help-0

[mortal trellis]

in fact, that fraction is a tiny bit less than 7, so x is more like 6.999999 or something

exact value doesnt matter

but therefore the quotient is 10^25-(nearly 7) = 99999....93.something

[waxen jacinth]

Really really thanks bro.

thanks

.close

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[dusk umbra]

The point of a vector field is to kind of get a feel for how a function affects each point in space right?

[dusk umbra]

For example, if I have a complex function to which I apply the transformation w = i*f(z)

Anywhere on the axes, i would draw a perpendicular vector, because at those points, the function is "pulling it" upwards while it tries to rotate the point by 90 degrees

Similarly, at some point say (1,1) on the complex plane, I would get a vector at that point thats kind of "diagonal"

Because the transformation w vector at that point has an x component and a y component

or an RE IM component

Is this interpretation correct?

.close

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[zinc haven]

this looks ill-defined

what exactly does D do to constant polynomials?

[mortal trellis]

(doesnt matter)

[zinc haven]

why?

[mortal trellis]

well you can find out something about what Dp for p constant could be. but it doesnt change anything about whether D is surjective or not

[tiny sky]

often, 0 is thought of as a polynomial with degree -infinity or -1 depending on the situation

[zinc haven]

hm

i'm thinking for this exercise i'd examine how D acts on P_m(R) and prove that it's surjective in P_m(R)

then somehow generalize it to P(R)

[mortal trellis]

idea is good

[quasi vector]

D isn't surjective in P_m(R)

D from P_m(R) to P_{m-1}(R) may be

[lone heart]

@zinc haven Has your question been resolved?

[zinc haven]

yes, typo

[zinc haven]

would induction work?

start with verifying m = 1, then work up to m = infty

if i suppose arbitrary m, then i struggle to prove im(D|_P) = P_(m-1)(R)

[mortal trellis]

you could phrase it as induction, sure. but where are you even using the induction hypothesis

remember, you only need to hit a spanning set

[zinc haven]

hit?

[mortal trellis]

the elements in the image are those that you "hit" with your function

[zinc haven]

oh wait, this might be simpler than i thought

by rank-nullity, m = dim(ker(D|_m)) + dim(im(D|_m))

but by definition, D only reduces p degree by 1, so dim(ker(D|_m)) is basically 1, which is the space of all constant polynomials

i think?

[mortal trellis]

not quite

[zinc haven]

just asking, if i want to prove im(D|m) = P(m-1)(R), then is double inclusion viable?

i can show subset, but i'm still thinking on supset

[mortal trellis]

yes thats viable

[zinc haven]

hm, i have an idea of the proof

but it would require that i can rewrite p in P(R) as a_0 + a_1x + a_2x^2 + ...

[mortal trellis]

how else would you write it?

or do you mean as an infinite sum?

thats yikes

[zinc haven]

yeah

axler still hasn't properly dealt with infdim spaces so i'm still very much shy

[mortal trellis]

polynomials are finite sum

forget infinite sums

what is D(x)

[zinc haven]

a basis of P_0(R)?

[mortal trellis]

yes

so what is D(span(x)) ?

[zinc haven]

whar

so like, D(ax)?

still a basis of P_0(R)

[mortal trellis]

span(x) is a set

D(set) should also be a set

D(set) means "apply D to every element in the set"

[zinc haven]

ok, then P_0(R)

[mortal trellis]

what is D(x^2) ?

[zinc haven]

a basis vector of P_1(R)

[mortal trellis]

more precise please

[zinc haven]

a member of a basis D(x^2), D(x) of P_1(R)

[mortal trellis]

why do D(x) and D(x^2) form a basis?

[zinc haven]

well if we write it as a sum, we have D(a_1x + a_2x^2), which is an arbitrary polynomial in P_2(R). we can prove that im(D) restricted to P_2(R) is precisely the space P_1(R)

[mortal trellis]

nono, thats what I want to prove using this

D(x) is a (nonzero) constant polynomial

D(x^2) is a linear polynomial

so they are two polynomials of distinct degrees

so they are lin independent

both living in P_1(R) which has dim 2

hence they are a basis

[zinc haven]

oh yeah..

ok so we can extend this to P_m(R)

what i'm worrying is how to extend this to P(R)

[mortal trellis]

you only need to hit a spanning set of P(R)

e.g. a list of polynomials of all degrees

or you can also phrase it differently, to show surjective you need to show that every polynomial is in the image

but every polynomial lives in one P_m(R)

so you are done

[zinc haven]

yes, i prefer the latter approach

axler doesn't really define a spanning set for infdim spaces

in fact, it's the definition of infdim spaces

[mortal trellis]

just because axler does it that way doesnt mean that you also have to do it that way. at some point you will encounter infdim spaces so it doesnt hurt to touch them from time to time

P(R) is barely even infdim, it still behaves very nicely

[zinc haven]

alright sure

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[worthy imp]

What is the turning point of the function: -x^4 - 2x^3+2x^2+4x
I got the derivative (-4x^3-6x^2+4x) then put it into chat gpt but it seems to have an answer inconsistent with the maximum of that curve in the function. Its answer was -1.41,0 but the maximum of the curve was different

[sharp panther]

Whats a turning point?

