precalculus

[sharp geyser]

True but its confusing

[red jewel]

Thats it

[tame shard]

whats confusing about it?

[red jewel]

Or such as 15π/2
You can do
15π = 15 × 180/2 = 1350°
Which is 270° (1350-3×360)

[sharp geyser]

But also i don’t understand how for example, why mostly all of the fractions on the outside are the same except they have different signs (ex: 1/2, sq root 3/2 then on pi/3) and (1/2, negative sq root 3/2) on 5pi/3

[red jewel]

Yeah the only basic thing you need to know is 1 circle = 2π = 360°
And 180° = π

[hushed sphinx]

Most of the details in the excessively complex diagram are not worth remembering piece for piece; instead figure out what the patterns are so you can remember them in case you need specific points.

[sharp geyser]

Ohh

[red jewel]

Oh those are value of those angles

[sharp geyser]

Okay 👍🏼

[red jewel]

You kinda need to memorize these
Best way is solving questions about these

[tame shard]

its flipped across the y axis

[sharp geyser]

OHH

[hushed sphinx]

That's the first important pattern to notice -- the points being shown have been selected exactly because both their coordinates and their angles have nice expressions, and the only coordinates that appear are 0, 1, 1/2, sqrt2/2, and sqrt3/2, with signs according to the quadrant they're in.

[sharp geyser]

Fair, you made me realize that

[red jewel]

such as = sin30 = 1/2
and sin60 = cos30 = √3/2

Cos²x + sin²x = 1 no matter what

Etc

[sharp geyser]

Ah okay

[hushed sphinx]

In particular the points that have such nice coordinates are the ones you get by dividing each right angle into either 2 or 3 equal parts.

[sharp geyser]

Ok ok

[sharp geyser]

Oh

Wait that makes more sense

[red jewel]

So, in nutshell
(x,y) (Such as, 60° is 1/2, √3/2)
first number means value of cosx, 2nd one means value of sinx

[round geyser]

Another trick for radians to degrees

[round geyser]

If you see like 2pi/3 radians think 2 (pi/3) and pi/3 is 60 degrees
2(60 degrees)= 120 degrees

[sharp geyser]

for example, im working on this topic, lets use example #4 for instance, and inside the box is the answer. Like, from -2 sq root 3/3, how did the example give the answer with fractions with different numbers?

[red jewel]

Secy means 1/cosy

[sharp geyser]

Oh, but then what happens after that?

[sharp geyser]

Oh okay got it

[round geyser]

So will it be 7(1/(cos y)= 1/(cos y) - 4sqrt(3)

[red jewel]

So, are you confused at y = 5π/6 + nπ, y = 7π/6 + nπ?

[round geyser]

After I did that I get cos y = -2sqrt(3)

[sharp geyser]

yes😅

[red jewel]

Oh
It is like
For example
Sin60 = sin120

5π/6 = 150°, 7π/6 means 210°

[round geyser]

Is there a triangle that has -2/sqrt(3)?

[sharp geyser]

Also sorry if im afk, i am not ignoring you guys, im thinking about what you all have said to me

[round geyser]

Oh yes nvm

[sharp geyser]

basically trying to comprehend

[red jewel]

Thats ok

Trig is hard at first

[round geyser]

-2/sqrt(3)

Draw 4 quadrants

Place A S T C

[sharp geyser]

thanks :>

[sharp geyser]

astc?

[red jewel]

1/√3/2

[round geyser]

QI and IV

[red jewel]

Which is, 1/sin60

[round geyser]

Wait

[red jewel]

Which is, csc60

[round geyser]

That’s how u remeber which functions are positive

[red jewel]

But there is a - there so it means it is on 3th or 4th area

[round geyser]

I use All student takes calculus

Also good news is

[red jewel]

What

[round geyser]

It’s a special right triangle

Its pi/6 radians

[sharp geyser]

alright 👍🏼

[round geyser]

I got 7pi/6 and 5pi/6

[sharp geyser]

oh wait i think i have that written out

[round geyser]

I had did that before

But I didn’t do “n pi” bc my teacher settled the domain as (0<=x>2pi)

[red jewel]

Do you forget what is negative or positive in certain areas

[sharp geyser]

yeah, because sometimes i seem to forget my signs or put the wrong sign

[sharp geyser]

ah i see

[round geyser]

I saw it wrong I did ur 4

But it should be 2

Number 2

[sharp geyser]

yeah, since it has the n pi

[round geyser]

I know it’s every odd numbers of pi/3

[red jewel]

Ok it is a weird analogy but
Think of it like rooms of house and functions as people
For example, area 1, living room, everyone can be (+) there
Area 2, is child room of sinus, only sinus can be (+) there
Area 3 is parent room, only cotx and tanx can be (+) there
Area 4 is child room of cosinus, only cos can be (+) there

[round geyser]

Or (odd numbers)(pi/3)

[round geyser]

I did “All student takes calculus”

[red jewel]

I have no clue what that is

So

[round geyser]

All means all positive
Student means sin is positive
Takes means tan is positive
Calculus means cos is positive

[sharp geyser]

Honestly its not it helped me just now, im flipping back through ASTC and writing this down but it does help!! thanks!

[red jewel]

Yw

[sharp geyser]

Okay to be fair im getting the idea of how the unit circle works and is being utilized in the work im doing now and other trig stuff, so thanks everyone:> i appreciate it a lot

[round geyser]

Okay

Yw

[winter osprey]

a lot of the times ppl say that if you memorize one quadrant and then "mentally" reflect the coordinates across the axes you can get all the values of the unit circle, but idk how this holds for everyone

i personally just brute forced my way into memorizing everything about the unit circle

[silver jungle]

same

[tribal kestrel]

hi what does surjective mean

[bronze geyser]

wut

[novel cradle]

surjective can mean something along the lines of either the exact thing as in a function wjere every possible output value is mapped to by atleast one input value orr along those lines i guess

it really depends im guessing you mean precalc

[hushed sphinx]

A function f: A -> B is surjective if every element in B is the output of f for at least one element of A.

