#geometry-and-trigonometry | Mathematics | Page 340

onyx cloud Jan 26, 2021, 11:42 PM

#

what calculator is it?

#

if you have a "2nd" button

#

then try 2nd and then hitting sin

strong star Jan 26, 2021, 11:42 PM

#

i am having a geometry problem

#

imagine u have a picture of 320x320 pixels

#

and i wanna put inside another picture (square) that can be anything randomly between 150 and 200

#

what are the coordinates (top left corner) to spawn the small square regarding the big picture?

#

idk if i explained myself

#

On X axe i though about
x = random.randint(50, dims[0] - s - 50)
the 50 is just a small offset
dims = [320, 320]
ands is the scale factor
s = random.randint(150, 200)
But on y axe i want the to spawn from the mid to the bottom
i dont want the on the top

strong star Jan 27, 2021, 12:14 AM

#

nvm, i did

neat relic Jan 27, 2021, 12:15 AM

#

Hey u free?

#

@strong star

strong star Jan 27, 2021, 12:15 AM

#

?

neat relic Jan 27, 2021, 12:16 AM

#

#help-0

upper karma Jan 27, 2021, 1:11 AM

#

📎 image0.jpg

#

Someone

#

Pls

#

Help

dark jacinth Jan 27, 2021, 3:28 AM

#

If I have the value of sin(theta), how would I use that to figure out cos(90-theta)?

wary frost Jan 27, 2021, 4:08 AM

#

@strong star

#

@zenith ember

zenith ember Jan 27, 2021, 4:10 AM

#

Yes?

wary frost Jan 27, 2021, 4:10 AM

#

Can you help a brotha out

grizzled lantern Jan 27, 2021, 4:10 AM

#

lol

zenith ember Jan 27, 2021, 4:10 AM

#

Read #❓how-to-get-help

#

And don't ping people directly.

grizzled lantern Jan 27, 2021, 4:11 AM

#

@wary frost post it

#

ill see if i can do it

wary frost Jan 27, 2021, 4:11 AM

#

📎 20210126_200514.jpg

grizzled lantern Jan 27, 2021, 4:13 AM

#

https://youtu.be/b4ORXN7a0-w

YouTube

Khan Academy

Dilation scale factor examples

▶ Play video

#

see if this can help you

wary frost Jan 27, 2021, 4:13 AM

#

Thank you

grizzled lantern Jan 27, 2021, 4:14 AM

#

yep

thorn wadi Jan 27, 2021, 11:41 AM

#

can anyone kindly help with this

📎 IMG_20210126_181758.jpg

half citrus Jan 27, 2021, 12:09 PM

#

whats the width if the length is 3x more than the width and the perimeter is 80

#

<@&286206848099549185>

#

it is 3 am and i have an headache

upper karma Jan 27, 2021, 1:44 PM

#

Maximo i'm sorry

#

The wifi was cutoff due to the snow 😦

upper karma Jan 27, 2021, 2:01 PM

#

thorn wadi can anyone kindly help with this

Find the area of the two sectors

#

and just subtract em

thorn wadi Jan 27, 2021, 2:01 PM

#

Oh, I already worked it out lol

#

forgot the equation for the area of a sector

#

wojakdeath

#

lmaooo

#

rookie error, but thanks

#

PepeCheers

upper karma Jan 27, 2021, 2:03 PM

#

$x \cdot \sqrt{2} = \frac{{18\sqrt{2}}{\sqrt{2}}}$

somber coyoteBOT Jan 27, 2021, 2:03 PM

#

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

📎 623220213399289858.png

upper karma Jan 27, 2021, 2:03 PM

#

Uhm

#

Can somebody help with that?

#

It is supposed to be 18 times the sqrt of 2

#

over the sqrt of 2

pure cape Jan 27, 2021, 2:04 PM

#

should be \frac{18\sqrt{2}}{\sqrt{2}}

upper karma Jan 27, 2021, 2:07 PM

#

Ah , so the suffix or /frac does not requre a bracket ?

#

Man , this reminds me of C++

#

$x \cdot \sqrt{2} = \frac{18\sqrt{2}}{\sqrt{2}}$

somber coyoteBOT Jan 27, 2021, 2:07 PM

#

Skyler

📎 623220213399289858.png

upper karma Jan 27, 2021, 2:08 PM

#

📎 unknown.png

#

too similar

#

Makes the brain bleed

pure cape Jan 27, 2021, 2:09 PM

#

i dont think that is readable 🤔

upper karma Jan 27, 2021, 2:09 PM

#

Oh noes

#

let me write something rq in html

#

<html>
<head>
<p style="text-color:red; font-family:times new roman;"> Maths(idk)</p>
<title>
Math
</title>
</head>
<body style="text-align:center; text-decoration:none; font-color:234211">
</body>
</html>```

#

Something like that ^

#

Ah more like this

📎 unknown.png

#

This is not mine , but it's in VSC

#

I use vsc

#

Oh sweet he is using LoRa

#

Wait ? I think that's an uno

#

hooked up to a LoRa gateway for long distance messages ???

karmic willow Jan 27, 2021, 6:32 PM

#

Hey, I got this question on a practice paper. I got x=360n where n is 0,1,2... and -1,-2... which seems to check out okay and satisfy the original equation. The official answer (in the image) includes this solution (apart from the negative values for n) but it also has two other solutions which don't even satisfy the original equation exactly. Have I done something wrong or would you say there was an error with the official answer?

📎 Desktop_Screenshot_2021.01.27_-_18.22.56.43.png

upper karma Jan 27, 2021, 6:52 PM

#

I need help

lusty slate Jan 27, 2021, 7:54 PM

#

Plz

#

Somebody

📎 image0.png

dire notch Jan 27, 2021, 8:16 PM

#

help PLEASE

📎 Screen_Shot_2021-01-27_at_3.png

wintry tundra Jan 27, 2021, 8:34 PM

#

@lusty slate that isnt geometry

#

@dire notch im guessing thats an equilateral triangle even tho it kinda sucks that they dont even tell u

#

so 3x-7 = x^2 - 17 = x+4

lusty slate Jan 27, 2021, 8:40 PM

#

📎 image0.png

#

Smbd

#

Gotta get this done before 4

wintry tundra Jan 27, 2021, 8:41 PM

#

u should tell us in hours since people live in different times

#

but whatever

lusty slate Jan 27, 2021, 8:41 PM

#

It’s 3:41

#

For me

wintry tundra Jan 27, 2021, 8:41 PM

#

oh ok

#

so what do u not understand

lusty slate Jan 27, 2021, 8:42 PM

#

All 5

wintry tundra Jan 27, 2021, 8:42 PM

#

hmm

#

well initial means starting value

#

so i mean what would the initial value be

#

@lusty slate u gotta answer me if u want this to get done

#

we are starting w number 2 btw

#

bc ill save number 1 for last

#

well ill save it for after number 3

lusty slate Jan 27, 2021, 8:46 PM

#

Alright

#

I did number 1

upper karma Jan 27, 2021, 8:51 PM

#

Hi

#

I need help in some geometry exercises

#

📎 unknown.png

#

Which theorem should I use?

wintry tundra Jan 27, 2021, 8:55 PM

#

so ur trying to say that angle 1 is angle 4?

upper karma Jan 27, 2021, 8:59 PM

#

yep

#

Can it be solved with respect to the opposite theorem by the vertex?

wintry tundra Jan 27, 2021, 9:03 PM

#

hmm

#

well u can problem solve thru tihs actually

#

so 3=4 right?

#

XW bisects VXY so 3 has to be 4

#

and since XS bisects RXT then 1=2

#

and 2=3 is a given

#

so if 2=3 then 1=3

#

and if 1=3 then 1=4

#

and boom there u go

upper karma Jan 27, 2021, 9:07 PM

#

wintry tundra XW bisects VXY so 3 has to be 4

Yes!

#

I was thinking about a problem my professor had done, where given the points of a vertex he starts demonstrating part by part until he gets to the angles I need to demonstrate.

wintry tundra Jan 27, 2021, 9:10 PM

#

yeah

#

the only theorem u need is bisector, rest is logic

upper karma Jan 27, 2021, 9:11 PM

#

Ok, I will check the theorem again.

upper karma Jan 27, 2021, 9:12 PM

#

wintry tundra the only theorem u need is bisector, rest is logic

there is another similar exercise I'll try to solve it.

wintry tundra Jan 27, 2021, 9:14 PM

#

ok

upper karma Jan 27, 2021, 9:19 PM

#

Hey i hope this chanel is free

#

Can somebody explain the steps how to simplify the expression ? I have no idea

📎 image0.png

wintry tundra Jan 27, 2021, 9:22 PM

#

hmm

#

i mean i would think its just sin^4(x) + sin^2(x) cos^2(x)

#

but there must be something else to it

upper karma Jan 27, 2021, 9:23 PM

#

Yeah thats what I thought too but apparently there’s gotta be neither way

#

Consider looking at both terms, notice that both have something in common

#

sigh

#

fck

#

cos

#

jajaja

#

@upper karma please don't give away answers. It doesn't help them learn.

#

I see, thank you😅😅

wintry tundra Jan 27, 2021, 9:24 PM

#

sin^2(x)(sin^2(x)+cos^2(x))?

upper karma Jan 27, 2021, 9:25 PM

#

wintry tundra sin^2(x)(sin^2(x)+cos^2(x))?

Yeh

#

my error F

#

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

#

Still not good.

#

That's nonsense

upper karma Jan 27, 2021, 9:45 PM

#

wintry tundra ok

📎 IMG_20210127_164503.jpg

#

Ooh, I'm an idiot point I II and III, is the thesis.

#

Would this be ok?

