help-13

[slow jewel]

Hmm that seems to be the only way then

[drowsy saddle]

so cost = rate x area

area = cost by rate

[slow jewel]

not times area

It's times length

Look at the units of rate

[drowsy saddle]

l * b?

l x b = cost by area?

[slow jewel]

No I meant here

Instead of area it's length

[drowsy saddle]

rate x length?

and we find length

and for area we again multiply length by breadth?

[slow jewel]

Yeah

[drowsy saddle]

oh

but isnt canvas of cone

a circle?

so how can we use lxb

[slow jewel]

It said it had a width

So I assumed it to be a rectangle

And the canvas is for the curved part of the cone

[drowsy saddle]

isnt width and diameter the same?

.close

[cedar kiln]

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[cedar kiln]

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[jovial gate]

Hi there, I need help on a certain thing, being Box Plots (And more information about it).

I'm confused about how to find the outlier in a set of numbers, for example, 1, 1, 2, 2, 2, 2, 3, 3, 5, 7, 9,

And I also forgot how to find the mean, mode, range, and median. But I know what the maximum and minimum are, the minimum being the lowest number, and the maximum being the highest in the set order of numbers

And I'm not entirely sure what the different Quadrilaterals mean. I think there are 3? At the top of my head.

So- yeah, my example question is the one above. 1, 1, 2, 2, 2, 2, 3, 3, 5, 7, 9.

I want to know how to find the Mean, Mode, Range, and the Median. As well as finding all the Quadrilaterals.

(And I need help making a box plot, using the example question above).

[jovial gate]

Because I'm struggling with this in math currently.

[royal loom]

I can give you a few quadrilaterals but idk if I can give u all of them

[jovial gate]

Tell me how to find them

Don't tell me the answer

[royal loom]

good point

quadrilaterals are shapes with four sides

[solid juniper]

[royal loom]

so to find some you can just start drawing a few like I did

[jovial gate]

For the numbers I said?

Bcs like

It's like box plot stuff

I might of worded it wrong

[royal loom]

oh do you mean quartiles?

[jovial gate]

yeah

mb

[royal loom]

post an example of a question you are confused about

[royal loom]

https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-data-statistics/cc-6th-box-whisker-plots/v/constructing-a-box-and-whisker-plot

Try watching this video for instructions

[jovial gate]

6th grade? man i'm dumb

[royal loom]

I don't think the 6th grade thing is accurate. No worries

[cedar kiln]

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[daring vault]

When I use pythagorean, one of the sides is 2+√3

[daring vault]

do I bracket (2+√3)^2 or not

or do I jus leave it as 2+√3^2

[clear umbra]

is your entire side 2+√3?

[daring vault]

yea

[clear umbra]

then yes

[daring vault]

so in my image im right?

[clear umbra]

yeah

[daring vault]

ok ty

.close

[cedar kiln]

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[daring vault]

.reopen

[cedar kiln]

✅

[daring vault]

btw

does it go like that

?

[clear umbra]

yes

[daring vault]

does this all make sense?

i havent rlly dealt with roots yet

xd

[clear umbra]

ok good

[daring vault]

and this?

[daring vault]

.

which one do I choose

the first one or the 2nd one

[clear umbra]

first

you cant break it up like that

[daring vault]

ok

and can I do something with the 4?

[clear umbra]

idk if this is denestable

[daring vault]

wut does that mean

[clear umbra]

means you dont have square root inside square root

everything only has one root

lemme try if its possible

[daring vault]

i mean this question is wut ive been workin on for the past 10 mins

maybe its my calculation or approach thats wrong?

[clear umbra]

oof i dont think i can denest it

[daring vault]

so I can jus leave x=√(8+4√3)?

[clear umbra]

yeah

[daring vault]

im watchin the solution, i dont quite get wuts going on

then he squared both sides

[clear umbra]

oh yeah this is also a possible way of denesting it

[daring vault]

and he expanded it?

[clear umbra]

yeah so you can compare it

like x + y = 8

[daring vault]

dam...

[clear umbra]

oh damn

i did think of denesting it that way but i didnt think it was possible

so i didnt try

[daring vault]

denesting is when you get the root out of a root right?

[daring vault]

.

[clear umbra]

yes

[daring vault]

can this work?

[clear umbra]

yes

[daring vault]

so if I have lets say, 2√3√4

it can js become 2√3*4?

[clear umbra]

yes

if you have multiplication/division you can split

but not addition/subtraction

[daring vault]

ok

so what he mentioned is one way of denesting it

[clear umbra]

yes

another way is to try writing it as (x+y)²

i tried that one because i thought it had a better chance of working

but apparently the other one was the way

[daring vault]

what grade lvl is denesting radicals?

[clear umbra]

idk

[daring vault]

i js looked up

ima js work on the denesting myself

then ill close the thread

[cedar kiln]

@daring vault Has your question been resolved?