Some books say "onto" instead of "surjective"; that means the same thing.

[tribal kestrel]

So what makes it different from injective is that it could have more than one x value for an output y

am I getting it right

[hushed sphinx]

Yes, and an injective function doesn't need to hit everything.

Surjective: "each thing in B must be hit at least once".
Injective: "each thing in B must be hit at most once".

[tribal kestrel]

Ah alright gotchu

[novel cradle]

did i do a good job defining it? heh.

[bronze geyser]

https://klipy.com/gifs/anya-forger-14--k01KS3XSXMQZXD8QHHDWE77BRS1

[novel cradle]

[bronze geyser]

https://klipy.com/gifs/jigglypuff-kick--k01KS3XWP2GV30HVG19W98P7AZ5

[tribal kestrel]

@hushed sphinx so when you mean that an injective function doesn't need to hit everything that's only after we defined the codomain and the range such that the range is only a subset of the codomain right?

[astral apex]

that’s right

[zinc lagoon]

real

i have a quizlet of it if anyone needs but tbh it wasn't that alone that got me to know it

[terse bolt]

DAMNNN

that one is good one

i recently learnt add sugar to coffee

but this is crazy

all students take calculus

[round ermine]

man i just visualize a unit circle and see whether the x or y axis is positive or negative i don’t have anything to memorize the signs

[tame shard]

If it works it works

[round geyser]

Yes

or take sin, cos, and tan in the unit circle

[buoyant abyss]

I learned Precalculus is one Day. Now i will jump to Advanced Calculus

[tame shard]

i doubt the advanced part.

[undone pumice]

Ummm sure…

[whole wing]

You mean Calculus 3? I believe it is best to start with Calculus 1 tbh

[exotic gull]

if A is the thing your mapping and B is where it gets mapped

surjective just means your map would hit all pts of B

so if you see something in B then there is definitely something in A aswell that gets mapped to that

by your function

[winter osprey]

i hope this upcoming semester i get to TA pre-calc, its lowk underrated

[tribal kestrel]

bro scrolled way up past everyone else just to help me🙏

thx unc

[round geyser]

How can I write change base form of 9 log_x 2 into log base 2

[astral apex]

the change of base formula is log_b a = (log_b' a)/(log_b' b)

so log_x 2 = (log_2 2)/(log_2 x)

[wise vine]

Yeah so 9logx(2) would be 9/log2(x)

[round geyser]

Bx I’m solving this very interesting equation that needs this

[wise vine]

I see, you know to to solve it after the base change right?

[round geyser]

Please check my work

[wise vine]

There was a small factorisation mistake on the end but you deleted it lmao

[round geyser]

I was doing ac method but forgot to divide by 2

Now I fixed it

[wise vine]

Now it should be right, you can always plug in the values of x in a calculator to check

[round geyser]

Btw these work

Bc I observe that their positive number

So they always work

[haughty crown]

https://tenor.com/view/lol-rdj-juggtok-giftok-tonystark-facials-gif-3958302751287261929

[bronze geyser]

https://klipy.com/gifs/deadpool-and-wolverine-jigglypuff--k01KS6QN1XKJR924KQSG71NNX60

[red jewel]

Really weird way
You just need to do ||log_2(x) = u
2u - 9/u = 3
2u² - 3u - 9
u² - 3u - 18 [[u/2]]
(u - 6)(u+3)
So the "real" roots are u = 3 and u = -1.5
therefore, log_2(x) = 3, so x = 8 and x = 2^(-3/2)
For anyone wondering why I did 2u² -3u -9 -> u² -3u - 18 and what is [[u/2]], I just did this by doing c×a/a, which is c/1, which follows the vieta rules (where I'm essentially doing c/a = ca/a², and [[u/2]] isnt an official term or smth, I just use it as a reminder, This is useful as if the roots are going to be fractals, it helps it to factorize faster and into integers||

[covert jasper]

taking it as u, makes it easy

[round geyser]

Good

[pine knot]

Hi guys im in algebra show me some average calc work i wanna see

[pine knot]

what is log ;-;

is it like the 10 thing

[round geyser]

Yes

log_10 = log

[round geyser]

It’s like
a^x=y

You make it a^y=x

log_a x =y

[tame shard]

definition of log :
$a^{\operatorname{log}_a(b)} = b$

[obsidian monolith]

∫ (larp larp) dx

[round geyser]

Larp

[round geyser]

I got a quadratic that has 27x^2-x-6=0

[round geyser]

^

[red jewel]

Uhhh let me help you

[round geyser]

My idea was

Make 1 = log_2 2

[red jewel]

You were correct until the step log_2[9(3x² - x/3)] = 1

Ok so

The step that caused this catastrophy is

Pretty simple

So good news is you had the core of the question

[round geyser]

Aced brought the 2 below 1 made it 2

I divided 9 to the 2 to make it 2/9

3x^2?

[red jewel]

No,

[round geyser]

I got x^2-(3/x)

Like this

[red jewel]

You did 9(x²-x/3) -> 9x² - x/3

[round geyser]

Oh

[red jewel]

Yeah

Thats where it exploded

[round geyser]

Should it be 3x

[red jewel]

Yeah

[round geyser]

9x^2-3x

[red jewel]

Yup

[pale oracle]

um

,iamnot studyig

[obsidian monolith]

No selfroles matching studyig.
See ,selfroles --list for the list of valid selfroles.

May 22, 2026, 3:41 PM

[pale oracle]

,iamnot studying

[obsidian monolith]

Removed the studying! role from you.

