geometry-and-trigonometry

[cedar tendon]

@upper karma i found it

the arc is 45

[upper karma]

good job

now convert it to radians

[cedar tendon]

i hadded help

ok

spider how do i convert a degree into a radian

[upper karma]

https://www.youtube.com/results?search_query=convert+angles+to+radians

[cedar tendon]

?

[upper karma]

^how to convert from angles to radians

[cedar tendon]

yea

[upper karma]

you really should change your nickname, you know

it only serves to demotivate you

change it to "I AM A MATH GOD"

and your outlook on it will improve

[cedar tendon]

i can't because math is so hard for me

[upper karma]

it's hard for everybody, man

nobody is born with math skills, lol

[cedar tendon]

yes their are people who born with math skills

[signal sky]

agreed

[cedar tendon]

their a lot people who were born that has math skills

[upper karma]

you really think that?

[cedar tendon]

yea

i do

[upper karma]

that they could derive equations straight from their mother's womb?

that they didn't have to learn that shit, like everybody else?

[cedar tendon]

because in my class we alway have an asian kid and he knows everything in math all the asian in my school knows everything in. math and they were born with it

such talent

[upper karma]

https://i.redd.it/gpy1brqrm0911.jpg

talent is overrated

[cedar tendon]

the asian were born with high intellgent math

math is alway been easy for the asian in my school

[upper karma]

you should stop comparing yourself to asian kids

you'll never be asian

Lol

[signal sky]

Not being racist or anything

but most asians work harder

thats why they are "smarter"

[upper karma]

no matter how hard you try, you'll never be asian

[cedar tendon]

i know

[upper karma]

not without some serious cosmetic surgery

[cedar tendon]

ok now help me solve this problem

so how do i convert it to a radian

[upper karma]

did you click any of the videos from here: https://www.youtube.com/results?search_query=convert+angles+to+radians

?

[cedar tendon]

i thought u will give me a video

like last time

[upper karma]

this time I'm giving you a choice

pick one

it's just a simple Youtube search, it's not that hard

Lul

[cedar tendon]

i got 0.78

@upper karma

[signal sky]

@upper karma How did you do that pfp?

By that pfp I mean no pfp

Or is it just grey?

[upper karma]

@signal sky Nah no pixels pfp

@cedar tendon it's not 0.78

it's 0.78539...

which rounds up to 0.79

[cedar tendon]

ok

[upper karma]

you keep reducing it to its lowest form until you eventually hit pi/4

then you just type it into a calculator

,w pi/4

[somber coyote]

Display results online and refine query

Requested by DSpider#6917

[upper karma]

0.7854 ≈ 0.79

[cedar tendon]

need help

[upper karma]

https://www.youtube.com/results?search_query=arc+length+of+a+circle

again, pick one

google and youtube should be your #1 resources from now on

if you don't find anything there (and I mean you looked at at least 3-4 videos), ask here for help

for example, if one video doesn't explain it very well, then go to the next one in the search list

[cedar tendon]

dude. but i forgot how to find the length of the arc

[upper karma]

that's ok, google it

[cedar tendon]

so can tell me

ok

[upper karma]

and: ||I'm pretty sure that "1.78" is a typo||

[cedar tendon]

45/360 * 2 * pie * 2.25 = 1.767145868 @upper karma

that what i got

[upper karma]

maybe they meant to type 1.8 instead

[cedar tendon]

ok

now i need help. with this

@upper karma

[upper karma]

weird question

[cedar tendon]

d spider you can help me right

yea

you can

[upper karma]

"arc measure of arc"

arc what?

[cedar tendon]

it told you the arc

it was given

the measure of the arc

[cedar tendon]

yea

[upper karma]

maybe refresh that web page?

[cedar tendon]

the arc is 3 radians

[upper karma]

or don't

[cedar tendon]

i did

and still the same

well up to the next question

[silent plank]

assume the diagram is somewhat to scale

[cedar tendon]

Ramonov now i need help with this

help

[eternal crag]

dude, are you ever going to post an attempt when asking for help

[dark sparrow]

no

[cedar tendon]

i need help

[eternal crag]

say i give you 1cm and ask you to convert that into meters, what fact do you need to know?

[cedar tendon]

i don't know

and i need help

[eternal crag]

do you know how to convert 1cm to m?

[cedar tendon]

nope

[wooden current]

you have to be joking

[eternal crag]

are you familiar with American currency? forget it i give up

[wooden current]

😦

[cedar tendon]

guys so how do i. convert a cm into meter?

[silent plank]

learn latin

[dark sparrow]

http://lmgtfy.com/?q=how+to+convert+centimeters+into+meters

[silent plank]

or at least the prefixes that are derived from latin

[dark sparrow]

did you get banned from using google or are you refusing to help yourself

[cedar tendon]

wait google can. ban people?

[wooden current]

😐

[dark sparrow]

so you're just refusing to help yourself then

[silent plank]

do you at least know how many centimetres are in a metre?

