help-39

[warm elbow]

oh mannn i hate this class so. much 😭

$ \
\begin{array}{cccccccc|c}
w_1 & w_2 & w_3 & w_4 & w_5 & w_6 & w_7 & M &\
-3 & 2 & 4 & 1 & 0 & 0 & 0& 0 & 1\
-1 & 4 & 3 & 0 & -1 & 0 & 1 & 0 & 3 \
-2 & 0 & -8 & 0 & 0 & 1 & 0 & 0 & 2\
\hline
-1 & 4 & 3 & 0 & -1 & 0 & 0 & 1 & 3\
\end{array}$

[jolly parrot]

toast

[warm elbow]

So i'm guessing here you still eliminate $w_2$

[jolly parrot]

toast

[warm elbow]

\textit{Initial Simplex Tableau} \
$
\begin{array}{ccccccc|c}
w_1 & w_2 & w_3 & w_4 & w_5 & w_6 & Z &\
-3 & 2 & 4 & 1 & 0 & 0 & 0 & 1\
-1 & 4 & 3 & 0 & -1 & 0 & 0 & 3 \
-2 & 0 & -8 & 0 & 0 & 1 & 0 & 2\
\hline
7 & 12 & 10 & 0 & 0 & 0 & 1 & 0\
\end{array}
$ \ \
Initial BFS: $w_1=0, w_2 = 0, w_3 = 0, w_4 = 1, w_5 = -3, w_6 = 2 \implies$ Infeasible. Introduce artificial variable $x_7$ \ \
\textit{Phase I} \
Minimize $M = w_7$\
Subject to \
$-3w_1 + 2w_2 + 4w_3 + w_4 = 1$ \
$-w_1 + 4w_2 +3w_3 -w_5 + w_7 =3$\
$-2w_1 - 8w_3 + w_6 = 2$\
$z + 7w_1 + 12w_2 + 10w_3 = 0$\
$w_1, w_2, w_3, w_4, w_5, w_6, w_7 \geq 0$
$ \
\begin{array}{cccccccc|c}
w_1 & w_2 & w_3 & w_4 & w_5 & w_6 & w_7 & M &\
-3 & 2 & 4 & 1 & 0 & 0 & 0& 0 & 1\
-1 & 4 & 3 & 0 & -1 & 0 & 1 & 0 & 3 \
-2 & 0 & -8 & 0 & 0 & 1 & 0 & 0 & 2\
\hline
0 & 0 & 0 & 0 & 0 & 0 & -1 & 1 & 0\
\end{array}$\
\ Eliminate $w_7$ from Tableau
$ \
\begin{array}{cccccccc|c}
w_1 & w_2 & w_3 & w_4 & w_5 & w_6 & w_7 & M &\
-3 & 2 & 4 & 1 & 0 & 0 & 0& 0 & 1\
-1 & 4 & 3 & 0 & -1 & 0 & 1 & 0 & 3 \
-2 & 0 & -8 & 0 & 0 & 1 & 0 & 0 & 2\
\hline
-1 & 4 & 3 & 0 & -1 & 0 & 0 & 1 & 3\
\end{array}$
\pagebreak
\
Introduce $w_2$ into the solution, Pivot around $a_{12}$
$ \
\begin{array}{cccccccc|c}
w_1 & w_2 & w_3 & w_4 & w_5 & w_6 & w_7 & M &\
-3 & 2 & 4 & 1 & 0 & 0 & 0& 0 & 1\
5 & 0 & -5 & -2 & -1 & 0 & 1 & 0 & 1 \
-2 & 0 & -8 & 0 & 0 & 1 & 0 & 0 & 2\
\hline
5 & 0 & -5 & -2 & -1 & 0 & 0 & 1 & 1\
\end{array}$
\ \
Introduce $w_1$ into the solutions, Pivot around $a_{21}$
$ \
\begin{array}{cccccccc|c}
w_1 & w_2 & w_3 & w_4 & w_5 & w_6 & w_7 & M &\
0 & 2 & 1 & -1/5 & -3/5 & 0 & 3/5& 0 & 8/5\
5 & 0 & -5 & -2 & -1 & 0 & 1 & 0 & 1 \
0 & 0 & -10 & -4/5 & -2/5 & 1 & 2/5 & 0 & 12/5\
\hline
0 & 0 & 0 & 0 & 0 & 0 & -1 & 1 & 0\
\end{array}$

[jolly parrot]

toast

[warm elbow]

@rough forge out of curiosity would this be optimal for the final tableau? since the cost = 0?

[rough forge]

i have no idea but i checked with an online calculator that got -11 as max

[warm elbow]

I just got -11 as the max as well

\textit{Phase II}\
Drop $w_7$ \
$
\begin{array}{ccccccc|c}
w_1 & w_2 & w_3 & w_4 & w_5 & w_6 & Z &\
0 & 2 & 1 & -1/5 & -3/5 & 0 & 0 & 8/5\
5 & 0 & -5 & -2 & -1 & 0 & 0 & 1 \
0 & 0 & -10 & -4/5 & -2/5 & 1 & 0 & 12/5\
\hline
7 & 12 & 10 & 0 & 0 & 0 & 1 & 0\
\end{array}$
\ \
Apply a correction by eliminating $w_1, w_2$
\
$
\begin{array}{ccccccc|c}
w_1 & w_2 & w_3 & w_4 & w_5 & w_6 & Z &\
0 & 2 & 1 & -1/5 & -3/5 & 0 & 0 & 8/5\
5 & 0 & -5 & -2 & -1 & 0 & 0 & 1 \
0 & 0 & -10 & -4/5 & -2/5 & 1 & 0 & 12/5\
\hline
0 & 0 & 11 & 4 & 5 & 0 & 1 & -11\
\end{array}$
\
We have reached an optimum solution of $Z = -11$ at $w_1 = 1/5, w_2 = 4/5, w_3 = 0$

[jolly parrot]

toast

[rough forge]

funny thing is i am also studiying optimization currently but there are so many ways how to do simplex so i cant follow

but it's prob right now

like butter on toast

[warm elbow]

yeah

okay thank you for confirming -11 though

i thought i was tripping hard 😭

[warm elbow]

its really painful

esp when theres a shit ton of fractions

[rough forge]

agree

[warm elbow]

we just learned simplex, big M, and 2 phase

and now we're doign somethign with strong duality theorem and CSp

which it was more theory based though i feel like im just bashing alot

[pearl pond]

@warm elbow Has your question been resolved?

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[grim fractal]

I think it's probably pi or something

[solid pier]

.close

[pearl pond]

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[cobalt hinge]

What are you doing?

[cobalt hinge]

The fuck is this for?

