help-17

[ornate ember]

try this instead

[tribal geode]

wait nvm i got it

[ornate ember]

excellent

[tribal geode]

Thank you

It was just the missing y'

.close

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[quaint flare]

how do i evaluate this since its not a quadrantal angle

[soft walrus]

$\cot\theta=\frac{\cos\theta}{\sin\theta}$

[twin meteor]

PajamaMamaLlama

[quaint flare]

i dont understand

i have to put it in terms of cot theta = x/y

[quaint flare]

?

[vast shale]

@quaint flare use the unit circle

[drifting jackal]

Use the unit circle and apply the above equation

[quaint flare]

thank you

[vocal sleet]

@quaint flare Has your question been resolved?

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[weary surge]

very confused on why it's not sin(x)^8/8cos(x) + C

[weary surge]

Although I do understand that I shouldn't have brought out 1/cos(x)

I just don't see how to get to the correct answer there

[graceful surge]

$\int_{ }^{ }u^{7}\cos\left(x\right)\cdot\frac{1}{\cos\left(x\right)}du$

[twin meteor]

Combustion

[weary surge]

ooooh did I drop out a cos(x) randomly

[graceful surge]

yeah and as you said you can't bring out 1/cos(x)

because it's not a constant

[weary surge]

so that becomes u^7 * cos(x)/cos(x) which is just 1, and then I just integrate u

awesome, thank you

[graceful surge]

[weary surge]

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[timber plaza]

Supposed to be using rational zero theorem to list all the rational zeros but i honestly dont know how

[raven owl]

Look up rational root theorem.

[timber plaza]

Oh wait its just synthetic division

[raven owl]

Yeah once you identify one of the zeros

[timber plaza]

How do i do that?

Cant x^3 just be one

Wait

Am i on the right path?

So confused

Nvm got it

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[timber marlin]

Anyone have any idea what they are asking here? The assingment is on addition and subtraction trig. Anyone know and videos I can learn this?

[mental falcon]

just better version :p

are you familiar with sum rule for tan function?

you could rewrite it this way then think about triangles and what tan(arcsin(x)) and tan(arccos(y)) would be

@timber marlin you alive there?

[timber marlin]

Yes apologies

I tried to eat some pizza.

Okay

Think about triangles

[mental falcon]

so if we know that the sin of some value is x, we can think of that as x/1 and use SOH CAH TOA to imagine what the tan value of that angle would be by associating $\frac{x}{1}=\frac{O}{H}$ if that makes sense

[twin meteor]

Soosh

[timber marlin]

Okay

So it would be like

O/H + A/H

Im sorry if I am being foolish. I have been struggling with identies this last 2 weeks

My exam is on monday 😦

[mental falcon]

so if we imagine triangle and some angle with side opposite of angle = x and hypotenuse 1 it would be like this

we can get 3rd side with pythagorean theorem $\sqrt{1-x^2}$

[twin meteor]

Soosh

[timber marlin]

OKay stop for a sec

[mental falcon]

so tan would be O/A = $\frac{x}{\sqrt{1-x^2}$

[timber marlin]

Is this just an identity I am expected to know?

[twin meteor]

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

[mental falcon]

i dunno, it's just thinking in terms of SOH CAH TOA

[timber marlin]

GotchaIh ijay

Ooops

[mental falcon]

if the sin is an angle is x and you know the value of sin is x then you can draw a triangle corresponding to that, at least thats how i'd approach this 🤷‍♂️

[timber marlin]

So

X^2 + Y^2 =1

so y side would be sqrt1-x^2

I think this makes sense

[mental falcon]

yes so we just got:
so $\tan(\arcsin(x)) = \frac{x}{1-x^2}$

[twin meteor]

Soosh

[mental falcon]

since that's tan, which is TOA and we do O/A

for that same picture

[timber marlin]

so x/sqrt?

1-x^2

But thats the same

[mental falcon]

hm?

[timber marlin]

Oh sorry I miss read

I thought you were asking me to answer

But you already did

[mental falcon]

so if we apply this thing

we basically just got tan a so far 😅

[timber marlin]

Jeeze

Okay

[mental falcon]

which is the tan(sin-1 x)

now we need tan(cos-1 y) also

[timber marlin]

yes

[mental falcon]

i wonder if theres an easier way to do this that im completely missing

there are often a lot of ways to approach trig problems, this is all i could think of

[timber marlin]

Im gonna guess not easily

Cause my prof said people always struggle with this chapter

Cause its a lot of memorizing and practice

[mental falcon]

ok, so can you draw a triangle for the second one, we know that the cos value is y, we can think of that again as y/1 and use CAH

i'll let you do it this time

[timber marlin]

so it would be a triangle with sides 1, Y, sqrt 1-Y^2

[mental falcon]

yes and make sure you draw them in the correct spots corresponding to the angle

[timber marlin]

how do you know which spot to put the angle?