[cedar tendon]

i know a centimeter is different from a meter

[silent plank]

or how to convert cents to dollars

real life skills: if you had a choice would you pay $1.50 or 161cents for the same thing

[ebon sundial]

DUDE just divide the length value by 100 @cedar tendon sheesh

[silent plank]

DUDE just divide the length value by 100

no

[cedar tendon]

what

divide?

[ebon sundial]

1 centmeter is equal to 0.01 meter

[cedar tendon]

Waht

@silent plank is this how to do it

[ebon sundial]

you supposed to know this already

[silent plank]

tldr: look up conversions and multiplication by 1

like seriously just google stuff

and read

[cedar tendon]

oh

so i just mutiply by 1\

[ebon sundial]

yea

that ame

same

also

[wooden current]

10*

[cedar tendon]

ok

[silent plank]

wdym by 10?

screenshot what you read and tell us what you don't understand about it.

[cedar tendon]

i got it all clearly

now time for the next question

i need help

[silent plank]

change your browser, it doesn't like that site

[wooden current]

wdym by 10?
what

[silent plank]

why did you type 10*

[cedar tendon]

ramonobv

ramonov

what i do?

[silent plank]

change your browser, it doesn't like that site

[cedar tendon]

ok

[silent plank]

and/or install some plugins

[wooden current]

i mean dont you multiply/divide by powers of ten when converting from one unit to another

[silent plank]

note a good description

you need to account for units etc

and not all conversions are in 10s

"multiplication by 1" requires no mentioning of units and can apply to all cases

[wooden current]

[cedar tendon]

oh i'm ficken stupid

it was the browser

this whole time

i. hate this

[ebon sundial]

welps what. did. i told you

you should be in algebra

#prealg-and-algebra channel

go to that channel you no business in geometry

[cedar tendon]

what

ok

[ebon sundial]

glad he left this channel

@silent plank i fixed your problem

you are good now

he gone to the prealg algebra

channel

[silent plank]

what problem?

[ebon sundial]

i told him to go to this channel #prealg-and-algebra

[silent plank]

why?

[ebon sundial]

because he a disappointment in geometry

so i tooken care of it

he just wants the answers

that all

he wants

he not puting any effort

[silent plank]

and how does switching channels solve that?

[ebon sundial]

he has been flooding up this channel

so i solve it

[silent plank]

@cedar tendon anyway where are you stuck now?

[cedar tendon]

ok

the first one

and the 2 one

[silent plank]

you should be able to do those based on what you've completed over the past few days

[cedar tendon]

What

ramonov

i forgot those

past days

[silent plank]

review them, and or do a quick search of the terms present in the question

screenshot what you read and tell us what you don't understand about it.

put in more effort

[cedar tendon]

but dude you don't understand this question is so hard for me

[silent plank]

read that first quoted line

[cedar tendon]

it said find the arc intercepted by each of the following angles?

[silent plank]

maybe search the terms: central angle, intercepted arc, etc and w/e terms your unfamiliar with

if you don't understand them

[ebon sundial]

Ramonov i just told you he going to flood this channel even more so we need to send him to algebra

channel

[versed river]

??????? why is this channel any worse than the other channel for him to be in?

[silent plank]

@ebon sundial if you have an issue then don't look here

[cedar tendon]

guys

i need help

[silent plank]

redirecting an issue, doesn't solve it

[cosmic pebble]

From

#❓how-to-get-help

Rule 7

When asking for help, do not insist on getting just the answer; we are here to help you learn, not cheat. Likewise, if you are providing help to others, try your best to explain and elaborate instead of simply giving away the answer.

[silent plank]

look, please

stop repeating:

i need help
i just need the answer

[cedar tendon]

ok

[silent plank]

i'm directing you towards finding resources to help yourself

[cedar tendon]

ok

[silent plank]

specifically

look up terms/concepts you don't understand, screenshot what you read and tell us what you don't understand about it
and I doubt anyone is willing to help you atm unless you do at least that

[cedar tendon]

ok

[cedar tendon]

the arc intercepted by DAE = 52
degree the arc intercepted DCE = 66
degree the arc intercepted GCF = 52
degree the arc intercepted GAF =  38 degree

@silent plank is thst right

that right?

[silent plank]

DAE and GCF are wrong

[cedar tendon]

ok

the arc intercepted by DAE = 66
the arc intercepted by GCF = 38

@silent plank is this right?

[silent plank]

yes

[cedar tendon]

DAE + GAF = 90 degree
DCE +  GCF =  118 degree

is that right

ramonov

@silent plank

[silent plank]

for question 2?
both wrong,

[cedar tendon]

ok

oh

ment 104

for dae

and gaf

[silent plank]

thats better

[cedar tendon]

@silent plank I'M finnally on the last question

this is the last one

[ebon sundial]

Ramonov lets just give him the answer so he can go

he on the last question anyways

[cedar tendon]

say what i can get the answer

[ebon sundial]

your on the last question right

[cedar tendon]

yes

[ebon sundial]

so lets just give him so he go

[silent plank]

do you know:
how to identify the slope of a line from the slope-intercept form?
find the slope of a line parallel to a line with a certain slope?
find the slope of a line perpendicular to a line with a certain slope?
find the equation of a line given the slope and a point?

