help-42

[hidden elbow]

so $\frac{(x+2)(x+3) + (x+1)(x+3) + (x+1)(x+2)}{(x+1)(x+2)(x+3)}$

[potent lotus]

Springsskateboard

[hidden elbow]

nice

[upper sparrow]

Now expand that out

[hidden elbow]

now wolfy simplify

[upper sparrow]

Or at least just get the constant term

[upper sparrow]

No, no wolfie

[hidden elbow]

constant term would be 11?

[upper sparrow]

Wolfie won’t save you this time

[hidden elbow]

LOL

wolfy bailed on us

[upper sparrow]

How did you get that, explain!

[hidden elbow]

multiplied and added

(2)(3) + (1)(3)+ (1)(2)

[upper sparrow]

Uh-huh, what do you notice about that sum?

[hidden elbow]

increasing?

[upper sparrow]

Well, that’s not what I wanted you to spot

[hidden elbow]

LOL

uhh

[upper sparrow]

Like you see how it’s kinda like we have (1)(2)(3) in each of them

But each time, one of the numbers is missing yea?

[hidden elbow]

yeah

[upper sparrow]

So can you figure out what the general case would look like?

(So much pain without sum and product notation )

[hidden elbow]

n(n+1) + n(n-1) + (n-1)(n+1)?

[upper sparrow]

Hmmm, not quite

[hidden elbow]

fk

[upper sparrow]

Say we have to k

[hidden elbow]

let’s do it

cool looking symbols

[upper sparrow]

We kind of have like k!/1 + k!/2 + k!/3 + … + k!/k

[hidden elbow]

wait

wha

[upper sparrow]

You agree with that one?

[hidden elbow]

I thought we had (1)(2) + (2)(3) + (1)(3)

[upper sparrow]

See what I said here

[hidden elbow]

yeah

[upper sparrow]

Put k=3 in that

What do you get?

[hidden elbow]

k=3?

[hidden elbow]

into this?

6+3 +2+…2

💩

[upper sparrow]

(Basically you just take these and then replace the k with 3)

[hidden elbow]

yeah

[upper sparrow]

Don’t worry about the …

As it doesn’t go that far out

[hidden elbow]

$\frac{3!}{1} + \frac{3!}{2} + \frac{3!}{3} +…\frac{3!}{3}$

[potent lotus]

Springsskateboard

[upper sparrow]

It’s just [shit] notation to explain you could have an arbitrary amount of terms

[hidden elbow]

ah

what’s the proper notation?

[upper sparrow]

Just these

[hidden elbow]

ah yes

what would the good notation be like

[upper sparrow]

Something like $\sum_{n=1}^{k} \frac{k!}{n}$ in this case

[potent lotus]

chartbit

[hidden elbow]

ah ok

[upper sparrow]

Put in n=1 and work that out, then put in n=2 and work that out, all the way up until n=k

Then add those all up together

Does that make sense? It’s kinda like a crash intro to the notation hahah

[hidden elbow]

wait what would the general notation be for this case

since we’re not just plugging in n=1

is it just n

then they would have laid out conditions like n is in the set of all reals or smth

$\sum_{n}^{k} \frac{k!}{n}$

[potent lotus]

Springsskateboard

[hidden elbow]

where $n\in{\mathbb{Z^+}$

[potent lotus]

Springsskateboard
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[hidden elbow]

OH WAIT

NAH

the top value is the last value n can take

[upper sparrow]

Yeah that’s it

It basically is like from the bottom value to the top value (in integer steps)

[hidden elbow]

ah ok

cool stuff

so now that we have the general sum what can we do

[hidden elbow]

sending for convenience sake

[upper sparrow]

Ah, sheiittt, notation makes this so much better

[hidden elbow]

yes let’s roll with notation then

I’ll understand along the way np

[upper sparrow]

We have that $A_{k} = \sum_{n=1}^{k} \frac{k!}{n}$, so $g_{k}(x) = \frac{A_k}{k!} = \sum_{n=1}^{k} \frac{1}{n}$

[potent lotus]

chartbit

[upper sparrow]

Oh shit they’re basically telling you to prove that series diverges

https://tenor.com/view/its-peak-peak-fiction-one-piece-gif-26139262

[hidden elbow]

w

t

f

[upper sparrow]

But are you happy with that?

[hidden elbow]

wait what’s Ak

[upper sparrow]

Oh I relabelled the A’s to A_k because they depend on k

Me being naughty, but I prefer showing they depend on k tbh

[hidden elbow]

ah so the A_k is f_k?

[upper sparrow]

It’s like not constant for everything so less confusion

[hidden elbow]

not as naughty as the person who made this

[upper sparrow]

It’s the constant term for the same f_k

[hidden elbow]

wait but where did it say $A = \sum_{n=1}^{k} \frac{k!}{n}$

[potent lotus]

Springsskateboard

[upper sparrow]

A is basically the constant term of the numerator of f_k

[hidden elbow]

I thought that was f’_k

[hidden elbow]

oh

[upper sparrow]

And that sum is how we expressed what it was

[hidden elbow]

OH

ok cool stuff

We have that $A_{k} = \sum_{n=1}^{k} \frac{k!}{n}$, so $g_{k}(x) = \frac{A_k}{k!} = \sum_{n=1}^{k} \frac{1}{n}$

[potent lotus]

Springsskateboard

[hidden elbow]

so you related $f’_k(x)$ and $g_k(x)$?

