help-0

[cloud nacelle]

yeah

[cosmic acorn]

what should I do now? S

[cloud nacelle]

youre asking if the answer's correct?

or you want to derivate it again

[cosmic acorn]

no, the computution must go further

simplify

[cloud nacelle]

oh simplify

[cosmic acorn]

Ye

s

[cloud nacelle]

its basically (2x^2 + 5) ^ 4/5

on the rhs

i mean its going to get complicated honestly

taking lcm n shit

[cosmic acorn]

seems like that

[thick lynx]

You aren't allowed to just ^4 numerator and denominator

You can only multiply (denominator^3) with the numerator and denominator

Then you will have denominator^4

[cosmic acorn]

Are you german?

[thick lynx]

Yes

[cosmic acorn]

geil

[thick lynx]

How did you know?

[cosmic acorn]

just asked tbh

aber

[thick lynx]

Das war tatsächlich sehr unwahrscheinlich aber hat geklappt

[cosmic acorn]

fr

[thick lynx]

Warte, lass mich prüfen ob die Ableitungen korrekt sind

[cosmic acorn]

okey

[thick lynx]

Bei der ersten Ableitung sollte es da (2x^2 + 5)^(4/5) sein

1/5 - 1 ist ja 4/5..

Du hast da aber nur ^(1/5) (fünfte Wurzel)

[cosmic acorn]

Habs in der zweiten ableitung korregiert

[thick lynx]

Du hast es in der zweiten Ableitung richtig gemacht, aber mit dem falschen ersten Ergebnis. Es wäre eigentlich [v = 5 \big(\sqrt[5]{2x^2 + 5}\big)^4]

[cosmic acorn]

Ich sehs

sooooo

klar

man man man

Tyyxxyxyyyyyyy

[thick lynx]

Warte, es sollte überall statt 8/5 eine 9/5 sein..

Und dann kannst du vereinfachen, wenn du zuerst beides auf den selben Nenner bringst (dazu mit (2x^2 + 5) multiplizieren)

[alpine sable]

Studierst du @thick lynx

[thick lynx]

Ja

[alpine sable]

Mathe?

[thick lynx]

Ja

[alpine sable]

Mach momentan auch Analysis II Kurs aber das so komisch irgendwie

[thick lynx]

Wieso ist das komisch?

[alpine sable]

Es ist irgendwie nicht wie ich es erwartet hatte, zu viel Formalität und wenig Intuition

[cosmic acorn]

wie alt?

[alpine sable]

Ist ein Schülerstudium

[thick lynx]

Bei mir auch!

[cosmic acorn]

oh

[alpine sable]

Bin noch 15

[alpine sable]

Schülerstudium?

[cosmic acorn]

ich bin 16 T-T und hocke in der schule

[thick lynx]

Ja

[cosmic acorn]

einfach Zwillinge

xD

[alpine sable]

Interessant wie alt biste

[thick lynx]

17

[alpine sable]

Nice

[thick lynx]

Bei uns geht das erst ab der 10. Klasse

[alpine sable]

Hier ab der 8 glaub ich

Aber unser Uni System ist auch bisschen schlecht organisiert

[cosmic acorn]

aus welchem bundesland wenn ich fragen darf

[alpine sable]

Nrw

[thick lynx]

[f''(x) = \frac{4}{(2x^2 + 5)^{\frac 4 5}} - \frac{16x^2}{(2x^2 + 5)^{\frac 9 5}}.] Dann kannst du den linken Bruch mit $(2x^2 + 5)$ multiplizieren, dann hast du den selben Nenner.

[cosmic acorn]

safe wuppertaler uni

[thick lynx]

Und dann vereinfachen.

[cosmic acorn]

Alles klar Chef

mak ich

[thick lynx]

Ah, bei mir BW

Bei uns braucht man auch 3 Leistungsfächer

[cosmic acorn]

kennst du die Essen-Duisburger uni?

[alpine sable]

Was hast du für Kurse gewählt

[thick lynx]

Mathe, Physik, Englisch

[alpine sable]

Nicht wirklich

[alpine sable]

Wir dürfen hier nur 1 glaub ich

[cosmic acorn]

ok

[alpine sable]

Weil du musst ja ganzen Schulstoff wiederholen das scheiße

[thick lynx]

Meinst du wegen dem Schülerstudium?

[alpine sable]

Ja

[thick lynx]

Bei uns ist das ganz anders. Wir müssen uns die Sachen selbst von einem Skript beibringen

Dann treffen wir uns jeden Sonntag um 10 Uhr

Und besprechen die Übungsaufgaben

[alpine sable]

Habt ihr keine Prüfungen?

[thick lynx]

(Also etwa 20 Seiten Skript jede Woche lesen)

[thick lynx]

An der Uni? Doch, aber noch nicht

[alpine sable]

Was for Module hast du schon gemacht oder machst du momentan

[thick lynx]

Lineare Algebra 1, leider darf ich jedes Semester nur eins belegen...

[alpine sable]

Ja bei uns auch

Ich hab direkt Analysis II gewählt 💀

Das wird richtig hart für mich

[thick lynx]

Kannst du Analysis 1 also?

[alpine sable]

Also es kommt auf das System an, ich kann eigentlich alle Grundprinzipien, integrieren, ableiten, extremwertaufgaben, Limes, etc.