[daring vault]

i dont get this part

how does he js get x=2, and x=6

pls heeeelp

[clear umbra]

quadratic formula

[daring vault]

x=-b±√b^2-4ac/2a?

[clear umbra]

yes

[snow pasture]

no you just caculate it after factoring

[clear umbra]

or factoring is easier

[snow pasture]

quadratic is good, but if you can solve it with factoring, use factoring

[daring vault]

i got to this

[snow pasture]

easier, and required less time

[daring vault]

and i dont know wut to do next

[snow pasture]

no, its x = 2 or x = 6

[daring vault]

huh?

[clear umbra]

x-2 = 0

x=2

[snow pasture]

do you see (x-2)(x-6)

yes

[clear umbra]

or x-6=0

x=6

[snow pasture]

this

[daring vault]

yes

[snow pasture]

so since x-2 = 0

x = 2

do you get this part?

[daring vault]

why is x-2=0

where does that come from?

this?

[snow pasture]

its rule thing like 1 + 1 = 2

yes

[daring vault]

so I split them up?

[snow pasture]

think about those simple quadratic fomula

0 = x^2 -8x +12

which (x-2)(x-6)

with this factoring

we can see x either can be 2 or 6

if this part doesnt make sense, substitude the number and check out the result

[daring vault]

i dont get after "with this factoring"

[clear umbra]

you got two answers right

[snow pasture]

ok let me do this step by step

you have quadratic function

which x^2 -8x + 12

you get this part?

[clear umbra]

(x-2)(x-6) being equal to 0 means x-2 or x-6 has to be 0

[daring vault]

yea

[clear umbra]

because multiplying 0 gives 0

[snow pasture]

now if you factor those, you get (x-2)(x-6) right?

[daring vault]

yea, its this part i dont get

[daring vault]

yea

[snow pasture]

you get this parT?

[daring vault]

yes factorisation

[snow pasture]

and you have hard time understading

why x= 2 or x= 6

yes?

[daring vault]

yea

[snow pasture]

its a rule thing its memorizing part tho

like + - / *

[daring vault]

plus minus divide times?

[snow pasture]

yes we dont trying to understand that sign

but we know what it means

the thing is 0 = x^2 -8x + 12
and 0 = (x-2)(x-6)

these are like diffrent diction, but the same meaning

[daring vault]

so I can jus split them up and both will equal to 0?

[clear umbra]

yes

[snow pasture]

yes

[clear umbra]

because one of them (or both) must equal 0

[snow pasture]

you get exactly same result if you use quadractic formula

[daring vault]

so wutever rule that is allows me to split them up or wut

[snow pasture]

yes

try out factoring first

if factoring doesnt work, then use quadractic

cuz factoring is lot lot easier

and shorten your process

if you need to test the theory, substitude the number

for this case, x = 2, x= 6

[daring vault]

goes like that

[snow pasture]

x^2 -8x + 12
4 - 16 + 12 = 0

[daring vault]

right?

[clear umbra]

yes

[snow pasture]

yes

[clear umbra]

now you go back to your second equation

x + y = 8

[snow pasture]

36 - 48 + 12 = 0

[snow pasture]

whats this

[clear umbra]

now put in when x = 2

[snow pasture]

it looks like Mayan civilization langauge

[daring vault]

i already substituted y with x-8

[clear umbra]

what do you get for y

[daring vault]

thats my mouse

[snow pasture]

oh

lmao

[daring vault]

im doing math on canva 💀

[snow pasture]

is thsi calculus one ?

[daring vault]

idk, their thing looks weird

[clear umbra]

you have two x values right, x = 2 and x = 6

[daring vault]

yes

[clear umbra]

now put x = 2 first into it and see what you get first

[snow pasture]

I think at this point you can solve it your own, but feel free to ask quesiton

[daring vault]

6

so y=6 or 2?

[clear umbra]

yes

x = 2 y = 6

or x = 6 y = 2

[clear umbra]

but since the order of x and y doesnt matter, you can just pick one

it can be √2 + √6

[snow pasture]

btw I wish someone ask questions about calculus 2

[clear umbra]

or √6 + √2

[snow pasture]

need to work on my calculus 2 proficiency

[daring vault]

i got it

that was tough

the denesting

[snow pasture]

ur good

keep working hard

everything is easy at the end

[daring vault]

ight

ill

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[burnt seal]

hello

[burnt seal]

i need help

im having trouble with part 2 of this problem

What do they mean by "free variables"

and how can i find the direction of rulings?

[cedar kiln]

@burnt seal Has your question been resolved?

[terse cobalt]

yo pls help how do I solve these 2d shapes idrk get it

[late lotus]

!occupied

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[heavy crypt]

if i want to calculate the volume by using an integral am i allowed to just put the x=6 into y=6 because you have to wallow it around the y-axis but the formula only goes for the x-axis

[crimson sedge]

what formula exactly

[livid hound]

there are different methods/integral setups depending on how you rotate

[heavy crypt]

.close

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[vapid lotus]

do the definable real numbers have the same cardinalty as the naturals?