[broken fossil]

wait why was it closed

[solid pier]

please stop being hostile

channel is closed

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[dapper kraken]

given positive integer $k\geq 2$, prove there exists an infinite amount of positive integer $n$ such that both
$$kn+1 \quad\quad (k+1)n+1$$ are squares

[jolly parrot]

skissue.in.a.teacup

[dapper kraken]

my best lead is contradiction with finite n?

but im not sure how to pull that off

[plush bramble]

that or just find the formulas for the squares

[dapper kraken]

formulas for the squares?

[plush bramble]

have you learned
https://en.wikipedia.org/wiki/Fermat's_theorem_on_sums_of_two_squares

special case of k=4

[dapper kraken]

no

lemme read

[plush bramble]

although it doesn't imply 5n+1 is square

[dapper kraken]

hm

[plush bramble]

maybe kn + 1 = s^2 and kn + n + 1 = n + s^2 = t^2 and then some algebra and probably induction

[dapper kraken]

im not sure how primes relate to the question with this?

[plush bramble]

i originally thought if you could prove it for k=4, it could be extended to any k

if p = 4m + 1 = x^2 + y^2 then you have 4m + 1 equal to a sum of squares.

but that was wrong, you want 4m + 1 and 4m + 4 + 1 are separately squares

[versed mica]

i’m not sure how to produce a larger one but i bet it would be useful to suppose it’s finite and thus has a maximum n then try to construct some n larger than it

probably the idea

[pearl pond]

@dapper kraken Has your question been resolved?

[dapper kraken]

if it matters at all the topic was pythagorean triples and pell's equations

[pearl pond]

@dapper kraken Has your question been resolved?

[dapper kraken]

.close gtg

[pearl pond]

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[steady wolf]

how come this is wrong

[pearl pond]

Please don't occupy multiple help channels.

[steady wolf]

oops

.close

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[shrewd rune]

b

what the duck do i do

[toxic lichen]

divide num and denom by x^(2/3)

that, or squint

top behaves like x^(2/3) and bottom is just x^1

[shrewd rune]

how

ohhh

can i firsly take out

cbrt(x^2) /x

and then divide it by x^2/3

[toxic lichen]

that is what i said

[shrewd rune]

but how do i divide

cbrt(x^2) by x^2/3

is the answer is 1

[toxic lichen]

idk what you mean by "how do i divide cbrt"

[shrewd rune]

and then x/(x^2/3)

[shrewd rune]

whats there

ohh ok i got it

.close

[pearl pond]

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[mortal geyser]

Hi im an algebra 2 student wondering if i could get an explanation on how to solve the following problem relating to imaginary numbers

[autumn fossil]

The best way to simplify that is probably multiplying the denominator by something called the conjugate

conjugate of a + bi is a - bi in general

so e.g. conjugate of 1 + i is 1 - i, conjugate of 5 - 4i is 5 + 4i etc...

oh and imporantly, you gotta multiply both the numerator and denominator, otherwise the fraction would be changed

do you know what the conjuagate of 3 - i is?

[mortal geyser]

3 + i

[autumn fossil]

$\frac{\left(6+2i\right)\left(3+i\right)}{\left(3-i\right)\left(3+i\right)}$

[jolly parrot]

MathIsAlwaysRight

[autumn fossil]

alright, then this is what you get when you multiply both the numerator and the denominator by that conjugate

now you may notice that the denominator (3 - i)(3 + i) looks exactly like (a - b)(a+b), which is a^2 - b^2. So (3 - i)(3+i) is just 3^2 - i^2 = 9 - (-1) = 10

$\frac{\left(6+2i\right)\left(3+i\right)}{10}$

[jolly parrot]

MathIsAlwaysRight

[autumn fossil]

so it becomes just this

now just multiply out the numerator and simplify

[autumn fossil]

and this will always happen when multiplying by the conjugate btw, that's why we do it in the first place

[mortal geyser]

so answer should look like this: 16/10 + 12/10i ?

[mortal geyser]

or actually wouldnt the denominater be 8 becasue (3 - i ) * (3 + i) = ( 9 + 3i - 3i - i^2) = (9 - i^2) and since i^2 is -1 it would go to (9 - 1) = 8

[autumn fossil]

i^2 is indeed -1

so it would become 9 - (-1)

which is 9 + 1

or 10

[mortal geyser]

oh right right

[autumn fossil]

(a + bi)(a - bi) will always be a^2 + b^2, so if you wanna save some time, you can just square both the coefficients and add them up

[mortal geyser]

alright

[mortal geyser]

and actually thats not an answer on my little work sheet here:

oh wait actually i see the answer now thank you for your help

.close

[pearl pond]

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[pearl pond]

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[daring bay]

yo

idk how to do the a)

let me translate

[file name] IMG_4930.jpeg
[file content begins]
21. Rescue Tunnel
In a mining accident, an attempt is made to reach the miners trapped in the shaft AB and the cavities ( H_1 ) and ( H_2 ) using six rescue drillings ( g_a ) originating from the tower T (4|6|0).

Data:
A (8|2|–2); B (15|16|–9)
( H_1 ) (22|6|–14); ( H_2 ) (12|16|–4)

[g_a : \vec{X} =
\begin{pmatrix}
4 \
6 \
0
\end{pmatrix} + r
\begin{pmatrix}
13-a \
a-4 \
a-11
\end{pmatrix}]

a = 0, 2, 4, 6, 8, 10

a) Is the shaft AB intersected by any of the drillings? If yes, where?
b) Determine whether the cavities ( H_1 ) and ( H_2 ) are reached.
c) Determine if any of the drillings point vertically downward.

[file content ends]

[jolly parrot]

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

[daring bay]

nice

yeah so i gotta check If the drill hits vecotr AB

but

how

can i just check with B-A and put into the equation etc

or?

<@&286206848099549185>

helll

help

[broken fossil]

what is thew(this new) notation bruh

[daring bay]

huh

[broken fossil]

ive never seen A(x|y|z) used for vectors or i atleast thats what i assume it means

[rough forge]

it's a point

[rough forge]

post mal the original

[daring bay]

here u go

[broken fossil]

oh alright

[rough forge]

a picture that you vorenthalten

[daring bay]

oh

but listen

im confused on the AB vector

was soll ich überhaupt machen bei der a digga

[rough forge]

AB ist eine Strecke, also eine Gerade tatsächlich

Also zB. x = A + t(B-A)

dann setzt x = g_a und schaust ob es für die Werte von a eine Lösung für t gibt, weil dann existiert ein Schnittpunkt

[daring bay]

hä aber

achso

also soll ich schauen ob die ein schnittpunkt haben

digga das dauert soooo lange

[rough forge]

Bei der b) ist vermutlich gefragt, ob die Punkte H1 und H2 auf der Geraden g_a liegen, für einen Wert von a. Also auch hier H1 = g_a und H2 = g_a und für a jeweils lösen

Bei der c) denke ich will man schauen, ob für ein a, die Gerade g_a den Richtungsvektor (0,0,-z) hat

[daring bay]

ja einfach punktüberprüfung

[rough forge]

also ein Vielfaches von (0,0,-1) weil der würde ja senkrecht nach unten zeigen

[daring bay]

hmm

[rough forge]

links ist ein Koordinatensystem zur Orientierung

[daring bay]

ich bin dir ehrlich ich glaub in einem test kommt nt so eine behinderte textaufgabe

[rough forge]

wieso behindert, du rettest hier menschenleben

[daring bay]

😭 😭

ich mach die a) dann gehe ich pennenoder so

[pearl pond]

@daring bay Has your question been resolved?