[cedar tendon]

what

ramonov

me

[silent plank]

yes

[cedar tendon]

nope i don't

[silent plank]

those are things you'll need to understand. search it up and read

[cedar tendon]

but dude were on the last question like mr.tai said

answ

[silent plank]

ignore him

[cedar tendon]

ok

ramonov for 1. how about slope intercept form

y = mx + b

[silent plank]

wdym by how about?

[cedar tendon]

ok

[upper karma]

y = 3x basically means that for every 1 unit you move on the x axis, you move 3 times that on the y axis

and the + 2 from y = 3x + 2 is saying that you don't start from 0 on the y axis

you start from (positive) 2

this is the "+ 2" part

so, how much further down (on the y axis) do you need to start from?

so that when you draw a parallel line you hit point A

because like I said, that line is "+ 2"

how much further down is that blue question mark?

count them

are you still following this?

I think it's not a very good method

it's so empirical

y = 3x makes perfect sense to me

how much do you need to move on the x axis? 12 units? ok, substitute "x" with 12

and you get a new "y" value

if you repeat this with an infinite number of values for "x", you get a line

because each value is basically a point

with their own x and y coordinates

and if you string along enough of them, you get a line

same thing with a circle

it's basically just a bunch of points

that are ALL the same distance from a central point (called an origin)

as for the second question, perpendicular looks like this:

but notice that this line isn't going up

it's tilted down

also, for every 3 units we move on the x axis, we move 1 unit down on the y

the perpendicular line would be y = -x/3

or -(1/3)x

same thing

but you still need to know the vertical starting point

so that it hits point A

https://www.desmos.com/calculator/q68idy8bun

play around with the v variable

if you wanna find its exact value, you need to think about how you would approach this

the v variable is your starting point on the y axis

in other words, the red question mark at the top

but to find that, first you need to find the red question mark at the bottom

and that boils down to...

once you have the length of side FG you can subtract it from the y coordinate of A

and you get your starting point on the y axis

I feel like I'm talking to myself, so I'll just leave

have fun

[silent plank]

I think it's not a very good method

[upper karma]

why not just say perpendicular is the negative reciprocal of the original slope?

yep

and desmos agrees: ||https://www.desmos.com/calculator/q5uduqeucu||

too bad he left, tho

[silent plank]

why overcomplicate things by trying to do / find the y-intercept graphically

[upper karma]

how would you do it?

[silent plank]

if you have the point, and the slope
why not just use the point-slope formula?

[upper karma]

$y=m(x-x_0)+y_0$

@upper karma

[somber coyote]

Al𝟛dium:

[silent plank]

it's literally screaming: use me

[upper karma]

^

||https://www.desmos.com/calculator/bpoqvtzpbm||

yep, that works, too 🙂

The point is that on a test time is gold, so you better don't overcomplicate it and solve this simple algebra in a minute

[green roost]

Wait people really find inttercepts graphically

[upper karma]

whats funny is that math i give up probably won't even learn anything from here

[green roost]

👁️ 👄 👁️

[upper karma]

🤣

[green roost]

You dont sit through hours of maths to learn to draw pictures use some algebra

[silent plank]

graphical methods get exponentially worse the less nice the values are

[green roost]

it honestly seems so much harder

[silent plank]

as demonstrated, the graphical method for find the y-int required contracting something outside the given plane

[green roost]

it would take me at least an extra 5 minutes to solve it graphically

like 5 if it was easy, just do some simultaneous equations and be done with it

I mean ive not actually read the question so it might be differnt but still i cant think of any case where it would be easier to solve graphically

[lavish glacier]

Hello

[ripe dove]

You didn’t post a question

[lavish glacier]

@ripe dove check again

[ripe dove]

3 rectangular faces, 2 triangular faces

You know how to find area of rectangles and right triangles?

@lavish glacier

[lavish glacier]

Yeah

[hard forge]

Just find the area of each figure

[ripe dove]

Then add them all up

[hard forge]

Yes

👍

[ripe dove]

Of course, only one of the possible solutions is even remotely plausible, but that’s because i could do some of those calculations in my head

[hard forge]

Mhm

[lavish glacier]

Its 1 right?

[hard forge]

890?

Can we see ur work

[lavish glacier]

I just found tbe area and added it up

[hard forge]

Okay, but can we see ur work

@lavish glacier I can walk u through it if u want?

[lavish glacier]

I got it

I realized i made a kistake and ot was d

[hard forge]

Yep

It’s 96

[lavish glacier]

Is ot just l× w× h?

[silent plank]

no

[silent plank]

that isn't a prism

[hard forge]

It’s a rectangular pyramid

[lavish glacier]

Waot how fo tou know it was a rectangular pyramid vs traingle prism?

