geometry-and-trigonometry

[dry phoenix]

oh i see what i wrote now

even if it was in correct order, how would you continue that

i dont know how to use interval [0,8] there

[supple abyss]

if you meant pi/4<x<3pi/4 that isn't correct

[dry phoenix]

Can you tell me how to do this inequation then?

[upper karma]

What are the transformations from 1/x to this graph? I assumed it was a verticle shift up 1, horizontal shift left 1, and a 1/2 dilation but that's not right.

[mental wharf]

order matters

[umbral snow]

@upper karma
Dialations are considered before shifts. You are right that the function is vertically stretched by 2, but THEN it is shifted up 2, left 1

Basically, whatever is on 0, stays on 0 during a stretch

[ornate lodge]

Is it true that r = A/s, where r is the inradius of a triangle?

[olive sinew]

Can anyone give me a hint on how to work this mess

Number 5

Semicircle measuring ?

[fossil ledge]

A protractor

That’s also roughly 60 degrees I think

[olive sinew]

I don’t have one

Damn

Let’s see if I can find an app or something

[fossil ledge]

Compass app technically works

[mystic shadow]

That is unnecessary

The instructions say that each angle theta is an integer

there are 2pi radians in an entire circle

they approximate pi as 3, so there is approximately 6 radians in one circle

[fossil ledge]

Okay look at 7 tho lol

That’s not 2 radians

[olive sinew]

Is like 170 degrees

Idk what to do

[mystic shadow]

pi is literally 3.14...

the answer for seven is 3

[olive sinew]

The book says that the answer for 7 is

6pi

I mean

Sorry you were right is 3

Ok I get it now thank you guys

[upper karma]

im taking a practice assessment and i dont have my graphing calculator but if someone could give me their steps and solution to 5b and 5c it would be appreciated

[fringe dirge]

wolframalpha.com

[upper karma]

not helpful to me

@fringe dirge if u can help me rn it'd be much appreciated cause i cant find a graphing calculator online that i can do this with and im not 100% sure i know how to do it correctly

[fringe dirge]

it is helpful to you

[upper karma]

no its not but ok

[mental wharf]

desmos

[fringe dirge]

https://www.wolframalpha.com/input/?i=graph+y+%3D+2x%2F(x-1)+-4+and+y+%3D+(6x)%2F(x+%2B+4)

"not helpful"

[upper karma]

I’ve been staring at this problem for like 15 minutes and still can’t figure it out despite watching numerous YouTube videos.. I would appreciate any help greatly

[devout shell]

,rotate

[fringe dirge]

You want to think about which side is bigger, the side labeled x + 5 or the side labeled 3x - 7

[upper karma]

It’s x+5 right?..

[fringe dirge]

but why?

[upper karma]

Cause 5 is greater than -7

[fringe dirge]

But what if x is 10

Then x + 5 is 15

And 3x - 7 is 23

[upper karma]

Oh I see

So it’s dependent on which side has like the potential to be larger?

[fringe dirge]

You have to use the picture and the geometry to decide which side must be larger

[upper karma]

Ok ill give it a try, thanks!

[upper karma]

i don't know if this is the right channel to ask but how would i do calculations with 2 angles in different dimensions? one angle rotating in the X axis and another rotating in an Y axis (it's 3D)

actually, let's make it simpler so i can understand some basic things

does anyone know how to draw dots in desmos.com?

nvm did it

how would i get the length and angle of the X dot compared to the Y dot? assuming 0° would be a flat line pointing to the right

as for the length, would i have to draw an imaginary triangle and use the pythagorean theorem to solve it?

which the answer would be √17

i don't know if it's right though

i still have to solve for the angle and i have no idea how to do it

is the angle in degrees or radians?

and what's opp and adj? opposite and adjacent?

[dark sparrow]

is the angle in degrees or radians?

whatever units your tan function takes inputs in

and what's opp and adj? opposite and adjacent?

yes

[upper karma]

thanks

but, uhh

i have to find the value of the angle, so, if tan(angle) is 1/4, what is angle?

how do i find that value?

[upper karma]

oh, inverse trig functions are a thing

[peak mesa]

@upper karma the lenth is:
sqrt(PointA.X*PointB.X + PointA.Y*PointB.Y)

the angle between this angles is:

diffY = PointA.Y - PointB.Y
angle = atan2(diffX,diffY)```

To get direction adding values:

Ydirection = cos(angle)```

Hope this helps 😃

[gentle meteor]

https://www.youtube.com/watch?v=8PcE_pzA9eU

https://www.youtube.com/watch?v=fBGOshZk94U

[upper karma]

@peak mesa omg thank you so much!!!

but why wouldn't i be able to get the lenght of it with the pythagorean theorem?

look at this graph for example

red is (2,1) and blue is (4,5)

using the sqrt(PointA.X*PointB.X + PointA.Y*PointB.Y) you gave me, it would be sqrt(2*4 + 1*5) which would be sqrt(8 + 5) thus sqrt(13)
but if i use the pythagorean theorem, i would have a triangle with the length of 2 on the bottom and a length of 4 on the right side, so it would be 2² + 4² = c², 4 + 16 = c², 20 = c², c = sqrt(20) which are different results. which one of them is the correct one and why?

