geometry-and-trigonometry

[shy shale]

Whys it similar

Like how

[crude kraken]

Mhm

[upper karma]

it is similar because all the angles are equal

[crude kraken]

Because the two sides are scalable and the angles are the same, it's similar

[upper karma]

and the sides follow a proportion

[shy shale]

It says if they are, write a similarity statement and give the scale factor

[upper karma]

if you multiply to sides of the small one by 1.2, you get the big one

scale factor = 1.2

[crude kraken]

(or 6/5)

(or for going from big to small it's 5/6)

[upper karma]

statement: Triangle VUW ~ Triangle SRT

[shy shale]

Ok thanks for explaining the whole process. And around that time got me confuse. So while you guys were explaining had to to take some notes and write something down on my paper.

[crude kraken]

=tex \frac{18}{15}=\frac{12}{10}\text{ and }32=32\therefore\Delta VUW \sim \Delta SRT\therefore\text{ The two triangles are similar.}

[shy shale]

Oh

Yeah now thats good more understanding clarification of mathbot

[crude kraken]

No problemo

btw, those three dots are therefore, you'll need to use it a lot so be warned :p

[shy shale]

Isnt the 3 dots look like a triangle?

[crude kraken]

Yea

[shy shale]

Oh ok

[crude kraken]

But it means therefore

And you get the scale factor, right?

[shy shale]

Thx for helping on number 7 and thanks to asian doge for helping me too. It helped alot the way you two explain well.

And I got another question I need help with too.

Ima send pic

[crude kraken]

mk

[shy shale]

I need help with number 8

Im too sure what the question is asking?

[crude kraken]

Me neither?

Oh, it's talking about the next question

or not...

[shy shale]

No for number 8 im asking

It doesnt answer to the next one

[upper karma]

@shy shale same technique my friend

[shy shale]

Thats different

[crude kraken]

🤷

I don't even get what it's asking for xP

[shy shale]

Wait so @upper karma how I do 8 again?

So basically for 8 its not similar?

[upper karma]

Find the ratio of corressponding sides

if they are equal they are similar

if not then no

i havent done the calculations yet

but that's the method

[shy shale]

Ok

So would it be equal?

[upper karma]

i dont know, did you calculate it?

[shy shale]

Im trying

[crude kraken]

OH

It's just asking you to prove the next two things are similar in the same manner as above ;p

[shy shale]

Oh

Oh ok same for 7

[crude kraken]

Same as 7 I think, do the math

[shy shale]

Ok

I think its similar?

Would it be?

[upper karma]

what do you have for the ratio?

no they are not similar

dude

[shy shale]

Whys it not similar?

[crude kraken]

Do you understand what it means for two shapes to be similar?

[shy shale]

Cause their the same shape that are similar meaning equal sides of the angles???

[crude kraken]

When two figures are similar, the ratios of the lengths of their corresponding sides are equal.

[shy shale]

Ok

[upper karma]

and corresponding angles are equal..

[shy shale]

Alright

[upper karma]

21.6/4.68 ≠ 9/1.8

[crude kraken]

If the corresponding sides are equal, the corresponding angles must also be equal

[upper karma]

alright?

so they are not equal

[shy shale]

So if its not equal not similar?

Then?

[upper karma]

yes

if the ratio is not equal

it is not similar

yes

[shy shale]

So for 8 I just say its not equal

Not similar?

[upper karma]

not similar

[crude kraken]

=tex \frac{QR}{VT}\neq\frac{RS}{TU}\therefore\Delta QRS \not\sim \Delta UVT

[upper karma]

the ratios are not consistent

[crude kraken]

except it uses squiggly lines instead of straight ones at the end, I just forget the symbol

[upper karma]

yeh there u go dromas

[shy shale]

Oh ok that helped.

I like mathbot though

[upper karma]

there u go mate

[crude kraken]

But I encourage you to think why this is actually so, because it will help a lot in later places.

[shy shale]

Ok thanks bro. Hope it actually does help for the near future to.

[upper karma]

too* lol 😂 😅

[crude kraken]

🤔

[shy shale]

Oh lol

[crude kraken]

I guess I'm going to the near future

or is the near future coming to me?

[upper karma]

lel

[crude kraken]

What if there's two near futures? 😮

[upper karma]

@crude kraken what maths are you in

[crude kraken]

This is too confusing D:

According to my transcript, pre-calc

but I'm studying stuff ahead :p

[shy shale]

Omg lol boi we are in the year of 2017

[crude kraken]

so I'm in some sort of mix of precal and cal

[shy shale]

So I guess the future is after 2020 somewhere around their or after the year of 2030?

Idk

[crude kraken]

*there

😛

[upper karma]

@crude kraken high-five im in precal too, what grade u in?

