precalculus

[undone pumice]

humm

let me think

[minor adder]

one of the last questions of our exams in uk😭

[round geyser]

Two Sides are the same

Ik

AC=AB

[undone pumice]

uhh

i got triangle ABC as sqrt3/4 * x^2

now im figuring out total area

[sick glacier]

Yo off topic but do my messages get shown as "likely from a spammer"

[minor adder]

yeah both r x

[undone pumice]

no (to marz)

[sick glacier]

Alr good

[minor adder]

yea

[round geyser]

And

[undone pumice]

wait nvm dont need the area

i read the question wrong lol

they're both equilateral

and similar

[minor adder]

i did it a diff way too without calculating area

[undone pumice]

umm

ratio of their heights

squared

[minor adder]

YES

wait hold on

[undone pumice]

sqrt3/2 x / (sqrt3+4)/2x

(sqrt3/sqrt3+4)^2

[undone pumice]

simplifying to uh... ||3:(19+8sqrt3)||.......?

[minor adder]

wow

yes

[undone pumice]

yay

xd i misread the problem at first

cus

im eating and typing lol

[minor adder]

i looked at some solutions on tiktok

[undone pumice]

trying to headsolve ts

[minor adder]

like same question js one square and look

[light tangle]

You should be asleep 😭

[undone pumice]

wow

thats u,

um

a lot of working

[minor adder]

crazy

im thinking why dont they js use the length scale factor square for area

[minor adder]

who😭

[undone pumice]

welp food finished and brain fuel restored yey

[minor adder]

what time is it for u

[undone pumice]

like

7

bought myself some food

[minor adder]

ooo

3am here

[undone pumice]

damn

haha wow

[minor adder]

that question

is like the hardest question from all papers we did

[undone pumice]

oh

wow

i see

[minor adder]

i found it pretty easy im js hoping

the real exams arent too bad

[undone pumice]

🙏

[minor adder]

1 month

[undone pumice]

you got this

i js have ap exams in a month to start to study for

then sat the month after

[minor adder]

i havent even revised bro

[undone pumice]

oop

um

nah its okay you'll cook trust

[minor adder]

2 weeks holidays done barely anything

[minor adder]

yes i hope

[undone pumice]

real i didn't do jack shi over spring break

[minor adder]

1st exam may 5th 😭

[undone pumice]

went shopping and stayed in my room like a bum 🥀 but trust we'll lock in until may

[round geyser]

Show us more

[undone pumice]

i have my first ap exam may 12th

[round geyser]

I like that kind of question

[undone pumice]

my second ap exam may 14th

[minor adder]

okay lemme find some of the hardest questions

[minor adder]

😭

yes

[minor adder]

damn

[round geyser]

As a trade take this SAT example question

[round geyser]

Take this

[minor adder]

oh gosh

[minor adder]

8?

[undone pumice]

i meannn considering the sat is calculator

the simplest solution would be to hop on desmos,

right?

[minor adder]

32 = 2(x-4)^2

divide 2 both sides

16 = (x-4)^2

sqrt

4 = x-4

8=x

im not sure😭😭😭

[undone pumice]

i js did ts icl

[minor adder]

omg its 8

[undone pumice]

i graphed and i got 8 so im pr sure

[minor adder]

yay

[undone pumice]

yay!

the entire math sat has a calculator btw so i js did that

[warm mauve]

I would kill to have that logic train of thought

[undone pumice]

its okay you'll learn to recognize patterns and improve problem solving

[minor adder]

ah right

[undone pumice]

yeah, quite a few questions can be solved with calculator (of course, there are also questions solved quicker by hand)

[minor adder]

Yeah

we have 3 papers in total

1 non calculator

[undone pumice]

ooh ic

[minor adder]

2 calculator

[undone pumice]

ah interesting

for our ap exam we have a non calculator multiple choice, calculator multiple choice

calculator free response and non calculator free response (x2 each)

[minor adder]

wow multiple choice

thats great no?

[minor adder]

ooo

[undone pumice]

well theres also like

written free response

[minor adder]

i mean our exams look easier than the questions yall be doing

[undone pumice]

on paper

[minor adder]

😭

[undone pumice]

the mcq is worth like

[minor adder]

ohh yes

[undone pumice]

2/3

and the written ones are worth like

1/3

of ur total score

[minor adder]

ohh

for top grade

we need about

73/80 each paper

[undone pumice]

icic

lemme google what u need to get the highest score

[minor adder]

alr

[undone pumice]

like...

hmm

lemme google the composite score

[minor adder]

okayy

[undone pumice]

oh

not bad

its a new ish exam so hopefully they dont ramp up the difficulty

theres prediction calculators for it lol

[minor adder]

ooo

oh wow

https://www.mathsgenie.co.uk/papers/1hmay2023.pdf

One of our papers not sure if ur able to go on it

[undone pumice]

ah it works

icic

what grade level is this for

[minor adder]

must be easy compared to what you do right

this is like

high skl

we do this at the end of high skl

[undone pumice]

oh

icic

im in high school rn

[minor adder]

same

abt to leave

2 months damn

ill miss it

[undone pumice]

oh

im in my first yr of hs

https://apcentral.collegeboard.org/media/pdf/ap-precalculus-practice-exam-multiple-choice-section.pdf heres like a section of the multiple choice on the ap exam im taking

which is the easiest math AP

(we have calculus with two levels)

AB and BC, and before that the class im taking Precalc

[minor adder]

wow don’t think id be able to do even one question from this

looks so hard😭

[minor adder]

oh

[undone pumice]

its supposed to be ah... 11th 12th grade level?

i think?

[minor adder]

what age roughly

[warm mauve]

😭

[undone pumice]

er...

idk

[warm mauve]

Imnso cooked if thats easy

Holy shit

[undone pumice]

17?