[hard forge]

It’s shape

[lavish glacier]

Im like a baby

[hard forge]

That’s a triangular prism

The shape u have is a rectangular pyramid

Cause the base is a rectangle

And it’s a pyramid

The formula for the volume of a rectangular pyramid is

$v=\frac{1}{3}bh$

[somber coyote]

Diesel Exhaust Cloud:

[lavish glacier]

So for a square pyramid, if i jave the height of 1 truangle is that the height of all pyramid?

[hard forge]

Where b is the area of ur base

U have the height of the pyramid as well

That is the height of the pyramid

[lavish glacier]

So i have 11 as base

And 14 as height of 1 triangle

I neee sa

[hard forge]

How is ur base 11?

[lavish glacier]

It's a different prob

[hard forge]

https://media.discordapp.net/attachments/326138757474680852/718519843157966858/20200605_134111.jpg

Aren’t u asking about this one?

[lavish glacier]

Different problem

No

[hard forge]

Well u need to post it for me to answer lol

[lavish glacier]

That is d

Thimo

Do i really need to?

[hard forge]

Yeah it would dispel confusion

Okay

[lavish glacier]

So 14 as hegight of 1 triangle

11 as base

[hard forge]

Since it’s a square all the sides of the square are congruent

So u just need to find the area of each side

And add them all up

Since it’s SA

[lavish glacier]

77 for triangle

121 for square

[hard forge]

,Calc (14*11)/2

[somber coyote]

Result:

77

[hard forge]

Yes

[lavish glacier]

77 times 4

[hard forge]

Since u have 4 squares yes

,Calc 77*4

[somber coyote]

Result:

308

[lavish glacier]

+121

[hard forge]

,Calc 308+121

[somber coyote]

Result:

429

[lavish glacier]

429

Ye0

[hard forge]

Yerp

[lavish glacier]

Yep

[hard forge]

That should be it

[upper karma]

solve for the radius, then plug that value for r in the in the surface area formula

[hard forge]

for a sphere, the ratio of surface area to volume is given by: S/V = 3/R.

Where “R” is the radius

[lavish glacier]

Ok

What is s? Sa?

[hard forge]

The approach rent free gave is also good

Yes

[upper karma]

i was helping this dude out yesterday and he dipped

[hard forge]

Oh

[lavish glacier]

Dipped meaning?

How do i iscolate for r?

[upper karma]

dipped meaning left

when i was helping you

you isolate for r by using multiplying the area with 3/4

then dividing it by pi^2

[lavish glacier]

Ok i got 136

Damn, my boys dipped😫

[upper karma]

@lavish glacier looks like a test, is it?

[lavish glacier]

Nah dawg, why yall always ask me that?

[upper karma]

BC ITS SEEMS LIKE SO

Okay

[lavish glacier]

Why?

Bc it's multiple choice?

[upper karma]

,ask volume of a triangular prism

Wait a sec

[lavish glacier]

I got that one now worries

[upper karma]

Got it?

Oh lol ok then

[upper karma]

@lavish glacier u need help

i here fo u

Nah he got it himself right after i offered him help lol

ye

[lavish glacier]

Nah man thx, y'all are so helpful@upper karma @upper karma

Yall are so devoted to helping guys like me, you guys should rewarded for your services.

[upper karma]

❤️

you make us who we are

we wouldn't be here without you

[lavish glacier]

Do u tutor irl?

[plucky marlin]

[upper karma]

🤔

[lavish glacier]

[upper karma]

no i don't tbh

i'm not good at explaining things tbh

[upper karma]

U very good

Consider becoming a teacher when u grow up

❤️

I dont tutor either

[lavish glacier]

@upper karma u also included in this

[upper karma]

👍

[cedar tendon]

Guys

[cedar tendon]

i got a question how do i find the equation

@upper karma

@upper karma

@silent plank

[upper karma]

bruh u showed this question this morning

[cedar tendon]

yea

ramonov told me to google search for it

and now i came back

and i found nothing

from google

[upper karma]

alright

for 1) you have to solve for b such that y = mx + b

for 2) you have to solve for b such that y = mx+b, where m is the negative reciprocal of m in the first equation

[cedar tendon]

so that how i found it both for question

the equation

thx

[glad gorge]

Does anyone know where the mistake is bcuz I can’t find it??

[silent plank]

reason 2

this has nothing to do with parallel lines

looking at that again, does it make any sense to use that here?

[analog hazel]

(75)2= 150+??+??=360 deg

[upper karma]

MO is 150 degrees

[analog hazel]

I thought 150 is mno

If you don’t mind could you explain a bit because I thought it’s all add up to 360

[upper karma]

Google inscribed angle theorem

[analog hazel]

alright

well thanks

[upper karma]

MNO is 75 degrees

Says in the question

[analog hazel]

Yea I thought I had to multiply by 2 because I thought it was already divided into half cus half diameter

[analog hazel]

I got GH is 164 but can’t find HG which is the answer for V

And for L I thought EF would just be 90 deg but I don’t think that’s right

oh wait 2nd one is central circle but still confusing

[upper karma]

HG is 360- 106(2) - 48

EF is 360 - 90 - 87.5(2)

[analog hazel]

Thank you so much

[upper karma]

@cedar tendon did you read what I posted yesterday?

https://discordapp.com/channels/268882317391429632/326138757474680852/718468861195845662

[delicate laurel]

The product to sum formula for cosine

is it possible to use it when

$\cos(2t)\cos(t)$

?