Ich hab eher das amerikanische System verfolgt als das deutsche

[thick lynx]

Hm, da gibt es eher kein Analysis 1 und 2, da ist es Calculus 1, 2, 3 und dann Real Analysis

[alpine sable]

Als ich dann in der Uni angekommen bin ich überrascht weil die Organisierung einfach komplett anders war

[alpine sable]

Genau aber bei uns gibt’s kein calculus 1, die stecken das in Analysis 1

Und die analysis 1 hier ist abstrakt

[thick lynx]

Aber zum Glück mit Beweisen, nicht wie bei Calculus wo es meistens nur computational ist

In der USA hat man das glaub ich erst ab Real Analysis wo man wirklich beweist

[alpine sable]

Ich finde die Beweise auch wichtig aber das heißt nicht dass man bei einer Vorlesung 2 Stunden lang nur Sätze und Beweise klatschen soll

Als unser Professor z.B. die Integralrechnung angefangen hat, hat er das nicht mal mit uns richtig geübt und hat dann in den Übungsblättern so ein gar nicht einfaches integral als Aufgabe gegeben

[thick lynx]

Habt ihr auch Übungsgruppen?

[alpine sable]

Wir bekommen jede Woche 2 übungsblätter, die sind nicht Pflicht aber wenn du die machst bekommst du Bonuspunkte

[thick lynx]

Wir sammeln mit den Übungsblättern Punkte und müssen 50% erreichen um die Scheinklausur schreiben zu können

[alpine sable]

Früher war das hier auch so aber jetzt nicht mehr

Wie viel Prozent braucht ihr um die Klausur zu bestehen?

[thick lynx]

Ich glaube auch 50%, wenn man die besteht, kann man die Modulprüfung schreiben

Und davon zählt dann wirklich die Note

[alpine sable]

Sind bei uns auch 50%

Dürft ihr Hilfsmittel mit in die Prüfung nehmen

[thick lynx]

Weiß ich noch nicht, bei der Linearen Algebra macht es eher nicht so viel Sinn..

[alpine sable]

Ja, wir dürfen jetzt keine mitnehmen, aber das auch unterschiedlich von Uni zu Uni

[thick lynx]

Das meiste werden ja Beweise, dann wahrscheinlich ein paar Matrixmanipulationen, Eigenwerte berechnen, ...

[alpine sable]

Ist es einfach?

Weil ich werde glaub ich lineare Algebra nach Analysis wählen

[thick lynx]

Ja, es ist nicht schwer

[alpine sable]

Meine Auswahl ist so komisch, ich schreib jetzt Ende März Kombi Klausur von Ana 1 und 2 und dann wähle ich lineare Algebra 1, wobei man eig lineare Algebra vor Ana 2 kennen sollte

Ana 2 ist aber schon herausfordernd

Weil da beschäftigst du dich mit mehrdimensionalen Funktionen und alles bekommt ein Stück komplizierter

[thick lynx]

@karmic rapids auf diesem Server ist auch deutsch und Frühstudent, aber er macht momentan Kurse vom Master

[alpine sable]

Wahnsinn

[karmic rapids]

hallo

[thick lynx]

Ah, da ist er

Hallo

[alpine sable]

Und wie alt ist er

Deutschland bietet eig schon gute Möglichkeiten in Bildung

Glaube nicht dass sowas in anderen Ländern häufig vorhanden ist

[thick lynx]

Du warst doch 15 oder 16, oder?

[alpine sable]

So jung und masterkurse?

Das echt krass

[thick lynx]

Ja, 15 (hab grad in der message history nachgeschaut)

[alpine sable]

Der nächste Maximilian haha

Der hatte doch auch irgendwie seinen Doctor mit 18 oder so

[heady bay]

Masterkurse mit 15?? 👁️👄👁️

[alpine sable]

Ja es gibt manche die schon auf diesem Level sind hahah

[heady bay]

Diese Selbstfolter muss man erstmal aushalten hooooly

[alpine sable]

Naja wenn man sich dafür interessiert ist es kein folter

[heady bay]

Ja klar

[alpine sable]

Manche interessieren sich schon in sehr jungen Alter für Dinge und die Eltern unterstützen sie zum Beispiel. Ich hab bedauerlicherweise erst mit 13 Jahren erfahren dass Mathe interessant sein kann hahah

[heady bay]

In der Schule wird nur die ganz trockene Mathematik gemacht leider

Uni-Mathe ist aber eher schon spaßiger

[lone heart]

@cosmic acorn Has your question been resolved?

[karmic rapids]

ich erst mit 14

[lone heart]

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[strong cloud]

idk where I got wrong

[vale wigeon]

it looks correct so far, just not yet finished...

[strong cloud]

(1+sqrt(17))/4 is > 1

[vale wigeon]

so it is.

[strong cloud]

then sin of whatever can not > 1

[vale wigeon]

that is correct.

[strong cloud]

so i got it wrong somewhere

[mossy coyote]

you got to take - from ±

[vale wigeon]

let's double-check that you solved the quadratic correctly

,w 2t^2 - t - 2 = 0

[ocean seal]

Results provided by WolframAlpha

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

Requested by ann_dec

[vale wigeon]

ok yeah you're good

[strong cloud]

so the + just wont work?

i just discard?