#

Hi, I am doing this exercise and it tells me that <1=<4.
I say it can't be the same, it should be <1= <2+<4
Am I right?

📎 unknown.png

wintry tundra Jan 27, 2021, 10:15 PM

#

1 isnt 4

#

it cant be 4

#

visually you can tell, and with the info given too u can also tell

upper karma Jan 27, 2021, 10:20 PM

#

"No, in fact, <1=<4.
You see, we have that, both angle BCH and EAB are flat, that is, they measure 180°, so, let's make <2=<3=a, <4=b and <1=c, so that <BCH=a+2c and <EAB=a+2b, but <BCH=<EAB, since both are flat, then:

a+2c=a+2b, then 2c=2b, so that:

c=b, that is, <1=<4."

I asked another person and he replied this.

#

according to what you told me and seeing the graph visually I do not see the equality either, the truth

wintry tundra Jan 27, 2021, 10:24 PM

#

he was wrong in BCH = a+2c

#

he means c+2a

#

if CD bisects FCH that is

upper karma Jan 27, 2021, 10:47 PM

#

📎 x6XcUK6hb07wAAAABJRU5ErkJggg.png

#

📎 623220213399289858.png

#

Is that correct ?

modern hemlock Jan 27, 2021, 10:58 PM

#

your answer in the end is correct

#

but what is $\sqrt(2)\cdot \sqrt(2)$?

somber coyoteBOT Jan 27, 2021, 10:59 PM

#

ScapeProf

📎 265853506664136705.png

modern hemlock Jan 27, 2021, 10:59 PM

#

its not 4

upper karma Jan 27, 2021, 11:04 PM

#

its 2 ?

#

so would it be

#

$x = \frac{18\sqrt{2}}{\sqrt{2}} = \frac{18\sqrt{2}}{2} = 18?$

somber coyoteBOT Jan 27, 2021, 11:05 PM

#

Skyler

📎 623220213399289858.png

upper karma Jan 27, 2021, 11:05 PM

#

uhm

modern hemlock Jan 27, 2021, 11:05 PM

#

no the answer in the end was correct

upper karma Jan 27, 2021, 11:06 PM

#

$x = \frac{18}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{18\sqrt{2}}{2} = 9\sqrt{2}$

somber coyoteBOT Jan 27, 2021, 11:06 PM

#

Skyler

📎 623220213399289858.png

modern hemlock Jan 27, 2021, 11:06 PM

#

yep

upper karma Jan 27, 2021, 11:06 PM

#

Thanks

#

Are you an ASDF potato ?

#

That reminds me of this crappy meme i used to know in like 3rd grade

upper karma Jan 27, 2021, 11:31 PM

#

@wintry tundra man

#

📎 IMG_20210127_183142.jpg 📎 IMG_20210127_183145_1.jpg

#

@wintry tundra Correct or not?

onyx cloud Jan 27, 2021, 11:36 PM

#

upper karma Hi, I am doing this exercise and it tells me that <1=<4. I say it can't be the ...

angles 1 and 4 are the same from the information provided

onyx cloud Jan 27, 2021, 11:38 PM

#

wintry tundra if CD bisects FCH that is

it says CF bisects BCD, which means his a+2c is correct

upper karma Jan 27, 2021, 11:42 PM

#

Maximo

#

Sorry to leave yesterday

#

the wifi cut out

#

Thanks for your help !

upper karma Jan 27, 2021, 11:44 PM

#

onyx cloud angles 1 and 4 are the same from the information provided

Ok thanks ❤️

onyx cloud Jan 27, 2021, 11:44 PM

#

upper karma Sorry to leave yesterday

no worries, np

upper karma Jan 27, 2021, 11:46 PM

#

🙂

wintry tundra Jan 27, 2021, 11:51 PM

#

sorry i have to do some stuff for robotics

#

i cant help u for now maximo but i bet u can find someone else

onyx cloud Jan 27, 2021, 11:52 PM

#

nono i dont need help

#

i was just lyk that the text was right

#

angle 1 equals 4

upper karma Jan 28, 2021, 12:00 AM

#

wintry tundra sorry i have to do some stuff for robotics

Thanks bro, no problem :''3 ❤️

upper karma Jan 28, 2021, 12:01 AM

#

onyx cloud nono i dont need help

I have just completed this exercise

📎 unknown.png

#

📎 IMG_20210127_185817.jpg

upper karma Jan 28, 2021, 12:41 AM

#

📎 Screen_Shot_2021-01-27_at_7.png

#

i need to use this formula to solve it

#

📎 triangle-45-45-90.png

#

help

onyx cloud Jan 28, 2021, 12:42 AM

#

@upper karma keep it to 1 channel thx, #help-5

upper karma Jan 28, 2021, 12:47 AM

#

Hi guys

#

one questions

#

Daum Equation Editor has just been closed, what other site do you recommend for my reports please !!!!

loud flare Jan 28, 2021, 2:56 AM

#

Can someone please help with 11?

📎 image0.jpg

modern estuary Jan 28, 2021, 3:04 AM

#

wtf is a triangle inequality

#

help

humble pulsar Jan 28, 2021, 3:06 AM

#

modern estuary help

Sum of any 2 sides of a triangle is greater or equal to the other side

modern estuary Jan 28, 2021, 3:07 AM

#

📎 image0.jpg

#

how would one complete this

modern estuary Jan 28, 2021, 3:45 AM

#

thanks guys preciate it

wintry tundra Jan 28, 2021, 4:54 AM

#

gotta mention helpers @modern estuary

#

not to mention ur name is annoying to mention

modern estuary Jan 28, 2021, 4:54 AM

#

oof

#

sorry

#

<@&286206848099549185>

#

🗣

wintry tundra Jan 28, 2021, 4:55 AM

#

i could help u tho

modern estuary Jan 28, 2021, 4:55 AM

#

well im going to bed rn

wintry tundra Jan 28, 2021, 4:55 AM

#

hmm

modern estuary Jan 28, 2021, 4:55 AM

#

🐸

wintry tundra Jan 28, 2021, 4:55 AM

#

F<A btw

modern estuary Jan 28, 2021, 4:55 AM

#

k thanks

wintry tundra Jan 28, 2021, 4:55 AM

#

for id say obvious reasons if you know geometry well enough

#

or know that 5>4

grizzled lantern Jan 28, 2021, 4:56 AM

#

these is isosceles
triangles

wintry tundra Jan 28, 2021, 4:57 AM

#

well yeah

#

but F has to be bigger than A since its isoceles and 5 is more than 4

jagged magnet Jan 28, 2021, 4:05 PM

#

📎 unknown.png

#

can anyone help me out

lusty slate Jan 28, 2021, 4:16 PM

#

Smbd rq

jagged magnet Jan 28, 2021, 4:17 PM

#

jagged magnet

atlest tell me what to do for it

jagged magnet Jan 28, 2021, 4:27 PM

#

lusty slate Smbd rq

yo do you know how to solve this

lusty slate Jan 28, 2021, 4:27 PM

#

Nah sadly bro

#

Use photo math

jagged magnet Jan 28, 2021, 4:29 PM

#

thanks man im not gonna lie that was great advice

prime jasper Jan 28, 2021, 4:34 PM

#

Anyone got any idea for 2a?

📎 unknown.png

#

nvm

#

nvm

alpine cliff Jan 28, 2021, 5:01 PM

#

jagged magnet

All of the triangles are similar

#

which means the proportions between corresponding parts are equal

nocturne thicket Jan 28, 2021, 6:07 PM

#

📎 unknown.png

#

Just making sure I'm right?

jolly summit Jan 28, 2021, 6:10 PM

#

📎 unknown.png

#

is this correct

grizzled lantern Jan 28, 2021, 6:14 PM

#

but can you explain?

#

why

jolly summit Jan 28, 2021, 6:18 PM

#

they have pair of congruents side and 2 pair sof congruent angles

#

one of them being b as a reflexdive

alpine cliff Jan 28, 2021, 9:50 PM

#

Isn't there more than one

deft rover Jan 28, 2021, 10:41 PM

#

Hey I really need help with this one, AD, BE and CF are medians in triangle ABC and they intersect in G. AD perpendicular BE. AC=10cm. ∠AFC=86°. CG=AB. I need to find the length of AF. I got stuck for a full day trying to do it

📎 unknown.png

sonic scroll Jan 28, 2021, 11:46 PM

#

How plss help

📎 image0.png

stuck dragon Jan 29, 2021, 12:14 AM

#

hint: diagonals of a rectangle are congruent and bisect each other

upper karma Jan 29, 2021, 1:12 AM

#

I Need To Solve For Each Missing Variable But I Don't Know The Formula Or Where To Start

📎 unknown.png

#

For Number 5

grizzled lantern Jan 29, 2021, 1:14 AM

#

@upper karma dont think of them as quadrilaterals

#

try and imagine if you could extend those parallel lines

upper karma Jan 29, 2021, 1:14 AM

#

Alright

#

I Still Don't Get It What Is The Formula

grizzled lantern Jan 29, 2021, 1:16 AM

#

📎 image005.png

upper karma Jan 29, 2021, 1:17 AM

#

So Would I Do 360=6x+10+2x+10

grizzled lantern Jan 29, 2021, 1:17 AM

#

is that what all angles add up to

upper karma Jan 29, 2021, 1:19 AM

#

Yeah

grizzled lantern Jan 29, 2021, 1:19 AM

#

i think you'd have to do 360= 2(6x+10) +2(2x+10)

upper karma Jan 29, 2021, 1:19 AM

#

Alright Thanks Couldn't Get Past This

grizzled lantern Jan 29, 2021, 1:22 AM

#

@upper karma id also check your answer for number six because

#

,calc 85*4

somber coyoteBOT Jan 29, 2021, 1:22 AM

#

Result:

grizzled lantern Jan 29, 2021, 1:22 AM

#

340 doesnt equal 360

upper karma Jan 29, 2021, 1:22 AM

#

OH SHOOT NOT ME BEING BLIND

#

Thanks

green tree Jan 29, 2021, 1:35 AM

#

https://i.imgur.com/AH5tLl0.png

Imgur

grizzled lantern Jan 29, 2021, 1:36 AM

#

@green tree is a trapezoid a quadrilateral?