[crude kraken]

11

[upper karma]

oh ok

lol

im a freshman

[crude kraken]

You're 12

oh

[shy shale]

Damn you one grade ahead of me

[crude kraken]

oh you're literally 12

[shy shale]

Im in 10th grade a sophomore

[upper karma]

no im not twelve

u shit

[shy shale]

In high school

[upper karma]

chill

no insults

hai senpai

sumimasen

Gomen gomen

[shy shale]

Oh also I need help with a ither question too

[crude kraken]

Alright

[shy shale]

Number 10

[upper karma]

@shy shale actually im 4 grades ahead of u, since u r a soph in geoemtry, im a freshman in precal

[shy shale]

Need help

[upper karma]

man samething

[shy shale]

Well what im guessing it is similar

Since the the triangles are vertical angles

Of the V and the teacher told me about it

So is it similar?

@upper karma @crude kraken

[crude kraken]

er

If my intuition is correct, yes

[upper karma]

they are similar yes

[crude kraken]

but if not, then not enough information given

[shy shale]

Oh yay was right it is similar

[upper karma]

there is enough info....

[crude kraken]

but I have absolutely no clue what theorem says "Two lines in a plane crossing are symmetrical goddamnit, and becasue the angles are the same across ways teh triangles must be similar"

[upper karma]

given 2 equal angles, u can figure the other one out by vertical angle

therefore similar by AA

[shy shale]

Also need to write a similarity statememt and postulate of theorem

Ok angle angle

@upper karma

[upper karma]

yeh angle angel

the statement just similar as above

i wont just give u the answer

[shy shale]

So how would I write it for the statement

[upper karma]

find corresponding angles

write it in similar order

put sqiggly line

[shy shale]

So is it HG~JK

.

[upper karma]

no goddamn it

you state the whole triangle

with triangle notation

[shy shale]

Like how would I state it

@upper karma

[upper karma]

.....

Triangle MKJ ~ Triangle MHG by AA

[shy shale]

And how do I state postulate or theorem?

@upper karma

Where everyone at?

[gritty flare]

🤔

@shy shale Like there is a theorem: a^n+b^n = c^n if n>2?

[shy shale]

Nvm dude someone is helping me in the other discord server on homework help

[gritty flare]

😦 ok

[shy shale]

I was just asking where was everyone at? Cause I was afraid this channel might go dead since no was talking.

[gritty flare]

this is quiet if Sev and Ann arent here

[shy shale]

Ik

I hate when channels like this go dead

[gritty flare]

where do you get help doe

:#

[upper karma]

@DromasGamerHD#3722 get you state it by either brackets or use the word "by"

do what your teacher ask u to do

[crude kraken]

Did I calculate this area right?

[thick coyote]

does anyone know any interesting uses of hyperbolic geometry in real life?

[quasi lion]

thermic transfer

I have it somewhere in my laptop

@thick coyote

[thick coyote]

thanks

[keen aspen]

@crude kraken Is that a Limacon?

[dark sparrow]

it's a cardioid

[upper karma]

This equation gives a hyperbola, right?

yes

k, thx. Thats all I needed. Was on my phone and maths came up.

ill disappear back into the shadows. bye.

[white swift]

=wolf plot y^2-x^2=1

[charred spear]

Query made by @white swift
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=plot+y^2-x^2%3D1
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

[white swift]

🤔

[thick coyote]

Three planar regions A,B,C partially overlap each other, with |A|=90, |B|=90, |C|=60 and |AuBuC|=100, where |•| denotes the area. Find the minimum possible |AnBnC|.

Can anyone help me with this

[upper karma]

The midpoints of the sides of a triangle are (5, 4), (8, 5), and (6, 0). Find the coordinates of teh vertices of the triangle.

[dark sparrow]

what have you tried so far?

[potent linden]

@thick coyote isn't it just be 40?

[thick coyote]

Yeah that's what I got, was just checking

How did you find it

[potent linden]

well, for the union to be 100, then the intersection of one 90 and the 60 has to be either 60 or 50, and 50 is a smaller number

so then you tack on the other 90, and align it so it drops it down to 40

[past mantle]

Is sin(x) the same as $$sin(\theta)$$?

[dark sparrow]

uh

no?

[past mantle]

Ok, I was just thinking that XD

[dark sparrow]

i mean they're the same function, just with different variables as arguments

[past mantle]

Ok, I see.

Variables don't make the function change

[dark sparrow]

the function is still sin()

[past mantle]

im confused on 34

[strange lintel]

Is it 16? OwO

Alright, let’s name them first.

One sec

[copper valve]

it is

[past mantle]

ik u cant read it

[strange lintel]

So, we take Triangle ADB and find AD

[copper valve]

u can use Pythagorean theorem for the right triangle 5-?-13

[strange lintel]

Pythagoras theorem

Then take the triangle ADC

[copper valve]

then use the ? to figure out the length of x in the x-?-20 triangle

[strange lintel]

¯_(ツ)_/¯

[past mantle]

i did $$13^2+(x+5)^2=20^2$$?

[copper valve]

oh be careful :o

[past mantle]

?

[copper valve]

do u have a right angle for that :o

[strange lintel]

Aight.

[past mantle]

no

[strange lintel]

It’s 11

:D

[copper valve]

o yea for some reason i thought x was DC not EC

[past mantle]

so u use Pythagorean theorem twice is what im getting

[copper valve]

ya

[past mantle]

ok

[past mantle]

seems easy enough

i did it so badly

[strange lintel]

Well, just write the given statements.

Draw the triangle.

And figure it out.

