precalculus

[daring osprey]

ah

[round geyser]

Yes

[neat sundial]

Oh okie so power rule !

I know basic formulas...

I rather need techniques..

Like they transform the whole integral to something 😭

[ancient inlet]

you mean indefinite integrals?

[chilly arrow]

i know, did u see what i said in #competition-math lol

and i didnt say it was impossible, just not really doable by normal methods

[round geyser]

Do they have period and amplitude

[neat sundial]

Mm yea

[ancient inlet]

that means what was your original eqn

[manic crest]

Calculas is the EASY CHAPTER

[round geyser]

I’m graphing csc and sec

It’s just inverses of sin and cos

[uncut mulch]

careful with the use of "inverse"

make sure to mention multiplicative inverse

[hushed sphinx]

"Reciprocals" is even better.

[dry verge]

Are linear algebra concepts like linear transformations usually taught in precalculus?

[tame shard]

Depends on where you learn it

[dry verge]

Ah, because I'm learning it on Khan Academy and they not only do vectors and what a matrix is (and how to solve systems of equations), but they're doing linear transformations.

It's like an "intro to linear algebra" course in a way, even if it's in precalculus.

[pale oracle]

its precalculus

[icy verge]

i got same in precalc book

[pale oracle]

yeah..cuz its precalc

[valid topaz]

<@&268886789983436800>

[round geyser]

I did was

Is there is amplitude in front of csc/sec x

Make a guide graph of sin/cos x

Place them there and draw asymptotes or poles

And draw “parabola”

Check my work

Mine got skewed

[undone pumice]

i dont like this wording, the pole is the x axis... you dont draw them

[round geyser]

I was told to draw the asymptote

[round geyser]

O

[undone pumice]

ye... draw the asymptote, not "poles".

[round geyser]

Okay

[undone pumice]

looks good... i think

[undone pumice]

do u need help?

[round geyser]

I’ll show u

Now I have period of pi and quarter period of pi/4

And then Horizontal shift is none

Where to start

[undone pumice]

yup (but remember that the period is over 2 set of asymptotes)

[round geyser]

0? then pi/4

[undone pumice]

first

think about how sec is the reciprocal of another trig function

[round geyser]

Yes

I graph the cos 2x as a guide graph

[undone pumice]

mmhm

(3cos2x, not cos2x)

[undone pumice]

based on that, where are the asymptotes?

[round geyser]

Like this

That’s what my teacher taught

[round geyser]

pi/4 and 3pi/4

[undone pumice]

yup

[round geyser]

Is it true that if I graph

[undone pumice]

ye

[round geyser]

I start with the horizontal shift

[undone pumice]

it touches so its useful to draw the cosine

[round geyser]

Then add quarter period

[undone pumice]

uh i dont know

i never learned all this quarter period bs

i just learned how to graph

[round geyser]

Period/4

[undone pumice]

ikk

i just dont like that

term

[round geyser]

Where do u start

[undone pumice]

for which function?

sin/cos or the ones with asymptotes

[round geyser]

csc or sec x

Evelyn what are some advice about graphing tan and cot

[undone pumice]

uh

[round geyser]

Since I’m learning that tmr

[undone pumice]

they look very similar just reflected versions moved differently

like

cot looks like tan

if it was reflected over the y axis

[round geyser]

It it’s sin x over cos x

[undone pumice]

and it was shifted pi/2 (cus cot=cos/sin and sin has diff zeroes)

[round geyser]

0 1 0 -1 0
1 0 1 0 1
0 und 0 und 0

[undone pumice]

ye there will be asymptotes

period of pi

and then

pi/4 and its other angles are useful(cus tanpi/4 is just 1, right?)

[round geyser]

the main I use is

[undone pumice]

so you can use 1, -1 for tan from -pi/4, pi/4

[round geyser]

0, pi/2, pi, 3pi/2, and 2pi

[undone pumice]

be careful

tangent has a period of pi

[round geyser]

Asymptotes of tan will be at pi/2 and 3pi/2

It’s asymptote of tan = (odd number)*(pi/2)

[undone pumice]

this is... an unusual way to put this

[round geyser]

also 5pi/2, 7pi/2, 9pi/2

[undone pumice]

my teacher taught us using

[round geyser]

Like I put
(radians)=(degrees) * pi/180
(degrees)=(radians)* 180/pi

Cool

[round geyser]

Then I did
Quarter period = period/4 or 1/4 * (period)
My teacher also wrote it that way

[undone pumice]

mm k

[round geyser]

Now ik

period = 4*quarter period

[undone pumice]

um

ye

[round geyser]

So on and so fourth

[undone pumice]

mm ic

so period shift but no horizontal dilation

vertical dilation

no vertical shift

[round geyser]

Do I start at pi/4

[round geyser]

^

Then add pi/2?

[undone pumice]

wdym by that

[neat path]

Im a 9th grader

And i came across this equation of making a heart shape

(X^2+y^2-1)^3 - x^2y^3=0

And i was wondering whether integrating from 0 to x and 0 to y will completely fill it up?

Im trying to learn all this and i think this is how you do it so can someone please explain it to me

[rancid mason]

do you have a diagram?

[rancid mason]

no first you must draw the function and

find its range

as shown the range is from -1 to 1

its possible but its not for this channel

[neat path]

Integrate from 0 to 1?

[rancid mason]

its very complex

you have to make the function to only have x

otherwise it wouldnt be possible

[neat path]

Why cant we integrate both x and y

[rancid mason]

thats to find like volumes and stuff

you need to intergrals then

which is very hard to do

[neat path]

Okayy

Thanks

[rancid mason]

so all you need to do is to make this equation into a function with y being the subject

and then intergrate it from -1 to 1

[regal tangle]

sqrt(x) does the limit exist at 0 ?

[tame shard]

pretty sure sqrt(0) itself exists

[hushed sphinx]

Yes, but some textbooks only define limits for functions that exist in an entire deleted neighborhood of the target point.

That's not any kind of deeper mathematical truth, though, only an attempt to save complexity because it takes a bit more explanation and notation to define them more generally..

What? Where?

[hushed sphinx]

Your equation does not define a function, as Cdkkpoke's plot shows, so it does not make sense to integrate it.

[hushed sphinx]

Yes, $\lim_{x\to 0} \sqrt{x}$ exists and is $0$, but some textbooks give simplified definitions that doesn't work in this case because $\sqrt{x}$ does not exist to the left of $0$.

[obsidian monolith]

Troposphere

[neat path]

So as he said, we create a function f(x) = that?

Could you please explain

[hushed sphinx]

Writing f(x) = (x^2+y^2-1)^3 - x^2y^3=0 would not make sense, because what you put to the right of "f(x) =" should be an expression rather than an equation.
You can write f(x) = (x^2+y^2-1)^3 - x^2y^3, but that still doesn't give you a function until you decide on a value for y.
You can write f(x,y) = (x^2+y^2-1)^3 - x^2y^3, but (a) that now is somewhat far removed from your heart-shaped curve, and (b) integrating a function of two variables is not a single thing -- that is there are several possible concepts it could mean.

[neat path]

Oh

So if we find an expression that gives us the heart shaped curve IN THAT X VARIABLE, then we can integrate it from 0 to x, filling the shape up?

[hushed sphinx]

It's also not quite clear to me what you mean by "filling the shape up".
I suspect perhaps instead of integration what you want might just be the inequality (x^2+y^2-1)^3 - x^2y^3 <= 0, which makes the solution set include the interior of the heart.

Or are you trying to compute the area of the heart shape?

[neat path]

I was maybe thinking of the same shape but a tiny width smaller than the last all the way until 0, which would make it not only 1 heart shape but "filled up" in the sense that it is completely coloured. And i figured that using integration could perhaps give that small width needed to complete the fill up?

[twin eagle]

where does number theory go?

[hushed sphinx]

#elementary-number-theory, or possibly #advanced-number-theory for techniques that go beyond standard undergrad topics.

[neat path]

Is this beyond undergrad?

[hushed sphinx]

No, I was replying to the other person with the weird name there.

[neat path]

Hes my friend from school

Yeah @hushed sphinx so my question wss

Was

Hope you understand

[hushed sphinx]

Is this not an answer to what you want?

[neat path]

It is an answer

But i want infinitesimally small hearts inscribed in each other

Like concentric circles perhaps, but with the width in between them almost 0

[hushed sphinx]

Okay, that is definitely not something where the concept of "integration" is relevant.

[neat path]

Oh, okay. Could you tell me how to do that though?

[round geyser]

Wow

[hushed sphinx]

You can get a family of curves that fill out the heart by setting the right-hand side of the equation to different negative values instead of 0. However, as this plot shows, they behave somewhat wonkily around the axis crossings -- In fact, I don't completely understand how the original manages to look nice and smooth around the x-axis crossing.
I don't immediately know how one would create neat equations for curves at a constant distance from the heart shape. That is in general a hard problem that cannot be done with polynomials.

[rancid mason]

bro you drew mickey mouse

[jolly karma]

Would anyone perchance know how to find the focus of a parabola?

[hushed sphinx]

What do you already know about the parabola? Equation, or some points on it, or a drawing in the plane and you have straightedge and compass?

[honest meadow]

is the parabola aligned to an axis?

[regal tangle]

well my course book says that limit exists if left hand limit = right hand limit.. but in this case since LHL doesn't exist so I'm a little confused

[hushed sphinx]

Yeah, that's one of the "simplified definition" patterns I'm talking about.

[regal tangle]

I see.. then what would be the absolute definition ?

[hushed sphinx]

It would be something like the limit of f(x) for x->c is a number L such that

for every epsilon>0 there exists a delta>0 such that for each x with 0<|x-c|<delta and f(x) defined we have |f(x)-L|<epsilon.
and then with a caveat that the concept doesn't make sense (because L would not be unique) unless c is a limit point of the domain of f.

[regal tangle]

hm.. i see

thanks for helping

[round geyser]

Hey I need help for doing y=-5 cot 8x

Amp: None
Period: pi/8
Quarter point: pi/32
Vertical shift: none
Horizontal shift:none

[hushed sphinx]

Or in fancier topological terms: "the preimage of every neighborhood of L is a (deleted) neighborhood of c, intersected with the domain of f".

[round geyser]

I need help graphing

[manic crest]

If 7x+2y=4, 128^x×4^y=?

SAT sneaky question

[tame shard]

Sneaky?

[manic crest]

Yes

[manic crest]

Wait your user/nickname is A EQUATION

[tame shard]

Take the god damn log2 of both terms

[manic crest]

Solve without logs. Solve with Jungle Method

[tame shard]

Wtf is a jungle method

You realize that 128^y = 2^7y, 4^x = 2^2x?

[manic crest]

An Approach to solve equations that can't be logged or NOT using natural/standard logs

[tame shard]

1 that doesnt explain what it is or how it works
2 this is clearly loggable

And standard logs?

[manic crest]

Log(4)(2)

[tame shard]

Fyn standard

[tame shard]

What do you mean by that?

[manic crest]

Standard means the classic log in precalculus

[tame shard]

You know logs arent constrained to base 10 or base e right

[manic crest]

I meant
Standard Logs: Log_4(5)
Natural Logs: In_4(5)

[tame shard]

Pardon me what does the log4(5) mean?

[manic crest]

$\log_4{5}$

[obsidian monolith]

ishantgaming67

[manic crest]

Bruh

[tame shard]

Define log 4 of x

Cuz im pretty sure we do not understand each other

[uncut mulch]

Natural Logs: In_4(5)
makes no sense

[tame shard]

And this

[manic crest]

We are cooked by Dagestan 2-4 years old

[tame shard]

<@&268886789983436800> dagestan joke

[manic crest]

It's not that bad joke as E word

[tame shard]

But thats still homophobic, so what?

[manic crest]

How?

Homophobic?

[tame shard]

since dagestan is a muslim region, they are homophobic there, getting send to dagestan implies you are going to get the gay beaten out of you

[uncut mulch]

Please be more mindful of what you post for reasons mentioned above.

[leaden portal]

u r not real

[tame shard]

Fym not real

You know plenty of memes out there are in fact bigoted

[leaden portal]

dis world not real

all ts fake

https://tenor.com/view/im-nothing-like-yall-nothing-like-yall-im-nothing-like-yall-fish-fish-meme-gif-7279072448778168854

[tame shard]

Sybau

[round geyser]

You mean
If 7x+2y=4, 128^x * 4^y=?

Easy

128^x = 2^(7x)
4^y = 2^(2y)
2^(7x) * 2 (2y)
2^(7x+2y)
2^4 ; (7x+2y)=4
128^x * 4^y = 16

[tame shard]

This

[round geyser]

Ye

[round geyser]

You use exponents properties of
a^m * a^n = a^(m+n)

[tame shard]

Yup

Idk what that guy was up to tho with the log_4 shit

And wth is a jungle method

Vro never explained that

[round geyser]

Can you check this please

[round geyser]

^

[round geyser]

Do I graph like this

[novel delta]

Please recheck your amplitude and period Remember the general structure for cosine is a*cos(k(x-d))+c.

[round geyser]

It’s not cos it’s cot

[round geyser]

Right

[novel delta]

I thought it was cosine, sorry then this is correct

for the second one i do not know about how your teacher structures things however I believe its more preferable to say pi/6 right rather than -pi/6 and for VS it is 1 unit up. But that is just communication other than that everything checks out and is correct.

[round geyser]

Okay

Then it’s pi/6 + the quarter period

[novel delta]

That should be where the next increment is located. the period + quater period , then plugging that in will give you the y

[round geyser]

I thought I start the HS + quarter period

[novel delta]

HS ± quarter-period gives you the key points for one full cycle (the one you actually sketch). Adding the period just moves you into the next repeated cycle, it gives the same type of points, just shifted right.

[round geyser]

I started with pi/6 and add pi/8?

Then it’s 14pi/48?

[novel delta]

Well, that will give you the first quarter-period point of the cycle. You can also subtract π/8 to get the symmetric one on the left.

[round geyser]

Ok

[round geyser]

So is that the asymptote

That’s what I got so far

[novel delta]

No that is not the asymptote that is the important points where the tangent hits a specific y value after the transformation the asymptote is at x = -pi/12 and 5pi/12

[round geyser]

Ohh

So I add pi/2

[novel delta]

Almost, you have to solve when the inside of the tan is equal +- pi/2

So solve 2(x-pi/6) = pi/2

that will give one of the asymptote

the general is 2(x-pi/6) = pi/2 +k*pi

[round geyser]

Okay

[round geyser]

x-pi/6 = pi/4

x=pi/6 + pi/4
x=10pi/24
x=5pi/12

Don’t laughs at my sketch

[scarlet cedar]

Why is it broken like this

[round geyser]

I didn’t have a steady hands

That’s why

[scarlet cedar]

No worries there

[round geyser]

But is that right

[novel delta]

That is correct that is one of the asymptotes

[round geyser]

Yay

[novel delta]

And other than the drawing the tangent graph seems correct

[round geyser]

Okay

I think graphing trig functions are interesting

Like doing sec and csc

It’s like a parabola

And cot is like cubic

[novel delta]

Well cot is -\tan\left(x-\frac{\pi}{2}\right)

-tan(x-pi/2) for cot(x)

So especially the identites they are interesting and can aid later on in more higher mathematical sectors as they help simplify many equations.

[round geyser]

Yep

[timber jasper]

Can someone give me some basic limits questions (infinity included) (please don’t include cubes conjugates and trig I’m trash in them)

[oak night]

Here are some basic ones without trig, conjugates, or cubes

1. lim x->2 (3x+1)

2. lim x->4 (x^2 - 16)/(x - 4)

3. lim x->3 (x^2 - 9)/(x - 3)

4. lim x->-1 (x^2 + 3x + 2)/(x + 1)

5. lim x->0 (5x)/(x + 2)

6. lim x->infinity (2x + 1)/(x + 4)

7. lim x->infinity (3x^2 + 2x)/(x^2 - 5)

8. lim x->infinity (4x + 7)/(2x^2 + 1)

9. lim x->infinity (5x^2 - 1)/(x + 3)

10. lim x->0+ 1/x

11. lim x->0- 1/x

12. lim x->2 1/(x - 2)^2

13. lim x->2+ 1/(x - 2)

14. lim x->2- 1/(x - 2)

15. lim x->infinity (7 - 3x)/(2x + 5)

[raw hill]

🎁

[obsidian monolith]

Civil Service Pigeon

[timber jasper]

Ty

I’ll save it with myself

[round geyser]

I have one too

[round geyser]

^

[tame shard]

lim x->4 of 5 is crazy

[round geyser]

Also lim x-) 8 (7)

[tame shard]

And what is 7

[round geyser]

sqrt(6)

[tame shard]

$\lim_{x \rightarrow -1} \sqrt{x^2+5}$

[obsidian monolith]

P(this user is larp larp) ≥ 0

[tame shard]

This is trivial

Just sub -1 in

[round geyser]

Yea that was the way I solve these

$\lim_{x \rightarrow -1} \sqrt{(-1)^2+5}$

[obsidian monolith]

Jake

[round geyser]

$\lim_{x \rightarrow -1} \sqrt{x^2+5}$=$sqrt{(-1)^2+5}$=$\lim_{x \rightarrow -1} \sqrt{6}$

[round geyser]

Look

[tame shard]

Chop the limit off in the second step

Since you are substituting x in anyways

[round geyser]

Can you show me please

[obsidian monolith]

Jake

[tame shard]

$\lim_{x \rightarrow -1} \sqrt{x^2+5}$ \
Substituting x = -1 in we have \
$\sqrt{(-1)^2+5} = \sqrt{6}$

[round geyser]

Thanks

[obsidian monolith]

P(this user is larp larp) ≥ 0

[little surge]

Cmon guys y'all can solve ur questions 🔥

[valid topaz]

uhh well if we rearrange and divide by x we can see that x=2 is a solution

there has to be a second one too by ivt

[round geyser]

How do u use lambert w function

[valid topaz]

uhh i dont think we can use that here

atleast i can't manipulate it into that form

[undone pumice]

uhhh

(x^x)/x = 2 for the integer solution. then for the other solution

take the natural log

but im pretty sure its um

a transcedental equation

from my knowledge

[round geyser]

x^x=2x
x=log_x 2x

[undone pumice]

Uhhh yeah so it ended up as like a transcedental equation type situation

where you can't reallllyyyy algebraically solve it

[round geyser]

lambert w

Then the only way is to plug and check

Ik exponential functions’s domain is x>0

[undone pumice]

*x>0

[round geyser]

I plug 2
(2)^2 - 2 =2
2=2

[undone pumice]

ye so 2 is the first solution

theres another

uhhh i guess if you rewrite it

[round geyser]

Idk what’s their exact form of 0.34632

[undone pumice]

as (e^lnx)^(x-1)=2

cus e^lnx=x

as such $e^{\left(x-1\right)\ln x}=2$

[obsidian monolith]

evelyn \\ HS 9

[round geyser]

Take ln

(x-1) ln x = e^2

[undone pumice]

i would do like substitituion maybe

cus lambert w wants ue^u structure

so like maybe if u = lnx

(e^u)(e^u-1)= e^(ue^u-u)=2

$e^{\left(ue^{u}-u\right)}=2$

[obsidian monolith]

evelyn \\ HS 9

[undone pumice]

ue^u-u=ln2

hmm nvm actually... lemme think of a diff way to force a lambert

[valid topaz]

apparently there exists smth called the r-lambert function

that might work

[undone pumice]

hmm

haven't heard of that one before

rn im attempting to force this silly equation to be a lambert

[valid topaz]

i don't think that's possible cuz it becomes (x-1)lnx=ln2

[undone pumice]

wait i think i got it i managed to squish it into the form where the solutions are 1/W(ln2) and 1/W subsript -1(x) ykw for the branch index of ln2

[valid topaz]

huh how

[undone pumice]

first i raised it by uhh

wait lemme grab some paper

[valid topaz]

nvm i have dinner

[undone pumice]

1/x-1

ah seeya

a bunch of substitution work and stuff mostly just weird manipulation

[valid topaz]

hmm

[undone pumice]

(ts is not fitting on the small paper i have-) okay so basically first i substituted y=lnx which gave (e^ye^y)-1=2 or e^ye^y-y=2

[valid topaz]

alr

[undone pumice]

Then I took the natural log

so ye^y-y=ln2

Then factor out the y

[valid topaz]

y(e^y-1)=ln(2)

[undone pumice]

sub in u=-y

and then mult both sides by -1

and then multiply both sides by e^u

[valid topaz]

u(1-e^u)=-e^uln(2)?

[undone pumice]

uhhh ye

i wrote it like that

then isolate the u

so add ue^u to both sides

[valid topaz]

ok

[undone pumice]

Lwk my Chromebook just died but here’s the next few steps

I didn’t know what letter to pick so uh ye I substuted in an a

After more manipulation got it into ae^a form which is lambert w

And then from the first substitution y=lnx it got kinda squished in the corner but

[valid topaz]

ok gimme a sec to read over this

[undone pumice]

Kk if I made a mistake lmk lwk tried like four approaches until I got a lambert form 🥀

[valid topaz]

so u tried to apply the lambert w to a*e^a=a/2+ln2/2?

there's a linear factor in the RHS

wait

so i tried back-subbing the final value in and it didn't check out

[round geyser]

Anyways

Is this correct

[raw hill]

For this case yeah

In the future you may want to be more careful about mixing signs on your shift

Using “left”/“right” also works

[round geyser]

It goes right pi/2

[raw hill]

mhm

Or left pi/2 since pi/2 - pi = -pi/2

[round geyser]

Ohh

[raw hill]

Hence why I said it was fine in this case even though you mixed negatives and positives

[round geyser]

Bc these are also a point

For graphing

My teacher wants us to make Horizontal shifts the same denominator as the Quarter period

[raw hill]

That’s not what I’m talking about

I’m talking about how you wrote a negative and dropped it

[round geyser]

Oh yea

I should’ve wrote “left”

Do I add pi/4

[raw hill]

adding pi/4 to what

[round geyser]

pi/4 + 2pi/4

[round geyser]

Is that right

[raw hill]

um I mean you can add anything you want to

What are you trying to accomplish

I’m not following

[round geyser]

I want the next point

[raw hill]

uh ok ig yeah

[karmic brook]

the amplitude is not "none" its 1

[round geyser]

Tan has no amplitude

[karmic brook]

oh yeah

whoops

for adding points to the graph. ecspecially with tan where its a bit wonky, plot the asymptotes first and then the x intercepts. and then the rest of the graph doesnt have to be exact just follow the shape of the function

[karmic brook]

so pi/2 is an x- intercept but so is pi/2 +kpi (k belonging to intergers), so you need to plot (pi/2, 0) but also anyother x intercepts that fall into the graphs domain

and then draw the behaviour around the nearby asymptotes

[round geyser]

It’s x-pi/2

So is it (pi/2,0) the point

I’m trying to add quarter points easier so I made the HS 2pi/4

[karmic brook]

yeah so every x- intercept on this graph is in the form (pi/2 +kpi , 0) letting k = 0 gives us our 'first' x intercept at (pi/2, 0)

[round geyser]

Ok

[karmic brook]

whats HS sorry?

[round geyser]

pi/2

Specifically right pi/2

[karmic brook]

no like what does it stand for ?

[round geyser]

Horizontal shifts

[karmic brook]

oh okay

[karmic brook]

yeah i mean if it works then sure, but looking at the graph you have above i would make the far right line pi (vertical asymptote) the middle line (on positive x axis) make pi/2 (x intercept), the zero line is just another vert asymptote and then do the negative

you dont really need to split it into quarter points

[round geyser]

It’s 2pi/4, 3pi/4, pi

What would this be

[karmic brook]

no, the zeros are pi/2

not 3pi/4

[undone pumice]

*right pi/2, not left. graph will look the same i beelive for left or right but...

Oh and also yeah. If the zeroes of the parent tangent function are -pi/2, pi/2, 3pi/2, shifting pi/2 right just makes it like 0, pi, etc

[karmic brook]

and again you dont need the quarter points, they dont really add much, since they dont show anything special about the graph

make sure to label your vertical asymptotes with the equation of the asymptote as well

[undone pumice]

yeah or just give the equation of the general placement of the asymptotes (like the set theory notation i showed u)

[karmic brook]

this ^^, also along with that if you do have working out on the side, right the set theory for the x intercepts as well

[undone pumice]

which in this case would be x=0/pi/2pi/whatever starting point + pi k (or n, or whatever variable you want) where it is defined as an integer...

[round geyser]

My teacher requires it

[undone pumice]

*right pi/2. graph looks the same right or left but if you want to get technical its shifted to the right pi/2 and if you're being graded on whether you wrote the right direction..

[undone pumice]

wait one more thing

3pi/2 is not in the middle of 0 and pi

label that one as pi/2

[karmic brook]

from that it looks like you have a vertical asymptote at 3pi/2, your behaviour of your graph around asymptotes needs to be shown (bring the line closer to the pi)

[undone pumice]

^^ yup what i was saying, move the "zero" to that mark and label it as pi/2

[round geyser]

I meant this

[undone pumice]

that is also incorrect

it crosses at the point in between 0 and pi

which would be where you have marked 3pi/2, which is incorrect (redraw your graph, and also relabel)

[undone pumice]

... well i guess that works too

[round geyser]

So I can have

3pi/2

As a point

It’s (odd number)*pi/2

[karmic brook]

technically right but the more correct way to write this is pi/2 +kpi (k belonging to intergers) or if you are really adament with that odd number write it as (2k+1)pi/2 (k beloginging to intergers)

and just search up the math symbol and set notation for beloging to intergers

its like a weird E into curly brackets

[undone pumice]

Ummm I showed him before

Yeah and that z symbol

[karmic brook]

..... glad to see it stuck

[undone pumice]

Sigh

[winter osprey]

Graphing trig functions that aren't sin and cos 🤢

[karmic brook]

techincally you are just graphing both :D.

[winter osprey]

i guess bro 😒

[round geyser]

Yes

[lost rivet]

How do u do no8? I did dy/dx to get the negative reciprocal for the normal gradient

[round geyser]

I was taught
Graphing sin and cos
Graphing csc and sec
Then
Graphing tan and cot

[raw hill]

#calculus?

[round ermine]

#calculus but here’s my solution anyway

[manic crest]

4^x=x?

[tame shard]

Ive a feeling this gon be some lambert W shi

,w 4^x=x

[obsidian monolith]

Results provided by WolframAlpha

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Requested by hexagon69_

[tame shard]

Yup

[tough flare]

Is it correct?

|S| : number of elements in S

[tame shard]

Err that sum looks off

1+2+3+...+n = n(n+1)/2

[tough flare]

What's correct for $|S \cross S|$?

[obsidian monolith]

Ujjawal Gupta

[tame shard]

|S|^2

[tough flare]

How?

[tame shard]

Err

In that one example

S has 2 elements

There is like 4 ways to pair 2 elements together

So 2^2

[tough flare]

What if it has n elements?

[tame shard]

Then S^2 has n^2 i think

(Looking at it, its nicely related to pascals triangle too)

[warm panther]

Should I use the quadratic formula to solve for x in quadratic equations?

[tame shard]

Yes

Like

Its literally made for that

[river beacon]

lmaoooooo

[echo sinew]

Absolutely not! You should use it only to solve 16th degree equations

[manic crest]

Try it in Desmos

Y=the equation, Y=X

[tame shard]

What

Desmos

Err

The lambert W function extends to complex solutions too

Desmos cant do that

[warm panther]

Yeah but like, aren't there faster ways?

[tame shard]

Yes, its called factoring

But almost all quadratics are not factorable

[warm panther]

Really

[tame shard]

Yes

[hybrid zodiac]

Is there a way to solve x^x=a without using lambert w?

Cause im not really accoustumed to using lambert w so im really unsure if what i did was correct or if there was a better way

[hushed sphinx]

That's pretty much what the W function is (up to some logarithms).

However, you may count numerical methods (such as bisection) for root finding as "a way to solve".

That's how the W function itself is computed, anyway.

[hybrid zodiac]

I understand,thanks.

[tame shard]

,w

[manic crest]

Desmos will show an exponential y that grows fast

[tame shard]

Err you are clearly not understanding what i meant

[raw hill]

Note that
$$\frac{1}{5+\sqrt{24}}=5-\sqrt{24}.$$

[obsidian monolith]

Civil Service Pigeon

[round geyser]

I plugged in x=2 it works

So -2

[raw hill]

well yes those two work, but I'm assuming you want a fully rigorous solution to show that there's no more

[round geyser]

I wonder how

I thought it’s (5+sqrt(24))^-1

[raw hill]

Yes, it's that as well

[round geyser]

If I’m correct

It’s -2<x>2

[round geyser]

As its domain

[raw hill]

I don't follow

Both terms on the left are exponential functions with positive base

meaning that both terms on the left are defined for all real x

[round geyser]

Okay

[round geyser]

When I graphed it it looks like a parabola

But with discriminate > 0

[raw hill]

Yeah that's how a^x + a^(-x) usually looks:

but it's not really a parabola

,w graph cosh x

[obsidian monolith]

Results provided by WolframAlpha

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[raw hill]

Rather it's a transformed version of the cosh graph (I'll let you figure out the actual transformations)

$$\cosh x=\frac{e^x+e^{-x}}{2}$$

[obsidian monolith]

Civil Service Pigeon

[round geyser]

Ik (1+1/x)^x=e

[raw hill]

the limit of that is e, yes

[umbral timber]

what does the h mean?

[round ermine]

hyperbolic cosine

[raw hill]

Yeah that's how the hyperbolic functions are notated. So sinh is hyperbolic sine, tanh is hyperbolic tangent, etc.

[round geyser]

Ohh

I only know to how to graph sin, cos, tan, csc, sec, cot

[round ermine]

i think hyperbolic trig is a little outside precalc level though

[round geyser]

I agree

[raw hill]

eh it's just linear combinations of exponentials which they should be familiar with, no?

so it should be fine up to the one piece of new terminology

[round geyser]

Cuz I just got to graphing tan and cot

[raw hill]

though the hyperbola background is a whole rabbit hole of its own

[round ermine]

and proving that representation is definitely outside precalculus level but it’s nice to know

[undone pumice]

oh what do those look like? i haven't worked with those before, only like regular hyperbolas and ellipses and stuff

[round ermine]

i only know that cosh creates catenaries which show up in hanging chain problems
other than that you can graph them but i'm not aware of any special names for them

[round ermine]

also here's why that equation is true

[round geyser]

Ohhh

Taking their conjugates

And difference of squares

[round ermine]

really useful for systematically solving the original problem

since of course that means you can represent one of the terms as the other term raised to the -1

[loud juniper]

am i able to learn pre without understanding algebra that well

[undone pumice]

precal is kinda a lot of alg review so ummm like

i would sayyyy its possible buttt i would not recommend it if you can take some time to review like alg2 stuff over summer or something like

[round geyser]

Look that’s what my teacher taught us

[undone pumice]

ye

[round geyser]

Like the one yesterday

The

p(x)=tan(x-(pi/2))

[undone pumice]

ah

[round geyser]

I meant to do 4 tick marks

[winter osprey]

aint no one tryna graph trig functions anymore

[round geyser]

I would

[round geyser]

Good luck graphing them

[winter osprey]

i always hated graphing in precalc and alg2

i mean i could do them but it was just tedious and annoying

[undone pumice]

🥀 the graph isn't even labeled properly

[round geyser]

So just use white and cover them

[undone pumice]

ik... but like...

the period is still pi so it shouldn't be in units of 1, 2, 3, 4 ...

i mean its not that bad

ill go graph it rq will see how long it takes

[winter osprey]

labeling with integers when doing trig funcs is so nasty 🤣

[undone pumice]

exactlyyy

kinda hard but just imagine at -pi/4, -3pi/4, pi/4, 3pi/4 the y is 1/3, cus its hard to make it look like that

[winter osprey]

i dont miss that shit one bit gang

[undone pumice]

🥀 my teacher made us graph em too

[undone pumice]

like last month

[undone pumice]

this is actually left pi/2

and you didn't shift it

the asymptotoes of parent tangent are at -pi/2 and pi/2

so when its shifted its supposed to move

and tangent is strictly increasing in between asymptotes

[round geyser]