[rough forge]

wild

[daring bay]

ja scheiß deazf

letzte frage

hier gibt es doch keinen wert von a

bei den doe geraden g und h parallel sind

weil der richtungsvektor muss ja der selbe sein

[rough forge]

oder ein vielfaches des anderen

r(12,a-9,-2a) = (2,-2,1)

Für identisch guckst du g_a = h

[daring bay]

achsoo

weil 12 vielfaches von 2 isr

ok

und a dann halt ausrechnen

aber wenn es kein vielfaches gibt dann kann ich es direkt sehen

[rough forge]

ne

solltest schon richtig prüfen

Wenn du es direkt sehen könntest damn wärst du krass weil dann könntest auf ein LGS gucken und einfach sagen obs lösbar ist oder nicht ohne zu rechnen

[daring bay]

ok lol

ja ich improvisiere einbisschien

[pearl pond]

@daring bay Has your question been resolved?

[broken fossil]

bruh

[pearl pond]

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[pearl pond]

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[tight loom]

What the heck is this now how do I solve this?

[tight loom]

B

[cursive wraith]

what have you tried?

[tight loom]

Wdym

I tried to just turn cos and ran into x and y so I can understand it a bit better but it didn’t help

[cursive wraith]

maybe you could try putting everything on one side

say the left hand side

[tight loom]

I did that

[cursive wraith]

and factor

[tight loom]

(X^2) (y)

[cursive wraith]

x^2?

[tight loom]

=0

Oh

No

Not that

Wait no

I HATE THIS

[cursive wraith]

let's try factoring together

[tight loom]

ITS SO STUPID I HATE IT

SO MUCH

THIS IS SO DUMB

[tight loom]

Okay

[cursive wraith]

we got $\cos \theta \tan \theta - \cos \theta = 0$

[jolly parrot]

rafilou is not not born in 2003

[cursive wraith]

how could we factor the left hand side?

[tight loom]

Take cos out of both

[cursive wraith]

yeah factor by cos

[tight loom]

And then cos (tan -1)

[cursive wraith]

$\cos \theta (\tan \theta - 1) = 0$

[jolly parrot]

rafilou is not not born in 2003

[cursive wraith]

now we have a product of two things equal to 0

[tight loom]

Yeah

But like what now

[cursive wraith]

so a * b = 0 type of thing

[tight loom]

There’s not even any numbers

Just -1

[cursive wraith]

when is a product of two real numbers 0?

[tight loom]

When they cancel out

[cursive wraith]

?

[tight loom]

I don’t know

[cursive wraith]

no

[tight loom]

I DONT KNOW

[cursive wraith]

ok, let's try out a few stuff

[tight loom]

okay😕

[cursive wraith]

first of all, something * 0 = ?

[tight loom]

0

[cursive wraith]

yes

0 * something = ?

[tight loom]

0

[cursive wraith]

alright

and if both a and b aren't 0

what can we say about ab?

[tight loom]

They don’t = 0

[cursive wraith]

ab is not 0 yes

so

it turns out

that we do have ab = 0

[tight loom]

Because cos is 0

And the brackets are 0

[cursive wraith]

both at the same time are 0?

[tight loom]

No

I don’t know

[cursive wraith]

only one needs to be zero

[tight loom]

1 divided by tan

[cursive wraith]

so

a is cos(theta)

b is tan(theta) - 1

we now wonder when a is 0

and when b is 0

[cursive wraith]

let's start with assuming this is 0

what angles theta make this so?

[tight loom]

Okay

Would I do second cos 0

Or just regular cos

90

1

[cursive wraith]

90° is one of the angles that work

you can give them in radians or degrees whatever you prefer

[tight loom]

Degree

[cursive wraith]

ok

[warm spindle]

is the range 0 ≤ theta ≤ 90 ?

[tight loom]

This really just doesn’t make any sense

[cursive wraith]

do you have a restriction on where you draw your angles from?

[tight loom]

What does second tan cos or sin even find

How is it different from the regular ones

[tight loom]

Yes

0 and 90

Must be between

[cursive wraith]

ok

[limpid lily]

What do you mean by second cos? Like on your calculator?

[tight loom]

Yes

The -1

Option

[cursive wraith]

that's arccos

cos^-1

you mean

[tight loom]

Yes

[cursive wraith]

that's not cosine function

[limpid lily]

That means arccos, which does the opposite of cos.

[cursive wraith]

that's the opposite

[tight loom]

Wdym

[tight loom]

okay what does it do tho

[cursive wraith]

if you're solving cos(x) = a on [0,90]

you input cos^-1(a)

and boom

that gives you the x

[limpid lily]

Well, let's say cos(45) = sqrt(2)/2.

cos takes an angle and gives you a number.

[tight loom]

45 is the angle

And what’s the number

[cursive wraith]

sqrt(2)/2

the output of cos

[limpid lily]

arccos takes that number and gives you (one of) the angles you put into cos to get sqrt(2)/2.

[tight loom]

But what is it for

[limpid lily]

It's to undo cosine.

Like cos(45) = sqrt(2)/2, right?

[cursive wraith]

cos(x) = a

x = cos^-1(a)

[tight loom]

And then I put the new number into cos-1 and I get 45

Allegedly

[limpid lily]

arccos(sqrt(2)/2) = 45.

Right.

[tight loom]

Oh my gosh

Wow

Alright

[cursive wraith]

this is the same with all the "^-1" functions

I wanna solve tan(x) = sqrt(3)

I write on my calculator "tan^-1(sqrt(3))"

[tight loom]

Sigh

[cursive wraith]

gives me the angle

[tight loom]

60

Yeah

I’m gonna try a different question

THIS IS SO DUMB

WHY CANT NOTHING JUST MAKE SENSE

[limpid lily]

Which question are you looking at?

[tight loom]

I GOT 1/2 AND -3 AND IT JUST DOESNT WORK

D

[mental vine]

what seems to be the problem here?