[vale wigeon]

yes you just discard it

perfectly normal situation

[strong cloud]

alr

[vale wigeon]

you are not even left without solutions in this case

because you still have (1 - sqrt(17))/4 as the other value of sin(x)

which is between -1 and 1

[strong cloud]

yep

[mossy coyote]

But wait a minute

how will he calculate arcsin

[vale wigeon]

with a calculator

problem says to give answers in degrees to 1dp anyway

[mossy coyote]

Or maybe taylor series

[vale wigeon]

no, definitely not

this problem looks similar to cambridge papers

if it's that, then a calculator is definitely expected to be used

[mossy coyote]

in india we don't use calculator

we calculate by hand

[strong cloud]

i am currently doing fpm igcse edexcel

[strong cloud]

lol

[vale wigeon]

ok yeah igcse

[vale wigeon]

ok tell me the indian way then, how would you calculate arcsin((1 - sqrt(17))/4) by hand?

i'm curious now

[flint belfry]

@me if there is actually a way 😭

[lament pagoda]

Construct triangle…

[vale wigeon]

let's go somewhere else for this, like #geometry-and-trigonometry

@strong cloud do you have anything else left to ask or can you continue your problem from here

[mossy coyote]

sqrt 17 is 4.1 approx

[strong cloud]

would this be correct?

[vale wigeon]

,rccw

[mossy coyote]

Magic

[vale wigeon]

correct so far but you also have -40.6 + 180 as a solution in your range [and actually -40.6 itself is out!]

so careful with that

remember sin(2θ+30°) is periodic with period 180°

[strong cloud]

i dont understand

[vale wigeon]

do you know that sine is a periodic function?

[strong cloud]

does that mean it repeating up and down same way every time?

[vale wigeon]

informally yes

to say that in a way that's more applicable to problem solving: sin(x + 360°) = sin(x)

in particular this means the solutions of any trig equation of the form sin(x) = a will form infinite families of points spaced 360° apart on the number line

because for any solution you can add or subtract 360° to it any whole number of times and get another solution

does that make sense to you?

[strong cloud]

is it because it is 2θ so it is 180 not 360?

[vale wigeon]

overuse of the word "it", but yes.

[strong cloud]

alr ty

[strong cloud]

[and actually -40.6 itself is out!]
what does this mean?

[vale wigeon]

-40.6 is not in the range 0 ≤ θ < 180 so you should not put it in the final answer

[strong cloud]

yes yes

ok tysm everyone

.close

[lone heart]

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[wet vector]

how do ik whether these two vectors r lin indepdent or not?

[lament pagoda]

You could see if you can get the zero vector with a linear combination of the two vectors

Where their coefficients aren’t just zero ofc

[lone heart]

@wet vector Has your question been resolved?

[lone heart]

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[wet vector]

.reopen

[lone heart]

✅

[wet vector]

this is written down in my notes

i dont really understand how sin and cos are linearly independent

[lost terrace]

when we talk about linear dependence, we are talking about elements of a vector space

In this case, the sapce is the set of continuous function from R to R, for example

two vectors f,g (functions) in this space are linearely independent if $af + bf = 0 \implies a,b = 0$

[ocean seal]

Eduardo

[wet vector]

ye

[lost terrace]

so, let $a,b$ be real numbers such that $a\sin + b\cos = 0$

[ocean seal]

Eduardo

[lost terrace]

and this is a equality of functions, so what this really means is that for all $x\in \mathbb R$, $a\sin(x) + b\cos(x) = 0$ as two functions are equal if and only if they output the same value for all inputs

[ocean seal]

Eduardo

[lost terrace]

if we input $x = 0$ we get that $a\sin(0) + b \cos(0) = 0 \implies b = 0$

[ocean seal]

Eduardo

[lost terrace]

and if we input $x = \frac{\pi}{2}$, we get $a = 0$

[ocean seal]

Eduardo

[lost terrace]

so we conclude that sin and cos are linearely independent in the vector space of continuous functions

Did this make sence to you @wet vector ?

[wet vector]

why did u pick pi/2 and 0

surely that just proves it for them numbers only

and not the entire set of R

[lost terrace]

The equality $a \sin(x) + b \cos(x) = 0$ is true for all real numbers x. We want to find the value of $a,b$, so we need to pick values for x that gives us some information about the constants

[ocean seal]

Eduardo

[lost terrace]

x = 0 makes the sin disappear and the cos be 1 , so we just get b on the left hand side

and the same with x = pi/2

[wet vector]

ye i see

ok kl

that makes sense

thanks

[lost terrace]

np 👍

[wet vector]

.close

[lone heart]

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[stable grotto]

i'm solving 15

above is my solution

but the correct answer is

this

what is wrong with my solution ?

i do get radius of convergence 5 though

[vale wigeon]

this minus sign shouldn't be there

gives you a (-1)^n that should not be there

other than that you're fine

[stable grotto]

ok tnx

.close

[lone heart]

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[lone heart]

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[formal ginkgo]

what if we used the y like x = (5y)/2 is that going to be correct ??

[mortal trellis]

that would also still work, yes

[formal ginkgo]

okay tnx

.close

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[wise beacon]

2sin2θ+1=0

for 0∘≤θ≤360

[wise beacon]

how do i do this?

do i subtract 1 to make it 2sin2θ=-1

then divide by 2

[vague mirage]

Yes

[wise beacon]

to make it sin2theta = -1/2

is that fine so far?

[vague mirage]

Yes

[wise beacon]

im tired so il just replace theta with x

so like sin2x =-1/2

so 2x= -30

so x = -15

[vague mirage]

Do you know the value of sine where it's 1/2 or -1/2

[bright scarab]

u can use the identity

[wise beacon]

?

yes

30

[bright scarab]

Sin^2theta + Cos^2theta = 1
(Sintheta + Costheta) = 0

this one

[vague mirage]

A small mistake

[wise beacon]

?

[vague mirage]

Like in which coordinate sin is negative?

In which angle sine function is negative?

