precalculus

[wary plover]

itll just be a bunch of lines going up to down with asymptotes?

[echo plaza]

no

[rocky bison]

[wary plover]

I don't know what "we" define this stuff as all I know is what i was taught

[rocky bison]

I'm literally showing you it

[wary plover]

I've never seen that graph before in my life, I have no context to judge

I'm thinking of this problems in the terms I was taught, nothing more or less. Clearly I need a more dynamic perspective

[rocky bison]

It's just the graph of arccos

cos(x)

goes on forever

of to infinity in both directions

arccos(x) does not however

[wary plover]

so what is the functional interval of that on the circle? I assume the top half?

and then itll just repeat the same graph?

for any other bit

because it doesnt matter which way it's being looked at, it returns the same answers

[rocky bison]

,$ -1\leq x\leq1\0\leq y\leq\pi

[obsidian monolith]

Pseudo:

[wary plover]

ok good

so what would the arctan be defined as

[rocky bison]

Same idea

But that will converge to two asymptotes

,w plot \arctan\left(x\right)

[obsidian monolith]

Display results online and refine query

Requested by Pseudo#2703

[wary plover]

so I interpret that on the left-hand interval?

like arcsin?

or

wait

hm

would that be the left or right of the circle?

I guess right

[echo plaza]

yeah

[wary plover]

only way it would make snese

cause it has to be 0

[kind pier]

$ \frac{2x+1}{x-5} < 3 $

[obsidian monolith]

Emil:

[kind pier]

if you have an equation that says to solve that would you just list the domain?

(-infin,5)U(16,infin) or am i suppose to just put one answer

[hard hornet]

solve for all possible x values, so list the domain

[fiery spruce]

How to do 6

[fading token]

Write y = x/(x + 1) and express x in terms of y

[rigid beacon]

Switch y and x, solve for y, die because algebra is hard

[fiery spruce]

So is it y = xy+x

[rigid beacon]

Not quite

Get that to be y in terms of x

So y = f(x)

[fading token]

No, x in terms of y lol

[rigid beacon]

Oh ye ye

Y = f(x) is y in terms of x yea my bad

[fading token]

x = g(y) where then g = f^(-1)

[kind pier]

$ x^5 - 3x^.25 - 4 = 0 $

[obsidian monolith]

Emil:

[willow bear]

$x^5 - 3x^{1/4} - 4 = 0$

[obsidian monolith]

Ann:

[willow bear]

did you mean this

[kind pier]

x to the 1/2

[willow bear]

1/2 isn't 0.25

[kind pier]

the first term

it was suppose to be .5

[willow bear]

see this is why you should never, EVER write 1/2 as .5

[kind pier]

Now I know how fractions work with the bot

[willow bear]

0.5 is ok if you want a decimal so badly

[kind pier]

lol I was only gonna do it to make it easier to do with the bot

I dont

[willow bear]

but not .5

that's just ugly

[kind pier]

ya 1/2 is fine I didnt know the command for fracations

[willow bear]

$x^{ \frac{1}{2} } - 3x^{ \frac{1}{4} } - 4 = 0$

[obsidian monolith]

Ann:

[willow bear]

\frac{num}{denom}

anyway ok what have you tried so far and where are you stuck

[kind pier]

can I use u sub there? x^1/2 = u^2 and x^1/4 = u?

[willow bear]

you can ALWAYS make a substitution as long as you clearly define it

whether it turns out to be useful is another question

[kind pier]

I thought there was a rule about the powers having to have something specific in common

[willow bear]

in this case, the substitution you suggest does turn out useful

[kind pier]

ok then the answer is 4^4 and 1

thanks

[fiery spruce]

Can someone run me through the steps of a and b

[patent beacon]

Take the derivative, check where its signs are positive and negative

Also find the zeroes of the derivative

[fiery spruce]

What do you mean by where its signs are positive and negative

[patent beacon]

f'(x) > 0 implies the function is increasing
f'(x) < 0 implies the function is decreasing

[fiery spruce]

So how would you get the relative extrema?

Its something like plugging x into the second derivative of the function, just dont know what x is

[willow bear]

no, the relative extrema are where the first derivative is zero and the second isn't

[fiery spruce]

Ohh ok

[willow bear]

also idk why this is in precalc and not #calculus lol

[viscid thistle]

how do i show

[hard hornet]

One way to show

is to just plug it in

[timber plinth]

smh

[willow bear]

yeah

[hard hornet]

$(\frac{\sqrt{3}}{2} + \frac{i}{2})^6 + 1 = 0$

[obsidian monolith]

MemesPlease:

[hard hornet]

is my latex gud enugh

ah yes

Though, you will have an easier time if you use the format $rcis(\theta)$ instead of what I posted up there

[obsidian monolith]

MemesPlease:

[wise kelp]

@viscid thistle I think you can do it using a^2-b^2=(a+b)(a-b)

Although that requires calculating the cube root of i

Which does seem to be sqrt(3)/2 + (1/2)i but I’m not sure how you find it without knowing what it is it beforehand

Lol why the reactions

[hard hornet]

That's some mega brain solution you got there

lolololol

[wise kelp]

Lol that’s just the first method that came to mind

Only because it says the solution they’re looking for is a complex number

How do you calculate the cube root of i btw? If anyone knows

[daring gale]

with the formula

probs still taylor series but i hate analysis so i don't know

[echo plaza]

just oil it

[wise kelp]

I suppose you could do it like this?

[hard hornet]

if thats the case, dont you also have to show that (a+bi)^3 = -i too?