[limpid lily]

OK, how did you get 1/2 and -3?

[tight loom]

I DID IT LIKE A QUADRATIC IGNROINH THE COS AND THEN AFTER I PLUGGED IN BOTH ANSWERS AND IT JUST DOESNT WORK

IM SO FRUSTERATED

[mental vine]

please calm down, show your full work

[limpid lily]

OK, so usually you have like 2x^2 + 5x - 3 = 0.

[cursive wraith]

you got cos(theta) = 1/2 or cos(theta) = -3?

[limpid lily]

And then you get x = one thing and x = another thing.

But here, you're not doing x.

You're doing cos(theta).

[cursive wraith]

just to be clear this is correct

but you're looking for theta

stopping at cos(theta) = ... doesn't say what theta is

[limpid lily]

So you get what rafilou said.

[limpid lily]

OK, but you had cos(theta) instead of x.

So, your factoring says that x = -1/2 or x = 3.

[cursive wraith]

watch the x here

[tight loom]

ugh

I SWEAR IT WAS JUST A - 5

[limpid lily]

But since the "variable" was cos(theta), you have cos(theta) = -1/2 or cos(theta) = 3.

[tight loom]

AN DI CHANGED UT BECAUSE I WANST RIGHT

AND I JUST CHANGED IT

EHNDJDBDBDBD

OH MY GOSH

IM SO UPSET

IM SP ANGRY

[limpid lily]

,w factor 2x^2 + 5x - 3

[jolly parrot]

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[limpid lily]

So, you get x = 1/2 or x = -3 like before.

[tight loom]

I was right the first time

Yes

[cursive wraith]

it's alright we're already on the correct path

[tight loom]

No we are not I’ll never get this

Ugh

Okay what’s next

[cursive wraith]

so remember that 'x' stands for cos(theta)

[tight loom]

I have my stupid answers

Okay

[cursive wraith]

so we're looking to solve for theta

[tight loom]

Okay

How

[cursive wraith]

so now, for each value of cos(theta) we found

[tight loom]

Okay -3 and 1/2

[cursive wraith]

we're just gonna try to look for the theta that works

[tight loom]

Okay

[cursive wraith]

yep

[tight loom]

So how would I even check that

[cursive wraith]

well let's start with cos(theta) = -3

[tight loom]

Because it’s just giving me the stupid same answer on my dumb calculator

[cursive wraith]

does it sound... alright?

[tight loom]

Yes

No idk

[cursive wraith]

cos(theta) = -3 doesn't look strange?

[tight loom]

It’s error

[cursive wraith]

do you know why?

[tight loom]

Cos-1 right

No

[limpid lily]

Your calculator is right.

[cursive wraith]

you never learnt the range of cos?

[tight loom]

No

My teacher didn’t teach it

He didn’t do anything

[limpid lily]

Sine and cosine can go all the way up to 1 and all the way down to -1.

[tight loom]

He just threw the stupid work at my head and hoped it would go into my brain

[tight loom]

Okay

[limpid lily]

But -3 is below that, so it's impossible to get.

[tight loom]

But this is supposed to be circles

[cursive wraith]

the cosine is some algebraic length you get from the trig circle yes

but then another way to see why it never goes out of [-1,1]

[tight loom]

Okay

[cursive wraith]

is that the trig circle's radius is 1

(so length of hypothenuse)

so the cosine length must be smaller than that

[tight loom]

Okay

So now what

[cursive wraith]

so does cos(theta) = -3 give us solutions?

[tight loom]

No

[cursive wraith]

alright

so now we turn to the other possible value

[tight loom]

1/2 gives us 60

Is that the answer

Is that the angle

[limpid lily]

That's what theta is.

[tight loom]

Thank God

[limpid lily]

The 60 you got is 60 degrees, so the degrees means an angle.

[tight loom]

Okay

Phew

I’m gonna try another one

Is this 90

Degrees

[cursive wraith]

c)?

yeah I have only theta = 90°

[tight loom]

Yes

Oh my gosh wow

Guys and this is standard of excellence

I’m so smart

Thanks guys

[limpid lily]

No problem.

[tight loom]

I got this🫂

.close

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[daring bay]

yo guys

[daring bay]

just a quick question

can i solve the linear equation system like this directly or do i have to move the numbers to the other sides first

[toxic lichen]

if you can think of some "direct" way to do it then nobody's stopping you from going for that

[daring bay]

or maybe express the greek thing by another combination

[toxic lichen]

there is no formal requirement to bring the system into this or that form

[toxic lichen]

"greek thing" ...

[daring bay]

but in this case ill most likely have to express lambda

[toxic lichen]

there are two greek things here

there is lambda and there is mu

[daring bay]

yeah

[toxic lichen]

again: no, you don't "have to" do anything.

what is this though

[daring bay]

yeah

[toxic lichen]

-2 + λ [???]

[daring bay]

k

paramter

[toxic lichen]

that is a k?

[daring bay]

like a unknown

[toxic lichen]

one very distorted k...

[daring bay]

variable

[toxic lichen]

the letter looks unrecognizable

[daring bay]

so

[toxic lichen]

!xy btw

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[daring bay]

yk sometimes

u have to express a variable through another

thats tricky yk

[pearl pond]

@daring bay Has your question been resolved?

[pearl pond]

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[pearl pond]

@subtle stream Has your question been resolved?

[summer imp]

One way you can interpret this is that with this fixed radius, your varying parameter is the angle between the two cuts made. To get the volume of the cone generated, you need :

• the circumference of the base that hasn't been cut away
• the height of the cone

If you express both of those in terms of the angle you start with, then you should end up with a formula for the volume of the cone solely depending on the angle. Using calculus you can then maximize this function.

[subtle stream]

.close

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[summer imp]

Actually I think I overcomplicated this

[pearl pond]

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[loud ember]

How to do part (c)? I need a brief explanation

[random ermine]

take lim N --> infinity for different values of x

the limit N --> infinity is the infinite sum

[loud ember]

mind showing your working with your answers?

I am just confirming something

[random ermine]

i didn't work it out

[loud ember]

well, do you mind?

[random ermine]

yes

[loud ember]

..

[random ermine]

what ru confirming

[loud ember]

the answer

[random ermine]

u can show ur working

i may check it

[loud ember]

How come x = 1 isn't convergent?

[random ermine]

x=1 should converge

but not x= -1

[loud ember]

bruh the answer scheme is wrong then

[loud ember]

hm

[loud ember]

So what'd be the set values of x

0 < x < 1 or something?