[bright scarab]

oh wait this is 2 sin2theta

[wise beacon]

yes

not squared

[vague mirage]

You know theta can't be less than 0 it's given in the question

[bright scarab]

ok then wat u can do is ask urself wheres the domain

[wise beacon]

oh

[bright scarab]

2theta = -1/2
2theta = -30 deg

[vague mirage]

In which quadrant the value of sin is negative

[wise beacon]

yea

thats what ive done

[bright scarab]

theta = -15 deg or 360 - 15

345 deg?

[wise beacon]

yh

[vague mirage]

Noo 0>theta>360

[bright scarab]

ah

[wise beacon]

im so confused

[bright scarab]

wait

i did it right

[wise beacon]

so theta cannot be < 0

[vague mirage]

Yup that's an answer

[bright scarab]

345

[vague mirage]

Yup

[wise beacon]

wait

so it cant be -15

[bright scarab]

make a unit circle

[wise beacon]

idk how

I just started as trig

[bright scarab]

take radius as 1

[wise beacon]

ok

[bright scarab]

then y coord is the sintheta

[wise beacon]

ok

[bright scarab]

x coord is cos theta

[wise beacon]

ok

[bright scarab]

half of radius is wat u want

[wise beacon]

ok

[vague mirage]

You can make the angle positive

[bright scarab]

so make a triangle with sides 1,1/2, random one

[wise beacon]

how?

[bright scarab]

no

[bright scarab]

in this case sin is negative

[wise beacon]

it has four solutions

for theta

[bright scarab]

wait wat

ok one more sol

is 180 + 15

[wise beacon]

yh

[bright scarab]

195

[wise beacon]

oh

so 345

195

what r the rest

I hate trig

[bright scarab]

bro i get only 2 sol

no others

[wise beacon]

yh

it doesnt make sense

il try the 2 solutions

[bright scarab]

ok

[wise beacon]

wth

uhm

[bright scarab]

lol wat

[wise beacon]

I hate doing maths online

its so anoying

[bright scarab]

try 165?

[wise beacon]

ok

still incorrect

dammit

[bright scarab]

bro lets take diff approach

[wise beacon]

ok

[bright scarab]

lets take 2sin2t as 2 * 2 * sint cost = -1

can we solve this?

[wise beacon]

I think so

[bright scarab]

but this only works if theres zero so no

[wise beacon]

oh

idk why my teachers give me the most random and arbitrary questions

[bright scarab]

bruh we made a silly error

[wise beacon]

fr

[bright scarab]

2 theta is 195 and stuff

not theta

[wise beacon]

OH FOR GOD SAKE

NOOOOOOOOOO

right

so half 345

and half 195

[bright scarab]

wait wrong hold up

[wise beacon]

ok

[bright scarab]

2theta is either 360 - 30 or 180 + 30

[wise beacon]

right

[bright scarab]

330/2 = 115

210/2 = 105

[wise beacon]

ok

[bright scarab]

maybe these?

[wise beacon]

ok

il try

no

dammit

[bright scarab]

bruh come on wats wrong

[wise beacon]

idk

[bright scarab]

we did it correctly

[wise beacon]

il try something else

[bright scarab]

sin2theta = sin330 or sin210

[wise beacon]

gimi 1 secx

sec*

yea i have no clue

[lone heart]

@wise beacon Has your question been resolved?

[limpid aurora]

2theta = 210 or 330

or well, 2 theta = 210, 570, 330 or 690

[wise beacon]

how did u get to that?

[lone heart]

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[wise beacon]

.reopen

[lone heart]

✅

[wise beacon]

two of the correct answers are 165 and 345

[lucid marten]

Where is the original question?

[wise beacon]

thats it

[lucid marten]

Let me give you a simple explanation on my end.

[wise beacon]

ok

thank u

[lucid marten]

But first you must understand some simple things:
The inequality you have is not the correct range. The inequality makes use of $\theta$ instead of $2 * \theta$. So we need to modify it to represent that

[ocean seal]

Pyp

[wise beacon]

ok

I realy just dotn get trig equations

[lucid marten]

So using the equation: 2 * $sin 2*\theta$ + 1. We rewrite it to look like this:
2 * $sin 2*\theta$ + 1 = 0 --> 2 * $sin 2*\theta$ = -1. Which becomes
$sin 2 * \theta$ = $\frac{-1}{2}$

[wise beacon]

is that mean to be like sin 2theta = -1 over 2

[ocean seal]

Pyp

[wise beacon]

yh

[lucid marten]

Yeah

[wise beacon]

ok

i get that

[lucid marten]

So with that we have the following:
Let us assume that $2\theta$ = $\alpha$
So sin $\alpha$ = $\frac{-1}{2}$

[ocean seal]

Pyp

[lucid marten]

Understood?

[wise beacon]

OH ITS THIS

YES

ah ok

[lucid marten]

So we need to get $\alpha$

[ocean seal]

Pyp

[wise beacon]

so alpha equals -30

[lucid marten]

Let me get there first

[wise beacon]

ok

sorry

[lucid marten]

$\alpha$ = $\arcsin(\frac{-1}{2})\newline$
But let us not consider the negative for now, assume it is invalid for now. So we have this:
$\alpha$ = $\arcsin(\frac{1}{2})\newline$
So we get the value to be $30$ degrees

[wise beacon]

ok

[lucid marten]

Chiil I check latex for sin inverse

[wise beacon]

?