[worthy imp]

The point in a curve where it goes from increasing to decreasing

Or id assume its the maxim of the curve

[sharp panther]

Oh yk, what's your problem

[worthy imp]

What the turning point of the first curve in the equation is

I got the derivative then failed to equate it to 0

[sharp panther]

failed to equate to 0 ? Why?

[worthy imp]

I couldn’t factor it

My brain immediately went to goo

[sharp panther]

What did you get as the derivative?

[sharp panther]

I think you missed a + 4 here ...

In the derivative

[worthy imp]

-4x^3-6x^2+4x+4

[worthy imp]

I thought y didn’t factor that

When finding the derivative

[lofty heart]

huh

derivative of 4x is 4

[worthy imp]

Oh

Maybe I’m thinking of just a constant?

[lofty heart]

constants go

[worthy imp]

Ye mb

[sharp panther]

I agree it's not going to be easy to factorise the derivative

[lofty heart]

do you get to use a calculator

[worthy imp]

Yes

[sharp panther]

Then it's done

Find the roots of the derivative

Theyre the turning points

[worthy imp]

Why does my calculator give a different turning point of -1.7 =x

Chatgbt insisted -1.41

Is that an error on chatgbt part?

[fierce cipher]

!nogpt

[lone heart]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[worthy imp]

Mb

It’s their any good ais

[worthy imp]

Oh I see

[lone heart]

@worthy imp Has your question been resolved?

[lone heart]

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[sick tangle]

do i just work out

lets suppose that the intersection point was R

of those two lines

do i just work out

MR and PR

or

AM AR
and

AP PR

[fierce cipher]

you can extend PQ so that it intersects BC at E and AD at F

then prove that CE = MF

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[wraith marlin]

help

[wraith marlin]

did i solve this right?

please help

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[cinder nymph]

Find y

[naive valley]

found it:

[opal jolt]

!original

[lone heart]

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

[pallid scarab]

consider the quadrilateral that contains your angle y

the sum of the angles inside that quadrilateral is ...

then, try to find each angle inside that quadrilateral

[cinder nymph]

360?

[pallid scarab]

yes

there should only be one angle that's not "obvious"

[cinder nymph]

I has 50 90 90 and got 120

Mb not 120

130

Idk what to do to find y tho

[cinder nymph]

Hal

P

[pallid scarab]

I don't understand how you get any angle except for 50

I'm talking about this "walking legs" quadrilateral

find each angle separately, or express it in terms of y

to help us get the names correctly, let's name the points of this quadrilateral

say A is the top point

B bottom left

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@cinder nymph Has your question been resolved?

[cinder nymph]

I gotta find the centre tho

To get it

[lone heart]

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[hot drift]

Just some simple geometry homework if someone can help, I also have a answer key (provided by teacher) if need be

[broken pivot]

Do you know about rotational symmetry?

[hot drift]

jog my memory

[broken pivot]

If you can rotate a figure around, and the rotated figure looks exactly like the original figure, the figure has rotational symmetry

^ only if you can keep rotating at a constant angle and end up with a 360-deg rotation

[hot drift]

yeah i believe she mentioned that

my teacher

[broken pivot]

Can you try solving #1?

[hot drift]

ive got no idea what its asking me

i missed todays class

[broken pivot]

Ok one sec

Since there are 5 equal parts to this figure that differ only by a rotation about the center of the figure, the figure can be rotated 1/5 of a full rotation

[hot drift]

yeah understandable

[broken pivot]

Can you try solving #1 similarly, by counting how many equal parts there are

[hot drift]

but then wouldnt the circle be able to be infinitively rotated

[broken pivot]

You can rotate a circle as little or as much as you want, and you'd end up with the same figure.

[broken pivot]

*infinitely, not infinitively

[hot drift]

ok

[hot drift]

but this time id rotate it 8 times because thats the amount of slices right

[broken pivot]

No

[broken pivot]

There are 8 equal slices

But how many times must you rotate the object to get an identical looking object?

[hot drift]

∞

[broken pivot]

Are we talking about the circle or the figure in #1

[hot drift]

circle

thats # 1

[broken pivot]

This is not a pure circle; it still has 8 line segments

[hot drift]

so id rotate it 8 times right

[broken pivot]

Not 8 times

It should be 1/8 rotations

[hot drift]

why

[broken pivot]

Since each slice is off from another by 1/8 of a rotation, you must perform 1/8 of a rotation to align the current slices with where the previous slices were

[hot drift]

OOOOH

i get it

what now

[broken pivot]

So put 1/8 rotation

And also find how many degrees is 1/8 of a rotation

[hot drift]

i could do that by dividing 360/8

[broken pivot]

✅

[hot drift]

is that it?

[broken pivot]

You havent written the smallest fraction though

[hot drift]

huh

[broken pivot]

You need to write 1/8 rotation AND 45 degrees

[hot drift]

this is what my answer key says

now i need help with this

<@&286206848099549185> i need hep nobody's responded in 15 minutes

[sly river]

What is this question?

[hot drift]

something something symmetry

[sly river]

Try graphing where the x cordinate is the sides and the y cord is the diagonals

[hot drift]

huh

[hot drift]

what

[sly river]

The question is asking for diagonal’s for certain shapes yeah, so if you graph the sides to the diagonals and find a relationship you will find the answer

[lone heart]

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[tacit acorn]

I know how to do Eulers method but now that im given a graph and not a equation im stuck

[tacit acorn]

.close

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[weak lichen]

is this how i should be doing this

[weak lichen]

also im stuck idk what to do next

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[tranquil quest]

Im trying to wrap my head around chemistry mole/gram/etc calculations but i am struggling, i know how to do the formula when its put in front of me, but knowing how to set it up from the raw data is where im struggling.