#

📎 material-bdwjp2se.png

past harbor Jan 29, 2021, 2:06 AM

#

could i please get some help with this problem

📎 image0.jpg

silent plank Jan 29, 2021, 2:09 AM

#

distance formula

#

choosing the side with the simplest values for less tedious calculations

#

you are told that the heptagon is regular

#

so you can multiply that value by 7 instead of applying the distance formula with crappy values 6 more times.

past harbor Jan 29, 2021, 2:11 AM

#

thank you 😁

lofty crypt Jan 29, 2021, 2:35 AM

#

need help with these 2.

📎 unknown.png

lofty crypt Jan 29, 2021, 2:56 AM

#

nevermind i figured out the first one, i got C. i just need help with the 2nd now.

silent plank Jan 29, 2021, 3:28 AM

#

the 2nd question is considerably easier than the first

lofty crypt Jan 29, 2021, 3:28 AM

#

nvm im watching youtube tutorial i did them both i just forgot the formula

silent plank Jan 29, 2021, 3:28 AM

#

well actually its about the same amount of work

lofty crypt Jan 29, 2021, 4:43 AM

#

silent plank the 2nd question is considerably easier than the first

📎 unknown.png

silent plank Jan 29, 2021, 4:45 AM

#

pls no direct pings in the future.
draw a diagram, apply the definition of bisection.

silk briar Jan 29, 2021, 6:48 AM

#

silent plank distance formula

can you divide it to seven triangles and get it from there?

silent plank Jan 29, 2021, 7:22 AM

#

why would you need to divide it into 7 triangles

silk briar Jan 29, 2021, 7:24 AM

#

because im just a beginner😂

#

i havent learned geometry

stuck dragon Jan 29, 2021, 7:28 AM

#

i mean

#

you can make 7 right triangles

#

with each side as the hypotenuse (longest side of the triangle)

#

and then solve using the pythagoras theorem

#

all of that I just said is just rephrasing the distance formula

deft rover Jan 29, 2021, 11:49 AM

#

deft rover Hey I really need help with this one, AD, BE and CF are medians in triangle ABC ...

so guys is there a chance that this is unsolveable?

upper karma Jan 29, 2021, 12:06 PM

#

📎 unknown.png

#

Did I do this correct?

stuck dragon Jan 29, 2021, 3:00 PM

#

yep

stuck dragon Jan 29, 2021, 3:00 PM

#

deft rover so guys is there a chance that this is unsolveable?

I dont think it is unsolvable

#

but im not sure where to start

upper karma Jan 29, 2021, 5:45 PM

#

I Don't Understand What I Have To Do For 2 Because When I Do 2x+15=X+15 I Get 0 Which Doesn't Work

📎 unknown.png

humble pulsar Jan 29, 2021, 5:52 PM

#

upper karma I Don't Understand What I Have To Do For 2 Because When I Do 2x+15=X+15 I Get 0 ...

yeah, x = 0 is right

upper karma Jan 29, 2021, 5:53 PM

#

But What Then ST Is 0 Which Doesn't Work If I'm Right?

#

WAIT

#

NEVER MIND DUMB BRAIN MOMMENT

echo chasm Jan 29, 2021, 9:17 PM

#

https://cdn.discordapp.com/attachments/387447700234174464/804788429656162304/IMG_20210129_200106.jpg
https://cdn.discordapp.com/attachments/387447700234174464/804788429924990976/IMG_20210129_200053.jpg

pseudo plaza Jan 29, 2021, 9:31 PM

#

,rotate https://cdn.discordapp.com/attachments/387447700234174464/804788429656162304/IMG_20210129_200106.jpg

somber coyoteBOT Jan 29, 2021, 9:31 PM

#

📎 114875468959842306.png

echo chasm Jan 29, 2021, 9:31 PM

#

the sum of the lengths of the height and the edge of the base of the hexagonal pyramid is 10, the axial section is 48, Find the area and volume

pseudo plaza Jan 29, 2021, 9:31 PM

#

,rotate https://cdn.discordapp.com/attachments/387447700234174464/804788429924990976/IMG_20210129_200053.jpg

somber coyoteBOT Jan 29, 2021, 9:31 PM

#

📎 114875468959842306.png

echo chasm Jan 29, 2021, 9:40 PM

#

📎 9c133084bfb63c9ad3712a2ef3ac9f9d.0.jpg

#

📎 IMG_20210125_104113.jpg

pseudo plaza Jan 29, 2021, 9:40 PM

#

,rotate https://cdn.discordapp.com/attachments/326138757474680852/804828355881861191/IMG_20210125_104113.jpg

somber coyoteBOT Jan 29, 2021, 9:40 PM

#

📎 114875468959842306.png

echo chasm Jan 29, 2021, 9:42 PM

#

📎 IMG_20210129_224156.jpg

pseudo plaza Jan 29, 2021, 9:42 PM

#

,rotate

somber coyoteBOT Jan 29, 2021, 9:42 PM

#

📎 114875468959842306.png

regal pond Jan 30, 2021, 1:21 AM

#

i need help

somber coyoteBOT Jan 30, 2021, 1:22 AM

#

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

silk briar Jan 30, 2021, 1:37 PM

#

i also need some help😅

#

i am pretty new to geometry

#

and i cant solve this question

dark sparrow Jan 30, 2021, 1:38 PM

#

what question

silent plank Jan 30, 2021, 1:44 PM

#

👻

dark sparrow Jan 30, 2021, 1:44 PM

#

@silk briar ?

silk briar Jan 30, 2021, 1:47 PM

#

im drawinh it on geogebra

#

📎 math.jpg

#

this is the picture

#

and i have to solve the are of the big half circle

dark sparrow Jan 30, 2021, 1:49 PM

#

the area of the big semicircle...

#

is anything else given other than that chord marked as "2"?

silk briar Jan 30, 2021, 1:50 PM

#

no 😬

dark sparrow Jan 30, 2021, 1:50 PM

#

then there is not enough information i'm afraid

silk briar Jan 30, 2021, 1:50 PM

#

our geometry teacher just drew this during the class and said tthat we have to solve it

#

oh he also said that the chord is in 90 degree angle with the diameter

dark sparrow Jan 30, 2021, 1:51 PM

#

i could see as much, given that it's tangent to the two smaller semicircles.

#

did he not say anything about those smaller semicircles by the way?

silk briar Jan 30, 2021, 1:52 PM

#

no

dark sparrow Jan 30, 2021, 1:52 PM

#

then this problem is impossible.

#

by varying the size of those semicircles i can make the big one have many different sizes.

silk briar Jan 30, 2021, 1:53 PM

#

let me check it once more

#

📎 Nayttokuva_2021-01-30_155806.jpg

dark sparrow Jan 30, 2021, 2:00 PM

#

oh!

#

this...

#

changes everything

#

wait.

#

is the vertical chord's length meant to be a+b?

silent plank Jan 30, 2021, 2:01 PM

#

is that what you constructed/assumed yourself?

dark sparrow Jan 30, 2021, 2:01 PM

#

and also 2?

#

orwhat

upper karma Jan 30, 2021, 2:01 PM

#

📎 unknown.png

#

Did i do this correct?

silk briar Jan 30, 2021, 2:01 PM

#

silent plank is that what you constructed/assumed yourself?

yes

silent plank Jan 30, 2021, 2:01 PM

#

because that would make the chord a radii which seems like its not supposed to be

#

perhaps you were actually asked to calculate the area of the red region

📎 unknown.png

silk briar Jan 30, 2021, 2:02 PM

#

yes

#

sorry if i didnt say that

silent plank Jan 30, 2021, 2:02 PM

#

which would actually be possible

dark sparrow Jan 30, 2021, 2:02 PM

#

catbruh

silent plank Jan 30, 2021, 2:02 PM

#

fml\

#

when people miscommunicate the problem

silk briar Jan 30, 2021, 2:02 PM

#

sorry for that

upper karma Jan 30, 2021, 2:02 PM

#

So was I correct?

#

📎 unknown.png

silent plank Jan 30, 2021, 2:04 PM

#

assign variables for the radii of the smaller semi-circles

#

express the radius of the larger circle in terms of those

#

then subtract areas of the smaller ones from the largest semi-circle

#

that would be the first part

#

you can get more information about your radii by applying something like intersecting chords theorem and/or otherwise

#

and you should end up with a constant

silk briar Jan 30, 2021, 2:06 PM

#

should i use proportion with it?

silent plank Jan 30, 2021, 2:07 PM

#

use a proportion how?

silk briar Jan 30, 2021, 2:08 PM

#

this would be the first step

📎 nspire.jpg

#

📎 nspire_q.jpg

#

then like this

#

which would be 4rq=4, then rq=1

silent plank Jan 30, 2021, 2:11 PM

#

yes

silk briar Jan 30, 2021, 2:11 PM

#

blobsweat

#

thank you satisfiedblob

silent plank Jan 30, 2021, 2:11 PM

#

and then you're 1 step away from your answer

silk briar Jan 30, 2021, 2:11 PM

#

yes

onyx cloud Jan 31, 2021, 3:08 AM

#

https://tasks.illustrativemathematics.org/content-standards/tasks/1825

Illustrative Mathematics

Providing instructional and assessment tasks, lesson plans, and other resources for teachers, assessment writers, and curriculum developers since 2011.

zenith garnet Jan 31, 2021, 3:25 AM

#

📎 Screen_Shot_2021-01-30_at_7.25.28_PM.png

obtuse tapir Jan 31, 2021, 7:10 AM

#

What

unkempt glen Jan 31, 2021, 7:29 AM

#

📎 image0.jpg

#

just went through geometry revisited, excited to start this

obtuse tapir Jan 31, 2021, 7:34 AM

#

ok

upper karma Jan 31, 2021, 4:12 PM

#

Hello, could someone help me with this exercise pls.

