow that condition

And there exists 2 values of m such that it satisfies

They r m1 and m2

[cursive swan]

Shouldn't it be, "Let m_1 and m_2 be two values of m so that p/q + q/p = 2/3 is satisfied."?

[tough shale]

Meh

[exotic stratus]

m_1 and m_2 are values of m, which when either of them are substituted in the original equation, you get p and q as the roots

[tough shale]

Doesn't matter

[cursive swan]

Matters.

It should be in an order.

[tough shale]

Then blame indian english for it

[cursive swan]

This is just incorrect.

[tough shale]

Technically, p and q r in terms of m

[cursive swan]

That's the way to solve it.

[tough shale]

So m does indirectly exist in it

So m does have something to do with that equation

[exotic stratus]

It's literally just that, except it has 3 words less.

and one word in a different form

[lost sphinx]

they in the same system of equations anyways

you know the flaw, you know the intended meaning, can you still solve it?

[vocal raft]

Find roots of equation using quadratic formula, in terms of m.
You will get p and q as roots in terms of m.

Then put the values in
p/q + q/p = 2/3
All are in terms of m so it will result is something like
m=... or m=...
Take them as m1 and m2.

Then substitute in (1/11)[m1/m2² + m2/m1²).

[exotic stratus]

You don't need to do that, but that would work

[cursive swan]

p/q + q/p is symmetric, so it can be expressed easily in terms of the sum of p and q and the product of p and q.

[exotic stratus]

If you know that, then why're you not solving the question?

[slow thorn]

whats the current issue then

since you understand the question now

i think the word satisfying kind of makes sense

but in an obscure sense

like 'to make sure'

[cursive swan]

It's like working like a robot. To take every info without any order and finding that the info is enough at a certain point of time in the "process" of the question-solving, "solving" it.

[slow thorn]

damn no need to get emotional over grammar

[lost sphinx]

i dont get what youre saying but to me youre the robot for demanding proper grammar before you solve the problem

thats only needed in proofs and professional settings, youre just solving classwork

[slow thorn]

anyways you can see that p and q are directly affected by m

[cursive swan]

i agree firmly with the latter

Thanks y'all.

[cursive swan]

yes

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[slow thorn]

internet tweaking

[tough shale]

You're demanding proper english grammer in a country whose native isn't english

They will be some inconsistencies

[midnight plank]

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[blissful ether]

how do i input a system of equation in wolfram alpha?

[grim vector]

With , between it iirc

[blissful ether]

,w a+b=54, a-b=3

[grand pond]

Results provided by WolframAlpha

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

Requested by masteroogway0599

[blissful ether]

thank you

.close

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[placid patio]

Hey can anyone help me with algebra ¿

[vast swan]

Yes

[lyric charm]

yes, but:

!occupied

[midnight plank]

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).

[midnight plank]

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[keen axle]

hi this is a simple question but i dont rlly understand the difference between a power law function f(x)= ax^n and an exponential function f(x)= ab^x. if someone could explain to me the concepts in a simple way as well as provide examples that would be rlly helpful 😭😭😭😭

[lyric charm]

graph y=x^2 and y=2^x and see if you can spot the difference

[sharp coral]

power law: x is the base
exponential: x is the exponent

you can go on desmos if you like and plot both of those with sliders for a and b, see how each of them looks

[keen axle]

yes im aware that theyre different like on a graph and the differences between them as equations but i dont rlly understand how you would apply them in real life i guess? the exponential one i kinda get like it could for ezample be used to describe a salary percent increase per year but the other one i dont get rlly

[night hawk]

Quadratics can be used to optimize functions.

[keen axle]

oh okay

i meant more of like an irl example

just so i have a clearer understandinh

cause that usually helps me

ykw nvm

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[mental hornet]

What do I do to get x

[midnight plank]

@mental hornet Has your question been resolved?

[dusty portal]

Cosine law

[midnight plank]

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[dusty portal]

.reopen

[midnight plank]

✅

@mental hornet Has your question been resolved?

