geometry-and-trigonometry

[rugged shuttle]

Wait @dark sparrow

How would I prove 21

If tan (pi/2) is undefined??

[dark sparrow]

tan(pi/2) is indeed undefined

what route did you try to go?

@rugged shuttle

pings me
doesnt respond

[rugged shuttle]

Oh sht mb

I just tried to use tangent subtraction formula

But quickly found out that it doesn’t work .-.

Also sorry

[dark sparrow]

the instruction does say to use the angle sum/diff formulas, but it does not specifically say to use the ones for tan.

[rugged shuttle]

Omg ur the best

[faint comet]

Does anyone know where I can find some trig proof questions and questions on general soloution

[spice hornet]

I was practice with the regular way and the inverted way. Finally got it.

I don't think the recipricals ones aren't too important

[faint comet]

Yo guys

[spice hornet]

Like I just have to flip a fraction. But is that an important question that I need to be like "oh I need to flip"? Besides a very obvious way of asking to

[faint comet]

Does this work for general soloution (ignore handwriting im high asf)

[rugged shuttle]

Honestly yeah lol besides graphing and identities that’s pretty much all the reciprocal functions do (I believe) until calc

[cyan latch]

Does trigonometry equations use radians or degrees more?

[obsidian harness]

Radians

You only use degrees until you know how to use radians

Cause sin(x) in degrees has some bad properties for if you are using trig functions in calculus

E.g sin(x)/x only approaches 1 as x approaches 0 in radians

[cyan latch]

Its harder right?

[obsidian harness]

And the derivative of sin(x) is cos(x) only in radians (in degrees you would need to multiply by pi/180)

[cyan latch]

Cause i can simply put degrees into a calculator

[obsidian harness]

Not really

[obsidian harness]

You can also put radians into a calc

pi radians = 180 degrees

remember the conversion and you're good

[cyan latch]

I guess so

[obsidian harness]

Yeah you should use a proper scientific calculator

[cyan latch]

I do

[obsidian harness]

Okok

[cyan latch]

I use a 25 euro scientific calculator

[obsidian harness]

cool

[cyan latch]

But its genuine garbage

Ill probably buy a new one

[cyan latch]

but for Geometry can you even use radians?

[dark sparrow]

radians are not something mystical

they are just another unit of angle

any angle can be expressed either in degrees or in radians, and it is a matter of convenience and contextual appropriateness which unit is to be used.

[solid needle]

100-4% is actually correct though

% = 1/100

100-4% = 100 - 4/100 = 100 - 0.04 = 99.96

i see no problems with that one

students should properly understand the arithmetic operations when handling percentages

you should do 100(1-4%) or 100(96%)

[cyan latch]

but i do like your style

[spice hornet]

Thank you for helping me!

[cyan latch]

precision is always good

[snow plinth]

Define a parallel rectangle P in R^n to be an n-fold Cartesian product of compact intervals, so its sides are parallel to the coordinate axes. Let an oblique rectangle O be a rectangle whose sides are not parallel to the coordinate axes; it is obtained as the image under an orthogonal transformation T of a parallel rectangle, so T(P)=O. An almost disjoint collection of sets means that the sets intersect each other at most along their boundary. v(.) is the volume of a rectangle, i.e. the product of the length of its sides (where length of say [a,b] is simply b-a). Note, that since an orthogonal transformation preserves lengths and angles, we have v(P)=v(O).

Consider then the following lemma in my lecture notes:

[somber coyote]

Philip

[snow plinth]

How hard is it to prove this lemma? Do you know a proof or a reference where this is proved? I'm grateful for anything! Prior to this, the author has proved some results about parallel rectangles which may be useful in proving the above lemma. If you want to know more, let me know and I'll post some more results the author has already proved.

[fickle cosmos]

I feel like I did this right .....

but it seems a bit off**...**

[rugged shuttle]

:)

[faint comet]

How come you did -3?

[upper karma]

Why wouldn’t it be -3?

[upper karma]

It is indeed a bit off

There’s no need to use pi in this case

[faint comet]

I don’t know I’m asking

[upper karma]

Oh, you can see that the function has 3 times the amplitude of a regular cosine function

[faint comet]

But why - not postive ?

[upper karma]

cos(0) evaluate to 1 but in the image you can see that y evaluates to -3 at 0

[faint comet]

Ahh ok

And if it went through postive 3 then it would be postive ?

[upper karma]

Yeah, If it’s 3 at 0 then it would be multiplied with positive 3

[twin python]

Can someone help me find x

[dark sparrow]

@twin python do you still need help with this? if yes, share all progress you've made so far.

[twin python]

Yes

I can't figure out tbh

I jolted the remaining angles but I still can't figure out

[dark sparrow]

jolted?

[twin python]

Noted

[dark sparrow]

ok

so let me ask you first a conceptual/vocabulary question

do you know the word that describes the placement of these angles relative to the polygon?

[twin python]

Exterior angle

[dark sparrow]

right.

now tell me: what do you know about the exterior angles of a polygon?

[twin python]

M not sure

[dark sparrow]

what do they all add up to?

[twin python]

360

[dark sparrow]

and you claimed not to know that...

but ok, now you've stated it.

[twin python]

Oh shoot I forgot sry

[dark sparrow]

the exterior angles add up to 360°.

as you yourself said.

do you see how to do your problem now?

[twin python]

Thank u very much

[upper karma]

Hello

Does anyone know how to simplify this

I tried using the double angle cosine formula but it didn't help at all

[dark sparrow]

what's the goal?

[upper karma]

Simplifying the expression

[dark sparrow]

why not look at the numerator

[light dock]

oh let's go this is my favorite trig identity

[upper karma]

I used photomath

It didnt solve it

[light dock]

you don't have to, you'll be fine without it

[dark sparrow]

that's your reason for not following my suggestion?

or i guess that wasn't meant as an answer to my question

so i'm going to reiterate:

look at the numerator

[upper karma]

Ok

Oh i got it

[light dock]

epic

[upper karma]

It is 1/2 tg 6a

Right?

[dark sparrow]

the whole thing simplifies to that, yes.

[upper karma]

Ok, thanks

[light dock]

i miss when they would put smiley faces with sunglasses when i got things right

unrelated to the problem itself

it's a good feeling

to me

[upper karma]

By the way

What do you mean by saying "trig identites"?

Does it mean trig formulas or functions?