[wise kelp]

What do you mean? @hard hornet

[hard hornet]

There are 6 solutions, and you only got 3

But it doesnt rly matter

Cuz it only asks you to show that root3/2 + i/2 is a solution

As easy as just plugging it in and showing its 0

[wary plover]

https://i.gyazo.com/870160b48e9f4461affb4197aeeb2844.png

what is the fastest way of determining this answer?

[hard hornet]

Write equation, foil

[wise kelp]

There are 6?

Wolfram Alpha only gives 3

[wary plover]

@hard hornet so reverse it? make the equation myself?

[hard hornet]

There is a super fast way, but i forgot what it was called, dont think youll need that

[wise kelp]

You know the zeroes

[wary plover]

I can definitely solve this problem easily, but the problem is I have only a minute per and this is more time confusing than most of the other ones

[wise kelp]

So you can write the equation as k(x-2)(x+1)(x+5)

Where k is some constant

Since the x’s are just going to multiply to x^3, k is also the leading coefficient

[wary plover]

so there's no faster than than foiling it out?

[wise kelp]

Can you use a calculator?

[wary plover]

no

[wise kelp]

You could just figure out what the x^2 term would be

Since that’s different for all the answers

[wary plover]

-8x^2

I believe

alright that should be fast enough with the time I save from the ones like graph identification

thanks

[wise kelp]

Np

[hard hornet]

(x-2)(x+1)(x+5)

equals 0

And then foil/distribute

Because each of those roots, 2,-1,-5 make that equation true

[wise kelp]

I just said this lol

[hard hornet]

There is a faster way

That involves i forgot wtf his name was

[viscid thistle]

$x^3+4x^2-7x-10$ foiling

[obsidian monolith]

GABOS:

[hard hornet]

a+b+c, ab+bc+ac, abc

I forgot what the name of theorem was

But thats all you gotta do

[wise kelp]

What’s the wolfram alpha command

[viscid thistle]

Hmm

elaborate MemesPlease

you use ,w

[wise kelp]

Ah

There are four terms though

[hard hornet]

x^3 - (a+b+c)x^2 + (ab + bc + ac)x - abc

[viscid thistle]

i mean

[hard hornet]

I FORGOT THE NAME OF THE THEOREM

FUCK

[wise kelp]

Oh right

[viscid thistle]

hm

[hard hornet]

But dont need to worry about it

[wise kelp]

,w cube roots of i

[hard hornet]

Just write it out, n foul

[obsidian monolith]

Display results online and refine query

Requested by GrandPiano#1855

[hard hornet]

Foil*

[viscid thistle]

x^3+4x^2-7x-10 this is the foiled version

[hard hornet]

Grand, check the original original question

[spring thunder]

@hard hornet viete's formulas

[hard hornet]

x^6+1=0

AHH YES

Thank you

[viscid thistle]

viete

or was it a bit different?

viett

[hard hornet]

I knew it as vieta's formula

[wise kelp]

Oh yeah

[hard hornet]

Powerful formula

[viscid thistle]

https://en.wikipedia.org/wiki/Vieta's_formulas

[hard hornet]

But its probably a bit too advanced for this problem

[wise kelp]

I’m just talking about cbrt(i) though

[hard hornet]

Overkill imo

[spring thunder]

yea it's just that he's french, and his french name is viete

so watever

i say it like that

[viscid thistle]

I do remember our high school teacher teaching us this

[hard hornet]

Icic

[viscid thistle]

but i never really got it

[wise kelp]

Because using a^2-b^2=(a+b)(a-b) for that problem requires taking the cube root of i

[viscid thistle]

this one

so

ab + ac + bc

[hard hornet]

Were you trying to solve x^6 + 1 = 0 @wise kelp

If not, then im mistaken

[wise kelp]

I was trying to solve the problem

Which only asks about a specific solution

[hard hornet]

You could just plug it in lmao

I thought you were trying to solve for all 6

[timber plinth]

just plug in to see if x makes it 0 smh

[hard hornet]

Madlad

[wise kelp]

Yeah you could lol

[hard hornet]

You nuts bro

Hahahaha

[wise kelp]

I don’t think that’s what they want you to do though?

[viscid thistle]

anyone else here ever mark whole server as read just to see what channels are active

[timber plinth]

@wise kelp it is

[wise kelp]

Oh really

That’s dumb

[hard hornet]

You scared me when you were trying to solve for all of em lololol

[wise kelp]

Well anyway I was trying to find a way to solve it without doing that

[hard hornet]

Damn you make things hard for yourself

[timber plinth]

just pull out the bring radical smh

[viscid thistle]

HALELUJA BROTHAS

CAN I GET AN AMEN

[hard hornet]

Guess n check

[timber plinth]

kek

[wise kelp]

Or rather I saw that it could be rewritten as an a^2-b^2 and wanted to see if it could be solved that way

On the bright side, I figured out a way to find the cube roots of an imaginary number using algebra

@timber plinth Bring radical?

[timber plinth]

a messy thing

it kinda breaks the rules

[hard hornet]

Use the reverse demoivres theorem instrad

I forgot the name too

[wise kelp]

Wait

You don’t have to make it a^2-b^2, you can just rearrange it to x^6=-1

[hard hornet]

Uhhh

Yea?

You somehow

Chose the hardest possible way

To solve for x

LOLOLOLOL

[wise kelp]

And then get x = cbrt(i)

Lol yeah

[hard hornet]

x=cbrt(-i)

Dont forget that too

[timber plinth]

nah, just get x = (-1)^1/6

ez

[wise kelp]

Oh yeah I guess so

[timber plinth]

,w (-1)^1/6

[hard hornet]

That too

[timber plinth]

since there's actually 6 solutions, not 3

[hard hornet]

Yuppis

[obsidian monolith]

Display results online and refine query

Requested by Darkrifts#3866

[wary plover]

https://i.gyazo.com/a02b07751336ceccf3b76cfee9644baa.png

not sure how to go about this one

[short sorrel]

theres a few different approaches

my first instinct would be to just expand both sides

and see what you can do from there

[wary plover]

multiple them out?