[random ermine]

what no

-1 < x <= 1

[random ermine]

-1 < x < 1 cuz of this

x = 1 a bit subtle

[loud ember]

hmm

ok

got it

ty

.close

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[random ermine]

ur welcome

[pearl pond]

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[iron sigil]

this is the information on how to conduct a binomial hypothesis testing and how to find p value or critical reigon. However, I dont understand wht does 'p' refer to

[iron sigil]

I also dont get the logic for finding critical value

[pearl pond]

@iron sigil Has your question been resolved?

[pearl pond]

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[iron sigil]

.reopen

[pearl pond]

✅

[iron sigil]

<@&286206848099549185>

[pearl pond]

@iron sigil Has your question been resolved?

[pearl pond]

@iron sigil Has your question been resolved?

[pearl pond]

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[sonic hound]

Find (5424904012659,690821867186)

[sonic hound]

,w GCD(5424904012659,690821867186)

[jolly parrot]

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[dapper kraken]

??

[sonic hound]

If it evaluates to 1, there must be a trick!

[dapper kraken]

try euclids division algorithm

[sonic hound]

(x,y) represents the GCD of x and y.

That's too long.

There must be a trick for the GCD is 1.

Yk when things cancel out, and there remains a simple expression

[dapper kraken]

thats the trick

idk what to tell you man

[sonic hound]

No, that's the most basic method

Who would not try to use that method?

[dapper kraken]

wait making sure euclids division algorithm is the one that constantly use gcd(a,b)=gcd(a-mb,b) right?

[sonic hound]

Yes,

a = bx + r
Then (a,b) = (b, r)

Where r < b

??

<@&286206848099549185>

[midnight haven]

idk use math

[fair creek]

what's the issue exactly?

[midnight haven]

i dont know math

[fair creek]

just a really long chain of

euclid's algorithm

[midnight haven]

ok my job here is done too

[fair creek]

dont engage if you're not gonna be helpful

long numbers though... why do you wanna compute it manually?

[sonic hound]

(2^100 - 1)/(2^50 + 1)/(2^50-1)

One would think to first calculate 2^50 + 1, then 2^100 - 1 and then 2^50 - 1, perform division and get the answer.

But no. There is a trick.

It simplifies to 1. Just like how this simplifies to 1, there must be a trick that simplifies the GCD to 1.

[toxic lichen]

wait, so what are those huge ass numbers

were they given to you exactly like this, as raw decimal representations?

or did you deliberately obfuscate them again?

[iron basin]

isn't this just (2^100-1)/(2^100-1)

[fair creek]

yes that might be "neat" but how're you gonna know that those long strings of numbers equal that

[iron basin]

i don't get what you mean

[fair creek]

feels like just overcomplicating things

[sonic hound]

Useless server.

.close

[pearl pond]

Channel closed

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[toxic lichen]

<@&268886789983436800>

[verbal whale]

There's no need of explanation @sonic hound

It's not even the first time you have this behaviour

[toxic lichen]

undue hostility...

[sonic hound]

what?

[radiant terrace]

It is fairly hostile to respond to peoples good faith attempts to help you by calling the whole server (and them by proxy) useless. You should not do this.

[sonic hound]

Oky.

[toxic lichen]

.reopen

[pearl pond]

✅

[verbal whale]

And you shall always try to provide the original context/exercise of your questions

Since you never do it...

[sonic hound]

Thanks for your help blurple_galaxy, parabolicinsanity, Ann, Alberto Z. and skissue.in.a.teacup.

I come up with questions.

[toxic lichen]

for the record, this is what it looks like if you rawdog euclid (with a bit of lubricant)

[radiant terrace]

Okay, that is cool. But people also cannot read your mind.

[toxic lichen]

absent a more elegant representation for either initial number, there cannot possibly be any trickery available.

[radiant terrace]

They do not know what you have tried or what other requirements you have for solving the problem.

[sonic hound]

sorry, I won't do it again

[dusky grove]

Hey guys I'm struggling with a question "5x/3-x-2/4=9/4-(x-2x-1/3)" can anyone help

[toxic lichen]

!occupied

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[radiant terrace]

If you had simply added "without using the euclidean algo" or something that clarifies that you wanted a solution that was less time consuming than that they could have accounted for your constraint.

[sonic hound]

!help

[twilit harbor]

​No Category:
  help Shows this message

Type !help command for more info on a command.
You can also type !help category for more info on a category.

[pearl pond]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[dusky grove]

Oh okay thanks

[toxic lichen]

@sonic hound so now we got through this,

can you answer my query regarding the form of the problem as presented to you

or alternatively, where on earth you got these numbers from

[sonic hound]

I got them randomly. I used Wolfram Alpha. It told GCD = 1. So...

[cursive wraith]

There's no big universal trick for showing gcd=1 in my knowledge

[toxic lichen]

ok then there is no trick you just have to rawdog

[cursive wraith]

So if you got them at random, chances are there's not a direct proof apart from euclid

Or prime factorization (which is worse like 90% of the time)

[toxic lichen]

this is an incorrect assertion

"most" (informally) pairs of big numbers are coprime

[sonic hound]

,w GCD(3242342868883543111, 2999111919299911)

[jolly parrot]

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Requested by forlorn1002

[sonic hound]

You are right.

Sad.

[toxic lichen]

what do you stand to gain by bashing GCDs of ex-recto 12+ digit numbers

[sonic hound]

Nothing.

Thanks for your help again "blurple_galaxy", "parabolicinsanity", "Ann", "Alberto Z.", "rafilou is not not born in 2003" and "skissue.in.a.teacup".

.close

[pearl pond]

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[patent pagoda]

Hi

[patent pagoda]

I need help with finding z

could i use exterior angle theorem? therefore it would be 57 + 75

so 132

oh i just complicated it

i could have just used angles on a straight line

but thanks

[dapper kraken]

did you solve the problem?

[patent pagoda]

yes

but i need help on this one pls

[verbal whale]

What's troubling you? What have you tried?

[patent pagoda]

i made 2 equations
x-y+2x+y+2x-y=22cm then simplified it to 5x-y = 22cm then i made another and its x-y=2x+y but when i simplify it it doesnt make sense it become -2y=x

[verbal whale]

Yeah all good

Why do you think it's making no sense?

Now you can substitute x = -2y in the first equation

[patent pagoda]

cause then i times 5x-y=22cm by 2 and the ys cancel out but then i cant add x+44

[verbal whale]

So that you can solve for y and then get the value of x as well

[verbal whale]

Huh? I don't understand what you're talking about honestly. It seems you're just overthinking it

[patent pagoda]

my bad so what should i do after i form those 2 equations

[verbal whale]

Read my messages above 😅

[patent pagoda]

5(−2y)−y so it would becpme like this?

[verbal whale]

= 22

Yeah

Just as simple as that, isn't it?