Its it the unit circle angl

e

[mental falcon]

just pick any spot that isn't the right angle to place the angle, in the sin one then i labeled x OPPOSITE of the angle because sin is SOH which is OPPOSITE / hypotenuse, so opposite side = x, hypotenuse = 1

with cosine it is CAH = adjacent side / hypotenuse side which corresponds with y / 1

you do know soh cah toa right?

[timber marlin]

Yes

[mental falcon]

did you draw it?

[timber marlin]

Sorry im still trying to visually

ok I thin,k I get it im gonna draw it

[mental falcon]

ok just draw a triangle same as one above, draw the squiggly of the angle which is the one you decide you are focusing on first

[timber marlin]

done

so 1 for y would be the oppisite

[mental falcon]

i guess we can throw that in since its a triangle

[timber marlin]

yes

[mental falcon]

what is cosine in SOH CAH TOA

[timber marlin]

ajacent over hypo

[mental falcon]

yeah so use it then, there is no O next to C

$\frac{y}{1}=\frac{A}{H}$

[twin meteor]

Soosh

[mental falcon]

y=A, 1 = H, it's simple, it just tells you what they are

[timber marlin]

OH thats handy

I like mthat

[mental falcon]

thats the whole point of SOH CAH TOA lol 😄

[timber marlin]

That actually helps a ton haha

[mental falcon]

aha finally clicked

[timber marlin]

I never looked at it like that

y/1 made it make way more sense

[mental falcon]

yes

[timber marlin]

so 1 is the hypo

[mental falcon]

ok now get the 3rd side with pythagorean theorem, and use TOA, tan = O / A to say what tan is

[timber marlin]

1-y^2

[mental falcon]

sqrt(1-y^2)

[timber marlin]

woops

yes

[mental falcon]

so tan = ?

[timber marlin]

Okay im starting to see how identities are useful

It would be

x/sqrt1 -X^2 all over y/sqrt(1-y^2)?

[mental falcon]

ok you need to keep looking at the triangle

we said the adjacent is y right

and the opposite is \sqrt{1-y^2}

[timber marlin]

yea

[mental falcon]

so TOA for the second triangle is sqrt{1-y^2} / y not the other way around, dont assume it is like the first one

Opposite over adjacent

[timber marlin]

ohhh righ

okay

[mental falcon]

so we got $\tan\arcsin x = \frac{x}{\sqrt{1-x^2}}$

[twin meteor]

Soosh

[timber marlin]

yea

[mental falcon]

$\tan \arccos y=\frac{\sqrt{1-y^2}}{y}$

[twin meteor]

Soosh

[timber marlin]

yup

So first over the secon d

Gross

[mental falcon]

now we need to plug those in for tan a and tan b respectively in this whole equation lol

[timber marlin]

oh

shoot

No

Identies

Okay

wow

[mental falcon]

i really doubt you can even plug this into the answer thing on the website lol

[timber marlin]

this is crazy

I think I am gonna need to spend time in the math lab this weekend

[mental falcon]

are you sure there wasnt any more context to this question other than what you posted, like some relation between x and y?

[timber marlin]

sec

[mental falcon]

i dont see the question you asked

[timber marlin]

That was before

sec

Thats the section

[mental falcon]

well i guess it would just be plugging stuff into that tan sum equation from what we got and then trying to type it all in, it's a bit too much latex for me to keep going in the chat here

[timber marlin]

No no its all good

[mental falcon]

give it a go and see if it works i guess, best of luck 😅

[timber marlin]

YOu have helped a lot

I will give it a try

Thank you!

[mental falcon]

hopefully at least it will help you understand some concept with the soh cah toa even if it isnt a good solution to this problem

it does seem like a hard one

[timber marlin]

Ill make sure I will!

Cheers!

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[rapid shore]

can i differentiate the tangent line to get c?

[outer warren]

no

[rapid shore]

yeah thought so

[outer warren]

you differentiate y'

(and then sub in the point)

[rapid shore]

is there any faster way to do this?

[outer warren]

not really

[rapid shore]

what if i move the denominator to the other side?

can i still differentiate?

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[unborn quail]

how do i do those sinx/x graphs

[unborn quail]

like what are some general rules to generate the graph asap

similar to how sqrt(fx), has the x-ints of f(x) turn into vertical tangents

and u wipe out the original graph below the x-axis

[somber portal]

wait hold up, wdym the x intercepts of f(x) are vertical tangents of sqrt(f(x))?

[unborn quail]

like i get questions like this

this graph is f(x)

[somber portal]

what graph

[unborn quail]

what would sqrt(f(x)) look like

those type of questiosn

shit liek this

the answer for this i A

[somber portal]

yeah but that has nothing to do with this statement, which is false

[unborn quail]

need help with this

[unborn quail]

they do tho

[somber portal]

no?

f(x) = x^2
sqrt(f(x)) = sqrt(x^2) = x
Does y=x have any vertical tangent?