[mental hornet]

Cah

Cosine

=

A/h

[keen crown]

not that cosine law, the one that says $$c^2 = a^2 + b^2 - 2 a b \cos{\theta}$$

[grand pond]

vhj

[mental hornet]

Y use theta

I use c

X

$$c^2 = a^2 + b^2 - 2 a b \cos{\x}$$

[keen crown]

same thing, just different variable names

wait no

in the cosine should be an angle, but x in your figure is a length

[mental hornet]

I need to find the length between the tips of jaws

What if I put a line down the middle

[keen crown]

i don't think that'll help

in the cosine law i gave you, use c = unknown side, a, b = known sides, θ = angle opposite unknown side

nvm line down the middle might be easier

[mental hornet]

Can you explain why we'd put a line down middle

So I know

[keen crown]

since the known sides are the same length, the line down the middle is both perpendicular to the unknown side and bisects the known angle

perpendicular -> right angle -> trig is simple

[mental hornet]

Now I do

Cah

?

[keen crown]

here you do Soh, not Cah

[mental hornet]

Soh

Sine

[keen crown]

yes

[mental hornet]

What is adjacent

The line i drew?

[keen crown]

yes

adjacent = between right angle and angle used for trig

[mental hornet]

How long is it

[keen crown]

what? the line you added is the adjacent side. either half of the unknown side is the opposite side. do trig

you don't have to know how long the adjacent side is

[dusty portal]

No cosine law?

[mental hornet]

Opposite

Oh nvm

U forgot I had the angle

90

[mental hornet]

No

Hmm

Somethings wrong

Definitely weong

[keen crown]

where did 90 come from?

[dusty portal]

??

[mental hornet]

Hmm

[dusty portal]

What are you doing

[mental hornet]

What am I doing

[dusty portal]

Are you just bashing numbers into a formula and praying for a sensible output

[mental hornet]

I thought i needed

The 90

[keen crown]

you need it to be present in the shape, but you don't use it in the equation

[dusty portal]

Okay, what are you talking about

In what context

[mental hornet]

What do I do with soh

[keen crown]

use sin(bisected angle at bottom) = (length of opposite side i.e. half of top edge) / (length of hypotenuse i.e. known side to the left or right)

sine, opposite, hypotenuse = soh

[mental hornet]

I don't have the top edge

[keen crown]

so you solve the equation to get it

you have the other variables in the equation

[mental hornet]

Sin27.8=opposite/75

U said opposite is half of top edge

I don't have the top edge

How do I half it

[keen crown]

solve the equation for opposite

[mental hornet]

Gimme a sec

No idea what im doing

I have 2 angles and a side

[keen crown]

you should be using one angle and one side to find another side

show what you tried

[mental hornet]

So I do sinx=27.8/75?

[keen crown]

no, you start from sin(27.8) = (x/2) / 75

sin(x) doesn't make sense. x is a length. sin takes an angle as input

[mental hornet]

Now sin=68.5/75?

[keen crown]

number's a bit off but method looks right

[mental hornet]

Huh

[midnight plank]

@mental hornet Has your question been resolved?

[midnight plank]

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[midnight plank]

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[hollow fractal]

(= lambda)
If a, b or c is zero then would your answer for "write this in Cartesian form" just omit that fraction? Also are there no mathematical consequences for that? (With the whole dividing by zero is undefined situation)

[tawdry ermine]

Yes

We can write it in Cartesian form by setting each fraction equal to lambda

So that would be?

[hollow fractal]

Yes but if a b or c is zero isn't there an issue with the fact that you can't divide by zero

[tawdry ermine]

No there's no consequence or problem at all

[hollow fractal]

Oh okay then

And it doesn't stop you from being able to find every term you need?

[tawdry ermine]

Because once we set each equation to be equal to lambda, it just makes the line parallel to the plane with the omitted variable

[hollow fractal]

OH

Right of course

Thanks!

!close

[tawdry ermine]

No worries

[hollow fractal]

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[rotund isle]

if i tan inverse both sides will it get rid of the arg ?

[rotund isle]

if so may i get a hint on what to do from there

[wind oxide]

not what I would try

[rotund isle]

should i split the arg uup instead

?