[short blade]

what do you mean by definable

if you're asking if R and N have the same cardinality, no

[vapid lotus]

numbers that you can write a definition for

[cedar kiln]

@vapid lotus Has your question been resolved?

[short blade]

can you elaborate @vapid lotus

what are your restrictions on the definitions. are you working with a specific language

[vapid lotus]

I have no idea

a undefinable number is a number you can't define

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

[short blade]

you'll need to specify what you're looking for to get a clearer answer

though i will say if you interpret definable as requiring a finite description, then it is immediately countable, yes

there is an mse post going into it more in depth
https://math.stackexchange.com/questions/988429/cardinality-of-the-set-of-all-numbers-that-modern-math-can-define

[cedar kiln]

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[vapid lotus]

.close

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[digital goblet]

Hey, trying to figure out the best way to solve this question. They allow you to use a ti84 graphing calculator. Tell me: Is it better to solve this by turning it into an equation like 100,000a + 10,000b + 1,000c... etc. or is it easier to simply list out the digits from 0 to 9 for each letter and then use process of elimination to find what the digits must be?

[digital goblet]

@wicked mantle

I solved this one a few days ago and used process of elimination but it took a while, added on to the fact that this new one is a bit more tricky to do it with. https://cdn.discordapp.com/attachments/903481109420048385/1217987302731812866/Screenshot_2024-03-06_235241.png?ex=66060612&is=65f39112&hm=ae6a8ba9f5ce5cda0aeb9bb1c92334daea8eb9dc2cb4d1ba4190cadaa69d023d&

[digital goblet]

Anyhow, the answer to this one is 818181

[main needle]

yea there's not much process of elimination for this one, you can write it out as 100,000a+... and try to see a way to reduce the number of variables

like something on the left side basically shows up in the same way as the right side

[digital goblet]

How so?

[raven shard]

whenever u see question like this. always set up equation(s)

[main needle]

specifically bcdef is on both sides

[raven shard]

“process of elimination” is very rarely approach u want to use for this kind of q

unless for some reason the answer is just rly obvious or smth

[digital goblet]

This was my process of elimination approach for the one a few days ago (Q13):

[main needle]

like a logic puzzle

[digital goblet]

Yeah pretty much

It was kinda fun but during the competition its 50 minutes for 20 problems so...

[main needle]

there's a shortcut yea

just think of bcdef as a digit almost

[digital goblet]

really?

[main needle]

like on the left there's 10000b+1000c+... and on the right there's that but times 10

if you can call the whole blob x you can make progress

[digital goblet]

I see

Are you saying to write 100,000a + 11,111x = 4.5 * (111,110x + a) ?

[main needle]

the numbers in front of the x are off

x is like this big 5 digit number

although I guess it's the same thing honestly

[digital goblet]

So 100,000a + x = 4.5 * (10x + a)

[main needle]

I'd just write it as that yea

[digital goblet]

After that then what?

[main needle]

oh I haven't thought about the next step

[digital goblet]

I get 99,999a = 44x

Or

No

99,995.5a = 44x

That's what I get

[main needle]

oh you said you get to use a calculator that's huge lmao

[digital goblet]

Yeah

[main needle]

a has to be even right, to get the .5 to work out

and decently big like 4,6,8

[digital goblet]

Yes

[main needle]

so just plug those in and pray

[digital goblet]

Wdym

How do I plug this in

[main needle]

like a=4 to find x

[digital goblet]

Oh

[main needle]

and x is most of the 6 digit number answer anyways, idk how you're going to tell which one's right but hopefully something will happen

[digital goblet]

I'm still confused

I get that the proper way is to represent groups of digits with x but I don't know how to go further than that

[main needle]

umm so the full solution is:
make bcdef x and rewrite the equation
simplify to 99,999a/44 = x
plug in a=4,6,8 (or whatever you've narrowed down a to)
you'll get that a=4 and 6 give a fraction for x, while a=8 gives x=18181

[cedar kiln]

@digital goblet Has your question been resolved?

[digital goblet]

Ohhhhhhhhhhhhhh

That's so smart

[cedar kiln]

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[pliant hamlet]

R=0

idk my calculation correct or wrong, anyone can help to check pls

[cedar kiln]

@pliant hamlet Has your question been resolved?

[final crag]

,w determinant of {{1,2,0,1},{0,2,-4,-2},{-1,-3,3,x+1},{3,c,0,5}}

[wraith dagger]

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[final crag]

if R is zero then the above us just -2c + 16

@pliant hamlet

,w determinant of {{1,2,0,1},{0,2,-4,-2},{-1,-3,3,1},{3,c,0,5}}

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[pliant hamlet]

that mean i am correct?