I pretty much need to be taught it in a percise way so i can actually do the problems.

[broken pivot]

,rcw

[broken pivot]

!1c

[lone heart]

Please stick to your channel.

[broken pivot]

#1336527646963535943

[lone heart]

@tranquil quest Has your question been resolved?

[lone heart]

@tranquil quest Has your question been resolved?

[lone heart]

@tranquil quest Has your question been resolved?

[coral night]

22.4L/mole of gas at stp, so if you have 4 moles you want to cancel the moles and turn to volume (litres), so 4moles*(22.4L/1mole)= 4x22.4L of gas occupied

[lone heart]

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).

[serene ridge]

nevermind you're answering the question im stupid

[coral night]

lol all good

when doing unit conversions you want an equal ratio, so 60sec/min, 2gramH2/mole H2, and you can flip that ratio to do what u wnat with it, as long as it remains equal. So like 5 minutes to seconds would be 5min*(60sec/1min), and notice how now on the bottom you have minutes, and at the top u have minutes, so they cancel and now youre just left with seconds as units, so 5x60sec

[tranquil quest]

i get that part i suppose, but i struggle with building the actual formula from the raw data

like

when its in front of me filled out

its easy

but like

taking from that world problem

[coral night]

what raw data?

[tranquil quest]

and turning it into a formula to solve is where i go brain dead

[coral night]

there are some ratios you have to memorize, like 22.4L/1mole, 6.02x10^23/1mole (i think i havnt used avagadros number in a minute)

[tranquil quest]

like see in 7. i stopped midway becuase i realized i didnt actually know EXACTLY what to do,

[tranquil quest]

6.02x10^23

im aware of stp being 22.4

and 6.02x10^23

i ust get messed up actually setting up the raw formula, for example

80gMethane = mol / g

would i put the base weight of methane as a whole (16.04) where grams is and 22.4 where mol is so

80g = 22.4mol / 16.04g

[coral night]

oh okay

[tranquil quest]

like that

[coral night]

one sec

whats the formula for methane?

[tranquil quest]

to actually set it up, i dont know what to do, but i know how to solve it, weird lol

Ch4

so its 16.04

in molar mass

CH4

[coral night]

okay yea so 16.04g/mole, so you have 80 grams of it, start with whatever you have 80gramsCH4 x 1mole/16gCH4 (now grams cancel, we have our same units opposite), so were left with moles of CH4 80/16 moles specifically

notice how the only units youre left with are moles

[tranquil quest]

ill look this up lol for sure

[coral night]

so you want the same units as what you have opposite (so top of fraction or bottom of fraction) from where you start

[tranquil quest]

then id multiply it by that so in a calc itd 80x1/16

and then id get my answer

[coral night]

yes, and then you get moles

[tranquil quest]

i see i see, ill just have to remeber actually setting it up, since i can do the equation in front of me, but the upcomking test and all our work has me make the equation before hand from word problems

which ill get caught up in things like

27gH20x1mol/18.015gx6.02x10^23molcules/1g

[coral night]

yea the important thing to remember is to always cancel your previous unit, and keep your ratios equal

[tranquil quest]

i ended up getting 16.02x10^20 which is wrong apparently

[coral night]

which question is this?

[tranquil quest]

its just outta frame but it ask for

How many moles of water molecules are present in a 27 gram sample of water

which were now adding more operations to the formula lol, so it makes it more complciated

[tranquil quest]

hence here where i got it wrong after coming up with this, from that word problem

[coral night]

actually thats just asking for moles, not molecules so no need to use avagadros

[tranquil quest]

ah okay okay

[coral night]

so just this i think

[tranquil quest]

oh 😭

[coral night]

if it did ask for molecules though (gimme one sec)

[tranquil quest]

sometimes there'll be equations of large units itll have conversion rates (1000 = 1) stuff taht comes before which i also dont entriely understand the preresuquite for it to be needed

[tranquil quest]

np

[coral night]

the important thing is to always convert to moles when it asks for a different unit

such as volume to mass

[tranquil quest]

the 6.02x10^23 is the molecules to make sure

[coral night]

so thats like 1kg=1000g right?

[tranquil quest]

so if it was asking for molecules i wouldve had the formula right 😭

[tranquil quest]

yessir\

or mL to L

which L is different

[coral night]

yes there are 6.02x10^23 molecules or atoms in a mole of an element or compound

[tranquil quest]

22.4 STP

ill just have to get more in the routine of using it ig, ill go to tutoring aswell

[coral night]

well theres 1000mL in one L, but a gas at stp takes up 22.4L

[tranquil quest]

mhm, instead of 6.02x10^23

[coral night]

yea i mean theres a lot of different pathways in this

well theres a lot of things that can equal one mole right, like 2grams of H2/1mole

[tranquil quest]

which is my problem i think, im usually good at math since its just memorization, but ive fallen flat for chemistry

[coral night]

the molar mass of an element is actually equal to the weight of one mole

[tranquil quest]

im aware, it changes, hence why we have 1 mol / (molar mass)

the word problems may be messin g me up bad so ill mess aroudn with formatting it better

im about to be going to bed here, but thanks for answeing

.close

[lone heart]

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[manic wyvern]

!status

[lone heart]

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

[dawn otter]

2. I have begun but got stuck midway.

[manic wyvern]

show your work then

[dawn otter]

3a + 5b - 7 = 0

a - 2b - 4 = 0

I stuck in this
because I don't know for this how to go

[manic wyvern]

do you know elimination method

[dawn otter]

yes
eliminate the variable

eliminating the variable when adding or subtracting.

