geometry-and-trigonometry

[little osprey]

Where does this come from?

[upper karma]

that's how I take my integrals @rigid thunder

it's called the chemist technique

[rigid thunder]

Blew my mind when I first read about it 😄

[upper karma]

Heron's formula be looking thicc tho

[little osprey]

Heron's formula be looking thicc tho
@upper karma no homo

yes hobo

[lusty quest]

thank you buds

👍

[little osprey]

no worries

[upper karma]

Hey how do I find the height of a triangle to one of its sides if I already know all the lengths of the sides of the triangle

,w heron's formula

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by polynomial#4757

[upper karma]

@upper karma this is what you're looking for

Oh ok thanks

[charred zephyr]

Ok that probably took too much time
@upper karma thanks for the help!

[upper karma]

Ask if you have any doubts

[solar shale]

How do I find the area

Ik how to solve

I just don’t know what to set it equal to at the end

[glacial haven]

area?

for what

[solar shale]

Like how they get 718

I understand the rest

[upper karma]

Where’d they get the height from

Maybe show full photo

[solar shale]

Here

[upper karma]

They gave you the surface area

so now add up all the faces and set it equal to 718

And then solve for a

S*

Make sense?

[solar shale]

Yeah

[upper karma]

You got it now?

and need any more help?

[solar shale]

Rn nah I’ll @ u when I do if that’s fine with you

[upper karma]

sure

👍

Or DM idc which

[solar shale]

Ok

[fading zinc]

if nobodys in the middle of a question

Id like ask one

I have no idea how to do this

[glacial haven]

just multiply by 5/2

[bitter sequoia]

Can anyone help me out with this?

Nvm i got it

[upper karma]

herro

I need some help

With dis

[glacial haven]

just use special triangle

[upper karma]

@glacial haven Wdym Special triangle?

Like I know how to get to the answer

But I don't know how to answer in terms of pi

<@&286206848099549185> Can you help pls?

[umbral snow]

Well, what is the answer you got?

[upper karma]

I didn't get any answer, because I don't know how to answer in terms of pi

[idle bloom]

what did you use to get the answer?

[upper karma]

But I got $$x=6\sqrt{3}

[somber coyote]

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

[upper karma]

I used Sine

[somber coyote]

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

[umbral snow]

Oh, haha you're correct, this isn't expressible in terms of π

[onyx cloud]

all those answers look wrong lol

[idle bloom]

^

I thought it was asking for the angle lol

[upper karma]

Yeah I had no clue how to express it in terms of pi XD

I was like "Can u...?"

[umbral snow]

Yes, I thought it was asking for a radian angle but it isn't

[idle bloom]

well you got the right answer for x

[umbral snow]

So... uh

[upper karma]

Another typo

[onyx cloud]

i think they meant to ask for the last angle maybe

[upper karma]

Bruh these practice exams have had so many typos.

It asked "find the value of x"

[idle bloom]

yea uh

none of those are right

[onyx cloud]

cause d would work as the answer if x was the last angle

[idle bloom]

cause we know x has to be > 9

and also the side length can't be expressed in pi as stated earlier

so the exam is just wack

[upper karma]

There is so many typos in this exam 🤦‍♂️

(its a practice exam btw)

[glacial haven]

30 60 90 triangle?

but the answers dont

seem

riught

[upper karma]

They’re all wrong

It’s a trick question

/ wrong question

[hasty citrus]

why is 10 *1 less than 5 * 5

or i mean 9 * 1

as in,
if there is a square, A, and rectangle, B, such that
the perimeter(A) = perimeter(B)
why is Area(A) >= Area(B)

[upper karma]

So

10(1)>9(1)?

In arithmetic it’s a basic and true statement

Geometrically you can understand that 10(1) > 9(1) if you construct some rectangles.

[hasty citrus]

Sorry that was unclear, i meant if we hold perimeter constant why does a square always have greater than or equal volume to a rectangle

[dark sparrow]

@hasty citrus you saw it for yourself

a 9 by 1 rectangle has area 9 while a 5 by 5 square has area 25

same perimeter does not mean same area

oh you're asking why it's ALWAYS true for rectangles and squares specifically

there are a bunch of ways to explain that

are you comfortable with completing the square?

[dark sparrow]

@hasty citrus?

[supple onyx]

No need to complete the square, if a square has sides x and area x^2 , then we can make a rectangle of the same perimeter with sides x-a and x+a. (x-a)(x+a) = x^2-a^2 which is always smaller than x^2 when a is non zero

[dark sparrow]

okay yeah true

[hasty citrus]

@supple onyx awesome thanks for the explanation, that makes sense but i still dont understand the intuition behind it, is there anything you would suggest

[silver spruce]

can someone teach me this?

we can go to a voice chat where it may be more usefull

*useful

[mossy narwhal]

Welcome pancake

[silver spruce]

wait no

help me too

haha lol

plz

[mossy narwhal]

lol I didnt help him and idk how to help 😬

[silver spruce]

haha lol

its okay

can i move this convo to a diff chat?

i think it may be more help there

[mossy narwhal]

Wait what

Ill stop talking here. I didnt mean to distract. Just saying welcome to a thank you.

[upper karma]

@silver spruce why exactly do you have a problem with almost 30 diff answers

[silver spruce]

haha lol

[upper karma]

that is a pretty extreme multiple choice question

[silver spruce]

its just some off brand free khan acadamey stuff

can you teach me how to solve it

[upper karma]

khan academy is also free?

anyway

yes i can

the function you're looking for is

[silver spruce]

are you okay with moving to a voice chat?

i think it may be easier that way

[upper karma]

$f(x) = a(x-(9+7i))(x-(9-7i))$

[somber coyote]

polynomial:

[upper karma]

no

this is all there is to it

now you expand

[silver spruce]

so i just have to plug it in?