[short sorrel]

yes

[wary plover]

wont that make it all jumbled? I need to be able to solve this within 2 minutes

[short sorrel]

and you can

you should get $xy -6y + 5x - 30 = xy + 5y - 3x - 15$

[wary plover]

I guess I could factor them out after that

[obsidian monolith]

Namington:

[short sorrel]

note that theres an xy on both sides

so we can clean that up

moving everything to right side, we get

$0 = 11y - 8x + 15$

[obsidian monolith]

Namington:

[short sorrel]

which we can then solve for y

$11y = 8x - 15$

[obsidian monolith]

Namington:

[short sorrel]

er oops, we're solving for x

[wary plover]

hehe

[short sorrel]

$8x = 11y + 15$

[obsidian monolith]

Namington:

[short sorrel]

and divide by 8

[wary plover]

neat

I suppose that's a decent enough way, thanks

hopefully I remember

[wary plover]

https://i.gyazo.com/34d0d9bce752fea514964ea84d4b4ea0.png

knowing that the hor asymptote is 3, how do I tell whether it crosses that asymptote?

[timber plinth]

Well, a trivial way is to graph it

Another way is to examine the behavior near the more vertical asymptotes @wary plover

[wary plover]

so there's no quick fix?

[short sorrel]

you can check if theres any value of x such that $3 = \frac{3x^2 + 2 + 8}{x^2 + 10x - 9}$

[obsidian monolith]

Namington:

[short sorrel]

which gives a solution of x = 37/30

so it crosses (or at least touches) the asymptote there

to verify that it actually crosses, we can, like, test 37/30 + 0.0001 and 37/30 - 0.0001 or smthn

if one is >3 and the other <3, it indeed crosses

this isnt super rigorous but its good enough for elementary mostly-continuous functions

[wary plover]

and for that algebra I just have to multiply it out?

[short sorrel]

multiply both sides by the denominator, yeah

just graphing it is far more convenient, but like

the algebraic approach works

[wary plover]

I dont have the time to graph it

[viscid thistle]

I believe you have to do sign analysis using 5 points

[echo hornet]

hello

[viscid thistle]

Hey

[proud flax]

Boys could i get some help solving simultaneous equations with complex numbers?

@ me if you can please

[heady jewel]

@proud flaxdo i just solve and send you solutions

[proud flax]

That would be helpful thanks

[willow bear]

no

NO

don't do that\

[heady jewel]

lul

didd you try?

[proud flax]

Ahaah

[heady jewel]

atleast

[proud flax]

Yeah i tried expanding and equating complex and real parts but it didnt work

[willow bear]

that actually isn't the best route to go here

like, you can turn this into a system of four equations in four real variables

but it's too much of a hassle imo

bc this can be considered as a system of two linear equations just with complex coefficients

[proud flax]

It gave me values of w that didnt agree with others

So what can I do instead?

[willow bear]

can you show your work? i'm sure it's all down to an arithmetic fuckup

as is often the case with these

[willow bear]

ok yeah so like alright

your mistake is that you're assuming z and w are both real

[proud flax]

Ohhhh

[willow bear]

which you shouldn't be

[proud flax]

So how should I set it out?

[willow bear]

well i can show you how i would do it, if you want

[proud flax]

Yes please

[willow bear]

ok so let me just copy down the equations for ease of reference: $\begin{cases} (1+i)z + (2-i)w = 3 + 4i \ iz + (3+i)w = -1+5i \end{cases}$

[proud flax]

Typo in first equation but thats cool

[willow bear]

from the second equation, subtract $(3+i)w$ from both sides to get $$iz = (-3-i)w + (-1+5i)$$ and then multiply both sides by $-i$ to get $$z = (-1+3i)w + (5 + i)$$

oh yse sorry

[obsidian monolith]

Ann:

Ann:

[willow bear]

ofc, feel free to stop me at any point if there's something that isn't clear to you

[proud flax]

Why do i multiply by -i?

[willow bear]

it's the same as dividing by i

$i \cdot (-i) = 1$

[obsidian monolith]

Ann:

[proud flax]

Oh right

[willow bear]

ok

so then plug that into the first equation to get $$(1+i)[(-1+3i)w + (5+i)] + (2-i)w = 3+4i$$

[obsidian monolith]

Ann:

[willow bear]

expand that out to $$(1+i)(-1+3i)w + (1+i)(5+i) + (2-i)w = 3+4i$$

[obsidian monolith]

Ann:

[heady jewel]

she just solved it

good

[willow bear]

not before asking for work

[proud flax]

Mmk i think i should be able to get it from here, thanks!

[willow bear]

alright cool

[proud flax]

@willow bear i managed to get the right answer from there, thanks heaps 😃

[heady jewel]

nice

[pallid burrow]

need help,
A function ​f is a rule that assigns to each element ​x in a set ​A exactly one element ​f(x) in a set ​B.

What is ​A called? And ​B? What is the range?

A real function is a function defined on a subset of the real numbers and taking real values.

Which of the following statements is NOT correct?
the answer is B The range of ​f is ​B
I dont understand why

[frozen needle]

for example with $\fun f\bbR{[-4,2]}x{\sin x}$, the range of $f$ isn't $[-4,2]$

[obsidian monolith]

Tuong:

[frozen needle]

you won't be able to find $x$ in $\bbR$ such that $f(x)=-3$ for example

[obsidian monolith]

Tuong:

[pallid burrow]

Because even though the elements may be assigned a value in the set B, it does not necessarily take up every value in B?