[minor adder]

ah

that would be college for us in uk then

[undone pumice]

precalc is normal track tgaken 12th grade

[minor adder]

a levels

[undone pumice]

like

idk

tho

[minor adder]

yeah

gcse maths is quite easy tbf

[undone pumice]

but a levels supposed to be harder

[minor adder]

im doing a level maths september this year

gl to me then😭😭😭😭

[undone pumice]

🙏

[undone pumice]

i mean.. its easy to me but we've been learning ts all year

[minor adder]

do u have further maths by any chance over there?

[undone pumice]

whats further maths

[minor adder]

like additional maths

idrk but

[undone pumice]

OH

we have calc

[minor adder]

we have it as a choice of a level

[undone pumice]

yeah igs calc

ill be takig calc next year

[minor adder]

ooo

[undone pumice]

so it matches with like

aps and stuff

which is supposed to be college level

makes sense

[minor adder]

wow bro

impressive

at 15

[undone pumice]

well not taking the one it says up ther (AP BC)

which si the a - level one (its harder)

just AB

(AB is easier than BC)

[minor adder]

ohh

ur still so smart wth

[undone pumice]

So taking BC when i'm like... 16., i'm 14 now

anyway off now, be back soon

[minor adder]

cya!

[round geyser]

Wow

[minor adder]

ill try this

[dry sierra]

Isn't there a calculus 1 channel?

[undone pumice]

#calculus

[minor adder]

x = 19 y =6?

[round geyser]

Yes

[warm mauve]

Where did the 25 come from?

Oh nvm

Jesus

I gotta lock in 😭

[undone pumice]

🥀

ack i should get back to khan academy sigh i hate optimization

[minor adder]

😭trust me keep practicing ull get better

[undone pumice]

yeah most math is just understanding the why and the how, then doing practice problems

[warm mauve]

I understand the hows normally but not the whys

[undone pumice]

I see. It’s important to remember why everything works too

[chilly arrow]

Damn

I don’t really know what matrices are

For me in Australia, we do complex numbers, proofs, 3d vectors, integration, mechanics in math ext 2

I think we get harder complex numbers questions than a level further maths

We only do separable DEs

[zenith dagger]

Does anyone have like a list of a ton of integrals at A level maths/fm level (17-18 yrs old)? Things like by parts, substitution reverse chain rule etc.

[compact bay]

calc in precalc...

#calculus

[zenith dagger]

bro im uk idk what pre-calc means surely everything is pre-calc

[smoky schooner]

I just finished a 7 lines equation

[limber junco]

hi uk 👋

you can probably search up stewarts calculus and find a lot of just integration technique problems

[lilac moat]

@round geyser me back gang

[fossil kindle]

can anyone confirm

i swear this question is wrong

this one too

[round geyser]

That’s the form y=a(x-h)^3+k

[fossil kindle]

its a cubic tho

which ig still applies

[round geyser]

h goes left and right and k goes up and down

[fossil kindle]

yeah but im pretty sure my answers were right and the question is wrong

[velvet wyvern]

you are right

[fossil kindle]

ok ty

[uncut mulch]

what's the colour scheme,
did you select the red or the green

[fossil kindle]

red

[round geyser]

What formula will we use

[uncut mulch]

we're dealing with compound interest

so you should use the compound interest formula (exponential growth)

[round geyser]

Okay

[round geyser]

Per year, n=12

[uncut mulch]

can you show the uncropped question

[round geyser]

The cut part is Peter and He

[uncut mulch]

why are you dividing by 12

i see nothing mentioning compouded monthly

[hushed sphinx]

On the other hand, if there is no explicit compounding period in the problem, we'll need to fill that in from usual banking practice in the location the problem is from ...

[round geyser]

Ik this equation

1/(x-4)?

[fossil kindle]

-12/(x-4)

[round geyser]

Thanks

[round geyser]

a=2,c=3,b=4, d=1/2

Correct

[lilac moat]

@round geyser ig u forgot me

[round geyser]

Wsp

[round geyser]

Look at this graph

[lilac moat]

Yooooo

[lilac moat]

Ye so ?

[round geyser]

Is this correct

[lilac moat]

Ye feels like I put some value feels like matching

[round geyser]

Yep

[nimble wolf]

I think this goes on #calculus

[robust pecan]

Uh

[robust pecan]

This channel wasnt on my list

Sorry

[honest meadow]

you need one of the joinable roles

undergraduate math i believe

[robust pecan]

Im not in uni

[winter comet]

you don't need to actually be in uni, its just supposed to be the role for people studying "undergraduate math"

[compact bay]

go to #calculus for calculus, integrals and all

[round geyser]

How would we derive it

[raw hill]

Horizontal asymptote: $y=\frac{a+\frac{b}{x}}{c+\frac{d}{x}}$. As $x \to \pm \infty$, $\frac{1}{x} \to 0 \implies y \to \frac{a}{c}$

[obsidian monolith]

Civil Service Pigeon

[raw hill]

,w solve c*x+d=0 for x

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[round geyser]

I never knew that

I only know HA is domain but x=

Ans VA is dn <=> dm

[muted plover]

Where is this expression for alpha as a rational function of beta from? I can verify that it's true, but I don't know how you'd arrive at this, and the solution I have in mind takes significantly longer than what this gives

Is there an obvious manipulation or a trick that gives this expression?

[raw hill]

$$\beta=-2\alpha^2+4\alpha \implies \alpha^2=2\alpha-\frac{\beta}{2}$$

[obsidian monolith]

Civil Service Pigeon

[raw hill]

$\alpha$ is a root of $x^3+2x+2$, so
$$\alpha \left(2\alpha-\frac{\beta}{2} \right)+2\alpha+2=0 \implies 2\alpha^2-\frac{\alpha \beta}{2}+2\alpha+2=0.$$

[muted plover]

Complete the square?