[patent pagoda]

−10y−y=22 then −11y=22 y=−2

then x=−2(−2)=4

[verbal whale]

Exactly

[patent pagoda]

ok thanks

[worldly peak]

@patent pagoda was the question resolved? Consider typing .solved to signal that the question is solved.

[patent pagoda]

.solved

[pearl pond]

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[pearl pond]

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[sonic hound]

Consider pqr = a typical term in a sum where
p, q, r are distinct positive integers and between 1 and n, including 1 and n. That is, the sum contains every term ABC, where A neq B neq C, and 1 <= A <= n, 1 <= B <= n and 1 <= C <= n
Consider S(n) = the sum of all the unique combinations of A, B and C --- (product ABC).

[sonic hound]

Find S(n) in terms of n.

For n = 5, the sum would be:
1 * 2 * 3 + 1 * 2 * 4 + 1 * 2 * 5 + 1 * 3 * 4 + 1 * 3 * 5 + 1 * 4 * 5 + 2 * 3 * 4 + 2 * 3 * 5 + 2 * 4 * 5 + 3 * 4 * 5.

,w 1 * 2 * 3 + 1 * 2 * 4 + 1 * 2 * 5 + 1 * 3 * 4 + 1 * 3 * 5 + 1 * 4 * 5 + 2 * 3 * 4 + 2 * 3 * 5 + 2 * 4 * 5 + 3 * 4 * 5

[jolly parrot]

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[sonic hound]

S(n) is the coefficient of x^(n - 3) in the expansion of (x-1)(x-2)(x-3)....(x-n).

[toxic lichen]

so it would appear

well, it's -1 times that

[sonic hound]

Okay.

S(n) = |coeff. x^(n-3)|

[toxic lichen]

if you wanna view this thru the lens of generating functions i think it might be more "cooking" to view S(n) as the coeff of x^3 in the product (1+x)(1+2x)...(1+nx)

but i do not yet see an elegant way to proceed from there

if you're willing to sit down and go through some sigma notation bullshit, then you can say $$S(n) = \sum_{p=1}^{n-2} \sum_{q=p+1}^{n-1} \sum_{r=q+1}^n pqr$$

[jolly parrot]

Ann

[toxic lichen]

which is mildly unpleasant (but even then, much less so than if you had gone to write all of your summations with dotdotdots spread like grains of sand)

[sonic hound]

This is a machinegun.

[toxic lichen]

?

i fail to see the metaphor

[sonic hound]

Nothing.

[tropic saddle]

I would probably introduce T(n) as the sum of products of two terms and U(n) as the sum of products of one term (easy). and then you can give a recursion and maybe try to solve that

actually yeah no, what Ann did is easy enough

[inland ivy]

Its a bit brute-forcey but you can just take the sum of all the (non-unique) products and then subtract the duplicates

There's not too many cases

[pearl pond]

@sonic hound Has your question been resolved?

[sonic hound]

okay, let me do it

[pearl pond]

@sonic hound Has your question been resolved?

[slow grove]

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

[toxic lichen]

it's quite ass

im having to sum a quartic

[iron basin]

summations from n = 5 to 10 for the curious

or even till 20

a very interesting series

[sonic hound]

S(n) = 1 * (2 * 3 + 2 * 4 ............) + 2(3 * 4 + 3 * 5..............) + k[(k+1)(k+2) + (k+1)(k+3)........ + (k + (n - k - 1))(k + (n - k)] = sum from k = 1 to k = n of
k[(k+1)(k+2) + (k+1)(k+3) + (k+1)(k+4) + ....................... + (k + (n-k-1))(k + (n-k))]
Consider this now:
(k+1)(k+2) + (k+1)(k+3) ................... + (k + (n-k-1))(k + (n-k)) = K
So S(n) = kK
Number of terms = (n-k)(n-k-1)/2
A typical term = (k + p)(k + q) = k^2 + kp + kq + pq
K = (n-k)(n-k-1)k^2/2 + (n-k-1)(n-k)(n-k+1)(3n - 3k + 2)/24 + k[(1+2) + (1 + 3) + (1 + 4) + ........ + ((n-k-1) + (n-k))]
Let [(1+2) + (1 + 3) + (1 + 4) + ........ + ((n-k-1) + (n-k))] = P
Then S(n) = k[(n-k)(n-k-1)k^2/2 + (n-k-1)(n-k)(n-k+1)(3n - 3k + 2)/24 + kP]

P = [(1 + 2) + (1 + 3) + (1 + 4) + ......... + (1 + m)] + [(2 + 3) + (2 + 4) + ..........+ (2+m] + [(3 + 4) + (3 + 5) + ............ (3 + n)]...................... + [((m-1) + m] =
sum from i = 1 to to (m-1) of
[(i + (i+1)) + (i + (i + 2)) + ............. + (i + (i + (m-i))]
The number of terms = m - i
So the expression is (m-i)(i)(2) + (m-i)(m-i+1)/2
The sum = (m-1)(m)(m+1)/2
P = (m-1)(m)(m+1)/2
m = n - k
So P = (n - k - 1)(n-k)(n-k+1)/2

S(n) = k[(n-k)(n-k-1)k^2/2 + (n-k-1)(n-k)(n-k+1)(3n - 3k + 2)/24 + kP] = k[(n-k)(n-k-1)k^2/2 + (n-k-1)(n-k)(n-k+1)(3n - 3k + 2)/24 + k(n-k-1)(n-k)(n-k+1)/2]
*the sum from k = 1 to k = n

 k[(n-k)(n-k-1)k^2/2 + (n-k-1)(n-k)(n-k+1)(3n - 3k + 2)/24 + kP] = k[(n-k)(n-k-1)k^2/2 + (n-k-1)(n-k)(n-k+1)(3n - 3k + 2)/24 + k(n-k-1)(n-k)(n-k+1)/2]
The sum from k = 1 to k = n```

[iron basin]

hmm

[tropic saddle]

why would you not start at n=1

[iron basin]

three numbers..

op started at 5

[tropic saddle]

well ok. n=3

[iron basin]

random

like i said

op started at n=5

and this series does match a diagonal in the sterling numbers

[iron basin]

you madman

anyways

we have a quintic sum now

🥳

have fun finding the summation of whatever this abomination is

[slow grove]

the answer can be found with the wikipedia page
its ||(n+1 choose 2) * (n+1 choose 4)||

[iron basin]

😭

explains the oddly high number of ten multiples

now i wonder where op got this problem from...

[slow grove]

its a pretty natural question, someone else asked something similar recently

[sonic hound]

wdym

[iron basin]

like if you got this from some examination or something

[iron basin]

it is natural in a sort of manner...

[sonic hound]

I am actually working on another problem.

It is. It is the absolute value of the coefficient of x^(n-3) in the expansion of (x-1)(x-2)(x-3)...(x-n).