[upper karma]

or even you can just do this

$f(0) = 390$

[silver spruce]

oh yea

lol

[somber coyote]

polynomial:

[upper karma]

then solve for a

[silver spruce]

i remeber now

[upper karma]

and that's it

[silver spruce]

thank you so much

haha lol

[upper karma]

np i guess

and @dark sparrow taught me to tell people to go to right channels

so next time, ask in #prealg-and-algebra since this is an alg question

[mossy narwhal]

Nvm poly said it

[hasty citrus]

@dark sparrow i’d be interested in seeing your completing the square argument as well if you want

[upper karma]

@hasty citrus what is your question

[dark sparrow]

take a square of side a (perimeter 4a, area a^2). take a rectangle with dimensions x by 2a-x (whose perimeter is, again, 4a)

the rectangle's area is x(2a-x), or -x^2 + 2ax

complete the square on this and see that its max value is a^2 and occurs at x=a

[silver spruce]

wait @upper karma what does a =?

[upper karma]

that's what you're supposed to solve for

[silver spruce]

its just like a thing you may need to multipy for so the y int is 390 rite?

[silver spruce]

i still can' t find the awnser

i got 314.6590 thats obviously not the right awnser

can someone move to voice chat with me and explain it

thank you so much if it is possible

should I list my steps here?

[upper karma]

what's with you

and voice chat

lol

,w a(-9-7i)(-9+7i) = 390

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by polynomial#4757

[upper karma]

amazing wolframalpha

,w calc a[(-9-7i)(-9+7i)] = 390

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by polynomial#4757

[upper karma]

fucking useless

[somber coyote]

Commander Vimes:

[upper karma]

vimes

what country are you from

[surreal bolt]

hi?

[azure reef]

i am from Ankh-Morpork

[silver spruce]

oh

lol im so sorry

thank you so much for your time

let me try again

wait

what was your awnser

i got X^2-18x+88=390

x^2-18x-302

1-18-302=319

[azure reef]

ok

so

can you show step by step how you got quadratic?

@silver spruce

[silver spruce]

okay

[azure reef]

because 319 is wrong answer

[silver spruce]

yea lol i think i realized

do you want the question too

i uploaded it before

[azure reef]

1-18-302=319
moreover here it shold be negative

[silver spruce]

oh yea

thats what i meant

[azure reef]

so ok

[silver spruce]

but like its not in the awnser key

[azure reef]

9-7i is one of the roots

[silver spruce]

or like the thingy thing

[azure reef]

second root obviously is 9+7i

[silver spruce]

yea basically he wants a quadratic with real coefficents and a y int of 390

let me try

[azure reef]

?

[somber coyote]

Commander Vimes:

[silver spruce]

so $(x-(9-yi)(x-(9+7i)

wait i did it wrong

$(x-(9-yi)(x-(9+7i)$

[somber coyote]

AZ.0629:

[azure reef]

why y

[silver spruce]

but then we solve to get $x^2-18x+88$

[somber coyote]

AZ.0629:

[silver spruce]

sorry i meant -7

[azure reef]

ok, so firs of all

you forgot a

[upper karma]

$y \approx 7$

[somber coyote]

polynomial:

[azure reef]

second 88 is wrong

[silver spruce]

is it?

[azure reef]

ye, there is no 88

[silver spruce]

thre is only two numbers

[azure reef]

do not ignore a also

it exists

you cannot set it 1

[silver spruce]

for the frist one there is the 99=81 and the 7i-7i=7

how do we figure out a?

[azure reef]

do you know how complex numbers are multiplied?

[silver spruce]

yea

its like i=squareroot -1

and then i^2=-1

i^4=1

[somber coyote]

Commander Vimes:

[silver spruce]

yea

[azure reef]

i mean to be precise

[upper karma]

the y-intercept occurs at x = 0.

[azure reef]

[silver spruce]

lol

[azure reef]

ok so AZ

[silver spruce]

yea i know

[upper karma]

good job deleting your messages

[silver spruce]

the ?

wait wut

im so confused to

[azure reef]

look

[silver spruce]

*too

[upper karma]

AZ

[azure reef]

AZ

multiply that

[silver spruce]

wait can commader vimes just help me?

[azure reef]

multiply inner parts

[silver spruce]

its just easer with one person

okay

[azure reef]

do not touch a yet

[silver spruce]

so its a(x-9-7i)(x-9+7i)

right?

[azure reef]

nvm

yes

[silver spruce]

okay imma solve the s stuff right now

[azure reef]

but prolly it is better to keep (9+7i) and (9-7i) bracketed

[upper karma]

i am not sure how you can't solve it when i give you the answer

you should review how to expand binomials

i am not trying to be mean

but seriously

https://www.youtube.com/watch?v=ZMLFfTX615w

[silver spruce]

$a(x^2-9x+7xi-9x+81-63i-7xi+63i+7)$

[upper karma]

and all you need to know is that sqrt(-1) = i

[somber coyote]

AZ.0629:

[silver spruce]

oh yea

lol

[azure reef]

ok so you are getting fucked up at that stage

[silver spruce]

its okay @upper karma , thanks for the help, but i'm pretty sure i think i get explanding

did i do anything wrong?

@azure reef did I do anything wrong?