[frozen needle]

Yep

[pallid burrow]

Also in the parabola x+(y-1)sqr=0, is it parabola inrespect to y? because if you solve for y its radical

how do you express the bottom half of that prabola equation with graphing?

with equations

how do i do this question

how do you even get this answer in b) for this question

[limber bone]

y = mx +b thats any linear function

for slope 2

y =2x+b thats a linear function with slope 2

thats a

b says find the equation such that f(2) = 1

so y = 2x+b

f(2) = 1

1=2(2)+b

4+b=1

b=-3

so y=2x-3

thats c

b or whatever

got it?

sketch several families is just ig choose some values for b

and sketch them

[pallid burrow]

what does y=mx+1-2m mean in geometric sense

okayso its just the expansion of slope form with m

generally how does m affect the position of the graph? like those reflection and 180 degree along some axis

[limber bone]

m is the slope

[pallid burrow]

by changing m

i remember something like multiplying by -2/3. how do i visualise the mapping of values

is it possible for this sort of rotation transformation of graph to be measured by change in m?

[gloomy shard]

Hey, a question about possibilities.
How do you solve the following:

Safe code consists of 3 different digits in ascending order. How many codes are possible?

[short sorrel]

Note that there's 3 different digits, so we're going to be selecting digits without replacement

So, we want to select 3 digits

But for each 3-digit combo

There's only 1 possible order

So we need to get rid of other orderings

Another way we can think of this: the unique orders don't matter

That makes this a combinations question

[gloomy shard]

1 possible order out of 3 digits

right....

[short sorrel]

How many different digits are there? From 0 to 9

[gloomy shard]

well it's a safe code, so it's 0-9 for each digit

3 digits in total

[short sorrel]

Right

Which means you have

10 options

And you're choosing 3

And need to get rid of duplicate orders

Can you take it from there?

[gloomy shard]

hopefully :D

[gloomy shard]

1 possible order out of 3 digits opened my eyes quite a bit, thanks @short sorrel

[short sorrel]

Right, it means that this acts as just a combination question

nCr(10,3)

[gloomy shard]

which means 120 :>

[short sorrel]

Right

[gloomy shard]

need to revise possibilities and combinations a lot :D

[short sorrel]

Another way to think of it:

10 options for the first digit × 9 for the second × 8 for the third

Then divide by 3! Different arrangements

You'll note this is the same as the formula for nCr

[gloomy shard]

that's what i thought at first but forgot about division

isn't it a permutation then? The order seems to matter in this case

Okay, that's a combination...

[short sorrel]

Permutations count each possible ordering

Combinations don't

@gloomy shard

We don't care about every possible ordering

We only care about 1: ascending order

[gloomy shard]

i see

[edgy forge]

in trigonometry, when people are given an angle in degrees, how are they so quick to point out what the coordinates of P are? for example, in this exercise, it seems they just expect you to know that the coordinates of P1 are (sqrt2)/2, (sqrt2)/2

[hard hornet]

$(rcos(\theta), rsin(\theta))$

[obsidian monolith]

MemesPlease:

[hard hornet]

If your radius is 1, and your angle is 45

in this case

your coordinate would become

$(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$

[obsidian monolith]

MemesPlease:

[edgy forge]

could you please explain how you came up with that answer

[hard hornet]

If you're given a radius of the unit circle

and the angle of the unit circle

then you can easily solve for any coordinate on the circle

given radius and angle

In your case, the radius of the circle is 1 (unit circle), and the angle is 45

using the formula i put above

you get the coordinates

$x = rcos(\theta), y = rsin(\theta)$

[obsidian monolith]

MemesPlease:

[edgy forge]

what i think i'm failing to understand is how to get cos and sin of theta

from radius and angle

[hard hornet]

do you know your trig functions?

[edgy forge]

i understand that x is cos, y is sin, tan is y/x, etc

[hard hornet]

oh so you're wondering how I got those two formulas above

[edgy forge]

i think so

[hard hornet]

mmm give me a second

let me pull up a picture

maybe this'll help you a bit

[edgy forge]

did you first make a triangle and compute the hypotenuse using s=r(theta)

ah yeah

[hard hornet]

think of cos as the x coordinate, and think of sin as the y coordinate

in reality

[edgy forge]

then used the pythagorean theorem?

[hard hornet]

$cos(\theta) = \frac{x}{r}$

[obsidian monolith]

MemesPlease:

[hard hornet]

$sin(\theta) = \frac{y}{r}$

agree?

[obsidian monolith]

MemesPlease:

[edgy forge]

agree

[hard hornet]

multiply both sides by r

and then you get the formula i posted above

the formula comes from drawing a triangle on the unit circle

and using trig ratio + algebra

to find your x + y coordinates

[edgy forge]

what i think i'm not understanding is how did you know cos(theta) was (sqrt2)/2

[hard hornet]

cos(theta) is NOT sqrt(2)/2

cos(45) IS sqrt(2)/2

You punch in the angle, and then evaluate the trig ratio

Oh how i knew cos(45) is sqrt(2)/2?

[edgy forge]

right

[hard hornet]

Lots and lots of practice

before that, lots of triangle drawing

and lots of ratio figuring out work

[edgy forge]

thanks for your help

[hard hornet]

If you were curious on all the common ratios

np

[kind pier]

How do you do greater than or equal to on the bot commands?

