geometry-and-trigonometry

[upper karma]

Pls help me

Someone

[upper steppe]

With what?

@upper karma

[upper karma]

Proofs

I’m about to cry

It’s so difficult

[upper steppe]

I hate proofs lol

What are you working on?

[upper karma]

Idk it’s jusf stupid proofs

[slate blade]

Diameter = 26, therefore circumference = 26pi.
y= aSin(bx-c)+d
b=2pi/26pi
b=1/13
The y transfer =+13

Triangle proofs or induction

[upper steppe]

The pinned tabs are proofs and ima do it last

@upper karma send me pic maybe I can help

[upper karma]

Ok

[upper steppe]

I’m still learning though

[slate blade]

@upper karma my best suggestion is just to get a piece of paper and “redraw” all of the figures with the given/derived information

[upper karma]

I fried

[upper steppe]

@upper karma can I see the choices?

Like click the box so I can see the options

[upper steppe]

What the actual huh?!?!

[slate blade]

First do AE is equal in magnitude to EC

[upper karma]

It’s impossible

[slate blade]

Since it is give. That DB bisects it

[upper steppe]

That’s geometry right?

[upper karma]

Ye

[slate blade]

Yeah. Trying to prove congruency using ASS or SSS formulae

Or any of the other combinations

[upper steppe]

The x and t threw me off

Y*

[slate blade]

X and y and stuff are there as placeholders

You’re meant to fill it in with the letters from the triangles

[upper karma]

Nvm

I got it

100%

[upper steppe]

Xyz is congruent to tuv

I think

[upper karma]

Less go

[slate blade]

Epic

[upper karma]

Gn

[upper steppe]

What was it?

[upper karma]

Not sure

I just did it

And it says it’s right

So pce

[upper steppe]

What was the answer tho...

[slate blade]

Lol

[upper steppe]

Please help

I’m to stupid and tired for this

I think it’s 38

It was 38 now I don’t get this one at all

[west basin]

sounds good

oh lol

well you are given angles on a line

so they add up to?

[upper steppe]

I’m sorry if I seem weird I been in school since 7:20 and been doing hw since I got out at 2:20 and now it’s almost 11 my brain is dead

[west basin]

its cool

[upper steppe]

180

[west basin]

yes

so 180 - 114

is the red angles measure

[upper steppe]

Huh?

[west basin]

angles on a line add up to 180

so the measure of the red angle will be 180 - 114

[upper steppe]

66

[west basin]

yes

so both red angles will be 66

so add them and subtract from 180

[upper steppe]

48?

[west basin]

yeah

[upper steppe]

Fml I got another one

And it feels like I’m about to faint

This is 60 right?

[west basin]

yeah

[upper steppe]

I have 168 questions estimated

Left then I’m done

[west basin]

sick

[tender prawn]

168 jeez

[west basin]

were you assigned them for one assigment or did you just slack off

and wait til the last minute

[upper steppe]

My dumbass slacked off

[west basin]

yeah there ya go

played yourself

[upper steppe]

This was assigned for last and this week

And it’s due Sunday

[west basin]

if you just did a small portion everyday it would of bene a lot better and your understanding would probably be better

[upper steppe]

Yup

I’m kicking myself rn

[tender prawn]

lol im doing that too

[upper steppe]

I can barely remember something for 2 seconds rn

[tender prawn]

i had 2 weeks to do these 8 problems but ive spent the whole entire day bashing them out

[upper steppe]

Bro I was studying for my algebra eoc instead

Cause I tested Tuesday

Instead of doing this

I should’ve done this instead of studies

[tender prawn]

ig thts fair

[upper steppe]

Brb ima to get water

I’m back

Can someone explain this to me?

[tender prawn]

just like

count the side lengths

u can use side angle side cuz u know one of them is a right triangle

[upper steppe]

They both are

[tender prawn]

oh yeah both oops

[upper steppe]

Your method is so much easier

I was trying to calculate the points and get the measurement like that

[tender prawn]

lol

[upper steppe]

Apparently I can’t count 😦

[tender prawn]

rip

[upper steppe]

Can you pls help me?

I been on it for a hour

And my smart score is 61 and I need to 90

[pure cape]

I dont think we are supposed to help you with passing test

[upper steppe]

It’s not a test....

[pure cape]

Then why should there be score

[upper steppe]

I don’t think a teacher would give a test at 11:49 pm

[upper steppe]

That’s the way the program goes

[pure cape]

And also, help you with what?

Could you post the problem?

[upper steppe]

Scroll up I need a better way to solve these

Cause for some reason even counting them i get it wrong

[pure cape]

You need help with the picture above right?

[upper steppe]

That’s the type of questions I’m getting yes

[pure cape]

Well so basically we need to prpve it by using sides, since we dont have any angles

The fastest way i could think of us sss

If you do intend tp use sas with the angle being the right angle, then you must prove that they are both right triangle

[upper steppe]

I don’t need fast

I need accurate

Can’t you just tell a triangle by the angles (looking no measurement)

[pure cape]

So do you know how to calculate the distance between two points on a graph?

[upper steppe]

Yeah distance formula

[pure cape]

Yes, by looking, but if you say by looking at a triangle i see this is a right triangle, not a lot of mathematicians will agree with you

Like you said, you need accurate not fast

[pure cape]

What is it?

[upper steppe]

D=(x2-x1)^2 + (y2-y1)^2

[pure cape]

Sth is missing innit?

[upper steppe]

Huh?