[ocean seal]

Pyp

[wise beacon]

ok

[lucid marten]

So we know that the value we are looking for is somehow related to 30 degrees, but its not 30. Now remember what I said about the inequality:
$0 <= \theta <= 360$ But remember we are working with $\alpha$ not $\theta$. So we need to modify the inequality to capture all values of the range and thus becomes: $0 <= 2 * \theta <= 720\newline$. By multiplying by 2 we reach alpha and thus we have the correct range: 0 <= $\alpha$ <= 720

[ocean seal]

Pyp

[wise beacon]

but dont u divide by 2

becuase its 2sin

not .5 sin

[lucid marten]

Remember I already divided by two make it $\frac{-1}{2}$

[ocean seal]

Pyp

[wise beacon]

yes but its

theres 2 2s

[lucid marten]

Yes, the 2 for $\theta$ allows us to create the correct range of the inequality so as to easily find all values instead of having to guess

[ocean seal]

Pyp

[wise beacon]

ok

OH OK

YEA

yea sorry

[lucid marten]

Thanks to see we are on the same page

[wise beacon]

u multiply the range by 2

to get 720

ah

[lucid marten]

Yes

[wise beacon]

yea

im a bit slow sorry

idk why for trig

[lucid marten]

Let me continue so that you could see where we need to be careful. Takes some practice and convincing yourself, I wont lie to you trig is very easy if you get the basic concepts rights which are very easy

[wise beacon]

ok

yea thank u once again i just suck at trig

[lucid marten]

I also used to suck but I did alot of practice, but far from that let us do the solution

[wise beacon]

ok

[lucid marten]

Having this

Let us form a solution

[wise beacon]

ok

[lucid marten]

The value we have, 30, is the value for alpha but remember we ignored the negative, Here is where it comes to place

[wise beacon]

ohok

[lucid marten]

Remember the unit circle

[wise beacon]

yes i know of it

idont know what it mean tho

[lucid marten]

It is a circle of a radius of 1, hence called a unit circle

[crystal thistle]

trig really does suck

[wise beacon]

ok

r = 1

got it

my teacher said not to worry about it

[lucid marten]

Know the following:
A --> All values of tan,sin,cosine are positive
S --> Only values of sin are positive
T --> Only values of tangent are positive
C --> Only values of cosine are positive

[wise beacon]

but she said it makes trig so much easier

[lucid marten]

Abbreviated as All Students Take Coffee

[wise beacon]

ok

[lucid marten]

It goes anticlockwise always

[wise beacon]

ok

[lucid marten]

Never forget that

So it looks like this

[wise beacon]

ima make a note card of that

[crystal thistle]

wow! i never knew of this mnemonic

[lucid marten]

The 1st quadrant(first of 4), 2nd quadrant(second of 4) and so forth

[wise beacon]

ok

[lucid marten]

Now remember, the 30 we have is not actually 30, but we said we ignore the negative first. The negative means that the value of that angle,$\alpha$ will give a negative value when read. According to the unit square, the sin function gives negative at only two places: the 3rd and 4th quadrant

[ocean seal]

Pyp

[wise beacon]

ok

3,4 sin pos so in 3,4

got it

[lucid marten]

To get the angle for any trig function at any quadrant we use a simple approach

[wise beacon]

ok

[wise beacon]

uhm

[lucid marten]

Where x is the acute angle you have, in this case ,30

[wise beacon]

ok

[lucid marten]

To get the angles:
3rd quadrant = 180 + 30 = 210
4th quadrant = 360 - 30 = 330

But we are not done yet

[wise beacon]

ok

[lucid marten]

Remember our range 0 <= $\alpha$ <= 720. Which means we have values for $\alpha$ above 360. Dont worry as it is a circle we just add 360 to the already found values until they surpass 720, and the last value we have before 720 is the end:
Above 360:$\newline$
210 + 360 = 570$\newline$
330 + 360 = 690$\newline$
570 + 360 = 930 --> We cannot choose this value as it is above 720 so we stop here$\newline$
So we have $\alpha$ = 210,330,570,690

[wise beacon]

ok

[ocean seal]

Pyp

[lucid marten]

But remember: $\alpha$ = $2\theta$. So:$\newline$
$2\theta$ = 210 , 330, 579, 690$\newline$
$\theta$ = 105, 165, 285, 345.$\newline$
And like that we solve the process

[ocean seal]

Pyp

[wise beacon]

WOW

wow

thank u

[lucid marten]

I know it seems a lot, but if you get used to it you can solve them faster and with less time.

[wise beacon]

YESSSSSSSSSSS

thank u so much

I have another one now

il try solve it

[lucid marten]

Yeah, try to as many as you can, the maximum time for such equations should always be like two minutes as the process is straightforward

[wise beacon]

ok

ty

[lucid marten]

Remember you must always have the correct range: Without this, you may miss some values which are on the negative side

[wise beacon]

ok

[lucid marten]

Like $\theta$ may be -30

[ocean seal]

Pyp

[wise beacon]

ok

[lucid marten]

Well hope you benefited

[wise beacon]

lmao yep immensly

you should be working at my school

not my teacher lmao

[lucid marten]

Am just a student like you, always looking for more knowledge and trying to find a simpler and nice method to present solutions

[wise beacon]

dam

this time the base values positive

and from -180 to 180

[lucid marten]

Can I see

[wise beacon]

yes

this one may be easier tho

can you just sub a into 1/2

then make a = arc tan 2/3

[lucid marten]

Yes, but what is a according to you?

[wise beacon]

1/2

so u times by 2

at the end

[lucid marten]

At the end, yes but remember the range

[wise beacon]

it only has one solution tho

[lucid marten]

You must have the correct range

[wise beacon]

oh yea

so do u half it

so -90 to 90

[lucid marten]

In this case yeah

[wise beacon]

oh ok

so the base value is the only solution

[lucid marten]

Exactly and then you make sure to start at -90

[wise beacon]

ok

so its q1

[lucid marten]

q1?