[spring thunder]

$\leq$

[obsidian monolith]

emeric75:

[spring thunder]

well $\geq$ wahtever

[obsidian monolith]

emeric75:

[kind pier]

ty

[wary plover]

https://i.gyazo.com/8e3a10b276853608c5926b2cb93a0fb4.png

how to go about this problem?

[hard hornet]

tbh I would probably distribute everything, combine all the terms, and see if you can factor something

Oh

I see something

Yes

distribute everything, and utilize pythag trig identity

Let me know how far you get

[rocky bison]

No spoilers @royal gull

[royal gull]

wut? How was that a spoiler? I guess the hardest part is to determine which ones are the correct solutions

after what I wrote before

[rocky bison]

Yes

[royal gull]

And what memesplease wrote won't really help without knowing that trick to use a substitution

[hard hornet]

You don't need a substitution

You can solve it just fine, without any substitutions

[royal gull]

how?

[hard hornet]

Il

I'll wait for Employee to do it, then I'll put my solution

[royal gull]

I mean, you cant just find delta with cos(x) as coefficient

and roots etc

but I guess you have a different method

[rocky bison]

You don't need a trick sub

Just expand

Then double angle

[hard hornet]

Yup

[wary plover]

https://i.gyazo.com/67d1942c314f6a3f53496eb1868b02ea.png

[rocky bison]

Yep

[hard hornet]

perfect

[rocky bison]

Now move the two square terms next to eachother 👀

[wary plover]

oh...

[rocky bison]

tell us what you see

keeping in mind

,$ \sin^2(x)+\cos^2(x)=1

[obsidian monolith]

Pseudo:

[wary plover]

yes yes I know

[rocky bison]

So what do you get from there?

[hard hornet]

what does your resulting equation look like

[royal gull]

oh wtf I read the parens wrongly lmfao

[wary plover]

60cos(x)sin(x) + 6

[rocky bison]

no 🤔

[hard hornet]

careful, check your math again

[wary plover]

6cos(x)sin(x) -30

[hard hornet]

60cos(x)sin(x) - 30

you forgot a 0

[wary plover]

mistype

[hard hornet]

now

[wary plover]

but yeah

[hard hornet]

$60cos(x)sin(x) - 30 = 0$

any idea on what to do next?

[obsidian monolith]

MemesPlease:

[hard hornet]

Hint: Use double angle formula

[wary plover]

yeah I know I had to look that one up

30(sin2x-1) ?

[rocky bison]

no

,$ \sin\left(2x\right)=2\sin\left(x\right)\cos\left(x\right)

[obsidian monolith]

Pseudo:

[wary plover]

Do you have to use variations of thonking face with every single message? It's really obnoxious and demeaning

[rocky bison]

That's your double angle formula

[hard hornet]

actually thats right

30(sin(2x) - 1) = 0

you can solve from here

[rocky bison]

wdym

Wait wot?

Oh lmao

Factored both thirties out

[hard hornet]

Yup

[rocky bison]

[hard hornet]

did something that confused me a bit too

but yes

[rocky bison]

I thought he was trying to us a reduction

[hard hornet]

$30(sin(2x) - 1)) = 0$

now, time to solve for x, do you think you can solve for x?

[obsidian monolith]

MemesPlease:

[wary plover]

well obviously the 30 is irrelevant

[rocky bison]

tru

[hard hornet]

nice observation

so

[wary plover]

sin2x = 1

[hard hornet]

yup

[rocky bison]

finish it

Also are we solving for acute solution?

Or generalized solutions?

[hard hornet]

there's only one solution

the question is multiple choice

[rocky bison]

Oh okei

[hard hornet]

can you repaste the question?

https://i.gyazo.com/8e3a10b276853608c5926b2cb93a0fb4.png

actually nvm i got it

right here

[rocky bison]

Values tho

so prolly has two of them

[hard hornet]

hmmm

i only see one solution on the multiple choice

[forest canopy]

same

[hard hornet]

sin(2x) = 1

[rocky bison]

i've not checked them lmao

[hard hornet]

funky

[wary plover]

pi over 4?

[hard hornet]

yup

you're right

tada

[wary plover]

that one was tough... idk if I can do that kind of problem in 2 minutes

[hard hornet]

practice makes perfect my friend

the hardest part is usually just "seeing it"

once i saw that there was a trig idnetity that cancels out nicely, it was easy rowing from there

25+11 = 36 if you know what i mean

[frank nebula]

Hey, knowing me I would've subbed and checked. At least you did better than me.

[hard hornet]

subbing in and check works

but im more used to solving for more general solutions

[wary plover]

what's cos2x again?

[hard hornet]

aka being presented the problem with no multiple choice

cos(2x) gives you three different identities

[wary plover]

oh god it does

I remember

[hard hornet]

$cos(2x) = cos^2(x) - sin^2(x)$

thats one of them

[obsidian monolith]

MemesPlease:

[frank nebula]

cos² - sin²
1 - 2 sin²
2 cos² - 1

I think?

[hard hornet]

and you get the other two by subbing in cos^2(x) or sin^2(x) with the pythagorean identity

Yup that's right

[frank nebula]

It's a pain in the anus region.

[wary plover]

I hope I dont get a problem like that

[forest canopy]

if you remember cos(2x) = cos²x - sin²x you can find the others from this one

[hard hornet]

learning how to derive it is good

[frank nebula]

Yeah. It's just manipulating the sin² + cos² = 1

[forest canopy]

azu, you made a mistake on one the formulas I think

2 cos² - 1 seems weird

[hard hornet]

thats fine

[forest canopy]

my bad

[frank nebula]

$cos^2x - sin^2(x) = cos^2(x) - (1 - cos^2(x))?$

[obsidian monolith]

Azuralytix:

[forest canopy]

you're right I posted directly after thinking something stupid sorry

[frank nebula]

Lmao it's fine.