[unborn quail]

say i have $\sqrt{f(x)}$

[twin meteor]

Big Chicken

[unborn quail]

take the derivative

i get $\frac{f'(x)}{2\sqrt{f(x)}}$

[twin meteor]

Big Chicken

[unborn quail]

so as f(x) approaches 0

the graidnet approached infinity

ie a vert tangent

[somber portal]

thats a particual case of a function, not a general rule

[unborn quail]

anyway can u help me wiht my question please?

[unborn quail]

.

[somber portal]

well, you would start with what information you have about sin(x), and about x

[unborn quail]

right

i so i know x = 0 is ana symptote

[somber portal]

it isnt

[unborn quail]

oh

[somber portal]

for it to be an asymptote, the limit has to be infinite in at least one side

have you proven that said limit is infinite?

this is precisely why i told you your statement was false

[unborn quail]

as the graph gets close to 0

it approached a vlue

the same value

from both sides

[somber portal]

yes. But that value is not +inf or -inf, so that function has not a vertical asymptote

[unborn quail]

bro what

that for sqrt(f(x)) graphs

[somber portal]

also not true

[unborn quail]

i thought i proved it

[somber portal]

and i did put a counterexample

[unborn quail]

byt taking the derivative

wait are we still taking about the same question

[somber portal]

if the function has an asymptote it's not differentiable at the point. So you cant use the derivative as a proof for vertical asymptote

yes

[unborn quail]

the sinx/x one right

[somber portal]

yes

[unborn quail]

ok

[somber portal]

look at your graphs

you got one graph where both lateral limits at x=0 are infinite. Either with equal or different signs

you got one graph where the limit of the function is 1

[unborn quail]

right

[somber portal]

and you got another graph where the limit of the function is +1 on one side, and -1 on the other

[unborn quail]

as it approaches 0 right?

[somber portal]

you can also see that the LEFT limit at x->0 is different in all 4 graphs. It can be either +inf (a), -inf (b), 1 (c) or -1 (d)

[unborn quail]

in my calc i did like sin(0.00001)/0.00001

[somber portal]

that's not a limit

[unborn quail]

and got like a decimenl

[somber portal]

you have to prove it analytically

[unborn quail]

hmm ok

[somber portal]

if you take the left limit of sinx/x at x->0, you will obtain one of the four values.

[unborn quail]

hmm

$\frac{\sin {0^{-}}}{0^{-}}}$

[twin meteor]

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

[unborn quail]

is this what u are saying

[somber portal]

yes. However that is an indeterminate form of the class 0/0, so you need to work through it

[unborn quail]

sorry what an indeteminate form

[somber portal]

an evaluation that does not give you enough information to obtain the value of the limit

[unborn quail]

so i have to get rid of x from the denominator

[somber portal]

you have to rewrite the function in a way that allows you to obtain a value instead of an indeterminate form

[unborn quail]

isnt sinx / x an identity

[somber portal]

no

[unborn quail]

oh nvm

uhm

how do i remove x from the denoninator again

[vocal sleet]

@unborn quail Has your question been resolved?

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[grizzled halo]

set some boundaries on a, b, c
e.g. they have to be integers

[vocal sleet]

@random sorrel Has your question been resolved?

[random sorrel]

wht do u mean and how?

it is an integer

[outer warren]

could a be something like 11?

[random sorrel]

umm their sum is 11

thts the answer

so a is not 11

[sweet birch]

could one of them be 10?

[random sorrel]

no.

cus their sum is 11

so i guess not

[sweet birch]

why not 10, 1, 0?

[random sorrel]

ummm they are digits of a 3digit no.

but the numerical values of those numbers is individually distinct

hence, a+b+c is sum of the digits of abc

also the sum is not given

[sweet birch]

okay, so we can say each of a, b, and c is...what limitations on the number

[random sorrel]

its the answer we have to find

[sweet birch]

because they are digits

[random sorrel]

ig so

[sweet birch]

yes, what does "they are digits" work out to in mathematical terms

[random sorrel]

wht do u mean?

abc is anumber and a,b,c are their digits

we have to find a+b+c ...

with the help of the given equation

[sweet birch]

what is the biggest number a can be?

[random sorrel]

9

but tht is not given to

us

in the question

so we cant say tht

[sweet birch]

does 64*9 work in the equation?

[random sorrel]

i just know the answer from the solution key

[sweet birch]

if a is its biggest possible value (9), then does the 64a + 8b + c = 403 work?
if a is its smallest possible value (0), does it work?

[random sorrel]

we dont know the biggest possible value

[sweet birch]

well we know a is a digit.

[random sorrel]

its not in the question

yeah

[sweet birch]

the biggest possible number a digit could be is...

[random sorrel]

9

oh snap thts right mb

so we just hit and trial right?

take 8 common and do tht

[sweet birch]

slightly smarter than trial and error - like we know 64*a has to be less than 403, because otherwise the equation is impossible to solve

[random sorrel]

c might be 3 just get 403-3=400

yeah yeah i

[sweet birch]

but it can't be too small because otherwise the 8b+c = cannot equal (the rest)

[random sorrel]

yep

got it thnx alot

a=6

b=2

c=3

.close

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[floral ledge]

not rlly a qn but just checking if the formula 1/2ab sin c works in right angled triangles?