[supple abyss]

second one

[upper karma]

i see

so it would be sqrt(x2 - x1)² + (y2 - y1)²)

[supple abyss]

(

yes

[upper karma]

thanks

[supple abyss]

np

[upper karma]

as for this one, is it correct too? i'm a really beginner at sin, cos and tan

[supple abyss]

depends which angle you're talking about

also what you mean by Xdirection

[bright parrot]

you're missing things

[upper karma]

that's what Splitrox said, if you look above

[bright parrot]

what are you trying to find

those formulas are incomplete

[upper karma]

just the length and angle between two points, but i guess i know it now

the angle would be arctan((x2 - x1)/(y2 - y1)) right?

[supple abyss]

which angle

[upper karma]

this angle

[bright parrot]

this is...a little more complicated than you're trying to make it

ok for one, generally the angle between two points is from origin

[upper karma]

so (0,0)?

[bright parrot]

what you have there is an angle in an arbitrary right triangle drawn from two points

yes, from (0,0)

when not specified, the convention is angle between two points as if there is a line drawn from origin to each of them

[supple abyss]

as chosen in the picture it would be arctan((y2-y1)/(x2-x1))

[upper karma]

y goes first?

[supple abyss]

yes

[upper karma]

oh, right

because y would be the opposite side of theta

makes sense

[bright parrot]

soh cah toa

[upper karma]

yep

i've started hearing soh cah toa yesterday

and i'm already getting it, i think

[bright parrot]

that's also why x and y values are not just sin or cos of an angle

[upper karma]

i see

also

there's a little problem now xd

[bright parrot]

like in your chosen triangle with your chosen theta, you have x, y, and h for hypoteneuse

if sin(theta) = y/h, then y = h*sin(theta)

[upper karma]

right

so

now i want to step it up

and bring the z coordinate

how would i calculate the length and angle between these two points?
(X,Y,Z)
(2,4,3)
(4,3,6)

[supple abyss]

a singular angle and length cannot describe it

[upper karma]

right, it has to have two angles

but one length cannot describe it?

[supple abyss]

either 3 lengths, 2 lengths and one angle or one length and 2 angles

[upper karma]

one length and two angles

[supple abyss]

https://en.wikipedia.org/wiki/Spherical_coordinate_system

[upper karma]

one angle would rotate in the Y axis with the X axis (the same as the 2d one) and one that would rotate in the X axis with the Z axis

oh

[bright parrot]

there's a distance formula for 3d space

it's the same as the distance formula in 2d, which is the pythagorean threorem, just extended one dimension

there's no need to change coordinate systems

[supple abyss]

I know that

[bright parrot]

THEY didn't

[supple abyss]

he asked to find the angle

[bright parrot]

lol yeah, but you still don't need to change systems like that

it would be easier explaining dot product

[upper karma]

oh

[peak mesa]

@upper karma you're welcome 😉

@bright parrot true

[upper karma]

so for the length between (2,3,4) and (5,6,9) would be 9 + 9 + 25 = c² which is sqrt(43)?

[peak mesa]

no

[upper karma]

it's in a 3d dimention

[peak mesa]

ah true

-.-

imagine

arctan((x2-x1),(y2-y1))

arctan((x2-x1),(z2-z1))

or use sin and cos

because an angle cant go to 3d

u need to have min 2 angle

[upper karma]

yeah

it is 2 angles

[bright parrot]

right so r you can see is just a single coordinate analog to pythagorean

[waxen python]

dudes

can u help me in math

[peak mesa]

xD

[bright parrot]

no, we are incapable

[ebon spear]

i sure cant

[upper karma]

ok splitrox

1 is theta or the other thingy?

[bright parrot]

@upper karma if you look at r and take either point it's the same as distance from (0,0,0)

[upper karma]

what?

[peak mesa]

u need to do this

https://cdn.discordapp.com/attachments/326138757474680852/593490754949611533/unknown.png

for a distance on space X Y its only sqrt(X^2+Y^2)

but u are working with more values

[upper karma]

yeah

there is also Z

[peak mesa]

so multiply with sqrt(pointA.XpointBX + pointA.YpointB.Y + pointA.ZpointB.Z)

...

is the same thing

sqrt(pointA.XpointBX + pointA.YpointB.Y) for X and Y

sqrt(pointA.XpointBX + pointA.YpointB.Y + pointA.ZpointB.Z) for X,Y and Z

..

[upper karma]

isn't it sqrt((x2 - x1)² + (y2 - y1)² + (z2 - z1)²)?

[peak mesa]

yes..

I was wrong

[upper karma]

also, theta is just arccos(z/r)?

[peak mesa]

yes

its simple

xD

[upper karma]

ok xD

[supple abyss]

also need to remember which planes you want the angle from

[upper karma]

wdym

[supple abyss]

this is taking the angle from the positive x axis for example

but you could just as easily ask for this angle

[peak mesa]

true

[upper karma]

oh

[peak mesa]

normally is that

dont worry

u want to get the angle

you're correct

[supple abyss]

if you choose the red angle then this becomes correct actually

[upper karma]

i hope i don't mess with the wrong angles xD

[peak mesa]

xD

[upper karma]

huh

wait

there are 2 symbols for the two angles, theta and what's the other one?