During an exam, I wrote $6 * 60 = 300$.

[obsidian monolith]

Azuralytix:

[frank nebula]

Idk how.

[hard hornet]

happens to the best of us

one time

i did

1+1 = 1

[forest canopy]

oof, nice one

I've already done e^1 = 1

[frank nebula]

Lost me 5 marks during my physics paper.

Because it was the first step of a 3-step qn.

[forest canopy]

ah, not lucky

[frank nebula]

Oops latex.

[wary plover]

so I took this and I tried to simplify it, this is what I have so far

[hard hornet]

why do you need to simplify it, out of curiosity

[wary plover]

that's the problem

just simplify it

can I simplify it any further than 2x+2 over the radical?

[hard hornet]

lemme see

[wary plover]

that's one of the multiple choices

but it might be a trick

[hard hornet]

what are your choices?

but yes, thats what i got too

[hard hornet]

i got d

[wary plover]

me too

it was just really easy and I dont trust things when they're easy lol

[hard hornet]

s a m e

[wary plover]

https://i.gyazo.com/5b3307a05a5b8cebbbc6ddf49ab6cc5a.png

hm

it has asymptotes right?

[hard hornet]

Are you familiar with sign chart?

[wary plover]

uuuuuh

[hard hornet]

sign chart is super omg wtf powerful for this problem

[wary plover]

show plse

[hard hornet]

search it up a bit, and lmk if its confusing

[wary plover]

k

no I dont want fucking astrology

google please

[hard hornet]

lmao

uhhh

math sign chart

LOL

[wary plover]

reading

oh these things were in my calculus class

[hard hornet]

yesss powerful

you'll need it for calculus too, so learn it now 😄

[wary plover]

we used them for second derivatives

to map the difference from the first

ok so how do I use these with this problem

[hard hornet]

set up a sign chart

find all your critical points

[wary plover]

I just use the numerator for critical points right?

[timber plinth]

The denominator has a critical point at 9

since the not-so-definedness is rather critical to the behavior

[wary plover]

or am I thinking of derivatives...

[timber plinth]

numerator gives you zeroes

critical points i'd call everywhere where either the thing is undefined or 0

[wary plover]

so at 9 it's undefined

clearly

[timber plinth]

yes

so it's a critical point to put on your sign thing

[wary plover]

and if it's -4 or 10 it will just multiply out to 0

ms paint get out here

do I just need to plug things in

[timber plinth]

nah

you can definitely do it without plugging anything in

[wary plover]

how do I find the signs then

[timber plinth]

find the signs of (x+4)(x-10)(x-9)

since putting it in the denominator (applying ^-1 to it) preserves the sign

[wary plover]

I dont follow

How do I find the signs for my chart without putting in numbers between those critical points

so I think I can do this kind of problem without help, but the answer I got isnt there

is it seriously none of the above?

oh wait

now I remember, you can factor out the signs

got it

[fading token]

🍮

[heady jewel]

why you guys even doing this easy shit

i mean start real calc

[short sorrel]

believe it or not, some people take classes.

i know, shocking

you dont have to be elitist

(Or if youre gonna be elitist, calculus isnt the hill to die on lmao)

[hard hornet]

wait til he sees advanced mathematics channel

[heady jewel]

^^

imagine this boi seeing those putnam integrals for the first time and their answers

@hard hornet

[short sorrel]

imagine thinking putnam integrals are mathematically anything more than a parlour trick

a parlour trick that can get you scholarships, mind, but thats about it

[hard hornet]

Eh im not interested in competitive math

I do it for fun n out or curiosity

If you think putnam makes a great mathematician, ive got bad news for ya

[heady jewel]

no

[hard hornet]

Its the other way around

A great mathematician means more than just being able to do putnam problems

[heady jewel]

putnam dosent make you a great one

problem solving is a mere part of math

but seriously are you planning to becone a mathematician? @hard hornet

[hard hornet]

I think what makes a great mathematician is one whose able to solve new problems using whats given, think differently, creatively

Eh dunno

If all you know how to do is replicate tricks

Then ive got bad news for ya

[short sorrel]

nah a great mathematician can multiply 4-digit numbers in their head

[hard hornet]

Lmao fudge odd

[short sorrel]

according to some of my family

[heady jewel]

....

lol

you all got science related families?

[hard hornet]

Some chinese kids can do 9 digit multiplication in seconds

[heady jewel]

thats not mathematics

[hard hornet]

It looks impressive

[heady jewel]

that dosent even shoe high iq

[hard hornet]

But it really isnt

[heady jewel]

show

[short sorrel]

well i mean

it is impressive

[heady jewel]

what grade are you in @short sorrel

[hard hornet]

Utilizes some algorithms n tricks

[short sorrel]

its just not really noteworthy besides as a neat trick

uhh

hold on

[hard hornet]

And memorizing abacus

[short sorrel]

i guess grade 20

[hard hornet]

Good point

LMAO WTF

phd?

[short sorrel]

yes

[heady jewel]

woahhhhh

[hard hornet]

Oh congrats

Whats your research topic on?

[short sorrel]

algebraic geometry

[heady jewel]

good luck

[hard hornet]

Fancy

[heady jewel]

mate im in grade 10

lol

idk what im doing here

[hard hornet]

How many more years do u have?