[outer warren]

the area formula?

capital C for the angle
works in all triangles

[vocal sleet]

@floral ledge Has your question been resolved?

[floral ledge]

ohh alright thanks!

[vocal sleet]

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[cobalt ocean]

!status

[vocal sleet]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[young jungle]

!status

[vocal sleet]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[young jungle]

1

Oops

[vocal sleet]

@thick raven Has your question been resolved?

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[vast shale]

Isn't it supposed to be 6?

There is no such anwser

[peak axle]

What makes you think six?

[vast shale]

Ohh i read it wrong

Sorry

.closr

.close

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[stable sigil]

How can I rigorously prove that {+, *} is not a functionally complete (boolean) set

[stable sigil]

Note: cant use 0-preserving or 1-preserving property right away, have to somehow get them if I were to use them

[vocal sleet]

@stable sigil Has your question been resolved?

[sullen moon]

try to prove there exists at least one Boolean function that cannot be expressed using only these two operations

Assume that { + , ∗ } {+,∗} is functionally complete. Then it should be able to express the not operation

then try to express the not operation using only + and *

then derive contradiction

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[hoary cypress]

In an acute triangle ABC, prove that secA + secB + secC >= 6

[hoary cypress]

can someone help me solve this question

[vocal sleet]

@hoary cypress Has your question been resolved?

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[rustic ridge]

i need help regarding how to use a sci cal to simplify a radical. II'm m not good at talking in english so bear with me

[dim quail]

post your quesiton

[drifting jackal]

It depends on the calculator, because not all calculators can simplify radicals

[rustic ridge]

i have this uhm fx-fx-82ES plus. I was able to simplify radicals with it with a fraction form answer but my sister borrowed it from me then ththe answers became decimals instead of frations

so i was wondering how can i make it a fration form

[drifting jackal]

If you ever have calculator problems, it is best if you look up the manual for it. But with what you showed, that might be a value that it has to show a decimal for

[rustic ridge]

uhm i actually solve it yesterday and this is what is showed (pic that i will send) i just check it now to see if my answer is correct and it is actually a hand me down calcu so i dont have the manual

[drifting jackal]

Google the manual

Mess with the settings if you press shift then mode

[rustic ridge]

ok copcopy

[drifting jackal]

So you are saying that if you typed in cos(22.5) into the calculator, it displays the fraction?

What if you typed that in again

[rustic ridge]

i am actually solving for a half identities

[drifting jackal]

.

[rustic ridge]

the same po

0.90.92

typo

0.90.9222

same to the pic i send earlier

[drifting jackal]

What I am asking is, if you typed in cos(22.5), did you see the fraction or decimal, the first time you did it

[rustic ridge]

oh i didn't type 22.5. I substitute the given to the main formula then i type it to calcu. Omg I'm sorry I'm such an idiot

the first pic that i send thaw was what i input on my first time i did it

I'm hopeless hshs why i am like this

btw, thanks for your effort and time

[drifting jackal]

I don't own that calculator myself so I don't know how it works

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[young aurora]

I need help with finding measurements of angles

[strong grove]

angles 2 and 1 lie on a straight line

and the property is that , angles on a straight line are equal to 180 degrees

so
<2 + <1 = 180

[young aurora]

So its 180?

[strong grove]

no

<1 is 30 degrees
<2 + <1 = 180
< 2 + 30 = 180

[young aurora]

if it aint 150 I'm bewildered

its been such a long time since I had to do this

How is 5 a 90 degree angle

[raven owl]

Angle 3 is 30 degrees, and there are 180 degrees in a triangle. 180-60-30=90

[young aurora]

yeah but 90 degree angles never look like that? I thought they alwyas held an L position

[raven owl]

It is not drawn to scale, it is drawn like that to make sure you dont guessed or measure the angle and actually use math

[young aurora]

so then its not actually a 90 degree angle because that isnt 90 degrees

[raven owl]

When people make figures of things it is not always possible to be accurate in the drawing. Thats why its more important to focus on measurements given rather than how its drawn\

But yes technically in the drawing its not 90 degrees

[young aurora]

sorry bro its been a while I just need a reminder of how to solve stuff like this... could you give like a quick rundown of how to solve these problems so I can do them on my own?

[young aurora]

ah ok

[raven owl]

Use stuff from this and that angles on a line add up to 180

[young aurora]

okay thank you

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[real loom]

I have a proof 0/0=1

[real loom]

Written in red

[vast shale]

What does x(1/1) even mean

[median crane]

x/x != x(1/1)

[wide sundial]

lol

0/0 = 0

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[orchid pier]

how do i turn the polar form (80/0) into rectangular form

[orchid pier]

or would it be easier to die the other way around

this is about complex numbers btw

[drifting jackal]

Sin and cos

Here's rectangluar to polar

[orchid pier]

and how do i divide 2 polar forms @drifting jackal

[drifting jackal]

For the angles, if you divide, you subtract the two

The numbers are just still division

Like this is multiplying so you add the angles

65 + -12 = 53

Division is just subtracting

[orchid pier]

and numbers multiply for real number

okay thank you nova!