📎 unknown.png

upper karma Jan 31, 2021, 4:55 PM

#

📎 image0.png

#

yeah that's a cool problem?

#

What about it? @upper karma

upper karma Jan 31, 2021, 5:21 PM

#

:))

#

nothing

silk patio Jan 31, 2021, 9:39 PM

#

x=40

pine brook Feb 1, 2021, 3:15 AM

#

hey guys
how do i calculate the trigonometric functons without information on other angles or sides?
in a question i got for a quiz the question tan1 + tan2 + tan 3 + tan4 ... + tan88 + tan89 appeared and i don't know how to solve those kinds of trig

#

pweese <@&286206848099549185>

pine brook Feb 1, 2021, 4:09 AM

#

PLEASE I NEED HELP <@&286206848099549185>

onyx cloud Feb 1, 2021, 4:28 AM

#

#❓how-to-get-help

#

do you know double angle formulas?

#

or tan(a+b)

silent plank Feb 1, 2021, 4:32 AM

#

that question seems to be too complicated for your current understanding of trig functions

hybrid solstice Feb 1, 2021, 2:56 PM

#

📎 Screenshot_56.png

#

can someone 1 help me

#

kind of just learniing this i think i understand it

upper karma Feb 1, 2021, 6:38 PM

#

Does anyone know how this is done

📎 20210201_123740.jpg

green tree Feb 1, 2021, 7:17 PM

#

https://i.imgur.com/wfz1gU6.png

Imgur

#

please seomone

#

please

#

<@&286206848099549185>

tropic shard Feb 1, 2021, 7:20 PM

#

Did you solve for AE?

#

@green tree @hybrid solstice

#

@upper karma That's solved by equality of proportions.

upper karma Feb 1, 2021, 7:26 PM

#

How do I solve volume of sphere without Calc

tropic shard Feb 1, 2021, 7:28 PM

#

Err, I wonder if that's possible.

#

You mean* deriving 4/3 πr³ from scratch, right?

#

I guess in the old days what they did was got a heavy ball, dumped it into water and saw how much the water level rose.

#

@upper karma I found an interesting site called gingersnapmath wordpress that shows a derivation of the volume of a sphere using the geometry of a cylinder and a cone. Might be what you're looking for.

upper karma Feb 1, 2021, 7:35 PM

#

@tropic shard --> isn't the formula there from Archimedes days? And he didn't know Calc

tropic shard Feb 1, 2021, 7:39 PM

#

It is, he's quite the smart, legendary philosopher

upper karma Feb 1, 2021, 8:25 PM

#

no

wide thunder Feb 1, 2021, 9:16 PM

#

is advanced trigonometry by durell a good book to self-study trigonometry?

#

i want a somewhat rigorous treatment if that is possible

oak citrus Feb 1, 2021, 11:10 PM

#

#book-recommendations

viscid garnet Feb 2, 2021, 12:07 AM

#

I know this is a parallelogram, but i am wondering if it could be a different shape, where could I start to find out

📎 unknown.png

copper dove Feb 2, 2021, 12:21 AM

#

Hey, when verifying trigonometric identities, is it possible to use 1 + tan x = sec x as a Pythagorean identity as opposed to 1 + tan^2 x = sec^2 X?

humble pulsar Feb 2, 2021, 12:38 AM

#

copper dove Hey, when verifying trigonometric identities, is it possible to use 1 + tan x = ...

No

#

cause 1+tan = sec and 1+tan^2 = sec^2 arent the same thing

copper dove Feb 2, 2021, 1:07 AM

#

Oh I see

#

Thank you

upper karma Feb 2, 2021, 2:26 AM

#

how do we determine that an angle bisector of the angle formed by two intersecting lines is an acute- or obtuse- angle bisector?

wintry tundra Feb 2, 2021, 2:52 AM

#

sqrt of 1+tan^2 isnt 1+tan

upper karma Feb 2, 2021, 3:05 AM

#

wintry tundra sqrt of 1+tan^2 isnt 1+tan

it's secant of that angle...

wintry tundra Feb 2, 2021, 3:11 AM

#

secant is hyp/opp right?

#

no sry hyp/adj

finite sky Feb 2, 2021, 6:40 AM

#

https://cdn.discordapp.com/attachments/546941040205234176/806047944125972490/Pythag2B2.png

sleek depot Feb 2, 2021, 8:28 AM

#

finite sky https://cdn.discordapp.com/attachments/546941040205234176/806047944125972490/Pyt...

Since the line starts at the origin, y intercept = c = 0

#

m = 2

#

Substitute that in y = mx + c to get y = 2x

#

Use distance formula for a point from the origin [sqrt(x^2 + y^2)]

#

So u have x^2 + y^2 = 2645

#

Now solve y = 2x and x^2 + y^2 = 2645

finite sky Feb 2, 2021, 8:31 AM

#

oh

#

i figured out now

#

soz i knew alreadi bu u can go on

sleek depot Feb 2, 2021, 8:32 AM

#

Consider only + values as graph is in quadrant 1

finite sky Feb 2, 2021, 8:32 AM

#

cause im no sure if im correc

sleek depot Feb 2, 2021, 8:32 AM

#

Ans is 23,46

reef parrot Feb 2, 2021, 8:33 AM

#

how do you find the sides of an isosceles triangle given 3 angles

wise pulsar Feb 2, 2021, 8:51 AM

#

why is radian part of trig

reef parrot Feb 2, 2021, 8:53 AM

#

<@&286206848099549185>

silent plank Feb 2, 2021, 8:57 AM

#

3 angles alone is insufficient to determine the sides of any triangle

reef parrot Feb 2, 2021, 8:59 AM

#

given an area also whixh is 36 sq m

silent plank Feb 2, 2021, 9:00 AM

#

apply trig formula for area of a triangle

wise pulsar Feb 2, 2021, 9:10 AM

#

How do I type this in my calculator

📎 image0.jpg

silent plank Feb 2, 2021, 9:12 AM

#

depending on your calculator, you may have a \
$\circ$ ' " button

somber coyoteBOT Feb 2, 2021, 9:12 AM

#

ℝamonov

📎 220517274845577216.png

wise pulsar Feb 2, 2021, 9:12 AM

#

So I just press it

#

Ok

silent plank Feb 2, 2021, 9:12 AM

#

its for degrees minutes and seconds

wise pulsar Feb 2, 2021, 9:13 AM

#

k

silent plank Feb 2, 2021, 9:13 AM

#

they'll appear as boxes,

#

cos (32 [button] 24 [button])

wise pulsar Feb 2, 2021, 9:15 AM

#

k ty

silent plank Feb 2, 2021, 9:17 AM

#

also make sure your calc is set to degrees

agile axle Feb 2, 2021, 9:37 AM

#

Is there something wrong with my calculator? I am new to Sin, Cos and tan and when I enter tan 45 it is supposed to be 1 but when I enter tan 45 in my calculater I get 1.61977... why is this?

silent plank Feb 2, 2021, 9:38 AM

#

your calc is currently set to radians mode

#

change it to degrees

agile axle Feb 2, 2021, 9:39 AM

#

I probably sound/am stupid but how do I do that?

#

Got a Casio fx-9750GII

silent plank Feb 2, 2021, 9:42 AM

#

not familiar with that model

#

google search maybe, or read your calculator manual

agile axle Feb 2, 2021, 9:43 AM

#

Will I have to change it back for normal calculations? Or can I keep it in degrees mode?

silent plank Feb 2, 2021, 9:44 AM

#

you can keep it in degrees mode

agile axle Feb 2, 2021, 9:44 AM

#

Thanks a lot 😄

silent plank Feb 2, 2021, 9:44 AM

#

just choose the correct mode depending on the units in the question

agile axle Feb 2, 2021, 9:45 AM

#

But it only effects it when you use sin, cos and tan?

short grail Feb 2, 2021, 4:36 PM

#

📎 image0.jpg

#

could someone help me on this

silent plank Feb 2, 2021, 4:37 PM

#

where are you stuck?

short grail Feb 2, 2021, 4:38 PM

#

x-2(9)= x-6(15)

#

i got -18=-90 after that

silent plank Feb 2, 2021, 4:44 PM

#

order of operations please

#

@short grail

#

assuming you recognised you have intersection chords and are attempting to apply the related theorem,\
$x-2(9)$ is NOT the product of $x-2$ and $9$ \
$x-6(15)$ is NOT the product of $x-6$ and $15$ \

somber coyoteBOT Feb 2, 2021, 4:48 PM

#

ℝamonov

📎 220517274845577216.png

upper karma Feb 2, 2021, 6:55 PM

#

can someone help me with this

📎 Screen_Shot_2021-02-02_at_10.55.22_AM.png

upper karma Feb 2, 2021, 8:23 PM

#

📎 unknown.png

placid jasper Feb 2, 2021, 10:20 PM

#

upper karma can someone help me with this

$\sin{4.76^{\circ}}=\frac{4}{\textsc{carpet length}}$

somber coyoteBOT Feb 2, 2021, 10:21 PM

#

jolimath

📎 783481419283300383.png

placid jasper Feb 2, 2021, 10:24 PM

#

upper karma

The 26-deg doesn’t matter. AC is twice the length of DF because it’s the segment joining the midpoints of AE and CE. The answer is 12.4

shrewd kettle Feb 2, 2021, 11:31 PM

#

honestly i just started geometry and im already so confused, i tried my best to pick things up but i cant really wrap my head around it, could anyone assist?