[final crag]

Seems you lost a factor of 2 somewhere

oh

its outside

yes you're correct

[pliant hamlet]

ok thx!

umm

one more question

[final crag]

yup!

[pliant hamlet]

Need some hints

R=0

[final crag]

So do you know the effect of transposes, switching, and shears (adding multiple of one row to another) on the determinant

[pliant hamlet]

I'm not sure how to switch the row

[final crag]

Note that to get from one to the other, it's just a matter of
transpose (doesn't affect determinant)
switching rows (changes det by -1 each time you do it)
You added 2*the ghi row to the def row which does nothing
You did -3 on the def row which multiplies it by -3

So the final answer will basically be plus or minus 6 depending on how the signs work out

[pliant hamlet]

like this?

[final crag]

yeah! whenever you swap two rows you just add a minus sign outside

[pliant hamlet]

that mean if i add the - in front then i can switch the row to any other row?

[final crag]

yup

[pliant hamlet]

okok thx

is that correct?

[pliant hamlet]

<@&268886789983436800>

[pliant hamlet]

.close

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[frozen wasp]

From 12 to 2541 scale range how to find out how much % is 257 ?

[cedar kiln]

Please don't occupy multiple help channels.

[nimble mulch]

shift the interval

then determine %

(257-12)/(2541-12)

[frozen wasp]

Okey thanks

.close

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[terse cobalt]

quick question why do they change the 3 to -3 and why did they now put it at the bottom instead of the top like in the first question

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[sand cradle]

Übrigens @nimble mulch würdest du sagen meine Lösungen hierzu sind korrekt?

[sand cradle]

Oder sollte ich was ändern?

[cedar kiln]

@sand cradle Has your question been resolved?

[crimson delta]

naja ich hätte mal angenommen dass sie was anderes hören wollen bei (a) anstatt einfach nur CnP subset T abzuschreiben

[sand cradle]

Ich glaube das war ein Beispiel

[sand cradle]

Aber ist z.B. (f) korrekt?

zwei mal : sieht irgendwie nicht so gut aus

[cedar kiln]

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[cedar kiln]

@sand cradle Has your question been resolved?

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[shell aspen]

yo i need help solving this

real analysis

[final crag]

So to show uniform continuity, you need to show that given e > 0, there exists d > 0 so that |x-y| < d implies |f(x) - f(y)| < e.

Then note that if |x-y| < d that means M|x-y|^a is at most Md^a. So you want e = Md^a or alternatively d = (e/M)^(1/a)

Now let's see if this works. Given e > 0, let d = (e/M)^(1/a)

Then |f(x) - f(y)| < M|x-y|^a = Md^a = M(e/M)^(1/a)^a = Me/M = e so this works!

You might want to think about where we used a > 0 amd M > 0 in this!

[cedar kiln]

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[glad bear]

how can a function itself be greater than zero. Take a function y = x, if y(x) is greater than zero then f(y) = 1?
Am I right?

[final crag]

why can't a function be greater than 0?

for y = x, if x > 0 then y > 0 too?

anyway here you have y' = 1 if y > 0, y' = -1 if y ≤ 0

locally y' = 1 has a pretty common solution i.e. dy/dt = 1 so y = t+C, similarly on the other side, you just need to see if given these conditions you can piece a function together

[glad bear]

I just don't understand what conditions are given

does it mean I have a derivative of a function that's 1 when the function is negative and -1 when the function is positive or equals 0?

[final crag]

you have a function whose derivative is 1 when the function is negative and -1 when its positive

[subtle harbor]

y'=f(y) is seperable

[glad bear]

would -abs(x) be a solution to this problem?

[subtle harbor]

<@&268886789983436800>

[final crag]

bingo!

[glad bear]

Yeah, but we got a 0

[final crag]

and? -abs(0) = 0

[glad bear]

I might be rusty with calculus, but I rember there're troubles when we try to take derivative of abs in 0

[final crag]

ah right

[glad bear]

a solution to a problem must be continuous and have a continuous derivative

[final crag]

then that's all you gotta say then! f(y) is not continuous as a function of y so there isn't a solution

[glad bear]

there isn't?

are there any other functions besides -abs that would satisfy the conditions?

[final crag]

any function you'll pick will have a non-continuous derivative cause f(y) is not continuous at 0

[solid juniper]

this is so weird...

why would they write it like that

[subtle harbor]

i mean i don't feel like they're chosing y to be both independent AND dependent variable

but maybe they are?

idfk

[solid juniper]

i'm pretty sure i know what it's trying to ask and it's... very weird to write it like that

[glad bear]

also we don't use y(0) = 0

[subtle harbor]

i think you're supposed to think about it as a seperable ode

then split it into cases of y>=0 and y<0

[solid juniper]

the y' = f condition forces y to look something like this (ignoring 0)

you can shift each component up and down

[subtle harbor]

huh

hmm

[solid juniper]

but it has to be of that form

[subtle harbor]

yeah

[solid juniper]

the y(0) condition forces it to look like -abs

[warm crescent]

so there wouldn't be any function of that style with a derivative at 0

[glad bear]

if we integrate both parts we just get abs

[solid juniper]

actually i don't quite like this drawing

let me make a new one

[subtle harbor]

okay, what does integrating y>=0 get us? what does integrating y<0 get us?