[somber coyote]

Commander Vimes:

[crimson smelt]

Why does everyone refer to khan academy

[silver spruce]

okay

[upper karma]

because i can't think of an even easier example

he explains at such a slow pace

[silver spruce]

so

we get

[upper karma]

surely there shouldn't be any unclear steps using khan academy

[crimson smelt]

A good ol textbook

[silver spruce]

$x^2-x(9+7i)-x(9-7i)+(9+7i)(9-7i)$

[somber coyote]

AZ.0629:

[azure reef]

yes

[silver spruce]

should we expland the x(9+7i) stuff now?

[azure reef]

so factor out x and see what happens near it

no

[silver spruce]

wait what do you mean?

[somber coyote]

Commander Vimes:

[silver spruce]

oh okay

[somber coyote]

Commander Vimes:

[silver spruce]

ookay

so

[crimson smelt]

Factor out x

[silver spruce]

$a[x^2-18x+[81-63i+63i+7]

[crimson smelt]

$a[x^2-18x+[81-63i+63i+7]$

[silver spruce]

then a(x^2-18x+81)

[somber coyote]

Yes:

[silver spruce]

yea

[azure reef]

again fucked up

[silver spruce]

so we get a(x^2 -18x+88)

[azure reef]

i mean coefficient near x is correct

[silver spruce]

wait whre

im so confused

81-63i+63i+7=88

[somber coyote]

Commander Vimes:

[crimson smelt]

Bruh youre litterally just multiplying stuff

[silver spruce]

am i wrong?

81+7=88 -63i+63i=0

im so confused

[upper karma]

will you review

the video i linked

or do you still want to stand strong with your "principles" and ego

[azure reef]

can you show exactly what you do here @silver spruce

[silver spruce]

okay

so

[azure reef]

show multiplication literally step by step as for childs

[silver spruce]

9*9=81

[azure reef]

true

[silver spruce]

9*-7i=-63i

[azure reef]

true

[crimson smelt]

$i = \sqrt{-1}$

[dark sparrow]

{}

[silver spruce]

7i*9=63i

[azure reef]

true

[upper karma]

you don't even need to multiply it out

it's a difference of squares

hello?

[somber coyote]

Yes:

[silver spruce]

7i*-7i=oh shoot

im so stupid lol

haha

49

[azure reef]

its alright so what is total product

yes

[silver spruce]

so 130

[azure reef]

yes

now

[silver spruce]

bro

i musthave made you so fustrated

[crimson smelt]

How do people get the honourable role btw?

[somber coyote]

Commander Vimes:

[silver spruce]

so x^2+18x+130

a=130

[azure reef]

no

[silver spruce]

yea *-18x

so then

[crimson smelt]

130a=390

[silver spruce]

we mulitpy everything by 3 to get 3x^2-54x+390

then add a+b+c

thanks so muych

[azure reef]

yw

[upper karma]

lol

1 hour and 4 mins ago

i gave you everything you needed to solve the problem

had you watched the khan academy video

this would've been done so much earlier

[silver spruce]

haha lol

its all good now

thanks @upper karma @azure reef @crimson smelt for helping me!

[crimson smelt]

[azure reef]

[silver spruce]

👍 👍 👍

[upper karma]

what do you mean @crimson smelt

what did he do lol

[crimson smelt]

Lol

I did say that all he needed to do was multiply

[tepid dragon]

If a pyramid is to a Pentatope, and a cube is to a tesseract, what is a regular icosahedron to?

[dark sparrow]

the 600-cell might be what you're looking for

[austere dragon]

this might sound extremely dumb, but why do we add exponent 2 and 3 for volumes and area?

[dark sparrow]

wym

[austere dragon]

like cm^2 cm^3 that stuff

does it represent a different type of unit or is it just an exponent?

[dark sparrow]

different unit lmao

[austere dragon]

why

[dark sparrow]

when you find the area of a rectangle you multiply its length (in meters) by its width (also in meters)

meter * meter = square meter

[austere dragon]

ooooooooOOOO

🤯

[little osprey]

Good question not a dumb one

[austere dragon]

i still cant differentiate length and width

[little osprey]

Length is usually the longer side

Is what I was told by my teacher

[arctic vortex]

can anyone help me with some geometrry homeworks??

[silent plank]

#❓how-to-get-help

[arctic vortex]

oh ok

can u help me with this one?

[dark sparrow]

@austere dragon wym

[little osprey]

What do you think @arctic vortex

[arctic vortex]

perpendicual to a point on the line

@little osprey

[austere dragon]

sorry i was just confused about the difference with length and width, i didnt know that length is actually the longest side instead of being named accordingly to the axis and stuff ^^^^

[dark sparrow]

meh

it's not as set in stone as you make it sound

[little osprey]

I forgot constructions with a compass

epic

[arctic vortex]

o

lol

[little osprey]

I think youre right though

[arctic vortex]

yay

would this be Z?

[little osprey]

https://www.youtube.com/watch?v=z-qdyuQ-JSw

would this be Z?
@arctic vortex I have no idea because I dont have the sheet of paper

[arctic vortex]

o

i dont know these ones

[little osprey]

well what do you think

[arctic vortex]

i think i know the first one

idk about the second one

do you know it @little osprey

[little osprey]

Ok so its really confusing

have you drawn it maybe?

the wording

@arctic vortex

[arctic vortex]

yes

i dont have a compass

[little osprey]

Then how will you draw a circle

[upper karma]

use a water bottle cap

[little osprey]

D Itsy Bitsy Spider

went up the water spout

[arctic vortex]

i just draw it with my hand

wait

[little osprey]

Ok fair

[arctic vortex]

this is important one

[little osprey]

as long as you know its a circle ig

[arctic vortex]

i know its either b or d

which would it be

[little osprey]

Wait

what happened to the last question

[arctic vortex]

oh nvm this is b

and yeah idk it

these two

[little osprey]

yeah its b

[arctic vortex]

yea

[little osprey]

The first one is based on the question you just posted

[arctic vortex]

yea

[little osprey]

So what do you think the first one is

The second one you have to draw and show me

I know the answer

But I want you to try and find out

[arctic vortex]

i think either a or c

[little osprey]

Why a and why c?