[round geyser]

Yay

[red jewel]

Yeah good job

Also you didnt have to rewrite (x-6)(x+3) as 3x-2 and 3x+1 after finding

But yeah

[round geyser]

If you mind correct this problem

[red jewel]

2log_5(x) + 1 = log_5(12x-4)

I dont have time rn so

[round geyser]

Ok

It’s x=2 or 5/2

Yay

In a few days I like to solve hard log questions

[red jewel]

Hard is really subjective

[red jewel]

For example, you can try solving this

Easy one among hard ones

[round geyser]

Is this “b” the one I circled

[round geyser]

^

[red jewel]

Yea

My b is like that

[round geyser]

Okay

Can I convert them into exponential form

[round geyser]

x=a^y ?

[red jewel]

Sure

As long as it is correct

[round geyser]

Okay

[round geyser]

So is it ab^x or (ab)^x

[red jewel]

What do you think

[round geyser]

Second choice

That doesn’t loook right

[round geyser]

My friend

[red jewel]

Hmmmm

There is a valid answer tho

[round geyser]

Okay

[red jewel]

So, what would the answer be

Answer is ||12|| btw

[round geyser]

This is tough

[red jewel]

It is

[red jewel]

Actually, youre pretty close , c = x^1/72

But you need to find ac

[round geyser]

It is x^(1/72)

But then I moved them back into both a and c

Took their reciprocal

But they want (ac)^y

[red jewel]

Have you tried writing a in terms of c

Eg a = c^u

[round geyser]

It’s 72/31

[round geyser]

Wait what’s the “u”

[red jewel]

A random one

I like u

[round geyser]

I used y

[red jewel]

Fair

[round geyser]

So a=c^y

Ik know

[red jewel]

Yeah so
||a⁹b⁹ = a⁸b⁸c⁸||

I hope i didnt ruin it

[round geyser]

I need help

Since I have no clue

[red jewel]

Try this

This should give you a huge clue

[round geyser]

c^8=ab

c=x^(8/9)

[red jewel]

And b¹⁸c¹⁸ = c⁷² (a⁸b⁸c⁸)

[round geyser]

I wonder why

[round geyser]

How come it’s c^72

[red jewel]

But you find b¹⁸ = c⁵⁴

So you find a

In terms of c

[round geyser]

I got a^9b^9(ab)

[red jewel]

a = c⁵

As ab = c⁸

If log(c⁸)x = 9, (aka log(ab) = 9

What would be log(c⁶)x

You did a great job tho

[round geyser]

Okay

Thanks

[round geyser]

7

[red jewel]

How about this

[round geyser]

Like (ab)^9=(abc)^8

[red jewel]

Aha

[round geyser]

Is it c^5 * c^3 * c^8

You add them up

[red jewel]

Just read the all of the colors

[round geyser]

It’s just 72/6

To get 12

[red jewel]

Yeah

Simple

[round geyser]

Thanks

[red jewel]

Yeah that was the type of top questions i solve usually

You did a good job

[round geyser]

Thanks

I don’t usually see this in my level

It’s just solving by condensing logs or expanding it

[round geyser]

I like this kind secondly

[red jewel]

I'm sure I saw a 15-75-90° triangle...

Well, whatever

[round geyser]

It’s in #competition-math

[round geyser]

I got

[red jewel]

Good job

[round geyser]

That’s the mainly question I love to solve

[red jewel]

Eh

[round geyser]

Yea

[uncut mulch]

Slight error
(x-3) - (x+2)
is 0-5, not 0-6

[round geyser]

Okay

[ebon lion]

i tried everything i can think of

its so simple

but it's ten at night and I'm pulling up blanks

[round ermine]

is it not just y=-2

[ebon lion]

im pulling my hair out i forgot to add the - and i realized as i pressed enter....

i need sleep...

[round ermine]

[round geyser]

I think of y=1/|x|

[ebon lion]

you would think but once you sketch the equation the math is pretty clear

[undone pumice]

kinda except this graph doesn't have a vertical asymptote

[round geyser]

y=e^|x|

[winter osprey]

fake parabola

fraud watch

[undone pumice]

lol

wym fake parabola its just a leaf

[round geyser]

Negative

It’s y=e^(-|x|)

[winter osprey]

yeah that is a diff graph than e^|x|

[round geyser]

Oh really

[winter osprey]

yeah

e^{-|x|} in green, e^{|x|} in purple

[round geyser]

Oh

[undone pumice]

cus this is basically 1/e^(|x|)

[round geyser]

It’s a rational function so it’s y=1/x

[undone pumice]

no

[round geyser]

What’s s in absx

[winter osprey]

the graph that blank put looks more like a variation of $e^{-x^2}$

[obsidian monolith]

TestTickler

[winter osprey]

but its obv not a rational function

[round ermine]

i have found the function of that graph to be (-2x^2)/(x^2+1)

[karmic crow]

Verify that the sum of three quantities x, y, z, whose product is
a constant k, is maximum when these three quantities are equal.anyone got any idea on how to slove this using calculus

[hushed sphinx]

That is not true. 5+5+5 is smaller than 1+1+125.

[valid topaz]

did he mean minimum?

[tame shard]

2+2+2 = 6 , 2^3 = 8 meanwhile 1+2+3 = 6 , 1*2*3 = 6 so both of them arent true i guess

looking back at it, that was not a counterexample

"Verify that the sum of three quantities x, y, z, whose product is a constant k, is minimum when these three quantities are equal"

hm....

[valid topaz]

i mean i used am-gm for this

[tame shard]

-2-2+2 = -2 , -2* -2 * 2 = 8 is a counterexample

lmfao

no

pfft

[valid topaz]

hmmm we allow negatives

[tame shard]

x,y,z implies they are real

=> negatives are allowed

[valid topaz]

well if we have xyz=k, then z=k/xy

then we can rewrite and take partial derivatives

[oblique current]

Well -2 ≠ 2

also hi

[karmic crow]

I am pretty sur that the book has a maximum in it

[hushed sphinx]

Then it's just not true.

[karmic crow]

What is your counter example?

[hushed sphinx]

This.