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[unique swan]

Is this setup right or no:
https://imgur.com/ILx7J0N

[storm lance]

the formula is a^2+b^2 -2bccosA so it is right but its not 48 its 10x14x2

then square root the answer

[unique swan]

I wanted guidance though, not direct answers. Otherwise, I will not learn as much. But thank you. I added instead of multiplying.

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[frail eagle]

so i wonder how to find the angle phi. theta is simply atan(y, x) i understand why. But from my understanding phi is atan( sqrt(x^2 + y^2), -z ). but somehow its the opposite why is that?

[frail eagle]

thats the formula but why?

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[hollow scaffold]

Help

[hollow scaffold]

can a quadratic function have the same range as an exponential

[brisk moss]

what do you think?

[vocal sleet]

@hollow scaffold Has your question been resolved?

[hollow scaffold]

No

[brisk moss]

why?

[hollow scaffold]

Is it yes

?

[brisk moss]

i won’t say yet haha

why do you think no?

[hollow scaffold]

O ok

Its bc

The quadratic would have the vertex point included

So like [0, infinity)

But exponential cant include

That point

Cus it doesnt touch

[brisk moss]

yea that’s right

[hollow scaffold]

O ok ty

[brisk moss]

there are only two flavors the range for a quadratic can come in

and two for an exponential

[a, infty) or (-infty, a]
vs
(a, infty) or (-infty, a)

[hollow scaffold]

Ahh okay

I also have a question

It was like luigi starts at a height of 8 ft and falls down with an initial velocity of 5 ft

Create an equation

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[vast shale]

can i get an explanation of this

is the NN backpropegated?

[sullen moon]

yes it is

backpropagation is needed to iteratively update the weight of the neural network to minimize the difference between the predicted Q values and the target Q values

@median crane

deep learning?

[median crane]

wth is dis

[sullen moon]

advanced stuff I will teach u in the future

[median crane]

Okay Master Chris

[sullen moon]

np

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[vast shale]

what does each layer represent

and what are the q-values?

[vocal sleet]

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[vast shale]

.open

what does each layer represent
and what are the q-values?
@sullen moon

[vast shale]

I need help with this

[tidal umbra]

@vast shale you have such a bad timing LMAO

bot's dead

[vast shale]

Lol

[dull bear]

[as a note, don't start channels with a message beginning with . because it won't get registered and you could lose your channel]

[sullen moon]

q values represent the quality of a particular action taken in a particular state

[vast shale]

it got autoclosed

Damn😂

[sullen moon]

it sort of tells the agent how good an action is in a given state

there are like input, hidden, and output layers

[vast shale]

is the q-values the actions

and the NN is the policy?

[sullen moon]

im in my spanish class, will get to u later

@median crane

[median crane]

Si

[sullen moon]

lol

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

?

[vocal sleet]

@vast shale Has your question been resolved?

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[vast shale]

Would you factor out a 4

[vast shale]

to make it x

how would you factor it out

my initial thought was make it 0.5sin(4(x-x/64))-5

not sure

if that's right

[pale perch]

to do what?

[vast shale]

to get it in the form $y=asin(k(x-p))+q$

[twin meteor]

RecRio

[pale perch]

you can factor out 1/4 within the argument yeah

[vast shale]

so k would be 4?

[pale perch]

1/4

wait

[vast shale]

1/4 x 1/4 is 1

[pale perch]

combine the x/4 and x/16 before you do anything

[vast shale]

can't

[pale perch]

you can

[vast shale]

it's my phase shift

[pale perch]

youll get your k by default

it isnt

a phase shift is a constant

that, is not

[vast shale]

ohhhh

lol

the x

flew over my head

thanks for clarifying

.close

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[vast shale]

.reopen

wait a minute

[vocal sleet]

✅

[vast shale]

@pale perch

it's supposed to be

theta over 16

not x

typo

[outer warren]

pi or theta

[vast shale]

this is the actual question

pi

😂

you can tell how out of it i am

but

in this instance

you would take out 4 correct?

wait no

1/4

(1/4(x-pi/4))

[pale perch]

yeah

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[exotic remnant]

How do i prove by induction that {a2n} (even subsequence) of a recurring converging sequence converges?

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[proper rampart]

y intercept is where the graph touches the y-axis

you can find slope by:\$\frac{0-120}{100-0} = -\frac{6}{5}$

[twin meteor]

Tangerine

[proper rampart]

equation of the line is expressed in the form of:\$y=mx+c$,\ where m is slope, and c is y-int

[twin meteor]

Tangerine

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[silk charm]

I made the common ratio be equivalent to r so i have 2196 = 4r^6 since 2196 is the last term and 4 being the first, and thats where I am stuck currently, so how do i solve for the 5th term?