[light dock]

by "identity" i mean "formulas that are true for any angle" essentially

[dark sparrow]

shit like sin^2(x) + cos^2(x) = 1

[upper karma]

Got it

[light dock]

for example, when you have an equation, say, sin(x) = 1, you can ask the question, "is this an identity?" and the answer would be no, since sin(x) is not always 1

but you can ask something like Ann said, like "is sin^2(x) + cos^2(x) = 1 an identity?" and then the answer would be yes

becuase we know that this is true no matter what x is

[upper karma]

Thanks. I understand it

[uncut totem]

Hello everyone. The condition of the task is to find the two sides of the isosceles trapezoid with the information given in the drawing. The correct answer should be 13√10, but somehow I got 13√8. I struggle to find my mistake

[dark sparrow]

1521 + 169 = ?

[uncut totem]

lol

[dark sparrow]

без да искаш си извадил числата вместо да събереш

[uncut totem]

не забелязах

благодаря

[lilac narwhal]

e^ipi = exp(ipi)

why do we not use the exp() form?

It seems simpler

[upper karma]

exp() is longer compared to e^

e^ix can be abbreviated to cis(x) anyways

[lilac narwhal]

Right

[upper karma]

And this isn’t really about #geometry-and-trigonometry

[lilac narwhal]

I kinda thought it was due to the cosx and isinx

Precalculas?

[upper karma]

Maybe

[weary haven]

Does anyone here study spherical trigonometry?

[upper karma]

Yeah but only the basics

[weary haven]

Nice

Have you found it useful

Or is the subject obsolete in terms of applications?

[north kindle]

Applications are for nerrrrrrds

[weary haven]

Yeah but I still wanna know if it is still used today

[dark sparrow]

i mean, navigation is a thing

would imagine that spherical trig has that as its biggest application

[upper karma]

It’s still useful in geography, astronomy and etc.

Celestial sphere calculations, stars transit time, celestial coordinates system conversion

As well as calculating great circle distances

[weary haven]

So geographers and astronomers still know this stuff

I just studied spherical trig for fun, not because it will be useful for anything

[upper karma]

Yeah and they used this as a base to many automatic computations

[grim notch]

i came across a super hard non euclidean geometry problem

in my hw

[tribal sequoia]

dang what's the problem

[faint comet]

How long did it take you guys to get good at trig proofs ?

[tribal sequoia]

around a week or so.

wdym by good

basic trig proofs or complicated onnes

[maiden brook]

pretty nice thing i saw: every convex quadrilateral can be disected into seven convex kites like this, for example

[mortal crow]

Um….I’ve only had proofs involving Pythagorean identities, complementary angle identities, and the most basic, reciprocal identities

[tall coral]

Assume $a^2+b^2=c^2
\newline c^2=b^2+2b(c-b)+(c-b)^2
\newline a^2 = 2b(c-b)+(c-b)^2
\newline a^2=2bc-2b^2+c^2-2bc+b^2
\newline a^2=-b^2+c^2
\newline a^2+b^2=c^2$

a little pythagoras theorem proof I made, not sure how to prove a^2 is equal to that expression though, of if I even need to.

[somber coyote]

KirbysGames

[wanton edge]

Did you just assume that a² + b² = c² to prove that a² + b² = c²? That looks like circular argument

[tall coral]

i have no clue how to prove a^2 though

(I'm also just not good at geomertry proofs ughh)

[tall coral]

or I guess it'd be better to say
\newline Let b and c be sides of a triangle such that c is greater than b
\newline $\exists a, b^2+a^2=c^2$

[somber coyote]

KirbysGames

[tall coral]

yeah, that's much nicer

[faint comet]

How do i Write a general solution for a trig equation

Like most of the time should I join the join angles together in 1 line or leave it seperate with angle +2piN

[dark sparrow]

it's fine to have separate families

like if you write {π/4 + 2πn, 3π/4 + 2πn | n ∈ Z} that's perfectly ok

[faint comet]

Ok, but like what if they ask you to write it as one is there a formula to do it or not really?

[dark sparrow]

gonna need to hear the exact formulation of that requirement.

[velvet bone]

from https://en.wikipedia.org/wiki/Delaunay_refinement

Chew's second algorithm
...
At each step, the circumcenter of a poor-quality triangle is inserted into the triangulation with one exception: If the circumcenter lies on the opposite side of an input segment as the poor quality triangle, the midpoint of the segment is inserted.

What do they mean by "input segment" here?

[obsidian harness]

Not an expert but

The circumcentre can lie outside the triangle

So it would be lying on the opposite side of say side YZ

[velvet bone]

so, that is what they meant? that the point is outside triangle?

[obsidian harness]

Hopefully

I think you'd have to read the papers linked in the references to understand what they mean

[velvet bone]

this is what I thought initially too but that would be a really weird to phrase it

[obsidian harness]

IDK

I hate when technical writing doesn't make any sense

[velvet bone]

my other assumption was that they talk about points that lie outside of constrained geometry, that would make more sense

not just triangle one single, but global boundary instead

well, still, what's input segment?

maybe they mean the longest edge of a triangle?

because it lies against worst angle in the triangle?

so, if the circumcenter is outside triangle, we divide this longest edge in half instead?

[obsidian harness]

Probably?

Yeah I guessed that's what they mean

[velvet bone]

is this algorithm good if I target edge size instead of angle?

[velvet bone]

btw they could've just said that circumcentre is used for accute triangles and midpoint of edge lying against obtuse angle for obtuse triangles

(well, and for right triangles both are the same location)

[obsidian harness]

Yeah why don't they communicate clearer...

[hallow flare]

Is this right

[clear anvil]

how would i find the diagonal?

[fallow sonnet]

how do i do this i have no idea

[spark tree]

Look form a right triangle for the 6 and 3 and get the hypotenuse same for the three and 4 . Then draw the other diagonal for the square. Then get the angle for the 3,4,5 triangle opp 4 adj 3 so tan4/3 + 90 then use the cosine rule to get the diagonal

[spark tree]

Both diagonals are equal anyways

[clear anvil]

yeah i considered that

but this is a geometry problem lol

ideally you don't need trig

[spark tree]

Huh?

You can’t solve it otherwise

[clear anvil]

you can

with pythag

[wanton edge]

yes, but there is a sneaky trick

[clear anvil]

what is it?

[spark tree]

Trust me you can’t

[clear anvil]

:V

You literally can

Smh

[spark tree]

Ok so how?