Is it pi and 0

[undone pumice]

yea

[undone pumice]

^^ i drew mines up here if u wanna see (mine is more stretched out but)

[round geyser]

I was look at this

[undone pumice]

thats cotangent

cotangent as i mentioned before looks like tangent but reflected horizontally and has different asymptotes

[round geyser]

I though I start at -pi/2

Then make it -2pi/4 and add pi/4

[undone pumice]

for the cot or the tan graph

[winter osprey]

did yall have to use the rational root theorem

at all in ur class

[undone pumice]

yes

dont get me started

it'd be like 24 and 2 or something

and my teacher made us guess and check every value with synthetic division

not in precal tho

[winter osprey]

teach gave us a 6th degree poly and told us factor it, the constant term as a 24 or something that had a bunch of options

[undone pumice]

in that other course, integrated 2

[undone pumice]

🥀 rrt made me hate numbers like 24 like wym this number has so many factors

couldn't have been 20 or something

[winter osprey]

fr

[round geyser]

Yes

I got that too

I love rat theorem

[round ermine]

used it like twice ever

[undone pumice]

🥀 its js annoying

for me

ts pmo

glad we didin't use it this year at least

[round ermine]

i was taught like some rules that helped eliminate certain terms but i forgot them all

[winter osprey]

same here

[round geyser]

My teacher used desmos

Which made it quicker

Get the two roots

Choose one do synthetic

And done

[undone pumice]

lucky

we had to just

guess

no calculator allowed

[winter osprey]

we were only ever allowed scientific calcs, never graphing

[undone pumice]

you still have it written down as "right pi/2". graph looks right, make sure when you label its clear which tick mark is that value because it kinda looks like the one above where it's supposed to be

[undone pumice]

ye... not until precal were we allowed graphing

learned RRT in integrated 2 so was just like... "is x-1 a root" no its not "okay lets try x-2" going off vibes until u got one

[round geyser]

Also that unit was the first unit in pre-calc to get an 100

[undone pumice]

nice

didn't go over rrt in precalc tho at least so

yay

[round ermine]

why is y=1/3 1 unit above the x-axis and y=-1/3 is 2 units below

[round geyser]

That’s what was my question

[round ermine]

they should be the same distance from the x-axis though assuming your scale represents each unit as a change of 1/3

[foggy steeple]

yo so like im working on sigmas but im lowkey clueless on how to find these darn formulas

i mean like it gives me a series or sequence and the answer key has something completely different than what i put

[bronze geyser]

https://klipy.com/gifs/sigma-64--k01KRW7EYH8DYDC0D86S08XKV9T

[foggy steeple]

https://cdn.discordapp.com/attachments/813481530872561715/1360027669651263661/togif.gif

[round ermine]

systematic solution
||rearrange: k^3-k^2+36=0. rational root theorem: roots could be ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36. test them, you get -3 is a solution: (-3)^3-(-3)^2+36=0 -> -27-9+36=0 -> 0=0. so (x+3) is a root. polynomial division: (k+3)(k^2-4k+12). solve for zeros of quadratic: k= 2±2i√2||

[round geyser]

And yes

[modern yarrow]

Can anyone tutor me? I'm working on derivatives in my precalc class

[ionic temple]

isnt it -36

[undone pumice]

oh derivatives in precalc? like actual derivatives or limit definition

im in precalc too, i can try to help u

[modern yarrow]

product rule, quotient rule, second derivatives, chain rule

[undone pumice]

oh wow i see

alr so which parts dont make sense

do you not understand how/when to apply the rules or is it like more you dont get the concept/intuition behind stuff

[modern yarrow]

well its not that they dont make sense, I was absent last week for some personal stuff and now I have to learn the whole unit before the test tomorrow

[undone pumice]

oh i see

uhhh

so do you know the rules?

if not ill go over them and try to give u some examples

[modern yarrow]

I think I got the product rule down, Im trying to learn the quotient rule atm

im just learning as I do my papers I missed

[undone pumice]

if you look at quotient rule its actually very similar to product rule

in the numerator

like if you have say f(x)/g(x) its the same thing except you have to put the derivative of f before g and theres a minus sign instead of plug

and then divide that by g(x)^2

[modern yarrow]

Okay, thank you!

[undone pumice]

you're welcome

if you want more help ill be lurking here while i play some brawl stars

[modern yarrow]

Okay!

[winter osprey]

make sure you do it in that order, because if you choose to do g'f - f'g it will switch the sign

for the numerator that is

[scarlet cedar]

It is derived from the product rule

[round geyser]

This one is 100% easy

What do u get

[small estuary]

3

[round geyser]

And hint use difference of cubes

lim x-> 1 ((x-1)(x^2+x+1))/(x-1)
lim x-> 1 ( x^2+x+1)
(1)^2+(1)+1= lim x-> 1 3
D

That’s my step

[small estuary]

yeah, thats the most logical solution

[round geyser]

I would say this is the basic of pre-calc

[small estuary]

ts is way too easy

but for instance the kinf od limits that include recursice functions are just pure labor

[round ermine]

our precalc class never taught limits to this degree

[small estuary]

wow

[round geyser]

Also I saw one when I see lim x-) inf and the one u can’t solve take the coefficient of biggest power

[round ermine]

we were just taught limits of polynomials as x->infinity or -infinity

[small estuary]

nah those are eay too

[small estuary]

well, yeah

but like did you learn undefined cases?

[round geyser]

Like this

It’s just -1

[round ermine]

oh that’s just figuring out horizontal asymptotes

[round geyser]

-x^2/x^2 = -1

[round geyser]

Loook

[small estuary]

graphically yes

[round geyser]

Is this the one u learn

[round ermine]

we did learn about continuity of a function by seeing 1. whether a limit from the left exists 2. whether a limit from the right exists 3. whether the function is defined at that point

[round geyser]

Yea that’s what I did

[round ermine]

but that was really early on and i don’t think anybody remembered anything about it

other than me

because it’s a little more relevant in calculus

[round geyser]

f(a) exist
lim x-> a f(x) defines
lim x-> a f(x)= f(a)

Check

[round ermine]

forgot that last step

whoops

[small estuary]

so you learned partial limits

[round geyser]

3^2x + 4^2x= 7^2x

[ancient inlet]

hmm

x = 1/2 ?

[round geyser]

Yes

It is 1/2

[dire otter]

anyone open for practice?

[round ermine]

god damn you are in a lot of gacha servers

[dire otter]

u are in a lot of gacha servers

[dire otter]

wanna do integrals?