[thorn monolith]

which variable are you trying to eliminate?

a or b

[dawn otter]

eliminate a first then eliminate b

[alpine sable]

can someone help me... i need 5 illustrative examples of angle sum theorem..

[thorn monolith]

eliminating one should be sufficient to find the other

[manic wyvern]

how should we do that?

[alpine sable]

!help

[lone heart]

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

[dawn otter]

my teacher say that need eliminate a first
after calculate then eliminate b

but I don't know how to do

[manic wyvern]

lets take a different example

x + 2y = 5
2x - 3y = -4

lets eliminate x first

so we can multiply the first eqn by 2
2x + 4y = 10

then subtract both the eqns

?

[dawn otter]

don't understand

[manic wyvern]

which step?

[dawn otter]

all

@crimson dawn
there ah?

i may need some chinese explaination or make it simple
let me easy understand

[manic wyvern]

oh sure

good luck!

[dawn otter]

@ocean seal
3a + 5b - 7 = 0

a - 2b - 4 = 0

[crimson dawn]

Yea I'm back mb

你看那两个方程的左边

是3a+5b-7

[dawn otter]

没事

[crimson dawn]

和
a-2b-4

[dawn otter]

我老师讲要eliminate a 还能拿到b
之后eliminate b 还能拿到 a

[crimson dawn]

这是对的

我们要把这个方程做一点儿改变

所以我们有
3a+5b-7
和
a-2b-4

a-2b-4 有 a
3a+5b-7 有3a

如果我们把a-2b-4程3

可以变a-2b-4成3a-6b-12

这让这个方程有3a

因为a-2b-4=0
3a-6b-12=0*3
3a-6b-12=0

因为两个方程有3a我们可以减掉那个3a

(3a-6b-12)-(3a+5b-7)=0-0

3a-6b-12-3a-5b+7=0

[crimson dawn]

那两个3a 被减掉

所以-6b-5b-5=0

-11b-5=0

从这应该可以把

@dawn otter

[dawn otter]

能简单吗
我很难理解

[crimson dawn]

我去吃饭先

[dawn otter]

OK

[lone heart]

@dawn otter Has your question been resolved?

[crimson dawn]

我们先看一个多容易的题

a-2b=20
a+b=25

这你可以做吗？

[dawn otter]

eliminate 什么？

[crimson dawn]

一个是比另外一个容易

你有看过一个number line吗？

[dawn otter]

有啊

[crimson dawn]

我们用number line 来解释

This is a on a number line

[dawn otter]

ok

[crimson dawn]

我们也写a+b和a-2b

我们知道a+b=25,a-2b=20
我们也写下去

[dawn otter]

我先吃晚餐

[crimson dawn]

ok

[dawn otter]

@lone heart
set mine not complete

[dawn otter]

回来了

@crimson dawn

[crimson dawn]

Ok

[crimson dawn]

这个绿色的长度

你可以用数字写吗？

@dawn otter

[dawn otter]

不能

这一题我会些
就是
3r+r
然后全部加r

[crimson dawn]

你选的是错的

[dawn otter]

wah
爆头咯

[crimson dawn]

你不可以在这问

#❓how-to-get-help

[crimson dawn]

我们回到这边先

@dawn otter ，那个绿色的东西is between 20 and 25

[dawn otter]

ok

[crimson dawn]

所以那个长度是25-20=5

[crimson dawn]

我们也可以看出绿色的长度是2b+b

[dawn otter]

ok

[crimson dawn]

所以2b+b=5

如果你把(a+b) - (a-2b)也可以拿到b+2b

[crimson dawn]

在这如果
3R-3S=-10
R-9S=2

把 R-9S=2 程三给我们
3R-27S=6

如果把(3R-3S)-(3R-27)会给24S

所以24S=-6-10

S=-16/24

哎这也很难解释😭

@dawn otter 你可以问你老师吗？

[dawn otter]

3a + 5b - 7 = 0
a - 2b - 4 = 0

这一个
可以教多一次吗
equation

[crimson dawn]

3a+5b-7=0(equation 1)
a-2b-4=0(equation 2)

3*(equation 2) :
3(a-2b-4)=0*3
3a-6b-12=0(equation 3)

(equation 3)-(equation 1):
(3a-6b-12)-(3a+5b-7)=0

3a-6b-12-3a-5b+7=0

-6b-5b-5=0

-11b-5=0

@dawn otter

[dawn otter]

okay

谢了

.close

[lone heart]

Channel closed

Closed by @dawn otter

Use .reopen if this was a mistake.

[lone heart]

Available help channel!

Send your question here to claim the channel.

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[proven ivy]

we did this before break and ngl i forgot everythign. this is a new problem tho

[sturdy haven]

do u guys have any polynomial functions that can graph a heart

im using geogebraaaa

[proven ivy]

find ur own channel.