[wise beacon]

yh

quadrant 1

[lucid marten]

What is arc tan 2/3

[wise beacon]

or do u not have to do that

21.14

[lucid marten]

So we need answers for positive values of 21.14 which means originally q1 and q3

[wise beacon]

oh ok

[lucid marten]

But q3 is beyond range so only q1

[wise beacon]

ok

so what do i do

now

[lucid marten]

If it was in -180 range we would consider q3 but the negative value

[wise beacon]

ok

[lucid marten]

This is because for negatives, we go clockwise in the unit circle

[wise beacon]

ok

wait

surely if a = 1/2

and 21.1 equals alpha

isnt it just 42.2

can u do that

[lucid marten]

No

[wise beacon]

oh

[lucid marten]

Trig functions are different

[wise beacon]

oh ok

[lucid marten]

Let me explain a little better

[wise beacon]

ok

[wise beacon]

my teacher did this as an example

can u do that

[lucid marten]

This is allowed as they got the arcsin of $\frac {\sqrt{3}}{2}$ and then went to solve for $\theta$

[ocean seal]

Pyp

[wise beacon]

oh

so we cant solve for theta

[lucid marten]

Am slightly confused, what is the main issue?

[wise beacon]

so if i did it the same way my teacher did

so + 2 and /3

to make tan 1/2 theta = 2/3

then cant u just substitute a for tan half theta

and make a equal arc tan of 2/3

[lucid marten]

You can totally do that, as a is not tan half theta but a = half theta instead

[wise beacon]

oh

[lucid marten]

You cannot absorb the trig function into a, this messes up a lot of logic

[wise beacon]

so would it be 42.3

ok

ah no

its wrong

[lucid marten]

Let me check

[wise beacon]

ok just go back to this

[lucid marten]

Dont go there

[wise beacon]

that makes more sense

oh

ok

ok

[lucid marten]

Let me check for you something, first

[wise beacon]

ok

ru joking

my calculator

was wrong

arc tan of 2/3 is 33.7

not 21.1

DAMMIT

[lucid marten]

Yeah

[lucid marten]

I wanted to confirm that

[wise beacon]

sorry for wasting ur time on that

im still gettinmg used to my graphing calculator

sorry

[lucid marten]

You use graphing calculator?

[wise beacon]

yh

for polynomials

and some exponential graph stuff

[lucid marten]

Us we used 82ms and went with it until we finished, but I preferred 100ms, gave me xtra functions. Made me powerful in math tests

[wise beacon]

fair enough

yh in the uk i use a casio cg 50

[lucid marten]

Not sure of the model, but I hope it has powerful functions like vectors,matrix,quadratic and cubic eq

[wise beacon]

yh

my phones dead

but it has aload of things

like 28 modes

[lucid marten]

Really powerful stuff then

[wise beacon]

yh

I coded chess onto it with my buddies

but thats besides the point

[lucid marten]

This is illegal in my school

[wise beacon]

fr

[lucid marten]

Yeah, we cant use programmable ones. If I had that then id be done in 1 hour

[wise beacon]

oh

yea theres an exam mode on it ho

[lucid marten]

I would simply basically bruteforce everything

[wise beacon]

lo

lol

yea they make us put it in exam mode though

[lucid marten]

So you got nerfed

[wise beacon]

so that removes everthing apart from graphs and marix mode

lol rep

yep

[lucid marten]

Unlucky, did my finals with and without my calc (100 ms). I almost did not finish the 2nd paper, but I did just on time

[wise beacon]

dam

[lucid marten]

But dont wanna clog the server, if you have any more questions just dm

[wise beacon]

ok

ty

cya

[lone heart]

@wise beacon Has your question been resolved?

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[kind violet]

k

[kind violet]

is this possible?

[alpine sable]

I don't understand what you wrote here, mind explaining?

[kind violet]

the maximum domain

of this function

[tacit arch]

Why can't -1 be in the domain

[kind violet]

bcs of x^2. the 2 makes it positive

[tacit arch]

And why is that not allowed

Is 2^(-1)/2 undefined?

[lone heart]

@kind violet Has your question been resolved?

[lone heart]

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• Show your work, and if possible, explain where you are stuck.
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[summer dirge]

?

[lone heart]

Channel closed

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as this action is irreversible, and this channel may abruptly lock.

[lone heart]

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[dusk ingot]

Un resorte está colocado en forma vertical. La masa del cuerpo es de 9 kg y la constante
del resorte es 25 N/m. Al inicio la masa se libera desde un punto que está a 4 cm arriba
de la posición de equilibrio imprimiéndole una velocidad hacia abajo de 2 m/s.

[lone heart]

@dusk ingot Has your question been resolved?

[dusk ingot]

<@&286206848099549185>

[alpine sable]

translate.

[dusk ingot]

A spring is placed vertically. The body mass is 9 kg and the constant
is 25 N/m. At the beginning, the dough is released from a point that is 4 cm above
from the equilibrium position, giving it a downward velocity of 2 m/s. What is the position, velocity, and acceleration of the mass at 10 seconds?

[lone heart]

@dusk ingot Has your question been resolved?

[regal dragon]

the force is weight

9.8*9

do u use kinematics to find spring amplitude?

[lone heart]

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[faint monolith]

I dont understand the last part of the solution

the part where it says since x and y are positive integers...

[fierce prairie]

hm what part specifically?