[iron basin]

😭

[sonic hound]

👍

Thanks y'all.

[slow grove]

<@&268886789983436800>

[sonic hound]

What

[slow grove]

some spammer

[sonic hound]

okay

i thought you called them because of me not closing the channel

lol

[slow grove]

lmao

[sonic hound]

,w substitute n = 10 in n^2(n-2)(n-1)(n+1)^2/48

[jolly parrot]

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[sonic hound]

This seems to be the simplified form.

,w substitute n = 5 in n^2(n-2)(n-1)(n+1)^2/48

[slow grove]

you did /4 not 48

[jolly parrot]

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[sonic hound]

Goodo.

Lol.

[sonic hound]

Good. It matches.

I shall close this now. I will now find this for the product of four terms.

Thanks.

.close

[pearl pond]

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[little swan]

<@&268886789983436800>

[pearl pond]

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[little swan]

<@&268886789983436800>

[pearl pond]

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[lone nimbus]

if you're trying to find the surface integral of a sphere, does your "r" (radius) have to be a constant?

i ask this because in physics (gauss's law), when finding the electric field at some variable distance "r", we put "r" as the radius of an imaginary gaussian sphere
and we integrate over that sphere

[tiny lintel]

Hello, if I remember correctly. The law of gauss says that the electric field is proporcional to the charge inside a closed surface. So we “vary” the surface in order to get the value of the electric field at every boundary. In the case where the surface is a sphere, we vary the radius of the sphere in order to get the “ successive” surfaces.

[lone nimbus]

if I have an insulating charged sphere with radius R, and I wanna find the electric field at some arbitrary distance r that is greater than R, when I would draw the gaussian sphere with radius "r", is my "r" here fixed or variable?

[flat cedar]

Why would the radius matter if it's given to be already greater than R? The amount of charged enclosed will remain same at any 'r'

[lone nimbus]

integral (E dot dS) = qenc/e_0

the flux will be the same, but not the E-field

[flat cedar]

If the electric field will change, so will the surface area

[lone nimbus]

actually it will have the same expression

kQ/r^2

it will depend on r, so r is a variable here

but when we draw the gaussian surface (to find the expression for the field), we put "r" as our radius, im confused whether we can have a variable radius

[midnight haven]

The radius of a sphere of a certain size is fixed. You dont want r to vary as you integrate. Im not sure I understand what your confusion is

[lone nimbus]

my confusion comes from the application of surface integrals in physics, namely, gauss's law, when we wanna find the field at some variable distance r, we create a gaussian surface with radius "r", which is a variable, when it should be a constant

[flat cedar]

Why should it be constant if you are finding it at any variable distance r?

[lone nimbus]

Because the surface integral of a sphere has a constant radius

we're integrating over the gaussian sphere

int(E dot dS) = qenc/e0

[midnight haven]

the variation in r is a product of the integration. If you set up the integral you are just integrating dA

[flat cedar]

I am not sure if I am understanding you correctly but, we find the electric field at 'any' distance 'r' from the centre. Now the formula thus formed will be in terms of this 'r' (which you would have integrated from 0 to r). Now you put in 'any' constant value for r.
It's more of 'any' than 'variable'

[midnight haven]

Spherical charge is nice because E will always be perpendicular to the surface. So if you set up the integral Then you are only integrating dA since E will be constant as well. the surface integral of dA is 4pi r^2. Then when you E in terms of r and Q you can solve for it at any radius

the radius of your gaussian sphere you drew is completely unimportant. Its just r. if its large or small you still get the same answer for your surface integral

so yes it is variable as in you can change it to test E at different distances away from the charged object. But in terms of solving for E you dont have to do anything fancy with it

[pearl pond]

@lone nimbus Has your question been resolved?

[pearl pond]

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[bitter herald]

<@&268886789983436800>

[obtuse relic]

NO MATH QUESTION?

[pearl pond]

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[mild lion]

<@&286206848099549185> i took a test today and my professor made a bonus question, i think it was like this

[ruby cargo]

!15min

[pearl pond]

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[mild lion]

k

[autumn fossil]

i dont get what the question is askign for

[mild lion]

a function that satisfies x^2 2x^2 3x^2 4x^2 and so on

[autumn fossil]

well, since it's a function, it can have only one y per x

so you'd have
y = x^2 = 2x^2 = 3x^2 = ...
which can only be true if x = 0

and then y = 0

so it would be function with domain {0} mapping the only element - 0 to 0

but i think youre just misremembering the question

wasnt it asking what the graph approaches perhaps?

in that case, it would be approaching the half-line from origin going up on the y-axis

[mild lion]

yeah but were trying to find a function that represents x^2,2x^2, not the final function

that was my first thought

so if x=1 the function gives out 1^2, 2(1^2), 3(1^2), and so on

[autumn fossil]

then it's not a function

function can't have that many y-values for a single x value

function only has 1 output per input

[mild lion]

so its impossible to represent this as a function

[mild lion]

he said this leads to a function

an actual function

[autumn fossil]

it doesnt

[mild lion]

wait

couldnt we replace 1x^2, 2x^2, 3x^2 as another varialbe

variable*

like a

ax^2

a=1 we get x^2

a=2 we get 2x^2

@autumn fossil couldnt this be a multivariable function?

[pearl pond]

@mild lion Has your question been resolved?

[autumn fossil]

uh i mean sure

you can do
f(x, n) = nx^2

but like, is that really what the professor wanted?

i think youre just misremembering the question

[pearl pond]

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[trail forum]

.close

nvm

[pearl pond]

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[hoary nacelle]

Help

[hoary nacelle]

Please help me with checking my proof of thm 5.1.1 which I did along though and a couple examples

They are exercises on the books

They are a lot of notations errors I know especially open covers ish thing maybe should be chain of open sets but I am not really not really a math student like for real so please point them out

@rough forge you’re there and you not talking it made me sad

[regal herald]

theyre likely just reading, give it a bit

[hoary nacelle]

I am joking with him 🥰

[pearl pond]

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[pearl pond]

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[fathom fulcrum]

E

[summer imp]

e

[fathom fulcrum]

Why does ts bot close channel 💔

[errant cedar]

please try drawing it yourself first

we are here to help not to directly give the answers

[errant cedar]

it closes them when they are inactive for too long

[pearl pond]

@fathom fulcrum Has your question been resolved?