[past mantle]

oh, i get it now

got it 😄

[strange lintel]

x=11

[past mantle]

i thought it way to much XD

i thought to use the equation up there

3 problems left and they only ask for perimeter and area XD

[strange lintel]

Nice.

[past mantle]

even though there is one side missing

should be easy

[strange lintel]

@copper valve That’s too accurate. Lol.

Nvm.

[copper valve]

?

[past mantle]

== sqrt(80)

[charred spear]

8.94427191

[past mantle]

im stumped for the area of this

i have $$A=8*4\sqrt(5)$$

[white swift]

sqrt5 ?

[past mantle]

ye

[white swift]

=tex \sqrt5

[past mantle]

oh

i dont understand how?

[white swift]

looks impossible to me

[dark sparrow]

it is possible

[white swift]

its equillateral 🤔

[dark sparrow]

no

[white swift]

what do u call them 😢

[dark sparrow]

isosceles

[white swift]

two-side-equal

[dark sparrow]

okay so if the top vertex is A, the base is BC from left to right, and that height is BH

[white swift]

that's the one

[past mantle]

is the base then 10 and 5?

[dark sparrow]

no

you can calculate AH

[past mantle]

oh

[dark sparrow]

using the pythagorean theorem

then HC

then BC

[sudden helm]

I have problem here. Triangle edge lenghts are
a=39
b=20
c=45
I have to calculate bisectrix(bisector)
what formula i have to use and could someone explain it please?

bisectrix faces to edge a

[sudden helm]

it's angle bisector

[tropic stirrup]

w/e, I was too lazy to give explanations

[copper valve]

nice chalkboard

so smooth

[sudden helm]

hey! thanks!

[tropic stirrup]

No problemo ^^;

[sudden helm]

@tropic stirrup one question. what are those lines above formula?

[tropic stirrup]

Oh it just means "segment"

in here the line segment itself and its length are both expressed by the lines above the letters

[sudden helm]

because i think that in my school we learn it somehow differently but i might be wrong

[tropic stirrup]

Nah it's just the difference of notation

[sudden helm]

because i have never heard of Stewarts theorem

[tropic stirrup]

Oh

It's good and useful to know that formula

[sudden helm]

okay i will keep that in mind

[tropic stirrup]

The proof of it can be searched easily through the internet

[sudden helm]

i know

[tropic stirrup]

Uses Pythagorean

[sudden helm]

which grade usualy learn this theorem if you know

[upper karma]

5th

[sudden helm]

in usa?!

[tropic stirrup]

Dunno, I just learnt it by myself

[sudden helm]

i go in 10th

[tropic stirrup]

And nah that's a lie over a cake

[sudden helm]

im from europe btw

[tropic stirrup]

American math is slow
Don't worry

[sudden helm]

ok

i see cos in one proof. is that normal?

[tropic stirrup]

Ye

Anyways I need to finish some stuffs

So I need to depart for now

[sudden helm]

ohh i remember something about this. i think teacher taught us this theorem but didn't call it steward's

ok bye

[tropic stirrup]

Pfft

Buhbye~

[upper karma]

How can you tell if a slope is positive or negative? Am I looking at the picture to determine that or the answer?

[umbral snow]

Is the slope going up or down? What was the original question

[upper karma]

Find the slope of (0,0) & (9,3)

Am I determing that from the origin?

[umbral snow]

You mean find the slope of the line connecting (0, 0) and (9, 3)?

[upper karma]

Yes

Can I just look at what quadrant it's in?

[umbral snow]

m = rise / run
m = difference in y / difference in x
m = (3 - 0) / (9 - 0)
m = 3/9 = 1/3

[upper karma]

Yes, I know how to find slope.

I just need to determine if that's a negative slope or a positive slope.

Am I looking at the actual answer to see if that's positive or negative?

[umbral snow]

Well, 1/3 is a positive number

[upper karma]

Or am I looking at the line on the graph?

[umbral snow]

So we have a positive slope

[upper karma]

ah thanks

So, for (-6, 0) and (0, -3), I would be getting a negative slope?

Since the slope is a negative number?

[copper valve]

ya

[rare ravine]

i get no answers in #help-1 so i'll ask here :
here's the pb :

You have to set 5 points to build an elliptic function of [-5;4] interval:

• at 0,0
• at 31,2
• with x at 127 and u have to find y
  The 2 other points so that the function has the 31,2 point as maxima (Those points will have a x<0 & y<0)
  And the sum of all entire x from 0 to 127 included has to be at -128
  What's the quadratic formula of this function ?

[dark sparrow]

christ, inserting notation right in the middle of a sentence

[upper karma]

I forgot what alt., con., and int. are short for

[dark sparrow]

alt and int are alternate and interior i think

idk what con might be short for

[upper karma]

corresponding

[dark sparrow]

con, not corr

[upper karma]

ohh

consecutive or something maybe?