[heady jewel]

wbu memes please

[hard hornet]

Gonna start my sophomore yr in college

[heady jewel]

what the fuck

im out

[devout raft]

[hard hornet]

Lul

[devout raft]

My students

[heady jewel]

,w differentiate 0

[obsidian monolith]

Display results online and refine query

Requested by Lionel#5497

[devout raft]

That's so evil

[heady jewel]

wasnt it undefined

wtf

[hard hornet]

Its 0

Taking the derivative of a horizontal line yields 0

Or you can treat 0 as a constant

And its still 0

[devout raft]

,w differentiate Lionel into 100^8990 pieces

[obsidian monolith]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Display results online and refine query

Requested by Krishna#2893

[devout raft]

Lul

[heady jewel]

lol

,w weather antartica

[obsidian monolith]

Display results online and refine query

Requested by Lionel#5497

[heady jewel]

lmao

[devout raft]

,w thug life

[obsidian monolith]

Display results online and refine query

Requested by Krishna#2893

[devout raft]

I am gonna delete my discord for some months

Bye frens

[formal wave]

1

should i always modulus the ln when i get the integral

like they did here

[torn swift]

the integral of 1/x is defined as ln|x| so yes, do that

[charred hull]

also, that's absolute value

not modulus

[willow bear]

@torn swift not "defined"

[viscid thistle]

This is what I got so far

Nvm got it

[spring palm]

can anyone help me find the integral of : sqrt( a^2-x^2)

[short sorrel]

dx? Is a a constant?

[spring palm]

ye

[short sorrel]

So you're gonna have to use a trig sub

What trig sub should you use here?

[spring palm]

is it possible without trig sub?

[echo plaza]

hyperbolic sub

[hard hornet]

Or noticing that it's the formula of a half circle

Dunno how that is going to help though

[rocky bison]

There's probably a bunch of pretty neat but complicated ways to do it tbh

[spring palm]

im trying to learn the proof for the area of a circle

[short sorrel]

Fair warning that this ends up being miserable

[hard hornet]

trig substitution is by far the most standard way for this problem

imo at least

[short sorrel]

,w integral of sqrt(a^2 - x^2) dx

[hard hornet]

If you want the area of a circle, you could just do polar

[obsidian monolith]

Failed to get a response from Wolfram Alpha. If the problem persists, please contact support.

[short sorrel]

..n

[rocky bison]

how are you proving the area of a circle tho

[echo plaza]

proven impossible

[short sorrel]

,w antiderivative of sqrt(a^2 - x^2) dx

[obsidian monolith]

Display results online and refine query

Requested by Namington#1983

[spring palm]

splitting it into quarters

[short sorrel]

It sucks.

[echo plaza]

[spring palm]

wao wtf

[rocky bison]

Because you could also use a polar integral

[spring palm]

alright thanks, teacher just made it seem like it was a viable option

[echo plaza]

that might just be wolfram using its bizarre list of favourite functions btw

if it can use arctan it normally will

even if it makes it uglier

[spring palm]

hmm i see

[hard hornet]

Would you still like to know how to solve it using trig substitution?

[echo plaza]

you can probably write that nicer with arcsin

[spring palm]

no im fine thanks i have the notes for that

[hard hornet]

alright

[spring palm]

thanks for the help frens

[echo plaza]

np

[molten flume]

Don't know how to start this question,.

[echo plaza]

$e^{qx}\equiv (e^{q})^x$

[obsidian monolith]

CaptainLightning:

[molten flume]

Thanks for the help! Easy problem but i over complicated it.

[echo plaza]

np

[narrow willow]

@spring palm you should factor the $a^2$ out of the square root then you'll see how that resembles $\cos{u}=\sqrt{1-\sin^2{u}}$

[obsidian monolith]

Sherwyn:

[timber plinth]

WA: exists
Everything but arctan: Am i a joke to you?

[echo plaza]

wolfram's goal is to use only arctan and artanh I swear

[proud raven]

what is the significance of a w2 vector

as in w2=u-w1

[willow bear]

what are u and w1

[proud raven]

it looks like its just the vector being projected onto rotated 90 degrees

they are uhh

[timber plinth]

Looks like the perpendicular component

[proud raven]

vector components of a vector orthogonal to another

[echo plaza]

[timber plinth]

Where w1 is the projection and w2 is the "rejection"

[proud raven]

lol i saw that @echo plaza

[willow bear]

...

is there a problem you're doing

[proud raven]

i dont really see what w2 has to do with a projection is what i mean

[timber plinth]

w1 is the parallel part, therefore w2 is what's left, creating a perpendicular part since none of it must be parallel @proud raven

[proud raven]

how do you mean 'whats left'

[timber plinth]

you know how projections work, yes?

[proud raven]

i know how to calculate one

[timber plinth]

smh

w2 is the vector rejection

where w1 is the projection, it's a sort of measure of how parallel it is to a vector I suppose

[proud raven]

oh

[willow bear]

"rejection"

[proud raven]

i think i just got what you mean

[timber plinth]

@willow bear sometimes it's actually called the vector rejection

[proud raven]

bizarre that it always ends up being orthogonal to v

[willow bear]

[timber plinth]

$v_{\perp u}$ or some shit

[obsidian monolith]

Darkrifts:

[timber plinth]

@proud raven it's obvious that w1 must be parallel by construction, so in order to reconstruct the original piece, you need a piece perpendicular

[proud raven]

does the original vector always have some specific angle to those two resultant ones

or it just lies between them

last question 😄

[timber plinth]

It lies between them

The specific angle clearly varies

[spring palm]

oooh thank you @narrow willow

[static vapor]

need help with delta

how does one recognise the need to use the "inclusion-exclusion principle"

or how do i not need to use it alltogether

[edgy flicker]

count how many cases there are in which the same letters do appear next to each other and subtract that

[viscid thistle]

with the determinant test its a hyperbola

but by graphing its y=-x

???