[arctic vortex]

because abd and cbd look congruent and they have an arc passing through the B

and ba = bc bc b is the common point of both

[little osprey]

do you know how to draw a bisector

of an angle

base your knowledge on that

[arctic vortex]

idk

[little osprey]

Hm

I think its both too

nvm

actually

it can only be a

[arctic vortex]

whys taht

that

[little osprey]

because A and C are not necessarily equidistant

They just serve as points on the line

[arctic vortex]

o

this would be false right @little osprey

since there is no right angle formed

[little osprey]

Im sorry but I dont know

Honestly I havent done compass and angle constructions in a really long time

[arctic vortex]

o

its fine

[austere dragon]

im pretty sure that means perpendicular which it is not

so yea false

[silent plank]

its unclear what that construction even represents

[little osprey]

^

[silent plank]

it would help if things were numbered

[arctic vortex]

idk

summer course geometry is really weird

what would tihs be?

[little osprey]

have you drawn it

and if you have

please send a photo

[arctic vortex]

think of a cricle

and a square in a cirlc

ee

[little osprey]

yes

[arctic vortex]

yes

[little osprey]

so what do you think

[arctic vortex]

and its asking for the measure of the space

[little osprey]

since the arcs will be congruent

[arctic vortex]

i think

probably like 90

[little osprey]

yes

[arctic vortex]

what

[little osprey]

wdym what

[arctic vortex]

its 90

[little osprey]

thats what I agreed to

[arctic vortex]

o

[upper karma]

@arctic vortex Did it solved?

[arctic vortex]

not this one

i dont understand this one

[upper karma]

okay sure lemme go

[little osprey]

Didnt we just

do this question

[arctic vortex]

i dont think its 90

tho

[silent plank]

have you drawn a diagram?
why don't you think its 90°?

[arctic vortex]

wai

tit is 90

because each measure is 0-

90

oops

[upper karma]

yup i think so its 90

@arctic vortex But why so messy language?

[arctic vortex]

?

wdym

sorry i just type fast lol

would tihs be false or true

[upper karma]

..

[arctic vortex]

i just have 1 more

after this one

then i am DONEEE

[upper karma]

okay go

[silent plank]

seriously nfi what that's asking

[arctic vortex]

o lol

[upper karma]

@silent plank

ask next?

[olive solar]

@arctic vortex which illustration

[arctic vortex]

L

wait no

it says D but i think its L

[olive solar]

if the angle ADH is the illustration why does it ask you to use point D

wait but it refers to an arc from the angle

like you have to draw an arc left of the arc in ADH

[arctic vortex]

o

[olive solar]

this question is incomprehensible

[silent plank]

its most likely a typo

mixed with some grammatical errors

[upper karma]

Its like a gud qn to pass ur Lockdown period

[arctic vortex]

this one then?

last one btw

[upper karma]

True

[arctic vortex]

how

im confused

[upper karma]

It seems to be like

[arctic vortex]

o

[upper karma]

no mathematically logical

Test ur eyesight

[arctic vortex]

i have poor eyesight

[silent plank]

there's no clear information relating the 2 angles

[upper karma]

No, they are asking to observe the figures

[arctic vortex]

im thinking its false

[silent plank]

if its unclear, the default response would be false and ignore it

[upper karma]

Cool @silent plank

[lofty adder]

Is it possible to convert a scalene triangle into an isosceles triangle?

Like is it possible if i construct a new 3rd side such that the median of the original triangle is the perpendicular bisector of the new side?

anyone?

this has been bothering me since 2 weeks

[paper vale]

What do u mean convert, like obviously if u just change a side then sure

[lofty adder]

im sorry i didnt mean convert

is it possible to construct an isosceles triangle on the existing scalene triangle

ABC is scalene

im trying to construct a side EF such that AF=AE and AD is the perpendicular bisector to EF

[paper vale]

Well all u need is 2 sides to be the same, so u can do that

AD is perpendicular to midpoint iff it is isosceles

[lofty adder]

why is it that if i multiply BC by cos(alpha), i get a value equal to EF

[paper vale]

Is the line other than AD defining alpha perpendicular to BC

[lofty adder]

i dont understand?

[paper vale]

Well that line could just be arbitrary so it it meant to be perpendicular to BC

[lofty adder]

yeah i get that

that line is perpendicular to BC

hence forming alpha with the median

how does cos(alpha) give a result equivalent to EF

[midnight hearth]

Looking for someone who'd be willing to spend a bit of time with me to explain me how to find the points of intersection of three spheres and implement it into code and vector arithmetic. Willing to pay

[solar shale]

X=10 is that right

[upper karma]

I think he might be but there’s a dispute so just send yes/no to clarify.

,w solve for x 30=1/2((5x+10)-3x)

No its not right.

[solar shale]

Ok

[upper karma]

Recall this theorem

@solar shale

[solar shale]

Thanks I understand now x=25

[upper karma]

Yep

Np

hm

Nice

[solar shale]

To solve this I would 12^2+16^2=16^2 right?

And the answer would be not tangent

[upper karma]

AO = 8 + radius

Now what’s the question

Looking for the angles?

Finishing the sides?

Area?

Area of the circle not including the triangle?