[supple abyss]

looks like phi

but I don't know with these weird fonts

$\phi$

[somber coyote]

CaptainLightning:

[upper karma]

φ

[supple abyss]

$\varphi$

[somber coyote]

CaptainLightning:

[upper karma]

huh

[supple abyss]

guess it's phi

[upper karma]

but hold on

on this image

https://media.discordapp.net/attachments/326138757474680852/593491130784415781/unknown.png?width=571&height=471

is 1 theta or phi

[supple abyss]

uh

depends which axis is which

[upper karma]

left one is Z, right one is X and the above one is Y

[supple abyss]

in that case those aren't the angles that get labelled

[upper karma]

phi is the azimuth angle?

[supple abyss]

phi is the angle from the positive x direction of the xz plane (in the picture)

[upper karma]

xz? it says xy there

[supple abyss]

the angle is in the xy plane but it's taken from the xz plane

[upper karma]

oh...

[supple abyss]

labelled the positive x direction in red and the xz plane in blue

[upper karma]

aaa i get it now, thanks

[supple abyss]

np

[mighty narwhal]

Jeez
3D polar coordinates

[lusty quest]

hey guys, i am really confused at what this is asking of me

so i have to find the values somewhere in the whole circle, so 3/4th of a circle is 270 degrees, the radian is 3pi/2

[covert orchid]

what is the usage of referential angles for those angles between '0 and 180 degrees' and '0 and 270 degrees' ?

[lusty quest]

are you asking me?

[covert orchid]

nope

i mean in general haha

[lusty quest]

its so you can figure out the trig function

and youll know whether to use the negative or positive

it also narrows the search result when youre trying to solve for one quadrant

[fossil ledge]

So let’s start

We have cos sq = 3/4

Let’s just root both sides

Cos s = rt3 / 2

So now, what values of s allow for this

Note, you’ll have two answers because the interval is the full circle

@lusty quest

[lusty quest]

this was what i got

ooh

ok i get it

[fossil ledge]

My bad, don’t forget that sqrt means +-, so 4 answers

Also interval was 0-2pi, no negatives here

[dark sparrow]

so i have to find the values somewhere in the whole circle, so 3/4th of a circle is 270 degrees, the radian is 3pi/2

no, you have completely misunderstood the problem

[upper karma]

if a dot is going at X speed at an Y angle in a plane, what would be the formula for the speed for both axis?

[dark sparrow]

your notation is weird

[upper karma]

uhh

[dark sparrow]

can i call its speed v and its angle θ

[upper karma]

ok

[dark sparrow]

the velocity along the x axis would be v cos(θ) and along the y axis would be v sin(θ)

[upper karma]

how do you know that?

i don't want just the solution and that's it, i want to understand how that solution was made and understand its concepts

i can't see triangles in this sort of situation 😅

[dark sparrow]

how familiar are you with the unit circle

specifically as the term is used in trig

[upper karma]

i'm kinda new at trig

i know the soh cah toa thing of the triangle

[upper karma]

oh wait, that's because sin is the opposite divided by the hypothenuse (i totally spelled it wrong), isn't it?

and the cos is adjacent divided by the hypothenuse

[fossil ledge]

Yes

So now we solve for opposite and adjacent from those equations

Our hypoténuse just being v in this case

[dark sparrow]

🥖?

[upper karma]

and how would i check if the dot got to X = 1 or Y = 1?

vsin(theta) = 1?

i can't write the theta

[fossil ledge]

If your angle is in Quadrant 1

And your v is positive

It will reach those values at some time

[upper karma]

well, basically i'm writing a script that needs that to run a lot of them and check for some stuff that is on the way xD

thanks for helping me^^

[upper karma]

i don't know what i did wrong but something wrong happened xD. i'm gonna explain in a sec

so i have a point in space, and it goes in an angle in the XZ plane (φ) and in another angle in the XY plane (θ) at speed v.
what i did to calculate the coordinates of the dot is:

x = cos(φ) * v
z = sin(φ) * v
y = sin(θ) * v

[cinder portal]

spherical coordinates?

This looks very VERY similar to spherical coordinates ngl

[upper karma]

ye

[cinder portal]

then you are a off

In spherical coordinates, there are two angles, theta, and phi

[upper karma]

yeah

those are the angles

[cinder portal]

theta is usually denoted for the XY plane, as in polar coordinates, and phi denotes the angle of the Z axis

where theta goes from 0 to 2pi, while phi goes from 0 to pi

[dark sparrow]

x = v cos(θ)cos(φ), z = v cos(θ)sin(φ)

[cinder portal]

understand that phi is really how much of an angle it makes with the Z axis

[upper karma]

wait, what...?

[cinder portal]

^^^

lemme pull up a pic

gimme a sec

[upper karma]

theta and phi don't have the same maximum value?

[cinder portal]

no thye dont

for phi

its how much of an angle it makes with z axis

and if you really think about it

the highest angle it can really make is, in the negative z direction, which would be 180 degrees

a cross section would be more like half a circle

[upper karma]

what about the other side?

if it can only go 180 degrees how would i go behind?

like

rotate behind

[cinder portal]

thats where the theta comes in

[supple abyss]

conventions on which angle is which vary

[cinder portal]

notice how you can take care of any direction, any point in space spherically

if you first choose your theta

and then you choose a phi

you choose which direction it goes on the XY plane, and then it takes care of it vertically

Think of a sphere

how would you take care of all the points in a sphere

first theta, then phi

the bounds $0\leq\theta\leq2\pi,0\leq\phi\leq\pi$

[somber coyote]

MemesPlease:

[cinder portal]

is more than enough

to take care of all the points

if you want an analogy, ill try my best

think of it as an orange, standing upright

theta determines which slice it is, and phi is the whole slice vertically

[upper karma]

oh wait

hold on hold on

you're telling me

[cinder portal]

sorry best analogy I can come up with LOL

[upper karma]

in order to rotate to the other side, you'll have to use theta to rotate 180 degrees?