[potent lotus]

Springsskateboard

[hidden elbow]

OH WAIT

nvm

I didn’t read the question LOL

ok cool so how do we show that as k increase g_k(x) —> infinity

[upper sparrow]

It happens, more than I’d like to admit

[hidden elbow]

LOL

[upper sparrow]

With what they said, I’m so tempted to just screenshot the proof I have in my notes and send it

[hidden elbow]

oooo you did this question before?

[upper sparrow]

But anyways, we turn our attention to $\sum_{n=1}^{k} \frac{1}{n} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{k}$

[hidden elbow]

wait

isn’t it until 1/k

[upper sparrow]

Well, not this exact question, but the last bit of it here I have, it’s part of series convergence/divergence

[potent lotus]

chartbit

[upper sparrow]

Sorry my bad!

[hidden elbow]

ah I see I see

aye let’s go im learning

np

wait but that sum can’t tend to infinity

[upper sparrow]

Ayo this shit surely is not in your syllabus

Anyways, we want to use this

[hidden elbow]

fk my sch math department

[upper sparrow]

It’s more that as you let k go to infinity, so you add even more terms of 1/k

[hidden elbow]

oo

OH

YEEE

I see

[upper sparrow]

So like if you add that infinitely many times, does the value of the sum go off to ♾️

[hidden elbow]

shouldn’t be

I thought

since it’s all reciprocals

oh wait

nvm

it’s the same thing added repeatedly

[upper sparrow]

https://tenor.com/view/it-does-yes-yeah-yup-jack-harlow-gif-16352674

[hidden elbow]

so eventually it goes to infinity

LOL yep

[upper sparrow]

You know what, I’m on mobile, too lazy for all the typesetting

[hidden elbow]

how do we prove it 🧐

[hidden elbow]

yea same LOL

hmmm

[upper sparrow]

https://youtu.be/5ejmgwXVSqQ

That guy can explain it for me

Basically search up “harmonic series” and a proof should come up haha

[hidden elbow]

O ok cool

let me watch

[hidden elbow]

so this is an example of harmonic series?

[upper sparrow]

It is the harmonic series

[hidden elbow]

idek what that is 😭

time to find out

[hidden elbow]

WATCHED

so we needa bring limits into the proof?

OH

wait nah

we can just show

Sn and S2n

so confirm it’ll diverge

[upper sparrow]

Pretty much the question they're asking is a limit question in disguise

[hidden elbow]

bruh

wait

so how would I prove it

show Sn and S2n?

[upper sparrow]

Basically yep, you'd do it in the way that they've done it in the video really!

[hidden elbow]

but how do I show the part where it tends to infinity regardless of x

[upper sparrow]

That's why they gave you the hint

[upper sparrow]

Note that g_{k}(x) doesn't depend on the value of x, it only depends on k

[hidden elbow]

oh and that was shown here?

cool

[upper sparrow]

Matter of fact, they're kind of dicks for labelling it as a function of x, when it clearly doesn't depend on x

[hidden elbow]

let me write down everything and try to present it correctly lol

[upper sparrow]

Massive dickheads

[hidden elbow]

LOL

why is it even a function of x then 😭

shouldn’t it be a function of k

[upper sparrow]

Exactly lmao, it really didn't need to be because they didn't add any dependencies on x

[hidden elbow]

ah ok

[upper sparrow]

It's basically a function of both, only that it's constant with respect to x

Like for a given fixed k, it doesn't change as x changes

[hidden elbow]

bruhhhh

😭

ok lemme write down

[upper sparrow]

Ayo, whoever set these questions... even satan is looking at them like "you're going way too far though, stop, that's enough"

[hidden elbow]

should pull them down with him

[hidden elbow]

so after I find S8

do I use S_(2^n) > 1+n/2

and hence proved?

since it’ll eventually tend to infinity

[upper sparrow]

Basically, yep, this is the key statement we want, from that point we can deduce that the sum goes off to ♾️ as k get's bigger, though be careful with the labels

[hidden elbow]

ah ok

where’s my label error

[upper sparrow]

Let me dig through my notes a moment

[hidden elbow]

noiceee

[upper sparrow]

Ah found it finally, looked in the wrong .pdf for this haha

[hidden elbow]

ayeee

[upper sparrow]

Actually I think you're good, I might be fried

I'm still in shock they've basically tortured you like this

[hidden elbow]

[hidden elbow]

wth is that 2^infinity

I see

[upper sparrow]

I mean, I guess it's doable, but it surely isn't trivial

[upper sparrow]

Ah, that's 2^n, poor resolution

[hidden elbow]

ah LOL

[hidden elbow]

[upper sparrow]

Hopefully this is clearer maybe

[hidden elbow]

aye nice

[upper sparrow]

Ah yes, better

[hidden elbow]

yea

wait

the first math line

when it’s 2

how is it 1/2 + 1/3?

isn’t it 1 + 1/2?