[faint monolith]

the last paragraph

i got the algebra part

then how do they just randomly know the values of x and y

[fierce prairie]

well, the problem gives that the sides are integers

so if (x-1)(y-1) = 2, and x and y are integers, x-1 and y-1 are also integers

and if two positive integers multiply to 2, they must be 1 and 2

[faint monolith]

it says 2 and 3 on there

also if its 1 and 2 the perimeter would be 10 not 14

[fierce prairie]

yeah, I was referring to x-1 and y-1

which makes x and y, 2 and 3

[faint monolith]

ok i understand thanks

[fierce prairie]

no problem, happy new year

[faint monolith]

wait

i have another question pls

i dont udnerstand this algebra

how did they go from the first line to the second line

and the second line to the third line

[fierce prairie]

difference of squares I guess?

p^2 - q^2 = (p + q)(p - q)

[faint monolith]

doesnt (10a+b)^2 turn into 100a^2 + 20ab + b^2

[fierce prairie]

yeah but you can avoid all that with the difference of squares

[faint monolith]

ohh i understand that

what abuot the third line though

[fierce prairie]

yeah that’s really strange..

is there any context to this problem?

[faint monolith]

this is the problem

[fierce prairie]

might be beyond my level, sorry

[faint monolith]

wait

ill show u the second part

of the solutions

Problem:

currently do not understand the third line of algebra

[lethal solstice]

second to third line?

[faint monolith]

yes idk how they turnedthe second line into the third line

[lethal solstice]

10a+b+10b+a = 11a + 11b

10a + b - 10b - a = 9a - 9b

[faint monolith]

ohhh NVM

thanks ik now

[lethal solstice]

skip in algebra 😭

[lethal solstice]

np

[faint monolith]

wait i also dont understand the part where they say a-b<=9

[lethal solstice]

a-b is positive because (a-b)(a+b) is positive

and a and b are positive

then since a and b are single digits, they range from 0~9

then the maximum difference is 9

[lone heart]

@faint monolith Has your question been resolved?

[faint monolith]

i have on emore question

it says

that a+b is a multiple of 11

then how do they add up to 11

because the multiples of 11 all dont add upto 11

like 44,55,66

4+4=8, 5+5=10 6+6=12

[lethal solstice]

since a and b are digits

their sum cant be over 18

so the only possible way for a+b to be a multiple of 11 is for the sum to be 11 or 0

and you dont want the sum to be 0 cuz then p wouldnt be a 2 digit number

[faint monolith]

huh

how is it a multiple of 11 if it sums to be 11

[lethal solstice]

11x1=11

[faint monolith]

??

sum means plus??

[lethal solstice]

we want the sum to be a multiple of 11

and the sum is restricted because each number a and b are from 0 to 9

so 0<=a+b<=18

but since we dont want a+b=0, the only way 11|(a+b) is if a+b = 11

[faint monolith]

ohh nvm i understand

thanks

[lone heart]

@faint monolith Has your question been resolved?

[lone heart]

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[alpine sable]

how to do this how many roots does the equations sin 6x = 1/3 have over the domains 0 <_ x < 2pi

[alpine sable]

.close

[lone heart]

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[indigo bramble]

in this problem, i tried doing this:

T cos 53 = mg

T = sin 53 mg

and sin 53 = 4/5

[real parrot]

alright

[indigo bramble]

so T = ( 4/5 ) * mg

but that is not present in the options

am i doing something wrong?

or are all the options wrong?

[real parrot]

no

you need the force exerted by tension on pully

not the tension itself

[real parrot]

tell me what is the angle between the string below and above the pully

[indigo bramble]

90 + 53 ?

[real parrot]

yes

[real parrot]

now

do you know the parallelogram law?

[indigo bramble]

of vector addition?

[real parrot]

yes

[indigo bramble]

yes i do

[real parrot]

ok so theta = 90 + 53 you have here

[indigo bramble]

yes

[real parrot]

the purple vector will be the resultant force by tension on the pulley

do you get it now?

[indigo bramble]

no

why would it be the purple vector?

[real parrot]

because the tension is tangent to the pulley

so although the string remains the same

the tension is still tangent

[indigo bramble]

okay that makes sense

[real parrot]

so we basically calculate the net force

by parallelogram law

$R^2 = 2T^2[1+cos(90 + 53)]$

[ocean seal]

Wither

[indigo bramble]

i cant work out where this is coming from

[real parrot]

parallelogram law?

do you need the derivation?

[indigo bramble]

if i cant then yes please

so lemme figure out what you did

[real parrot]

alright

[indigo bramble]

parallelogram law is for addition of two vectors right

[real parrot]

yes

[indigo bramble]

so you added two vectors to get that?

[real parrot]

yes

[indigo bramble]

so thats the value of the purple vector?

[real parrot]

yes R is the value

[indigo bramble]

so the purple vector is derived from the addition of the downward mg and the horizontal component of the force?

[real parrot]

no

its derived from both the tension forces +

[indigo bramble]

hmm

yea can you tell me how its derived?

[real parrot]

alright

you might know this

[indigo bramble]

yea

[real parrot]

so the orange vector is the sum of the two vectos

[indigo bramble]

yes

[real parrot]

alright

[real parrot]

okay so the sum of indivisual components of A and B will be the resultant as well?

[indigo bramble]

yea

[real parrot]

ok tell me what are the vertical components of A and B

[indigo bramble]

A sin theta and B sin theta?

[real parrot]

okay but for A, theta is not the angle between A and B

[indigo bramble]

but in this diagram theta is the angle between A and B vectors right?

[real parrot]

ok let me give you a simpler explaination

A will have no component in vertical direction

[sour shore]

does zutto know cosine rule?

[indigo bramble]

why not?

[indigo bramble]

nope

[real parrot]

ok

the vertical component of A will be Acos90

[indigo bramble]

why A cos 90?