If you mean squared

I don’t have that symbol

On my@ketboard

[pure cape]

Your formula is missing something

Yes

[upper steppe]

Keyboard*

[pure cape]

There should be a square root

Ok so can you calculate the distance between point E and F for me?

[upper steppe]

8-9

[pure cape]

What?

Use the formula

[upper steppe]

I did

[pure cape]

Point E(0;8)
Point F(9;-3)

So whats the distance

[upper steppe]

8.94

Round of to 9

[pure cape]

You did something wrong

Try again

Tell me x1 and x2

And y1 y2

[pure cape]

Whats 11^2

What does it equal

[upper steppe]

121

[pure cape]

So 81 + 121 is?

[upper steppe]

202

[pure cape]

For the record (-11)^2 isnt -1

Do you know why?

[upper steppe]

Yeah I know why

[pure cape]

And there should be a square around the -3 -8 on the second line

[upper steppe]

Just ain’t think right I been doing school since 7:20 am

Gotta go slower with me

[pure cape]

Sure

So (x1 - x2)^2 + (y1 - y2)^2 = 202

Right?

Now take the square root of that

Which is sqrt(202)

[upper steppe]

14.21

[pure cape]

Which is arppox. 14.2

But next time when doing math, try not to approximate

Just keep it as sqrt(202)

Ok so now we know that EF = sqrt(202)

Can you find BC?

Write out x1 x2 y1 and y2

Then plug it in the formula

Show me your work ok?

[upper steppe]

Same thing

[pure cape]

Good

[pure cape]

Ok so we proved ef = bc

Now can you do that for the other two sides?

With the same method?

[upper steppe]

Yes thank you

[pure cape]

Yw

[upper steppe]

@pure cape just finished that assignment

Now it’s time for the most hate proofs

Oops sorry

Didn’t mean to tag

[pure cape]

its ok

[remote sierra]

I need help with a problem

[green osprey]

hey i need help with some geometry

[dark sparrow]

y'all should post your questions right away instead of waiting for someone to show up

[green osprey]

oh

okay so what is a perpendicular bisector and how do i calculate one with two perpendicular bisectors

[silent plank]

a perpendicular bisector as the name suggests is a line that is perpendicular to another line segment and bisects it

the latter part of the question is unclear

perhaps post the question exactly as worded

[green osprey]

the segments are in a triangle

the question is In Triangle ABC DZ, EZ and FZ are perpendicular bisectors. If AZ = 76 and BD = 51 and ZE = 23, what is the length of AB and ZC.

this is the triangle

[placid bone]

ah

you can use vectors for this if you know how to

[green osprey]

dont know

[placid bone]

it's quite simple if you do use vectors

ah

in that case

[silent plank]

DZ being a perpendicular bisector here means that it bisects AB,
i.e. BD = DA

[upper karma]

Are calculus proves based on geometry

[placid bone]

wait sike

you can't use vectors for this

you can prove that the altitudes of a triangle are concurrent with vectors stho

[green osprey]

idk man

we havent went over vectors yet

[upper karma]

no

[silent plank]

ignore the stuff about vectors

[green osprey]

i still dont know what ZC is lol

[silent plank]

apply pythagoras a few times

[green osprey]

tried that and i ended up with irrational numbers

[silent plank]

and why would that be an issue?

are you implying that lengths can't be irrational?

[green osprey]

its on delta math and usually its supposed to be rational

[upper karma]

...

[silent plank]

supposed to be...

[green osprey]

it should work

[upper karma]

show calculations

[silent plank]

perhaps show your work

[green osprey]

ok

[silent plank]

you could even just apply sas congruency

[green osprey]

wait im stupid

its 76

wait what

so far i figured out bz = 71and AD= 51 and i know az = 76, ZE = 23, and BD = 51

[silent plank]

how are you getting BZ as 71

[silent plank]

i can't read that second line

[green osprey]

76^2 = 51^2 + DZ^2

DZ = 5*sqrt(127)

sqrt((5*sqrt(127))^2+51^2)=BZ

[silent plank]

and you won't get 71 from that

[green osprey]

oop my bad 76

[silent plank]

i also mentioned a bit later that you could also just apply SAS congruency

[green osprey]

yea but then i would need to know ZE and EC

[silent plank]

not really no

you're told certain segments are perp bisectors

[green osprey]

yeah

[silent plank]

DZ is common,
DB = DA since DZ is a perp bis
<BDZ = <ADZ = 90°
for congruency of ADZ and BDZ (by SAS) hence AZ = BZ
similar idea for the other triangles

[green osprey]

wait what

how can db = ab

[silent plank]

whops typo

[green osprey]

.-.

i knew that

they both were 76

i need to find zc

[silent plank]

similar idea for the other triangles

do exactly the same thing

[green osprey]

i dont have ze though

or ec

or BC in general

[silent plank]

your told what ZE is not that it really matters

its a common side in the bottom two triangles

its also a perpendicular bisector

and congruency can be proven in the exact same way as the other triangles earlier

[green osprey]

but i need to find either ZE and EC or FC and ZF too prove SAS right

sorry if i sound stupid i still havent slept

[silent plank]

use exactly the same ideas i did for the two triangles above

ZE is common in tri BEZ and CEZE

BE = EC since ZE is a perp bis

right angles etc

congruency SAS etc

[green osprey]

ohhhhh

i get it

so 76?