📎 unknown.png

#

ive tried looking at some online resources but i dont really pick things up the best with those types of resources and would rather have someone right here to help me out

#

its been quite a while since ive had a math class, it was a year and a half ago when i had algebra and i forget a lot of terms easily

#

I just started geometry with the new semester

upper karma Feb 2, 2021, 11:35 PM

#

can someone also help me

📎 image0.png

shrewd kettle Feb 2, 2021, 11:43 PM

#

scratch that figured out the first portions ill use the help channels if i have questions

past harbor Feb 3, 2021, 1:33 AM

#

can someone please help me with this

📎 image0.jpg

upper karma Feb 3, 2021, 1:46 AM

#

📎 unknown.png

#

just tell me if it is congruent or not

west basin Feb 3, 2021, 2:01 AM

#

They are

#

9x - 31 = 5x + 13

upper karma Feb 3, 2021, 2:06 AM

#

👏

#

thanks

#

📎 unknown.png

#

what is the way to setup the equation

#

The area of a trapezoid is 60. The length of one base is 7 units greater than the other base, and the height of the trapezoid is 5. Find the length of the median of the trapezoid.

#

i dont understand anything lmao

west basin Feb 3, 2021, 3:02 AM

#

use the given to find the lengths of the bases

#

and then use that to find the median

upper karma Feb 3, 2021, 3:05 AM

#

📎 unknown.png

west basin Feb 3, 2021, 3:07 AM

#

take distances between each point

#

then you can narrow it down a bit

crystal quail Feb 3, 2021, 7:14 AM

#

Would it be fair to say that, since "two lines can intersect at one point at most," directly implies that "two planes can intersect along one line at most"? I can't see why it wouldn't be true but wanted to get some thoughts before I settle on that answer

#

reasoning just being that planes can be thought of as linear extensions of lines into another dimension, so the path they trace through space forms a plane, and the point they intersect traces the path of a line

upper karma Feb 3, 2021, 9:05 AM

#

i have a question....

green frost Feb 3, 2021, 9:19 AM

#

damn fr

unreal patrol Feb 3, 2021, 12:47 PM

#

slide help

#

📎 unknown.png

storm portal Feb 3, 2021, 12:57 PM

#

unreal patrol

📎 800622149903056927.png

upper karma Feb 3, 2021, 12:59 PM

#

@unreal patrol you can use the hint above.

unreal patrol Feb 3, 2021, 12:59 PM

#

thank you

#

i dont get this one

#

📎 unknown.png

upper karma Feb 3, 2021, 1:05 PM

#

Have you tried anything so far?

unreal patrol Feb 3, 2021, 1:21 PM

#

nop

upper karma Feb 3, 2021, 1:44 PM

#

Well, start by trying something out considering you have all the trig identities you can use to manipulate sec(A/2) into something you can find easily.

#

If you get stuck let me know.

upper karma Feb 3, 2021, 2:24 PM

#

It says "test" on the title.

#

<@&268886789983436800>

#

@worn folio hi. Delete dis

#

ty

#

@worn folio well I did it on your stead

#

given that you were doxing yourself by asking help for that test

#

never ask for test solutions

#

@upper karma ty as well for the notification

upper karma Feb 3, 2021, 2:44 PM

#

How do

📎 image0.jpg

worn folio Feb 3, 2021, 3:04 PM

#

@upper karma Im in Cp geomotry, and my teacher uses General geometry tests for our hw lol.

#

I dont want to have to wait to consult my tutor after school

#

cause then it turns into an hour of studying spanish after the one part of math

hybrid solstice Feb 3, 2021, 3:08 PM

#

📎 Screenshot_61.png

#

my brain hurts can some one help me

#

i just need the equation

#

this is proportions

silent plank Feb 3, 2021, 3:09 PM

#

apply angle bisector theorem VaP$

#

@upper karma your question is cut off

hybrid solstice Feb 3, 2021, 3:11 PM

#

alright

#

yeah but 21 wouldnt equal y

silent plank Feb 3, 2021, 3:12 PM

#

wdym

#

why would 21 equal y

hybrid solstice Feb 3, 2021, 3:12 PM

#

it wouldnt

silent plank Feb 3, 2021, 3:13 PM

#

why did that statement warrant a "but"

hybrid solstice Feb 3, 2021, 3:14 PM

#

wait hold up i think i get it

#

i was getting the imagery wrong

arctic path Feb 3, 2021, 3:33 PM

#

Hello, may i ask some help about solid geometry? frustum of a cone to be exact.

obtuse tapir Feb 3, 2021, 3:44 PM

#

Ok

#

Go ahead

humble pulsar Feb 3, 2021, 3:49 PM

#

"can I ask something?"
doesnt ask

arctic path Feb 3, 2021, 3:53 PM

#

obtuse tapir Ok

if you were given a product of radius of bases in a frustum, how do you get each radius if the given is height, volume and the product of radius?

arctic path Feb 3, 2021, 3:57 PM

#

arctic path if you were given a product of radius of bases in a frustum, how do you get each...

I mean the height of frustum, volume of frustum, and the product of the radius of the bases of frustum...

obtuse tapir Feb 3, 2021, 3:57 PM

#

Apply the formula

#

The volume formula

#

then substitute

#

the product

#

so if the product is xy

#

suppose 50

#

x=50/y

#

then plug this in volume formula

arctic path Feb 3, 2021, 3:59 PM

#

ohh i get it

#

thank youu

obtuse tapir Feb 3, 2021, 3:59 PM

#

Good

upper karma Feb 3, 2021, 4:46 PM

#

📎 Screen_Shot_2021-02-03_at_11.43.04_AM.png

#

someone explain what the heck a kite is i dont full under stand

exotic plinth Feb 3, 2021, 5:23 PM

#

can someone help with this please

#

https://cdn.discordapp.com/attachments/387447700234174464/806575414658990150/Screen_Shot_2021-02-03_at_12.22.39_PM.png

#

<@&286206848099549185>

scenic glacier Feb 3, 2021, 5:40 PM

#

📎 unknown.png

#

hey guys does anyone know how to use the unit circle

#

i havent payed attention during the last few months n we are doing a quiz review

grizzled lantern Feb 3, 2021, 6:11 PM

#

📎 unit-circle7_43215_lg.png

#

@scenic glacier

#

for number one its asking what angle makes cos = -1

#

think of cos as your left and right and your sin at top and bottom

#

or you can think of cos as your x and your sin as your y

worn folio Feb 3, 2021, 6:21 PM

#

📎 D9ilPTmleQWPAAAAABJRU5ErkJggg.png

#

Need help asap

#

i literally cant solve it ive looked at every theorm and proof

worn folio Feb 3, 2021, 7:55 PM

#

@graceful totem when you have a sec do you think you could help me with this?

languid lark Feb 3, 2021, 11:17 PM

#

Hello

sacred geyser Feb 3, 2021, 11:28 PM

#

can someone help me plz

📎 Screenshot_2021-02-03_at_5.27.52_PM.png

upper karma Feb 3, 2021, 11:29 PM

#

sure

sacred geyser Feb 3, 2021, 11:29 PM

#

ok thx i need it

upper karma Feb 3, 2021, 11:30 PM

#

hmm

#

im not too good at geometry

#

but i think it might be talking ratios?

#

maybe

#

ok @sacred geyser so this is a ratio thing

#

the geometric mean is the ratio of the shorter leg (KN) to the longer leg (LN)

#

and if you broke that triangle into two triangles,

#

their altitudes would be the same

past harbor Feb 4, 2021, 12:43 AM

#

could someone please help me with this problem

📎 image0.jpg

viscid garnet Feb 4, 2021, 1:48 AM

#

i do not really know where to go from this

📎 unknown.png

#

i made some isoceles traingles and found that x/2 = 90 - <OCB

exotic plinth Feb 4, 2021, 2:05 AM

#

what is NP and NL

📎 image0.jpg

fleet hill Feb 4, 2021, 2:12 AM

#

worn folio

AD=(1/2)(BD)

#

@worn folio use the triangle midsegment theorem

#

@worn folio google the triangle midsegment theorem and use that

upper karma Feb 4, 2021, 2:32 AM

#

📎 image0.png

#

Dunno why I am getting really off answers someone double check for me and tell me if it is a parallelogram (this doesn’t cheat I gotta show work)

#

<@&286206848099549185>

#

Desperate deadline midnight for progress reports

idle bobcat Feb 4, 2021, 3:34 AM

#

Hi, can someone explain the intuition here behind dividing every term by cosacosb?

📎 unknown.png

silk patio Feb 4, 2021, 3:50 AM

#

To turn everything into a tan term

idle bobcat Feb 4, 2021, 4:14 AM

#

ah gotcha

#

How do I derive the right hand side?

📎 unknown.png

#

deriving cos(2x) = cos^2x-sin^2x is pretty straight forward

#

dont know about the one in the picture

#

📎 unknown.png

#

same with this one

#

the lecture notes didnt actually explain how they got these

#

they just told us that these are also identities

versed river Feb 4, 2021, 4:23 AM

#

have you seen the pythagorean identity $\sin^2(x)+\cos^2(x)=1$?

somber coyoteBOT Feb 4, 2021, 4:23 AM

#

Sneaky

📎 329749386315956225.png

idle bobcat Feb 4, 2021, 4:24 AM

#

yeah I know that

#

oh

#

ok wow

#

LOL

versed river Feb 4, 2021, 4:24 AM

#

:)

idle bobcat Feb 4, 2021, 4:24 AM

#

❤️

#

so I guess there's no easy way of deriving 1-2sin^2x directly from cos^2x-sin^2x right?