@solar shale

[solar shale]

Its asking if AB is tangent to circle O

[upper karma]

Hmm

You need to figure out if AB is perpendicular to BO

Meaning if there’s a 90* angle there

[charred zephyr]

pythagoras

👏

[upper karma]

^

If it checks out, it’s a tangent

[charred zephyr]

AO = 8 + radius
@upper karma use this to find AO

then apply pythagoras

[upper karma]

I agree completely

12^2 + 16^2 ≠ 16^2

[solar shale]

Ok got it

[charred zephyr]

what? i actually got it's a right triangle

[upper karma]

perhaps

[cloud stump]

do pythagoras and then if BO is equal to the missing part then it's tangent

wait

never mind

[upper karma]

just apply it for the triangle as a whole

That way if it works you know it’s right

[cloud stump]

AO is 20

[upper karma]

And clearly the only possibly right angle on that thing is where it intersects AO

er

AB

[solar shale]

So I can’t just do 12^2 + 16^2 ≠ 16^2

[charred zephyr]

no

[solar shale]

I would have to use pythogoras

Aight

[charred zephyr]

u need to check if
AB² + OB² is equal to AO²

[upper karma]

Where’d you pull those from

@charred zephyr yes

But why the equation 12^2 + 16^2 = 16^2 @solar shale

[charred zephyr]

u have AB and OB, u need AO and it's easy to find since AO = 8 + r and radius is already pointed out by OB

[upper karma]

Where’d the other 16^2 come from

[cloud stump]

ok let's say the point where AO intersects the circumference is X then XO is 12 because it's a radius. That directly implies OBA is 90 degrees

[upper karma]

sex degrees what

[cloud stump]

sexagesimal

[upper karma]

Mmm yes

[solar shale]

I got the other 16^2 because half of one side was 8 so I added 8 now (other side) I have 16^2 now so I used 12^2 + 16^2 ≠ 16^2

@upper karma

[upper karma]

Half of one side is not 8, that is part of one side though

Half of that side would be 10 in fact

But good try

[charred zephyr]

are u done @solar shale

[solar shale]

AB² + OB² is equal to AO² ok so 16^2+12^2=AO^2

[cloud stump]

that's waste of time

[solar shale]

One side would be 400=AO^2

[cloud stump]

just translate the radius

the radius is 12 everywhere which means AO = 8+12

20

[charred zephyr]

whats the waste of time?

[solar shale]

I’m lost now

Ok I I get it so radius is 12 so I can do 12+8=20 and that is AO=20 so 16^2+12^2=20^2

400=400

[charred zephyr]

Step 1: Finding AO
AO = 8 + r (r is the radius of the circle)
Since OB is the radius of the circle, and OB = 12
AO = 8 + 12 = 20

Step 2: Applying the reciprocal of pythagoras
You have sides AB = 16, OB = 12 and AO = 20.
AO is the longest side.
Check if AO² is equal to AB² + OB² and deduce the nature of triangle OBA.

Step 3: Angle ABO
After deducing the nature of triangle OBA, conclude the measure of angle ABO.
Using a famous circle tangent propetry (hint: perpendicular lines), is AB tangent to the circle?

[solar shale]

Yes it’s tangent

400=400

[charred zephyr]

gr8

[solar shale]

Someone verify that?

[cloud stump]

2x*6=(x+3) *2

12x=2x+6

10x=6

x=0,6

it's ok

@solar shale

[solar shale]

Ok thanks

[solar shale]

I got 2870 for an answer can someone verify that

[runic beacon]

What is the question? To find area?

[glacial haven]

pi?

what do you need pi

for

tf

huh

theres no circle

[solar shale]

Oh wait nvm I confused two problems ignore the pi

Feel free to @ me once someone verifies that if 2870 is the answer

[glacial haven]

@solar shale yes sir

[cloud stump]

the volume?

find the volume of the parallelepiped and divide by 2

o wait

never mind

it's not perpendicular

[silent plank]

@solar shale show work

[novel flax]

how do i prove cos20-cos40-cos80=0

without calculator

[upper karma]

cos20=cos40+cos80

[somber coyote]

HoboSas:

[upper karma]

For RHS

Then some manipulation

[novel flax]

ok i will try

$\cos (40)+\cos (80)=2\cos (60) 2\cos (20)
= 2\left(\frac{1}{2}\right) \cos(20) = \cos (20)$

[somber coyote]

Orange905:

[novel flax]

wow thanks @upper karma

[upper karma]

Np

[upper karma]

if i have two line segments coming out of the middle of a circle and they have two points on the edge of the circle, how do i find the minor arc between the points

Middle of a circle?

Do you mean center?

yes

There's a formula

If you know the angle between the 2 segments

o i think i got it

[next fable]

The rules for rotating a point in a plane are something to the sort of 180 degree rotation counter clockwise (positive) or clockwise (negative) as "-x. -y", but what about rotations such as 130, 120 etc is there any rule for these or do they just have to be done using a protractor?

[next fable]

<@&286206848099549185>

[weary drift]

rotation in the plane by any angle t can be viewed as a linear map on R^2 and can be represented by this matrix R_t @next fable

$R_t=\m{\cos(t)&-\sin(t)\\sin(t)&\cos(t)}$

[somber coyote]

RokettoJanpu:

[weary drift]

a rotation by $\pi$: the image of $(x,y)$ under $R_\pi$ is
$$R_\pi\m{x\y}=\m{\cos(\pi)&-\sin(\pi)\sin(\pi)&\cos(\pi)}\m{x\y}=\m{-1&0\0&-1}\m{x\y}=\m{-x\-y}$$

[somber coyote]

RokettoJanpu:

[solar shale]

@ me

[hollow mica]

Form is perfect dude

And calculations are ok

To check yourself : put your answer where "r" is at in the volume equation

[umbral snow]

Easy way to check, what's the volume of a 4.5 cuft radius sphere?