[somber coyote]

175 braces:

[delicate laurel]

$\cos(2t)\cos(t) = \frac{1}{2}\left(cos((2+1)t)+\cos((2-1)t)\right)$

?

[somber coyote]

175 braces:

[silent plank]

what's making you doubt whether you can use it?

[upper karma]

Everything

[somber coyote]

staticjpl:
Compile Error! Click the reaction for details. (You may edit your message)

staticjpl:

[solemn warren]

How would I find the length of ? without using sine
Sorry for the crap drawing

sine(pi/4) = o/h
sine(pi/4) = ?/1
sine(pi/4) = ?

But idk how to find it without sine.

[dark sparrow]

you could notice that the triangle thus formed is an isosceles right triangle

[solemn warren]

Right, I tried to draw more accurately, but I think it looks worse.

Tho it looks much more like an isosceles now

Tho are u sure its an isosceles?

[dark sparrow]

also, you can hold shift while using the line tool in MS paint to make it snap to the nearest horizontal, vertical or 45° diagonal line

just sayin'

[solemn warren]

Ah right, lol, i forgot that feature existed thanks

[dark sparrow]

anyway, does the fact that this triangle is isosceles need more explanation?

[solemn warren]

I thought isosceles needed 2 sides with same length?

Or am i confusing it with equilateral

[dark sparrow]

yes, an isosceles triangle is a triangle where two sides are the same length

this is equivalent to having two angles be the same

all isosceles triangles have matching angles at their bases, and all triangles with a matching pair of angles are isosceles

[solemn warren]

So I feel like its not isosceles because of this?

Tho, I guess the angles satisfy it?

Oh shit nvm, i see it now.

Idk why I thought one of the blue and red lines were 1.

Instead, blue = red

Making it an isosceles...

[dark sparrow]

BRUH

[solemn warren]

Hmm still not sure how I can find the actual length without using sine tho.

[solemn warren]

Yeah

[dark sparrow]

this is a right triangle

[solemn warren]

sqrt(x + x) = 1?

[dark sparrow]

what

[solemn warren]

oh lol

sorry

sqrt(x^2 + x^2) = 1

[dark sparrow]

ok i guess you can say it like that if you wish

[solemn warren]

Well, idk, is there another better way?

I was thinking going:
sqrt(2*x^2) = 1
sqrt(2) * x = 1
x = 1/sqrt(2)

[dark sparrow]

i mean idk

i'd rather have seen you write x^2 + x^2 = 1^2 in accordance with how the pythagorean theorem is usually stated

but w/e

[solemn warren]

Oh right lol yea

Either way, it worked so, thanks heaps!

[solemn warren]

damn, i never realized tan has range (-inf, inf)
idk why but i never really thought about or looked at it properly.

[wooden current]

😳

[fading carbon]

Evening, could anyone help me understand trigonometric identities? I particularly am unsure which side of the equation to tackle ( this is in general not specific to one identity problem) any tips/ insight you can give would be great

[upper karma]

@solemn warren without using sin:

@fading carbon https://www.khanacademy.org/math/trigonometry/trig-equations-and-identities

[fading carbon]

Thankyou🙏

[solemn warren]

@upper karma thats cheating lol, using cos

cos is basically sine

[upper karma]

https://tenor.com/view/marvel-is-it-though-thor-bruce-banner-gif-14242312

[solemn warren]

yes

cos = sin(x + pi/2) or something

and tan is too (sin = cos(x - pi/2) = tan * cos)

[upper karma]

errbody sine up in here

[solemn warren]

lel

[silent plank]

usually, start with the side that feels more complicated

[upper karma]

cos(x) = sin(x + pi/2)
sin(x) = cos(x - pi/2)

I didn’t think of this until now but it is epic

There are so many trig facts and identities I learn all the time.

[upper karma]

well, yes

[eternal crag]

1. terrible meme
2. not the right channel

[upper karma]

well, if it helps you learn...

[upper karma]

Ahahhhhhhhhh

Nice one

[woeful acorn]

The terminal side of $\theta$, an angle in standard position, intersects the unit circle at $P(\frac{-1}{3},\frac{-\sqrt{8}}{3})$. What is the value of $\sec{\theta}$

[somber coyote]

Nitrousoxide:

[woeful acorn]

Is the answer to this -3?

[silent plank]

what's making you doubt it?

[woeful acorn]

I havent done trig in months

And on top of that I find trig to be a little annoying so I tend to worry that I've screwed up somewhere

[silent plank]

show your reasoning

[woeful acorn]

As in how I got to my answer?

[silent plank]

yes

[woeful acorn]

I did $\arctan{\frac{y}{x}}$ then got the cos of that, then got the reciprocal

[somber coyote]

Nitrousoxide:

[woeful acorn]

my calculator doesnt support sec

[silent plank]

if you adjust signs properly, that would work.