No

[obsidian monolith]

Civil Service Pigeon

[raw hill]

Again,
$$\beta=-2\alpha^2+4\alpha \implies 2\alpha^2=4\alpha-\beta,$$
and so
$$4\alpha-\beta-\frac{\alpha \beta}{2}+2\alpha+2=0.$$

[obsidian monolith]

Civil Service Pigeon

[raw hill]

,w solve for a if 4a-b-1/2 a*b + 2a + 2 = 0

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[muted plover]

Would you call this "ad hoc"?

[raw hill]

The idea here of “reducing powers” like this is quite standard in high school comp math, so probably not.

[muted plover]

Yeah ironically I'm doing something kind of like this in part (i) and couldn't figure out that the grader did the same thing in (ii)

[raw hill]

[muted plover]

Thanks!

[raw hill]

[lucid ledge]

what do you guys recommend me to do when I stumble onto something that I needed previous knowledge to do it?

[muted plover]

If the problem statement makes sense to you, but you're not sure how to start, play around with it for a little bit. (If the question doesn't make sense look over your notes or ask for help). Write down what you know and what you don't, write down all the ways you know the things you know are related to each other, try to think about how you can relate the givens and the goal. If you've done this and you're still stuck, look over your notes or ask for help

[warm mauve]

For problems is there some prespective shift or Line of thinking one could follow for solving problems? My main issue is not knowing what "tool" to use for the job

[round geyser]

How would we solve this

[round geyser]

Just do long divison

[muted plover]

With what divisor and dividend?

[round geyser]

I saw it wrong

I thought that beta was also alpha

[muted plover]

x=8 is one solution. I think it shouldn't have any others if you look at the derivative

[raw hill]

Divide both sides by $4^x$ and you end up with $(x/4)^{4(x/4)}=256$

[obsidian monolith]

Civil Service Pigeon

[round geyser]

Also when I graphed it

It never touch

[thick blade]

x^x=2^(8+2x) at x=7 when analyzed graphically

[raw hill]

🤨

The left hand side is then odd while the right hand side is even

[round geyser]

How do you know

[hardy venture]

different bases of exponents tho, I don't think there is any issue

[hardy venture]

when this happens, a rule of thumb is to check for complex x

[round geyser]

That had to be complex solutions

[hushed sphinx]

The point was that x=7 cannot be a solution to x^x = 2^(8+2x) because 7^7 is an odd integer, but 2^(8+2·7) is an even integer.

(You can compute exactly which odd and even integers if you're curious, but it is certain even without computing that they won't be the same number).

((Though for full disclosure it took me a few minutes to realize that was what what CSP meant, because I was thinking about odd and even functions at first.))

[main bone]

I don't even know what I tried to do here... I felt like I needed to share this thing

[warm mauve]

Oh lord

Someday i hope i can solve that

What even are imaginary numbers 😭

[round geyser]

Solve for x

[raw hill]

Divide both sides by 2 and see what happens. Presumably, b is not 2 or -2.

[round geyser]

That’s what I thought of

(2^(ax-1))/2 = b^(ax-2)

[raw hill]

mhm

[round geyser]

There is a exponent rule

x^a/x^b = x^(a-b)

2^(ax-1)-(1) = b^(ax-2)

[raw hill]

You should put parentheses around your entire exponent on the left. But yes.

[round geyser]

I’m just doing that show we can subtract

[round geyser]

2^(ax-2) = b^(ax-2)

b=2 I assume

[raw hill]

My point is that in a literal sense, you wrote $2^{ax-1}-1$.

[obsidian monolith]

Civil Service Pigeon

[raw hill]

But I digress

[round geyser]

Oh

[raw hill]

That’s one possibility

But then you can’t really solve for x (why?)

[round geyser]

Cancel base

Set exponents the same

[raw hill]

yeah

This is why I said b is presumably not 2 or -2

[round geyser]

ax-2=ax-2

[round geyser]

Oh yea

[potent chasm]

Quick tip for exponential equations like 2^(ax-1) = b^(ax-2):

Always check first if the bases can be made equal.
If yes → set exponents equal directly.
If no → take ln() of both sides and solve.

The contradiction trap (like getting -1 = -2) usually means
your assumed value for the base is wrong. Good discussion! 👍

[round geyser]

Okay

The ans is take log_2

[hushed sphinx]

<@&268886789983436800> mrbeast scam

[round geyser]

What would I do

[quaint stirrup]

If u are facing any issue regarding maths I can help you

[rare ferry]

So what is e used for? As far as I can understand, it’s the exponential growth as it approaches the limits of like a bunch of intervals. Idk how or why it’s used though.

[fossil kindle]

the "perfect constant for exponentials"

in terms of limits its like the compounded interest

[echo sinew]

<@&268886789983436800>

[round geyser]

This is a good one

[fossil kindle]

what do i do with the rate

i think its 17000e^20 times something

[round geyser]

Hint

A=Pe^rt

[round geyser]

Set that over 100

[fossil kindle]

ohh

[fossil kindle]

17000×e^(.0104×20) ?

[round geyser]

It’s 0.04

[fossil kindle]

oh

oopsies accidentally added 1 to it

ok

[round geyser]

Lmk what did u get

While that compare this to this

[fossil kindle]

37834.20

[round geyser]

Here

That 12 is exact

[fossil kindle]

mine was 37834.195

huh

[round geyser]

195 can be 200

Rounded as 200

[fossil kindle]

equation is 17000(1+1/12)^240

[round geyser]

You mean 0.04

0.04/12

[fossil kindle]

ok i keep forgetting the rate

[round geyser]

Yea

Lmk what did u get

[fossil kindle]

37783.8954?