[glad bear]

Like this

Oh, I forgot +C

[subtle harbor]

then you can figure out what the c is with the y(0)=0 condition :)

[glad bear]

Oh wait we have y(0) = 0

[glad bear]

Just remembered

[subtle harbor]

technically two different cs but y(0)=0 will force them to be the same

[solid juniper]

even just y needing to be continuous forces them to be the same

which is why i don't love my drawing anymore

[warm crescent]

ok quick question off topic and sorry for typing in this channel what would be the derivative at extremes of an closed interval? f(x) = x^2 with x in [0, 5]. Does f'(0) would exist? there is no left limit... But I think I've read you can redefine it to be one. But what for a piecewise function?? ok -abs(x) has derivative -1 at 0?? lmao I'm lost

[final crag]

side note: this is not uncommon! this is the definition of an autonomous differential equation. y' = y whose solution is y = e^x is the classic example!

[solid juniper]

y' = y is fine lol

y' = f(y) here though....

[subtle harbor]

i guess its just the fact that y is being used as both independent and dependent variable, but its in different contexts so I suppose its fine

if they defined f(t)=(...) and then did y'=f(y) maybe it would be more clear

[final crag]

I know but just making a sidenote that y' = y is a prototypical example of autonomous diff eq's which are defined by y' = f(y). It's not so much saying that y is both independent and dependent as much as saying the expression dy/dt only has terms in y and not in t!

[final crag]

hence why its called autonomous (not depending on t)

f is kind of irrelevant and is just being used to say dy/dt only has y terms in it

anyway sorry to digress!

[glad bear]

It's as though the only candidate for a solution is abs(x) but there are no derivative in 0

[solid juniper]

one candidate and the candidate is not a solution

[glad bear]

can we throw a 0 out and say it's a solution on (-inf;0)?

same goes for (0;+inf)

[solid juniper]

that kinda removes a lot of conditions from the problem

[glad bear]

it does?

[solid juniper]

what about y(0) = 0

[glad bear]

ohhh, we MUST include 0

I can throw out any x's I want except for 0 right?

[solid juniper]

wdym throw out?

[glad bear]

I mean what if we threw 0 out for example and said that this abs is a solution on (-inf;0) but we can't throw 0 out specifically

because y is defined in 0

[solid juniper]

y=-abs(x) with domain R\{0} is a solution to y'(x) = f(x) where f is defined similarly to the question just without 0, so f(x) = 1 if x<0 and f(x) = -1 if x>0

R\{0} being all real numbers other than 0

there are more solutions in that case though

this would also be a solution

[glad bear]

so, with the condition y(0) = 0 there is no solutions since only abs satisfies all the conditions but it can't be a solution due to the lack of derivative in 0 (i.e abs is not in C^1)?

[solid juniper]

yes

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[delicate elm]

Try drawing out the triangle, with A+(B-A)=B

[cedar kiln]

@fringe rain Has your question been resolved?

[fair geyser]

it's not clear what's the contradiction

B-A equals [2,-4]
AB equals [2,-4]
what you were told says that would happen

oh the signs

B-A equals [−2,4]
AB equals [−2,4]

[fringe rain]

is AB not [2,-4]?

[fringe rain]

these are different for me

[fair geyser]

you said i have a vector B [2,5] and vector A [4,1]

AB is [2, -4] if it's A = [2,5]

[fringe rain]

oh yeah wait i plotted that wrong

but does the formula terminal - starting always apply?

to find some vector A to another vector B

regardless of the direction

[fair geyser]

sure why not

it applies always

[fringe rain]

is it also called a linear combination

or is that something different

[fair geyser]

linear combination is when multiply 2 vectors by 2 different numbers and add

if i;m not mistaken

[fringe rain]

oh wait its coming back to me now

okay thanks

.close

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[frigid sinew]

im stuck at this integral of 1

does it just become teta?

[brazen kestrel]

$\int{dx}=x$ in general

[wraith dagger]

Hat MSC2020 82B10

[brazen kestrel]

So yes

[frigid sinew]

but how do i deal with it then

like how can i find what teta is

[brazen kestrel]

What do you mean find what theta is…the anti derivative would just be theta

[frigid sinew]

like how do i transform teta back to x is what i meant

[brazen kestrel]

Oh wait ignore the x there it was just a general variable

You can replace it with theta

Or any variable

[frigid sinew]

but it was an indifinite int

[frigid sinew]

like this

so i have to return back to x no?

how do i calculate it

back otherwise

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@frigid sinew Has your question been resolved?

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[timber violet]

Is this a question from a summer program pset?