[willow bear]

"the determinant test"?

also, no, this isn't y = -x; the graph actually consists of two lines, the second line being the y-axis.

[viscid thistle]

oh

so would it be pair of intersecting lines

the test where B^2-4AC > 0 hyperbola, <0 ellipse and =0 parabola

[willow bear]

what are A, B and C

[viscid thistle]

a=1, b=1 c=0

[willow bear]

yeah i mean ok the two lines is kinda like a degenerate case of a hyperbola

[fading token]

,rotate

[proud raven]

this is gonna sound really stupid i think but

is there a less painful way of finding a unit tangent vector than finding the slope then the tangent line at a point and then a vector along that line and then the unit vector of that vector

[echo plaza]

you don't need to find the tangent line

[proud raven]

i found some ways to do it with parametric equations but cant suss it out in my head if were just talking about like x^2=y

[echo plaza]

otherwise no

[proud raven]

alright

[stiff rune]

@patent beacon

were the points

1,1

and -1,1

that passed through a tangent line with points 0,-1

[patent beacon]

Yes I believe they are the points

[stiff rune]

niceee

[patent beacon]

Did you search for them, or did you find them algebraically? You may be asked to do the latter on a test

[stiff rune]

i used the gradient of 2x

or of 2 rather

and just found them by going across 1

up 2

both ways

[patent beacon]

Here's a way that will always find them for you, let's say y = x²:

Take the tangent line at a general point x = a

This tangent line is y = 2a(x - a) + a²

The line should pass through (0, -1), plug that into the tangent line:
-1 = 2a(0 - a) + a²

-1 = -a²

a = ±1

[stiff rune]

oh ok

thats good to know

[patent beacon]

It's kind of like solving the problem backwards, idunno

Use an unknown as the point, and solve for it

[stiff rune]

well im getting my answers right on the textbook now 😃

and i know how to do every other thing for the test so should be good

[rocky bison]

@patent beacon That's actually a pretty neat problem tbh

I don't think I've seen things like that before

[echo plaza]

you get ones with cubics and stuff

[rocky bison]

Yeah I imagine

I've just never actually seen a problem presented in that way

They're always the other way around

i.e. find the equation for the tangent/norm at x=...

[echo plaza]

I remember doing a lot of them

[rocky bison]

Ye

[proud raven]

can i ask for help on a problem

a) find all points of intersection of the two equations

b) find the unit tangent vectors to each curve at intersections

equations are y=x^2 and y=x^(1/3)

I can get the points of intersection, and i can kind of intuit what the unit tangent vectors are

but im having a hard time showing part b on paper

nvm i think i figured it out

i lied

part c is find the angles between the curves at the point of intersection

is that just equal to the angle between the tangents? or a different value

[echo plaza]

well there are technically 2 angles

but yeah the angles between the tangents

[rocky bison]

@patent beacon https://www.desmos.com/calculator/uf41mszecp

[echo plaza]

if you want one with cubics, find the equations of the lines that pass through the origin and are tangent to the curve y=x³-3x²-1

[rocky bison]

,time

[obsidian monolith]

The current time for Pseudo is 7:05 AM (BST(+0100)) on Sun, 09/06/2019

[echo plaza]

,time

[obsidian monolith]

The current time for Foodness is 6:05 AM (GMT(+0000)) on Sun, 09/06/2019

[rocky bison]

I'll try to figure how to do that later

[echo plaza]

how to set it to BST

[rocky bison]

I set mine to london

so hopefully it'll update itself

,time --set London

[echo plaza]

,ti --set London

[obsidian monolith]

Your timezone has been set to Europe/London

[proud raven]

okay, thanks!

[echo plaza]

also this one

You are given the curve y=ax²+bx+c, (2,4) is on this curve and the tangent to the curve at x=-1 has the equation y-7x=-1. Find a,b and c

someone asked that one about a month ago although it came with a lot of padding lol

[proud raven]

man

cross products are bizarre holy cow

thats sort of the last thing i have to cover to catch up with vectors in my class

[fading token]

Cross products are really just 3x3 determinants

[autumn sonnet]

if I have log_(1/5)(x^2-3) >= log_(1/5)(x-1) can I cancel the logarithms?

[fading token]

Depends on which base the logarithm has

[autumn sonnet]

Oh sorry the base is 1/5, I missed the _

[fading token]

Oh, then yeah, bc logarithm functions are monotonous increasing

Just be careful that the argument of the log must be greater than 0

[autumn sonnet]

When could I not cancel the logarithms?

[willow bear]

ah careful

if you want to undo the log_(1/5) you gotta flip the sign

bc log_(1/5)(t) is monotone decreasing

[fading token]

ahh yep

[viscid thistle]

stop saying that cross products are just 3x3 determinants to people who haven’t taken linear algebra
it’s confusing to them and will keep them from understanding it just as much as giving them a simpler formula would be
if someone isn’t at the level where they can understand the derivation for a formula they need, make it simple and clear that it’s just a formula and there’s nothing super fun going on here, it’s simply a tool, not part of the puzzle itself

[civic plaza]

Learning how to do the 3x3 determinant is simpler than learning a long af formula but that's just me

A lot of people , including me, were taught the cross product in a class like newtonian mechanics before being formally taught in linear algebra

It's not bad especially if you consider the "diagonals shortcut" to computing it

[deft crow]

why is (ln9)/ 2 the same as ln3

[willow bear]

$\frac{1}{2} \ln(9) = \ln(9^{\frac12})$

[obsidian monolith]

Ann: