help-39

[meager trellis]

but for instance the function that takes a turing machine and outputs whatever the turing machine outputs, and isn't defined if it runs forever, is partial recursive

you can basically just take the input machine and simulate it

[lament knot]

Right

[meager trellis]

and tbh i think being able to state this kind of result is a big part of why partial recursive functions are interesting

and also more generally that a turing machine just isn't always a total function, but you can always view it as a partial function

[lament knot]

But like this makes me a bit worried because we say Turing machines and partial recursive functions are equpotent models of computation right?

[meager trellis]

yes

[lament knot]

Then when we look at the turing machine for the halting problem

We can't point to a partial recursive function corresponding to it?

[meager trellis]

well that's because there isn't a turing machine for that either

[lament knot]

There is a Turing machine for that, but it just doesn't halt for all inouts

Inputs

[meager trellis]

that means it doesn't compute the halting problem

because it doesn't compute a total function, and the halting problem is a total function

[meager trellis]

if you mean this thing then this computes a very different function that isn't total, that does also have a partial recursive function that computes it

[lament knot]

So if it's a total function then it's corresponding turing machine must halt on all inputs

[meager trellis]

yes

[lament knot]

And if its a partial recursive function then the machine is allowed to loop on only those inputs which are undefined

[meager trellis]

yes

and on the inputs where the function isn't defined it also has to run forever

(otherwise it's instead computing a different partial function that is defined at the inputs where it halts)

[lament knot]

This is confusing me a little though, because why can't we say it's simply a Turing machine which tells if a {tm, input} will halt but loops if not, like it's just a acceptor machine?

Does this mean its a partial recursive function where the domain is the subset(halting pairs) among the superset(all pairs) of Turing machines, inputs?

[meager trellis]

well that's a thing you can talk about, but it's different to "a machine which outputs 'halts' if the machine halts and 'doesn't halt' if it doesn't halt" which is what people normally mean by "the halting problem"

[lament knot]

Isn't that just a hypothetical machine where the assumption is "halting problem is decidable"?

[meager trellis]

yes, a turing machine that halts on only the turing machines that halt computes a partial function and its domain is the set of turing machines that halt

[lament knot]

Yes! That would make a lot of sense

[meager trellis]

well yes, it's something that you mainly talk about while proving it doesn't exist, but still

[lament knot]

Yea I think much of my doubts cleared now

Thanks a lot!

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[lament knot]

.reooen

.reopen

[pearl pond]

✅

[lament knot]

Here why are we saying the domain of definition is unknowable?

Since we can as we mentioned earlier, list out all the pairs which halt

@meager trellis sorry for the ping but kinda again got confused

[meager trellis]

i think they're being kind of informal there by calling it "unknowable"

but also like, there's no computable function that can decide if something is in the domain or not, that seems reasonably "unknowable"

[lament knot]

Hmm right

We can have a computable function which starts listing them all the pair out, but ofc it never finishes

Yeahh

Makes sense

Thanks

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[toxic lichen]

how did a=5 happen

this would make the parabola open upward 💀

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[dense cargo]

Not really understanding whats going on, I included the markscheme

[regal herald]

theyre finding the angle between OB and OQ then using the arc length formula

and they find the radius with pythagoras

[dense cargo]

but how

[regal herald]

the angle?

they use soh cah toa

to find BOA

[dense cargo]

ok so they split the rectangle?

[regal herald]

yeah draw a line from O to B

then you have a right triangle

[dense cargo]

ok so after you get the right triangle you do Pythagorean theorm right?

wait

[regal herald]

for the radius yes

[dense cargo]

after what do we do

[regal herald]

we want the angle BOQ

so we need to know angle BOA

[regal herald]

^

[dense cargo]

what angle on BOA?

aint it a right trianglew

[regal herald]

yes?

you have an angle at B, an angle at O and a 90 angle at A

im talking about the angle BOA

as in the angle between BO and OA

[dense cargo]

i apolgize for my lack of intelligence as im still not getting it

so we got this right now right

A = 90 B = 45 O = 45 right because they are just right angles in half

[regal herald]

the angles arent cut in half

that would only be true if BA=OA

the angle youre trying to find the the one at O in the upper triangle

we call it angle BOA

[dense cargo]

so the angles are not 45 and we are currently trying to find out the angle at O. ok we do this with the SOH CAH TOA

[regal herald]

yup exactly

[dense cargo]

so according to the markscheme it would be 8/5

after we just add that to the 90

from BOQ?

[regal herald]

right again

[dense cargo]

then we use the lenght of an arc fomual to find the lenght of the arc

[regal herald]

right once more

[dense cargo]

wow thank you good sir and wish you very well in life

god bless you

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[hoary ridge]

mathematica agrees

@hearty flame

[hearty flame]

i used cylindrical coordinates

actually nvm its right

[pearl pond]

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[hallow zinc]

i dont understand standard score

also does third quartile mean 196< or <119

[toxic lichen]

3rd quartile means 196, the right edge of the box

[toxic lichen]

as in "i don't get what it means conceptually" or "i don't get how to compute it"

[hallow zinc]

standard score is her score - mean / standard deviation right

[toxic lichen]

brackets!

[hallow zinc]

ok

[toxic lichen]

standard score = (raw score - mean)/stdev

do i understand correctly that it's the concept of it that you're missing

yes or no

[hallow zinc]

actually not really im just confused how to get the answer

say 1 use (197 - 135) / 62

aint the score 1

since the mean can be in a large range

[toxic lichen]

that's the lowest(ish) possible score

[hallow zinc]

i used to most extreme values to calculate 0.9 - 1.2something

which rounds to 1.3

[toxic lichen]

(the mean need not be an integer)

[hallow zinc]

im not sure if im misisng something

is the question expecting an exact standard score i guess

anyways thank u very much

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[wind lagoon]

What is the strategy to solve this exercise? I am stuck with figuring out how a general element of V* looks like, so as to prove that f1,f2 and f3 span V*.

[toxic lichen]

a general element will look like f(x,y,z) = ax + by + cz

[wind lagoon]

ah, right. a=f(e1), b=f(e2) and so on, correct?

[toxic lichen]

yes

[gray hinge]

chat gpt failing to do simple math

[dapper kraken]

wow who guessed

[remote aurora]

always been like that

[wind lagoon]

That we are dealing with functions and not points in space as vectors confuses me. So we know f(x,y,z) = ax + by + cz is a general element in V*. What would be the next step in showing f1,f2 and f3 are a basis?

[toxic lichen]

!occupied

[pearl pond]

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

[toxic lichen]

prove that for any a, b, c in R there exist constants u1, u2, u3 in R such that u1 f1 + u2 f2 + u3 f3 = f

[wind lagoon]

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[chilly vector]

could you solve this?

[chilly vector]

A parallelogram is divided into 4 regions in which regions A, B, C have areas of 45 cm2; 30 cm2; 60 cm2 respectively. Calculate region D.

[white estuary]

Area of D?

[chilly vector]

yes

[unique ibex]

Consider the heights of the parallelograms

[chilly vector]

and?

[vapid cave]

observe the ratios in area and deduce where the region boundaries must be

[chilly vector]

wait

this is my solution

lets call them h1 and h2

A and B have h1

C and D have h2

[iron basin]

yes

[chilly vector]

then A/C = B/D

[iron basin]

correct

[chilly vector]

A/C = 45/60 = 3/4

the same to B/D

[iron basin]

yes

[chilly vector]

then D = 30/3x4 = 40

is it right?

[iron basin]

yea

you got it

[chilly vector]

alright thank you

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[mystic wraith]

can someone tell me how to do them

[keen quail]

what do you know about transformation of functions

[mystic wraith]

well i know

that like

if its f(x-3) then its 3 units to the right

so horizontal shift

and f(x)+3 its horizontal shift of 3 units up

3f(x) means its been stretched right?

-f(-x) means its reflected y and x axis

[keen quail]

you mixed up vertical and horizontal but yes you're right

[mystic wraith]

f(3x) means theres something done to it by 1/3

[mystic wraith]

oh

wait

oh ye

[keen quail]

where are you facing problem applying this

[mystic wraith]

i did some questions related to them like wait

like these

[keen quail]

stretched horizontally by a factor of 3

[mystic wraith]

but for this part i dont understand what im tryna do

[keen quail]

you know that the transformation will be 3 units to the right and 4 units down right

for the first one

[mystic wraith]

hm why did our teacher say smth like its a horizontal dialtion scale factor by 1/ |b|

oh

wait

yes move toright by 3 and 4 down

[keen quail]

she's right, she said what i said but more sophisticatedly

[keen quail]

yes, simply apply this to the point given

that's it

[mystic wraith]

ok wait uh

apply 3,-4 to -2,-5?

wait what

[keen quail]

yepp

[mystic wraith]

how do i do that

[keen quail]

add 3 to the x-coordinate and subtract 4 from the y-coordinate

[mystic wraith]

wait why are we doing that

[keen quail]

cause you know that horizontal change will apply to the x-coord and vertical change will apply to the y-coordinate

[mystic wraith]

yes

wait so we add them so then it becomes g(x)=f(x+1)-9

?

[keen quail]

nonono

the questions asks you to find the coordinates

after this transformation occurs

if (-2,-5) lies on f(x)

[mystic wraith]

yes

[keen quail]

we need to find what this point will be on g(x)

[mystic wraith]

oh so the change in f(x) to g(x) is 3 units right and 4 down

so we add that

to the previous coordinate

[keen quail]

yeppp

[mystic wraith]

to find the new place

[keen quail]

yess

[mystic wraith]

so (1,-9)

right

[keen quail]

yep

[mystic wraith]

what abt 7

wym as in say the axes of those

and

[keen quail]

can you write the points given in the form (a,b) for me

[mystic wraith]

hm i dont rlly understnad wording

[keen quail]

all three points

[mystic wraith]

(-3,0) (4,0) (0,2)

[keen quail]

yep, now g(x)=f(x)-3
so we know it's 3 units down
can you determine how this will change the given points

[mystic wraith]

hmmm

uh

that means the x int will change

all the intercepts will change

[keen quail]

yep

[mystic wraith]

because hm as it goes down like the graph basically becomes wider when they intercept

[keen quail]

i think they just want you to comment if the intercept will be larger or smaller, do you have an answer key?

[mystic wraith]

lemme check

how do u figure out the intercepts and stuff

whats inconclusive?

the graph never touches any axis?

the question so u dont need to keep looking back up

[keen quail]

ok the a part they only want you to judge the change in y int which is easy, and b is similarly easy too

just apply the transformation like before

let me see c

[mystic wraith]

also why does c have a ? at the end

i didnt know that was a sign in math

ahh i see so for a we cant determin the exact value of x as it doeesnt tell us but it does say the graph moved 3 units down so -3+2 which is -1?

[keen quail]

c is inconclusive because we don't have the function given, and transformations are being applied to both the x and y coordinates

[keen quail]

yess

[mystic wraith]

dont have the function given?

wym

[keen quail]

so you will be able to tell what the points are after transformation

but not what the intercept of the function will be

[mystic wraith]

wait but isnt the interecpts before the function (-3,0) (4,0) (0,2) and since we could do like juding the change between part a and b can we not do that for part c as it tells us grpah moves 2 units left and 4 down

[keen quail]

we've just been given f(x) and g(x) we need to know what f(x) and g(x) are to determine the intercepts in some cases

[keen quail]

intercepts are strictly when one coordinate is zero, do you agree?

[mystic wraith]

yes

[keen quail]

try the transformation and see if you get the points with coordinates that are zero

if not, then it's not feasible to determine the intercept

[mystic wraith]

ok hm i dont rlly understand as like

i know that int means x or y must be 0

but i dont understand like why is this inconclusive and what happens if transformations are both are applied to both x and y or "we've just been given f(x) and g(x) we need to know what f(x) and g(x) are to determine the intercepts in some cases" for that wym as in we need to know because hasnt it told us like other equations

is it because the sign are f(x) = f(x+2) -4

[keen quail]

did you understand why we only found the y intercept in the a part

[mystic wraith]

f(x)=f(x)

[keen quail]

and only the x intercepts in the b part

[mystic wraith]

because we cant determin the change of the x int

[keen quail]

exactly

[mystic wraith]

but in part c isnt change in x 2 to the left and y is 4 down so we just put that in so -2-3 and -2+4 for x

and then -4+2 for y

hm

[keen quail]

(-3,0) (4,0) (0,2)
(-5,-4) (2,-4) (-2,-2)

the line below is the points after transforming

we can determine this much from the information given

[mystic wraith]

yes

ohh wait

ohhhhhHHH

i get it we dont see the intercepts

[keen quail]

yeppp

[mystic wraith]

but for the others we assume that their x is 0 when we given y and vice versa right

because only 1 has changed

[keen quail]

yes

[mystic wraith]

ahhh

i get it

[keen quail]

yes perfect

[mystic wraith]

alr

tysm

ima go do more questions

[keen quail]

[mystic wraith]

cya

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[boreal raptor]

why are we able to integrate from 0 to 1 of 1/sqrt(1-x^2) when it's undefined at x=1?
isn't there an asymptote that tends towards infinite at x = 1-
meaning that there should be an infinite area?

[pearl pond]

Please don't occupy multiple help channels.

[oak ivy]

You could ask in #help-3

[boreal raptor]

oh i cant open multiple? :/

.close

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[plush sundial]

idk how to draw the figure

[plush sundial]

mb this is the ques

anyone?

just tell me the figure

<@&286206848099549185>

[pearl pond]

@plush sundial Has your question been resolved?

[pearl pond]

@plush sundial Has your question been resolved?

[pearl pond]

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[gray bluff]

Hi there. I just need some guidance on how to re-do this topic. I mostly know how to do it. I just need some help to make sure i'm doing it right. This is the question i'm currently doing.

[shrewd mist]

what have you done?

[gray bluff]

Currently nothing. But I think the perpendicular part would be.. -1/5? Thats the part thats throwing me off

[quasi sundial]

Almost correct

Do you know the relation between the slopes of two perpendicular lines?

[gray bluff]

I'm not sure what you mean by relation

[toxic lichen]

ok just, what do you know about the slopes of a pair of perpendicular lines

if you know that two lines are perpendicular to each other, what can you say about their slopes, and/or how can you find the slope of one knowing the slope of the other?

[gray bluff]

I know that its basically the opposite of a parallel line. And thats pretty much it.

[shrewd mist]

Probably a more direct question. If a line has slope "a" (where a is a number, like 2), what is the slope of a line perpendicular to it?

[gray bluff]

I mean... I know that if the problem was like. 2/3 then the perpendicular line would be like. -3/2 but the singular problems like these are trickier for me

[shrewd mist]

Fair enough, but can't you just write -5 as -5/1?

[gray bluff]

I think so? That somewhat sounds right.

[shrewd mist]

-5 divided by 1 is just -5... so -5 = -5/1

That way you can have it in a more workable form

So a line perpendicular to one with slope -5/1 would have what slope?

[gray bluff]

Okay and then you write the equation as -3=-5/1(2)+B to solve to find the equation of B?

[shrewd mist]

Wait a sec

We didn't get the perpendicular slope yet

[gray bluff]

ah. Okay thats the parallel then?

[shrewd mist]

Yeah

[quasi sundial]

Yep

[gray bluff]

okay okay my bad

[shrewd mist]

For the perpendicular, you do the same thing, but just do the "perprendicular" operation on the slope -5/1

[gray bluff]

so then you do the opposite? Which I think is 1/5? Ahhh i dont know

[shrewd mist]

exactly

you flip the numerator and denominator, then add a "-" (which cancels out the existing one)

[gray bluff]

okay that makes sense.

[shrewd mist]

Formally, if a line has slope "a", a line perpendicular would have slope "-1/a"

[gray bluff]

Understood

So I got this for the parallel. But i'm not sure if its right. y=-5/1x-13/1

oh wait hang on thats wrong

Okay. So I got:
Parallel: y=-5/1x+7/1
Perpen: y=1/5x-17/5

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[proper flare]

Prove that the angle bisectors of the angles of a triangle are concurrent.

[proper flare]

Just tell what theorems are to be used here.

I need to stop doing Maths all day.

I have a family 🥲 .

<@&286206848099549185>

[toxic lichen]

yes you do

but also the locus thing i said in your last channel does the job handily

[dark rock]

prove the smaller triangles congruent

you also need to do a construction

[proper flare]

That is a very vague statement dawg.

[dark rock]

wait do you mean the vector one ?

cus the one im thinking of is basic ash

[proper flare]

You said "Prove the smaller triangle congruent."
You didn't specify which "small triangles" you're referring to.

[dark rock]

you

probabl

[proper flare]

wdym

[dark rock]

y

oh well

let me just go ahead with what i meant

[dark rock]

what you can do is uhh draw three perpendiculars on three sides of the triangle

[dark rock]

and then you draw the angle bisectors asw

[proper flare]

Like this?

[dark rock]

uhh sorta yes

go around proving uhh

first prove ADE and

wait a sec

[dark rock]

you havent drawn one of the angle bisectors

i see

you needa prove ADE and ADG as two congruent triangles

[proper flare]

I have drawn the angle bisectors AD and CD.
Any point on the angle bisector of an angle is equidistant from the rays of the angle. So GD = DE.
and DF = DE.
So DF = DG.
Join D and B. DG is perpendicular to AB and DF is perpendicular to BC.
By RHS, triangle BGD is congruent to triangle BFD.
So angle GBD = angle DBF.
So BD is the angle bisector.

[dark rock]

basically you have EAD = DAG, AD as the common side and if you draw the perpendiculars you will get DEA = DGA

[proper flare]

Is this correct?

No way this is as simple as this.

[dark rock]

looks correct to me

yeah it is pretty simple

the method i was going ahead with is easy as well

just do 3 sets of congruencies and you would be done

[proper flare]

Because when I searched for the proof of this theorem on the internet it gave me proofs using Ceva's theorem and the converse of the angle bisector theorem.

[proper flare]

Yeah, that's the same thing

[dark rock]

yh

[proper flare]

Okay thank you

.close

[pearl pond]

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[proper flare]

I am leaving this server.

[minor meteor]

Byee

[toxic lichen]

what is this sacred geometry shit

[pearl pond]

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[agile valley]

A home improvement contractor is painting the walls and ceiling of a rectangular room. The volume of the room is 998.25 cubic feet. The cost of wall paint is $0.12 per square foot and the cost of ceiling paint is $0.18 per square foot. Find the room dimensions that result in a minimum cost for the paint.

[agile valley]

here's what i've done so far
letting the length, width, and depth of the room be x, y, and z, we have the following relations

xyz = 998.25
and
the cost C(x, y, z) is 0.12(2xz + 2yz) + 0.18xy
= 0.24(xz+yz) + 0.18xy

trying to eliminate z from the equation, we have z=998.25/xy
and
substituting into C(x, y, z)

=C(x, y)
= 0.24(x+y)(998.25/xy) + 0.18xy

taking partial derivatives to find the minimum

[tulip ore]

(here, you should simplify (x+y)/xy to be 1/x + 1/y so that the partial derivatives are easier)

[agile valley]

i get

[jolly parrot]

2derpy

[agile valley]

and here i don't know how to solve for x and y such that both become 0

[tulip ore]

you can solve the bottom equation for y, then pluyg that into the top

[agile valley]

oh right

yeah i forgot how to do algebra 💀

[tulip ore]

nw

[jolly parrot]

2derpy
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

2derpy
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[agile valley]

x = y = 11
z = 8.25

and C_min = C(11, 11, 8.25) = 65.34

thanks!

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[quiet vessel]

In the first attatchment, consider the following question:
How do we improve the leading question in Section 4.2, so "satisfying" is defined in Section 3.2 and the expected value of fᵣ* (where fᵣ* converges to an explicit, everywhere surjective function) is finite?
If you want evidence to make sure I know what I write, see the second attatchment.

[quiet vessel]

Here is the link to the first attatchment: https://www.researchgate.net/publication/389499633_Defining_a_Unique_Satisfying_Expected_Value_From_Chosen_Sequences_of_Bounded_Functions_Converging_to_an_Everywhere_Surjective_Function

Here is the link to the second attatchment: https://www.researchgate.net/publication/389005047_Defining_a_Satisfying_Expected_Value_From_Chosen_Sequences_of_Bounded_Functions_Converging_to_Pathological_Functions_v2

If you can't asnswer the question directly, is there a research paper that can help?

[spare lark]

You might find better help in the advanced channel

Here its mostly for pre-uni undergrad math

[pearl pond]

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[bitter pasture]

For this NFA, can it be simplified such that it still describes the same language but has fewer states overall?

[bitter pasture]

I also came up with a regex for this: 1(000* + (100*011)*), I wonder if it is correct

[pearl pond]

@bitter pasture Has your question been resolved?

[pearl pond]

@bitter pasture Has your question been resolved?

[pearl pond]

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[lucid hazel]

@regal herald srry for ping but cuz u were helping me before

i tried ur statefy

strategy

but it iddnt work

a^[x-h] -1 = y

[regal herald]

si

[lucid hazel]

a^0 - 1

1-1 = 0

so that means no horizontal shift?

[regal herald]

when y=0, x=-3

[lucid hazel]

imma gonna cry fr

can u show me how

[regal herald]

you got y=0 for x+h=0 si?

[lucid hazel]

so i always set x+h = 0

i wanna learn how u do it

[regal herald]

yeah, otherwise we have an a in there

a^0=1 so that takes care of it

[lucid hazel]

can u show me how u do it

like step by step pls

i rlly wanna understand

x+h = 1

x =0

so h =1?

[regal herald]

no
x+h=0

[lucid hazel]

ok then what do i do

[regal herald]

then we have y=a^0-1=1-1=0

[lucid hazel]

so no translation?

[regal herald]

thats not what that means

we just have y=0 at x-h=0

now look for the point on the graph where y=0

because that is where x-h=0

[lucid hazel]

-3

oh i think i get it

[regal herald]

so x=-3

-3-h=0

so h=-3

x-h=x-(-3)=x+3

shift left of 3

[lucid hazel]

thanks bro

[regal herald]

nw

[lucid hazel]

i understand bro

i love u bro

[pearl pond]

@lucid hazel Has your question been resolved?

[pearl pond]

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[junior vapor]

is partial fractions the fastest way to do this?

[fair creek]

appears so

its not that long though since all the partial fractions have constant numerators

[junior vapor]

Kk thank you

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[junior vapor]

kinda related to my other question, but is the partial fraction meant to be ax+b/x^2 + cx+d/(1+x^2)

.reopen

[pearl pond]

✅

[wild fable]

x^2 should be a repeating factor

[junior vapor]

Oh so it's A/x + B/x^2 + (Cx+D)/1+x^2?

[wild fable]

yea should be

[junior vapor]

kk thank you

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[cobalt patrol]

does mass a accelerate?

Determine the tension developed in the cords attached to each block and the
accelerations of the blocks. Neglect the mass of the pulleys and cords

this is the prompt

[pearl pond]

@cobalt patrol Has your question been resolved?

[rough stream]

There's a common way to analyze mass-pulley systems, by writing equations for the length of each string

Okay I see you did get that far

You differentiate those equations to get equations for the pulleys' accelerations

@cobalt patrol

For example, 2a_a - a_c = 0

[pearl pond]

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[pale atlas]

what does it even mean by one root tends to infinity??

[toxic lichen]

you tell us lmao

the hell does that mean

[pale atlas]

💀

[placid geyser]

Doesn't it just mean a tends to 0?

[toxic lichen]

does it?

[pale atlas]

my teacher ghosted me when i sent him this 😭

[placid geyser]

Lol

[toxic lichen]

how much is at stake with this problem

cause if you dont get a clarification and can skip, i would

[pale atlas]

yeah i should ig

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[idle bramble]

i need to find m and i have no idea how to solve polinoms

[toxic lichen]

find m such that what happens

[idle bramble]

its in turkish wait lemme translate the question

[toxic lichen]

post the original Turkish too just in case

[idle bramble]

What is the sum of the possible values ​​of m?

please help i have exam in two days and people say that this chanel helps a lot 🙏

[junior anchor]

A polynomial is defined such that the exponents of x in all terms have to be integers

So basically we have to make sure all exponents are integers

And find all possible "m"s such that the exponents are integers

Non negative integers actually

Does that make sense

[idle bramble]

so do i leave the negative ones

[junior anchor]

So whatever m you take, both "16/(m+2)" and "m-3" need to be non negative integers

[idle bramble]

okay

[junior anchor]

If you look at 16/(m+2) that really leaves you with a limited set of outcomes don't you think?

[idle bramble]

yes so maybe a 3?

[junior anchor]

What all non negative integer values do you think 16/(m+2) can take

[idle bramble]

do i need for 16 to be devided completely?

[junior anchor]

Yes

[idle bramble]

ok

does 14 wrok

[junior anchor]

Yes

What else works

[idle bramble]

İDK

[junior anchor]

What do u think of m = 6

[idle bramble]

then itsll equal to 2

16 isnt completely gone

[junior anchor]

Yes which is a non negative integer right?

[idle bramble]

yea

[junior anchor]

We want 16/(m+2) to be a non negative integer

2 is a non negative integer

[idle bramble]

so i can use any numbers which can devide 16 right

[junior anchor]

Yes

[idle bramble]

oh then its 2, 6 and 14

[junior anchor]

Correct

Now

We need m-3 to also be

Non negative integer

So what possibilities of m remain

[idle bramble]

2 and 6?

will 14 work?

[junior anchor]

Do you think 2 works

[idle bramble]

no my bad

[junior anchor]

Do you think 14 works

[idle bramble]

ohhh its -3x

okay

[junior anchor]

?

What do you mean -3x

[idle bramble]

like the final answer of it will depend on its exponent number (language barier 😭 )

thats why

then 14 wouldnt work since the answer would be negative

[junior anchor]

14 works

Because

14-3 is still a positive integer

We just want the exponents to be non negative integers

The terms themselves can be negatuve

Negative

[idle bramble]

oh okay

then its just 14?

[junior anchor]

What do you think of 6

[idle bramble]

it will work too

[junior anchor]

Actually you should really start with defining what a polynomial is

That's what this question is really asking you to do

[junior anchor]

Yes

[junior anchor]

This is the heart of this question

[idle bramble]

okay then the answer is D

[junior anchor]

Defining what a polynomial is

[junior anchor]

💯

[idle bramble]

thank you profesor 🙏

[junior anchor]

Make sure you get the definition of a polynomial tho

[junior anchor]

I'm not a prof 😭

I'm barely a person

[idle bramble]

im gonna try to speedrun whole polynom chapter tomorrow

[idle bramble]

still thankyou

althrough do yall know imagination numbers

wait let me find an question example

nvm i gave up 🔥

.close

[pearl pond]

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[midnight haven]

how do i differentiate help i genuinely forgot

[glacial sequoia]

you can write it as 1 - 1/(x + 1)

and differentiate it from there

[midnight haven]

what

[glacial sequoia]

or apply the quotient rule

[midnight haven]

my attempt was

x(x+1)^-1

[glacial sequoia]

yeah that works too

what did you get

[midnight haven]

that ws my only step

😭

[glacial sequoia]

do you know the product rule?

[midnight haven]

yes

[glacial sequoia]

yep go ahead and use it

[midnight haven]

nx^n-1

[glacial sequoia]

that's the power rule

[midnight haven]

oh wait

is that different

then idk whats a product rule

[jolly parrot]

neon

[glacial sequoia]

does this look familiar

[midnight haven]

ohh yes

i know this

[glacial sequoia]

yep so use it

[midnight haven]

oh

[glacial sequoia]

u is x and v is the other thing

[midnight haven]

ok

u=x v=(x+1)^-1 = 1/(x+1)
u'=1 v'=-(x+1)^-2 = -1/(x+1)^2

dy/dx=uv'+vu'
= x(-1/(x+1)^2) + 1/(x+1) (1)

idk

[glacial sequoia]

yep that's right

[turbid sequoia]

you can simplify to 1/(x+1)^2

[glacial sequoia]

or directly achieve it by doing this

[turbid sequoia]

also directly acheive it by doing the quotient rule

[midnight haven]

😭

oh

is this the final answe

r

[turbid sequoia]

should be

[midnight haven]

i hope so

thanks

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[desert thorn]

I've done this question myself but am wondering anothher persons opinion on working out A

to calculate the dip estimate

[pearl pond]

@desert thorn Has your question been resolved?

[cobalt patrol]

did i not end up getting the right answer?

[cobalt patrol]

but i wasn't given any velocity or displacment values

[pearl pond]

@desert thorn Has your question been resolved?

[placid geyser]

Around 0.48? Idk what shape to consider except a cuboid

[desert thorn]

Awesome I got 0.47 so good to know I'm not way off

[pearl pond]

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[compact ridge]

correct, then you need to bound y

which two lines y = .... bound the region?

you already have x = 0 and x = 1 remember

well y = x is also correct

but not y = 1

exactly

the bottom line is....

the bottom line?

you're overthinking this

yep! you just needed that

yes

so if you now start evaluating $\int_{x = 0}^{x = 1} \int_{y = 0}^{y = x} 1 \ dy \ dx$

[jolly parrot]

south

[compact ridge]

note how the bounds match with the dy then dx

you'll just get $\int_0^1 [x - 0] \ dx$

[jolly parrot]

south

[compact ridge]

so it checks out

your inner bound is a function of x

but that's the principle

replace 1 with a different function if you want the integral of that function over that region

1 actually just gives you the area of the region

np!

[pearl pond]

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[compact ridge]

definitely!

[pearl pond]

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[compact ridge]

those are common textbook exercises

[compact ridge]

so you can find formulas for the volume of a cone, a sphere, and so many more shapes using triple integrals also

sadge

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[cursive wraith]

you can definitely transform finding the extrema of $g(x,y) = x\log(1+x^2+y^2)$ on $D$ into finding the extrema of $h(r,\theta) = r\cos(\theta) \log(1+r^2)$ on $[0,1]\times \bR$

[jolly parrot]

rafilou is not not born in 2003

[cursive wraith]

by arguing that first, $(r,\theta)\in [0,1]\times \bR \Longleftrightarrow (r\cos(\theta),r\sin(\theta)) \in D$

[jolly parrot]

rafilou is not not born in 2003

[cursive wraith]

and then that for $(r,\theta)\in [0,1]\times \bR$, $h(r,\theta) = g(r\cos(\theta),r\sin(\theta))$

[jolly parrot]

rafilou is not not born in 2003

[cursive wraith]

but now I'm realizing that the question is phrased a bit weird

are you supposed to find the global extrema of g defined on D

or are you supposed to find the global extrema of g defined on R^2, that lie inside D?

[calm wing]

what is the difference?

[cursive wraith]

pretty huge difference

if g is defined on D, a compact set

it will always have global minimum and maximum

if g is defined on R^2

then some extrema may not exist

and who's to say they even lie on D

continuous*

[calm wing]

so why use the word "global" here

[cursive wraith]

global as in "on the entire domain"

?

[calm wing]

oh the question writes that

yeah that's weird

[cursive wraith]

that's not the weird part

"global" means "on the entire domain"

so if the domain is D

there's no shame in calling the extrema on D "global"

the thing that trips me is they specify afterwards that the closed unit disk is "in the Euclidian plane"

but

[calm wing]

i think we are overthinking it

[cursive wraith]

if g were defined on R^2

they wouldn't have written that

[calm wing]

it's this one

according to the solution

[cursive wraith]

yes

g will always have global minimum and maximum

you just need to find where they are in D

[calm wing]

solution

[cursive wraith]

yeah sure

$\nabla g (x,y) = \begin{pmatrix}\log(1+x^2+y^2)+\frac{2x^2}{1+x^2+y^2}\ \frac{2xy}{1+x^2+y^2}\end{pmatrix}$

[jolly parrot]

rafilou is not not born in 2003

[cursive wraith]

if it's 0 inside the interior of D, then x = y = 0

and g(0,0) = 0

compare to g on the boundary

which is g(x,y) = log(2)x

we get -log(2) on (-1,0) and log(2) on (1,0)

[pearl pond]

@naive marlin Has your question been resolved?

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[mellow idol]

any ideas on how to solve this?

[mellow idol]

<@&286206848099549185>

[cinder thistle]

what does R ^0 mean?

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[mellow idol]

I think it's supposed to be just R

[cinder thistle]

well, firstly
a^2 + b^2 + c^2 + (k-1)(a^2b/a+b + b^2c/b+c + c^2a/c+a) >= ab+bc+ca + (k-1)(ab+bc+ca)/2

(a-b)^2+(b-c)^2+(c-a)^2 >= (k-1) ( (ab + bc + ca) - 2(a^2b/a+b) - 2(b^2c/b+c) - 2(c^2a/(a+c))

(b-a)^2+(c-b)^2+(a-c)^2 >= (k-1) ((b-a)(ab/a+b) + (c-b)(bc/b+c) + (a-c)(ac/a+c))

[mellow idol]

ye I got this too

[pearl pond]

@mellow idol Has your question been resolved?

[mellow idol]

<@&286206848099549185>

[frozen lantern]

thats

hm

interestin

you think am gm might help?

[mellow idol]

probably not

in most oly inequalities, am-gm helps usually only after simplification

[mellow idol]

I got stuck beyond this

[cinder thistle]

hm, lemme try somethings
also, i am going to talk about "t" where t = k-1, as maximising t is the same as maximising k

[mellow idol]

yup

only thing I got is k is at least 1

[cinder thistle]

yeh, because the first part is trivially atleast 0

[mellow idol]

yeah

is there any way to upper bound the expression on the right

lower bounding will probably not work since we need an upper bound for k

[cinder thistle]

hm

(b-a)^2(ab/b^2-a^2)+(c-b)^2(bc/c^2-b^2)+(a-c)^2(ca/a^2-c^2)

[mellow idol]

since it's homogenous i'm gonna try assigning some sum for a,b,c and hope it gives me something

[cinder thistle]

hm

[mellow idol]

yup

<@&286206848099549185>

[pearl pond]

@mellow idol Has your question been resolved?

[pearl pond]

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[orchid token]

Acceleration due to gravity is approximately 10 m/s2 pointed down. Mr. Husic throws an eraser vertically into the air, from the ground level. The eraser lands 10 seconds later. (4, 2 marks)

can someone explain this to me

[orchid token]

i've so far said a = -10m/s/s

so v(t) = -10t + Vo

from there im not sure what to do

[rare burrow]

this isn't a question 0_o

[compact bison]

what do u need to find

[orchid token]

oh wait sorry lol

it says to find the initial velocity of the eraser

[compact bison]

0

[rare burrow]

no

[compact bison]

oh nvm

sry sry

i misread

[orchid token]

we are supposed to do with math btw

i mean not SUVAT

like calc

[compact bison]

u need to use this formula T = (2 * u * sin(θ)) / g

[orchid token]

💀

[compact bison]

here theta is 90 degrees

[orchid token]

no i dont think so

[rare burrow]

use $v_f=v_i+at$

[jolly parrot]

00100000

[orchid token]

its math so like thats not

[rare burrow]

what??!?

[orchid token]

i can't use physics stuff

[toxic lichen]

oh yes let's drown OP in black boxes

[rare burrow]

ok what in the world is that supposed to mean

you can't use any kinematics formulas?

[compact bison]

I am sry😭

[orchid token]

like

heres the solution

[rare burrow]

that's a really arbitrary restriction

[rare burrow]

and also, this isn't a math problem

you physically can't do it without physics things

[compact bison]

those are kinematics formulas tho

[orchid token]

its part of my math course and we aren't expected to do it with a detailed knowlege of kin stuff

[rare burrow]

arguably, he's using as much physics as I am when I suggest using $v_f=v_i+at$...

[jolly parrot]

00100000

[orchid token]

i might've not been clear, like we only need to know that acceleration is the deriv of velocity which is teh deriv of dispalacement

[rare burrow]

what?!?! why would you use a differential equation for this

[orchid token]

like we should be able to solve with js that knowledge and the rest being math

[rare burrow]

that's so overkill lol

[orchid token]

its not differential is it?

i haven't even learned that

[toxic lichen]

well then like
h''(t) = -10, h(0)=h(10)=0

[rare burrow]

you're using derivatives...

[rare burrow]

lmao let's take the characteristic polynomial now

and then find an annihilator to solve the IVP

[orchid token]

ok wait heres my working so far

[rare burrow]

well ok this is actually not a diffeq but...

[orchid token]

i kind of got confused

as you can see

[rare burrow]

ok you lost me at using an integral

why would you use an integral for this

[turbid sequoia]

i mean regardless it probably still has eigenvalues if you can split it into two lin. indep. 2 element vectors somehow lol

[orchid token]

to find velocity

its a math course i have to use math things otherwise i can't get marks

[toxic lichen]

what characteristic polynomial

[jovial kiln]

v(t) can be expressed by integrating a(t) wrt t

[rare burrow]

there's none, I'm just being dumb lol

[supple mural]

do a = 10 = dv/dt

[rare burrow]

but... why would you do that for this question that requires no calculus?!

[orchid token]

bru

its my math course

i have to use math 😭

[toxic lichen]

-10

[orchid token]

yeah ive done that

[orchid token]

also why is it -10

[toxic lichen]

bc downwards

[orchid token]

i js thought its + 10 but idk

ok

[toxic lichen]

and presumably you want to run your height axis up

[supple mural]

mb

[toxic lichen]

not down

[turbid sequoia]

perfectly fine way to do it tbh with the integral

[jovial kiln]

you can define g=+10 if you stay consistent with your coordinate system and define up to be negative

but often times it is just convinient to define down negative

[orchid token]

yeah i can js flip my answer at the end the teacher said if its unreasonable

but anyway

what am i supposed to do after v(t)

cuz i tried finding the slope but thats js going to give me the equation of v(t) again

[jovial kiln]

do you have s(t)?

[orchid token]

no but i can't figure out why i'd need it

i can find it rn

[jovial kiln]

max height is given by max(s(t)) which is given by v(t) = 0

[rare burrow]

ok can i just say, he's using no calculus in the model solution here

[orchid token]

he gets the equations using calc

[orchid token]

is this for initial velocity?

[supple mural]

why dont you use kinematics equations
just derive it first then use it directly

[rare burrow]

well, isn't it a standard that in a math class, when your professor derives a result on the board, you're allowed to use it on hw/exams?

if he derived them in class, they're fair game

[turbid sequoia]

he actually does he just uses the formula for the integral of acceleration and velocity

because if your accel is a constant it always comes out to look like that

[rare burrow]

oh, you mean from step 1 to 2?

[orchid token]

well idk i can use kinematics equations but i dont know them well enough and i js want to understand it from the math perspective

[turbid sequoia]

and 2 to 3

[rare burrow]

I honestly read that as him just using $v_f=v_i+at$ originally

[jolly parrot]

00100000

[turbid sequoia]

integral of 10t plus Vo is certainly 5t^2 +v0t +C

[rare burrow]

but I see how it can be interpreted as him taking the integral

[orchid token]

what do i do next

[jovial kiln]

you find s(t) to find the initial velocity, yes

[rare burrow]

from a math perspective, an integral is just the limit of riemann sums (well, the riemann integral at least). from a strictly "math perspective," you understand it already without applying it to physics

[orchid token]

can you explain this agian, i really dind't understand what you mean by saying the max height is given when s(t) is maximized cuz in my head thats related to finding the max height

[rare burrow]

I see I see. I'm convinced he's using calculus now. sorry for being annoying 😅

[turbid sequoia]

we know the position function is maximized when V(t) = 0 and a(t) has a negative sign due to the first and second derivative test

[jovial kiln]

we have that the eraser falls back down after 10 seconds. We also have that
s(t) = -5t^2+ut where u is the initial velocity (found by integrating v(t))

we know that s(10) = 0 since at 10 seconds, the eraser falls back down to its initial position (s is defined as change in position aka displacement)
this lets us find u and then we can solve max(s(t)) using v(t) = 0

[rare burrow]

when you integrate $a(t)$ to get an expression $v(t)$, you have some constant of integration $v_0(t)$ left over. you know $t$, so now you need $v_0(t)$ to solve for $v(t)$ at the relevant $t$

[jolly parrot]

00100000

[rare burrow]

I guess to understand it intuitively, you can think about it like this. If you have some acceleration, $\frac{d}{dt}(vt+v_0)$, it only tell you what rate at which the velocity changes. That's why the derivative will be $v$, and the "information" of the initial speed you were traveling at, $v_0$ is "lost"

that's why the integration constant $C$ exists, and it can be interpreted as $v_0$ here

[jolly parrot]

00100000

[orchid token]

okok

i see

[rare burrow]

hopefully that helps a bit with understanding why the equations have their respective physical interpretation 😅

[orchid token]

right no that makes sense now

thank you

it was confusing for me why we needed distance to find velocity

[jolly parrot]

00100000

[orchid token]

.close

[pearl pond]

Channel closed

Closed by @orchid token

Use .reopen if this was a mistake.

[rare burrow]

sorry, the way I worded this earlier made it sound like $vt+v_0$ was the acceleration

[jolly parrot]

00100000

[rare burrow]

acutally, my notation is really shitty and confusing 💀 I am using $v$ for the acceleration whoops

[jolly parrot]

00100000

[rare burrow]

sorry about that

[orchid token]

lol nw

thank you all

.close

[rare burrow]

dw, it's closed already; the bot just hasn't kicked us out yet

[pearl pond]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[carmine storm]

How would I solve this without using l'Hopital's rule?

[cobalt apex]

theta-pi/2 = x

try that

[carmine storm]

The denominator simplifies to cos(theta) as cos(2pi-x)=cos(x)

[autumn fossil]

and also some trig equalities
(e.g. cos(2pi - x) = cos(-x), because the period is 2pi so you can just subtract it)

[carmine storm]

Yes

[autumn fossil]

later, this should make it turn into sin of something

[cobalt apex]

instead of that

write cos(2(pi-theta/2)

and subsitute x in it

2x/cos2x

where x is tending to =

cos 0 =1

x=0

final answer you get 0

is that the answer

[carmine storm]

No

It's -2

[cobalt apex]

hmm

were any of my steps wrong

[carmine storm]

Okay

I have worked this out

[
\lim_{\theta \to \pi/2} \frac{2\theta - \pi}{\cos(2\pi - \theta)}
]

[jolly parrot]

waffle

[cobalt apex]

its not even 0/0 form

why are you applying l'hospitals

oh wait

nvm

actually

wait

it its not

[carmine storm]

[
\lim_{x \to 0} \frac{2x}{\cos(x+\frac{\pi}{2})}
]

for $x=\theta-\frac{\pi}{2}$

[jolly parrot]

waffle

waffle

[carmine storm]

[
=\lim_{x \to 0} \frac{2x}{-\sin{x}}
]

[jolly parrot]

waffle

[carmine storm]

This is -2

[cobalt apex]

let me see

1 sec

yeah

that should be it

[carmine storm]

.close

[pearl pond]

Channel closed

Closed by @carmine storm

Use .reopen if this was a mistake.

[pearl pond]

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

problem 2

[spare lark]

Why is problem 1 and 2 from two completely different level

[pearl pond]

@balmy oracle Has your question been resolved?

[balmy oracle]

dont know

its kind of a hybrid of field theory and complex analysis

[pearl pond]

@balmy oracle Has your question been resolved?

[pearl pond]

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[buoyant crater]

im completely stuck, no idea what to do, asked chatgpt but they dont understand my question

[fringe robin]

What do you know about all the triangles?

[buoyant crater]

literally nothing

equilateral has all equal sides

thats all i know

oh WAIT

is it D

@fringe robin

[nimble socket]

someone asked this earlier today...?

anyways