Also, be sure to put cuft on your answer

Units and all that

[solar shale]

Oh ok I’ll check using that thanks

I’ll fix that

[hollow mica]

Well, just feet right? It's raidus

[solar shale]

Just feet

[umbral snow]

Oh haha yes.

[hollow mica]

Why did you drop the pi*r^2 before that part of the equation for the surface area of a cone?

[solar shale]

I’m not sure what you mean like where that go

[hollow mica]

Sorry let me show you the eq I was thinking of

But if your assignment defines it as whatever you just showed me as then I'd continue using that

From what I saw in your screenshot your math was correct

[solar shale]

Wait I’m not sure honestly

I’ll just solve it and give my final answer and then someone could verify that

[hollow mica]

Stuck?

Show me your work m80

Is the computer you're doing your hw on the same one with discord ?

[upper karma]

How do I find the side length of an equiangular triangle if I know the radius of the triangle

[dark sparrow]

wym by radius

[austere dragon]

what do people use as an alternative to pi in a job or career somethin like that? do they use calculators? 22/7? 3.14? or just leave the pi symbol?

[dark sparrow]

wym

what do you mean by alternative

[austere dragon]

i mean, there has to be a convention or something, calculating with pi would be impossible wouldnt it? and theres a whole lotta method out there

[glacial haven]

nah they use pi

[austere dragon]

but pi is infinite?? it has to be estimated

[dark sparrow]

pi isn't infinite

don't misuse the word infinite like this

[austere dragon]

what

[dark sparrow]

pi is a finite value. it's less than 4.

[austere dragon]

oh i mean, it uh.. hmm idk what word to use

[dark sparrow]

its decimal expansion is infinite but so what

[austere dragon]

its uh, infinite?

wha?

[dark sparrow]

by your logic, so's the number 1/3.

[austere dragon]

shit

how does that even work

[dark sparrow]

yes pi is irrational yes its decimal expansion is infinitely long and non-repeating but again so what

[austere dragon]

you cant calculate it if the decimals never ends!! it's like multiplying a number you dont know the exact value of, you need to estimate it

[dark sparrow]

...

if you're doing any calculations with computers you aren't storing anything with infinite precision anyway

you can't calculate 1/3 exactly as a terminating decimal either so what

[austere dragon]

[dark sparrow]

if you want an approximation then take as many decimal places as you need, and i can hardly imagine needing more than 40, tops

as far as symbol manipulation is concerned, pi is just pi

[earnest sphinx]

How can I approach this?

[twilit zenith]

I could be mistaken.
Since you are required to perform a projection, it would be more convenient to transform your current basis (the standard basis with e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1)) to an orthogonal basis where (1, 2, 2) is one of the vectors.
Then, you can convert your vector x from the standard basis to the new basis, and then you can apply the "projection matrix" on the "converted" vector.

More formally, you begin with a matrix $P_{B}^{E}$ that converts from the standard basis $E = { \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix}, \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix}, \begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} }$ to an orthogonal basis where $b = \begin{pmatrix} 1 \ 2 \ 2 \end{pmatrix}$ is one of the basis vectors, and then you apply $T_{B}^{B}$ which is the projection on a vector whose components are given in the new orthogonal basis $B$, and in the end, the desired transformation is $T_{B}^{T} \circ P_{B}^{E}$.

[somber coyote]

RoiKadmon:

[twilit zenith]

I think that was a blob of text with no real explanation. I'll try to give a simpler example: $\$
Let's say that you're interested in finding the projection of a vector to the subspace spanned by $e_1 = \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix}$. In this case, you need to find an orthogonal basis where $e_1$ is one of the basis vectors. Fortunately, the standard basis, $E = {e_1, e_2, e_3}$ is already such a basis, and thus, in order to perform the projection operation, we must only consider what the projection $T_{E}^{E}$ does on each of the basis vectors. Indeed, since the projection only considers the component of the first vector, keeps it as is and "ignores" the other vectors, our desired transformation is $T_{E}^{E} = \begin{pmatrix} 1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix}$. $\$
Indeed, for an arbitrary vector $q = \begin{pmatrix} a \ b \ c \end{pmatrix} = a \cdot e_1 + b \cdot e_2 + c \cdot e_3$, the result of the projection is $T(q) = \begin{pmatrix} a \ 0 \ 0 \end{pmatrix} = a \cdot e_1 + 0 \cdot e_2 + 0 \cdot e_3$, and if we compute the matrix, we get: $\$
$T(q) = \begin{pmatrix} 1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} a \ b \ c \end{pmatrix} = \begin{pmatrix} a \ 0 \ 0 \end{pmatrix}$.

[somber coyote]

RoiKadmon:

[earnest sphinx]

Great! Very useful! Thanks!!

[livid sequoia]

is this right?

[near trail]

How can I calculate how fast to rotate the corner piece texture around it's axis, in order to make it match the texture movement for the joining straight pieces.

Picture for clarification. Any tips are appreciated!

[upper karma]

Establish an angular velocity

Every pixel will have the same angular velocity

The linear velocity is just the distance from the center of the circle times the angular velocity

@livid sequoia can you provide more context?

[livid sequoia]

identify if those are perpindeicular bisectors or angle bisectors

@upper karma

[livid sequoia]

nvm I figured it out

[upper karma]

Oh I c

Nice

[pearl lava]

How do I write a proof for this?
https://gyazo.com/e8e35a545a5b3eb672e1d2c53a67cf0b

[upper karma]

"How do I write?" or "how do I prove this?"

[pearl lava]

where do I start

[upper karma]

Consider triangle HIJ

What can you say about it?

[pearl lava]

it's an isosceles triangle

[upper karma]

Damn right

HI=IJ and IF=IG

what follows?

[pearl lava]

FH=GJ?

[upper karma]

Yes

What next

[pearl lava]

HK=KJ

[upper karma]

Yes

Then

[pearl lava]

do I use SAS?

[upper karma]

Yes you got it

[pearl lava]

ok

thanks

[upper karma]

Np

[paper vale]

FIK is congruent to GIK is another way

[upper karma]

Yep

Because IK is both median and angle bisector

[upper karma]

help please :)

i think this is where SOHCAHTOA could apply..?

i'd recommend looking into SOHCAHTOA to solve this problem and other similar problems

[prisma cape]

Hint: what trig function is opp/hyp

[upper karma]

@prisma cape sin

[prisma cape]

That would be correct. @upper karma

[twilit zenith]

In this specific case, you are given that the two dimensions of the rectangle are equal, so it's a square pyramid.
However, assuming you have 3 right angles (meaning this is a rectangle) and you aren't told that the dimensions are equal, that it's a square or any other information that would indicate that this is a square, I think you can only assume it's a rectangular pyramid.

[solar shale]

Ok so if 2 sides of a square are the same then I can say that is a square pyramid as a rectangle would be different numbers @twilit zenith

[twilit zenith]

Indeed. (assuming the two sides are next to each other. The opposite sides are equal in any rectangle.)

Also, just a tiny correction:
A rectangle in which all the sides are equal (i.e. a square) is still a rectangle, so it's probably more accurate to say "a non-square rectangle would have different numbers" as opposed to "a rectangle would have different numbers".

[idle bloom]

#❓how-to-get-help

[upper jetty]

sorry. [edit : I deleted the question and ping. (see #help-2 ) ]

[solar shale]

Can someone verify that?

[dark sparrow]

total surface area or lateral surface area?

[solar shale]

Total

[dark sparrow]

ok

checks out

[hidden phoenix]

So I just popped on to have a quick question about a video I saw

So in this one video

https://www.youtube.com/watch?v=1JDt5P5n75E&t=79s

It claims there is a "chigorchuk group"

[dark sparrow]

hyperrogue

[hidden phoenix]

but it looks like {8,3} schläfti thing

[dark sparrow]

hmm

[hidden phoenix]

and then it goes through the vertexes and fills it

is that really ghigorchuk group?

[dark sparrow]

you mean grigorchuk

[hidden phoenix]

yes grigorchuk

my name memory is horrible

[dark sparrow]

https://en.wikipedia.org/wiki/Grigorchuk_group found this

[hidden phoenix]

Can someone help me with this group

I dont get it

[dark sparrow]

@strong sage mind explaining? i don't seem to get it either but you may have a better idea since you made the vid

[hidden phoenix]

wow he is here

[hidden phoenix]

@strong sage So isnt the "grigorchuk set" featured in the video same as if you take {8,3} schläfti tiling and fill it though the edges

[strong sage]

@hidden phoenix generally this is a visualization of the (index 2) subgroup generated by ac, ca, and b. Each of these generators corresponds to a move (isometry) in the hyperbolic plane. You have (ac ac)^4 = (ca b)^4 = (ac b)^4, which is exactly the same as in {3,8}, so the three moves are the moves in {3,8} which makes these relations hold. There are of course also other relations in the Grigorchuk group (which do not hold in {3,8}) -- these are represented by using a hyperbolic manifold (with its isometry group = Grigorchuk group) rather than the hyperbolic plane, so you can take a path in this space which would move you back to the starting point, while in {3,8} the same path would end in a different spot. (The gray lines at the end of the video visualize this.) The implementation and some more comments are in https://github.com/zenorogue/hyperrogue/blob/master/rogueviz/grigorchuk.cpp [I must update it to yesterday's changes in HyperRogue engine, it does not compile at the moment]

[hidden phoenix]

ok

[dark sparrow]

so what you're saying is that in the video we're looking at a quotient?

[strong sage]

quotient space of the hyperbolic plane, yes (although this quotient is a bit too large, so repeated images are only in faraway, small, unseen regions of the Poincare disk)

[upper karma]

when i mention derivatives i get yelled at

when people talking about groups and hyperbolic planes in #geometry-and-trigonometry it is fine

Still geometry

[strong sage]

geometric group theory

[dim pulsar]

analytical geometry

[hidden phoenix]

Hyperbolic geometry is geometry

[livid sequoia]

Can someone check if I got this right?

[paper vale]

it is 14.2

[supple onyx]

Hyperbolic geometry might be geometry but it aint pre university stuff

That's what #point-set-topology is fo

For*

[hidden phoenix]

oophe

i did not read the channel class

[radiant nimbus]

how many 0.2 km^3 cubes would fit in a 1 km^3 cube would it be 125?

like if a 1 km^3 cube was scored by increments of 0.2 in each dimension, that would be 5 per dimension so 5^3?

[idle bobcat]

Hi, how come the answer isn't x=8?

[radiant nimbus]

sqrt(5^2+4^2)

[next jackal]

Pythagoras

[idle bobcat]

My logic being that since we can separate the triangle into two 30 60 90 triangles, the hypotenuse would be double the shortest angle

yeah I did the pythagorean theorem to solve

but the more i thought about it if the 30 60 90 proportions check out the answer should be 8

[radiant nimbus]

its sqrt 41

the length 5 stays same, 8 is divided

so 5^2+4^2=r^2

to get r you square root both sides

[idle bobcat]

but if you split the entire triangle into two equal triangles you get two 30 60 90 triangles

[radiant nimbus]

no

[idle bobcat]

and therefore shouldnt I be able to use the 30 60 90 proportions

[radiant nimbus]

you get 2 45 45 90s

the original is a 30 60 90

when you split it down the middle it becomes a perfect right triangle

well 2 of them anyway

[idle bobcat]

im confused

how is the original triangle 30 60 90

the base angles are equal

its isosceles

[radiant nimbus]

well im assuming since it's not listed

ah well

in that case

i stand corrected but splitting it wouldn't form 2 30 60 90s, i know that for certain

[idle bobcat]

ah you're right it wouldnt..

idk why I thought it did

im tired

thanks though

[radiant nimbus]

well

actually you might be right lol

eyeballing it it looks like all 3 sides could be different angles

but

the important thing is that you are dividing the bottom length in half

so it becomes 4^2

5 stays the same

since both triangles share it

[idle bobcat]

its definitely not 30 60 90 otherwise x would be 8 lol

yeah

it actually just makes another isosceles triangle

now that i think about it

wait no it doesnt lol

ok im confusing my self im gonna stop

[radiant nimbus]

lol it's all good. were you using the angles to solve?

you don't really have to in this case you just need the lengths

you would only need angle info if you weren't given info about lengths, then you could solve for a sides length using sin, cos etc

[idle bobcat]

nah my first instinct was just use pythag

same way you were describing

[radiant nimbus]

yeah that's all it is

[idle bobcat]

but I just tried to disect it more and thought I had a 30 60 90

and then we end up in this rabbit hole

lol

[radiant nimbus]

lol

i think it is a 30 60 90 though

but

that would make the length 16 i think

[idle bobcat]

if it were 30 60 90 wouldnt it make x = 8

[primal flicker]

I did the math

[radiant nimbus]

well

[idle bobcat]

since shortest side is 4

[radiant nimbus]

yeah

[primal flicker]

Left/right angle is 51.34

So top one is 180-51.34*2

[idle bobcat]

w0w

how did you figure that out?

[primal flicker]

Calling the left (or right) angle a

tan(a)= 5/4

[radiant nimbus]

well the hypotenuse of a 30 60 90 is 2x

[primal flicker]

so a=arctan(5/4)

and using a calculator i got 51.34

[idle bobcat]

ok that is giberish to me i havent gotten to trig yet lol

cool tho

[primal flicker]

Yeah i dont think there is an easier way

But you dont actually need them

[idle bobcat]

fair enough

thanks for the help guys

[radiant nimbus]

so

pythag for the win lol

I wish I was still doing geometry. I'm taking differential equations next semester and I'm a little nervous about that

[idle bobcat]

im 21 and im only just finished pre algebra recently lol

didnt care for math in highschool

[radiant nimbus]

nothing wrong with that. I'm in my 30s and going back to college lol

[idle bobcat]

went onto community college for programming

got an internship doing full stack development

[radiant nimbus]

Working on a B.S. in Engineering

nice

[idle bobcat]

now im trying to learn math from the ground up

ive been using khan academy for 6 hours a day everyday the past month

lol

its a grind but im enjoying it

[radiant nimbus]

a few things that will help you be a better dev, learn some linear algebra (mostly basic matrix ops) and learn some discrete mathematics/ stats/probability

[idle bobcat]

yeah thats the plan, i wanna get to a level where I can understand the core machine learning algorithms and start implementing my own neural nets

and yeah I heard discrete math was a big part of computer science

[radiant nimbus]

its pretty important. you can probably mostly ignore the stuff with proofs. but boolean algebra is a godsend for writing algorithms

learn how to write karnaugh maps too, most people find that easier than simplifying/reducting the boolean

[idle bobcat]

dont know what any of that means but 🙂

[radiant nimbus]

if you're already doing full stack stuff, chances are you've already been doing some form of boolean algebra and didn't realize it lol

like your statements like say if (this):

then this

etc

or if(not this and not that or this) stuff like that is boolean expressions

and the stuff we come up with is often redundant so you can use boolean algebra to get it in its simplest yet equivalent form to your original expression

helps make programs more efficient basically

[idle bobcat]

ahh I see what you mean

hmm, still a bit difficult to imagine what a whole field of math dedicated to booleans could entail

im intrigued

[radiant nimbus]

well discrete math is more than just boolean algebra. but boolean algebra in general will be really useful to you. boolean algebra is less of its own field and more of a set of tenets or rules for handling expressions that only deal in trues of falses, or 1's and 0's

like if I gave you an expression if(A or B) you could write a truth table that would show there is only one state that can exist for that expression where the output would be false, any other time and it's a true result

and with boolean algebra you can do cool things that are basically equivalent expressions like if I said (Not A and Not B) it is logically equal to (A or B)

[idle bobcat]

man that sounds it would be incredibly useful for lower level development

when do you typically start learning discrete math

[radiant nimbus]

its like

probably a sophmore level class at uni

maybe freshmen if you started higher in your math classes

some people can skip a level or 2 of calculus

I think you need at least Calculus 1 to take it

but really, you might be able to get by without Calculus if you are studying it on your own

though knowing basic derivatives and integrals is always useful

[idle bobcat]

I don't think I'd be skipping calculus lol

seems interesting to learn

[radiant nimbus]

It's pretty useful lol

[idle bobcat]

ye

[radiant nimbus]

some of it is pretty boring tbh, but I thought it was fun for the most part

[idle bobcat]

my only fear is that by self learning on khan academy I'll be missing out on a lot more core mathematical concepts

[radiant nimbus]

A lot of Calculus, especially Multivariate is really more critical thinking and less "plug and chug" numbers

[idle bobcat]

that you'd really get only from a real textbook or curriculum

i feel like khan academy is a supplement for in school learning

some people on here suggested some website but I forgot the name