[cinder portal]

what do you mean as rotate to the other side

[upper karma]

like

you said that phi can only rotate 180 degrees

[cinder portal]

yes

[upper karma]

so if i want to go more, i have to rotate theta + or - 180 degrees and work with it like that?

[cinder portal]

phi only takes care of the "verticality" of it

there is no plus or minus

its 0 to 180

thats it

if it helps

think of it like this

theta is for the XY plane, a 2D plane

while phi

is the z axis

uhhh

gimme another second

[upper karma]

i think i understand your point but there's a misconnection somewhere xD

[cinder portal]

oh different notation on that one

they swapped theta and phi

[devout shell]

This diagram has different notation lol

Just what we needed

[cinder portal]

FUK

[supple abyss]

happened to me lol

[cinder portal]

agAAAHH

ok we takei tslow

in 2 dimensions

we have just theta

theta takes care of all the points

right

0 to 2 pi

[upper karma]

[cinder portal]

right/

in 2 dimensions

theta takes care of all the points

in a circle

actually to be more specific

r and theta

with just r and theta, you can find ANY point

how long it is from the origin, and what angle it makes with the x axis

right?

[upper karma]

uh...

do you mind going on voice chat? xD

[devout shell]

Lol I want to hear this then

[cinder portal]

oh god

nooo not like this

[devout shell]

Let’s hear what you got then

[cinder portal]

can i not

ok here lemme give it one more shot

[supple abyss]

gif incoming

[upper karma]

oh my xD

that's... what

[cinder portal]

YES
THANK YOU

[devout shell]

I would do voice chat but I have to whisper since everyone is sleeping right now lol

[cinder portal]

That animation is FUCKIN AWESOME

[upper karma]

now pls explain memes

[cinder portal]

ok so do u understand

in 2 dimensions

[supple abyss]

a picture tells a thousand words and a gif is a thousand pictures

[cinder portal]

all you really need is r and tehta

to take care of ANY POINT we wanted

you probably learned this

if we wanted to figure out what (1,1) was in polar

it would be

pi/4 and r = root(2)

[upper karma]

polar?

[cinder portal]

r is root(2) from the origin, makes an angle of pi/4

yes

[upper karma]

what?

[cinder portal]

in 2 dimensions, polar utilizes r and theta

to take care of all points

[upper karma]

i've never seen polar

[cinder portal]

uhh

wA

👀

[devout shell]

Lol you have to tell him about that now

[cinder portal]

ah SHIET

No wonder it was a bit confusing at first

Ok let's start from the bottom

[upper karma]

oh boy

[devout shell]

I don’t think a diagram is what we want to jump to right away

[cinder portal]

In normal graphing, we use (x,y)

[upper karma]

yeah i know

[cinder portal]

would you agree that (x,y) takes care of any point on a graph

just choose an x, and choose a y

now, here is some next level black magic

instead of doing cartesian, or casual (x,y) stuff

we can do something called polar

instead of finding a horizontal + vertical point

what we can now do instead, is find the distance from the origin, and find the angle it makes with the x axis

[devout shell]

give motivation for using polar coordinates

[cinder portal]

lightning put a decent picture up there

instead of x and y

we use r and theta, r being the distance the point is from the origin

and theta being the angle it makes with the positive x axis

[upper karma]

oh

[cinder portal]

makes sense?

[upper karma]

so instead of using Y it uses a line with a specific length and the angle of the line?

[cinder portal]

angle the "line" makes with the positive x axis

[devout shell]

Not quite there yet, perhaps we should introduce polar coordinates using a familiar example first

[cinder portal]

like i said before, lets say we wanted to convert (1,1) to polar

what would the distance be from the origin (r), and what angle would that (r) make with the x axis?

[upper karma]

uhhh

wait wait

don't say it

i'ma try

[cinder portal]

hint try pythagorean theorem

LOL

[upper karma]

i was doing that

the length is sqrt(2) and the angle is 45°

[cinder portal]

yup

perfecto

now

[upper karma]

yay

[cinder portal]

lets do another example

[upper karma]

i'm smort

[cinder portal]

(0,1)

[upper karma]

the second one is y right?

[cinder portal]

yea

[upper karma]

ok so

[cinder portal]

find r and theta basically

[upper karma]

r is just 1 and the rotation is pi/2?

[cinder portal]

yup

[upper karma]

i'm starting to get it ,yay

[cinder portal]

now

[upper karma]

so the Z thing uses polar instead of the normal coordinates?