[pale perch]

find r

[silk charm]

Ye but if you divide 2196 by 4 then take its 6th root it becomes irrational

[pale perch]

,w (2196/4)^(1/6)

[twin meteor]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by aldrnari_

[pale perch]

so what

theres no rule r cant be irrational

as long as its real in this context, its fine

[twin meteor]

Result:

549

[outer warren]

2916 not 2196

[pale perch]

aha

[silk charm]

Ahh

3

Zamn

Then i just take it to the power of 5 right?

Or 4

[outer warren]

what's the expression for the 5th term in terms of r?

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[tired copper]

Prove that sina/cosb + cosa/sinb = 2cos(a - b)/sin2b

[gaunt sparrow]

Put the LHS on a common denominator. It should be pretty simple now to use the angle sum formulas to get the right numerator and the double angle formulas to get the right denominator.

[tired copper]

I made it into one fraction to get (sinasinb+cosacosb)/sinbcosb = cos(a-b)/sinbcosb

[tired copper]

i was thinking about using sin2a but i'm not sure

for the denominator

[tired copper]

so sin2b = 2sinbcosb but i feel like i'm close but 'im missing smth

[gaunt sparrow]

It is that identity

What's sinbcosb then?

[tired copper]

sin2b/2

[gaunt sparrow]

Ok and then

[tired copper]

can i multiply with it then?

[gaunt sparrow]

Just plug it in no?

You have it in the denominator already

[tired copper]

oh yea alr

got it thanks

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[light lodge]

999= -1 (mod 1000)

[light lodge]

help

why do i have a negative nubmer

[cyan shadow]

if you take away 1000 from 999 you have -1 left

[light lodge]

we are dividng

not subtracting

[lilac plaza]

mod = remainder?

[cyan shadow]

(we went through this in another channel, Brian needs an intro to mods lecture/content)

[outer warren]

$999 \equiv 999 + 1000k \ (\mod 1000), \text{for}\ k \in \Z$

[twin meteor]

ℝαμΩℕωⅤ

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[north anvil]

If w is a non-real root of z^5 = 1, show that the other non-real roots can be expressed as w^2, w^-1, and w^-2

I've went with w^4 + w^3 + w^2 + w + 1 = 0

[north anvil]

then $\bar w + w + \bar{w}^2 + w^2 + 1 = 0$

[twin meteor]

魔法の🌙kitty!

[north anvil]

but am stuck from there

though i just realised z bar = z^-1
so if i just convert it to w^-1 + w + w^-2 + w^2 + 1 = 0
does that sufficiently show that they are hence roots?

wait can u send it again even if its not relevant i wanna save it somewhere ☠️

[acoustic terrace]

so the solutions of $z^5$ is $z_k =e ^{i \theta_k}$ where $\theta_k= \frac{2 \pi k}{5}$ for $k=0,1,2,3,4$

[north anvil]

mmhmm?

[twin meteor]

Zaitzer

[acoustic terrace]

now lets see, for what $(e ^{\frac{2 \pi i k }{5}})^2=1$ where $k\neq 0$

[twin meteor]

Zaitzer

[acoustic terrace]

$(e ^{\frac{2 \pi i k }{5}})^2 = $(e ^{\frac{2 \pi i (2k) }{5}})$

[twin meteor]

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

[acoustic terrace]

you now need to check if w is a solutions is w^2 a solution etc

ignore the other stuff i sent

[north anvil]

tyty

.close

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[wide sundial]

I kinda remember the answer from birthday paradox, but how was i supposed to do this without that memory?

[wide sundial]

It's like 23 or something right so i guess the answer is 24?

But without that memory, this question is not doable right (especially with a basic calculator)?

,w (365-n)!/(365)^n = 0.5

[twin meteor]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by kanna22

[wide sundial]

even wolfram struggles finding n

[dark kiln]

trial and error

[wide sundial]

till 24?

[dark kiln]

binary search

,calc (365-30)!/(365)^30

[twin meteor]

Result:

Infinity

[wide sundial]

lol

wait is my equation wrong or something

[dark kiln]

the equation is weird yeah

[dark kiln]

1 minute

[wide sundial]

1 - that result i guess idk

,calc 1- ( (365-30)!/(365)^30)

[twin meteor]

Result:

-Infinity

[wide sundial]

lol

that doesn't make sense

the denominator is bigger than the numerator

[dark kiln]

https://www.wolframalpha.com/input?i=(product+k+%3D1+to+20%2C+366-k)+%2F365^20

[wide sundial]

o

why does my notation not work

,w (n! * (365 choose n))/(365^n) when n = 20

[twin meteor]

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Requested by kanna22

[wide sundial]

okay but this looks like it works

was the (365-n)! part wrong

[dark kiln]

with just a calculator, you'd have to search until 24, should be less than a minute

[wide sundial]

no calculator allowed ig

[dark kiln]

kinda slow and humiliating but barely doable

[wide sundial]

only basic one that can't display more than 8 digits

i'll just skip the questison ig

seems contrived

[dark kiln]

i mean basic one

[dark kiln]

that's P(365,n), (365-n)! is not P(365,n)

[vocal sleet]

@wide sundial Has your question been resolved?