#

its mainly just a matter of knowing the pythagorean identity?

versed river Feb 4, 2021, 4:26 AM

#

not without pythag identity lol

#

but pythagorean identity is everywhere

#

you won't be forgetting it

idle bobcat Feb 4, 2021, 4:26 AM

#

yeah thats fair

#

thank youu

versed river Feb 4, 2021, 4:26 AM

#

no worries

trim nexus Feb 4, 2021, 5:41 AM

#

📎 unknown.png

trim nexus Feb 4, 2021, 6:02 AM

#

📎 unknown.png

#

📎 unknown.png

#

📎 unknown.png

#

@everyone pls help

carmine sundial Feb 4, 2021, 9:03 AM

#

📎 IMG_20210204_165537.jpg

#

What would best describe ray DE? Line tangent?

silent plank Feb 4, 2021, 9:06 AM

#

tangent would be enough

carmine sundial Feb 4, 2021, 9:07 AM

#

Thank you!

silent plank Feb 4, 2021, 9:07 AM

#

you'd need to fix #2

carmine sundial Feb 4, 2021, 9:40 AM

#

What would it be? Equidistant?

#

Or just radii

silent plank Feb 4, 2021, 9:48 AM

#

just points on the circumference

ebon star Feb 4, 2021, 1:29 PM

#

Is it possible for someone to help me with 10 pages of unit 6, lessons 1-5 of geometry? The unit is called relationships with triangles.

storm portal Feb 4, 2021, 1:38 PM

#

ebon star Is it possible for someone to help me with 10 pages of unit 6, lessons 1-5 of ge...

#info

📎 Screen_Shot_2021-02-04_at_8.38.13_AM.png

ebon star Feb 4, 2021, 1:41 PM

#

storm portal <#525440994242461702>

is that better

storm portal Feb 4, 2021, 1:44 PM

#

The only problem is that people will likely not want to do 10 pages of work

ebon star Feb 4, 2021, 1:46 PM

#

understood

pseudo bison Feb 4, 2021, 2:31 PM

#

📎 unknown.png

#

<@&286206848099549185> How do I do this?

cosmic pebbleBOT Feb 4, 2021, 2:32 PM

#

From

#❓how-to-get-help

Rule 4

If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once. Please do not try to bump your question using this ping unnecessarily. Do not abuse this ping. Do not individually ping users with the Helpers tag without their express permission.

devout harbor Feb 4, 2021, 4:51 PM

#

@pseudo bison do you still need help?

upper karma Feb 4, 2021, 5:14 PM

#

📎 image0.jpg

#

Anyone get this?

#

Nvm got it

obtuse tapir Feb 4, 2021, 5:24 PM

#

note: blah blah
Me: draw perpendicular 😎

upper karma Feb 4, 2021, 7:31 PM

#

📎 unknown.png

obtuse tapir Feb 4, 2021, 7:36 PM

#

Acute

#

Right angled

upper karma Feb 4, 2021, 7:43 PM

#

flushEDP

placid jasper Feb 4, 2021, 8:38 PM

#

Have you heard about arcsine?

upper karma Feb 4, 2021, 8:48 PM

#

no

hollow raven Feb 4, 2021, 9:06 PM

#

Well here's one hint

#

As the two shorter sides form an angle that approaches 180 degrees, the third side's value approaches the addition of the two shorter sides

#

So essentially, if the third side is greater or equal to the two shorter sides, then a triangle cannot be made, because the triangle's value would essentially need to be > 180 degrees, which is of course impossible

#

and that even if side one and two were to be parallel to each other, they would not be able to connect with the third side end by end

cloud meteor Feb 4, 2021, 10:06 PM

#

@upper karma

#

can you repost it here?

upper karma Feb 4, 2021, 10:06 PM

#

alrighty

#

📎 unknown.png

cloud meteor Feb 4, 2021, 10:06 PM

#

I'll start off by stating some theorems

#

The sum of the areas inside a triangle is always 180º

#

therefore if you know two angles

#

you can find the other angle

upper karma Feb 4, 2021, 10:08 PM

#

thanks man

cloud meteor Feb 4, 2021, 10:08 PM

#

Anyway, so

upper karma Feb 4, 2021, 10:08 PM

#

so for the first one

cloud meteor Feb 4, 2021, 10:08 PM

#

wait mb

upper karma Feb 4, 2021, 10:09 PM

#

i do 120, minus 108 + 29 which is 137

#

and then that's the angle?

cloud meteor Feb 4, 2021, 10:09 PM

#

no no no

#

you do 180 - 108 - 29

upper karma Feb 4, 2021, 10:09 PM

#

mb i meant 180 instead of 120

cloud meteor Feb 4, 2021, 10:09 PM

#

you do 180 - (108 + 29)

#

= 180 - 137 = 43º

#

📎 unknown.png

#

so the angle under D should be 43º

#

Alright. So the next theorem is

viral magnet Feb 4, 2021, 10:11 PM

#

how the hell do you do that.

cloud meteor Feb 4, 2021, 10:11 PM

#

geogebra

viral magnet Feb 4, 2021, 10:11 PM

#

HAHA

cloud meteor Feb 4, 2021, 10:11 PM

#

if two points on the circumference of a circle

📎 unknown.png

#

have a common chord

#

they're the exact same angle

#

which means that ......?

#

@upper karma

upper karma Feb 4, 2021, 10:12 PM

#

ur mom ?

#

bruh idek man

#

im 12 and this shits so hard

cloud meteor Feb 4, 2021, 10:14 PM

#

those angles are the same

#

D = B = x

#

therefore x = how many degrees

fallen ivy Feb 4, 2021, 10:38 PM

#

TOS banhammer

finite comet Feb 4, 2021, 10:53 PM

#

📎 unknown.png

placid jasper Feb 4, 2021, 10:54 PM

#

cloud meteor The sum of the areas inside a triangle is always 180º

*angles

opaque violet Feb 5, 2021, 2:32 AM

#

📎 unknown.png

zealous geyser Feb 5, 2021, 3:05 AM

#

hey can anyone help me with a problem

#

📎 unknown.png

idle bobcat Feb 5, 2021, 3:18 AM

#

Hi, how do I find out if this is a cosine or sine function?

📎 unknown.png

idle bobcat Feb 5, 2021, 4:31 AM

#

Also for this question

📎 unknown.png

#

is this correct?

📎 unknown.png

unique flower Feb 5, 2021, 5:25 AM

#

yes, although you could simplify it to 2+sqrt(3) by rationalizing the denominator

obtuse tapir Feb 5, 2021, 10:44 AM

#

@cloud meteor same arc bro

worn folio Feb 5, 2021, 1:08 PM

#

Need some help

📎 D9ilPTmleQWPAAAAABJRU5ErkJggg.png

storm portal Feb 5, 2021, 1:15 PM

#

idle bobcat Also for this question

You can check using desmos

#

Just make sure you're in radians

📎 Screen_Shot_2021-02-05_at_8.15.51_AM.png

lusty slate Feb 5, 2021, 2:34 PM

#

📎 image0.png

#

Smbd rq

obtuse tapir Feb 5, 2021, 5:52 PM

#

Good question

#

📎 image0.jpg

#

basically the diagram

#

does it click now?

#

📎 image0.jpg

#

shift the perpendicular to the side

#

the radius of the smaller circle adjacent to the bigger circle’s radius

#

shown in dotted lines

#

now pythagoras theorem

#

The values are given

#

Create an equation

#

nice done

blazing pond Feb 5, 2021, 7:19 PM

#

📎 unknown.png

#

plz help

humble pulsar Feb 5, 2021, 7:19 PM

#

what have you tried?

blazing pond Feb 5, 2021, 7:19 PM

#

everthing

#

and thats the answer I got

#

is it correct?

humble pulsar Feb 5, 2021, 7:20 PM

#

no

#

did you draw a diagram?

blazing pond Feb 5, 2021, 7:20 PM

#

yea

#

is this one correct

#

📎 unknown.png

humble pulsar Feb 5, 2021, 7:21 PM

#

no

blazing pond Feb 5, 2021, 7:22 PM

#

wow I am terrible

humble pulsar Feb 5, 2021, 7:22 PM

#

humble pulsar did you draw a diagram?

going back to the one I was talking about. . . what trig ratio did you use?

blazing pond Feb 5, 2021, 7:22 PM

#

I got a new answer for the first one

#

I got 18 ft

#

17.976

humble pulsar Feb 5, 2021, 7:22 PM

#

what trig ratio did you use

blazing pond Feb 5, 2021, 7:22 PM

#

sin

humble pulsar Feb 5, 2021, 7:23 PM

#

right, 17.976 is right

blazing pond Feb 5, 2021, 7:23 PM

#

ok thx

#

how abt the second one

humble pulsar Feb 5, 2021, 7:23 PM

#

did you draw a diagram?

blazing pond Feb 5, 2021, 7:23 PM

#

yes

humble pulsar Feb 5, 2021, 7:24 PM

#

what trig ratio do you use?

blazing pond Feb 5, 2021, 7:24 PM

#

tan?

humble pulsar Feb 5, 2021, 7:24 PM

#

no you dont know adjacent

blazing pond Feb 5, 2021, 7:24 PM

#

oh so sin?

humble pulsar Feb 5, 2021, 7:25 PM

#

yes, then add on how high the triangle would be off the ground

blazing pond Feb 5, 2021, 7:25 PM

#

and I would put 540 in as the hypotneuse right?

humble pulsar Feb 5, 2021, 7:25 PM

#

yes

blazing pond Feb 5, 2021, 7:26 PM

#

wouldit be 536?