[karmic crow]

Wait a sec

Here is what the question is asking you the sum neeed to be at minimum and the product of the sum need sto be maximum

[tame shard]

the point is that since -2 != 2 it disproves
"the sum of three quantities x, y, z, whose product is a constant k, is minimum when these three quantities are equal"

[karmic crow]

Wait i am still in a bit of a mess

I will also implement another condition, the numbers need to be of equal signs

I g7ess now ive completed the question

[tame shard]

congratulations you have discovered ad hoc additions

[karmic crow]

What does that even mean

[tame shard]

https://en.wikipedia.org/wiki/Ad_hoc

[karmic crow]

Alr😅

[light tangle]

Can someone explain to me why 0/0 is undefined?

Shouldn’t it be 1 because x/x = 1?

[valid topaz]

that rule is written so that it strictly doesn't include 0

if it did, then we could prove that 1=2

[light tangle]

How so?

[valid topaz]

so say like we know that 6 x 0=0 and 3 x 0=0

[light tangle]

Yeah

[valid topaz]

if we divide both sides by 0, we get (6 x 0)/0=(3 x 0)/0, which is also equal to 6 x (0/0)=3 x (0/0)

[bronze geyser]

Harharhar

[valid topaz]

and because we set 0/0 to be 1, then 6 x 1= 3 x 1

so 6=3

[bronze geyser]

ik ts

[light tangle]

Oh, okay 👌

So you can never divide by 0?

[bronze geyser]

hes

yee

[valid topaz]

well in like standard algebra and stuff like that, no

[bronze geyser]

what about in boolean algebras

[valid topaz]

(insert larping)

[light tangle]

There’s non standard algebra?

[valid topaz]

uhh yes

but im not qualified at all to speak on that

[bronze geyser]

no i have a phd in algebraic topology

[valid topaz]

ok

[bronze geyser]

https://klipy.com/gifs/ragebait-monkey

[light tangle]

Fair enough, thank you for the explanation 😄

[valid topaz]

np

[undone pumice]

<@&268886789983436800>

[bronze geyser]

mods sleeping

<@&268886789983436800> oi oi

crypto scam

[plush violet]

Theres no need to ping a second time

[bronze geyser]

oops sorry mod

[dry verge]

I finally found out why you have to change the matrix in a specific way before multiplying it by the determinant^-1.

a and d are swapped because they represent the horizontal and vertical shifts, so it's no different from swapping x and y. b and c are made negative because you're reversing the direction of the shears.

[reef star]

wait why do we have to swap x and y to inverse tho

[night socket]

in inverse functions?

[reef star]

in a matrix

[reef star]

based on this

[night socket]

oh ok

because a and d sort of act as the scaling factor for the x axis and y axis

think of it like x and y

x controls the x axis

and y controls the y axis

when we swap a and d

[reef star]

yea

[night socket]

you invert the transformation over the x and y axes

[reef star]

if we scale the x axis by x then by y its just scaled by xy no?

wait actually that makes sense

[night socket]

was that your question?

[reef star]

yes

ty

[night socket]

np

[round geyser]

Hey i was working on this

[red jewel]

I still cannot understand why are we increasing the degree

[tired sedge]

It looks like a Binomial expansion

[orchid urchin]

why is it to the 15th power?

[round geyser]

It was in my mind

I was bored so i wanted a 15th power or degree question

[orchid urchin]

makes sense, you know the formula for the rows for pascal triangle?

[round geyser]

Yea

[round geyser]

I got this x^4 power

[orchid urchin]

what program u use to draw

[round geyser]

Apple screenshot

And i can edit

[orchid urchin]

try concepts drawing software, it has a dark mode

[round geyser]

Ok

[orchid urchin]

there is an app too

if ur not pc

for each drawing you make you have an infinite canvas

[round geyser]

I’m on iPhone

[orchid urchin]

concepts is free

[round geyser]

K

[round geyser]

I’m on app and don’t see that app

Is there any other app

[orchid urchin]

send pic

this the icon and is titled "Concepts"

[orchid urchin]

oh remove the 'drawing software' part

[orchid urchin]

u est time i see

[round geyser]

I see it now

[orchid urchin]

what u think of it?

[round geyser]

I like it

[orchid urchin]

replaces apple screenshot?

[round geyser]

Sure

[orchid urchin]

thats a w

with perspective grids and shading u can make good 3d stuff if you wanted too

I mainly just use it for when I gotta do equations and I don't want to write on anything

or model specific things

[round geyser]

Okay

[iron egret]

guys quick question how do u say mew

the sign that looks like a u

as in when ur typing it

is it mew or meu

[orchid urchin]

u mean mu

the greek letter

[iron egret]

yea

[orchid urchin]

just say it like it is

mu

[iron egret]

ok

[sage trellis]

wait i say mew but spell it as mu

its pronounced like the pokemon i think

[reef star]

yea its pronounced mew not moo

[round geyser]

Find x,y,z,u,v

[raw hill]

add the equations

[tired sedge]

I think this is fairly ez

U just

Do it simultaneously

[round geyser]

ngl it’s easy

[tired sedge]

Over and over

And over

[tired sedge]

Ye

[round geyser]

I do from top to bottom

[tired sedge]

Same

Solving it rn

[round geyser]

z=1-x+y
y-(1-x+y)+u=2
y-1+x-y+u=2
-1+x+u=2
x+u=3

[tired sedge]

Yessir

U just

Substitute

Now

[round geyser]

u=3-x
(1-x+y)-(3-x)+v=5
1-x+y-3+x+v=5
y+v=7

[round geyser]

I notice that
x-y+z=1 that x,y,z has to be zero

[raw hill]

||Me when I add all the equations to get x+y+z+u+v=17. Then add two at a time to get x+u=3, y+v=7, z+x=7, u+y=9. So z=17-(x+u)-(y+v)=7. Thus x=0, u=3, y=6, v=1 as well.||

[exotic barn]

Yes that's you def

[raw hill]

Yeah I do not miss the rapid-fire testing where all your effort was condensed onto a one-page answer sheet

I'm still mad for you making me figure out Wheatstone on my own in exam grrrrrr

[round geyser]

You did was you add x+u=3 into second equation

With z-u+v=5

[raw hill]

I did this substitution ig

[raw hill]

I never dealt with anything like this

[solemn peak]

can someone help me with an arithmetic series question?