[woeful acorn]

That's my main concern lol

[silent plank]

instead try doing it without a calculator

and consider what cos(theta) represents in relation to the unit circle

[woeful acorn]

The only reason I picked -3 is because +3 wasnt an option.

I would've thought more about it if they both were options tho

It represents the x value at which that angle intersects the circle

[silent plank]

you've mixed things up

[woeful acorn]

Sorry I meant x

[silent plank]

yes, so here you can get
cos(theta) = -1/3 directly

information is lost when you apply arctan like that

[woeful acorn]

that why we have atan2 in programming

[silent plank]

yeh

you could achieve the same result by applying appropriate shifts

[woeful acorn]

Wdym

[silent plank]

atan2( -sqrt(8)/3 , -1/3) = pi + arctan( (-sqrt(8)/3)/(-1/3))

[woeful acorn]

atan2 has to require some logic though?

I mean I've always assumed it did

[silent plank]

mainly properties of the unit circle

applies shifts of pi based on the quadrant

[woeful acorn]

So it does require logic then

[silent plank]

ig

[potent mango]

is the volume of a parallelipiped still base*height even if they're not 90 degree's

wouldn't it be base*length of the slant

[upper karma]

Volume=∥a×b∥ ∥c∥ |cosϕ|=|(a×b)⋅c|.

this is the volume of a parallelepipid

[potent mango]

so

(a cross product b) multiplied by (magnitude of C (cos(phi)) ?

so just base times height...

[upper karma]

Yah basically

[formal citrus]

incenter is only for triangles? can I say that the meeting point of this polygon is incenter? (sorry for my bad english)

[formal citrus]

<@&286206848099549185>

[paper glacier]

the incenter is the point where all the bisectors meet

for a triangle, it will always exist

for an n-gon, if it is regular, then it will exist

otherwise, it can not exist

[ocean axle]

I want a point equidistant from all vertexes of a triangle. What do I need.

1) Orthocenter
2) Circumcenter
3) Centroid
4) Incenter

[dark sparrow]

what do you think you need

[pastel anchor]

any help?

i need to find area btw

[eternal crag]

you can first find the area of the right triangles

[pastel anchor]

like the small triangles?

[eternal crag]

(there are probably better methods)

yes

[pastel anchor]

but then what do i do with the area of the right rirnagles

[eternal crag]

are you trying to find the area of the center triangle?

[pastel anchor]

yes

triangle FBD

[eternal crag]

area of triangle FBD = area of the entire rectangles - area of three right triangles

[pastel anchor]

do u have any idea how to find the area of the right triangles? the squares are cut off

[eternal crag]

i've labeled two side lengths of that upper right triangle

use those two numbers, you can find its area

[pastel anchor]

ah so u count it fully?

[eternal crag]

heh?

[silent plank]

apply formula for area of a triangle

[pastel anchor]

ah ok. do i add up all the areas of the right triangles?

[silent plank]

area of triangle FBD = area of the entire rectangle - area of three right triangles

[urban egret]

Cool problem

[upper karma]

does anyone here know why we use general form for circles

im learning geometry rn and i dont see why general form is useful

[urban egret]

Are you referencing the x^2+ y^2=r^2 equation

[upper karma]

no

its like

[silent plank]

x^2 + y^2 +ax + by +c = 0 ?

[upper karma]

$x^2+y^2+Ax+By+E=0$

yeah

[somber coyote]

w h e r e i s t h e b r o t h:

[silent plank]

its not useful

noone says it is

[upper karma]

then why do we have it?

[silent plank]

"general form" is just a term used to describe certain equations in the form stuff = 0

it just exists because it can

[upper karma]

oh okay

[silent plank]

i mean you can still determine the center from it without too much effort

[upper karma]

how do you do that?

hm

wait

is it just

$x^2+y^2+Ax+By+E=0$

[somber coyote]

w h e r e i s t h e b r o t h:

[upper karma]

where A is X coord of center and B is y coord?

[silent plank]

no

[upper karma]

do you just convert it to standard form?

[silent plank]

the center/standard form: (x-h)^2 + (y-k)^2 = r^2 is preferred.
where: h = -A/2 and k = -B/2

and those coordinates can be obtained without going through the whole process of completing the square

(but ideally you would anyway)

[upper karma]

okay thanks

eh, but to find the radius i would assume you have to convert to standard anyways

[silent plank]

yeh

[ocean axle]

what do you think you need
@dark sparrow I think I need Centroid.

[dark sparrow]

why do you think so

[ocean axle]

I've looked up google images and it looks like it's the same length from vertexes.

[dark sparrow]

looks like ≠ is

[leaden bridge]

where do you guys learn all the tricky trigonometric identities from?

Abhijeet just helped me verify an identity using $\cos(2\alpha)=2\cos^2(\alpha)-1$

[somber coyote]

lucidorio/sparklester:

[upper karma]

High school...