[round ermine]

i am currently going to class

[chilly arrow]

yea same

but then they put it in the exam 😭

did you do sinx/x

[round ermine]

i learned it on my own and i know it equals 1 but no

[round geyser]

Hey is zero also an asymptote specially vertical

[dire otter]

hmmm i cant tell with the hand writing on

[round geyser]

Graphing tan function

That’s should’ve been

[tame shard]

Why does this look like lambert W material

,w 49^x=9^x + 16^x

No lambert w suprisingly

[obsidian monolith]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by hexagon69_

[round ermine]

i mean if you simplify it down to 3^(2x)+4^(2x)=7^(2x) you can recognize that 3+4=7 so 2x has to equal 1

[tame shard]

Yea

I know

[round ermine]

i’m not entirely sure how you would more rigorously solve it though

[tame shard]

Logs perhaps?

Or lambert w function but idk

[round ermine]

i don’t know how logs would help

[tame shard]

Yea...

But like log(x^y) = y log (x), but the damn addition term

[dire otter]

https://docs.google.com/document/d/1wREEmT8ngiGGJGhOoUBWXcK75PYvkqeCjhlN-FhbUAQ/edit?usp=sharing
can someone do my homework

[tame shard]

no.

[dire otter]

why

[tame shard]

we can only help you with the questions, not outright give you the solution or do your hw

[dire otter]

well can u help me

[dire otter]

whats the fastest solution

[dire otter]

well ill just look for someone who can answer it

[tame shard]

also this is for like #multivariable-calculus

[dire otter]

its just a calculus with hard problem

[tame shard]

well if you think so then go to #calculus , or read #❓how-to-get-help to get help

[dire otter]

problem 3 and 4 uses traditional integration techniques

[dire otter]

so can u solve it or not

[tame shard]

no, but if you ask it in #calculus someone might be able to help you. Please note, we can only help you with solving the questions, not outright give you the solution or do your homework for you.

[dire otter]

so u will only give out solutions if its not homework

[tame shard]

no

we dont give out solutions in general

where did you even get that conclusion from?

[dire otter]

hmm so this chanel doesnt give out solutions in general

[tame shard]

this entire server*

you would know if you read #rules

[jolly karma]

How do I determine which value the foci is attached to when working with Ellipses?

Value as in x or y

[round ermine]

the foci will always be on the major axis
in other words you’ll need to determine whether the ellipse is oriented horizontally or vertically and from there you can determine where the foci are located (y for vertical, x for horizontal)

[round geyser]

What’s this mean

[raw hill]

Rectangle stuff.

[round geyser]

Thanks

[round geyser]

What’s the asymptote for parent graphs for tan x and cot x

[raw hill]

,w asymptotes of tan x

[obsidian monolith]

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[raw hill]

,w asymptotes of cot x

[obsidian monolith]

Results provided by WolframAlpha

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Requested by civil_service_pigeon

[raw hill]

odd way to write it but ok

$y=\tan(x)$ has asymptotes at $x=\frac{\pi}{2}+n\pi$ and $y=\cot(x)$ has asymptotes at $x=n\pi$ ($n$ is an integer.)

[obsidian monolith]

Civil Service Pigeon

[raw hill]

This is immediate from noting that $\tan x=\frac{\sin x}{\cos x}$ and $\cot x=\frac{\cos x}{\sin x}$ and considering where the denominators are equal to $0$

[obsidian monolith]

Civil Service Pigeon

[raw hill]

@round geyser

[round geyser]

Yes

Between 0 and 2pi

Is it gonna be pi/2 and 3pi/2

While cot x is 0 and pi/2?

[raw hill]

https://www.desmos.com/calculator/v5q3pyzje7

[round geyser]

x=0

[raw hill]

eh depends on how precise/colloquial you're being

"between" usually means you exclude the endpoints

[round geyser]

And x=pi

[raw hill]

but if you're talking about the closed interval $[0,2\pi]$, then $y=\cot x$ has vertical asymptotes at $x=0$, $x=\pi$, $x=2\pi$.

[obsidian monolith]

Civil Service Pigeon

[raw hill]

If you're talking about $(0,2\pi)$, then just $x=\pi$

[round geyser]

Since my teacher adds in quarter period

[obsidian monolith]

Civil Service Pigeon

[round geyser]

And horizontal shift

I would start at horizontal shift then add in quarter period

[orchid urchin]

hardest part of precalc?

[undone pumice]

wdym by this

[orchid urchin]

what was/is the hardest part of precalc for you specifically

[undone pumice]

oh

uhhh

matrices?

[orchid urchin]

thats an interesting one ngl

[undone pumice]

i understood most of what we learned from prior courses

[undone pumice]

and vectors and stuff

its just like... mostly cus i didn't really learn them before. having to learn stuff like unit vectors and stuff is just new and a bit hard to understand

[orchid urchin]

yeah I see how, personally for me its mostly probability and statistics

[undone pumice]

oh

we aren't learnng that in our precal course

i only know like probability up to normal distributions and stuff

bell curve thingy

[orchid urchin]

dam, its good because it shows how set theory becomes practical in probability

[undone pumice]

from prior course

[orchid urchin]

used to be just stuff I learned for proofs

[undone pumice]

oh

[orchid urchin]

that too, because it comes back up in calculus, you eventually learn why the bell curve has that formula

[undone pumice]

we only have 4 units in my class. two are more algebra ish focused, 1 is trig focused, 1 is matrices/parametric equations/vectors

[undone pumice]

mm i see

i didn't learn that in precal tho

a course called integrated math 3 that i took last yr

[orchid urchin]

yeah imo statistics is like the physics of math if that makes sense, you learn just like a base foundation and then way later you build on with way more advanced stuff (calculus)

[undone pumice]

mm icic

interesting

[orchid urchin]

well depending on the major

that is

[undone pumice]

ah

my teacher told us we should take ap stats or calc next year depending on our major but i didn't really know so i js picked calc

[orchid urchin]

yeah calc easily because calc is everything, calculus and linear algebra are top fucking tier, especially once you see how they are interconnected a bit, it be crazy enlightening in differential equations

[undone pumice]

ah

interesting

[winter osprey]

calc is good

i didnt like AP stats but i still got an A in it

[round geyser]

I would say limits

[round ermine]

stats is that one math subject i’m never interested in whenever it’s taught in classes

[chilly arrow]

is limits even precalc and not calc

[chilly arrow]

fair enough

[round ermine]

i think an introductory to limits can be precalc

[chilly arrow]

i feel like u just need some interesting questions

[round ermine]

but calculus dives much more deeply into limits

[chilly arrow]

right

[round geyser]

Wrong section

[dry verge]

Should I assume that the vectors' angles equal 0, 90, 180, and 270, respectively?