[spark tree]

Ok? Now what?

[wanton edge]

that is the clue

[clear anvil]

hard to draw with a sketchpad but yeah you didn't need trig at all for this

[spark tree]

Oh wow good job dude

Well i could have gotten it if i drew but am not home lmao sorry for doubting you 💀

[crystal sand]

i can just ask for help here righy

[wanton edge]

just post a question with full context

[trail tendon]

yes

[ember grove]

At first when i saw addition formulas and proofs i was like " why don't you just add the output of sines and that's it" but then I forgot that we are talking about functions not just simple direct numbers. I don't understand completely why but i guess due to the fact that a function follow a certain system of correlation like in the case of trig functions. They are not just direct values but fractions. What do you think?

[trail tendon]

what progress have you made on it?

[crystal sand]

i think for the value it goes with would 5pi/4

[trail tendon]

for what value?

[crystal sand]

-1

[trail tendon]

um

i mean its true but are there any other values?

wait its not true actually

💀 mb

[trail tendon]

what's your thought process for finding the 'value'

[crystal sand]

you would need to find -1 on the unit circle

like the one that matches up with it

and then you would need to find theta

[trail tendon]

not exactly

i'm going to call the value x ok?

like x = theta/2 + pi/4

right

so like tan(x) = -1

are you following or nah xD

[crystal sand]

yea

[trail tendon]

ok

tanx = sinx/cosx right?

[crystal sand]

ok

[trail tendon]

im just gonna get a picture of the unit circle

you're looking for a point where sinx/cosx = -1

cos is the x value

sin is the y value

of each point

right?

if this doesnt make sense at any time lmk lol

[crystal sand]

yea this is hard part for me

[trail tendon]

ok

look at 60 degrees

60 degrees = pi/3

the (x,y) coordinates here are (1/2 , (sqrt(3) )/2)

the 60 degrees or pi/3 are the inputs of a trig function

like sin(60 degrees)

or sin(pi/3)

sin(60 degrees) means the y value at the "60 degree" point

i'm not explaining this well 😭

[crystal sand]

uh

[trail tendon]

do you see the (x,y) points?

[crystal sand]

yea

[trail tendon]

ok

[crystal sand]

but if u do sinx/cosx

itll give me sqrt 3

[trail tendon]

thats if x is 60 degrees

so you kinda want to find where the degree gives a point (x,y) such that y/x = -1

[crystal sand]

i know that much

[trail tendon]

ok

[crystal sand]

is it just to a point of memorization?

[trail tendon]

kind of

but also

you're dividing two numbers to equal -1

[crystal sand]

yea but am i just supposed to guess to figure out which two point equal to -1

like the angle

[trail tendon]

um yes...

but it isn't too hard

because you know if you divide two numbers and you get 1

they have to be the same number

so what happens if you divide two numbers and u get -1

[crystal sand]

so sqrt2/2. sqrt2/2?

[trail tendon]

that would be if tan(x) = 1

but we're looking for tan(x) = -1

[crystal sand]

so one of them is negative

[trail tendon]

yes

either one, but not both

[crystal sand]

both would be positve?

[trail tendon]

like the x can be negative, or the y value can be negative, but the x and y cannot both be negative, or the x and y cannot both be positive

[crystal sand]

3pi/4 and 7pi/4

[trail tendon]

yeah

and keep in mind you could keep adding pi, but they limited you to only be from 0 to 2pi

like you could get 11pi/4 and its the same as 3pi/4

but its out of the domain

[crystal sand]

ye

[trail tendon]

so anyway

x = theta/2 + pi/4
x = 3pi/4 and 7pi/4

do u know how to continue

from there

[crystal sand]

it has something to do with 3pi/4+2pik

i would move the pi/4 over

amd then mult by *2 to get the theta alone

or x

[trail tendon]

you'd set x = 3pi/4 and solve for one solution of theta, then set x = 7pi/4 to solve for the other soltuion of theta

yeah

[crystal sand]

would i multiply everything by *2?

[trail tendon]

sure that would work

to get theta alone

[crystal sand]

i got pi+4pik

[dark sparrow]

!redir

[lime crown]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[gusty crane]

how do i find the surface area for problem 13

[maiden brook]

how do u do this 😭 i got that BC is 7 and drew a diagram but then what?

there are also similar triangles but u need more

[dark sparrow]

can you show us your diagram

[dark sparrow]

aight lemme see...

triangles AZY and ABC are similar

could do something with that i think

[maiden brook]

yeah but what else?

u need more

[dark sparrow]

let β := angle B (which is equal to angle AYZ of course)

then angle AZY = 120° - β

and angle BZX = β

so it would appear triangle BXZ turns out isosceles

[maiden brook]

yeah

and then what?

[dark sparrow]

XYC is isosceles as well

with some more angle chasing

mark off in both triangles which sides are equal

[fervent halo]

can someone help me find l6?

i can pay

[maiden brook]

oh yeah so it’s 7/2 then?

[dark sparrow]

so it would appear

[fervent halo]

shhh

[dark sparrow]

<@&268886789983436800> we got a payment solicitor

[obtuse field]

@fervent halo please stick to asking your question in one place in the server, and don't offer to pay for help

[fervent halo]

did i ask anywhere else?

i thought it was fitting to ask here since its a trigonometry chat

sorry

[obtuse field]

yeah you asked both here and in a bunch of help channels

please just ask in one place

it is on-topic, but we'd rather not have the same question posted in a bunch of different places in the server

[fervent halo]

mb

[fervent halo]

How do I solve the Riemann hypothesis?

[north kindle]

Step one: assume the reimann hypothesis is true
Step two: if the reimann hypothesis is false, this contradicts the assumption that the reimann hypothesis is true
Therefore, the reimann hypothesis must be true
Q.E.D. /s

[raven dove]

is this riight wat i did?

[maiden brook]

what is that last step

the reciprocal isn’t the same?