[silent plank]

yes

[green osprey]

hopefully i dont fuck this up

im almost at a c

POGGERS

the homework automatically grades

[earnest fjord]

Show that the vertices A(-2,0), B(0, 2sqrt3), C(2,0) are vertices of an equilateral triangle

[dusky surge]

What have you tried?

[earnest fjord]

My current best guess is starting to find the lengths of the sides

[dusky surge]

Good start!

[earnest fjord]

And then compare AB = AC = BC

[dusky surge]

That's actually a very good approach.

Do you know how to find the distance between 2 points?

[earnest fjord]

Yes, formula

[dusky surge]

Then it's going to be easy as you just sub the points into the formula

[earnest fjord]

I'll try that

[dusky surge]

Good luck!

[earnest fjord]

Thank you

The length of AC is 12 cm. CD is tangent to the circle at D. The radius is 6cm. What is the area of the shaded region?

With this one I'm quite lost as it's one of the harder things for me

[earnest echo]

Do you know that tangent makes right angle with radius??

[earnest fjord]

here?

[earnest echo]

Yes

[earnest fjord]

I have a pretty good idea about how this would work now

[earnest echo]

👍

[earnest fjord]

Find sector first and then the entire triangle

and subtract sector from the triangle area

to get shaded region

[earnest echo]

Yes

[earnest fjord]

Wait so

Because that's right angled we can just use pythagora's to find the third side?

[earnest echo]

Third angle??

[earnest fjord]

third side*

[earnest echo]

Yes

[hybrid quiver]

Help?

I've tried using the 15 tangent formula and then doing Poly transforms

Is this optimal?

[gray whale]

Could someone be so kind to explain D) I don’t know how to add them to get the result in 2nd image

Please @ me

If you respond. Thanks in advance 🙂

[main lintel]

@gray whale Can you write the question here as text with substituted values

I have one solution. It seems kind of obvious that you have to try to split S2 into two parts, by the sin(a-b) = sinacosb - cosasinb rule

This gives you S2 = 2*sin200t*cos(0.2) - 2 *cos200t*sin(0.2)

Now ST= S1+S2 = (3+2cos(0.2))*sin200t - 2sin(0.2)*cos(200t) [1]

Where 5*cos(0,2) and 3+2sin(0.2) are some plain numbers.

Now, by the same formula above, just in reverse you have that these two numbers:
5cos(0.2) and 3+2*sin(0.2)
can be transformed in such a way that they form the terms of an analogue sinusiode:
A*sin(x-y) = Asinxcosy - Acosxsiny

Comparing this with equation [1] above we have that:

x = 200t
Acosy = 5*cos(0.2)
Asiny = 3+2*sin(0.2)

You can substitute A from first into the second to get: tan y = some number, which gives the basic solution for y as 0.08 and an infinite number of them y=0.08 + k*PI

Put that into the first equation.
You get two sets of solutions,
A=4.916 phi=0.08+2kPI
A=-4.916 phi=0.08+PI+2kPI
Because the negative amplitude just shifts the phi by PI, the final equation is the same for both solutions
4.916*sin(200t-0.08+2kPI)
For some reason they just took the solution with k=0 so
4.916*sin(200(t - 0.08/200))

[upper karma]

Explain why the distance point P travels is the same as the angle of revolution (measured in radians) of this wheel.

[main lintel]

P is travelling along the arc of this circle, so the distance it travels when circle rotates by angle of revolution is equal to the length of the arc respective to that angle.

The length of the arc is length = radius*angle_in_radians=1 [ft]*angle_in_radians= angle_in_radians [ft]

@upper karma

[upper karma]

thanks

[main lintel]

First, I think there are 7 solutions strictly between 0 and pi/2. 7 solutions between -pi/2 and 0, and 0 as the 15th solution.

There might even be 8 intersections on each side plus 0, so 17 in total from -pi/2 till pi/2

[hybrid quiver]

@main lintel https://awwapp.com/b/umeekivy5f9ua/

[main lintel]

I don't get it what you did there

What is the Poly treansform? @hybrid quiver

[hybrid quiver]

k=sqrt(1/m) @main lintel

[main lintel]

You mean something like this: https://trans4mind.com/personal_development/mathematics/trigonometry/multipleAnglesTangent.htm#Observations_from_the_Table

[hybrid quiver]

@main lintel im changing the polynomial such that it has roots of 1/tan^2(x)

[main lintel]

@hybrid quiver How you got the formula for tg 15x, have you manually did the steps or you have some table?

[hybrid quiver]

the coefficients are 15 choose n @main lintel

[main lintel]

Aha, I see this

@hybrid quiver After plugging in that, you got a polynom of 7th degree

[hybrid quiver]

Correct, it cancels out

[main lintel]

And you concluded that the wanted sum is the "second term coefficient/leading term coefficient" of the polynom

by Viete's formula

[hybrid quiver]

yessir

[main lintel]

It is a pretty interesting and very hard problem

[hybrid quiver]

5/78 is incorrect!?

@main lintel can you recheck my work? it's wrong!

[main lintel]

I'm still not sure about the sqrt 1/k swap.

Other things seem reasonable

[hybrid quiver]

It's by polynomial transformations

the inverse function 1/k^2

[main lintel]

I mean, seems reasonable, but too late to comprehend it 100% 😄

[hybrid quiver]

https://www.youtube.com/watch?v=VomStUJ8BT0

Here's what I watched

[main lintel]

After the first step you have a polynom equation
P(u) = 0 where u=tan(x).