[sturdy haven]

oker

[tough relic]

Since they're rotating, think of both of them as functions of sine and cosine.

[proven ivy]

yup

[tough relic]

Area would be length × width

[proven ivy]

i think they want me to find the distance or smthing?

[tough relic]

Area of the quadrilateral ABB'A'

[proven ivy]

it would be the distance formula or smth right

[proven ivy]

how would that be possible

[tough relic]

I'm thinking A(t) = l(t) × w(t)

l(t) can be in terms of sin and w(t) in terms of cosine.

[proven ivy]

width of the quadrilateral shown would be 12 right?

like the "defualt" one

[tough relic]

Think of the two dots as oscillating points.

[tough relic]

Yes

[proven ivy]

ok

[tough relic]

$l(t)=6\cos(\omega t)\$
$w(t)=6\sin(\omega t)$

[ocean seal]

GoldBarley

[proven ivy]

so the center=the midline right

[tough relic]

?

Centre of what?

[proven ivy]

wait i ithought the radius=the amplitude? why is it 6 instead of three

[proven ivy]

center of the circle? or is that only for problems with one circle

[tough relic]

But there are two circles.

[proven ivy]

so u times 2 because its 2 circles?

[tough relic]

Yea, and considering the signs of the trigonometric functions in each quadrant of a unit circle, things should be fine.

[proven ivy]

what is the omega part?

[tough relic]

The angular velocity.

[proven ivy]

the what?

i havent learned that yet 😭

im a sophomore in high school

[tough relic]

$\omega=\frac{2\pi}{time period}$

[ocean seal]

GoldBarley

[proven ivy]

ohhh ok

[tough relic]

Oh 😅
I thought thinking in terms of oscillations rather than two rotating circles would make things a little easier. Sorry.

[proven ivy]

yeah its ok. only reason i sorta know oscillations and sine waves is cuz i play around with synthesizers in my free time. so i know the sound but not the visual

[tough relic]

When calculating omega, keep the units in mind. Because it says you'll have to graph A(t) as well.

[proven ivy]

so is it just 2π over 8

π over 4

[tough relic]

In radians/minute

That should give you A(t) I think.

[tough relic]

^

[proven ivy]

and wouldnt it be +3 cuz point a is at 0,3

at the end of each function?

[tough relic]

Huh, I didn't think about that. But I think the length function remains unchanged.

I considered length to be the horizontal displacement and the width to be the vertical. So you can add +3 to the width function.

[proven ivy]

so if its 2π over 8 minutes itd be π over 4 in a minute?

wnat abt phase shift/horizontal shift

[tough relic]

I thought of them as independent oscillations (the two circles).

You can add -π to the oscillations of the first circle because they're rotating clockwise.

That will only change the signs of the trig functions, ± π doesn't change the functions themselves.

[proven ivy]

so in the form of y=asin(bx-c)+d there is no c?

[tough relic]

There is a phase difference between the oscillation of both circles.

The first circle is π radians behind the 2nd one.

[proven ivy]

but what would that look like for l(t) and w(t)

[tough relic]

I really think it's best to keep it unchanged 😅

That causes some trouble because then both functions become 0.

Since a phase difference of pi doesn't change the functions themselves, i think it's simpler if we just keep it as it is.

Perhaps l(t)=-6cos(...)
w(t) = well this is getting difficult

w(t)=-6sin(...)

So I think the area remains unchanged.

[proven ivy]

hmm ok. Ill ask the teacher abt it tmr

u mind if i friend u to dm if i have any questions before class tmr? I needa go to bed but imma attemp the problem on the bus @tough relic

[tough relic]

Sure

This is probably what A(t) should look like. I'm not too sure now.

(t in minutes)

[proven ivy]

says ur not accepting requests. Plz friend me

[tough relic]

Oh yes

[tough relic]

Sorry about that, I forgot about it.

[proven ivy]

ur good

.close

[lone heart]

Channel closed

Closed by @proven ivy

Use .reopen if this was a mistake.

[lone heart]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[waxen turtle]

Idk how to do it like at all even when i use a percentile calculator it still gets it wrong

[lone heart]

@waxen turtle Has your question been resolved?

[lone heart]

@waxen turtle Has your question been resolved?

[winged crescent]

wsg

[glad pier]

hello
does anyone have trigonometry questions??????

[lone heart]

@waxen turtle Has your question been resolved?

[lone heart]

@waxen turtle Has your question been resolved?

[mossy pike]

can somebody help me with finding the x values in equations like this: x^2+10x=-12

[proven ivy]

i think u move the 12 to the other side and factor

[twin comet]

quadratic equation :>

[reef roost]

wow

[lone heart]

@waxen turtle Has your question been resolved?

[vast root]

can you send the raw data set

like in a txt file

actually you dont even need to do allat

find the position with (P/100) * (n+1),
where P would be 79, for 79th percentile

whatever number it is, round up and down to find out what positions your value is inbetween

and interpolate

(lower value) + (the decimal part of your original value) * (upper value - lower value)

round to one decimal place

and thats your number

[fading willow]

is 0.999... + 0.000...1 = 1?

[chrome tiger]

What is even 0.000...1

How many zeroes are there?

[fading willow]

infinitely many

[chrome tiger]

And how would you define such a number?

For example, 0.37 = 3/10 + 7/100

0.1337 = 1/10 + 3/100 + 3/1000 + 7/10000

So you basically add up fractions

A number with infinitely many digits is represented as series

Series for 0.999... is 9/10 + 9/100 + 9/1000 + ... + 9/10^k + ... = 1

(sum of geometric progression)

[chrome tiger]

And now hopefully this question makes sense ^

[fading willow]

is there latex here?

[chrome tiger]

yes

[fading willow]

\sum

[chrome tiger]

$\sum$

[ocean seal]

spindle

[fading willow]

$\sum_{n=1}^\infty \frac{1}{10^n}$

[ocean seal]

ye91

[chrome tiger]

Yeah that's how you define 0.111...

[fading willow]

$\lim_{n \to \infty} 10^{-n}$

[ocean seal]

ye91

[chrome tiger]

zero

[fading willow]

approaches zero

never 0

[chrome tiger]

the limit is zero

[fading willow]

yes the limit is 0

but It can never be 0

yea

[chrome tiger]

Which object are you refering to?

[fading willow]

?

[chrome tiger]

$\lim_{n \rightarrow \infty}10^{-n}$ is a well-defined real number. In fact, it's $0$

[ocean seal]

spindle

[fading willow]

the limit is zero

it's never 0

if we say that that's zero

what's the point of limits

infinity is a concept

it doesn't work with real manipulations

[chrome tiger]

Let's make it clear: $10^{-n}$ is never zero for any natural $n$. However, $\lim_{n\rightarrow\infty}10^{-n}$ is zero (literally and mathematically)

[ocean seal]

spindle

[fading willow]

just because some mathematicians said so doesn't make it true literally

its basic logic

[chrome tiger]

I think you misunderstand the concept of definitions in mathematics

Limit of a sequence has a strict definition

[fading willow]

there's a reason we only estimate the area under the graph

we define it by limits

[chrome tiger]

$a \in \mbb{R}$ is the limit of a sequence ${a_n}$ iif for any $\varepsilon > 0$ there exists $N \in \mbb{N}$ such that for all $n > N$ the following holds: $|a - a_n| < \varepsilon$

[ocean seal]

spindle

[fading willow]

because we will always get bigger numbers

[chrome tiger]

Verfiy this for a=0, a_n = 10^{-n}

[fading willow]

if we don't use that then the area under the graph will just be equal to the limits

which is not true

[chrome tiger]

Alright at this point I think you're just trolling

[fading willow]

I'm not

I don't like definitions that exist simply because that's how they look

it's not intelligent

[chrome tiger]

wdym

[fading willow]

and could be wrong

[chrome tiger]

How a definition can be wrong? As long as the object being defined exists, the definition is correct

[fading willow]

its like saying because we fall gravity exists

but you can't explain it

[fading willow]

well infinity is not defined

[chrome tiger]

In mathematics we have a collection of certain axioms. We then build more stuff on top of those axioms by introducing definitions and proving theorems

The entire thing is purely logical

[chrome tiger]

I'm not even mentioning infinity anywhere, look at this definition of the limit: #help-0 message

[fading willow]

is 1 = 0.999...?

[chrome tiger]

Yes

[fading willow]

ok well now your playing with infinity

[chrome tiger]

0.999... really means "the limit of the sequence 0.9, 0.99, 0.999, ..."

The sequence is well-defined, the limit is well-defined

[fading willow]

it's a limit

a limit to an infinite series

which only approaches the limit

which is why it's called a limit

which why it'll never reach the limit

well it's numbers

numbers are infinite

it's like saying that the expansion of the universe is equal to some value

the universe is infinite

it will never reach a limit

sorry I meant space in the universe, but yea I think the rate of expansion is also increasing

lol I just had an argument with a kid at school about 0.999... equals 1 so I came on here

I mean it's inherently simple logic

[serene ridge]

0.9 = 1

proof by "just round it up"

[chrome tiger]

Again, of course the elements of the sequence don't reach the limit, but we're talking about the limit itself

We define an infinite decimal fraction as the limit of a certain sequence

So $0.999...$ is really just a shortcut for $\lim_{n\rightarrow\infty}\sum_{k=1}^n \frac{9}{10^k}$

[ocean seal]

spindle

[chrome tiger]

And by definition of the limit this number is 1

Again, I know that $\left{\sum_{k=1}^n\frac{9}{10^k}\right}_{n\in\mbb{N}}$ doesn't contain $1$, but the limit of the corresponding sequence is $1$, and that's what we assign to $0.999...$

[ocean seal]

spindle

[tardy stag]

why are yall even talking about this in here

this is inverted's channel

[chrome tiger]

My bad, should've redirected them

[fading willow]

Do you believe in the -1/12

Theory

[chrome tiger]

We can proceed in a separate help channel if you want
(see #❓how-to-get-help)

[lone heart]

@waxen turtle Has your question been resolved?

[robust geode]

Hello, sorry for the time it took to get some help, I'll be giving you some right away

Since you needed to get the 79th percentile, k will be 79 here

Since there is 117 numbers in the data set, as stated in the exercise, we will have n=117

The result from this calculation will normally give you a number which will be looking similar to the numbers in the dataset. More precisely, it will either be exactly like a number in the data set, or it will be in between two of them

If it's the same, then you just found your percentile

If it's not, then you will be able to calculate your percentile with the two values besides the one you found with the previous calculation with this :

With d, being the number minus the floor of that number (we also call this the decimal part)

And your previous number being in between $x_i$ and $x_{i+1}$

[ocean seal]

Epsilia aka Epe

[robust geode]

Hope you'll be able to move forward from this help

[lone heart]

@waxen turtle Has your question been resolved?