[raven dove]

ye yu right

i figured it out just now

[fathom stone]

I need to find the radius of the circle with an equilateral triangle inscribed in it but for the sides I have 2 lengths that can be the side length of the triangle since everything I did led to ⊾BAC=⊾BCD=90

this is the question

[maiden brook]

yep

[final vale]

System: tg(x)=-1 and cosx<0. Help please

I predict answer is x=-Pi/4+Pn, where n€Z

[sick vessel]

est ce que vous pouvez parler francais

[foggy tartan]

x=(3pi/4)+2kpi k in Z

ouii o

oui je parle fr

[sick vessel]

tu pe mexpliquer le tehoreme de tales

[foggy tartan]

celle de triangle

oui

[sick vessel]

oui stp

[foggy tartan]

d acc . mais ou tu veux je t ecris

[sick vessel]

au pire laisse tomber merci quand meme

[faint comet]

What’s argument angle ? Is it angle the angle that is the closest to the postive x-axis from your angle

[scenic merlin]

to find the negatives, do we ALWAYS go clockwise
im so confuse rn when doing other practice questions

all this arrow directions got me confused back when hs physics had the hand signs

[obsidian harness]

Yes

Angles are measured anticlockwise from the positive x-axis, by definition

[obsidian harness]

Oh what's this about

The right-hand rule?

[scenic merlin]

Man I miss physics tbh

[scenic merlin]

Thanks

[obsidian harness]

Interesting that you're not doing physics now, oh well

[scenic merlin]

I still ask myself that question why

[foggy lintel]

I need more difficult problems in the field of analytic geometry with solutions

[jaunty sail]

ABCA1B1C1 is a regular triangular prism, the sides of the base and the side edge have a height of 6. A secant plane is drawn through the base of the edges AC and BB1 and the vertex A1 of the prism. Find the cross-sectional area of ​​the prism of this plane.

How to solve it?

[tidal tulip]

Tbf mate I don’t know

[trail tendon]

i can't visualize it 💀

[dark sparrow]

did you make a diagram

[sudden shell]

I forgot how to find EA and FB Can someone help me out

[lime dune]

not enough information, show us the entire problem statement

[halcyon sierra]

I'm guessing E is the midpoint of AC and F is the midpoint of BC

[sudden shell]

Yes

[halcyon sierra]

then it's 20

AE=20

[sudden shell]

Can you explain

[halcyon sierra]

Since E and F are midpoints, AE =BF=20

all sides are equal

40/2 you get the midpoint

[sudden shell]

40/2=EF

[halcyon sierra]

yea

[sudden shell]

So how you know EA if we have nothing on CA

[halcyon sierra]

the sides are all equal

[sudden shell]

Oh I’m so stupid

[sudden shell]

Thank you

[halcyon sierra]

no problem

[strong vault]

Heyy does anyone know how the perimeter of e is 46 cm

[dark sparrow]

do you think the perimeter should be something else?

[pseudo jackal]

we have

• 10+7+9 = 26
• the three equal sides are each 3 = 9, and the 7 side is equal to a side as well so
  26+9+7=42
• the difference of lengths of the 10 side and 7 side is 3, using this and the other equal 7 side, the side directly below it should be 7-3=4
  42+4=46
  P = 46cm

[jaunty sail]

No

[dark sparrow]

then make a diagram lol

[jaunty sail]

lol

[dense garnet]

Nvm that wasn’t trig

[crisp yarrow]

can someone explain how the answer is C?
I worked it out and I got -cos(x)-2 in contrast to the answer 2-cos(x)

[rich scroll]

Im confused in number 18, letters a and b seems promising to me.

Number 17 as well

[lime dune]

a and b are both incorrect for 18

you cannot apply "isosceles triangle theorem" when the equal angles are in two different triangles

||c|| is the answer to 18, to see this ||consider the two angle equalities you have and the fact that the angles of any triangle sum to 180 degrees||

[scenic merlin]

There’s 4 answers in this question
What am I supposed to do for the remaining 3? Or did I do something wrong? This is as far as I can remember what to do when referencing to my lecture video in its examples.

[dark sparrow]

so your point of confusion is that this equation has 4 solutions...?

[scenic merlin]

Yea but how do I get the other 3

[dark sparrow]

hm

[crisp yarrow]

you only got -pi/3

what about -2pi/3?

thats one as well

[dark sparrow]

i would say that you would do well to first get the general solution.

[crisp yarrow]

no thats a waste of time

[dark sparrow]

sin(3y/2) = -sqrt(3)/2

y = -π/3 + 2πk, y = -2π/3 + 2πk, for k in Z

[dark sparrow]

au contraire

it will help you not fuck up which solutions are included and which are not

so it's anything but

[crisp yarrow]

I would just get the first two values of -root3/2

which is -pi/3 -2pi/3

then minus 2 pi from both

[dark sparrow]

from here, find the integer values of k with make your values fall in the desired range

[crisp yarrow]

then boom u have ur 4 solutions

[dark sparrow]

i mean sure you can also step up/down by 2pi

until you are out of the interval

amounts to the same shit

[dark sparrow]

this has an error

those should actually be 3y/2 not y

i forgor to include the 3/2 factor

so the general solutions for y would be
y = -pi + 4πk/3, y = -2pi + 4πk/3

[scenic merlin]

The issue with this method, my lecturer never showed this way from what I remember unless I forgot somewhere in those multiple lecture recordings - so it just confuses me because I don’t get it

I’ll get back to you because I had to rush outside for a moment

[crisp yarrow]

No need for general solution lol

@scenic merlin

[torpid tinsel]

helloo

[river falcon]

A right angled triangle has one 90° angle

It obeys the pythagoras theorem and its area is the easiest to find

[torpid tinsel]

does it also obey trig ratios?

[river falcon]

Equal to the (area of a rectangle with the same side lengths)÷2

[river falcon]

Yes. For example, if one of the two non 90° angles inside is x then sin x is the opposite side/hypotenuse

[torpid tinsel]

ohhhhh, it's starting to make more sense now

[torpid tinsel]

i didnt know this!

[river falcon]

Think about it

A rectangle cut in half diagonally is a right angled triangle

[torpid tinsel]

ur so right wth

HAHA sounds simple, i'm just super slow 😭

[river falcon]

That's about it idk what else to tell you ask any doubts

[torpid tinsel]

all i needed to know for now really

thank you for your help!

[river falcon]

You're welcome

[twilit surge]

can someone help me find the area of a kite

[hexed crater]

you can split a kite into two triangles and find the area of each and add them

depends on what youve been given tho, send a photo of the question

[twilit surge]

ok

[hexed crater]

okay so

look at the top triangle

its base is the long horizontal line, whihc is 40+3 = 43ft.

height of the top triangle is 8 ft.

area of a triangle = base x height x 0.5

so area of top triangle = 0.5 x 43 x 8 = 172 ft^2

bottom triangle is also 172 ft^2, same procedure

so total is 172+172 = 344 ft^2

[twilit surge]

imma see if it's right rq

[hexed crater]

yeah check

[twilit surge]

Yeah it was right

thank you 🙏

[hexed crater]

there u go then, np

theres also a general formula for area of a kite

but u can just do this, more intuitive

[dark sparrow]

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[dark sparrow]

you ended up doing the entire thing for him

[hexed crater]

ah okay sorry

[swift light]

huh???