[hybrid quiver]

@main lintel did you watch the poly video?

[main lintel]

No

Polynom is just a function

[hybrid quiver]

yes

@main lintel this was what was wrong -_-

thanks to the legendary David Altizio

[west basin]

light theme

[hybrid quiver]

@west basin 🤣

[main lintel]

If we have f(x) and we do x=g(y), and if g is bijection on the observed range, we have f(g(y)). Then the numbers x1, ..., xn that are the nullpoints of the old function have the respective y1=g-1(x1), ..., yn=g-1(xn) that are the nullpoints of the new function.
Because f(g(g-1(x1))=f(x1)=0

[hybrid quiver]

no, it's just inverted dark theme!

🇩🇯 🦋 👨

djmothman

aka djmathman

aka David Altizio

[upper karma]

A carousel moves in a counterclockwise direction. It makes one complete revolution every 16 seconds. Han is sitting 11 feet from the center of the carousel. Here is a view of the carousel from above.

Based on the view from above, write an equation describing Han's vertical position v, in feet, relative to the center of the carousel t seconds after the carousel starts moving.

[mild anvil]

can someone help with my trig hw

[slate blade]

@upper karma you get a cos eqn

[
v=-11 \cos{(\frac{\pi}{8} \times t)}
]

[somber coyote]

Doodaide

[upper karma]

k thanks!

[slate blade]

I think at least. I'm not that smart either

@mild anvil you get 2 solutions from this right

[mild anvil]

i thought it was 4

i tried 2 solutions but it kept saying to was wrong

[slate blade]

Well do you have your working?

[
x=\frac{\pi}{3}, \frac{2\pi}{3}
]

[somber coyote]

Doodaide

[mild anvil]

this is what it keeps saying

[slate blade]

oh shit u right double angles my bad

yeah it's 4 solutions

[mild anvil]

i’m rlly struggling with double and half angle identities bc our teacher didn’t teach us 😢

[slate blade]

ok so for your answer, it's 4 answers with the following

basically all variations of 60 degrees

so pi/3, 2pi/3, 4pi/3, 5pi/3

I think if I converted right sorry I'm kinda rusty as well

but here I'll give you a crash course on double angles if you want

[mild anvil]

omg thank you i love you

[slate blade]

but was the thing correct?

[mild anvil]

yes

[slate blade]

boom shakalaka

ok so now onto double angles

[mild anvil]

i’ve been stuck on it for so long 😭

[civic crown]

hi can i have help

[mild anvil]

can you help me with the rest of my hw?

[slate blade]

you basically know the three basic angle identities sin cos and tan

lol sure

[mild anvil]

yayyy

[slate blade]

let's finish the hw before the magic lesson

[mild anvil]

okie

everyone in this server is rlly smart i’m jelly

[slate blade]

empty set I think

[mild anvil]

i tried it it was wrong

[slate blade]

bruh what

[mild anvil]

yea i thought it would be that too but it’s not

[slate blade]

hmMMMMM could it be a typo?

oh wait no it's 3tan not 30

sorry but could you send screenshots instead of images its kinda hard to tell

[mild anvil]

yea gimmie one sec

[slate blade]

ok so here are the answers:

pi/6, 5pi/6, 3pi/2

[mild anvil]

dang i got it wrong cause it wnted it in degrees 😦

[slate blade]

oh lol whoops my bad

[mild anvil]

its ok

[slate blade]

sorry too used to workin gin rads.

[mild anvil]

its giving me another chance

[slate blade]

it's 30, 270, 150

hopefully

[mild anvil]

no its not that

[slate blade]

bruhhhh

wait I just realised solving for sin is not equal to secant

I am so sorry

I ded today

[mild anvil]

its ok you can still give me the right answer for it. its giving me another chance

if you want to you dont have to

[slate blade]

0, 120, 240

Interval wise, [x,y) means that it doens't include y right

[mild anvil]

wow you're so smart

i wish i had the talent

[slate blade]

lmao you're overestimating me

but let's complete the quesitons

[mild anvil]

okie

[slate blade]

uhhhhhhh have you tried empty set on this one

from memory I did one similar and it didn't have an exact answer

[mild anvil]

yea its wrong

did you leave 😦

[pure cape]

try to do it slowly

[slate blade]

no I'm still alive I'm just confuse d

[pure cape]

divide by 5 on both sides, and consider a formula for theta

[mild anvil]

oh ok 🙂

[slate blade]

shouldn't there be multiple answers is what im considering

[mild anvil]

yes because its a set

[pure cape]

eventually you get something like (pi/4 + kpi)/3

for all integers k

[slate blade]

cuz if so, it's just pi/4 and someother thing

[pure cape]

and now apply that to the interval [0:2pi)

[slate blade]

[0:2pi[

[mild anvil]

it gave me 6 different answers

[pure cape]

yes

there should be 6 different answers

for k from 0 to 5

[placid bone]

yep

solid

[slate blade]

lit

[mild anvil]

like i plugged in 2 answers but it said that was wrong and gave me 6

[pure cape]

since when k is 6, it exceeds the condition [0;2pi)

[mild anvil]

but i got it wrong

[pure cape]

what are your answers

and show me how you got it

[placid bone]

if you just think about it, tan3x just compresses the graph by a factor of 3

[mild anvil]

it went away

[pure cape]

what went away?

how can your answers go away

[slate blade]

they don't love you anymore

ok did you get

pi/12, 5pi/12, 3pi/4, 13pi/12, 17pi/12

wait I can use greek keyboard

[mild anvil]

like the correct answers went away because i clicked give me a similar one cause i got it wrong

yes rhats what i got but i got it wrong so it gave me another problem

[slate blade]

βρθη

[pure cape]

you're missing an answer

a solution, to be more exact

[slate blade]

isn't the next one 25pi/12? so out of its domain?