[wind oxide]

yep

arg(z1/z2) = arg(z1)-arg(z2)

[rotund isle]

i did that first but i had no idea what to do from there

coz i couldnt get rid of the args

[wind oxide]

u dont need to get rid of them

[rotund isle]

oh?

[wind oxide]

have u been taught exponential form of complex numbers

[rotund isle]

yeah

like re^itheta

[wind oxide]

yep so try using that

[lethal path]

lmao this exact q was asked recently

it's probably from some Australian syllabus

[rotund isle]

yea it is

[lethal path]

HSC? or wrong state

[rotund isle]

hsc

[polar star]

if argument of a complex number is -π/2 and the modulus is r, what would this turn into?

[rotund isle]

re^/-i(pi/2)

[polar star]

okay and how else does one express e^(-π/2)

[rotund isle]

oh wait is that like a rotation by -i or some

[polar star]

convert it to rectangular form

and yeah, it is -i

[wind oxide]

err good point i could have done that directly

[polar star]

so you get -ir

z/w = -ir

now work with that and see if u can come up with something

[rotund isle]

hmm ok lemme try

oh ok i got it thanks everyone

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[viral dagger]

given $x<y$ are prime, if the solution for $x^3+y^3+2018=30y^2-600y+3018$ is $(a,b)$, find $a$

[grand pond]

skissue.in.a.teacup

[viral dagger]

mod 6 gave me $x^3+y^3=4 \mod 6$

[grand pond]

skissue.in.a.teacup

[viral dagger]

which means $x\equiv y\equiv -1 \mod 6$

[grand pond]

skissue.in.a.teacup

[viral dagger]

and i dont know how to continue it

oh wait actually nvm

[peak walrus]

Oops

[viral dagger]

i forgot to check the fact x can be 3 (and it works)

.solved

[midnight plank]

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[molten bay]

Integration ( \frac{\sin x}{\cos(x - a)} )

[grand pond]

Andy

[obtuse basin]

!status

[midnight plank]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[molten bay]

Cos(x-a)

X-a=t

[obtuse basin]

Right, that gets you sin(t+a)/cos(t)

Now the numerator can be expanded

Can you recall how?

[midnight plank]

@molten bay Has your question been resolved?

[molten bay]

Sint.cosa+costsina/cost

tant.cosa+sina

a is constant so cosa is constant

Thanks

.close

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[midnight plank]

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[narrow forge]

Anyone know how to prove this?

[dreamy lichen]

,rccw

[dreamy lichen]

maybe start calculating all the angles

you can start with the angles in triangle RPS

express them in terms of x

thats for 27, i didnt notice the 26

[narrow forge]

Don’t worry about 26

I already did that

[dreamy lichen]

oh alr then

[dreamy lichen]

for 27, this works

just hunt down all the angles you can, even if you arent sure you need them

[narrow forge]

Alright I’ll try

Since RV bisect PRS, does PRV and VRS should be called x? Or something else

[dreamy lichen]

nope, dont call them something else

and dont introduce more variables

focus on the triangle PRS though

you know that the angle P is 4x

can you determine angles R and S?

in terms of x

use the fact that sum of angles in triangle is 180

[narrow forge]

R and S are the same

Cuz it isosceles triangle

Oh wait

Could u try to help me later

I got school now

.close

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[lament sigil]

Hello 🙂 I am struggling to understand this part of my lecture notes on Greens Theorem:

[lament sigil]

The right line/path of the rectangle on the far right uses (x + delta x), even though over that line the changing variable is y as the line is vertical?

(Similarly, the top line/path uses (y + delta y) when it is x changing along that path as it is a horizontal line)

[midnight plank]

@lament sigil Has your question been resolved?

[azure oracle]

it doesnt matter?

[lament sigil]

Hey 🙂 Doesn't it? 😮

Fair enough haha

[azure oracle]

like, who cares what label you give the axes

[lament sigil]

Someone who isn't good at math >.< (me) haha

[azure oracle]

no prob at all

[lament sigil]

So I can just change the delta y on the right line to delta x?

To make v(x + delta x)*delta x ?

Or v(y +delta y)* delta x ?