[cinder portal]

no

we going slow first

stop thinking ahead

[upper karma]

😅

i'm an anxious boi

[cinder portal]

k

now heres one

find

(3,9)

[upper karma]

oh wow

i actually have no idea

[devout shell]

Normal as in rectangular, that’s the better way to describe (x, y, z)

[cinder portal]

what did you do last time

do the same thing

figure out the distance from the origin (r) and figure out the angle it makes with the x axis

draw a picture

[upper karma]

i just knew the angles 90° and 45°

[cinder portal]

if it helps

[upper karma]

OH

[cinder portal]

oh the angle is a bit disgusting in this one

[upper karma]

i got this

[cinder portal]

u might have to use a calculator

[upper karma]

ye

might take a while hold on xd

[cinder portal]

ye np np

[devout shell]

Give in terms of an arc function and we’ll know if you had the right idea or not

[cinder portal]

naw i used wolfram alpha in bot to figure out the angle LOL

o shit dont look in bots

[upper karma]

i won't

[cinder portal]

in terms of arc function works

[devout shell]

Smh, you could have just written in terms of an arc function and that would do just fine

[cinder portal]

in case mofu punched in an actual angle, i would know

but any works

[devout shell]

There’s a very natural arc function to use in this case

[upper karma]

r = sqrt(90)
angle = 71.57°

[devout shell]

How did you calculate that angle?

[cinder portal]

yup that is right

[upper karma]

arctan(9/3)

[devout shell]

Good, he used arctan(x)

[upper karma]

lol

someone taught me that yesterday

[cinder portal]

Now

(378482,7252)

Naw jkjk

[upper karma]

WTF

LMFAO

[cinder portal]

But you get the idea right?

[upper karma]

ye

[cinder portal]

Thats the power of polar

[devout shell]

Talked about polar and never once said anything about a circle of any kind

[cinder portal]

circle helps understand

[upper karma]

r = 2 * sqrt(35825303957)
radius = 1.098°

[cinder portal]

holy fuck

is that actually it?

[upper karma]

ye

[cinder portal]

omg

i told u I WAS JOKING

[upper karma]

BUT I DID IT ANYWAY

[cinder portal]

LUL

10/10

[devout shell]

It’s just simple application of the THM and arctan

[upper karma]

^

[cinder portal]

alright

you think you got polar down now?

r and theta

[upper karma]

takes time with head but easy for wolfram

[cinder portal]

is all u need

[upper karma]

mhm

you're one of the best teachers ❤

better than irl teachers

[cinder portal]

aight

lets move on now

instead of 2D

now we bring ing

3 DIMENSISONSSSS

ok dont get too comfortable yet

[upper karma]

woaaehaweaaeaw

[cinder portal]

ok actually lets back track a tad bit

[upper karma]

oof y

[cinder portal]

in two dimensions, polar coordinates is usually associated with a circle

when r stays constant

[upper karma]

i gotta go in 30 minutes

[cinder portal]

you can vary your angle to whatever the fuck you want

so that its a circle

right?

[upper karma]

yeah

[cinder portal]

now, what I want you to think about now, is in 3 dimensions

its a sphere instead

just think about a sphere

[upper karma]

mhm

[cinder portal]

that's all i want you to do

[upper karma]

i am

[cinder portal]

so a sphere is centered at the origin

[upper karma]

yeah

[cinder portal]

and now, we introduce something called phi

[upper karma]

is phi related to pi?

[cinder portal]

back in 2 dimensions, we only worked with r and theta to take care of all possible points

no

phi is a greek symbol itself

[upper karma]

oh ok

[cinder portal]

now in 3 dimensions

try to think logically about htis

is there a way to take care all the points

by introducing a phi?

Hint Think about the verticality

[upper karma]

uhh

idk

i gotta go now

[cinder portal]

ah shiet

[upper karma]

can i ping you when i come back?

[cinder portal]

ye

[upper karma]

thanks ❤

[cinder portal]

ye np np

[cinder portal]

oh my god i just realized dogegirl is amphy

.-.

[devout shell]

Yes lol

[peak mesa]

@supple abyss

where did u find that

i need more like that for a test

[cinder portal]

he probably animated it himself

[peak mesa]

what rly?

[cinder portal]

PROBABLY

idk

[peak mesa]

do u know the program?

[cinder portal]

I need to learn this skill too, too many times I try to help others without providing visuals and it cripples ppl

nop

[supple abyss]

google

[peak mesa]

@supple abyss keywords

[supple abyss]

spherical coordinates gif

[fossil ledge]

3blu1brown released his software I believe, you could learn that

[upper karma]

@cinder portal hi teacher xD

[devout shell]

asking about polar and spherical still? @upper karma

[upper karma]

ye

i had to go

[devout shell]

you think it will be fine if I give me take on it then?

[upper karma]

if you don't mind

[devout shell]

voice chat is fine if you want as well lol, but if you don't want, then I'll start with polar again

[upper karma]

ye vc is fne

hold on a sec tho

k i'm ready

@devout shell

[devout shell]

https://quantummechanics.ucsd.edu/ph130a/130_notes/node216.html

[devout shell]

measure off all angles in a triangle all sum to...?

[hallow token]

360?

[devout shell]

no

that's the central angle of a circle

but a triangles angles all sum to...?

[hallow token]

180

[devout shell]

so then what should you do in this problem?

[hallow token]

Add all of them up to get a sum of 180

[devout shell]

we know that m1 + m2 + m3 = 180 then

[hallow token]

So it should be C

[devout shell]

I didn't work through the algebra but I assume you did then

and answer choice C has all angles summing to 180 so then that seems correct then

well all answer choices sum to 180 lol

but I'll assume you did the algebra correctly then

[hallow token]

Welp thats a yikes

[devout shell]

what did you get for x then?

[hallow token]

I have no idea how to do this. I failed geometry hard.