[pure cape]

what happens if k=5?

doesnt seem like it goes out of the domain

[slate blade]

wait no I forgot to type it lol

7pi/4

I just skipped it my b

alright what's the next question

[slate blade]

same panic different disco

I'll type it correctly this time watch me

π/12, 5π/12, 3π/4, 13π/12, 17π/12, 7π/4

I hope that's right

[pure cape]

basically same stuff, i think you should let luna do it first before giving out the answers

or try to explain

[slate blade]

good point

[pure cape]

giving out the answers is more of harming the person than actually helping them

[mild anvil]

it was correct thank you

[slate blade]

do you understadn how it is don

e

loona

[mild anvil]

no i dont 😦

[placid bone]

eek

when is tan 3x=1

[mild anvil]

could you explain

[pure cape]

like i said

[slate blade]

that ain't good

[pure cape]

its better to explain

[placid bone]

that's essentially what it's asking

[mild anvil]

like ive never learned it my teacher never taught us

[placid bone]

alright

so

[slate blade]

trig time

[placid bone]

when is tan x=1

from 0 to 6pi

[pure cape]

then why did he give you this?

[placid bone]

and then divide your answers by 3

[pure cape]

thats a bad teacher

[mild anvil]

i know she sucks i hate her

[pure cape]

oh well

[slate blade]

or is it they taught you but did it really badly

[mild anvil]

but i cant blame her its because of corona

[upper karma]

Which angles are represented by the same point on the unit circle as \small \frac{\pi}{4}? Select all that apply.

[mild anvil]

no like she never taught us

[upper karma]

really need help with Trigonometric Functions

[pure cape]

this channel is occupied

try a different channel down at math-help

[slate blade]

lmao trig is one of the trippiest topics if you aren't taught well

[upper karma]

Virtual school while doing math isn't helping

[placid bone]

luna do you need the answers now?

[slate blade]

@upper karma try adding and subtracting pi, 2pi, etc. and doing pi/ (pi/4) etc. You'll get a ton of answers

[upper karma]

alr

[mild anvil]

she was like ok we dont have enough time so just try to learn it on your own

and she didnt give us notes or anythong

[placid bone]

@mild anvil what you basically do is just find when tan x=3 from [0, 6pi], then divide answers by 3

[slate blade]

and you change the domain to 6pi because it's 3x and 3*2pi = 6pi

orrrrrr you can also cheese the problem

by graphing it

[mild anvil]

i tried graphing it but like idk what im supposed to be looking at

[slate blade]

u use desmos?

[mild anvil]

no i use mathway

[slate blade]

that's not a tan function m8

[mild anvil]

yea im on a new problem

[slate blade]

ah ok can I see the question?

[mild anvil]

yea sure

[slate blade]

so I'll teach u the graphing way first, then the normal way

[mild anvil]

ok thank you

[slate blade]

take a look at your graph right, you're technically graphing 2 functions.

y=cos(2theta)

and

y=-sqrt(3)/2

just that they equal at points y right

[desert vortex]

Arent you suppose to do your own homework

[mild anvil]

im confused

[slate blade]

you basically have to click on the intercepts and that is where the functions equal eachother

pshhh it's lurning

[mild anvil]

yes im trying to learn

i dont want to fail

[slate blade]

uh, basically when you say that cos(2x) = -sqrt(3)/2

[placid bone]

we're not doing it for them

[slate blade]

you're equating 2 functions together.

2 graphs are intersecting at certain points so to say

loona are u still alive

anyhow

[mild anvil]

yes im still alive

[slate blade]

so to find where they equal oneanother, you look at the intercepts of the graphs between the domain indicators

does mathway allow you to analyze intercepts?

[mild anvil]

im just trying to comprehend

no maybe i should use desmos

[slate blade]

https://www.desmos.com/calculator/sz29crbbwx

click this

[desert vortex]

🇧 🇷 🇺 🇭

[slate blade]

I already have like 7 graphs named stuff so I needed a creative name

[mild anvil]

OMG

I MUST BE BLIND

WTF

I DIDNT SEE THE PURPLE LINE

[slate blade]

...

we all have our days

today apparently isn't any of ours

[mild anvil]

thank you for understanding

ive had a rough day today

[slate blade]

F well

alright you can also solve this through normal "analysis"

just by solving $\arccos{\frac{-\sqrt{3}}{2}}$

[somber coyote]

Doodaide

[slate blade]

and make that = 2x

expand the domain and badabing badaboom

that wasn't an ideal explanation was it lol

[mild anvil]

no im sorry im just dumb

[slate blade]

anyways do you get how to solve it graphically?

[mild anvil]

kinda

[pure cape]

easiest way is to just remember the special angles

and the values their trig funcs compute

there arent much to start with

[slate blade]

any other questions?

[mild anvil]

i do have more problems but im trying to solve this

[slate blade]

k msg if you have issues

[mild anvil]

okie

youve been a lot of help thank you and sorry for bothering you

all of you actually

[upper karma]

The point P lies on the unit circle, making an angle of measure \small \theta with the positive \small x-axis.