[teal fjord]

correct

[rugged shuttle]

What’s the product to sum formulas for (cos a)(sin b) and (sin a)(cos b)? Or what’s the difference?

[trail tendon]

sin(a+b) = sin(a)cos(b) + cos(a)sin(b)
sin(a-b) = sin(a)cos(b) - cos(a)sin(b)

[rugged shuttle]

Isn’t that add/subtract identities

Oh nvm I found it

[trail tendon]

oh sorry i didnt see product to sum 💀

[trail tendon]

ok cool mb 💀

[rugged shuttle]

Lol it’s fine

,help

[somber coyote]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[rugged shuttle]

,tex sin $$8x$$

[somber coyote]

Cyberr

[rugged shuttle]

,tex sin 8x

[somber coyote]

Cyberr

[rugged shuttle]

,tex sin 8x = 2(sin 4x)(cos 4x)

[somber coyote]

Cyberr

[rugged shuttle]

Yo I did it

@empty yew wanna try this problem (verify identity)

[jagged wyvern]

Could someone tell me if i'm doing this right? Im not really good at word problems, but i tried to visualize it out

I assumed kite was 300m away, so it's HYP
And 53 degrees is greater than the other angle so it's going to the longer side
Sin(53) = x/300
Sin(53)(300) = x
x = approx 239.6

And since my hand is 4 feet above the ground, I add 4 to 239.6?

So 243.6? Someone pls confirm

[jagged wyvern]

<@&286206848099549185> 15 minutes

[gentle haven]

I believe you are right

Are you able to check your answer?

I'm assuming you are using bim

[trail tendon]

that is correct

[jagged wyvern]

This is a practice quiz cant

[jagged wyvern]

No I assumed both visualizations where either 300 was the HYP or the height, how do I know Im right

[trail tendon]

they should have specified the distance between what and what, but i think its safe to assume they meant the distance from your hand to the kite

and if that is the case, then yes it is the hypotenuse of the right triangle

otherwise there would be no trigonometry 🤣

that's not why i came to that conclusion, but it is funny lol

[jagged wyvern]

my bad

it's just the word problems that mess me up

[trail tendon]

i agree

[jagged wyvern]

like which values are the HYP, and the height, because you have to visualize the stuff

[trail tendon]

that word problems mess you up

XD jk

word problems are hard for me too lol

[jagged wyvern]

Ye..

[trail tendon]

yeah

[jagged wyvern]

Ty for the help

[rugged shuttle]

,tex tan(x/2) + csc x = (2-cos x)/sin x

[somber coyote]

Cyberr

[rugged shuttle]

,tex tan 3x = (3tan x-tan^3x)/(1-3tan^2x)

[somber coyote]