[round geyser]

Yes

Or 37783.90

Now compare

[fossil kindle]

yeah

[round geyser]

Continuously would have the most

It’s like money compounding in seconds

[fossil kindle]

smth like (1+1/inf)^inf i think

from the limits of compound interest

[round geyser]

Guess what

Do u know Nerdwallet

They had that compound interest calculator

If u are using that

Set contributions as 0

[fossil kindle]

interesting

[round geyser]

Did u use it yet

You notice the amount is the same as what me and u got

[round geyser]

^

[round geyser]

From this

[fossil kindle]

uh

still downloading

[round geyser]

Okay

[round geyser]

Just use their website

[round geyser]

https://www.nerdwallet.com/banking/calculators/compound-interest-calculator

Here

[fossil kindle]

yeah i got 37783.9

[granite hedge]

Can someone help me with this problem

I translated it from gpt so idk if you can read it properly

[valid topaz]

<@&268886789983436800>

[lilac moat]

It's easy so he telling point B reflection is point C wrt A that means point A,B,C on same line and point A is mid point

If I said smthing wrong plse @round geyser correct me

[split summit]

Int. 1+sin(logx)/1+cos(logx) dx

[granite hedge]

Oh ok

I think it has something similar like linear equations because I remember something similar

[undone pumice]

#calculus

This not precalc if it has integrals

[limber junco]

Bro sent it in a buncha channels

💀

[undone pumice]

Wild

🥀

[ruby jungle]

you literally directed him here to post (from the #competition-math channel)

[woeful geyser]

Integration isn't precalc?

[brisk raft]

The prefix pre- means "before," "prior to," "in advance of," or "in front of". Derived from Latin prae-, it is used to indicate that an action or state occurs earlier than a specified time or position. It is used to form verbs, nouns, and adjectives.

[swift shuttle]

Brother

That's irrelevant

[brisk raft]

First, consider the definition "pre-". Next, consider what subject integration is primarily taught in. Finally, deduce why integration is not usually regarded as a topic in "pre-calculus".

[undone pumice]

integration is calculus, precalculus is more the buildup to teaching calculus like teacching concavity secant lines limit definition etc.

bridging algebra and calc

[swift shuttle]

I don't think that's quite sufficient to deduce that

[brisk raft]

on my world it is

[dense zealot]

wdym you guys didnt do 3D integrals in pre-calc

[valid topaz]

uhhh

[limber junco]

No I told him to post his second thing to precalculus

[chilly arrow]

daily integral i see

[chilly arrow]

i mean its because people learn it in high school, and they have the pre-university role so they can't see the calc channel

i feel like this seems to happen a lot

[brisk raft]

most people learn what an integral is in their calc class, not a precalc one

not knowing there's a calc channel doesn't really change tat

[split summit]

[split summit]

Can anyone gimme that role

[uncut mulch]

id:customize

self assignable,
undergrad

[river drift]

you can view and message in the #calculus channel even with a pre-university role

you may need to go to id:browse to enable the channel

[undone pumice]

yeah i didn't give myself early uni role i just cherry picked the ones i wanted

[lone temple]

can someone help me with this?

[woeful geyser]

I see

In my textbook

It's different

[swift shuttle]

Have you not studied differentiation?

[velvet wyvern]

!status

tf

!status

[limber junco]

what conditions are needed for the function to be strictly decreasing on an interval?
(hint: think about the derivative)

[velvet wyvern]

since there is modulus you need to deal cases,

if the functions is h(x)= |g(x)|
h*x) increasing if g(x) >0 and g(x) is increasing

if g(x) < 0 and g(x) is increasing h(x) is decreasing because of the modulus,
if g(x)<0 and g(x) is decreasing h(x) is increasing

[spice sandal]

is this a cubic graph ?

[undone pumice]

Sorta but it’s a first derivative

[raw hill]

This belongs in #calculus, and there's online resources on this like Khan Academy as well.

[wispy plume]

oh okay got it that you!

[hushed sphinx]

<@&268886789983436800> mrbeast

[round geyser]

Can we just subtract 2548 to the LHS

Making it x^12-x^8-2548=0

[hushed sphinx]

That's one thing you can do, though it doesn't necessarily help a lot.

(Not that I can immediately see what would help a lot).

[round geyser]

Can we substitute

u= x^4

[hushed sphinx]

Yes we can, but we're still left with a cubic in u.

[valid topaz]

<@&268886789983436800>

[raw hill]

(Fortunately, you can guess a root by testing integers)

[hushed sphinx]

Ah, so we can -- I mistook x^8 for x^4 when I tried initially, and that looks hopeless.

[flat remnant]

I think you can change the variable 2 times then use Delta

I don't know if this is the good approach, it seems hard.

[round geyser]

Which is

u^3-u^2-2548=0

Check it out

[round geyser]

Is this correct

[raw hill]

dropped negative

[round geyser]

I thought there is 12 solutions

[raw hill]

You had a negative on the 13 in your numerator on the left

then dropped it when copying it over to the right

[round geyser]

So that’s the 12 solutions

[raw hill]

? You only listed four solutions rn

[round geyser]

Is that it

[raw hill]

2 from each of these

[round geyser]

Also

• • and - +

[raw hill]

Note that i^4 = 1

[round geyser]

So what’s it gonna be

[raw hill]

,texsp ||$$\implies x^4=a^4 \implies x=a, -a, ia, -ia$$||

[obsidian monolith]

Civil Service Pigeon

[raw hill]

oh texsp fail

whatevr

[round geyser]

Oh

It’s

,texsp $$\implies x^4=14^4 \implies x=14, -14, 14i, -14i$$

[obsidian monolith]

Jake

[round geyser]

Why is it 14^4

[raw hill]

It's not

I picked a^4 for notational convenience since writing roots is annoying, just pick a appropriately or replace a^4 with a

[round geyser]

Ok

x^4-14=0

You use sqrt(14)

[raw hill]

,w (sqrt 14)^4 - 14

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[round geyser]

(x+14)(x-14)(x^2+sqrt(14))

Is this right

[raw hill]

(x-14) being a factor implies x=14 is a root

is that true?