[leaden bridge]

can confirm not at my high school @upper karma

[upper karma]

Wow

Welp in my country yes, idk on yours

[leaden bridge]

what country did you study in

[upper karma]

Spain

[leaden bridge]

dude I'm currently in high school in Spain

how long ago did you go to it

[leaden bridge]

@upper karma

[upper karma]

Dms

[upper karma]

does anyone know how to do factors?

[leaden bridge]

yea

but you cannot factor it straight on

[urban egret]

@upper karma are you familiar with grouping

Ohhhhh your helping nevermind

[leaden bridge]

not really go on topcat

[urban egret]

@upper karma ok so you can’t factor straight on cause you have a 3 that is the coefficient of the leading degree but you can use the same concept. So you multiply the constant by the 3. Find the factors of 48 that adds up to -14x. Then use the concept of grouping to solve.

I got 48 from 16 times 3

So you should have 3x^2 -6x-8x+16. Then use grouping to find a common expression. So I will factor out 3x and 8 to get 3x(x-2)-8(x-2). Notice (x-2). You can rewrite it as (3x-8)(x-2) hope that helps

Of course just use the foil method to check to see if your answer is correct.

[leaden bridge]

TopCat how efficient is grouping when compared to solving the quadrating equation formed by the polynomial? What I would have done is solving the quadratic equation, getting x=8/3 and x=2, which can be written out as (x-8/3)(x-2). But then, since the coefficient of the x² is 3, I would take that and multiply any of the groups, so going from 3(x-8/3)(x-2) to (3x-8)(x-2)

It seems p fast, but is it correct?

[urban egret]

Well of course grouping is not a concept that will be used every problem I used the long way for him so he can use solve his problems. Honestly knowing that 3 is a prime number and seeing 16 as a positive number we know we will get something like this (3x-)(x-). You can just test numbers

[upper karma]

Just factor out 3 and solve for the quadratic equation

[urban egret]

Or could’ve done that as well^

@leaden bridge you can check to see if your final answer is correct by using the foil method

[leaden bridge]

Yeah, factoring out the 3 and getting the quadratic solutions is essentially what I'm doing

[urban egret]

When I would TA for math I noticed students preferred patterns so to get them use to analyzing differences I get them use this so they can become comfortable solving for quartic expressions

You would mainly need grouping when the leading degree is greater than 3.

Unless your familiar with synthetic division, roots, zeros, etc.

[leaden bridge]

makes sense

alright thank you topcat

[ebon sundial]

@upper karma @upper karma im so glad that guy isn't comming

this server is peace ful

noq

now

@silent plank im glad this is peace

[cedar tendon]

I NEED HELP

[ebon sundial]

OH NOO

HES BACK

WE ALL ARE DOOM

[cedar tendon]

HELP ME

i need help

[upper karma]

you know, it's my fault he's back

i was searching for his name bc i haven't seen him post in this server for a while

[cedar tendon]

help me

SOMEONE

I need help

HELP ME

someone

[upper karma]

k

what are the options

[upper karma]

how do you not know what AC is?

have you ever looked at a circle before?

[cedar tendon]

I NEED HELP

what ac

[upper karma]

it's radius

[sinful anvil]

3 is a sector pretty sure

[cedar tendon]

what ab

is ab arc length?

[sinful anvil]

2 is arc length

[cedar tendon]

ok

what 3

[sinful anvil]

3 is sector

So Radius, Arc length and lastly the sector

[cedar tendon]

need help with that

[upper karma]

google it

[sinful anvil]

It's D

[cedar tendon]

what D?

[sinful anvil]

The formula

It's A= Pi x R^2

[umbral snow]

Dude literally Google

[upper karma]

^

kaynex knows

[umbral snow]

There's questions we're good for. This ain't it

[sinful anvil]

I already said the answer

[sinful anvil]

oh

[upper karma]

google

[cedar tendon]

What is it

?

[sinful anvil]

It's A pretty sure

[cedar tendon]

ok

[upper karma]

its an American multinational technology company that specializes in Internet-related services and products, which include online advertising technologies, a search engine, cloud computing, software, and hardware

use it to help solve your math issues

[sinful anvil]

LOL

[paper glacier]

how is this guy still on this server smh

[sinful anvil]

Idk

[cedar tendon]

HELP ME

i need help

[weak shoal]

Should we summon the mods?

[sinful anvil]

Well if it's necessary, do it

[cedar tendon]

i need help

[weak shoal]

Read #❓how-to-get-help

the answer to your problem is in there

[cedar tendon]

im asking for help

[upper karma]

@cedar tendon wasup

[cedar tendon]

xi

[upper karma]

remember me

ill help

[cedar tendon]

ok

[upper karma]

just learn the basic technique of finding the area of a sector

remember what the other guy said

look at youtube or google

[cedar tendon]

what he said?

ok

[upper karma]

look at youutubee or googleee

[cedar tendon]

xi is this how (sector angle / 360 ) * (pi *r2)

[upper karma]

yep

ur doing great

and do u know why do u have to multiply them

do u know what pi r2 is?

its pi r^2 btw

@cedar tendon Do u understand why

or did u just look up the formula

understanding why is an important fundamental in math

it helps

[cedar tendon]

i just look up the formula

[sinful anvil]

For area of the whole circle?