[dark sparrow]

this is all

full of overly-fancy terms and notation

[upper karma]

exactly why i need help with it lol

I mean, i get the logic behind the math and everything and can do it when its in like basic terms

but not this

you get the hang of the terms by practicing

[dark sparrow]

except these are overly fancy

[west dune]

Hey guys,

DC = 12 cm
AB = 6 cm
AD  = 8 cm
ABCD is a trapeze
AB || DC
And the distance between AB to DC is 8 cm

* I proved that triangle AEB similar to triangle DEC

I need to find ABCD's size, how to?
I don't know that AD is height

[final prairie]

למה אתה מתכוון בגודל של ABCD?

שטח?

@west dune

[west dune]

@final prairie כן

[final prairie]

יש נוסחא לשטח של טרפז

הגובה כפול סכום הבסיסים חלקי שתיים

אז הגובה כן הוא 8 כי ידוע שזה טרפז

הבסיסים הם 6 ו-12

[west dune]

לא ממש לא

אתה לא יודע ש8 זה הגובה

תאורתית זה כן

אבל מה המשפט

שאני אגיד
AD גובה

[final prairie]

אם ABCD טרפז ו AB||DC אז AD הוא גובה

זה בהכרח

אה רגע

הבנתי מה אתה אומרר

[west dune]

אז? 😅 @final prairie

[final prairie]

אין עוד נתונים?

[west dune]

אני אשלח לך כי אתה יודע עברית דקה

[final prairie]

@west dune אה הבנתי עכשיו

[west dune]

אוקיי

[final prairie]

״המרחק בין הבסיסים״ זה בעצם הגובה

אבל לא בקשו שטח של הטרפז

[west dune]

יש לך משפט לזה?

[final prairie]

כשאומרים ״מרחק״ בין שתי ישרים זה בעצם המרחק הכי קצר, שזה הגובה

אין משפט זה הגדרה

זה כבר יותר הבנת הנקרא אפילו

אז שטח הטרפז זה שתי שניות

אבל יותר קשה זה שטח המשולשים

[west dune]

כן אבל אני לא יכול להוכיח ש

AD 
DCאנך ל```
בלי משפט

[final prairie]

נכון

אתה גם לא צריך

אתה רק צריך את הגובה

והוא נתון

אז גם תמחק את הסימון שלך ש AD=8

כי אתה לא יודע שהוא גובה

[west dune]

?

[final prairie]

סימנת AD=8cm

[west dune]

אז?

[final prairie]

זה לא נכון

כי אתה יודע שהגובה הוא 8

אתה לא יודע שAD גובה

[west dune]

אוקיי

[final prairie]

אז שטח הטרפז הוא

8(12+6)/2=4*18=72

[west dune]

=tex 8(12+6)/2=4*18=72

[west dune]

כן

ומה המשפט שלי?

[final prairie]

איזה משפט?

זה הנוסחא לשטח טרפז

[west dune]

?מרחק בין מקבילים הוא גם הגובה

[final prairie]

זה הבנת הנקרא, המילה ״מרחק״ זה פשוט אורך הגובה

מיותר לציין

תכתוב ״המרחק הוא גם הגובה, שנתון שהוא 8״

[west dune]

שנינו יודעים שמתמטיקה זה נושא מתחכם אז גם שאתה מבקש עזרה צריך להתחכם בהתאם

איאפשר סתם לרשום דברים בהוכחות

אבל תודה, הבנתי אותך

אני אצטרך עזרה בעוד משהו אחרי זה

[final prairie]

המרחק בין שתי ישרים הוא אורך הגובה

הניסוח שלהם קצת חרא פשוט, זה סתם בשביל לבלבל

אתה עכשיו גם צריך לחשב את שטחי המשולשים

[west dune]

Oriented shit
Like we like to call it when programming 😄

[final prairie]

בין שתי ישרים מקבילים **

חח

חישבת את יחס הדמיון במשולשים?

[west dune]

כן, יצא לי חצי

ויחס שטחים זה

=tex (0.5)^2

[final prairie]

בדיוק כן

אתה תסתדר עם השטחים?

[west dune]

'דקה, אני רק סיימתי את א

אני אוהב להבין איך לפתור הכל ולהבין את השאלה מאשר לרוץ לפתור

[final prairie]

Take your time :p

[west dune]

:}

[final prairie]

חח

אני הולך לכמה דקות

אז @ אם אתה צריך עוד משהו

[west dune]

👍

cya

[hard atlas]

Wait, Hebrew is also written right to left? :o

[west dune]

Yes, It's the first language in histroy to be written RTL.

[hard atlas]

Now I did not know that

[west dune]

A day without learning something new is a wasted day @hard atlas

[hard atlas]

True 😀

[west dune]

@final prairie חזרת אני רואה

חחח

[final prairie]

כן חח

Stalker :p

[west dune]

לא... בדיוק בדקתי אם התחברת

😂

יש לי טיימינג טוב

[final prairie]

חח

[west dune]

שכח מהשאלה למעלה אני צריך עזרה במשהו אחר 😬

[final prairie]

הסתדרת? חח

[west dune]

ב

[final prairie]

jesus fuck

[west dune]

'אני לא מצליח את ב

?'יש לך רעיון איך לעשות את ב

[final prairie]

אממ

[west dune]

זה היה קל אם זה היה
EOC = EDO אבל זה לא

[final prairie]

EDC=90º

[west dune]

נכון

[final prairie]

אוקי אז ככה

נכון יש משפט שאומר שאם שתי ישרים מקבילים אז כל שתי משולשים עם אותו בסיס על אחד מהם ונקודה על השני הם שווי שטח?