[upper sparrow]

It is, but for the purposes of showing it diverges [or converges] you can ignore the first term[s]

[hidden elbow]

ohh Damnn

but we can still do it with 1+1/2 instead right

same thing

[upper sparrow]

As in if it converged without that 1, then it would converge with it, and similar if it diverges with it, it would diverge with it

[hidden elbow]

ah I see I see

[upper sparrow]

Yea you can include it if you wanted, just modify the argument as appropriate

[hidden elbow]

nice stuff

thanks a bunchh!

[upper sparrow]

No worries, was an interesting question! Somewhat surprised it went down the path of harmonic series in the end, but I really should have seen that coming

But hopefully all of that makes sense!

[I mean, I couldn't see myself doing this at 16 tbh ]

I think, a better way of stating this would be that you have $g_{2^{m}}(x) = \sum_{n= 1}^{2^{m}} \frac{1}{n} > \frac{m - 1}{2}$ via similar arguments as before

[potent lotus]

chartbit

[upper sparrow]

As that explicitly involves the function they stated, but it's like a tiny point hehe

And as a final point, the $\Pi$ product notation is basically the equivalent for $\sum$ but for products, so $\Pi_{m =1}^{n} a_{m} = a_{1} \times a_{2} \times \ldots \times a_{n}$

[potent lotus]

chartbit

[hidden elbow]

ahhh okok

NICE I’m learning notations

[upper sparrow]

E.g. you can represent the factorial $n! = \Pi_{m = 1}^{n} m$

[potent lotus]

chartbit

[upper sparrow]

Yep it actually makes life so much easier, that's how I instantly spotted it would be a harmonic series question

• there's less ambiguity as well

[hidden elbow]

yea HAHA the thing is we didn’t learn these notations

so idk what my teacher is smoking cuz he expects us to try these

[upper sparrow]

So e.g.
$$
\ln\left( \Pi_{n=1}^{k} (x + n)\right) = \sum_{n=1}^{k} \ln(x+n)
$$

[potent lotus]

chartbit

[upper sparrow]

Is how I'd have written the first part of the question if it were me doing it

[upper sparrow]

Whatever it is, me no wan it

[hidden elbow]

AH so

OHH okok I see see

YES

[hidden elbow]

LOL

[upper sparrow]

https://tenor.com/view/drake-nah-gif-20825086

[hidden elbow]

https://tenor.com/view/headache-stupid-people-oh-my-god-omg-oh-my-gosh-gif-22026951

[upper sparrow]

But yeah, there you go, done quite a bit of math here hope that's been all good!

[hidden elbow]

YEA it’s been amazing

gna rub this in my teachers face

he’s gonna think I cheated

technically it’s not officially an assignment

he just asked us to try

I still learnt new stuff though 🤙

[upper sparrow]

I mean, I think quite a lot of it you would have made progress on, you could have done the differentiation and common denominatoring, it's just knowing how to write that constant term [which might have taken time to spot]

The general case is a bit to visualise [hence do small examples and try to generalise from there]

But we done made it now 🥳

[hidden elbow]

we made it bros

🙌

[upper sparrow]

https://tenor.com/view/yeah-asher-half-brothers-oh-yeah-heck-yeah-gif-18780717

It always do be the small bits tho as well

But yeeee, been a good one

[hidden elbow]

YEA

[hidden elbow]

wait for this

what is m?

[upper sparrow]

It's supposed to be some random integer, but to be fair you can replace that m with any other letter of your choice [except n, as we used that as our summation variable, and that would be really bad form ]

[hidden elbow]

ahhh

so I can just use m

but I need to define it

[upper sparrow]

You can just say "for some integer m", it should be clear from the context what you're doing - like you can say "if we take the sum up to 2^m, then our sum becomes..." then clarify that m is a positive integer if you wanted

[hidden elbow]

oh so

[hidden elbow]

$g_{2^{m}}(x) = \sum_{n= 1}^{2^{m}} \frac{1}{n} > \frac{m - 1}{2}$ for some $m \in \mathbb{Z^{+}}$?

[potent lotus]

Springsskateboard

[upper sparrow]

More the other way round: let $m \in \mathbb{Z^{+}}$ be given, then $g{2^{m}}(x) = \sum_{n= 1}^{2^{m}} \frac{1}{n} > \frac{m - 1}{2}$

[potent lotus]

chartbit

[hidden elbow]

ah the order matters?

oh danggg

then I finish off with

hence proved

[upper sparrow]

That's something you'll hear about in e.g. classes in logic, the order of statements and the wording really matters

[upper sparrow]

Yep that'll do

[hidden elbow]

oh sheesh

what’s QED?