Use the cosine and sine functions to express the exact coordinates of P in terms of angle \small \theta.

[jaunty marsh]

Draw a right triangle and label sides x and y for point P with coords (x, y)

Use opp/hyp and adj/hyp for sine and cosine

I think

[umbral snow]

What are the definitions for sinθ and cosθ? If you know that, this question is immediate

AV is correct that doing the above will get the right answer, but knowing what the unit circle is is important

[slate blade]

sinθ and cos theta are special ratios between the sides of a right angle triangle

that's the "best" way I can think of defining it

[umbral snow]

After the unit circle is introduced, there's a different definition

[slate blade]

true but in the end, it's still the same "idea" isn't it. within the unit circle, it is still a ratio between the traingle, its jsut that one side is 1

[upper karma]

thanks

My teacher didn't teach us any of that

[slate blade]

what is with bad teachers lmao

[umbral snow]

It isn't, it fails for negative lengths

[white cradle]

hmm

[slate blade]

what is your proposed definition then?

[umbral snow]

Okay, I'll give the answer.
Take the terminal arm with angle θ. Where that intersects the unit circle is (cosθ, sinθ)

[slate blade]

so just a coordinate system?

@upper karma they are known as "similar angles" I think. Basically, it is just variations of the angle pi/4, and you must consider the CAST thing

wait @umbral snow sorry to disturb but why does it not work for "negative length"

cuz the length's direction is just down isn't it? which is still possible

[umbral snow]

What's "down" for the side of a right triangle?

[upper karma]

[slate blade]

well I'm thinking of this from a physics POV but can't you arbitrarily define it?

[silent plank]

negative length...

[slate blade]

like if you went Up and Right are positive, Down and Left are negative

doens't that work for vectors and forces?

[placid bone]

mmm

so length is known as a scalar quantity

[slate blade]

sorry I'm still in High school and am still barely into university pants yet

[placid bone]

and scalar quantities, cannot be negative

this is still high school maths

if you do physics it's grade 11 physics

if not, it's grade 12 maths

[slate blade]

I am grade 11 lmao

[umbral snow]

"Ratio of a triangle's sides"
Works until something like cos(135)

Then it doesn't and you have to lean on unit circle def

[placid bone]

so basicallly, things like speed, number of things, length are known as a scalar quantity

cannot be negative

[slate blade]

ah ok make sense

I'm just dumb then

thanks for the new definition

[placid bone]

but stuff like velocity can be negative

[slate blade]

vector quantity right

[placid bone]

since they're known as vector quantities

yeah

the negativity can represent direction

so in essence, scalar can't be negative, vector can

[slate blade]

ah ok ty

[white cradle]

scalars can be negative too, its just that the examples you mentioned have physical meaning that doesnt make sense for negative

for example electric charge maybe

[slate blade]

Numerous, what's the formula for period

$$period = \frac{2\pi}{b}$$

[somber coyote]

Doodaide

[slate blade]

use this and try to figure it out

[placid bone]

scalars only have magnitude, which is given by |x|, meaning that it's always positive/0, so i'm pretty sure they can't be negative

a quick google search seems to corroborate with my information too...

[white cradle]

thats only the specific scalars you define as the norm of something

[placid bone]

i mean as far as high school maths goes, you don't get negative scalars

[white cradle]

what about this then

[slate blade]

I think that's just due to the "reference point of 0"

[placid bone]

doesn't negativity indicate direction though?

[white cradle]

not for charge

[placid bone]

so negative charge means that electricity is moving away? idk how charge works

[slate blade]

like for temperature, a scalar quantity, reference fo 0 degrees celcius

[white cradle]

yes for temperature that argument works

not for charge

[white cradle]

nah, it has nothing to do with direction, its just a purely scalar quantity

but i guess its just semantics at this point

since you could just shift the reference like doodaide said

[slate blade]

Yeah. to us it's more of an assumption that we've made

cuz Charge Positve or negative is really just an arbitrary definition we've created

that purely describes how the two dipoles interact

They don't really have any meaning beyond describing how they interact with each-ther

[white cradle]

how about this, we say dot product of two vectors results in a scalar

and theres certainly cases where that results in a negative value

[slate blade]

yeah I guess let's just go with that

[placid bone]

well ig my definition is flawed in that aspect i guess XD

so a more refined definition for scalar is any quantity without a direction?

[white cradle]

yeah thats all

[slate blade]

more like it is completely described by a single positive quantity

[white cradle]

but i guess you can always 'shift your scale' to make it positive if you want

[slate blade]

bruh this is why we need servers like this

[steep temple]

how do parabolas exist as conic sections if the cone is infinite? wouldnt the plane cutting into the cone form a hyperbola?

[dusky surge]

The plane of a parabola should be parallel to the corresponding tangent plane of the corresponding cone.
Therefore there will be not interception between the two planes.

[dark sparrow]

@steep temple hyperbolas are conic sections too

but if you angle the cutting plane in such a way that it's parallel to one of the generating lines of the cone, you get a parabola. and thats the only way to get a parabola.

any other orientation gives you either an ellipse or hyperbola depending on if the plane goes through one or both parts of your bicone

[cosmic bear]

I'd recommend working backwards on this one. Try to find a formula for getting the area in terms of the sidelength, and then use that

[placid bone]

ah

herons formula

[patent berry]

Ty

[placid bone]

huh

wait you were talking to thecaptainyaya

lol don't take my suggestions seriously

i was messing around LMAO

[steep temple]

@dark sparrow somehow i am having a really hard time trying to form a parallel plane to this cone

[dark sparrow]

yeah cause you have to be exact

and also not have the plane pass through the apex, ftr

[steep temple]

ive pretty much manually bruteforced all the combinations and i dont get any parabola

its either a hyperbola which would exist if i imagined the cone is infinite, or an ellipse which for some reason never evolves to a parabola

[dark sparrow]

say, what even are those parameters you're controlling?