[celest marsh]

never heard of that in my life lol

[timber violet]

Where is it from?

The question

[celest marsh]

why

[cedar kiln]

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[crimson sedge]

koshan time

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[celest seal]

.close

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[lone lily]

is the answer 9?

[late lotus]

nope

[lone lily]

you cant simplify 6 root 6 as 6?

[tepid blade]

No

[lone lily]

okay

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[coral matrix]

Hey, I was watching a trigonometry video on youtube. The topic was trig identities. The person in the video said that: sec(x) - sin(x) * tan(x) = cos(x) is an identity. Solving algebraically it seems correct, but I noticed that sec(x) and tan(x) are undefined for x = π/2 + κπ, whereas cos(π/2) is just 0. Does that still mean that the above equation is an identity?

[cedar kiln]

@coral matrix Has your question been resolved?

[coral matrix]

<@&286206848099549185>should I have asked my question somewhere else? Sorry, I'm new to this discord server.

[hollow trail]

a trigonometric identity is an equation that's true for every angle where both sides are defined

[coral matrix]

but isn't cos(π/2) defined?

i thought it's just 0

[hollow trail]

if one of the two sides is not defined then it's not true that "both sides are defined"

[coral matrix]

so it's not an identity?

[hollow trail]

by this definition it is an identity because it's true for every angle where both sides are defined

[coral matrix]

I thought that cos(π/2) is defined as just 0, whereas sec(π/2) is undefined. Which would mean that the equation is not an identity. Am I wrong in the sense that cos(π/2) is defined? Or am I missing something else?

alright I got it, I misunderstood, thanks!

.close

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[shut fiber]

i need to see if my answer is right or not please check it

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No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[late lotus]

,rotate

[tepid blade]

No

[late lotus]

@shut fiber do you know about variables?

[shut fiber]

somewhat

[late lotus]

alright so you think of the missing number as n

[shut fiber]

okii

[late lotus]

then n = 3/4 * 8/5

$n = \frac{3}{4} \times \frac{8}{5}$

[wraith dagger]

flurry

[shut fiber]

wait u did resiprocale right?

[late lotus]

yeah

can you do the rest?

[shut fiber]

yeah

so the answer would be

n=15/32

right-

[tepid blade]

No

[late lotus]

how?

[floral mantle]

answert this one pls

[tepid blade]

Just multiply this

[late lotus]

!occupied

[cedar kiln]

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).

[shut fiber]

i did-

4 x8

and

[tepid blade]

No

[shut fiber]

Oh-

[tepid blade]

Just normal multiplication,

3x8

[shut fiber]

multiply the tops n bottoms to each other?

[tepid blade]

And 4x5

[shut fiber]

ohkkk

[floral mantle]

pls reply

[tepid blade]

Yess

[late lotus]

!help

[cedar kiln]

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

[shut fiber]

alrr tysm!,

[late lotus]

@floral mantle

[tepid blade]

Your welcome

[cedar kiln]

If you are done with this channel, please mark your problem as solved by typing .close

[shut fiber]

24/20

[late lotus]

reduce that

[shut fiber]

alrr

6/5

[late lotus]

yep

you got it

[shut fiber]

tysm!!!

.close

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[gloomy hound]

how do i do (b)? a bit stuck on where to start. thanks

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@gloomy hound Has your question been resolved?

[gloomy hound]

<@&286206848099549185>

[nimble veldt]

assume they are not linear independet. then you could express one of them as a linear combination of some others. show that this is a contradiction.

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[modest sable]

how?

[brisk dirge]

divide both sides by 2pi and then square both sides

[modest sable]

do we sqaure the square root too?

[tepid blade]

Yes, doing that will remove the rt

[modest sable]

am i correct

i have another quewstion

this

[tepid blade]

Yes, it’s correct

[modest sable]

how do i do this

[tepid blade]

Take cube, both sides

[modest sable]

ohh

ok thanks i undertsna

what about this i squared both sides so it cancels then i to get rid of a I multiplyed w by a and the fraction by a so the a can cance

then im lost

yes

yep thats whati did

yep

a and a

OHH THEN WE DIVIDE THE FRACTION BY (W SQUARED +1 AND X

so a= x/w squared +1

ohhh

thankss

i get it now

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[undone star]

I need someone to verify how fraudulent my math is

[undone star]

This seems like a rather trivial inequality but somehow I came up with contradiction

Simply dividing (3) by (4) will yield the opposite

I feel like I’m missing something really obvious

And I’m trying to avoid using cs wherever possible since it kinda ruins the purpose of these questions

[crimson sedge]

i dont think you can divide an inequality like that

you can't conclude anything by dividing the inequality like that

if we know that
a>b and a>c
this doesn't tell us whether b>c or b<c

[undone star]

Yeah that’s fair

Massive oversight on my part and idk how I missed that

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[west turret]

Would this be correct?