[round geyser]

No

It’s 4th root of 14

[raw hill]

mhm

$x^4=14 \implies x=\sqrt[4]{14}, -\sqrt[4]{14}, i \sqrt[4]{14}, -i \sqrt[4]{14}$

[obsidian monolith]

Civil Service Pigeon

[round geyser]

Ohh

[raw hill]

Ig you can see this from
$$x^4-1=(x^2-1)(x^2+1)=(x^2-1^2)(x^2-i^2)=(x-1)(x+1)(x-i)(x+i)$$

[obsidian monolith]

Civil Service Pigeon

[raw hill]

perhaps I should've mentioned that

idk

[round geyser]

Where is 8 more solutions

We have 4 now

[raw hill]

I think you can answer this yourself

look back at your working

[round geyser]

Look

[raw hill]

what

also

what

[round geyser]

This is complicated

[raw hill]

You had $x^4=\frac{-13 \pm i \sqrt{559}}{2}$

[round geyser]

You noticed

[obsidian monolith]

Civil Service Pigeon

[raw hill]

And then you somehow restricted yourself to only working with $x^4=\frac{-13 +\sqrt{559}}{2} \dots$?

[obsidian monolith]

Civil Service Pigeon

[round geyser]

…

[round geyser]

How would we do this

Like reuse the equation

[raw hill]

My point is that you just used $x^4=\text{a completely different thing}$

[obsidian monolith]

Civil Service Pigeon

[round geyser]

So do I need to do two cases

A positive and a negative

[raw hill]

Not really because what you do is the same for both

||you just take the fourth root and tack on \pm 1 or \pm i||

[round geyser]

$x^4-\frac{-13 \pm\ sqrt{559}}{2} $?

This

[obsidian monolith]

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

[round geyser]

^

,w x^4-\frac{-13 \pm i \sqrt{559}}{2}=0

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by tofu5477

[raw hill]

My point was that you could say
$$x^4=\frac{-13 \pm i\sqrt{559}}{2} \implies x=\pm \sqrt[4]{\frac{-13 \pm i\sqrt{559}}{2}}, \pm i \sqrt[4]{\frac{-13 \pm i\sqrt{559}}{2}}.$$

[obsidian monolith]

Civil Service Pigeon

[raw hill]

feel free to verify this with a calculator if you'd like

[round geyser]

Ohh

Yea

That’s the remaining 8 solutions

$x= \pm\sqrt[4]{14}, \pm i \sqrt[4]{14},\pm \sqrt[4]{\frac{-13 \pm i\sqrt{559}}{2}}, \pm i \sqrt[4]{\frac{-13 \pm i\sqrt{559}}{2}}.$$ $

Like that

[obsidian monolith]

Jake

[round geyser]

Thanks

[dreamy seal]

hello, why is the answer D and not C? is it important that the resultant points in a direction different to arrows A and B

[fossil kindle]

looks like its both c and d

[dreamy seal]

i concluded that, but i don't exactly get the difference of swapping the head and tail of the resultant

[fossil kindle]

the key is prolly wrong if thats what you're using

[fossil kindle]

i thought abt direction but uh

the direction for d is not the same either

[dreamy seal]

it happens to be my tutors solution hisself, but i tried to make sense of it since maybe the resultant's direction should be "unique" (?)

this question sucks but i want to learn

[fossil kindle]

d looks wrong acc

[dreamy seal]

hm, i guess i should trust the answer i chose then?

(these are past paper questions which is why i'm hellbent on this)

[velvet wyvern]

which one did you choose

[dreamy seal]

i chose C

[dense zealot]

<@&268886789983436800>

whys there so many scammers

[round geyser]

Synthetic or long divison

[valid topaz]

both work

synthetic makes u looks sophisticated tho

[round geyser]

Bc it’s 3/2

[round geyser]

I’ll do two ways

[warm mauve]

Show ur process when ur done

[warm mauve]

https://tenor.com/view/anime-i-see-gif-25052382

Bet

[round geyser]

Good

[foggy forum]

fm gcse aqa l2?

[round geyser]

Wdym

[foggy forum]

whered you get that question from

[bronze geyser]

ts handwriting gave me a seizure

[round geyser]

What

[bronze geyser]

jk

[foggy forum]

??

jake

[round geyser]

Insta

[undone pumice]

oh... interesting, didn't expect ig

[bronze geyser]

lmao

[undone pumice]

my ig feed is just food lol no math-

[bronze geyser]

your bio says you like food and math

and noone posts shi abt math

[valid topaz]

3b1b 🥺

[bronze geyser]

what

[valid topaz]

nvm

[round geyser]

Me too

[undone pumice]

lmao

[round geyser]

I really like to solve like the last time with x^12

Also made me thought of solving x^3=50653

[thorny remnant]

Hi, just joined! I'm 14yo and just finished precalculus!

[hardy venture]

$\sin(15^\circ)$

[obsidian monolith]

YeetusDeletus5

[thorny remnant]

About 0.6502878 or so

[round geyser]

Or

[round geyser]

What’s the exact value

[thorny remnant]

{√6-√2}/4

[undone pumice]

nicee

[orchid urchin]

for a second i thought that said I pee

[undone pumice]

you can do it with the sin difference formula

[thorny remnant]

That's what I used

[coral agate]

Hi guys

I'm really lost can someone [please help

[raw hill]

The polar part is completely superfluous — this is a standard average rate of change exercise. And since $\pi/12$ is halfway between $0$ and $\pi/6$, you can just say that using the average rate of change gives you $f(\pi/12) \approx \frac{f(0)+f(\pi/6)}{2}$. (Why?)

[obsidian monolith]

Civil Service Pigeon

[orchid urchin]

what you mean why?

[raw hill]

It's common to use "(why?)" as a stand-in for "this is something that you should be able to justify, but make sure you actually go through the exercise of justifying it"

[round geyser]

That made me thought of

sqrt((3.14-pi)^2)) =

[thorny remnant]

Unless 3.14 is not a rounded number, isn't the answer zero?