#

or 537

#

yeaaaa it would right, cuz last time I subtracted 36 instead of adding it

#

thx a lot @humble pulsar

graceful trail Feb 5, 2021, 7:44 PM

#

For trig, is the ratio of tangent cos/sin or sin/cos? I forgot

humble pulsar Feb 5, 2021, 7:45 PM

#

graceful trail For trig, is the ratio of tangent cos/sin or sin/cos? I forgot

sin/cos

#

cot is cos/sin

graceful trail Feb 5, 2021, 7:45 PM

#

omg ty 🙏

upper karma Feb 5, 2021, 7:56 PM

#

📎 unknown.png

#

The 1/2x part is very confusing for me

silent plank Feb 5, 2021, 7:57 PM

#

1/2 x is just half of x

upper karma Feb 5, 2021, 7:57 PM

#

Yeah but how does that affect me while I'm solving

#

Cuz I still have to solve for X

silent plank Feb 5, 2021, 7:57 PM

#

1/2 is just a number

upper karma Feb 5, 2021, 7:58 PM

#

so do I disregard it?

silent plank Feb 5, 2021, 7:58 PM

#

no

#

its just a number

#

treat it like you would any other number

upper karma Feb 5, 2021, 7:58 PM

#

Sorry I'm still not following 😅

silent plank Feb 5, 2021, 7:59 PM

#

1/2 x is 1/2 multiples of x, (1/2)x, x/2 or however you want to view it
just like how 2x is 2 multiples of x
3x is 3 multiples of x
etc

upper karma Feb 5, 2021, 8:00 PM

#

So it's just 1?

silent plank Feb 5, 2021, 8:00 PM

#

no

#

1/2 is 1/2

upper karma Feb 5, 2021, 8:00 PM

#

Ok I understand that

#

but can I do anything with it?

silent plank Feb 5, 2021, 8:01 PM

#

wdym

upper karma Feb 5, 2021, 8:01 PM

#

Subtract add etc.

silent plank Feb 5, 2021, 8:01 PM

#

1/2 is just a number

#

don't overthink it

upper karma Feb 5, 2021, 8:01 PM

#

OK

#

Can we start over?

#

I'm kinda all over the place with it

silent plank Feb 5, 2021, 8:02 PM

#

ok.

upper karma Feb 5, 2021, 8:02 PM

#

Ok so

#

📎 unknown.png

#

This is the equation

#

I'm confused on what to do with 1/2x

silent plank Feb 5, 2021, 8:03 PM

#

well technically you don't even need to apply it because the question is over specified,
but this seems like a real issue you need to get past

#

what does it mean for a triangle to be equilangular?

upper karma Feb 5, 2021, 8:03 PM

#

All sides are the same

#

so like

#

60+60+60 = 180

silent plank Feb 5, 2021, 8:04 PM

#

those are angles,
but yes as a result all the sides would be the same

#

and apply that

#

for the purposes of dealing with 1/2 x

#

set XZ equal to either of your other 2 sides

#

e.g XY

#

and you'll get the equation:

#

$2x+10 = \frac{x}{2} + 28$

somber coyoteBOT Feb 5, 2021, 8:06 PM

#

ℝamonov

📎 220517274845577216.png

silent plank Feb 5, 2021, 8:06 PM

#

and solve that like any other equation

upper karma Feb 5, 2021, 8:06 PM

#

O

#

Ok thank you

#

Basically set

#

XY and YZ to be the same

#

which makes solving XZ easier

silent plank Feb 5, 2021, 8:08 PM

#

you can set any pair of sides to be equal and solve for x

#

XY = YZ will result in the easiest calculations

#

you can sub your value of x into all 3 sides to double check your answer and the question is actually valid

upper karma Feb 5, 2021, 8:09 PM

#

That makes sense

#

Thank You

#

I have another one

#

mnp_flushedsex

#

📎 unknown.png

silent plank Feb 5, 2021, 8:15 PM

#

identify corresponding vertices and apply distance formula

upper karma Feb 5, 2021, 8:17 PM

#

How would I go about doing that

silent plank Feb 5, 2021, 8:18 PM

#

be familiar with congruence notation

upper karma Feb 5, 2021, 8:48 PM

#

Guys

#

how could I know if the norm of any of these vectors would be equal to

📎 unknown.png

#

the norm of this vector

📎 unknown.png

#

I know the norm of the vector v is 1, but don't know how to see the norms of the others vectors

idle bobcat Feb 5, 2021, 10:13 PM

#

Hi, how do I know if I should write this as a cosine or sine function?

📎 unknown.png

#

This is what I have so far, I'm just not sure if I should write it in terms of cos or sin

📎 unknown.png

weary drift Feb 5, 2021, 11:36 PM

#

@upper karma what's e1,e2,e3?

torpid belfry Feb 6, 2021, 12:57 AM

#

heyy i got some questions and i think this is the right channel for this type of math maybe

lusty slate Feb 6, 2021, 1:58 AM

#

📎 image0.png

fleet hill Feb 6, 2021, 2:51 AM

#

@lusty slate 4m = 4000 mm
300 mm = 30 cm
234 mm = 2.34 dm

storm portal Feb 6, 2021, 3:44 AM

#

@lusty slate look up metric conversions

idle bobcat Feb 6, 2021, 6:33 AM

#

A sinusoidal function with an amplitude of 15 units, a period of pi/2, a phase shift of pi/7 radians to the right, axis at y =33
This is my answer: y= 15f[4(x-π/7)]+33
Is f cosine or sine? How do I know?
Thanks

upper karma Feb 6, 2021, 8:23 AM

#

@idle bobcat i'm guessing that "axis at y=33" means it touches the y-axis at y=33, then consider what occurs when both functions cos and sin are at x=0

meager garden Feb 6, 2021, 12:09 PM

#

How would I get the radius of the inner circle? I know the perimeter would be 6R

📎 unknown.png

#

would it be sqrt(3)/2 * R ?

#

I guessed that from the options I got

obtuse tapir Feb 6, 2021, 1:21 PM

#

Good question

#

I’ll give some hints

#

You can divide the hexagons into isosceles triangles

#

Then the median

#

Would become the radius

#

and the pattern continues

idle bobcat Feb 6, 2021, 5:01 PM

#

upper karma <@301519660531187712> i'm guessing that "axis at y=33" means it touches the y-ax...

axis just means the midline

#

I'm just not sure if I'm given a question like this how I determine whether to use cos or sin for my function

idle bobcat Feb 6, 2021, 5:20 PM

#

usually, it tells me whether its sine or cosine, and I just need to put together the rest

hasty ferry Feb 6, 2021, 5:24 PM

#

idle bobcat A sinusoidal function with an amplitude of 15 units, a period of pi/2, a phase s...

i think the answer should be (4x-pi/7). Not 4(x-pi/7)

hasty ferry Feb 6, 2021, 5:25 PM

#

idle bobcat usually, it tells me whether its sine or cosine, and I just need to put together...

technically, cosine is just sine with a phase shift of pi/2. so i think u should answer with sine

#

this is the answer imo: 15 sin[4x-π/7]+33

idle bobcat Feb 6, 2021, 5:29 PM

#

hasty ferry i think the answer should be (4x-pi/7). Not 4(x-pi/7)

can u elaborate on why?

#

because it says the period is pi/2

#

and 2pi/(pi/2) means our k value is 4

#

wouldnt you need to distribute that value to pi/7 if you wanted to express it the way you did

idle bobcat Feb 6, 2021, 5:33 PM

#

hasty ferry technically, cosine is just sine with a phase shift of pi/2. so i think u should...

but the thing I'm confused about with that is sine can also just be a phase shift of pi/2 to the left of cos cant it? so how do I know if the function is a base cos function shifted pi/7 to the right, or if it's a base sin function shifted pi/7 to the left?

#

usually in these problems they either explicitly tell me to write it in terms of sin or cos, or they atleast give me hints in terms of what the y axis intersects, or some other related hint

hasty ferry Feb 6, 2021, 5:35 PM

#

idle bobcat because it says the period is pi/2

yes. even my answer has a period of pi/2. The only difference btw my ans ans urs is the phase shift

#

by convention, sin(ax+b) has a phase shift of b

#

so if u want a phase shift of pi/7, b must be pi/7

idle bobcat Feb 6, 2021, 5:36 PM

#

yeah but in your equation you brought the k = 4 value inside the brackets

#

which means you would need to distribute

hasty ferry Feb 6, 2021, 5:37 PM

#

idle bobcat but the thing I'm confused about with that is sine can also just be a phase shif...

yea you're right there. it can be cos also. the question must have mentioned clearly

hasty ferry Feb 6, 2021, 5:37 PM

#

idle bobcat which means you would need to distribute

yes, the period doesnt change after u distribute

#

try plotting both in desmos

idle bobcat Feb 6, 2021, 5:38 PM

#

y=af[k(x−d)]+c this is the general form I was taught to use

#

so if you bring the k into the brackets, you would need to distribute the k to the d

hasty ferry Feb 6, 2021, 5:39 PM

#

idle bobcat so if you bring the k into the brackets, you would need to distribute the k to t...

yea you would. but the period doesnt change bcos of that. the coefficient of x is the **only **determinant of period

idle bobcat Feb 6, 2021, 5:39 PM

#

15 sin[4x-28pi/7]+33

#

oh im not saying the period would change

hasty ferry Feb 6, 2021, 5:40 PM

#

yeah bro but the period is still pi/2

idle bobcat Feb 6, 2021, 5:40 PM

#

im saying the d value would change

hasty ferry Feb 6, 2021, 5:40 PM

#

ok i misunderstood

hasty ferry Feb 6, 2021, 5:41 PM

#

hasty ferry by convention, sin(ax+b) has a phase shift of b

.