[wide sundial]

how can a basic one even do all of that

it's like the windows one

but can render like 8 digits

so way worse

[dark kiln]

you enter 364/365

then you do *363/365

24 times, ~2 seconds per step

[wide sundial]

that seems crazy lol

but okay okay

I have another question actually

how to find the IQR of numbers that are like smaller than 4?

I mean IQR of a set of numbers that have less than 4 elements?

i should probably start a new channel cuz new question

thanks @dark kiln

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[lime kraken]

i have no idea to where even start with thisd

[broken nimbus]

Have you heard of multiplying by conjugate?

Or L'Hospitals Rule?

[lime kraken]

i havent this

[lime kraken]

this yes actually

[broken nimbus]

Alright, we see we have a radical here.

[lime kraken]

so the way to go is to multiply by conjugate

[broken nimbus]

Try it out and see what happens.

[lime kraken]

oka ytu

okauy i got it i forgot for some reason ty

.close

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[broken nimbus]

u betcha.

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[mellow echo]

for graphing logs, when finding the x variable we take 1/k as in my notes to find the transformation of x.
can someone explain to me in words why it is 1/k * x and not just k * x so i can maybe understand / memorize this conecpt better

[broken nimbus]

k simply represents a factor

if we want to horizontally compress the function we would do k * x

Conversely, if we want to horizontally expand the graph its simply 1/k where k just represents how much we have expanded it by in this instance. Ex. 1/2 we have horizontally expanded the function 2 times horizontally. It would take 2 times the amount of effort to get the same result than having the factor be simply just 1.

I guess your probable thinking was if k just represents a factor at the end of the day, then what is the need for 1/k as we can choose anything we want for k?

[mellow echo]

maybe im thinking why we need to change k into 1/k to find the value of x and not just k*x

to find the new value*

[broken nimbus]

I see.

This one?

[mellow echo]

yeah

[broken nimbus]

Well remember what I said about horizontal reflections and expansions.

In our scenario 1/2 on the log function horizontal reflection means that its going to take 2 times the amount effort to get the same output for the x values.

[mellow echo]

2x vs the original graph for every +1 change in x value

[broken nimbus]

mhmmmm.

Also do remember that we horizontally translated it afterwards.

But remember the order of transformations

R - Reflection
V/H/Exp/Comp
Translation

I believe for the top 2 it doesn't matter if we swap the order, but abiding this order is easier from personal experience.

[mellow echo]

.close

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[spice garden]

hi

[spice garden]

does someone know how to factorise this

so i get a lower degree

[vast shale]

Oh it’s hard

[spice garden]

yeah

<@&286206848099549185>

[tepid beacon]

does lambda=2 give you a root

looks like it does

then factor theorem

[spice garden]

how u see that

[median crane]

look up rational root theorem and then its just a matter of trying a few numbers

[tepid beacon]

all the coefficients are powers of 2 so it felt like a natural first guess

[spice garden]

ow

okay i will try

[vast shale]

You can use long division and from there it’s easy

[spice garden]

owww ye you're right

i forgot all about that

wait so what divides what how do u write it up

i know how to do the division tho

[median crane]

you original expression divded by lambda-2

because 2 is a root

[spice garden]

like so?

[vocal sleet]

@spice garden Has your question been resolved?

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[cedar kernel]

whats wrong

[vast shale]

!status

[vocal sleet]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[indigo karma]

I get here but I am not sure how to get it to the form it is asking for

[cedar kernel]

you can simplify more

-1 looks alone dont you think?

[cedar kernel]

and the thing they asking you to compare with does not have a "-1"

so..

[cedar kernel]

what

why did you equate it?

huh

[indigo karma]

Am I not meant to?

[cedar kernel]

no?

just take in the -1

like make the base the same

for -1 too

[vocal sleet]

@indigo karma Has your question been resolved?

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[dapper cedar]

So

[dapper cedar]

lim x goes to pi/2 of sin(x)/cos^2(x) - tg^2(x) and WITHOUT LHOPITAL

[flat whale]

Why do you even need l'hopital's

Is this your problem
$\lim_{x \to \pi/2} \frac{\sin(x)}{\cos^2(x) - \tan^2(x)}$

[twin meteor]

riemann

[flat whale]

^

[dapper cedar]

No

[vast shale]

It makes sense if tan^2(x) isn't in the denominator

[dapper cedar]

The tan^2 x is alone

Yes

[outer warren]

combine into single fraction,
factorisation/conjugates/pythagorean trig identity

[dapper cedar]

Yes

I understand

Sinx - sin^2(x) nominator, cos^2x deniminator

What next?

[outer warren]

factorisation

[dapper cedar]

sinx(1-sinx) / 1 - sin^2(x)

Denominator is just formula for cos^2 x

What next

[outer warren]

1 - sin^2(x)
can be factorised

[dapper cedar]

? How

[outer warren]

1 = 1^2

[dapper cedar]

Wait what

[outer warren]

difference of two squares

[dapper cedar]

Can you write exacgly

Oh I see

1-sinx x 1 + sinx

Right

.solved

!solved

[tidal umbra]

.closed

.close*

do this

[median crane]

.close

[vocal sleet]

Channel closed

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[jaunty sandal]

If all sides of a cube are extended, the space is divided into a number of areas. How many?

[magic wasp]

As many as a cube has faces, edges, and vertices

[jaunty sandal]

3^3?

If you extend all the sides of a cube, you create a total of 27 areas. This is because each side of the cube has three different faces (front and back, top and bottom, and left and right). By extending these side surfaces, 3^3 = 27 areas are created in space. is this correct?

[magic wasp]

If you count the inside of the cube, yes

[jaunty sandal]

Okay thanks

.close

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[young jungle]

https://gyazo.com/f262b5cc1b63f62295d52a652041db25
would the co domain for all functions be set S?

[young jungle]

and the range would be set S for f(n), h(n) but for g(n) it would be set S excluding 4,5

[tepid plinth]

range is the set of all values the function takes

wait

oh okay

[tepid plinth]

u listed them in different order, my bad

i think i need a lesson on what is range and what is co domain

[young jungle]

range is what the actual values the function outputs whereas co domain are the possible values it can output

i was just wondering if the co domain would be S as it states "possible"

[tepid plinth]

by definition then co domain can be R the set of reals 😆

[young jungle]

and the range would be set S for all functions as every x in (x,y) contains the set S

i just want to know if my thinking is correct

!status

[vocal sleet]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[young jungle]

4

[tepid plinth]

since all the functions are from S to S, both domain and co domain of them will be S

so you're correct

[young jungle]

ok so the range it will also be set S for f(n) and h(n) but for g(n) it would be set S excluding 4,5 as min(3,n) will make it so y values in function g(n) that exceed 3 will always be 3 i.e: {(4,3), (5,3)} ?

[tepid plinth]

yes

[young jungle]

Ok thx

.close

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Channel closed

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[vocal sleet]

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[opal dragon]

!help

[vocal sleet]

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

[opal dragon]

Is this right

[ornate ember]

channel is occupied, can you open a new one in #❓how-to-get-help ?

[ornate ember]

can you show all of your work here? It's cut off in the screen shot

[tidal umbra]

wait doesnt this channel belong to naashe?

[ornate ember]

oh wait

[vast shale]

technically steel shld be the one to open a new one

[ornate ember]

yeah you're right

[tidal umbra]

LMAO

[opal dragon]

but it says my name on the thing lol

[tidal umbra]

idk if the calculation is right but yea ur logic is right

[ornate ember]

weird, Steelcrowe's screenshot was first

[opal dragon]

wait is it not right

[vast shale]

ah

[opal dragon]

because i found the volume of the cylinder then volume of the hemisphere and added them all up

[tidal umbra]

i didnt say ur not right

[opal dragon]

is it not right

or right

[ornate ember]

ok sorry for the confusion, thread belongs to Naashe. Can you open a new one Steelcrow? Also, can you provide a screenshot where your work isn't cut off? It looks like it's on the right track but I can't tell after a point

[cedar kernel]

bot somehow registered naashe's message first?

damn

[tidal umbra]

ur radius is wrong i think

[opal dragon]

lemme see

im sure its right its 6.59-4.78

[tidal umbra]

dont you think ur radius should be 1.81/2?

because 1.81 is the combined length of radius of first hemisphere and second hemisphere

[opal dragon]

Nah

cause its a hemisphere

I can see that I made a mistake nwo

[tidal umbra]

yea good

[opal dragon]

.close

[vocal sleet]

Channel closed

Closed by @opal dragon

Use .reopen if this was a mistake.

[cedar kernel]

they both sent the message at the same time
steelcrow message :1175485961497477191 : 11/18/2023, 10:51:32 PM IST
naashe message : 1175485961900134400:11/18/2023, 10:51:32 PM IST

steelcrows message however was millisecond faster like 0000000000096 unix millisecond faster

so bot got confused ig , also explains why diff things at our side of screen

[opal dragon]

its not that deep

[vocal sleet]

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[lime kraken]

For the first question

To replace x with -1

Should I consider x to be x or (x)

[lime gorge]

(x) ofc

[lime kraken]

Is it like that

Everytime it’s a variable?

Also why

Or is it just like that

Because yes

[median lake]

It literally said channel closed but ok whatever

[lime gorge]

Because yes

[vocal sleet]

@lime kraken Has your question been resolved?

[vocal sleet]

Channel closed

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[vocal sleet]

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[languid dune]

is this E or F

[languid dune]

could be another points just that i think its prob one of those

[hushed pewter]

The problem is kind of annoying because no exact point is labeled in the image, just a letter there but where is the actual point?

[pallid zenith]

i think there are multiple

[hushed pewter]

E and F both look reasonable. I see two more that looks reasonable too