[devout shell]

ok well we know that m1 + m2 + m3 = 180

now put in what m1, m2, and m3 are

[upper karma]

I can’t seem to get the right answer for number 12. Can anyone help me? My finals are tomorrow 😭

[fossil ledge]

Do we know SOH CAH TOA?

[upper karma]

@hallow token add all three sums together and set it equal to 180. So it should be 40z-27+25-2x+26+x=180. Than solve for x and z and plug it back in

[fossil ledge]

Actually Pythagorean should be enough

[upper karma]

I briefly learned about soh cash toa last year but my teachers disregarded it as trig

[fossil ledge]

Pythagorean good tho?

[upper karma]

I don’t really remember what soh cah toa is for. I don’t think u use it unless your dealing with sin cos and tan. I think Pythagorean is for finding the missing side

Don’t quote me on that tho xd

[fossil ledge]

okay let’s do this

Let’s start with using Pythagorean on the right triangle

So SR^2 + QS^2 = 12^2

So we dont know SR (which is what we are looking for) and we dont know QS.

so lets use the other triangle to find QS

QS^2 + PS^2 = 30^2

but look at the diagram, PS = 40 - SR

so QS^2 + (40-SR)^2 = 30^2

SR^2 + QS^2 = 12^2

QS^2 + (40-SR)^2 = 30^2

So we have two unknowns and two equations, do you know enough algebra to take it from here?

[upper karma]

Hmmmm

I assume we factor something?..

[fossil ledge]

not quite

[upper karma]

Square root something?..

[fossil ledge]

not yet :P

lets solve for QS^2 in one of the equations

your choice to pick

[upper karma]

Numero dos

[fossil ledge]

okay so QS^2 + (40-SR)^2 = 30^2

QS^2 = 30^2 - (40-SR)^2

now, lets plug this into the other equation

SR^2 + QS^2 = 12^2

SR^2 + 30^2 - (40-SR)^2 = 12^2

now all you have to do is solve for SR

which is left as an exercise to the reader

[upper karma]

alright I’ll give it a try

Thank you!

[fossil ledge]

no worries, @green coral if you get lost :)

sorry me

[devout shell]

lol accidental ping

gets 'em every time

[tawdry pivot]

@austere forum

[devout shell]

lol

[sonic yew]

Find shaded area help please

[crystal rain]

ok

[olive sinew]

Is there a better and more accurate way to determine whether the function is positive or negative with radians values

Without a calculator of course

[fossil ledge]

unit circle

[olive sinew]

ok

[crystal rain]

kek

[olive sinew]

Ok but if you are given cos2 for example how can you determine that is in quadrant 2

I am brand new to this, help me understand please

[fossil ledge]

2 what

2 degrees?

[olive sinew]

Radians

[fossil ledge]

pi/2 is roughly 1.5ish

so each quadrant is roughly 1.5ish

[olive sinew]

Oooohhh

I get it now

[fossil ledge]

but just use the conversion

[forest hazel]

How do I prove the midpt theorem for this

[dark sparrow]

,rotate -90

[vestal bronze]

Not sure if super legitimate, but if you make a mark in the middle of line XY call it Z, and draw two lines to the line CD and specify that where it intersects is equal distance on each side of P at point N and M, then you have ZA=ZB, and BN=AM which means that AB and NM must be parallel

And by the same logic but using point P instead of Z, you can say AB is parallel to XY

so your tangent line CD must also be parallel b transitive property of equality

[spark mortar]

Need some private help with vectors as I don't understand anything so far I need a patient person to help me understand and solve my stuff before I fail completely please, quite desparate 😢 dm me

[upper pebble]

L1 and L2 are parallel lines. What's the value of alpha?

I got 45 degrees but i'm not sure if its correct

[vestal bronze]

do you know any of the other angle values?

wait nevermind

i agree

cause like if you expand the line from the 3a it makes a parallelogram and so consecutive sides sum to 180 right?

then you get 4a=180

i also might not know any geometry

[olive sinew]

Can anyone provide me a hint?

For 87

[devout shell]

the arc length is given by what formula?

[olive sinew]

S=theta • r

[devout shell]

and what is r in this case?

[olive sinew]

It’s looks like is 60dg

[devout shell]

r?

[olive sinew]

Or pi/3

[devout shell]

r is the radius of the circle

[olive sinew]

I mean

[devout shell]

so the radius is...?

[olive sinew]

1

2.35

[devout shell]

so know we know that S = θ

and we have an (x, y) point, and what do we know about the unit circle and how x and y are related to trig functions?

[olive sinew]

Cos s = x and sin s = y

[devout shell]

and tan(θ)=y/x, so it's really just your preference as to which arc function you want to use to get the angle

make sure the output is in radians

that will give you the answer

I'll let you determine the shortest length for 88, as this was concept check, I shouldn't have given too much away lol

but at least you know now

[stoic wraith]

https://www.sebz.xyz/i/firefox_June_28_2019_BlandConure.png

I'm not sure how they want my to represent my answer

Nevermind I'm stupid...

They want the tangent of 41pi/4 which is literally the tangent of pi/4 which is 1

[spark mortar]

@upper karmathe answer is 32.5

I don't see why you do root 5 by 4

It's just a 5 by 4 factor

So what you do is 26 times 5/4

[spark stag]

When the area is enlarged

The scale factor is the square

Of the factors the side lengths were enlarged by

Hence the square roots.

So yeah, your method doesn't work.