idle bobcat Feb 6, 2021, 5:41 PM

#

hasty ferry yea you're right there. it can be cos also. the question must have mentioned cle...

nope, they gave no hints or anything

hasty ferry Feb 6, 2021, 5:41 PM

#

what im telling is, if u want a phase shift of pi/7, k*d must be pi/7

#

and not d=pi/7

#

idk if they taught u that d is phase shift

#

the general convention is k*d=phase shift

hasty ferry Feb 6, 2021, 5:43 PM

#

idle bobcat nope, they gave no hints or anything

then dont worry about f. it can be both sin or cos

idle bobcat Feb 6, 2021, 5:44 PM

#

📎 unknown.png

#

most websites im reading doesnt say k*d is the phase shift

#

even in my textbook

hasty ferry Feb 6, 2021, 5:45 PM

#

ok if this is what they taught u, go with it

hasty ferry Feb 6, 2021, 5:45 PM

#

idle bobcat most websites im reading doesnt say k*d is the phase shift

in electrical engineering, k*d is phase shift

idle bobcat Feb 6, 2021, 5:45 PM

#

hasty ferry then dont worry about f. it can be both sin or cos

ok yeah so anyways, how can it be both sin and cos?

hasty ferry Feb 6, 2021, 5:45 PM

#

but go with your textbook

idle bobcat Feb 6, 2021, 5:46 PM

#

because they both give different graphs

hasty ferry Feb 6, 2021, 5:46 PM

#

yea but the question hasnt mentioned it clearly right

idle bobcat Feb 6, 2021, 5:46 PM

#

yeah

hasty ferry Feb 6, 2021, 5:46 PM

#

so both should be accepted

idle bobcat Feb 6, 2021, 5:46 PM

#

ok so I guess I was right that there wasn't enough information

#

in order to determine whether they want sin or cos

hasty ferry Feb 6, 2021, 5:47 PM

#

yepp

idle bobcat Feb 6, 2021, 5:47 PM

#

ok yeah I was confused for so long about this question..

upper karma Feb 6, 2021, 8:07 PM

#

📎 unknown.png

humble pulsar Feb 6, 2021, 8:11 PM

#

upper karma

do you know the simplest 30-60-90 side lengths?

upper karma Feb 6, 2021, 8:11 PM

#

No

humble pulsar Feb 6, 2021, 8:12 PM

#

Ok draw a 30 60 90 with hypotenuse 16

#

then use trig to find the legs

upper karma Feb 6, 2021, 8:12 PM

#

I have no idea how to do that 😅

humble pulsar Feb 6, 2021, 8:13 PM

#

have you tried anything?

upper karma Feb 6, 2021, 8:13 PM

#

Yes

#

and I've failed misreably

humble pulsar Feb 6, 2021, 8:14 PM

#

ok well can I see what you have tried..?

upper karma Feb 6, 2021, 8:15 PM

#

Can't access it anymore

#

Do you know what this type of math is called

#

GOnna see what I can find on Khan academy

humble pulsar Feb 6, 2021, 8:16 PM

#

Trig...

scarlet pawn Feb 6, 2021, 10:15 PM

#

upper karma

this is a 30 60 90 triangle

📎 Screenshot_29.png

#

it's a triangle that has one 60º angle, a 30º angle and a 90º angle

#

don't mind the thingys on the sides of it

#

draw one of these

#

and that bigger side

#

the hypotenuse

#

make it 16 cm long

#

then

#

you see the side that's the opposite of the 30 degree angle?

#

it's the opposite side to the 30 degree angle

#

and the adjacent side to the 60 degree one

#

you'll need that, but height first

#

that you can solve by doing this:
sin 60= height/ 16

#

so height will be sin 60 * 16

#

then the base

#

is

#

sin 30= base/16

#

so base= sin30* 16

#

then to calculate the area you do

#

base*height all dividing by two

#

there u go

#

@upper karma

#

i'm learning trig now, this was the easiest way i found to explain it, excuse me if it ends up being wrong 4

upper karma Feb 7, 2021, 1:08 AM

#

Sorry just saw this

#

@scarlet pawn Thank You

acoustic notch Feb 7, 2021, 1:57 PM

#

So uh I was thinking about how the graph of sinh looks nothing like sin, but does somewhat resemble tan. It's kinda like a stretched out version of it

📎 tansinh.png

#

The same can be said about cosh and sec

📎 seccosh.png

#

And not only are they both stretched out, but they're stretched out by the same amount

#

So if sinh(x) = tan(y), then cosh(x) = sec(y)

#

This is not so hard to see given the identities cosh^2(x) - sinh^2(x) = 1 and sec^2(x) - tan^2(x) = 1

#

What I noticed is that every hyperbolic function can be viewed as a regular trig function cut off at +/- π/2 and stretched out to infinity (by the same factor)

#

tanh is sin stretched out

📎 sintanh.png

#

sech is cos stretched out

📎 cossech.png

#

csch - tan

📎 cotcsch.png

#

coth - csc

📎 csccoth.png

upper karma Feb 7, 2021, 2:22 PM

#

And they hv identities similar to trig functions

#

Maybe this is the reason for the similarities

acoustic notch Feb 7, 2021, 2:24 PM

#

All identities that don't change the argument are analogous as far as I've seen

#

if sinh(x) = tan(y), then cosh(x) = sec(y)
And this holds true for all pairs

#

Though signs are a bit tricky with the even functions

upper karma Feb 7, 2021, 2:26 PM

#

👍

acoustic notch Feb 7, 2021, 2:30 PM

#

To take it a step further, since e^x can be written as sinh(x)+cosh(x), it's "squished" version is tan(x)+sec(x)

📎 ex.png

#

Which means that the squished counterpart of x is ln(tan(x)+sec(x))

#

So we can squish any f(x) function by taking f(ln(tanx+secx))

supple onyx Feb 7, 2021, 2:34 PM

#

are you aware of the complex identity relating sin, cos, cosh, and sinh?

acoustic notch Feb 7, 2021, 2:35 PM

#

I'd say I have a surface level understanding

#

cosh(x) = cos(ix)
sinh(x) = -isin(ix)

supple onyx Feb 7, 2021, 2:35 PM

#

maybe something to dig in that direction? Not too sure

#

i need to reread what you said

acoustic notch Feb 7, 2021, 2:37 PM

#

Well with the complex identity the hyperbolic counterpart of sin is sinh

#

But from this perspective sin's counterpart is tanh

upper karma Feb 7, 2021, 2:38 PM

#

supple onyx i need to reread what you said

Darpleon is relating trig and hyperbolic functions

supple onyx Feb 7, 2021, 2:39 PM

#

yeah ik

#

just rereading to try and see if a hint might come to mind as to where to look for this relation

upper karma Feb 7, 2021, 2:39 PM

#

Oh ok

acoustic notch Feb 7, 2021, 2:40 PM

#

Well I assume if you just do the algebra with the exponential forms it'll all check out

#

But that might not provide any additional insight

supple onyx Feb 7, 2021, 2:55 PM

#

hmm interesting, for the sinh-tan example, computing sinh(ln(tanx + secx)) does give you something similar to the tan function, but it's missing every second period

#

same with sech - cos, graph disappears for pi/2 + 2pi * k < x < 3pi/2 + 2pi * k

#

right okay of course it would, those are intervals where tanx + secx is negative

acoustic notch Feb 7, 2021, 2:59 PM

#

yes you have to use absolute value

#

And with cos and sec you're not gonna get negative values

supple onyx Feb 7, 2021, 3:03 PM

#

i think i may have maybe some sort of path that could be worth looking at

#

consider the cos to sech example

#

cos denotes x coordinates of a circle

#

while cosh denotes x coords of a hyperbola

#

considering that a hyperbola is in some sense a sort of inverted circle, i guess it makes sense that 1/cosh and cos should be related?

#

super loose but maybe worth digging into

acoustic notch Feb 7, 2021, 3:05 PM

#

supple onyx while cosh denotes x coords of a hyperbola

technically sec and tan also describe a hyperbola

#

Due to them having that same identity

supple onyx Feb 7, 2021, 3:05 PM

#

yeah

exotic plinth Feb 7, 2021, 3:15 PM

#

📎 unknown.png

devout harbor Feb 7, 2021, 3:17 PM

#

area of a circle is pi r^2

#

so for circle A, you have 225pi = pi (r_a)^2

#

do the same for circle B

acoustic notch Feb 7, 2021, 3:29 PM

#

$sq(x)=\ln(\tan{x}+\sec{x}) \ e^{sq}=\tan{x}+\sec{x} \ e^{-sq}=\frac{1}{\tan{x}+\sec{x}} \ = \frac{\sec{x}-\tan{x}}{\sec^2{x}-\tan^2{x}} \ e^{-sq}=\sec{x}-\tan{x}$

somber coyoteBOT Feb 7, 2021, 3:33 PM

#

Darpleon

📎 503601869570965524.png

acoustic notch Feb 7, 2021, 3:36 PM

#

$e^{sq}+e^{-sq}=2\sec{x} \ e^{sq}-e^{-sq}=2\tan{x}$

somber coyoteBOT Feb 7, 2021, 3:36 PM

#

Darpleon

📎 503601869570965524.png

acoustic notch Feb 7, 2021, 3:38 PM

#

$\cosh{sq}=\sec{x} \ \sinh{sq}=\tan{x} \ \tanh{sq} =\sin{x}$

somber coyoteBOT Feb 7, 2021, 3:38 PM

#

Darpleon

📎 503601869570965524.png