[fathom fulcrum]

i get it
but like i know how to solve this with distance formula mid points and blah blah

what im looking for
is solving this
with scaler vector
can u help w/ that?

basically we know
M here is midpoint
which must be 2,2
and AC = AM + MC
BD = BM + MD

[errant cedar]

yep i just finished to solve it on my side

i think

[fathom fulcrum]

what do i next? ( i have to leave my house within 20 mins)

[fathom fulcrum]

what answers did u get
let me match them first

[errant cedar]

for me the next step was to construct a perpendicular vector to AM

[fathom fulcrum]

reversing i and j comp?
yeah about that its kinda using coordinate geom stuff

[errant cedar]

using a little trick in R^2 if you have (a,b) then (-b,a) is perpendicular to the first and of the same length

[fathom fulcrum]

yeah :(( i dont want to use this
im looking for one specific way

[errant cedar]

since the intersection of the diagonal of a square cut in half the diagonals then can you see how to get OB of OD?

i mean that’s the solution i got if you find out something else you’re on your own unless you get another helper

[fathom fulcrum]

yeah this way i know as i already told u at start

tho thanks a lot for your effort <33

[errant cedar]

anyhow at the end i got the area of 40 (0,6) for B

[fathom fulcrum]

correct D ( 4, -2 )

[errant cedar]

yep

[errant cedar]

I mean you can find the area by the pythagorean theorem alone and the length of the diagonal AC because the sides of a square have the same length, but to get the other diagonal without using coords idk

[fathom fulcrum]

Lengths isnt issue
My concern is just B and D coords

[errant cedar]

tbh it feels wrong not to use coords the anwser is coordinates

there is this construction in red from euclidiean geometry to construct the perpendicular bisector

you could try to find the 2 intersection between the 2 red circles

then scale it by 1\sqrt(2) but getting theses intersections that’s still using coords

[errant cedar]

ngl it feels a bit hopeless

[pearl pond]

@fathom fulcrum Has your question been resolved?

[pearl pond]

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[hollow hound]

ρcosϕ=1 how do i describe this surface

[hollow hound]

this is calculus 3/ multivariable calculus

nevermind i forgot that ρcosϕ = z

i always forget these important formulas bro 😭

thanks

.close

[pearl pond]

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[faint sky]

Hellooo!

[faint sky]

What did I do wrong here

Wait hold up

This one has the plotted points

🙂

[dapper kraken]

it looks like you plotted y>0

[faint sky]

Yes is that not right?

[dapper kraken]

its asking for y>-1?

[faint sky]

Wait shouldn’t it be a horizontal line

[dapper kraken]

no thars correct

[faint sky]

Bcuz

Oh ok

[dapper kraken]

your only mistake was the red is y>0, when its asking for y>-1

[faint sky]

So is it the shaded area that’s wrong or

I’m sorry I’m slow lol

[dapper kraken]

yes

the red graph is incorrect

[faint sky]

Oh okay gotcha tyvm

I understand now

.close

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[manic cape]

I need help with this

[manic cape]

Idk how to solve for x and y

[plush bramble]

angles that make up a straight line add up to 180 degrees

And the sum of angles inside a triangle also sum to 180

The unlabeled angle inside the triangle can be expressed in terms of x

[manic cape]

how do i solve tho

from where do i start

[plush bramble]

Use this to get an equation with x and y

[manic cape]

wait but that's not interior that's exterior

[plush bramble]

If you didn't learn what I said then I don't know how you can do the problem

https://www.mathsisfun.com/angle180.html
Specifically

[manic cape]

wait but when it has those lines isn't 360 ?

[plush bramble]

What are you saying equals 360

[manic cape]

the sum

[plush bramble]

Use equations rather than "it"

What sum are you talking about

[manic cape]

someone said that to me before u prob the same person

[manic cape]

exterior

[plush bramble]

Said what

[drowsy adder]

dude explain what youre trying to say

[plush bramble]

Exterior what

[drowsy adder]

dont just say sum 360 exterior and expect us to understand

[manic cape]

just look at the triangle

[plush bramble]

Lol

[manic cape]

listen

The triangle in the picture is it interior or exterior angle ?

Cause the way I solved the first part is by starting with z= 180

but I think I should have used z=360

you tell me to explain and then you leave

<@&286206848099549185>

I need help with this question

[pine temple]

Z’s wrong

Perpendicular angles are 2 angles that add up to 180

[plush bramble]

Chill out and learn patience

[manic cape]

So is it supposed to be Z= 360 - 117 -51 ?

[pine temple]

No

[limpid lily]

z and 51 degrees are two angles that touch.

Combined, they make a straight line.

[pine temple]

Z + 51 = 180

[limpid lily]

If some angles touch and combine to make a straight line, you can add them together and you'll get 180 degrees.

So, like z and 51 degrees add up to 180 degrees.

x and y add up to 180 degrees.

What else on your diagram combines to make a straight line?

[manic cape]

but in my notes it says ''The sum of the exterior angles of a triangle is 360''

That's what I'm confused about

[limpid lily]

You need the exterior angles to be sort of in the same direction.

[pine temple]

Well, z + 117 + y = 360

[limpid lily]

Like angle z and y are sort of tilting clockwise.

[manic cape]

then why did you tell me its 180 digree

[limpid lily]

But 117 degrees is tilting counterclockwise.

[obsidian egret]

guys

[manic cape]

wdym

[limpid lily]

Well, look at the vertices of your triangle.

Like the lower left vertex.

[obsidian egret]

I need help in a ITF (inverse trignometric function) question

[limpid lily]

!occupied

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[limpid lily]

The angle inside the lower left vertex can be split into two.

[pine temple]

That was supposed to be rhetorical

No offense

[limpid lily]

Then if you continue to draw the splitting line outside the triangle, it'll look like this:

See how z is a little bit clockwise from that red line?

[pine temple]

…what?

[manic cape]

I don't get what chai is saying

[limpid lily]

The green line there is a little bit clockwise from the red line.

[manic cape]

pls don't confuse me more

[limpid lily]

OK.

[manic cape]

What is the sum of the triangle i sended ?

is it 180 or 360

[pine temple]

Anyways

Two angles based off a straight line will add to 180

So z + 51 = 180

[manic cape]

what about x and y

[pine temple]

Let’s hold off on that until we can find x

[manic cape]

ok

...

[pine temple]

But in the meantime, Z + 51 = 180

[manic cape]

so Z =129 ?

@pine temple

[pine temple]

Sorry!

That is correct

[manic cape]

so now 180-129-51

[pine temple]

= 0

[manic cape]

yup im not doing this right

[pine temple]

Eh, it’s ok

I’m still here

Now, we want to find the value of the unmarked interior angle

[manic cape]

ok

[pine temple]

If we can find that, we can find x

[manic cape]

ok

[pine temple]

So the missing angle equals 180 - 117

[manic cape]

are you solving for x ?

[pine temple]

No

Not yet

[manic cape]

what are you solving for

[golden fjord]

oh you want to solve for z?

[pine temple]

We already did

[pine temple]

Here

[golden fjord]

oh these are old ntoes

sorry