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[rough zinc]

Hellow guys can someone tell me where it went wrong

[runic garnet]

whats the numerator of the original integral

oh now i see it

thats an x not a pair of parentheses

[peak minnow]

i think it can be a u sub

[rough zinc]

Thanks guys I figured it out

Thank uuu !!

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[north tide]

hi

[north tide]

i got no idea how to approach this

[jagged nexus]

yes soo ... first to make f(x) upside down or to make any curve upside down what you do?

[north tide]

uh

idk

[jagged nexus]

ok so to make specifically x^2 's graph upside down you add a - sign infront of it so it give all negative values of y, which is upside down...

[north tide]

mhm

[jagged nexus]

soo , -x^2 is first step towards g(x)

try plotting..

[north tide]

alr

@jagged nexus

okay I got the same thing but reversed at x=0

[jagged nexus]

yes...

then to shift it on x axis to 4 units ... what you do ..??

[north tide]

dont I add or subtract for that

[jagged nexus]

you add 4 in x ... so -(x)^2 becomes -(x+4)^2..

[north tide]

its cuz its not on the f ( x − h )

oh

[jagged nexus]

adding in x cuz we are shifting on x

[north tide]

oh and if I added it outside the parentesis

it would change the Y

[jagged nexus]

yes as f(x) = -x^2 (or y=-x^2) adding outside gives y= -x^2 +4 which is y+4

[north tide]

is there any way to set up a equation

using the points]

[jagged nexus]

umm how ?

[north tide]

like a (x,y) point

from the first graph and one for the second one

to try and find the equation of transformation

[jagged nexus]

no...

[north tide]

idk im just wondering if theres any easier way

dam

alr thanks for ur help tho

[jagged nexus]

this is easiest tho

[north tide]

.close

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[jagged nexus]

your welcome gentleman

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[tropic pecan]

I got 2 values for alpha

Do I use the smallest one

[wraith jolt]

yeah

[tropic pecan]

thx

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[cobalt plover]

I once again need help with this system

[nimble veldt]

posting the same question again and again - and then never responding. well done.

[cedar kiln]

@cobalt plover Has your question been resolved?

[cobalt plover]

Sorry man

Im posting it while working on other problems

And checking whenever i take a break

Just to find out it got timed out

I tried DMing you but you've got them disabled

Damn thats fucking crazy

@nimble veldt

Wait, youre not a mod

@wanton sail

@floral sinew

[cobalt plover]

<@&286206848099549185>

[cedar kiln]

@cobalt plover Has your question been resolved?

[main needle]

the constants are so annoying that you might as well just write them all as constants a,b,c,d... and solve the system as a matrix equation

[cedar kiln]

@cobalt plover Has your question been resolved?

[cobalt plover]

Hmh

Mhm

[main needle]

here's an example

you just apply the inverse matrix to the numbers on the right side of the equation

[cedar kiln]

@cobalt plover Has your question been resolved?

[cedar kiln]

@cobalt plover Has your question been resolved?

[raven shard]

@cobalt plover has been many many many times u have done this

(and u just did it again)

[raven shard]

though i see in ur work u have this

hint but there's not much to be said i suppose: ||once u get here there's very little u need to do||

[cedar kiln]

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[hearty junco]

I did the arc length

and got 300

but I'm not sure what to do after that

[cedar kiln]

@hearty junco Has your question been resolved?

[hearty junco]

nvm I figured it out

[cedar kiln]

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[prisma cairn]

Hello would anyone be able to help me. I’m unsure what to do next when im doing polynomial long division using complex numbers. Thanks!! Also the question is asking me to factorise p(x)

[prisma cairn]

Ive tried two ways too i dont understand tho

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[pseudo merlin]

how do i do this?

[pseudo merlin]

Part a

[cedar cypress]

Compare the RHS with the definition of nCr, and also you can't just write n! there like that

[pseudo merlin]

y?

oh

[cedar cypress]

n! Is not equal to n(n-1)

[pseudo merlin]

iot is not n-1!

[cedar cypress]

You need the numbers below 98 as well

All the way upto 1

So essentially this is something like

(n)(n-1)98!/[3!97!]

After multiplying 97! in the numerator and denominator

[pseudo merlin]

what............

[cedar cypress]

What is the definition of n!

[pseudo merlin]

uh

n factorial

idk..

[short blade]

what is 3!

[cedar cypress]

n!=product of all natural numbers from 1 all the way to n

[pseudo merlin]

\\\

[pseudo merlin]

wait

howd i t turn to 98!

[cedar cypress]

Did this make sense?

[pseudo merlin]

yes

[cedar cypress]

Cool then 98*97!

What should that equal

n! = n(n-1)(n-2)......1

=n (n-1)!

[pseudo merlin]

ye

[cedar cypress]

Do you see that

[pseudo merlin]

98 x 97! =98?

[cedar cypress]

98!