[tender quest]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[solemn peak]

Can I ask about this question?

[raw hill]

ah formula bashing my favorite...

Do you remember the formula for the sum of an arithmetic series

[solemn peak]

yeah

it was sn = n/2 (t1 + tn) or (2t1 + tn)

[raw hill]

$s_n=\frac{n}{2} (t_1+t_n)$, this follows from a pairing argument (pair the first and last terms, the second and second-to-last terms, etc.)

[obsidian monolith]

Civil Service Pigeon

[raw hill]

The reason I bring this up is because you're given two pieces of information about the sum of the series, so you can apply this to both of those pieces of information

[solemn peak]

alr

we get the sum of up to the 3rd term and 8th term correct?

[raw hill]

correct

[solemn peak]

where would I go from here, I've just been trying to plug in the variables

[raw hill]

Show me what you have so far and explain what specifically is tripping you up

[solemn peak]

alr

[raw hill]

oh you used the more optimal $S_n=\frac{n}{2}(2t_1+(n-1)d)$ that's good

[obsidian monolith]

Civil Service Pigeon

[solemn peak]

yeah

[raw hill]

this is fine

[solemn peak]

got it

[raw hill]

notice how there's two variables in one equation and no other conditions

so that's probably why you're having issues

but like we've discussed, you also have info about the sum when n=8

so try to deduce something similar from that

[solemn peak]

so plug it into the same formula?

[raw hill]

Yes, you can use $S_n=\frac{n}{2}(2t_1+(n-1)d)$ again (since it holds for \textit{any} number of terms).

[solemn peak]

ok thank you

[obsidian monolith]

Civil Service Pigeon

[raw hill]

[solemn peak]

I'm stuck again

this part

[raw hill]

transcription error

[solemn peak]

oh man

didn't realize that sorry

[raw hill]

ur good

[solemn peak]

I've got up to here, I don't know where to go again

[raw hill]

Remember you also have $-1=t_1+d$ from earlier

[obsidian monolith]

Civil Service Pigeon

[solemn peak]

oh, so 2t1 would be -1?

[raw hill]

why?

[solemn peak]

I made a mistake

I confused it for

t1 + t1, its t1 multiplied by t1

[raw hill]

huh

what is "it"

[solemn peak]

im referring to 2 t1

I thought I would plug in the -1 from earlier into there

[raw hill]

So you're saying $2t_1$ is the same as multiplying $t_1$ by $t_1$?

[obsidian monolith]

Civil Service Pigeon

[solemn peak]

yes

nvm thats wrong

[raw hill]

yeah I was gonna say lol

[raw hill]

but I'm seeing some kind of system of equation idea here, which is good

[solemn peak]

I don't know what to do with the -1 from earlier though

It tells me what term 2 is

[raw hill]

All you have a system of two equations in two variables, namely $t_1$ and $d$

[obsidian monolith]

Civil Service Pigeon

[raw hill]

so you can solve it as such

Ex. (This is not the only way to do this) You can say that
$$-1=t_1+d \implies t_1=-1-d,$$
and substitute that into $33=2t_1+7d$ to obtain
$$33=2(-1-d)+7d.$$

[obsidian monolith]

Civil Service Pigeon

[solemn peak]

okay

[raw hill]

[solemn peak]

from that point on when you get 33 = 2(-1 - d) + 7d, you would solve for the difference now?

[raw hill]

yeah you can solve for d from that

[solemn peak]

okay I'll do that

[raw hill]

because now the equation is only one variable

that's the point of solving a system of equations by substitution

[solemn peak]

alright

I got the answer man, thank you so much

[raw hill]

Does that mean you can finish from here?

[solemn peak]

Yeah

[raw hill]

[solemn peak]

much appreciated

[hushed sphinx]

Another way to get this result with barely any formula bashing:
The sum of an arithmetic sequence with an odd number of terms is the middle term times the number of terms. So from t1+t2+t3 = -3 we know 3·t2 = -3, so t2 = -1.
Similarly, t4+t5+t6+t7+t8 = 132-(-3) = 135, so t6 = 135/5 = 27.
The difference between t2 and t6 is 4d, but we now know it is also 27-(-1) = 18, so d = 28/4 = 7.

[raw hill]

(Note this is the pairing argument I alluded to up here)

[hushed sphinx]

They're equivalent when you squint enough, but writing it that way gives us two relations between t1, t3, and t8 which take more work to get useful information out of than focusing on the single term in the middle of each range.

[solemn peak]

This is very useful

I don't have a good grasp on how it works yet though

[hushed sphinx]

Here's a sketch for five terms that might be enlightening if you can stomach the layout:

t4+t5+t6+t7+t8 =

      t6               t6                 t6
=  +t5  +t7  =  +(t6-d)   +(t6+d)   =  +t6  +t6  -d+d    =  5×t6
   +t4  +t8     +(t6-2d)  +(t6+2d)     +t6  +t6 -2d+2d

[solemn peak]

The number of terms times the middle term = the sum of an odd number of terms?

[solemn peak]

ok

[solemn peak]

I think I understood the concept wrong here

[hushed sphinx]

Or perhaps it's just my weird ASCII art rendering that's more confusing than enlightening. If so, apologies.

[solemn peak]

No it's completely fine, I'm just trying to grasp what = what x what right for the concept you showed

[hushed sphinx]

In t4+t5+t6+t7+t8, the number of terms is 5, and the middle term is t6, so t4+t5+t6+t7+t8 = 5·t6.
(Note that this works only when t4,t5,t6,t7,t8 is an arithmetic progression!)