[hushed sphinx]

The context wouldn't make sense unless "3.14" means the rational number 314/100.

[thorny remnant]

Certainly!

[bronze geyser]

Ts is too advanced

[fossil kindle]

how do you even find that out

how do you make a 15-75-90 triangle

[thorny remnant]

You can use the half angle formula

[hasty mica]

or angle difference

[rough leaf]

What was that again

Isn’t it uh

Sqrt((1-cosx)/2)

Smth like that right

[timber jasper]

Yall im learning basic limits cuz im jobless as a 13 year old and i got the question

lim (x → 0) of (√(x + 1) − 1) / x

How tf would I do that bro 😭✌️

[uncut mulch]

rationalize numerator using conjugates

[round geyser]

What about |pi-3.14|

[rough leaf]

Lhop

The limit

Or rationalize

[low mountain]

Rationalise

[potent slate]

L’Hôpital

[thorny remnant]

Rationalize the numerator

[hushed sphinx]

Which Ramonov already said immediately under the question.
(Though I wonder whether that description is too pithy for the OP to get anything out of -- but he hasn't asked for clarification, so perhaps).

[swift shuttle]

That's just annoying

Not hard tho

[solar bough]

is that the damn exponent

lord have mercy

[swift shuttle]

Well for starters log√6 (216) is just 6

[timber jasper]

Well wtf if rationalise

[low yoke]

you can separate by exponents like: $$ {6^{1/2}}^{2+\sqrt[3]{\sqrt{2x+22}}} = 6^3$$

[solar bough]

um the exponent without the 2 is equal to 4

its hard to work out when ur in bed at 6AM without working out paper

[swift shuttle]

You just get that the cubic root part is equal to 4

Then you simplify

[round geyser]

I agree

[solar bough]

root(2x) = 42

[swift shuttle]

√2x +22 = 4³
√2x = 4³ - 22
x = (4³-22)² / 2

[solar bough]

ithink

[swift shuttle]

And you evaluate that

[round geyser]

I would just use √x = x^(1/2)

[solar bough]

x is 882 i think

i dunno maybe not

[obsidian monolith]

S0S4

[low yoke]

now log in base 6

$$\frac{1}{2}(2+\sqrt[3]{\sqrt{2x+22}}) = 3$$

[obsidian monolith]

S0S4

[low yoke]

and well, now you solve

[swift shuttle]

Brother

[low yoke]

tell me

[round geyser]

Wow

[warm mauve]

What

[solar bough]

uh

dont think thats right

[thorny remnant]

Omg this is nuts

[undone pumice]

Damn I’ll try to solve it

I set the exponent to 6 and solved from there got x = ||1922||

[bronze geyser]

aura

[brittle adder]

Guys I learned some of trigonometry in precalculus so idk if I need to talk about it in trigonometry, since I wanna learn trigonometry for Calculus

[orchid urchin]

i feel like you shoul be able to talk in both

since precalculus covers trig

[brittle adder]

Yeah

Can you help me?

[orchid urchin]

yea

[brittle adder]

YEEEEEES

You had a calculus class?

[orchid urchin]

im self study i never officially taken a calculus class but I am studying calculus, linear algebra, and how to prove it by daniel j velleman

[brittle adder]

Ooooo

You know derivatives and limits?

[orchid urchin]

yea

[brittle adder]

I need some help with trig

It’s a topic that made me fail my calc class

[orchid urchin]

what parts of trig?

what topic i mean

[brittle adder]

Everything

:((

Everything related to sine cosine

I don’t understand how to solve something that has sine or cosine

All of that

[orchid urchin]

Hmm, that sounds tough. When you learned trigonometry would you say your teacher said "memorize this" or did your teacher actually teach understanding

i ask because sometimes trig teachers lowk suck, unless its college or something

[brittle adder]

So the professor was crazy because went over everything without stoping

So never got the chance to actually understand

And somehow I passed precalc but I failed in calc

[bronze geyser]

similar triangles

[orchid urchin]

What do you already know about trigonometry or feel comfortable with?

including sine and cosine

[brittle adder]

There is no vc in here?

[brittle adder]

So rn nothing but it’s not like I wouldn’t be able to understand the first things fast

I kinda know some of the stuff but since I never used it don’t really know how to apply it

[orchid urchin]

Okay thats good honestly, to be precise I learned trig a little oddly so I don't know if I could help very well. However it does start with the unit circle, most things in trig come all the way back to the unit circle, if you master that then you can master trig because its the foundation.

What do you know about the unit circle?

[brittle adder]

Yeah I’ve seen that and I really don’t understand it

You think I can call you text book? I don’t want you to give me a class maybe just to tell me how to start

Or maybe just tell me 🙁

Because first I don’t understand the purpose of it

And when I see it in some square roots and stuff I don’t know what to do

[orchid urchin]

That is okay, I show you. You should have learned in precalculus that a circle is (x-h)^2 + (y-k^2) = r^2, the unit circle is defined as a circle centered at the origin with radius 1, so that gives you the following x^2 + y^2 = 1, that should be the easiest part to understand, next is SOHCAHTOA, assuming you know that I will give you a problem, and just say if you can answer it or not, if you can then do so, in the image provided lets say between the x axis and the line that has length 1 has angle theta, if sin is opp/hyp then what is the opposite side and the hypotenuse if you can understand this part then you should go through this lightning fast because in reality trig is really simple.

@brittle adder I prefer not to call however I recommend a good teacher on youtube, his name is professor leonard and he goes through trig in his precalc section thoroughly.

like REALLY thoroughly

[brittle adder]

So he teaches you really well?