[steep temple]

from a 2d perspective, the y parameter is shifting the plane on the x axis and the z parameter is shifting the plane on the y axis

the angle controls the tilt of the plane

https://www.intmath.com/plane-analytic-geometry/conic-sections-summary-interactive.php

the simulation is available here, please do let me know of the combination which results in a parabola

it's not like the simulation does not allow for a parabola - the website mentions you can construct parabolas but i cant seem to manage

[inland canyon]

Line x is tangent to the circle. what is its length?

man imma be honest my teacher is kinda bad

<@&286206848099549185> could use some assistance

would i use the pythagorean theorem

[upper karma]

Do you have a better pic

[inland canyon]

well i mean i fucking drew it bruh

i could like

try to draw it better lmao

[upper karma]

Or... take a pic of the question as stated

[inland canyon]

bam

perfect circle

[upper karma]

Okay

Is 12 the whole line

Or just the chord

[inland canyon]

oh the whole line

prob should have done a big line

[upper karma]

Ok then yeah you can use pythag

[inland canyon]

would that be like

[upper karma]

Bc you are given the line is tangent

Try yourself first

[inland canyon]

12=9.6^2+x^2

or 12^=9.6^2+x^2

breh

@upper karma did u die homie

[upper karma]

...no?

What kind of question is that

[inland canyon]

like u went silent for a sec

[upper karma]

You are not my only focus rn

[inland canyon]

was a bruh moment

fair enough

[upper karma]

And yeah that's correct

[inland canyon]

wait which one

theres 2 of them

[upper karma]

12²=9.6²+x²

[inland canyon]

ight

i may need more help soon so like

is it ok if i ping u if i do

[upper karma]

The hypothenuse needs to be squared, otherwise you are not using the pythag theorem correctly

Yeah ping me if needed

[inland canyon]

thanks

[inland canyon]

@upper karma i know the theorem of like getting an angle if its inscribed on the outside which is like measure1 = 1/2 (x-y)
if im not wrong

but wouldnt i need atleast like

2 pieces of information

[jaunty marsh]

You can use the fact that a radius at the point of tangency is perpendicular to a tangent

[inland canyon]

u mean the fact that its like

90 degrees

oh yeah i can like

do 60 + 90 + 90 + x = 360

120

nice

@jaunty marsh do you know how to like do this too

im assuming u would do something with the 15

[jaunty marsh]

Power of a point: 15^2=9(x+9)

https://artofproblemsolving.com/wiki/index.php/Power_of_a_Point_Theorem

Very useful for circles problems

[inland canyon]

may i ask what power of a point is

[jaunty marsh]

It’s a set of theorems that have to do with the lengths of secants, tangents and chords
Check out the link above

[obtuse tapir]

These are just extensions of the BPT

[jaunty marsh]

BPT?

[obtuse tapir]

Proportionality theorem

Thales theorem for some

Not the 90 degree arc triangle one

This must be some olympiad thingy

[jaunty marsh]

The Triangle Proportionality Theorem states that if a line is parallel to one side of a triangle and it intersects the other two sides, then it divides those sides proportionally.
How is this related?

[obtuse tapir]

Yes

It connects

[jaunty marsh]

But how?

[obtuse tapir]

Look at the proofs lmao

Rearrange AE*CE

And you can see the similarity

[inland canyon]

wait

bruh

https://cdn.discordapp.com/attachments/326138757474680852/785239980842025010/unknown.png

9 ix just like

the outside part

of the

triangle

wait

so is like

the uhh

radius

the 15^2

[obtuse tapir]

What is this

Ok

Use pythagorean theorem

15^2+x^2=(x+9)^2

solve

[jaunty marsh]

Could you also use power of a point?
9(x+9)=15^2

[inland canyon]

but isnt that just gonna tell me like

what 9 =

not the whole side of the triangle

[obtuse tapir]

Olympiad tricks are always faster

[inland canyon]

also is 8

[obtuse tapir]

What are you saying

[inland canyon]

u said solve 15^2+x^2=(x+9)^2

its 8

x = 8

[obtuse tapir]

Cool

[inland canyon]

wait

so x = 8

thats the answer?

wait wait wait wait wait

how did we use

pythogrean

when like

[green osprey]

speaking of proofs can someone help me after

[inland canyon]

9 only = like the half outside of the circle

right

so how the fuck

did we use pythogorean

[obtuse tapir]

The whole triangle is a right angled triangle

[jaunty marsh]

Angle between 15 and radius is a right angle

[inland canyon]

mhm mhm

[obtuse tapir]

15 is a tangent

And tangents are perpendicular to the center

[inland canyon]

ok so

15 is tangent

that means 15 is perpendicular to the center?

[obtuse tapir]

The line of length 15

[inland canyon]

like the line

[obtuse tapir]

is perpendicular

Yes

[inland canyon]

ok

but

but h o w

like

like i get the angle

is 90 degrees

[obtuse tapir]

The proof for why tangents are perpendicular?