What you can do is use the scale factor to find side lengths.

Please don't answer questions without knowing what you're talking about

@upper karma I'm assuming the second image you linked is the answer key solution?

If so, consider:

Scale up a 3x2 rectangle by multiplying each side by 2

Then, you get a 6x4 triangle, right?

The scale factor is 6/3 or 4/2 = 2, which makes sense

As we scaled up each side by 2

But what happened to the area

It went from 3*2 = 6

To 6*4 = 24

The area increased by a factor of 4

You'll note that this increase is 2^2

And this applies no matter what the shape was

The scale factor of a change in area

Is the square of the scale factor of a change in side lengths

(and perimeter is just side lengths, so)

Please disregard entwine's reply, it's incorrect

We can check our work, too: P_2/P_1 = 29/26 = 1.115

Sqrt(5/4) = 1.118

Accounting for slight error due to rounding, this checks out

And we can observe the original area is 5*8 = 40

And if we multiply 5*1.115 and 8*1.115, the scale factor for length

You'll note the area of that shape is 5/4 times the area of the original

Which is 50

(well, 49.729 due to rounding inaccuracy)

Does that explanation make sense?

We can mathematically see where this relation comes from, too

Let's let a and b be side lengths

So the area is A_1 = a*b

If we scale up a and b by some value k

That means we multiply a and b by k

So the new area is

A_2 = (a*k)(b*k) = abk^2

Note the k^2

This demonstrates that the scale factor for area is the square of the scale factor for length

in other words, A_2 / A_1 = (L_2 / L_1)^2, where L is a side length (or a perimeter)

[spark mortar]

Question 7 saying that the area of a 5 by 8 rectangle is enlarged by a factor of 5/4 which is basically 25% increase. The area is 26 plus 25% gives 32.5. I don't really get where you get a1 and a2 it only mentions one rectangle that gets bigger by a factor

Sorry if I don't get it but it seems simple as that as I read the question

[spark stag]

a_1 is the original area, a_2 is the image's

yes, it's a 25% increase in area

but a 25% increase in area

is not a 25% increase in side length

otherwise a 2x2 rectangle enlarged to a 4x4 rectangle would have an area of 8

which obviously doesnt make sense

if the sides increase by a factor of k, then the area increases by a factor of k^2.

proof for rectangles:

let a rectangle have side lengths a and b

then, the area of the recftangle is ab

I'll call this original area A_1

A_1 = ab

if we multiply each side length by k

then a becomes a*k

and b becomes b*k

so the new area, let's call it A_2

is

A_2 = (a*k)(b*k) = ab*k^2

so the side lengths increased by a factor of k

but the area increased by a factor of k^2

that's essentially what's happening here, but in reverse

we're given the factor the area increased by - 5/4

so 5/4 = k^2

so to find k, which is the ratio of the perimeters

thats the sqrt(5/4)

and again, we can confirm this

if each side is scaled up by sqrt(5/4)

the original rectangle is 5x8

so the area is 5*8 = 40

we multiply each side length by sqrt(5/4)

5*sqrt(5/4) * 8*sqrt(5/4) gives us our new area

which is 5*8*5/4 when simplified, which is 50.

and, of course, 50 is 5/4 of 40 ("25% more")

you were making the assumption that scale factor of area = scale factor of side lengths

but this assumption is not true.

and that can be shown by thinking about what that'd mean for a bit, or just playing around with examples

[spark mortar]

Yeah I think I simplified the question quite badly sorry about that

Well that makes sense cause area is sides multiplied yeah?

So 5 times 8 times 5/4 still gives 50 😂 so you don't need the square root 😅

[spark stag]

when we talk about 5*8*5/4, thats the area * 5/4

which is what 5/4 represents

when we want to talk about perimeter

we need to think about how sides are scaled, not area

and sides are always scaled by the square root of the amount area is scaled by

$(k_{sides})^2 = k_{area} \ k_{sides} = \sqrt{k_{area}}$

[somber coyote]

Namington:

[spark mortar]

Think is he doesn't need all that in the question? It says area of rectangle 5x8 which is 40 is enlarged by 5/4 which is 50. Question done. If you want the perimeter you can do simply p = 2xa + 2xb

Which is not needed ecause he has both sides

Normally if he had one side and the area he would do A = a times b then substitute to find the second side

And then find the perimeter

[spark stag]

"you can do simply p = 2xa + 2xb"

uhh

what are a and b?

[spark mortar]

The 2 sides

5 and 8

[spark stag]

but those sides were enlarged

because the shape was enlarged.

[spark mortar]

But it gives you the sides

By 5/4

[spark stag]

it gives you the sides of the original

[spark mortar]

Just multiply it

[spark stag]

you need to find the side lengths of the image

[spark mortar]

Done

[spark stag]

no

you cant do that

[spark mortar]

That's all it asks

[spark stag]

no

no

if you do that

then the side 5

becomes 5 * 5/4 = 6.25

the side 8 becomes

8 * 5/4 = 10

but whats the area

of a 6.25 by 10 shape

62.5

thats not 50.