[obsidian monolith]

Civil Service Pigeon

[raw hill]

(ok I'll fuck off now lol)

[hushed sphinx]

Uh yeah, that's definitely not how it's supposed to look ...

[solemn peak]

I'm having trouble on the very next question, is this a mistake?

already gave me the answer

[undone pumice]

Wym

[solemn peak]

I calculated n = 10, but it already gave me the answer of 10 in the question

[undone pumice]

10 is different in the question

I thought

Cus it’s just saying each section has 10 rows

But it’s asking for the number of sections

Unless I’m reading the problem wrong

[ancient inlet]

isnt this the sum of a Arithmetic progression?

[solemn peak]

I honestly don't even know

I interpreted it as asking for the number of terms which was the rows I think

[tame shard]

It is i think, each section gets 6$ cheaper and by the definition of arithmetic progression

[solemn peak]

alright

[undone pumice]

idk i interpreted it as like each term is each section since its one seat per section but lowk whoever wrote this problem uhhh 😭

[turbid saddle]

Tisk Tisk Tisk

[round ermine]

it says number of rows
my assumption is that you find the amount of sections
since each section’s ticket is $6 cheaper than the previous section, you use the given info to find the # of sections
then multiply that by 10 to get the total amount of rows in the concert

[undone pumice]

Oh ye

[dry yoke]

if n!! means 1×3×5×7×9×11...×n
would n!!! mean
1×4×7×10×13...n
?

[tired sedge]

Uhm. I don't think so

I've never seen multiple factorials before

The "!!!"

[tired sedge]

I think IF multiple factorials were a thing, it would be smth like

n!!=(n!)x! Like 5!!= 120!

[simple bramble]

Found this, don't know how reliable it is tho

[tired sedge]

Oh

Nvm

[dry yoke]

tgis formula below suggests 20!! as 20×18×16×...2
instead of
19×17×15×13...1

[simple bramble]

Isn't that what double factorial means, like it's the multiplication of the same type of numbers less than or equal to(odd or even) the one we're doing the double factorial off?

[dry yoke]

nah iirc !! multiply odd numbers

so even!! = (odd before it)!!

[hushed sphinx]

That's not what the notation usually means.

[dry yoke]

troposphere is typing

[hushed sphinx]

See e.g. https://en.wikipedia.org/wiki/Double_factorial.

[dry yoke]

ah so i'm wrong then

[red jewel]

Just why

Like why do we have that

[dry yoke]

multidimensional hypersphere hypervolume

uses !!

[dry yoke]

but however this thing iirc only uses the odd terms

as in odd!!

4d hypersphere and 5d hypersphere shares same non-pi denominator

[dry yoke]

eh, wrong again

[hushed sphinx]

I'm not exactly sure; intuitively it does look like the odd and even values don't mesh particularly well. But the convention is what it is.

[red jewel]

I wonder, do they teach archimedes parabola formula in uni?

Or american high schools?

It is extremely quick and simple

[hushed sphinx]

What is that?

[red jewel]

Basically,

You put a triangle

For "outsider part"

You do 2/3

4/3 for inner side

Works on every parabola

(yeah the guy proved this using exhaustation method in ~200BCE)

[hushed sphinx]

What is the purpose it works for?

[red jewel]

Uhh, parabolas in integral

[hushed sphinx]

I think what is usually taught is just a general power rule. It's relatively rare to need to integrate a homogeneous parabola like that, relative how often you encounter other polynomials.

[tired sedge]

Ye idk what this is

Tbh

Tryna understand it

[hushed sphinx]

Well, it's not something that belongs in PREcalculus, anyway.

[red jewel]

So for example, 4 - x² parabola
Instead of doing (from -2 to 2) ∫ 4-x² dx = 4x - x³/3 => 8-8/3 - (-8 +8/3)

You do 4 × 4/2 × 4/3 = 32/3

[tired sedge]

Ye I do this usually

[tired sedge]

Ohhh

That's neat ig

Don't know how it can be useful when u alr have power rule

Tho

:/

[red jewel]

Speed

And efficiency

[tired sedge]

It only works for quadratics right?

[red jewel]

Also y = √x

[tired sedge]

Oh

Noice

But like

I feel power rule is way, way more general

[red jewel]

Basically anything that is parabola shaped

[tired sedge]

I wouldn't mind taking like 10 seconds longer to integrate smth rather than learning a new rule

[tired sedge]

Ye

[red jewel]

Well i try to use Simpsons in 3rd anyway

They don't even ask 3rd here

Usually 2

Well At least i know how to prove that one... (Hopefully)

[fossil kindle]

w precalc

[round geyser]

Ik power rule in log

And exponents

[dry verge]

Programming a bot that solves systems of equations (two variables and two equations only) is really funny now that I can solve Algebra 1 worksheets I’ve found online with my own calculator.

All I had to do was translate multiplying the matrix by the B vector into Python and it was good.

[round geyser]

Do this for me

[dry verge]

I cannot translate that to the bot because x and y are multiplied in the second equation.

Also, if it’s homework, I can’t solve it.

[round geyser]

It’s not hw

It’s just a random question

[round geyser]

Here I’ll remake it

xy=10 -) 5x+10y=10

Now will u do it

x-y=3
5x+10y=10

[dry verge]

Got it, hold on.

[round geyser]

Thanks

[dry verge]

Is this correct?

Let me graph it to make sure.

Yep, outside of that 5 and 7 (which I have no idea where they came from), this is correct!

[tame shard]

Use vietas formula i think

[tame shard]

Rounding

[round geyser]

Yes

It’s correct

[round geyser]

I gave him an example problem to test the bot she created

[tame shard]

her*

[round geyser]

Oh

I see

[dry verge]

That said, I love how this was given because every programmer has a nightmare like this.

[round geyser]

Thanks

[dry verge]

No, thank you! /gen

Because I know someone will try to do that and break the bot.