יודע מה אפילו יותר פשוט

[west dune]

נפלת לזה גם נראה לי

[final prairie]

אז זהו תראה ככה

ED

הוא גובה במשולש EOC

[west dune]

רוצים את אלו

[final prairie]

אני יודע אבל יש לי טריק

[west dune]

אוקיי

[final prairie]

EOC

ו

EOD

בעלי אותו שטח

[west dune]

כן

[final prairie]

כי יש להם בסיס וגובה משותף

אז זהו

תחסיר S∆EOP

משתיהם

וקיבלת OPC=EPO

[west dune]

חיסור גדלים שווים מגדלים שווים נותן גדלים שווים?

רגע מה

[final prairie]

שטחים

ברור..

[west dune]

אוייי נכון

תודה רבה לךךך

[final prairie]

=tex a=b\implies a-c=b-c

[final prairie]

Np

[west dune]

מזל שיש את השרת הזה עם אנשים שמוכנים לעזור

[final prairie]

Yep

יש לך עוד גיאומטריה?

כי אני לא סובל גיאומטריה חח

[west dune]

אבל איך אני מוכיח ש
EOC = EOD
תאורתית אני יודע איך אבל מעשית בהוכחה מה אני אצטרך לרשום

?מישהו כן סובל גיאומטריה בימינו

[final prairie]

יש להם בסיס משותף

EO

ויש להם את אותו הגובה

לכן יש להם אותו השטח

הגובה של שתיהם ED

[west dune]

אז אני פשוט ארשום במשפט
משולשים בעלי בסיס משותף וגובה זהה הם שווי שטח

[final prairie]

טוב אני הולך להכין אוכל :p

[west dune]

ביי, תודה

[final prairie]

כן זה בסדר

אין בעיה

[west dune]

בתאבון

[final prairie]

תודה חח

[west dune]

😉

חח לא אמרו ש
DC || EO
אבל ניתן להוכיח זאת בעזרת
שני ישרים המקבילים לישר שלישי מקבילים בינהם

[final prairie]

בסדר כן

טרנזיטיביות קוראים לזה דרך אגב

If $$AB||CD$$ and $$CD||EF$$ then $$AB||EF$$

[sudden helm]

Right triangle sides are a=18 b=24 c=30
And i have to calculate narrow-angle bisector lenght.
what formula do i have to use, if i have to?

looks like this

[gritty flare]

is the 18/30 angle split in half?

[sudden helm]

yes and 24/30

[gritty flare]

angle bisectrix theorem

:/

maybe

[sudden helm]

there is such thing?

wait

[gritty flare]

yes of course

[sudden helm]

maybe

maybe you're right

[gritty flare]

oh

i think i got it

narrow angle bisector splits 18 into 2 halfs

[sudden helm]

are you sure?

[gritty flare]

judging by it

:/

[sudden helm]

i have answers but only answers

do you want to know?

[gritty flare]

tell me the answer

[sudden helm]

ok one is 9√5 and other one is 8√10

[upper karma]

I got 8 root 10

[sudden helm]

you got it?

[gritty flare]

square root of 657 is what i got 😦

[upper karma]

And I used cosine formula

[sudden helm]

i have to find that formula then probs.

[gritty flare]

got ittt

😄

[sudden helm]

cool

[gritty flare]

so

[shy shale]

hey I need help

[gritty flare]

wait Dromas

[gritty flare]

omg

dud

[shy shale]

I need 3 more questions to finish

[gritty flare]

angle d = 47

[upper karma]

The cosine rule. We can use thecosine formula to find the length of a side or size of an angle. For a triangle with sides a,b and c and angles A, B and C the cosine rule can be written as: a2 = b2 + c2 - 2bc cos A.

@sudden helm

[gritty flare]

WHAT?

COSINE FORMULA?

WE DONT HAVE ANGLE DOE

we can inverse it

fuck

i forgot

[sudden helm]

@upper karma thanks!

[upper karma]

We don't need the angle

[shy shale]

ok someone dm me I need help though

I dont want to get confused

[upper karma]

Post the question dromas

[shy shale]

here

Idk how should I start this off

[sudden helm]

x is 90

if im not wrong

[shy shale]

yeah x should be 90

cause its a triangle being split in half

I think

[sudden helm]

y = 180 - 90 - 47

[gritty flare]

i told him

[shy shale]

ok

[gritty flare]

as it is that triangle

[sudden helm]

do you understand why?

[gritty flare]

i forgot the name

ugh

[sudden helm]

dromas

[gritty flare]

so angle C is parralel to Angle B

[gritty flare]

so angle C = angle B

[shy shale]

ok I have that

[gritty flare]

thus being equal to 47

😄

yay

gj

[shy shale]

so its not 43?

[mild cargo]

It is 43

[shy shale]

hello?

ok

good

ok got 2 more questions and im done

[mild cargo]

However, angle D is not 47

I think you meant angle B

[shy shale]

35 I need help

[gritty flare]

can u zoom

;=;

[shy shale]

you going to have to read it

[gritty flare]

i cant see it

[mild cargo]

What have you tried?

[shy shale]

ok

[gritty flare]

i cant see a damn word in 45

[shy shale]

can you read 35

[gritty flare]

35*

[shy shale]

35

thats clear

[mild cargo]

What have you tried so far?

[shy shale]

trying to think

[gritty flare]

omg cant you take picture up close

I think that is Law of Sin word problem

[gritty flare]

there

[shy shale]

their took a picture side ways

like can you explain how to get to the answer, cause I don't understand how to do number 35

[mild cargo]

Yeah

Working on it

You know about similarity right?

[upper karma]

a. ✓10?

[shy shale]

no

[mild cargo]

I will show you

[shy shale]

ok

[mild cargo]

Now you can work out the equation

[upper karma]

U know basic trigonometry? Dromas?

[mild cargo]

You don't need trigonometry

[gritty flare]

you redrew it wrong

or did you

[mild cargo]

How?

[gritty flare]

am i seeing it incorrectly

[mild cargo]

I did it correctly

[gritty flare]

am i blind

no one will ever know

[upper karma]

Ik but I'm using trigonometry and that's easier that way IMO

[shy shale]

so when you input the numbers I do cross multiply to find my answer?

[mild cargo]

You need to imput values first

[shy shale]

ok

[mild cargo]

What is length DC and what is length BC?

[upper karma]

Hence , ✓10.....🙃

[mild cargo]

But he does not know trigonometry

[gritty flare]

archytext

[mild cargo]

It is better to do it this way

[gritty flare]

look

[mild cargo]

yeah

[shy shale]

i this far

[mild cargo]

Call the unknown value x

[gritty flare]

I dont know trig law that can actually compute that

:/

[upper karma]

x^2 = 10

[shy shale]

like that?

square root 10?

[mild cargo]

No @upper karma

Oh nevermind

Yes

It is correct

[upper karma]

Also , I used tantheta and u used similarity , same idea on both our approaches , almax

[mild cargo]

Yeah

So you get sqrt(10) as the answer

[shy shale]

when I square root 10 I get 3.16227766

[upper karma]

I don't remember similarity now , so used tantheta

[mild cargo]

Yeah

I haven't learned tantheta hehe

[shy shale]

so the decimal is my answer?

or no

[upper karma]

Yes

[mild cargo]

you can just write it as sqrt(10)

[upper karma]

^

[shy shale]

oh ok

ok last question I got

[gritty flare]

Anarchetect

one question

which angles tangent did you use

[upper karma]

The angle at home

[shy shale]

This last question I need help

[mild cargo]

Do you know the formula for volume of sphere?

[upper karma]

And the angle at football field , both are complementary

4/3 pi r^3

[gritty flare]

PiR^2 is volume

[mild cargo]

pir^2 is area

[gritty flare]

no

thats area

english math confusing

[gritty flare]

Volume = V

right

[mild cargo]

=tex 4/3pir^3

[shy shale]

I think its that

[mild cargo]

shit

[shy shale]

so how I do it

[mild cargo]

=tex \frac{4}{3}\pir^3

[charred spear]

Rendering failed. Check your code. You can edit your existing message if needed.

[gritty flare]

=tex \frac {4π}{3}4^3

[mild cargo]

=tex \frac{4}{3}\pi r^3

[upper karma]

Dromas's pic of the formula is correct

Ahh finally :D

[shy shale]

yeah Ik

my teacher taught it

[mild cargo]

Having a hard time with latex

You just plug in the radius into the formula

And solve

[gritty flare]

=calc (4pi/3) * 4^3

[charred spear]

Failed to parse equation: Invalid syntax at position 2

(4π/3) * 4^3
  ^

[shy shale]

like that?

[mild cargo]

Yeah

[gritty flare]

=calc (4pi/3)*64

[charred spear]

Failed to parse equation: Invalid syntax at position 2

(4pi/3)*64
  ^

[upper karma]

You'll get the answer in cm^3 . Just remember

[gritty flare]

dam

[shy shale]

for 4 to the third I get 64

[gritty flare]

:/

[shy shale]

for 64 Do I times 4 and divide it by 3 and times pie?

[gritty flare]

268 cm^3

4 pi devided by 3 times 64

[upper karma]

No Alex that conversion is wrong

It's cm^3 not cm

[gritty flare]

:/

ye

[shy shale]

oh ok I got got it

[gritty flare]

0.268dm^3 right?

:/

0.00268m^3 :/

[upper karma]

Ahha I don't remember conversions

[shy shale]

I this so far

[gritty flare]

1 m = 10dm = 100 cm

[shy shale]

had to calculate it

[gritty flare]

1m^3 = 1000dm = 1000000cm^3

[shy shale]

pretty sure im correct

[gritty flare]

that 0

bugs me

thats why i said app. 268

[shy shale]

yo thank you guys for helping me the 3 of the questions it helped

thanks

[gritty flare]

i learned a lot

thanks to you to

😄

[crude kraken]

Can someone explain the difference between a parabola and a hyperbola

[dark sparrow]

a hyperbola has asymptotes

a parabola doesn't

but also: a parabola is the locus of all points equidistant from a focus and a directrix (line)

a hyperbola is the locus of all points such that the difference in distances from each point on the hyperbola to the two foci is constant

[crude kraken]

Uh huh...

So how by looking at one can you tell the difference?

[dark sparrow]

a parabola has no asymptotes and only has one arm

[crude kraken]

By that image, the hyperbola does as well

[dark sparrow]

no

that secant plane intersects the other half of the conical surface

[rare ravine]

Hi, i found some solution to my pb

but i need ur help

Here's my function : A abs((B ln(C x + D)) / (x + E)) - F
I get this kind of courb :

how to make the maxima a little more at right plz ?

(A to F are constants)

[upper karma]

You want to shift the graph to the right?

Or exactly what do you mean?

[rare ravine]

i want to shift the courb to the right

but if i do it directly, i get no values at little abscissas

[upper karma]

uhmmm does A abs((B ln(C (x-1) + D)) / ((x-1) + E)) - F

not work?

or something like that?

[rare ravine]

x+1, not -1

[upper karma]

oh ok

[rare ravine]

but no

it doens't resolve

[upper karma]

I dont know what is happenin in that case then 🤷

[rare ravine]

dw

someone ? :/

[rare ravine]

it's resolved now ^^

[past mantle]

i dont understand number 9

[white swift]

what is radical form 🤔

[past mantle]

=tex \sqrt(x)

[white swift]

o yes right

radical as in sqrt

its isoceles 🤔

have you done trigonometry? 🤔

[past mantle]

no, im not in trig yet

[white swift]

how the fuck do you do them then lol

have you done 10

o pythagorean?

[dark sparrow]

=tex \sqrt{x}

[dark sparrow]

@past mantle

anyway

these are all isosceles right triangles

their sides are in the ratio 1 : 1 : sqrt(2)

[past mantle]

maybe since its a 45-45-90 triangle, maybe $$15 = l\sqrt{x}$$

[dark sparrow]

\sqrt{x}

[past mantle]

x should be 2

in the square root

[dark sparrow]

well then write \sqrt{2}

lol

[past mantle]

i got $$\frac{15\sqrt{2}}{2}$$

[dark sparrow]

which one?

[past mantle]

is that right?

[dark sparrow]

9?

[past mantle]

for 9

[dark sparrow]

yeah

[past mantle]

😄

i did in my work $$\frac{15}{\sqrt{2}}*\frac{\sqrt{2}}{\sqrt{2}}$$

...

[dark sparrow]

backslashes!

[tranquil ridge]

https://cdn.discordapp.com/attachments/269573202018041856/385865763832987648/image.jpg

I'll try here

Can any1 help me with this?

[west dune]

We don't know what's the data which were given

[tranquil ridge]

I need to count the area of A

[west dune]

Again... give us the question's data

[tranquil ridge]

uhm

do u mean i should translate the question

sorry im not a native speaker but I can translate it if thats what you mean

[west dune]

So translate please

[tranquil ridge]

So the quadrilateral is divided to four pieces by drawing diagonals. Three areas of the pieces have been marked. Count the area of the fourth piece A

Tell me if you cant understand

[modern sonnet]

@tranquil ridge

Yo

[tranquil ridge]

hey

[modern sonnet]

I’m here

But firsy

*first

I need your help to label all vertices on that diagram

Including the intersection of the diagonals

Just to make it easier to communicate

[tranquil ridge]

Uhm

[modern sonnet]

Label whatever way you want

[tranquil ridge]

just a sec

[modern sonnet]

Take your time

[tranquil ridge]

So I need to name 4

right

[modern sonnet]

5

There’re 5 points to name

[tranquil ridge]

Whats the fifth

[modern sonnet]

I said vertices =.=

Hang on

[tranquil ridge]

Im sorry english isnt my native language xd

[modern sonnet]

I’ll do it myself

Give me 2 minutes

[tranquil ridge]

Oh you meant edges

Well Im in no hurry I just want to solve this

[modern sonnet]

No not edges.

Points

[tranquil ridge]

Ah sorry

[modern sonnet]

So

You wanna find the area of DEC

Okay

[tranquil ridge]

Yes

[modern sonnet]

First, have a look at triangles ABE and CBE

[tranquil ridge]

mkay

[modern sonnet]

They share the same perpendicular height dropped from B

[tranquil ridge]

yep

wait doesnt perpendicular line form a right angle

[modern sonnet]

See the line in blue?

[tranquil ridge]

yes

[modern sonnet]

(It’s not the same line as BC by the way. Let’s call it BH)

[tranquil ridge]

yeah

[modern sonnet]

Then area = 1/2×height×base

So Area of ABE = 1/2×AE×BH

Area of CBE = 1/2×CE×BH

In other words,

[tranquil ridge]

Im still with you 😄

[modern sonnet]

=tex \frac{\text{Area of ABE}}{\text{Area of CBE}} = \frac{AE}{CE}

[modern sonnet]

Oh hey it worked

[tranquil ridge]

thats noice

[modern sonnet]

So now you can find AE/CE

The ratio

Now, by the same logic,

=tex \frac{\text{Area of ADE}}{\text{Area of CDE}} = \frac{AE}{CE}

[modern sonnet]

In other words,

=tex \frac{\text{Area of ABE}}{\text{Area of CBE}} = \frac{\text{Area of ADE}}{\text{Area of CDE}}

[modern sonnet]

And so you can find the area of CDE because you already know the other 3

Done!

[tranquil ridge]

ill try to get the answer

[modern sonnet]

It’s just calculator work

[tranquil ridge]

yep

I got it!

Thank you so much

[modern sonnet]

Is it 9.6

I’m trying out quick mental mafs

[tranquil ridge]

It is

[modern sonnet]

2+2 is 4

Oh wait that meme is banned

Okay!

[tranquil ridge]

dont say it 😄

[modern sonnet]

Glad I could help

[tranquil ridge]

yeah thanks a ton

never could have solved that by myself

[shy shale]

So I went to tutoring today after school and the teacher was helping me for geometry before the test

And it helped

[past mantle]

If there is a 45-45-90 triangle, is the hyponeuse $$l\sqrt{2}$$

[upper karma]

if anyone needs like help in physics, chemistry, biology, etc lmk

[red citrus]

hi guys

im a freshman and im taking Honors Geometry

can any1 help me understand rotations on a coordinate plane and how a-(a-c)=2a-c

[umbral snow]

a - (a - c)
= a - a + c
= c
It doesn't

[red citrus]

wtf

my teacher made like 99 problems like that

[bronze harbor]

@red citrus ur a freshman and ur 10

xd

[red citrus]

yes man imma genius

@umbral snow

[bronze harbor]

lies

[red citrus]

look

bottom problem

[bronze harbor]

i hate geometry so much ngl lol

[red citrus]

imagine taking honors

thats like geometry mixed with algebra II

[bronze harbor]

i do well in algebra

but when it comes to graph and geometry

🤦

[red citrus]

lol

[native night]

I like how your grader drew clouds over the incorrect parts

[red citrus]

Lol

[upper karma]

random question: when you're adding segments are you supposed to put the bar thingy on top?

if anyone needs help in calculus/or math in general i can help(edited)
im on a lot these days
but please PM me tho, im in so many channels i forget to come here
just wanna help since my projects are finished 😃

but if someone can answer for them that works for you but if you see me online shoot me a question

[dark sparrow]

@Greenie#4023 adding segments?

@upper karma

[upper karma]

@dark sparrow like Segment AB=Segment AC+CB

Do you do AB=AC+CB or actually put the line on top

[dark sparrow]

do you mean vectors?

bc if you're talking about segment lengths then that's only true when C is between A and B

[upper karma]

yeah

but notation

[dark sparrow]

anyway... notation

ffhh

i'd prob put an arrow over them

it's a bit of a bitch to TeX but

=tex \overrightarrow{AC} + \overrightarrow{CB} = \overrightarrow{AB}

[upper karma]

Oh okay

thank you

[jade breach]

@dark sparrow Isn't that for rays tho?

[dark sparrow]

uh

no?

i've seen [AB) for rays

but arrows are definitely vectors

[jade breach]

I meant the arrow thingy above the letters

[dark sparrow]

yes

those denote vectors pretty much always

[jade breach]

ok

[opaque kayak]

I came up with my own question:

Show that the sum of two gradients A and B, is: https://prnt.sc/hi9tde
for example, shown in this picture: https://prnt.sc/hi965n
You can see that the gradient of the original blue line is 1/2 and of the red line is 1/3
You can see that the new blue line has a gradient of 1 which is (1/2 + 1/3)/(1-(1/2)(1/3))

Am I overlooking any simple solutions to this?

[dark sparrow]

=tex \tan(x+y) = \frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}

[opaque kayak]

that's the intended solution, yes, by converting the gradients to angles such that tan(x) = A and tan(y) = B so x and y are the angles so tan(x + y) gives you the gradient of the new line

but is there anything simpler, eg no trig?

[surreal bolt]

Let complex numbers P, Q be P =a + bi, Q = c+di. (P)(Q) will have angle = sum of the angles of P and Q.
PQ = (ac -bd) + (ad + bc)i

Slope of PQ = (ad+bc)/(ac-bd) = (d/c + b/a)/(1-(b/a)(d/c)) Voila.

=tex P = a + bi, Q = c + di
\ (P)(Q) = (ac - bd) + (ad + bc)i
\ m = \frac{ad+bc}{ac-bd} = \frac{ad+bc}{ac-bd}\cdot\frac{1/ac}{1/ac}
\ = \frac{d/c+b/a}{1-(b/a)(d/c)} = \frac{b/a + d/c}{1-(b/a)(d/c)}

[past mantle]

i dont understand this problem from the new chapter

number 30

[thorn talon]

Well

You can determine the internal angle

Since all internal angles are equal and angle sum

And then x is 180 - internal angle

Since angles on a straight line

Though

I'm pretty sure you can also just divide 360 by 5

[minor quest]

@thorn talon I thought he said 30

[thorn talon]

i can't read

[minor quest]

Xd

[thorn talon]

whoops