[indigo bramble]

where is 90 coming from in this diagram?

[real parrot]

the angle between y axis and A is 90 is this fine?

[indigo bramble]

how

[indigo bramble]

the angle here isnt 90 right?

or is it not drawn to scale?

OH WAIT

SORRY

[sour shore]

in this case we have set A to be completely horizontal by convention

[indigo bramble]

i was thinkin A is another vector

A is just the x axis so its 90 degrees to y yes

i was thinking A is the side vector my bad

[real parrot]

yes, if we set them to an angle it might be difficult for @indigo bramble to understand

[real parrot]

yes

[real parrot]

A is a vector on the x axis

which has 0 angle with x axis

so Asin0 = Acos90 = its vertical component

ok so @indigo bramble what is the final vertical component of those vectors?

[sour shore]

so our strategy is to find horizontal and vertical components of both A and B, form a right-angled triangle with them, and then apply pythagoras

[real parrot]

exactly

[indigo bramble]

just B cos theta?

[real parrot]

no, vertical

[indigo bramble]

i still dont get this tho

[real parrot]

vertical will be $Bsin\theta$

[ocean seal]

Wither

[real parrot]

leave it, the angle between A and B is $\theta$

[ocean seal]

Wither

[real parrot]

ok?

[indigo bramble]

yes

[indigo bramble]

yes

[real parrot]

ok now tell me the horizontal one's

[indigo bramble]

sorry i got confused there:
vertical component - only B sin theta
hoirontal component - A cos theta + B cos theta

[real parrot]

ok, but isnt A just colinear with the x axis?

[indigo bramble]

yes so it would be just A?

[real parrot]

yes

[indigo bramble]

ok

[real parrot]

so vertical components are $Bsin\theta$ and horizontal are $A + Bcos\theta$

[ocean seal]

Wither

[indigo bramble]

yes

[real parrot]

now the resultant can be calculated using pythagorus theorem

$R^2 = (Bsin\theta)^2 + (A + Bcos\theta)^2$

[ocean seal]

Wither

[indigo bramble]

yea

[real parrot]

now solve this

and you will get the expression for resultant

[indigo bramble]

the resultant will be the square root of the expression to the right

[real parrot]

yes

open the brackets

[indigo bramble]

ok lemme work it out

[indigo bramble]

$R^2 = (A)^2 + (B)^2 + (2ABcos\theta)$

[ocean seal]

Zutto

[real parrot]

yo

you got it!

[indigo bramble]

yay

[indigo bramble]

is that related to this

[sour shore]

T=A=B, then simplify

[indigo bramble]

$R^2 = 2T^2(1 + cos\theta)$

[ocean seal]

Zutto

[indigo bramble]

and theta is 90+53

how would this further solve the problem?

@real parrot

[real parrot]

put theta = 90 + 53

[indigo bramble]

-sin 53 = - 4/5

?

[real parrot]

yes

[indigo bramble]

$R^2 = 2/5T^2$

[ocean seal]

Zutto

[indigo bramble]

so R = sqrt(2/5)T

@real parrot

but sqrt(2/5)T != mg right?

[real parrot]

?

[indigo bramble]

how would i equate that to the mg component?

im still now sure how to get the answer 😓

[real parrot]

T = mgsin57

[indigo bramble]

57?

where did you get that from?

[real parrot]

53

[real parrot]

e

[indigo bramble]

ok so thats not the answer

@real parrot so all the options have mg as a component

[real parrot]

t = mgsin53

[real parrot]

put it in this

[indigo bramble]

r = sqrt(2/5) * (4/5) * mg

[indigo bramble]

@real parrot this isnt one of the answers tho

did i do something wrong?

@sour shore can you help?

[sour shore]

lemme see

[sour shore]

ok give me thes eqn just after you've evaluated T=mg and cos(143 deg)

[indigo bramble]

@sour shore this one?

[sour shore]

mm don't simplify just yet

[indigo bramble]

okay

[indigo bramble]

@sour shore this one?

[sour shore]

not yet

just substitute in

[indigo bramble]

@sour shore so this one?

[sour shore]

yeah, then substitute in

[indigo bramble]

r^2 = 2 * (sin 53 * mg) ^ 2 * ( 1 + cos 143)

[sour shore]

where does the sin 53 mg come from?

[indigo bramble]

tension on the string T = sin 53 * mg?

since its going to be in the opposite direction to the weight thats being pulled by gravity

so the force counteracting that, since its stationary so it must be equal, is sin 53 * mg

?

[sour shore]

ahh, that's not true

let me draw you an example

[indigo bramble]

okie

[sour shore]

What would your reasoning say for this pulley system?

[indigo bramble]

its at rest

there is tension in the string

[sour shore]

what's the tensin on the left, and wnat's the tension on the right?

[indigo bramble]

both are the same, mg

[sour shore]

why is the right not mg sin 180 deg = -mg?

[indigo bramble]

T - mg = 0

T = mg

[sour shore]

Also consider this system; by your formula/reasoning T1 has no tension since it is mg sin 90 deg

[indigo bramble]

but this system isnt at rest currently, right? its not a stable system

theres no weight on T1 but there is on T2 so itll just slide down till T1 is no more

[sour shore]

it's the same system as this but with a different angle

oops

yeah I mean if the rhs is tied to a wall

[indigo bramble]

oh okay yea then theres tension there

[fast basin]

does anyone need help

[sour shore]

what's the tension?

[fast basin]

in the original set up conianing 53° ?

[indigo bramble]

same as T2

[fast basin]

it would be mg

[sour shore]

and so in the original too