[round geyser]

Yw

Can you check this out

Btw 0,1=0.1

Different country has its own way to write decimals

[tame shard]

Substitute y = 3-x

[round geyser]

Okay

y=3-x
x(3-x)=10
3x-x^2=10
-x^2+3x-10=0
x^2-3x-10=0
(x-5)(x+2)=0
y=3-5=-2
(5,-2)
y=3-(-2)=5
(-2,-5)

I got this

[round geyser]

^

[tame shard]

Cool

[round geyser]

Thanks

[round geyser]

Is this correct, larp

log_a a^x = x

Is the rule i used

[tired sedge]

log base 0.1?

thats intersting

didnt know logs can have fractional bases

[round geyser]

Yep

Also yes

[tame shard]

Logs can have any real base

[tired sedge]

even negative?

[tame shard]

(Nonnegative)

[tired sedge]

im pretty sure

they cant

[tired sedge]

ohh

[tame shard]

Wait a minute

[tired sedge]

can logs have complex bases?

[tame shard]

Idk

But log base 0 doesnt exist for sure, as 0^anything = 0 (besides 0^0,which is either undefined or 1)

[tired sedge]

same with 1

btw

log base 1

is undefined

[round geyser]

Right

y>1

Is log’s range

log x
D: All real numbers
R: y>1

[tame shard]

Yea, since log_a(b) = ln(b)/ln(a) , log_a(1) is always equal to 0 for any a

[round geyser]

Okay

[tired sedge]

logs having complex bases would be cool tho

icl

[round geyser]

Can we have a negative log

[tired sedge]

ye

[tired sedge]

?

no

doesnt make sense

[round geyser]

Like log -x

[tame shard]

That would require complex numbers

Which is another whole rabbit hole

[round geyser]

Ik - log x

I graphed that before

What’s their range now

[round geyser]

y<1?

[tired sedge]

same as range of log

x

but

negative

[round geyser]

y>1

[tame shard]

log(y) for y<0 would require complex values

[round geyser]

Okay

I like this one

[tired sedge]

can logs have log bases?

like log base log7

?

lol

log base ln

would be lowk cool

[round geyser]

y= log (θ)

I like this one

[tired sedge]

thats just the same as 10^y=theta right?

[tame shard]

Logs can have any constant base(that isnt 0 or 1)/ variable base

[round geyser]

yea

[tired sedge]

ooo

[round geyser]

Yep

[round geyser]

Larp is y=log (θ) valid

[tame shard]

Depends on wtf is theta

Cuz idk

[tired sedge]

an angle?

maybe

[tame shard]

I barely know logs

[tired sedge]

idk

[round geyser]

Yea theta is for angles

Like y=sin (θ)

[tame shard]

I mean there is nothing wrong with taking the logarithm of the measure of an angle, but idk if that would even be meaningful

[round geyser]

I agree

[tired sedge]

exactly

[round geyser]

sin (θ)= o/h

I did from triangle ratios to graphing

In pre-calc

[tired sedge]

does this mean that arcsin theta is h/o

?

[round geyser]

Yea

[round geyser]

I call arcsin as csc

[tired sedge]

csc is normally denoted

for

cosec

sooo

[round geyser]

Or cosecant

[tired sedge]

ye

[round geyser]

I graphed that too

I graphed sine, cosine, cosecant, secant, tangent, and cotangent

[tired sedge]

noice

[narrow vine]

i just tried to argue with my friend about her formula being wrong because it gave that the distance of two lines that intercept is 0

its 8 days till my uni acces exams

[tired sedge]

thorfinn pfp

🥹

based

[tired sedge]

?

😭

[red jewel]

Usually written in radians

[round geyser]

Like what’s log(pi/2) like that

[red jewel]

Yeah

[round geyser]

It did work with log(pi/2)

I got 0.196

[red jewel]

Because

[round geyser]

Interesting interesting

[red jewel]

Log(π/2) = x,
10^x = π/2
10^x ≈ 1,67 I guess?

Etc

[round geyser]

What country are u in

Bc I write it like
10^x = 1**.**67

In the U.S.

[tired sedge]

i think in eruope its with commas

and everywhere else

its periods

[hushed sphinx]

It's generally a per-language convention, not per-continent.

[red jewel]

Well whatever. Same thing

[tired sedge]

oh

ig so

[narrow vine]

yeah

[round geyser]

Cool

That’s what i like

[valid topaz]

use difference of squares for the numerator and difference of cubes for the denominator

[round geyser]

lim x-> 2 ((x^2-4)(x^2+4)/(x-2)(x^2+2x+4)
lim x-> 2 ((x+2)(x-2)(x^2+4))/(x-2)(x^2+2x+4)
lim x-> 2 ((x+2)(x^2+4))/(x^2+2x+4)
(2+2)((2)^2+4)/((2)^2+2(2)+4)

[valid topaz]

ye

[round geyser]

8/3

[round geyser]

If i were graphing this there will be a hole at (2,8/3)

Am i right

[echo sinew]

Yes, you are

[round geyser]

Also

It’s not like the usual rational function

It’s just a straight line that touches (-2,0) and (0,2)

[robust junco]

what was lim again

[round geyser]

limit

limit as x approaches 2

[echo sinew]

Huh? No, it isn't a straight line

[round geyser]

What’s it like then

[echo sinew]

,w plot y = (x^4−16)/(x^3−8)

[obsidian monolith]

Results provided by WolframAlpha

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

Requested by zuanon.alberto

[red jewel]

What

[red jewel]

4x³/3x²

Oh

[round geyser]

L hospital rule

[undone pumice]

r u trying to like lhopital?