[leaden thunder]

,w define qed

useless wolfie

[upper sparrow]

https://tenor.com/view/spit-take-laugh-lmao-gif-9271200

[hidden elbow]

BRUH

[leaden thunder]

https://mathworld.wolfram.com/QED.html

[hidden elbow]

we broke wolfie

with our first question

so QED = hence proved

is that Latin

wait wtf did I ask

it literally says Latin

[upper sparrow]

They basically asked this guy to prove divergence of the harmonic series

[leaden thunder]

what a roundabout way to prove divergence of harmonic series

[upper sparrow]

Honestly the setup of the question didn't help one bit, why on earth did they need to define that function as one in x when it clearly doesn't depend on it

Swear they just want @hidden elbow to suffer

[hidden elbow]

SEND HELP

[leaden thunder]

indeed

[hidden elbow]

wait I asked my friend and she said it’s not conclusive cuz I only did it for 3 cases?

eh oops wrong ping

@upper sparrow

[upper sparrow]

Hmmm, as in you'd show it in general what it would look like, as per here, that should be enough really I would think

[hidden elbow]

ah I see

so n=1 to n=3 is enough?

[upper sparrow]

As in you should do something like this one here too

[hidden elbow]

AH

[hidden elbow]

I see I see

[upper sparrow]

Yep, once you've shown that pattern like that, you should be good imo!

[hidden elbow]

TY!

so I just show the pattern of 2^m

[upper sparrow]

Yep

[calm coral]

@hidden elbow Has your question been resolved?

[upper sparrow]

W H A T

[hidden elbow]

LMAO

that’s why I was prepared for a comedy show

[upper sparrow]

That's just outright disrespectful there whoever did that, honestly, is an opp

[calm coral]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[hidden elbow]

.reopen

[calm coral]

✅

[hidden elbow]

wait

@upper sparrow

this look alright?

[upper sparrow]

that's what I would like to see

[hidden elbow]

NOICE

[hidden elbow]

so then after this I do 3 cases

and show the pattern

[upper sparrow]

Actually wait a moment

[hidden elbow]

uh oh

[upper sparrow]

Group them like this

[hidden elbow]

oh wait why

[upper sparrow]

As you want to replace 1/3 by 1/4, all of the 1/5, 1/6, 1/7 by 1/8 and so on so forth

[hidden elbow]

but the picture grouped them like 1/2 + 1/3 etc

[upper sparrow]

Ah yeah, they did, but got an even better way to do it now

[hidden elbow]

aye cool alr

[upper sparrow]

This will make use of the hint they gave

[hidden elbow]

wait how about the 2^3 pattern

[upper sparrow]

Same idea

[hidden elbow]

wait nvm yep isee

wait

[upper sparrow]

Actually give me a second, I'll write it on paper for you

[hidden elbow]

1/2^n (n-1) times right

ah ok

[calm coral]

@hidden elbow Has your question been resolved?

[hidden elbow]

wait if 1/2^n n-1 times, then when n=2, shouldn’t it only be 1/4 1 time?

😭

[hidden elbow]

it’s been 3 hours on this problem ☺️

[upper sparrow]

Been quite some time! It's the wording of the last proof that's the nuts bit

[upper sparrow]

Try this one out, hopefully I've counted right

[hidden elbow]

ayeee

[hidden elbow]

wait what does the orange part say

[upper sparrow]

"there are n of these terms in total"

[hidden elbow]

ahh

[upper sparrow]

Ah shit it should be n+1 of them, fuck

Actually shit I've confused my poor brain

No it's n of them

...I think

[hidden elbow]

nw I’ve been cooked since the first question

wait btw

1/2^n repeated n-1 times right

then let’s say n=2

1/4

it should only appear once no?

why does it appear twice

[upper sparrow]

Oh yeah was gonna say, they done fuqd up there

Actually no hang on, I don't think so

No yes they did

That should be $2^{n-1}$ times

[potent lotus]

chartbit

[upper sparrow]

They wrote it wrong!

Assholes

[hidden elbow]

BRUH WHA

THE QN WRONG OMG

OMG YE 2^(n-1) times

[upper sparrow]

They have the nerve to set that shit and fuck up the quoting, kmt

[hidden elbow]

lmaoo is snow here also

@ancient thistle

wait so assuming they typed 2^(n-1) there,

I find the pattern for 2^k

wait I don’t have to show 3 cases?

[hidden elbow]

here they said using the estimate

what does that mean

[upper sparrow]

What they're saying is that they're going to bound the sum below by replacing all the terms with something smaller

So like in that one, all the terms will be less than 1/2n, so replace them all with 1/2n and you get a smaller sum

[hidden elbow]

wait so for our terms

since we’re grouping different from theirs

what would our estimate be

[upper sparrow]

We're basically here replacing all the terms with something smaller, the next 1/2^n up, so like the blue is replaced with 1/4, the purples are replaced with 1/8, the greens with 1/16 and so on

It's a similar idea to what they have, but we're trying to change our sum into one based on 1/2^n looking thingys

[hidden elbow]

ah

wait how would it look 😭

[upper sparrow]

Kinda like in this picture here!

[hidden elbow]

AH ok

[hidden elbow]

wait what does “the sum is bounded below by 1 + n/2” mean?

[upper sparrow]

Basically at the end because we replaced the OG sum with something smaller, we know the sum must be at least [bigger than] 1 + n/2

[hidden elbow]

oh we can prove that way?

[upper sparrow]

Yep, prove the sum is at least something which ends up going to infinity and you’re done!

[hidden elbow]

wait but how does it show the sun ends up going to infinity

[upper sparrow]

It's, like, trivial

[upper sparrow]

Think about it: you are at least something, and that something gets bigger, so you must get bigger too

[hidden elbow]

AH

so with that I’m done?

at last?

[upper sparrow]

Yep

[hidden elbow]

LETS GOOOO

4 HOURS

[upper sparrow]

Damn, for a whole 5 mark question

[upper sparrow]

But yep, all that all good now?

[calm coral]

@hidden elbow Has your question been resolved?

[calm coral]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[calm coral]

Available help channel!

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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

[remote mural]

Calculus methods not allowed

I.e. taking the derivative

Pretty lost on this

[somber finch]

Someone said before that it was already solved

[remote mural]

Yes I know them lol

But I don't know the answer

[velvet osprey]

how are we supposed to find tangents if not with calculus

[remote mural]

This was a question for 9th grade surprisingly haha

[velvet osprey]

how are 9th graders supposed to know what a tangent is

[remote mural]

Ask miss @ancient thistle herself haha

She administered it to her students lmfao

I have no clue though

[velvet osprey]

@ancient thistle

[ancient thistle]

[velvet osprey]

finger got pointed at you

[remote mural]

Name dropping snow

[ancient thistle]

the tangent and the conic are supposed to intersect at a quadratic root

so you solve for discriminant = 0

[remote mural]

Wait conic?

[ancient thistle]

f is a hyperbola

[remote mural]

Yeah

But a quadratic root? How in the hell

Are you supposed to see that

[ancient thistle]

if you have a linear root then the line will cut through the graph

[dreamy tangle]

so you can calculate the k for every x_0 by \delta=0

[remote mural]

Still a bit at a loss haha

I will give it a think more

[somber finch]

You mentioned loss?

[ancient thistle]

loss

[dreamy tangle]

and you can find two spots which satisfies x_1 = 1/x_2 and k_1*k_2 = 0

[graceful dust]

who let ms snow near those poor 9th graders

[somber finch]

I tried that, but the roots are not that easy

[remote mural]

She got like 5 people in vc trying to solve this

[somber finch]

$y = m\cdot\left[x-a\right]+f(a)$

[potent lotus]

sopinha

[somber finch]

$m\cdot\left[x-a\right]+f(a) = \dfrac{1}{x}+x \iff mx^2-max +f(a) \cdot x = x^2+1$

[night torrent]

she

[potent lotus]

sopinha

[remote mural]

It's meant to be a secret

[dreamy tangle]

i put that to practice and got a super long equation

your mind just keep you from doing this after you learned a advanced method

[remote mural]

I FIGURED IT OUT

THIS IS SO DUMB

I HATE NINTH GRADERS

.close

[calm coral]

Channel closed

Closed by @dawn spruce

Use .reopen if this was a mistake.

[calm coral]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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

[dull salmon]

I got an answer of f(g(x))= 9x^2 + 6x - 11, but the marking scheme says its f(g(x))= 9x^2 + 3x - 12, can someone help where i've went wrong?

[dull salmon]

wait nevermind

spotted it

.close

[calm coral]

Channel closed

Closed by @dull salmon

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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

[mint cloud]

how do i convert radians to angles/degrees

how do i find in what quadrant and what angle these are refferring to

my course completely jumped over that

[carmine sorrel]

multiply the radian by 180/pi

[mint cloud]

do i use a calculator or am i expected to do it manually

when it comes to pi

[carmine sorrel]

in the examples you have the pi will cancel out immediately

and generally, you'll be looking at examples like that

[mint cloud]

bare with me, can u walk me through converting this example

so
3pi/4 x 180/pi?

[carmine sorrel]

yes

[narrow spear]

you know that $2pir = 360$

[potent lotus]

phoestaclies

[mint cloud]

so 540pi
/ 4pi

or just 540
over 4?

[narrow spear]

$2*\pi*r = 360$ so you can have r = something

[potent lotus]

phoestaclies

[carmine sorrel]

no? 2pi(r) is the circumference of a circle. not a degree measure. unless you're using r as rad, which is very confusing.

[mint cloud]

i do not know that 2x pi x r =360. i thought 2pi equalled 360

[narrow spear]

oh no sorry i said nothing forgot

[mint cloud]

ok so 540pi/4pi

135 degrees

[carmine sorrel]

yes

[mint cloud]

ohhh

myyyy

goooodnesss

which would put it in the second quadrant

[carmine sorrel]

yes

[mint cloud]

with a theta of 45?

maybe i used that word wrong

i guess the theta is 135

whats the inner angle called

inside the triangle

[carmine sorrel]

yeah, it would be equivalent to +/-sin(45) and +/-cos(45) depending on the quadrant

[mint cloud]

whats with the +/-

i keep seeing that in trig

isnt it clear if its + or - based on the quadrant

[carmine sorrel]

because sin(x) and cos(x) change signs depending on the quadrant. For short hand +/-sin(x) just means "positive or negative sin(x)"

[mint cloud]

also last question, do i always multiply radians by 180. or does that change to 90. 270. 360 ever?

i suppose its 180 because that equals pi right

[carmine sorrel]

for example, in the first quadrant cosine is positive, in the second quadrant cosine is negative

[mint cloud]

ok yes

but when i know it lands in the second quadrant

do i still need the +/-

[carmine sorrel]

you need to decide which one to use based on the quadrant

[mint cloud]

ok wonderful i think im getting it

thank u very much for your time

[carmine sorrel]

np

[mint cloud]

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[remote mural]

so do you see what type of triangle triangle ABC is?

[ashen solstice]

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[main remnant]

How would i simplify this?

[leaden thunder]

What's f

[main remnant]

Oh right

I would just plug in i/2 in the function?

[leaden thunder]

Yes. Whatever f is

[calm coral]

@main remnant Has your question been resolved?

[calm coral]

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[viscid silo]

do you know how exponents work

[craggy oar]

i think it should 3^8

i mean im learning exponents now

its showing up as wrong though

6561 is it solved, but thats also wrong

[viscid silo]

its correct?

,w (3^-2)^(-4)

[potent lotus]

Results provided by WolframAlpha

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

Requested by DerpZ#9141

[viscid silo]

yeah its correct

ignore the fact that i just wolframalphaed something like this

[craggy oar]

my friend tried 6561

he said it didnt work either

[subtle osprey]

try commas

6,561

[prisma mango]

Post screenshot of the entire page? Maybe there are specific instructions

[craggy oar]

number 6

i just tried with comma

i havent submitted yet

im still working on 14

[subtle osprey]

,w (3^-2)^-4=6561

[potent lotus]

Results provided by WolframAlpha

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

Requested by Jash#2503

[craggy oar]

I figured it out, thanks for the help

my friend put a comma originally, and I didnt realize

it is 6561

.close

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[burnt cobalt]

quick question about propositions

[burnt cobalt]

is 5+7=13 a proposition?

it is, right? And it would be false?

[steel grail]

It is a false proposition

Because you can prove that 5 + 7 = 12

and that 12 is not equal to 13

[burnt cobalt]

got it, thank you!

.close

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[fiery torrent]

Hi

Can I get any hints please?

[sterile seal]

Drop a perp from D to BC and a perp from A to BC

[fiery torrent]

Oki

Similarity doesn't work here

@sterile seal

[sterile seal]

Let BD be x

[fiery torrent]

Let it be

And?

[sterile seal]

Write everything in terms of x

What is ous

[fiery torrent]

What do you mean ous?

Everything is written in terms of x @sterile seal

@sterile seal look what I did

I found

I did

IT

It's Square root of 2

.close

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[terse pine]

$cos\theta=\frac{1}{\sqrt{2}}$

[calm coral]

Please don't occupy multiple help channels.

[potent lotus]

Zyme><SOL

[terse pine]

yeah that's it

Need to solve for $0^{\circ}\leq\theta\leq360^{\circ}$

[mellow crater]

$1^{\circ}$

[potent lotus]

Modus

[sterile seal]

What do you need help with?

[rigid kettle]

\leq

\cos

(latex improvements)

[potent lotus]

Zyme><SOL

[rigid kettle]

What have you tried

[terse pine]

I tried to find a degree

using a handy dandy table

but none of the degrees for sin cos and tan had $\frac{1}{\sqrt{2}}$

[potent lotus]

Zyme><SOL

[rigid kettle]

its definitely a nice one

can you show us the table

[terse pine]

unless I modified $\frac{1}{\sqrt{2}}$

[potent lotus]

Zyme><SOL

[sterile seal]

If I draw a right triangle with leg 1 and hypotenuse sqrt2, what do you notice about it

[rigid kettle]

That aside, a table isn't a good method anyways

[terse pine]

I use a hand trick

but that didn't work so I resorted to a table

[rigid kettle]

The best way to remember trig angles related to 30, 45, 60 is to remember 2 triangles

[terse pine]

something like this

wait a second

[sterile seal]

look at the table

[rigid kettle]

1/rt2 is litearlly there

anyways

[terse pine]

is $\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$?

[potent lotus]

Zyme><SOL

[rigid kettle]

yes prove it

[sterile seal]

Isn’t the only method to calculate using the complex form of them

[rigid kettle]

what can you multiply by top and bottom

[terse pine]

because in mines it had sqrt2/2 as the values for that

[rigid kettle]

1/sqrt(2)

what can you multiply by top and bottom

[sterile seal]

Rationalize the denominator

[rigid kettle]

to rationalize the denominator yes

[terse pine]

sqrt 2

that's right

[rigid kettle]

yes

also, for remembering trig

https://i.imgur.com/fJcnnUg.png
https://i.imgur.com/YY1loIz.png

you only need these 2 triangles

[sterile seal]

Really?

[terse pine]

sqrt2*2 is 1?

[rigid kettle]

draw them neater on a piece of paper and you will find it does the job

yh for 30 60 45

[rigid kettle]

??? what

[sterile seal]

Oh I thought you were talking about for all triangles

[rigid kettle]

$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2}\sqrt{2}} = \frac{\sqrt{2}}{2}$

[rigid kettle]

yes?

[terse pine]

$\frac{\sqrt{2} * \sqrt{2}}{2 *\sqrt{2}}$

[potent lotus]

Zyme><SOL

[rigid kettle]

first step multiply by sqrt2 / sqrt2 = 1

[rigid kettle]

???

[rigid kettle]

no no, here

starting from 1/sqrt2

[terse pine]

bro what I'm lost

how do you change sqrt2/2 into 1/sqrt 2

divide both sides by sqrt 2?

why would that occur to you though

what's the trigger that would say yes, divide both sides by sqrt 2

[terse pine]

I'll try remembering this but I need to understand it at base value

[rigid kettle]

sure

i went the other way round though

[sterile seal]

$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2}^2}=\boxed{\frac{\sqrt{2}}{2}}$

[terse pine]

backwards would be dividing by sqrt 2 right

[potent lotus]

kappa

[sterile seal]

This might be easier to see

[terse pine]

yeah I get that

[sterile seal]

What are you having trouble with?

[terse pine]

this

[sterile seal]

It’s a common technique to make computation easier

For example, if sqrt2 is approximately 1.41, then what is 1/sqrt2?

[terse pine]

1/1.41

[sterile seal]

And that’s hard to find right

[terse pine]

yeah

[sterile seal]

But if we rationalize the denominator and find that 1/sqrt2=sqrt2/2

We can simply divide by 2 and find that 1.41/2=0.705

[terse pine]

oh alright, do you happen to know anything about transformations of the parent function sqrt x

[sterile seal]

Maybe??

[terse pine]

I posted a question in #help-32

if you could take a look at it

would be appreciated

.close

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[uneven saffron]

Hey

[uneven saffron]

Need help with a q

[swift marsh]

https://dontasktoask.com/

[uneven saffron]

"Lethna buys some boxes of pens and some packets of pencils
There are 40 pens in each box
There are 12 Pencils each box

Lethna givesone pen and one pencil to each student in Year 7 of her school

When she has done this she has no pens and no pencils left

Work our how many boxes of pens and pencils Lethna could have bought"

[vagrant oak]

Do you know least common multiple?

[uneven saffron]

idk howto find it

okay

i relearned how to find it

its 120

cause 2 x 2 x 2 x 3 x 5 = 120

What now

used prime factor decomposition to get this btw

alr

i got it

3 boxes of pens and 10 boxes of pencils

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[terse pine]

Just needed some help remembering what sides are on both the 30-60-90 triangle and the 45-45-90 triangle

[terse pine]

if there's a trick or something handy

easy to memorzie

[upper sparrow]

For the 45-45-90, you have a [right angle] triangle whose non-hypotenuse sides are equal

For the 30-60-90, take an equilateral triangle and fold it in half

[prisma mango]

Google

[rigid kettle]

https://i.imgur.com/fJcnnUg.png
https://i.imgur.com/YY1loIz.png

remember these 2 triangles @terse pine

[terse pine]

I'm looking through my lesson pages if your wondering

[rigid kettle]

the information in them is apparent

you just need to draw an equilateral and split it in half

and a right isos

[terse pine]

there was an example where I wrote down the exact side length associated with the angle

[rigid kettle]

Wait sorry misread your question

you wanted the sides???

It doesnt really matter what the sides are

[terse pine]

for example 30 60 90
1 sqrt3 2

example**

[rigid kettle]

Its the same scaled up or down

You get the same fraction but the numerator/denominator is multiplied by a constant

[terse pine]

is the 30 60 90 triangle the sine triangle or the cosine

or am I just making things up now

[prisma mango]

You use sine and cosine to solve for the sides

[terse pine]

that's the purpose of my question

[prisma mango]

It's not restricted to just one of those

[terse pine]

so sin 30 for the 30 angle?

which case would you use either

it gives different results

[prisma mango]

Do you know the mnemonic SOH CAH TOA?

[terse pine]

yup

[prisma mango]

So if you did $\sin(\theta)$ that's opp/hyp

[potent lotus]

dldh06

[prisma mango]

If theta = 30, then you are using the sides opp and hyp, if you are applying sine

If you are using cosine, that's adj/hyp

So you are using both trig functions to find opp and adj

[calm coral]

@terse pine Has your question been resolved?

[calm coral]

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[marsh canyon]

please tag me otherwise i wont see the message

nvm i did it

.close

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[toxic igloo]

i need help drawing this out

[calm coral]

@toxic igloo Has your question been resolved?

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[toxic igloo]

.reopen

[calm coral]

✅

[toxic igloo]

<@&286206848099549185>

to find the bearing, i found the angle of about 41 degrees but on the answer key it says the bearing is S27W

thats my work

never mind

i figured it out

i just had to subtract 41-14

.close

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[tribal kernel]

Hello i need help on a math problem

[tribal kernel]

i have to determine the equation of the parabola

but i dunno where to start

[calm coral]

@tribal kernel Has your question been resolved?

[tribal kernel]

@fading jacinth-role

<@&286206848099549185>

??

[tribal kernel]

<@&286206848099549185>

[calm coral]

@tribal kernel Has your question been resolved?

[calm coral]

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[wintry glacier]

whenever i write that something is true because [fancy A] is a sigma algebra

[wintry glacier]

can i just write by the sigmaness of [fancy A]

sounds cooler

[calm coral]

@wintry glacier Has your question been resolved?

[wintry glacier]

.close

[calm coral]

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[west fossil]

wasnt paying enough attention in class lmao

[strange lichen]

factor out an x

[west fossil]

1 x or 0.1x

[strange lichen]

x

[west fossil]

ok thx

wait

how do you factor out x from -0.8x

wait no nvm

i see now

whoops

[calm coral]

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[vivid idol]

Help

[vivid idol]

Need help finding the antiderivative of this

[prisma mango]

You don't need to close and open new channels if no one answers you, just wait patiently for help

[vivid idol]

Sorry.

I just have very little time.

[strange lichen]

So that’s x^(-3/2)

To find antiderivative, +1 to exponent and then multiply the whole thing by 1/(the number u got in the exponent)

[vivid idol]

does x^-3/2 go on the top

with the 1?

if ur counting it as a negative

[strange lichen]

Yes, if we change it to a negative exponent, it has to go to the numerator

[vivid idol]

OH

WAIT

2x^1/2

?

[strange lichen]

1. Use parentheses, 2. Show ur work. 3. No

[vivid idol]

oh

its -1/2

[strange lichen]

Ye in the exponent

[vivid idol]

Back to the original problem

[strange lichen]

Start from here

[vivid idol]

wdym

oh

No need for another integral notation?

[strange lichen]

Once u integrate, u don’t draw another

Also, don’t forget dx

[vivid idol]

Where do I put dx

Bc m

[strange lichen]

Just the way you’ve been doing it

[vivid idol]

Nvm

This better?

,rotate

[vivid idol]

or still something wrong

[strange lichen]

Ur good

Actually, the last line u did

The limits are from 1 to 3

Not 3 to 1

Remember we changed that alrdy

[vivid idol]

what?

We changed from 1 to 3 to 3 to 1

[strange lichen]

It’s read bottom to top

[vivid idol]

what?

[strange lichen]

The limits

[vivid idol]

the first second or third one

[strange lichen]

The original problem had 3 to 1 right?

[vivid idol]

No

It was 1 to 3

Originally

[strange lichen]

That’s not what it says on ur paper

Remember we read from bottom to top

[vivid idol]

Wdym

[strange lichen]

Bottom limit to top limit

[vivid idol]

,rotate

[vivid idol]

this is the original problem

[strange lichen]

Just read this lol

I’ve said it a few times

[vivid idol]

sorry i dont understand u

right now

[strange lichen]

The integral was from 3 to 1 originally, we read from the bottom number to the top number

[vivid idol]

ok

yep

[strange lichen]

…

[vivid idol]

but flipped it

[strange lichen]

Yes

[vivid idol]

so??/

i flipped it

[strange lichen]

Why’d u flip it back here…

[vivid idol]

ok im sorry

im just

very tired.

[strange lichen]

Ok I understand

[vivid idol]

ok i will

evaluate and send u end result

ok how do

i do

this

[strange lichen]

Don’t put +C for definite integrals btw

It doesn’t make sense

[vivid idol]

not sure how to subtract this mess.

oh nvm

i got it

multiply both by

sqrt

num/dem

den*

done

,rotate

[vivid idol]

look good?

[strange lichen]

Lemme see

Yep

Nice job

[vivid idol]

ty

can i show u some other simple problems

that i just finished?

that i asked u a while back

[strange lichen]

Can u wait like 5 mins

[vivid idol]

kk

lmk

[strange lichen]

aight sup

rdy

@vivid idol

[vivid idol]

kk

thes 3

,rotate

[strange lichen]

all 3?

[vivid idol]

mhm

pretty easy ones.

[strange lichen]

u got em all right

[vivid idol]

YAY

ok

,rotate

[strange lichen]

ye thats right

[vivid idol]

,rotate

[strange lichen]

btw if u just wanna check ur answers, search up integral calculator

[vivid idol]

last 1

and alright.

[strange lichen]

yea thats right

i only checked the answers btw

[vivid idol]

ye

im done btwbut like

do u wanna help on one more problem

before i head out

[strange lichen]

ok

[vivid idol]

u dont got to

a bit hard

[strange lichen]

nah its fine

[strange lichen]

ok

first flip the limits of the second integral

anyways

think abt it like this

the area under the curve from 1 to 7 is -5

from 4 to 7 its -7

[vivid idol]

should we combine?

and find 4 to 1?

given 7 and 1

but flip 4/7

[strange lichen]

not sure what u mean

[vivid idol]

me neither

im lost

[strange lichen]

actually lets do this

[vivid idol]

we should

[strange lichen]

break that last integral up into 2 integrals

[vivid idol]

flip the integral

[strange lichen]

well yea do that to the limits of the 2nd integral

[vivid idol]

should i do it like this

basicalyl

conviently