[azure oracle]

well my brain doesnt fnction on this time of day, but for these stuff you don't need to look at the symbols, just say oh good rectangle go there deltax go there deltay and go back

[lament sigil]

Yeah fair, its late lol

[azure oracle]

yea

these diagrams help you understand

so if you get the general gist your fine

[lament sigil]

I wish I was more relaxed like u, then I wouldn't stress all the time about math 😛

tyty

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[forest shoal]

is this correct?

[forest shoal]

isnt this column supposed to be -e^-2t

[blissful pier]

a_12 entry should be

yeah there we go

to me that appears correct now :)

,w inv({{2e^t,e^(2t)},{e^t,e^(2t)}})

[grand pond]

Results provided by WolframAlpha

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Requested by xxmrfancyu2xx

[forest shoal]

okay appreciate it

.close

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[rancid vigil]

The last one doesn't make sens

[rancid vigil]

,, \text{it should be} \ \pi \cdot r

[grand pond]

<rajel />

[cursive swan]

It makes.

[rancid vigil]

That's the circumference of a half circle

[gaunt jetty]

Remember the straight line at the bottom that you'd have if you sliced up the circle

[rancid vigil]

Huh ?

If I'm not wrong , perimeter is just circumference

[wary mulch]

The bottom of the semicircle isn't "open"

[gaunt jetty]

Well yes, but you're only considering the arc of the semicircle in what you wrote

There is the line segment of the semicircle that you're forgetting to consider

[rancid vigil]

Do you have any graph or smth

[gaunt jetty]

Try to draw a semicircle first

Colour the arc of the semicircle as well

[rancid vigil]

Oh I see now

[gaunt jetty]

Nice

[rancid vigil]

I was trying to imagine a line in the bottom but in the entire circle lol

[gaunt jetty]

Ah no

[rancid vigil]

Now I see it in the semi circle

[gaunt jetty]

Yeah

[rancid vigil]

Thx

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[odd lagoon]

Which one goes to infinity faster out of ln(x^x) and x^4?

[odd lagoon]

specifically positive infinity

or do they get there simultaneously?

[lyric charm]

ln(x^x) = x ln(x)

[odd lagoon]

right

[lyric charm]

ln(x) is weaker than any power function

[odd lagoon]

right

and a product of functions

doesn't help them get there faster?

only the fastest factor matters?

[lyric charm]

well x ln(x) is thus weaker than x*x

[odd lagoon]

but what about x^4?

[lyric charm]

who do you think is stronger at infinity: x^4 or x^2?

[odd lagoon]

x^4

[lyric charm]

x ln(x) << x^2 << x^4

[odd lagoon]

okay wait i think i get it

lnx is slower than just x too

and x^4 is just x times itself 4 times

x times a slower function than x can't keep up with x times itself 4 times

that makes sense

thanks Ann

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[umbral sparrow]

am i right with C? or is it B here

[barren venture]

do you remember what 1-cos x is?

If yes just put it in the second option and check whether it gives tan x/2 or not

[umbral sparrow]

i did

i think B is right now

[barren venture]

yup

[strong sun]

yo guys pppllllllssssssss dm me help me to understand powers must talk frensh but its ok if its english

[umbral sparrow]

tysm

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[umbral sparrow]

am i right with A here?

[midnight plank]

Please don't occupy multiple help channels.

[grim vector]

close the old one

argP = -2pi/3 and d(P) = 6

so P = (6*cos(-2pi/3), 6* sin(-2pi/3) )

seems good

yep

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@umbral sparrow Has your question been resolved?

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[pliant sorrel]

i already solved part i

=1

but im stuck on part ii

[midnight plank]

@pliant sorrel Has your question been resolved?

[fresh sparrow]

notice that when you find the derivative of s_n(x), it is a sum of 1+1+1+...+1+1+(-1)+(-1)+...+(-1)

where the sign changes from 1 to -1 depends on your x

[fresh sparrow]

5/2 isn't a random value in task 1: in fact, it's above the sign change which happens at x=2

s_3 '(x) > 0 when x > 2
s_3 '(x) < 0 when x < 2

the minimum is at 2

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[split oasis]

how to do this

[split oasis]

i mean how to do b

i don’t understand the wording of the question

i tried just solving it using the arranged dot product way

[hard pewter]

show what you tried

[split oasis]

ok one sec

when it says with each of the coordinate axes

are u gonna do

xz, xy, yz?

assume for each, that one is 0 while the rest are the numbers they are

find magnitude based on that?

so you’d have three answers?