[robust geode]

Bruh

Please tell me what's not being understood

[waxen turtle]

I CANT DO IT RN

😭

IN A BIT

[lone heart]

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

[deep condor]

bruv

[robust geode]

Ow okay lmao very sorry, no pressure then !

[lone heart]

@waxen turtle Has your question been resolved?

[waxen turtle]

OK IM HERE

[waxen turtle]

Idk how

<@&286206848099549185>

I tried this but it was wrong

[waxen turtle]

<@&286206848099549185>

[twin nimbus]

@waxen turtle 64th percentile means 64% of all people have less than X score. How many people are 64% of this population?

[waxen turtle]

like 26

I think

[twin nimbus]

25.6

So what is a score such that 25 people are less than this score, but not 26?

[waxen turtle]

Omg i got 26.24

[twin nimbus]

The population is 40, and so 64% is 40 * 0.64 = 25.6

[waxen turtle]

okkk

[twin nimbus]

Unfortunately, fiddly bits about how to handle the 0.6 person exactly, and whether it's less than or equal or strictly less than depends on the exact definitions that your text uses, so I would ask that you give your text's definitions

[waxen turtle]

i think it is like a decimal number idk

[waxen turtle]

Wat do I do

[twin nimbus]

The essence of percentile is this idea

@waxen turtle

[waxen turtle]

is it like 63.6

Idk

[twin nimbus]

I'd honestly say the answer is just "64" or "63" depending if you're using < or ≤ in the definition

But I'd have to see your definition of percentile to tell you for certain

[waxen turtle]

Should I try this

@twin nimbus

[twin nimbus]

I'd have to see your definition of percentile to tell you for certain

[waxen turtle]

Idk whatever i do is wrong

So 8.16 is like where it's at right

@twin nimbus

[twin nimbus]

I'd have to see your definition of percentile to tell you for certain

[waxen turtle]

Idek waht it is

[twin nimbus]

It should be in your book

[shy umbra]

heeh

[alpine sable]

Is it 23.76?

[sinful grove]

Guys what's the Taylor Lagrange formula in order 2 of tanx

Around 0

<@&268886789983436800>

[robust geode]

"Your previous number being in between $x_i$ and $x_{i+1}$" as I said. And it was 26.24, so it is in between 26 and 27. So here it would have been $x_i=26$ and $x_{i+1}=27$

[ocean seal]

Epsilia aka Epe

[robust geode]

Other than htat the rest was good

You can see how there is multiple ways to approach things like these. Although I'm surprised you haven't had any sort of definition, like a proper course ??

[marsh wedge]

who needs help

[compact sandal]

me

[sullen niche]

@thick shard

[marsh wedge]

wut wiz

[tepid drum]

#❓how-to-get-help

[urban swan]

just the next term

dont memorize

[waxen turtle]

It was the midway point

OK

.close

[lone heart]

Channel closed

Closed by @waxen turtle

Use .reopen if this was a mistake.

[full night]

what-

[dusk granite]

Bruh

[dusk granite]

bro

it is not

[dusk granite]

💀

Bruh

ill try

.close

Bruh

[coral anvil]

.close

[dusk granite]

dead chat lol

[dusk granite]

Bruh

<@&268886789983436800>

Reclose plz

[dusk granite]

Because what is that lol

.close

.done

[clever folio]

Give the bot time to do things

[dusk granite]

Ok

[tepid drum]

You need to type .coose not .close in this case

[dusk granite]

.coose

[alpine sable]

xd

[dusk granite]

XD

[halcyon walrus]

Why is a Hilbert space?

[summer dirge]

[mental python]

bot still hasnt closed this?

[chrome tiger]

!occupied

[lone heart]

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).

[chrome tiger]

But the short answer is that it's a closed subspace of a hilbert space

[dusk granite]

Yeah

[dusk granite]

Bruh

Scam

[halcyon walrus]

but I can show that this space is closed in the space 𝐻^1 (0,1).

[twin nimbus]

.reopen

[lone heart]

✅

[twin nimbus]

.close

[lone heart]

Channel closed

Closed by @twin nimbus

Use .reopen if this was a mistake.

[twin nimbus]

Double tapping

[waxen turtle]

No

[lone heart]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[modern skiff]

the answer is supposed to be $\frac{1}{2}log|tan{\frac{x}{2}+\frac{\pi}{6}|$

[ocean seal]

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

[wild umbra]

why tf r u doing integrals swerriee

[wild umbra]

that's such a mess

[wind cloak]

try converting the log into its inverse tanh equivalent

or vice versa

[wild umbra]

try to rewrite the denominator using the identity

[wind cloak]

it's probably the same integral

[wild umbra]

$a\sin x +b\cos x= R\sin(x+a)$

[ocean seal]

yajat

[wild umbra]

where $R=\sqrt{a^2+b^2}$ and $\alpha=\arctan \frac{b}{a}$

[ocean seal]

yajat

[desert raptor]

Multiply and divide by 1/2

Hi swer

[desert raptor]

You don't have to do allat

[wild umbra]

once it becomes a function of sin, convert it into csc and use $$\int \csc x \text{d} x = \ln |\tan\frac{x}{2}|$$

[ocean seal]

yajat

[wild umbra]

yea that's just #help-0 message

[modern skiff]

why shouldn't I lol

[wild umbra]

you're a commerce student, aren't u?