Cyberr

tan 3x = (3tan x-tan^3x)/(1-3tan^2x)
```Compilation error:```! Missing $ inserted.
<inserted text> 
                $
l.49 tan 3x = (3tan x-tan^
                          3x)/(1-3tan^2x)
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.

LaTeX Font Info:    Trying to load font information for OT1+lmr on input line 4
9.
(/usr/local/texlive/2023/texmf-dist/tex/latex/lm/ot1lmr.fd```

[rugged shuttle]

,tex $$tan \frac {x}{2} + csc x = \frac {2-cos x}{sin x}$$

[somber coyote]

Cyberr

[rugged shuttle]

,tex $$tan 3x = \frac {3tanx-tan^{3}}{1-3tan^{2}x}$$

[somber coyote]

Cyberr

[rugged shuttle]

Can someone help me prove these 2 identities?

Here is my work

[dark sparrow]

any reason why you are writing so small?

[dark sparrow]

don't forget \tan, \sin, \cos etc. in tex

[dark sparrow]

also in #80, how did $\frac{2-\cos(x)}{\sin(x)}$ transmogrify into $\frac{4-4\cos(x)+\cos^2(x)}{\sin^2(x)}$?

[somber coyote]

Ann

[dark sparrow]

oh wait, it looks like you tried to square both sides but then fell victim to freshman's dream on the left.

also in the tan(3x) one you didn't seem to fuck up (yet), all that remains on the LHS is to multiply outer num and denom by (1-tan^2(x)) as per standard nested fraction clearing procedure

[rugged shuttle]

Yo I got 87

It became clear when I put everything over 1-tan^2 x

I write small cuz I try to fit everything in 1 page

I’m new to tex so…

Hehe

Bruh no…

@dark sparrow I got $\frac {{1-\cos x}^{2}}{1-\cos^{2} x} = \frac {2-\cos x}^{2}{1-\cos^{2} x}

Bruh

$\frac {{1-\cos x}^{2}}{1-\cos^{2} x} = \frac {2-\cos x}^{2}{1-\cos^{2} x}$

[somber coyote]

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

[rugged shuttle]

BRUHHHHH

[dark sparrow]

bad habit!

you would do well to write a little bigger so that both you and whoever's reading your work don't have to use a microscope.

[dark sparrow]

it sounds like you did not fix the issue i pointed out

you cannot go from
tan(x/2) + 1/sin(x) = (2 - cos(x))/sin(x)
to
tan^2(x/2) + 1/sin^2(x) = [(2-cos(x))/sin(x)]^2

do you understand why? yes or no.

[rugged shuttle]

:0 yes :(

I forgot

[dark sparrow]

there's this lovely identity

$\tan(x/2) = \frac{1 - \cos(x)}{\sin(x)} = \frac{\sin(x)}{1 + \cos(x)}$

[somber coyote]

Ann

[dark sparrow]

the first half of which kills your problem

without any need for squaring

[rugged shuttle]

…

I have never seen this identity before in my life

Ok let me try the problem again

BRO I DID IN LITERALLY 3 steps

@dark sparrow how was this identity derived

Anyways I gtg so thx !

[spice hornet]

How does one get a root 3 /3

[dark sparrow]

are you asking how tan(30°) = sqrt(3)/3?

[spice hornet]

Yes

[spice hornet]

?

[spice hornet]

It was a video from Organic Chemistry Tutor

Where I was looking he was teaching how to find missing sides for various triangles

[dark sparrow]

i gave you a diagram to explain why tan(30°) = sqrt(3)/3

ochem tutor assumed that you know this value and used it to find some side lengths in his problem

[spice hornet]

Well can you explain this imagine for a beginner to this concept?

Or what can I look for to learn this

[spice hornet]

and is there another way to learn this as well?

[dark sparrow]

look up "SOH-CAH-TOA" and "trig values of special angles" maybe??

i am referring directly to the defn of trig functions

[spice hornet]

I do see one video kind of similar. I'll check that.

YT keeps showing me the right triangles. Which I actually did got the hang of them. But I want to keep learning more

[wicked tartan]

This is all I got from the teacher I need to find the distance of BD

[upper karma]

Looks trigonometric to me.

[wicked tartan]

Im not understanding this that well

[empty yew]

Hey @rugged shuttle how to do problems in the degrees? I don't know🫤

[digital egret]

yikes

[upper karma]

This is the cutest math I have ever seen.

Bro if this comes on my exam, I would rather fail

[upper karma]

yeah sorry mate, I messed up there.

[upper karma]

Like messed up real bad, like completely disregarded smth

Dude

The thought that smth like this even exists

Itself is frightening.

[upper karma]

What the mistake?

Am in grade 9 man 😭

Through I would still interested to see what went wrong

[upper karma]

Haha, no.

Nvm dude

I can't even solve simple side change maths

I mean… I know I messed up, but like this thing only required like grade 7 geometry and thats it. But problems can get a bit difficult in geometry, like this one which was meant for some high schoolers:

[winter garden]

can u bring back the old message

i was looking at it

please

[winter garden]

@upper karma so I used what ever I can remember from your old message, assuming you got that all the 3 area of the 3 triangles are the same, that is true and correct for all right triangle, or all triangle i am not sure only did it with right triangle

Why was that a reply message idk

Accident

[upper karma]

Oh essentially you have a triangle where the heights touch opposite sides at some points which form a triangle with lengths 5 12 and 13 and you need to find the area of the original triangle

[winter garden]

4 X 30 = 120

Why is it cap

Nvm gotta go rethink

[rugged shuttle]

Nvm it wasn’t that interesting of a problem you just had to use a certain identity to find the answer

[upper karma]

Oi oi oi

What's going on in trigonometry???

Speaking of trigonometry

I am very bad at trigonometry and geometry in general

[arctic vector]

Could someone help me with geometry?

[dark sparrow]

!da2a

[lime crown]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[past geyser]

sure lil bro

just post the question, no need to ask for someone to help you

[arctic vector]

Idk how to translate

Let me translate ir

Based on the drawing, calculate the area of ​​the shaded quadrilateral

@past geyser

[past geyser]

have you heard of the pythagorean theorem

@arctic vector

[arctic vector]

Yrd

yes

But im confused

Where to pit then umbers

And how to do the calculations

[dark sparrow]

from what language?

[arctic vector]

Lithuanian

[dark sparrow]

ok nevermind

right so you want the area of this, ok

let's maybe start by giving names to all the relevant points here.

gimme a moment to reproduce this shape.

[arctic vector]

Its asking fir the area

So yeah

[dark sparrow]

ok so here is the thing

i gave names to all relevant points and reproduced all of the markings from the diagram.

[arctic vector]

Alright

I i olny know where the rectanglr is

That at the top and botton should be 2

[dark sparrow]

the shape consists of 3 pieces, and we will proceed by calculating the area of each one:

• triangle CED
• rectangle BCEF
• triangle ABF

[dark sparrow]

can you tell me what segments, precisely, have length 2 cm?

i gave names to everything, specifically to avoid these vague terms of location.

[arctic vector]

If im right

It should be like that

[dark sparrow]

ok but can you name these two segments?

[arctic vector]

Wdym

[dark sparrow]

also wow, black bands.

[dark sparrow]

give them names.

as in

[arctic vector]

Umm

[dark sparrow]

they do have names, i want you to say their names.

[arctic vector]

Hard choicr

[arctic vector]

Idk

[dark sparrow]

"hard choice" means you either don't know or don't remember what i might be talking about.

[arctic vector]

Im terrible at math

[dark sparrow]

if a segment has one end named P and another end named Q, what is the name for the segment itself?