[orchid urchin]

Yeah, the videos are very long but if u already understand something you can just skip over a section

[orchid urchin]

He doesn't provide formal proofs but for just learning trig you don't need that, however he does teach identities imo a bit late, and tells you to memorize the sin/cos difference formula or sum formula, rather than showing you where it comes from, but to be FAIR understanding where that formula comes from requires a good amount of knowledge of circle theorems and triangle theorems, or even ptolemy's theroem which is about a inscribed quadrilateral.

[brittle adder]

If I’m not wrong and maybe I am, I think the opp is the left side and the hypotenuse might be the horizontal line but I’m not sure about that

I meant right side

Right line

[brittle adder]

Damn then maybe is better to memorize it

[orchid urchin]

oh wait let me draw theta mb, this should make more sense, and your correct the side opposite to theta is the right side, thats easy, and the hyp is the longest side which is 1, why 1? because it is the radius of the circle. Now if sin = y/1 then what does the right side (vertical line) equate to?

its not a trick question it is very simple

[brittle adder]

Y

?

Im dumb

[orchid urchin]

yeah and what is y equal to?

[brittle adder]

Im not sure about that

Sorry I’m dumb

[orchid urchin]

That is okay think about the opp/hyp and how it sin produces y/1

actually i just told u the answer

woops

its sin

[brittle adder]

Jajajaj

[orchid urchin]

because sin = y/1

Now when it comes to math its good to collect information and deduce, with that being said try to solve for x using cos.

recall cos is just adjacent/hyp

plug in the values for adjacent and hyp and you should get your answer

[brittle adder]

Okay let me see

X/1?

[orchid urchin]

yeah, and x/1 = x = cos

that make sense?

[brittle adder]

Not really sorry

[orchid urchin]

ill draw it, you know about SOHCAHTOA?

[brittle adder]

Yes

Basically the abbreviation to remember

How to do each one

Like soh means sine opp/hyp

[orchid urchin]

yeah, sine is defined as opp/hyp and cosine is defined as adjacent/hyp and tangent is defined as opp/adjacent

I say defined because you don't derive it, thats just what the functions are defined as

[brittle adder]

Okay

[orchid urchin]

srry if my drawing bad

basically you shoudl understand from that, that sin(theta) = y

geometrically the vertical side is y, and the hyp is 1 because that is the radius, 1 is the hyp because we draw a triangle where we drop a line from the y value of that point on the circle down to the x axis, and then go left from there

or however u want to think about it

(x,y) is a arbitrary point on the circle

now with that being said cos is adjacent side of theta over the hyp, which is x/1

like you said before

thats why cos(theta) = x geometrically

and sin(theta) = y

does that make more sense?

[brittle adder]

Wait sorry I had to do something

Im back

[orchid urchin]

all g

[brittle adder]

Why sin theta equals y

[orchid urchin]

lets say I start at the origin and I move 3 units to the right and 4 units up, what would my coordinate be?

[brittle adder]

(3,4)?

[orchid urchin]

yeah, i put two pictures so u can understand geometrically

these diagrams are useful because they tell you important things

notice how especially in the 2nd picture the blue is move 3 right, then that must mean the blue has a length of 3 units

and green 4 units

[orchid urchin]

now lets circle back, that same logic applies to this, sin(theta) = opp/hyp, we know the hyp is 1 because thats the radius but what is the opp? well if our point is (x,y) then we are moving y units up therefore our opposite side (vertical) is y.

y is the length of that vertical side if our point is (x,y) and the horizontal is x

that should make sense

if it does, then the answer to why sin equals y is because we just plug in the length of the opp side and hyp side into the formula opp/hyp = sin

and we will get sin = y/1 = y

that same process is done for cosine except is adjacent/hyp, so we plug in the adjacent's length which is x and then the hyp's length which is 1 and we get cos = x/1 = x

is that clearer?

[brittle adder]

Oooo got you

Is because y/1 is basically y

[orchid urchin]

yeah and any number over 1 is just itself unless its zero, which im sure you know, 6/1 = 6, 1888/1 = 1888, so y/1 = y

[brittle adder]

Yeah yeah I didn’t think you meant that

Im dumb

But yeah I know that

[orchid urchin]

all g, its common to use a slash (/) to mean division. Okay now that you got that we can deduce new information from the information we just deduced

we have point (x, y) but we know x and y so we can plug those in

and we get (cos(theta), sin(theta))

that make sense?

[brittle adder]

Yes

[orchid urchin]

okay good now this is important because it will tie everything in especially for trig equations, basically since we know a point (x, y) is (cos(theta), sin(theta)) that means any point on the unit circle is (cos(theta), sin(theta)) meaning the point is determined by an angle, now the unit circle looks like this if you search it up on the internet:

but why

well thats actually really simple

oh wait i just forgot you do need to know what radians are

forget that for now

look at the first angle that is 30 degrees, notice how its kind of a weird point

u got a square root and a fraction, not very pretty

WELL it turns out, that we don't need to do anything fancy to figure out why that point is the way it is

let me draw a picture to show you what I mean

that 30 degree angle paints this out

[orchid urchin]

also ima note this is also a right triangle

but anyway onto the thing I just drew

we know the other angle is going to be 60 because a triangle adds to 180

and 90 + 30 = 120, so 180 - 120 = 60

that means we have a 30-60-90 special triangle

I can explain why the values of a 30-60-90 right triangle are the way they are if you want, but here is what it is:

that should be familiar no?

@brittle adder

[brittle adder]

Sorry I’m trying to understand

[orchid urchin]

no problem

[brittle adder]

That is not familiar

[orchid urchin]

okay thats okay, ill show you something thats kind of niche and I think should be taught to more ppl, and its very simple. You know how an equilateral triangle has all the sides the same length with all the angles being the same as well which is 60?

[brittle adder]

Not really

[jolly karma]

Is anyone here decent at multi angle stuff?