[upper karma]

@cedar tendon Ya cool, if thats how u wanna do it

do it ur way

just apply that formula to the problem

and boom solved

[cedar tendon]

so the sector area is 90 right

[sinful anvil]

90*

Theta is 90

the Radius is 12

[upper karma]

yep

90 is the angle

[sinful anvil]

The angle is 90*, radius is 12

[upper karma]

@cedar tendon U confused my boi

i believe in u

👍

[sinful anvil]

I had 113.04 as the area in degrees

Is it asking for it in radians specifically?

[cedar tendon]

@upper karma it give me 113.097

[sinful anvil]

I used the full area formula

then distributed it into 4

[upper karma]

what r we talking about

[sinful anvil]

452.16 was the area of the circle @cedar tendon

And I divided it by 4

[cedar tendon]

i just divied 360 with 90

[sinful anvil]

oof

[upper karma]

the answer still has to be with pi

[cedar tendon]

and mutiply pi r12

and i got 113.09

[sinful anvil]

I had 113.04

[upper karma]

k cool

[cedar tendon]

xi

so what the answer

[upper karma]

hold on

[sinful anvil]

lol

In radians? Correct

We need the answer in radians

[upper karma]

the answer is 36 pi

what r u talking about

[cedar tendon]

ok

[upper karma]

@sinful anvil what problem r we doing

this one right

it didnt say in radians

@cedar tendon Do u understand why its 36 pi

[sinful anvil]

6^2 is 36

times that by pi

and it's 113.04

[cedar tendon]

idk how you got 36

[upper karma]

the radius isnt 6

i dont understand anymore

what problem r we doing

someone answer me

[sinful anvil]

lol

[upper karma]

isnt the radius 12

[sinful anvil]

yeah

I was mixed up with another server

Since I'm helping someone else with the same topic

and the radius was 6 for them, my bad

Yes, it's 12

[cedar tendon]

ok

[upper karma]

@cedar tendon Area of a cirlce pi r^2 right

[sinful anvil]

Theta is 90

[cedar tendon]

yea

[upper karma]

so 144 pi * 90/360

and u get 36 pi

[cedar tendon]

OH

ISEE

[sinful anvil]

lol

[upper karma]

sure u do

[sinful anvil]

I'm dumb

[sinful anvil]

I used a completely different formula

[cedar tendon]

ok

[upper karma]

@cedar tendon https://youtu.be/Qz_7T0GpCFU

watch this

hopefully that will help

https://www.youtube.com/results?search_query=fidning+sector+area or any video in here

[cedar tendon]

@upper karma i got 461pie

is that right?

[upper karma]

how tf did u spell pi incorrect?

[cedar tendon]

noo

pie is the right spell

i eat pie

don't tell me pie is the right spell

because i eat pie

[upper karma]

@upper karma Lmao this dude

[cedar tendon]

hes trying to mess with my food

xi

i got 461

pie

is that right

[sinful anvil]

lol

I'm lost

[cedar tendon]

xi

is that right?

[upper karma]

ok this guy is a troll

[cedar tendon]

xi

[sinful anvil]

Then call the mod

Simple as that

[upper karma]

damn

big brain

[cedar tendon]

math is hard

and ineed hep

help

[upper karma]

u need to learn

u need to try

[cedar tendon]

xi is it 461

[upper karma]

whats 9 + 10

[cedar tendon]

wait

what

[upper karma]

lol

[ebon sundial]

@cedar tendon yea what 9 + 10

[cedar tendon]

who

who you guys asking it to

[upper karma]

@cedar tendon Yea man whats 9+10

u

[cedar tendon]

guys i need help with the problem im currently on

[sinful anvil]

send a screenshot

Is it another "find the area of the sector" one?

[cedar tendon]

i need

help

to solve this one

xi is the answer 461pie?

is it?

[upper karma]

the answer is 9 + 10

[cedar tendon]

i need help

[upper karma]

u need jesus

whats 9 + 10

[sinful anvil]

You're also a troll, aren't you lol

[ebon sundial]

imagine this guy doesn't even know what 9 + 10

[cedar tendon]

if i answer the question would you help me

[ebon sundial]

yea

i would help you

out

[sinful anvil]

lol

[ebon sundial]

if you can answer what 9 +10

[cedar tendon]

ok

BUT THAT QUESTION is hard

it is impossible to get the answer

no one knows it

[sinful anvil]

9 + 10?

[cedar tendon]

that is the hardest insanely problems i. have seen

[upper karma]

whats 9 + 10?

[ebon sundial]

just give a answer

[sinful anvil]

Do you mean it as an outdated joke? or?

[ebon sundial]

we are serious

it not a joke

[upper karma]

^

[ebon sundial]

i wanna see if this dude knows the answer

[upper karma]

whats 9 + 10

[cedar tendon]

YOU GUYS don't understand that problem is impossible to solve

no one has ever solve that

[upper karma]

no one has ever solved 9 + 10 before either

so