[arctic vector]

I cant tell it in english

I dont know hoe its called in english

How

[dark sparrow]

it's language-independent, i would say.

[arctic vector]

yes

[dark sparrow]

such a segment would have the name PQ.

[arctic vector]

Oh

[dark sparrow]

segments are named after their endpoints in just this manner.

always.

[arctic vector]

Ohh

[dark sparrow]

this is why we give names to things.

names to points more specifically.

so that we can refer to segments, angles, polygons etc. unambiguously via their names

and everybody can tell from the diagram what lies where.

understood?

[arctic vector]

50/50

Let me translate it

[dark sparrow]

too bad i don't speak lithuanian.

[arctic vector]

Ohh

Yes

I understand

[dark sparrow]

alrigt

alright*

so yes, BC = FE = 2 cm, you are right about this.

[arctic vector]

Just nor sure where to get the orher 2

[dark sparrow]

now for what comes next, i'll need you to follow my instructions closely.

[arctic vector]

okay

[dark sparrow]

and i also need you to NOT do something i didn't ask about.

[arctic vector]

Ok

[dark sparrow]

ok

look at this diagram now, and focus specifically on triangle CED.
you will be applying the Pythagorean theorem to this triangle.

i want you to fill in each blank space in the following formula with one of the sides of this triangle (CD, DE, and CE):

____^2 + ____^2 = ____^2

• You must reply with the entire formula, with the blanks filled in.
• I will reject an answer that consists of just the three sides listed with commas between them; for example, the answer DE, CD, CE or anything similar will be rejected.
• I will reject any answer that contains any numbers (aside from the 2 exponents); for example, the answer 2^2 + 3^2 = CD^2 or anything similar will be rejected.

[arctic vector]

Okay

[dark sparrow]

ok, awaiting response.

[arctic vector]

So wait

Oh nvm

Got it

[dark sparrow]

awaiting response!

[arctic vector]

I dont have a calculator by me so ik doing it in website

[dark sparrow]

no

you do not need a calculator for this.

do NOT do anything with the equation, except what i am asking.

[arctic vector]

Yes i know

[dark sparrow]

you can and should write it here in plain text.

^

if you need this symbol

it is most likely shift+6 on your keyboard.

[arctic vector]

Im on phone

Difficult for me haha

[dark sparrow]

i believe in you.

[arctic vector]

So if it’s not √ 2^2 plus 3^3 = ?? then how should it be?

Thats how my math teacher does it

[dark sparrow]

sounds like you didn't read my instructions!!!

I will reject any answer that contains any numbers (aside from the 2 exponents); for example, the answer 2^2 + 3^2 = CD^2 or anything similar will be rejected.

[arctic vector]

Im confused about that

Do i have to take out the square root too or what

[dark sparrow]

no, no square roots.

ok, let's go back one step.

since you aren't great with english mathematical terminology (which is perfectly understandable),

[arctic vector]

Its 13

[dark sparrow]

no, forget the numbers!

[arctic vector]

Yes

[dark sparrow]

since you aren't great with english mathematical terminology (which is perfectly understandable),
let us go over the relevant words for our scenario.

first,

do you know what the word "right angle" means?

[arctic vector]

I think so

[dark sparrow]

ok.

so you know that a right triangle is a triangle in which there is a 90° angle.

[arctic vector]

Yes

[dark sparrow]

(and there can be only one. there cannot be two.)

ok.

now

[arctic vector]

Yes

[dark sparrow]

two of the sides of a right triangle are touching the right angle, while the third side is across from it.

do you know what the side across from the right angle is called?

it's ok if you don't, i will tell you. but i want to see if you know.

[arctic vector]

Let me translate

I dont know.. half of the angle?

[dark sparrow]

no.

it's called the hypotenuse.

the side across from the right angle is called the hypotenuse.

[arctic vector]

Oh

[dark sparrow]

and i'm also going to say that the other two sides, which are touching the right angle, are called the legs.

[arctic vector]

Ooh

[arctic vector]

Yeah

[dark sparrow]

i can look up the translations of these into lithuanian if you want.

[arctic vector]

Inunderstand

[dark sparrow]

ok, right.

now that we that in order,

the Pythagorean theorem states: the square of the hypotenuse equals the sum of the squares of the legs.

in other words, (leg)^2 + (other leg)^2 = (hypotenuse)^2.

these might be named by different letters, so i am avoiding the usual presentation as a^2 + b^2 = c^2, and on purpose.

[arctic vector]

Oh

[dark sparrow]

so coming back to this

applying the pythagorean theorem to triangle CED

you would have:

CE^2 + ED^2 = CD^2

understand?

[arctic vector]

Yes

[dark sparrow]

ok.

so now we can fill in the numbers that we know

namely that CD = 3 and ED = 2 (as given to us directly)

which gets us:

CE^2 + 2^2 = 3^2

can you find CE now?

[arctic vector]

Pythagorean?

[dark sparrow]

i don't understand your question.

[arctic vector]

Ce by pythagorean?

[dark sparrow]

with all due respect, have you been listening to all the things i've been saying?

[arctic vector]

Tes

Yes

[zinc reef]

@arctic vector

[dark sparrow]

you have the equation

CE^2 + 2^2 = 3^2
yes it did come from the pythagorean theorem.

[arctic vector]

Yes

[dark sparrow]

but now it does not matter.

you just solve it like any other equation.

for that, its origin in the theorem doesn't matter.

[zinc reef]

The Pythagorean Theorem was created by Pythagoras

[arctic vector]

Oh

[dark sparrow]

do not overthink. just do what i tell you.

[arctic vector]

Maybe im like extremely sleepy thats why its hard for me to focus

[zinc reef]

It can solve world problems

[dark sparrow]

ok then the question is

why are you doing math while sleepy?

[zinc reef]

a2 + b2 = c2

[dark sparrow]

are you doing homework for tomorrow morning?

[dark sparrow]

with all due respect, you aren't helping.

[zinc reef]

No

[arctic vector]

Yes ive been trying alk day

[zinc reef]

I study pre university math

[arctic vector]

Because nobody is online when i need help

[zinc reef]

I literally do math for fun

[arctic vector]

Most people are only obline late

[dark sparrow]

are you in grade 5?

are you over or under 13 years of age?

[zinc reef]

You are acute

[dark sparrow]

i repeat @tame gale are you over or under 13 years of age?

ok

anyway

@arctic vector do you want to continue?

[arctic vector]

Mhm

[zinc reef]

@dark sparrow Can I join

[arctic vector]

What the

[upper karma]

Umm

[dark sparrow]

discord's TOS forbid users under 13 years old.

[upper karma]

Kiarny

That was a bit too much.

[dark sparrow]

<@&268886789983436800> user kiarny was seen posting a decidedly inappropriate gif.

which has since been deleted.

[upper karma]

Nice joke dude

I was 14 2 weeks ago

[arctic vector]

But can we speed up things? I really want to sleep and i’d try to learn from it in the morning

[upper karma]

Ahh

You know abbreviations such as "idk"

But don't know English.

Very interesting.

[zinc reef]

@arctic vector Can I join whatever you and Ann were doing, I'm certified Mathlete

[arctic vector]

mhm

[dark sparrow]

ok, let us try.

again

you have the equation

CE^2 + 4 = 9
in front of you

find CE.

[dark sparrow]

no.

[zinc reef]

Why

[dark sparrow]