[inland canyon]

but how does that help figure out the radius

[obtuse tapir]

Algebra

[inland canyon]

ok karl lets dumb this down a bit

[jaunty marsh]

X is the radius

You solved for x

[obtuse tapir]

The triangle connects algebraic expressions

[inland canyon]

ok lemme reask this

how did u write up the pythogorean theorem if u didnt know what the whole side was (radius + 9) or (x+9)

how ever u wanna write it

[obtuse tapir]

I dont understand

What do you mean

[inland canyon]

bruh

so like

[obtuse tapir]

The pythagorean theorem works for right angled triangles

[inland canyon]

mhm

mhm

b u t

like

what i was thinking was like

(9+x)^2 = x^2 + 15^2

oh

thats

thats litterly how you write it out

man

im dumb

[obtuse tapir]

Nice

[inland canyon]

nice

[obtuse tapir]

speaking of proofs can someone help me after
@green osprey what is it

[green osprey]

uh nevermind

i figured it out

[obtuse tapir]

Good

[low scaffold]

Hi, I’m quite stuck on this question. I feel I maybe need to do something with the circumference equation . Idk
https://cdn.discordapp.com/attachments/628061094287769621/785247294387388416/image0.jpg

[jaunty marsh]

Maybe use the fact that angle ACB is right angle

[low scaffold]

how would I know for sure?

[jaunty marsh]

It’s a theorem, an angle that sub tends a diameter is right angle

And then set the sides of the equilateral triangles to variables a, b, c
Then a^2+b^2=c^2 by Pythagorean theorem

[low scaffold]

that sound familiar actually

[jaunty marsh]

BC^2+AC^2=AB^2
Then express the areas of the equilateral triangles using formula $s^2*sqrt(3)/4$

$\frac[s^2*\sqrt(3)][4]$

$\frac{s^2*\sqrt(3)}{4}$

[somber coyote]

AV8312

[jaunty marsh]

There, that’s it

[low scaffold]

s being a side

[jaunty marsh]

Yeah

[low scaffold]

Ok thanks i'll try it

[low scaffold]

Thanks I got it

[jaunty marsh]

👍

[green osprey]

can someone work through this problom with me

its really confusing

[jaunty marsh]

||Use vertical angles||
||Then triangle congruency through SAS||
||Then CPCTC to prove parallel lines through angles||

Try the hints one at a time

I don’t have much time to explain it more than that sorry

[green osprey]

ok

ty

[old jasper]

can someone help me with these equations

[upper karma]

What have you tried

And which one out of all

[old jasper]

i can't get any of those 3 mainly b

i am just stuck in all of them don't know where to go

[upper karma]

Ok our goal on a and c is to get to the form of log(something)=something

Are you able to try at least with a once, knowing the your log laws $\ \log(a)+\log(b)=\log(ab)\ \log(a)-\log(b)=\log(\frac{a}{b})$

[somber coyote]

Al𝟛dium

[upper karma]

Uh

Are you able to try at least with a once, knowing the your log laws $\ \log(a)+\log(b)=\log(ab)\ \log(a)-\log(b)=\log(\frac{a}{b})$

[somber coyote]

Al𝟛dium

[upper karma]

@old jasper

...

[inland canyon]

bruh

Whats the value of x

wouldnt all the degrees add up to 360 bruh

cuz then it would be like uhh 22

222*

but the angle is acute

or would it be like

180

it adds up to 180 doesnt it

😔

[autumn bloom]

Ain't it 138 because a triangle has to add up to 180, 360 would be a rectangle if I remember correctly

[inland canyon]

ye but like

acute angle

pretty sure its 42

[autumn bloom]

Wait no Im also dumb

I added the both up

Yea it's 42 not 138

[low scaffold]

Every triangle's angles add up to 180 degrees

yea 42

[inland canyon]

ight c00l

man fuck math

who invented math

yo fuck greek people all my homies hate greek people

[autumn bloom]

It's so confusing, too many numbers 😔

Once you know it, it can be fun tho

[low scaffold]

^

[inland canyon]

man

this image

explains my pain

also cannedfruit i like ur pfp homie

[autumn bloom]

Thank you 😎

[inland canyon]

welcome

[low scaffold]

Anyone able to help with this? I’ve tried to differentiate it but now I’m stuck

[pure cape]

Well you were almost there

You know that the derivative at that point is also the slope of the tangent line

so you got the slope of the linear equation y= mx + c where m is the slope

So now you have to find c

You can do that by substituting x=a into the linear equation

[low scaffold]

Ohhhh

Ok

Like this?

[astral bluff]

Hi guys can someone help me with a hw question?

[pure cape]

Uh no, you did it wrong

Your derivative is right but your subsitution is wrong

[low scaffold]

How should it be?

[pure cape]

right can you tell me what m might be?

[low scaffold]

Using the value of a I got:

[pure cape]

Yes which is 4e^4a right?

[pure cape]

Oops missed a bracket there

Oh yeah, this is geometry

Lol didnt even notice

[low scaffold]

I thought because of the tangent line it should be here

[upper karma]

It's fine ig, this comes from a geo problem

[low scaffold]

Where should it be?

ok

[pure cape]

I mean its quite on the calculus side, but doesnt matter

Yeah, so we got the slope

[upper karma]

It's fine here ig, i just didn't see that this problem was originally a geo problem

[pure cape]

So our equation now is 4e^(4a)x + c = y

[low scaffold]

ok

[pure cape]

You also know that the line crosses thru (a;e^(4a))