[astral apex]

what are the vectors you’re saying have those angles?

[echo sinew]

And?

[round geyser]

My first idea was combine from left to right

Like combine ln 3 -ln (x+5)

Info ln(3/(x+5)

[echo sinew]

Yup correct move 👍

[round geyser]

But then it’s getting stacked into a fraction tower

So my second idea was

[echo sinew]

Make sure, though, to check the domain of the whole equation (better if you do it at the beginning)

[round geyser]

y > 0

Now is 0 > y > -5

[echo sinew]

Yeah. If you don't like this, you can avoid that, by bringing the two negative terms to the right side

[round geyser]

If I’m correct

[round geyser]

Bring “ln x” to ur RHS

Now is ln 3 - ln (x+5) = ln x

[round geyser]

ln(3/(x+5)) = ln x

[round geyser]

ln cancels
3/(x+5) = x

[round geyser]

Will u get x^2 +5x-3=0

[echo sinew]

Yes

Which unfortunately doesn't have nice solutions at all

[round geyser]

I agree

It’s 0.54138

Or (-5 + sqrt(37))/2

[round geyser]

I like this kind of question

a=22.41(b+c)
b=0.83c

“What percent of b is a” can be a is what percent of b

Is that right

[echo sinew]

Yes

[echo sinew]

22.41 actually but yeah

Who tf did come up with that huge percentage lol

[round geyser]

So, my idea is sub in 0.83c
a=22.42((0.83c)+c)

a=22.42(1.83c)

Solving for c in b=0.83c which is c=0.83/b

a=(22.42 x 1.83 x 0.83)/b

[round geyser]

Check

[chilly arrow]

Trivial if you have a calculator but would be annoying to do manually

[round geyser]

Right

[round geyser]

But is that right

[prime bear]

Hi I’m a bit confused on how to take the derivative of this function (don’t know where to start) (12)

[echo sinew]

Do you agree it's a fraction, don't you? Therefore, you'll need to apply the formula for the derivative of a quotient

[prime bear]

yes

Do we use the power rule

To take the derivative of the numerator

[echo sinew]

Yup (together with chain rule)

[prime bear]

Ohh chain as well

Ok thanks

[undone pumice]

fyi this belongs in #calculus (if you dont have access, id:customize ) cus precal is just like alg trig limits and such. if you ever see anything inside a sqrt that would be chain

[prime bear]

Oops sorry we learning this in precalc so I forgot 😭

[undone pumice]

ah. is okay! (wow ur precal class is ahead hmm..)

[prime bear]

Help I guess it is prep for calc bc. Also thanks for reminder of chain

[undone pumice]

ah icic im only doing ab next year prolly

np

[dry verge]

The eastward direction ones are 0, the northward direction ones are 90, etc.

[fossil kindle]

chain rule, power rule, quotient rule

i think

i js did t(x) = sqrt((x³-1)/(4x²+4x+1))

and then f(x) = sqrt(x) and g(x) = (x³-1)/(4x²+4x+1) and t(x) = f(g(x))

prolly overkill tho

[fierce remnant]

I have 8 weeks to learn precalc

how tough will it be?

[undone pumice]

umm

how are your algebra and trig foundations

[fierce remnant]

I got 3.0 on algebra

[undone pumice]

3.0?

[fierce remnant]

yup for the smester but I slacked on the final

[undone pumice]

wait

3.0 out of like

what

[fierce remnant]

3.0 or B-

[undone pumice]

OH like gpa wise got it

uhhh

in my experience (im in precalc rn) precalc is just alg2 stuff with some new concepts, plus trig (unit circle and reciprocal) and that stuff, plus limits maybe

[fierce remnant]

I am older student and I want to get through this to get further in my CIS degree

[undone pumice]

I see

[fierce remnant]

I just bought precalc by sullivan

[undone pumice]

Ah

[lilac verge]

rewrite the numerator as (x^3 - 1)^(1/2) since a square root is a 1/2 power, then use quotient rule along with power rule and chain rule

[haughty crown]

I’m helping my friend through h algebra II and they’re doing trig and unit circle 💀

[fierce remnant]

Lord fang believes in me

[undone pumice]

ah

i did it in integrated 3 at my school

well not at mu school in middle school but ye

so lwk it wasn't that bad cus it was basically js review for me

but the kids who skipped int3 and were taking this class as sophmores lowk struggled with unit circle

[lost rivet]

How do u find the max of s

I’ve found dt/ds

[astral apex]

I think you have an exponent 1/2 in your derivative where you meant to have a -1/2

The second exponent 1/2 in the numerator

[lost rivet]

Wait oh yes

Y is my denominator 1 now bro

N I hv the (2t+100)^-1 still

[fickle gulch]

Hello

[tough flare]

Now I get it, it's $\sum_{i = 1}^{n} n = n^{2}$

[obsidian monolith]

Ujjawal Gupta

[tame shard]

I mean yea

Thats is just adding n to itself n times

[tough flare]

*adding

[fossil kindle]

no you need to use pythagorean theorem

[tame shard]

Oh yea

[tough flare]

There's also method of completing the square

[tame shard]

And factoring

[round ermine]

why did you re-add the ^(1/2) to the denominator on the second step there

[gaunt moat]

just use property of sigma

n*n times

[sharp geyser]

Does anyone recommend any good videos that clearly and effectively explain the unit circle and how to apply it? Im having difficulty understanding it and using it for my school work 😓

[red jewel]

what part of unit circle

[sharp geyser]

the whole thing😅

[hushed sphinx]

Can you explain more specifically what confuses you? The concept of a circle centered on (0,0) with radius 1 shouldn't take a video to explain.

[red jewel]

No like
There are basic stuff (such as the x axis is always cosx, y axis is always sinx)
Or trigonometric transformations? (Such sin(90-x) = cosx}

[sharp geyser]

Yeah, nah not that that’s obviously easy, but like the degrees given in the circle and the fractions with pi and the fractions on the outside

[hushed sphinx]

In some educational contexts, however, "unit circle" refers not just to the circle, but to an excessively complex diagram full of coordinates and angles for a dozen+ particular points.

[red jewel]

Ah like, 2π/5 = x°?

[sharp geyser]

Im sorry if i can’t really explain well 😓

[red jewel]

No its ok

[sharp geyser]

Yeah

[red jewel]

The only thing you need to in those stuff is

π = 180°