[red jewel]

No i kinda thought f(2) would be unknown as it goes 0/0

I know we do L'hopital on 0/0 and limit but

[undone pumice]

well its 0/0 cus its a hole

[red jewel]

This is the first time I see graphic of 0/0 thats why

[undone pumice]

oh

ic

[winter osprey]

Omg bruh I found my unit circle from like 5-6 yrs ago

[sage trellis]

unit circle never goes away

[winter osprey]

I brought it with me to college but forgot about it ngl

Just found it while cleaning my shit before moving back home

[red jewel]

I honestly need to digitalize 3 of my notebooks

Before I go uni

But that will take a Lot of effort

[winter osprey]

I still handwrite all my things

Hws, notes, etc

[sage trellis]

personally i cannot use digital notes

[red jewel]

Why not

[red jewel]

I mean thats what i'm gonna do

[sage trellis]

idk ive tried with like an ipad

[red jewel]

Probably rip papers one by one and make them a pdf

[sage trellis]

but i feel like paper just makes it click better ig

[winter osprey]

You can use LaTeX if u know how to to pdf it, it's what im doing rn for my qualifying exam practices

Makes it look nice and you can make a table of contents to see what u need asap, etc

[red jewel]

But I'm horrible on uhhh

Digital screens

[sage trellis]

why do u need them digitized

[red jewel]

Well

Easier to carry

Easier to find

As the topics are scattered

Also there are 3 notebooks

That has physics, math, and chem

So

[sage trellis]

o

[raw hill]

Surely this is just a recall exercise?

[round geyser]

I got A=$7444.31

[raw hill]

how is the balance smaller than what you started with

[round geyser]

I made a desmo built in

I putted P=5000, r=0.04, n=4, and t=10

[round geyser]

^

[raw hill]

that doesn't answer my question

[round geyser]

I read it wrong

[raw hill]

rip

[round geyser]

Actually that’s my ans too

[raw hill]

[round geyser]

I like Desmos

Bc i can built in any formulas i wanted

[round geyser]

Is it true that ab=(a)(b)=a•b

[raw hill]

Assuming that you're not using a dot product or anything like that, sure ig

[round geyser]

i got this kind of question

[round geyser]

Is it log 3 + 3 log 2

Since I’m expanding the log

[raw hill]

Yes

[round geyser]

Thanks

[raw hill]

[round geyser]

I’d seen questions where x is in the base of log

[raw hill]

that's fine

[round geyser]

Like this

[uneven matrix]

you can use change of base formula

[round geyser]

log_?

[uneven matrix]

$\log_b(a)=\frac{\log(a)}{\log(b)}$

[obsidian monolith]

Dirac Spinor

[uneven matrix]

Then, $\log_{x+3}(x^2+3)=2 \implies \frac{\ln(x^2+3)}{\ln(x+3)}=2$

[obsidian monolith]

Dirac Spinor

[round geyser]

Then cross multiply

[uneven matrix]

yup

then what?

$ln(x^2+3)=2ln(x+3)$

[round geyser]

ln(x^2+3)=2 ln (x+3)

[obsidian monolith]

Dirac Spinor

[uneven matrix]

yup

[round geyser]

$ln(x^2+3)=ln(x+3)^2$

[obsidian monolith]

Jake

[uneven matrix]

yup

[round geyser]

$(x^2+3)=(x+3)^2$ ->
$x^2+3=x^2+6x+9$ ->
$3=3(2x+3)$ ->
$1=2x+3$ ->
$-2=2x$ ->
$x=-1$

[obsidian monolith]

Jake

[round geyser]

^

[uneven matrix]

got it

[round geyser]

Like that my third types of problems i like to solve

First one is solving (ax+b)^n=(cx+d)^n and the Pythagorean theorem by polynomial

[barren horizon]

$log_b$

[uneven matrix]

another way to solve is to just ask "x+3 raised to what power gives x^2+3" then the answer to that would be 2, so then you would have $(x+3)^2=x^2+3$

[obsidian monolith]

emiya

[barren horizon]

What does log_b mean

[round geyser]

log base of b

[round geyser]

^

[vale swift]

who wants to find my limit

[brave briar]

How would one solve this?

[round geyser]

Me

[raw hill]

I'd start by factoring out the constants:
$$\sum^{4}{b=2} (a^2b-ab)=\sum^{5}{c=3} (ac+6) \implies (a^2-a) \sum^{4}{b=2} b=a 6 \sum^{5}{c=3} c+\sum^{5}_{c=3} 1,$$
which makes computing the sums more straightforward. And now you just have a standard quadratic.

[obsidian monolith]

Civil Service Pigeon

[last stream]

ln(x^2+3)=2 ln (x+3)

[sage trellis]

2ln(x+3) = ln((x+3)^2)

[sage trellis]

wait u need help doing this right

[last stream]

yes

[sage trellis]

ok yeah

[sage trellis]

so you can sub this into the rhs

ln(x^2 + 3) = ln((x+3)^2)

[willow bear]

so this is a thing now

i guess things like trig belong here

[faint acorn]

nice

[ocean cove]

woohoo

[viscid thistle]

ye !

[main flume]

Sixth

[calm thicket]

Noice

[viscid thistle]

last

hey

http://prntscr.com/gra2as

i'm confused on the equation of this line

y=4 x= -8?

[willow bear]

y = 4, x = -8 describes a single point

which isn't even on your line

[viscid thistle]

how do I write the equation?

y=mx+b

[willow bear]

you're given two points on the line

can you find the line's slope?

@viscid thistle

[viscid thistle]

-12?

[willow bear]

how did you get that?

[viscid thistle]

i counted from -8 to 4

[willow bear]

do you know how to find the slope of a line given two points on it?

[viscid thistle]

i thought i just did

lol

😣

[willow bear]

given two points (x1, y1) and (x2, y2), what is the slope of the line passing through them?

=tex m = \frac{y_2 - y_1}{x_2 - x_1}

[willow bear]

does this look familiar?

@viscid thistle

[viscid thistle]

i'm trying to figure it out but honestly i'm lost

[willow bear]

figure what out?

[viscid thistle]

i thought the slope was the difference

[willow bear]

it's not just a difference

[viscid thistle]

the rise

and the run

ohhhh

[willow bear]

slope = rise/run, by definition

[viscid thistle]

-12 and 4.5?

5*

[willow bear]

those are the rise and run respectively, yes

so