[brittle adder]

Im sorry I’m really new to this :((

[orchid urchin]

its okay, this picture is a equilateral triangle

the tick mark means all the sides are the same

[brittle adder]

Maybe text book

[jolly karma]

You’re good no worries

I get it

[brittle adder]

Okay got it

[jolly karma]

Why not go grab a help channel?

[orchid urchin]

can we do that?

how i do that

[jolly karma]

Just pick one of the open channels and send a message

[orchid urchin]

no command?

[jolly karma]

Nope!

Just send a a message :)

[orchid urchin]

thx

[jolly karma]

Ofc!

[orchid urchin]

@brittle adder type in #help-21｜아리스킨충1

or something

something from the math help category thats available

[orchid urchin]

u had a question?

[jolly karma]

Can you do multi-angle stuff?

[orchid urchin]

in trig context yes

[jolly karma]

Mind helping me in my channel?

[orchid urchin]

sure

[warm panther]

I don't think my textbooks answer this question: What is the concept of quadratics?

[solar bough]

its algebra with a variable term exponent of 2

no?

[low yoke]

Yeah, a quadatric equation is an equation with one variable that has a exponent of 2

$$ x^2 +2x +1$$

[obsidian monolith]

S0S4

[warm panther]

I guess it's just that

[sharp karma]

like ax²+bx+c, a≠0

[low yoke]

Yeah xd

[round geyser]

Is that golden ratio

,w x^2 +2x +1

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by tofu5477

[round geyser]

Oh nvm it’s (x+1)^2

[green star]

X= -b/2a is how you get the vertex of a quadratic right? If so then ehat do I do if I need to make a table and I can only use whole numbers

When it’s 13/-8 for the vertex

[raw hill]

You should technically put parentheses around the denominator, but yes. Just use the nearest integers (so -2 and -1 in this case) ig.

[green star]

Kk thx

[raw hill]

[round geyser]

Who thought of x=-b/2a

[orchid urchin]

what you mean, you want to know where it came from?

[round geyser]

Yea

[orchid urchin]

x = -b/2a is the axis of symmetry of a parabola, and you find it with nothing special its just finding the midpoint between the parabola's roots. The quadratic formula gives u the roots or the two x intercepts, and the vertex lies EXACTLY halfway between those points, if you average them then it simplifies to -b/2a

[bronze geyser]

its also found in vertext form

[round geyser]

or m+n/2

[orchid urchin]

also lets say you have the form y = ax^2 + bx + c, if you complete the square and put it in vertex form you will see that h is -b/2a

[round geyser]

And y=a(x-m)(x-n)

[bronze geyser]

yes

[bronze geyser]

just factor ig

quadratic formula to find zeros

then write in that form

[orchid urchin]

I wouldn't say anyone thought of it though, it appears naturally on its own

[bronze geyser]

yeah

[outer peak]

How do I simplify this more ?

I am lost for steps

[valid topaz]

||rearrange sin^2x+cos^2x=1||

[outer peak]

So this

Right ?

[raw hill]

Yes, barring the abuses in notation.

,w true or false sin x cos^2 x - sin x = -sin^3 x

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by civil_service_pigeon

[raw hill]

you can check like this in #bots btw

[outer peak]

cool thx

[raw hill]

[bronze geyser]

did you take a picture with a potato or smth

[outer peak]

my phone it is about 3 or 4 years old

[bronze geyser]

ts looks like the first photographs ever taken

[outer peak]

lol

[warm panther]

Why is intersection called “and” and union called “or”?

[round geyser]

How would i do 7

[undone pumice]

heres a very basic picture (80+x = height of building b)

how do you think you could find x?

[round geyser]

It’s a rectangle

[undone pumice]

um

well

lets look at the red

triangle

[round geyser]

Take tan

tan 22 =x/240

[undone pumice]

Yup

[round geyser]

x= 240*tan 22=2.124 ft -) 2 ft

[undone pumice]

Mmhm

So what’s x

[round geyser]

I mean

x=97 ft

My phone must been using radians

[round geyser]

In exact is 96.9662 ft

[undone pumice]

Mm ok

[round geyser]

That’s tall

Then 80 + 97 =177 ft

[undone pumice]

Yup

[round geyser]

Is that it

177 ft

[undone pumice]

I think so

[warm panther]

.

[river drift]

given an object x and two sets A and B, x is in the union of A and B if it is in either (x is in A or x is in B) and x is in the intersection of A and B if it is in both (x is in A and x is in B)

[round geyser]

Need help on

[hallow sluice]

Could we use the Maclaurin series expansion to help like yk solve or somewhat explain non elementary integrals like the integral of cosx/x?

[undone pumice]

This goes in calc and I think so… not sure

[formal frost]

hi

first draw it

divide it in 2 triangles with a square angle

[round geyser]

Thanks

[round geyser]

If you don’t mind draw it

[formal frost]

k

dm

lil bro send me dm

@round geyser

[round geyser]

Okay

[thorny remnant]

Why is it that I finished precalculus and I've never heard of this Maclaurian series??? 😭😭😭😭😭

[hallow sluice]

Idek lol. Im still in basic calculus yet i have to study about the taylor and maclaurin series. Though it is incredibly useful for proofs and stuff

[hushed sphinx]

Maclaurin series don't sound like a topic that would often belong in "precalculus" at all.

They can be spoken of without derivatives, so I suppose it's possible, but much of what there is to say about them would require calculus anyway.

[hallow sluice]

Yeah ur right. However where i study they're making us do this in calc 1

[hushed sphinx]

... which is explicitly not "precalculus", right?

[hallow sluice]

Yep

Guess i came to the wrong channel

[hushed sphinx]

There's #calculus.

[hallow sluice]

The Maclaurin series is basically just a special case of the Taylor series. You'll cross that bridge when u get there

[hallow sluice]

Cheers

[round geyser]

Doing bearing is easy

[thorny remnant]

I probably have to look into other curriculums, I'm self taught so I just buy textbooks and work books from the store.