hence, the amount you multiply the area by

is not the same as

the amount you multiply the sides by

to find the amount you multiply the sides by

we find the scale factor of area

A_2 / A_1

and take the square root

ive explained this twice already, it's essentially because length is 1-dimensional, area is 2-dimensiona

consider a rectangle with side lengths a and b

then its area is

A_1 = a*b

now if we multiply a and b by some value

k

its area becomes

A_2 = (a*k) * (b*k) = a*b * k^2

note that the area went from a*b to a*b*k^2

in other words, when we multiply sides by k

we multiply area by k^2

now, this question is in reverse

we know the area was multiplied by 5/4

so the sides must've been multipled by the square root of that

of course, the perimeter is made up of sides

so the perimeter must've been multiplied by the square root of that

the perimeter is originally 26, it's multiplied by sqrt(5/4)

26 * sqrt(5/4) is about 29

[spark mortar]

I suck at geometry :(

Well I'm stuck with vectors so

I cri

[worthy root]

I suck at algebra

[buoyant oasis]

nobody gets good without practice

[worthy root]

I do lot of practice but still suck

[upper karma]

,rotate -90

[upper karma]

The answer should be pi or pi/2?

I mean I know period=2pi/B

But B is 4 or 2?

<@&286206848099549185>

ping me

[chrome fiber]

that's awkward notation, but likely 4

[upper karma]

Oh, alright

So, do we always multiply the outer numbers?

[chrome fiber]

im assuming its all an argument for the sine function, so yea

[upper karma]

Ah, I see

Thank you!!

[olive solar]

sin2(2x...) looks more like a typo to me

[upper karma]

Oh, maybe...

[cedar sparrow]

Notation for sine squared maybe?

[upper karma]

Mhm, I guess

[olive solar]

no, i think that they do mean sin(2(2x+(pi/4))), just rather awkward

[cedar sparrow]

I can't imagine they'd forget parantheses.

But still make it look like that.

[upper karma]

If we make it look like sin(2(2x+(pi/4))) then B will be 4 right?

[cedar sparrow]

Yes.

[upper karma]

Okay

[cedar sparrow]

In which case the period would be pi/2

[upper karma]

Ah, I see

[cedar sparrow]

Personally, I think the answer is B. I've rarely seen "None of these" as an answer.

But that's just my opinion.

[upper karma]

Oh, alright

Thank you!

[worthy root]

Is 0 a integer

,w is 0 a integer

[somber coyote]

Display results online and refine query

Requested by Krishna🤠(A Nerd bye nature)#4971

[worthy root]

[dark sparrow]

...of course 0 is an integer...

[subtle gate]

also why do you ask that here to wolfram

[worthy root]

[minor arch]

https://i.imgur.com/TnXJSGF.png

Im a lost how they came to this conclusion.

Can anyone help me out?

[stoic wraith]

Can anyone simply explain to me why this is a fact? In my mind the inverse would put the 2 over the root 3 which would make the final result sine of 2 root 3 over 3

https://www.sebz.xyz/i/firefox_June_30_2019_ShadowySkunk.png

Thanks in advance

[spark stag]

inverse sine is not the same thing as raising something to ^-1

inverse sine is simply the function that undoes sine

more precisely:

$x = \sin(\sin^{-1} (x)) = \sin^{-1} (\sin (x))$

[somber coyote]

Namington:

[spark stag]

(within a certain domain)

[hard shard]

It's why I angrily argue that arcsin should be the standard notation

[spark stag]

the notation is a bit confusing, but thats all inverse sine is

the function that undoes sine

so when we have

$\theta = \sin^{-1} \left(\frac{\sqrt{3}}{2}\right)$

[somber coyote]

Namington:

[spark stag]

we can take the sine of both sides

$\sin\theta = \sin\left( \sin^{-1} \left(\frac{\sqrt{3}}{2}\right)\right)$

[somber coyote]

Namington:

[spark stag]

then the sin and the sin^-1 on the right side cancel

giving us

$\sin \theta = \frac{\sqrt{3}}{2}$

[somber coyote]

Namington:

[spark stag]

make sense? @stoic wraith

again, inverse sine has no relation to raising something to the -1th power

its just a notational thing

we use the ^-1 as a shorthand for "inverse" - it doesnt actually mean raising something to -1

[vagrant elk]

The -1 tells you how many times to do it

[spark stag]

...

[fossil ledge]

wait does it

waitno it doesnt

[vagrant elk]

it tells you to undo the function

[hard shard]

sin^-2 is the same as sin

[fossil ledge]

no it isnt

[vagrant elk]

no it isn't

[hard shard]

oh wait this is no joke region

yes sin^-2 is not the same as sin

[fossil ledge]

I was tricked

[vagrant elk]

But uh goat

have you ever seen something like $(f \circ g)(x) = f(g(x))$?

[somber coyote]

Darkrifts:

[fossil ledge]

nope

[vagrant elk]

damn

[fossil ledge]

never seen anyone abuse notation like that

[hard shard]

wait what

is that not

whta

[fossil ledge]

wait thats fine actually

[spark stag]

thats not notation abuse

[vagrant elk]

Yeah it's fine

[spark stag]

tf

[vagrant elk]

abuse
smfh

[hard shard]

isn't that wat it means

[spark stag]

thats the definition of function composition

[fossil ledge]

lmao

[hard shard]

im so confused

[spark stag]

🤔

[fossil ledge]

functions is probably my weaker part of math

I just do f(g(x))

[vagrant elk]

But uh, so all you gotta do now is say $(f \circ f)(x) = f^2(x)$

[somber coyote]

Darkrifts:

[spark stag]

although i have seen some authors use (g o f)(x) to mean f(g(x))