[hard pewter]

yes three different answers

three coordinate axes represent three different vectors

you need the angle made by b with each of those 3 vectors

[split oasis]

can you show me how to do it

so for yz vector, you’d do the magnitude formula for [0,-1,-2]?

and then you’d do the same for the other two, by putting 0 for y, and 0 for z?

[hard pewter]

wdym by the yz vector

we have the vector along x axis, i
along y axis, j
along z axis, k

so to get the angle between p and i

you would take the dot product of p and i and divide it by their magnitudes

[split oasis]

can u show me

i can’t visualize what you’re saying

[hard pewter]

i am just telling you to apply this

where the a vector is
in one case, i
in one case, j
and in one case k

[hard pewter]

and since you had written this yourself i assume you know how it works

[split oasis]

ok thank you

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[wraith creek]

Hello

[wraith creek]

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[dusty portal]

Did you have a question?

[midnight plank]

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[dry acorn]

Slightly confused on where to start on this, I'm sure it's stupidly easy but I'm just not seeing it

[dusty portal]

Parallel lines have equal slope

Set up an equation relating the slopes of each lines in terms of p

[modern sapphire]

Or even better, use similar triangles

[dry acorn]

Ah I see thank you guys

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[wet meteor]

Why doesn't this work?

Solution would be 5,485

[lyric charm]

you need to find h?

[wet meteor]

Yes

[lyric charm]

who crossed this out?

[wet meteor]

Me

[lyric charm]

and why so

how did you get b = 6.1?

[wet meteor]

To show u how I calculated

[wet meteor]

Yes

[lyric charm]

that wasn't a yes/no question...

also i can see the calculations -- why did you cross them out?

[wet meteor]

Bc they somehow arent right

But so u can see what I did

[lyric charm]

i am not getting 5.485 here

can you show the original problem and the answer key that says 5.485

my value of b also matches yours

[wet meteor]

It's german

[lyric charm]

ok, send it anyway

[wet meteor]

Nr 8

[lyric charm]

beta is 55.6°

actually i think you just took alpha and beta from #9 by accident

or no

just beta

anyway thats your problem

beta is not 24.3 but 55.6

[wet meteor]

Lol now i see

Thx

[wet meteor]

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[wary ferry]

i dont know where i went wrong

[lavish venture]

2^2 ≠ 2

[lavish venture]

0 * 2 - 1 =-1

[wary ferry]

Oops

somehting is still wrong

[lavish venture]

yea it said 2 wrong i didn’t check beyond the first two

[wary ferry]

ok

i did simple math wrong

again

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[woeful turret]

f(x) is a function whose domain is [-5,7]. Let g(x) = |2x+5| then domain of fog(x) is?

[woeful turret]

wont domain of f(x) be range of g(x)?

[dawn dagger]

you have to make sure the range of g lies in [-5,7]

[woeful turret]

how? how can g(x) be negative

[dawn dagger]

it can't but -5 <= g(x) <= 7 implies 0 <= g(x) <= 7 to consider, so one thing less to worry

[woeful turret]

right ao |2(7)+5| = 19

no wait

oh

shit

one second

[woeful turret]

my bad bro i was substituting 7 as x for some reason

thanks

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[woeful turret]

y=x^6-3x^5+8x^3+5
how to tell if this is onto or not?
f:R--->R

[prime hornet]

think about the limit as x tends to +- infinity

[woeful turret]

it will go to infinity

[prime hornet]

mhm, for both limits, x tends to +infinity

so there's a global minimum

[woeful turret]

ohh ok

[prime hornet]

you can find it by taking the derivative, locating all the extremum points, and finding which is the smallest

[woeful turret]

how do we find this?

[prime hornet]

but the point is, there exists one

so the function cannot be surjective

[woeful turret]

i took the derivative but that is a 5 degree eqn

how to find extremum from that?

[astral canyon]

Do you know the definition of a surjective function? At least in layman’s terms?