[pseudo merlin]

oh yeah

ok thanks

do u know how they got

20

how did it turn to

30C30-a

[cedar cypress]

nCr=nCn-r

[pseudo merlin]

shouldnt it just be 10

[pseudo merlin]

why

[cedar cypress]

Because you aren't changing the denominator essentially

(n-r)!r!

You're just switching these two up

[pseudo merlin]

uhhhhhhhhhhhhh

so

(30-a)!a!

[cedar cypress]

You really should work on the basics before seeing problems

[pseudo merlin]

sswtiching what up

[cedar cypress]

Do you know how we define nCr

[pseudo merlin]

choose r from n

[cedar cypress]

Ok that should do as well

Choosing r from n is equivalent to not choosing n-r from n

Soo if I say I have 10 apples

And ask you the number of ways to choose 6

It should be equal to the number of ways to choose 4

Because if you choose 6 you're leaving 4 out

[pseudo merlin]

is this from pascals triangle?

[cedar cypress]

And if you choose 4 you're leaving 6 out

So it's the same thing

[reef sedge]

how u doing

[pseudo merlin]

it is so bad

[cedar cypress]

Mathematically nCr is n!/[r!(n-r)!]

[pseudo merlin]

wait

[cedar cypress]

In that expression go ahead and replace r with n-r

You'll see the value won't change

[pseudo merlin]

is it like

this

wait

[cedar cypress]

You don't 'need' pascal's triangle to understand this

[pseudo merlin]

like this

[cedar cypress]

The definition of nCr should do

[pseudo merlin]

oh pascals is easier for me because im also doing pascals triangle

[cedar cypress]

Alright sure

If that's easier you can view it like that

Anyhow, that result should be remembered

I really recommend you going over your notes before attempting problems tho

[pseudo merlin]

oh i dont haven otes for nCr=nCn-r

[late lotus]

its an identity

[pseudo merlin]

i have this

[late lotus]

try proving how ncr = ncn-r

using factorials

[cedar cypress]

Yeah use that thing in the purple box

All you need to do is replace r with n-r

To see the result

[late lotus]

its basically the same thing though

since $\frac{n!}{(n-r)!r!} = \frac{n!}{n-(n-r)!}{n-r}!$

[pseudo merlin]

.close

[cedar kiln]

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[pseudo merlin]

thxthx

.reopen

[cedar kiln]

✅

[kindred agate]

?

[late lotus]

what happened

[pseudo merlin]

loike this?

[late lotus]

yeah

[cedar cypress]

Yes

[pseudo merlin]

.close

[cedar kiln]

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[cedar kiln]

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[rose trench]

i dont understand my statistics assignment. pls help guys

[rose trench]

<@&286206848099549185>

[formal cobalt]

That is a lot of reading.

[rose trench]

mhm

[cedar kiln]

@rose trench Has your question been resolved?

[rose trench]

<@&286206848099549185> pls help

[cedar kiln]

@rose trench Has your question been resolved?

[rose trench]

nah

.close

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[blissful wigeon]

I'm stuck and I've been searching an answer since january (#help-19 message)

kinda killing me so if any genius could help me i'd be down

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[umbral dew]

pretty dumb question

[umbral dew]

but we can't cancel this right?

[cedar kiln]

@umbral dew Has your question been resolved?

[umbral dew]

<@&286206848099549185>

[glossy wren]

Yes

[umbral dew]

hello help me im syl

[glossy wren]

@umbral dew What do you need help with ?

[umbral dew]

we cant cancel this right?

delta V vector

i've marked it out

[glossy wren]

I'm back

What do you need help with ?

@umbral dew

[umbral dew]

bruh i've literally written it like 2 times

[glossy wren]

We can't cancel it out

[umbral dew]

alr

thanks

.close

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[stiff acorn]

Hi

[dense locust]

Is there a way of doing this, without magically guessing its dy/dx of sin^3 x

[earnest socket]

u = sin(x)

[dense locust]

ok yea i know how to calcualte it

but how can you guess its sin^3 x

Why not sincosttancotcosecsecx

Is it just more of a memorization thingy

[earnest socket]

huh

I'm saying you can substitute

no need to guess sin^3

[dense locust]

Integration

Ur supposed to integrate it

Oh

[earnest socket]

after u = sin(x) you get int 3u^2 du

[dense locust]

no im saying from 3 cosx sin^2

[earnest socket]

do you know how to evaluate an integral using substitution

[dense locust]

no

Not yet

thats right after this topuc

Does the bottom mean anything

bc i dont hables this

[earnest socket]

oh then it's just that bottom rule

[dense locust]

ive been doing this before^

But not first image

[dense locust]

Ok ill try understanding it

First so u dont waste time

ty!

ill let you know

Ok i understood

this is such a bs rule tho

my gosh

.close

[cedar kiln]

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[cedar kiln]

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[nimble ibex]

so for part i I get v=35

EKE (start)=GPE end +EKE end

[rich field]

you can find v² by energy conservation too