you will obstruct more than you can help.

[arctic vector]

😭😭😭😭😭

Howwwwww

[zinc reef]

I'm a thirteen year old prodigy in highschool and I study university math and physics

[dark sparrow]

i do not need your assistance here, thank you.

[dark sparrow]

x^2 + 4 = 9, find x

i cannot tell you this in any other way except by doing it for you.

[zinc reef]

Not like assistance, it's that I'm lonely and no one wants to be my friend

[arctic vector]

13 x^2??

[dark sparrow]

if you just want to chat, there's #chill and #discussion. go there. @zinc reef

[dark sparrow]

.....

no, not even close.

[arctic vector]

This is how bad i am

[zinc reef]

Not chat talk about math

[dark sparrow]

#math-discussion .

but here you are just getting in my way.

[arctic vector]

I need tons of examples just for me to understand

[dark sparrow]

you need to review how to solve equations.

[arctic vector]

Yea

[dark sparrow]

i think it will be very difficult for us to progress here.

are you able to come back to this problem tomorrow?

[zinc reef]

No one pays attention to me because they think I'm crazy because I'm trying to solve Raymon's Hypothesis

[arctic vector]

yes

[dark sparrow]

go to #math-discussion.

or #discussion, or #chill.

[zinc reef]

No one pays attention to me there because they think I'm crazy because I'm trying to solve Raymon's Hypothesis

[dark sparrow]

<@&268886789983436800> we got a refusal to comply with redirection here

[zinc reef]

Okay fine

I'll go away

I'm sorry

[arctic vector]

@dark sparrow do we move this to tmrw

What time tho

[dark sparrow]

you're the one with the problem

come back here when it's convenient for you

[arctic vector]

Ok thx

[dark sparrow]

i can't and won't guarantee that i will be available

[arctic vector]

Yes ofc

Ok bye

[zinc reef]

@dark sparrow

[dark sparrow]

yes?

[zinc reef]

Why are you so mean

[dark sparrow]

in what way was i mean? by redirecting you while you were interrupting a convo with somebody else?

[zinc reef]

I said I wanted to join your conversation because I was curious

[dark sparrow]

the way you inserted yourself into the conversation was by repeating things i said earlier.

[zinc reef]

On what math equation you were talking about

[dark sparrow]

the conversation also had a purpose -- helping mm2swifty with his geometry problem -- and you got in the way of that.

i do apologize for being rather crass in telling you off.

[zinc reef]

I accept

[dark sparrow]

you try walking up to someone irl who's having a convo and interrupting them though -- i imagine they won't be very happy about that

[zinc reef]

I do imagine a lot of crazy things like in physics I imagined a blackhole curving the spce and time around it like Einstein

[dark sparrow]

seriously... #math-discussion is the place for that.

[zinc reef]

They think I'm crazy

[upper karma]

Bro maybe you weren't cooking food rather, smth else.

[zinc reef]

What do you mean

[upper karma]

It's a joke

[upper karma]

.

[glacial dawn]

how much earned

[glacial dawn]

ye

[dark sparrow]

uhh

does it tho.

[dark sparrow]

i think not. let me double check.

yeah, no, it doesn't.

[blazing veldt]

Why doesn't it? If I have a function $f\left(x\right)=x^{2}, \f\left(2x\right)=\left(2x\right)^{2}$

[somber coyote]

Primordial

[dark sparrow]

but you aren't saying the equality tan(A) = 3tan(A)/(1-tan(A)) - 1 holds for all A, are you?

(which indeed it doesn't)

[dark sparrow]

and that introduces f by saying f(x) is to be equal to x^2 for all x. so it is a different scenario entirely.

[blazing veldt]

I understand now. Thank you for clearing that up!

[forest blade]

Are there any good videos that go over double angles and half angles and explain what is happening well so you understand the concepts?

I just had an exam and feel like I'm struggling to figure out anything involving half angles

And I hate continuing to the following sections when I didn't feel good about the previous sections. I feel like it just builds a bad foundation :/

[rugged shuttle]

Professor Leonard (aka Superman)

[forest blade]

I'll take a look. I think I saw it pop up in a google search before. Thanks!

[jagged wyvern]

Can someone tell me why this is 10.1, and not 9.9?

I used Tan^-1(3.5/20)

3.5 is the height (O)
20 is the base length (A)

[cunning lion]

are you sure 20 ft is the base length?

[jagged wyvern]

Uhh

Ohh i see it now

Sin^-1 was suppose to be used because it says surface of ramp so hypothenuse

🤦‍♂️

[deft sail]

The line length in 20

I see 4 of 1/4 circles

I’m confused

[empty yew]

Which identity?

[oblique wadi]

what's your questoin?

[rugged shuttle]

Product to sum formula for right hand side

[faint comet]

Yo guys

For the equation like this without the pi/4 the y inter is 0 but since it has the pi/4 I know the y inter is going to be higher how do I calculate the exact vaule of that

[dark sparrow]

do you know in general how to calculate the y-intercept of a curve, trigonometric or otherwise?

[arctic vector]

@dark sparrow hi

[faint comet]

Yes bro read my question

[oblique wadi]

read their question

what does it mean to have a y-intercept in the xy-plane?

[faint comet]

My question is I know how to calculate the y-inter for sin which is represented by D, so I am asking how would calcite the y inter for sin when their is a horizontal transformation.

[oblique wadi]

answer this precisely, "A y-intercept of a function f(x) is ___"

look

you can set $f(x)=\sin (x+\frac \pi4)$

[somber coyote]

Kakaka

[faint comet]

D

[oblique wadi]

what is D?

do you know what a y-intercept is?

for a function

[faint comet]

Bro just answer my question lol idk why your being a prick

[oblique wadi]

a y-intercept is the value of f(x) when x=0

in your original questoin

why does it matter what the horizontal shift is?

$f(x)=\sqrt2 \sin (\pi/4)$

[somber coyote]

Kakaka

[dark sparrow]

don't call me bro.

please edit that word out of your message.

[oblique wadi]

was expecting that

xD

[dark sparrow]

f(0) not f(x)

[oblique wadi]

oops yes

[faint comet]

How is that = to 1

[oblique wadi]

sin π/4 = √2/2

from unit circle

[faint comet]

Bc without the shit y-inter is 0 but with the shift it is now 1

[oblique wadi]

I think I know what's going on

you don't actually understand what a y-intercept is

hence your confusion

[faint comet]

It’s not = to 1 tho

[faint comet]

You being condescending as hell when all I’m asking is a simple explanation says a lot about you as a person

[oblique wadi]

do you see your problem here

[dark sparrow]

@faint comet please edit the word "bro" out of your earlier message.

[oblique wadi]

I initially tried to direct you to think about the precise definition of the y-intercept to help resolve your confusion

[dark sparrow]

from here specifically. you have not yet done so.

[oblique wadi]

but you ignored me

[oblique wadi]

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

[somber coyote]

Kakaka

[faint comet]

That’s all I need bro thanks

[dark sparrow]

@faint comet i repeat: please edit the word "bro" out of your earlier message as i asked you 3 times now (incl. this one)

[faint comet]

Bro that’s so mean

[dark sparrow]

<@&268886789983436800> we got a deliberate misgenderer here

maybe 2 if we also count detakro

[faint comet]

What did I do I’m sticking up for her

[granite mortar]

i think detakro was referring to archer