[desert raptor]

commerce with maths exists

just in case you forgot

[wild umbra]

there's no way they teach this level of integrals in commerce

[modern skiff]

whta

[desert raptor]

this is ncert pal

stop being so egotistic

[modern skiff]

...mpc students and mec students study from the same book.

[desert raptor]

OH WELL GUYS MY FINAL INTEGRATION RESULT MIGHT BE WRONG

I ALWAYS FORGET INT SEC X DX

[wild umbra]

how is that being egoistic?

[modern skiff]

wow

[desert raptor]

,w int sec x dx\

[wild umbra]

r u serious lmfao

[wild umbra]

what's that

mpc, mec 🤔

[desert raptor]

pcm and commerce eco maths

[desert raptor]

yea im sorry js ignore the last part im sure u can handle that

[modern skiff]

I'm trying

[desert raptor]

do you know how to derive int sec x dx?

[modern skiff]

Log(secx+tanx)?

[desert raptor]

yes, and don't forget the +C

[modern skiff]

yeah

[desert raptor]

this should be that

i messed up, mb

[modern skiff]

but it's still not the same is it

[desert raptor]

it is a correct answer

ur teacher will mark it as correct

[modern skiff]

they told js to get the same answer whatsoever

I'm sure they used that formula

um a min

yes they did

or did they

[desert raptor]

in what world do you need to use this for that problem

it's just overcomplicating stuff

,w int 1/(sinx + √3 cosx) dx

[ocean seal]

Results provided by WolframAlpha

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

Requested by _nofluxgiven

[modern skiff]

what if I converg it into sin(a+b) n trh

[desert raptor]

well then you'd have 1/2 int 1/(cos(pi/3)sin(x) + sin(pi/3)cos(x)) = 1/2 int 1/(sin(x+pi/3) dx

1/2 int csc(x+pi/3) dx

[modern skiff]

I GOT IT

[desert raptor]

i mean it's still the same function, you'd just get the answer in different forms

[modern skiff]

yesssss
and then int of cosecx is log(tanx/2)

thank you!!

.close

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[smoky marten]

Is this correct?

[lone heart]

@smoky marten Has your question been resolved?

[heavy moon]

you need help with question b ?

[smoky marten]

Yes

[smoky marten]

I do this and i dont know if it is correct

[heavy moon]

i’ll try and see if i get the same value give me a minute

i tried to make the steps as clear as possible lmk if there’s something you don’t get

[heavy moon]

@smoky marten

[smoky marten]

I think this is incorrect

Thanks you but i need to go to my school

[lone heart]

@smoky marten Has your question been resolved?

[lone heart]

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tulip violet]

can someone help me with this.
you have to find the shaded area

[tulip violet]

nothing 😭😭

im a bit dumb.. so i need this explained

my teacher isnt much help

[mental python]

can you try finding x?

(using pyhtagoras)

[tulip violet]

ok

lemme try

wait

i dont have my calculator

brb

[mental python]

can you first try writing down the equation?

[tulip violet]

here..?

???

wait

a=3

b=3

???

[deep condor]

Nope

What does pythagorean theorem tell us here?

[tulip violet]

😭😭😭

ummmmm

honestly idk

my brain has switched off

[mental python]

have you seen this before?

[tulip violet]

yea

wait wait wait

im not at that level

ik all that

just like

ummmm

help

[mental python]

for any triangle that has a right angle (90 degrees)

[tulip violet]

yes

ik pythagoras

[mental python]

okay

[mental python]

here, what is a b and c?

[tulip violet]

smth..

[mental python]

any guesses?

[tulip violet]

4.357995321

[mental python]

hm?

whats that

[tulip violet]

idk

i guessed

i can feel my math teacher watching me rn

[mental python]

dont guess

i asked you what a b and c are

answer that first

[tulip violet]

apart of a triangle..?

the right angle triangle 😀

[mental python]

of this triangle

[tulip violet]

b and c are equal..?

[mental python]

no

c is the diagonal

called the hypotenuse

[tulip violet]

oh.

IK IK

😭😭

a and b are equal 👍😝

[mental python]

indeed

theyre both x

now what is c?

[tulip violet]

yay

[tulip violet]

6

cm

[mental python]

yup

[tulip violet]

omg

[mental python]

so fill that into a^2+b^2=c^2

[tulip violet]

i feel smart

isnt that what i did…

..

[tulip violet]

in this.

[mental python]

yes, but you didnt fill in a and b

[tulip violet]

because i dont knwo.. 😭😭😭

do i fill it in with x2..?

[mental python]

a=x
b=x

[tulip violet]

yes.

[mental python]

from this

so, now solve $x^2+x^2=6^2$

[ocean seal]

Bonk

[mental python]

(keep it in exact form)

[tulip violet]

..

can i

get a hint..

on how to

[mental python]

first

[tulip violet]

solve that

[mental python]

write x^2+x^2 as 2x^2

[tulip violet]

yes

now

algebra..?

or

smth

EQUATIONS..?

do i solve this like an equation

[mental python]

are you writing along?

[tulip violet]

yes

[mental python]

good

so, we have $2x^2=36$

[ocean seal]

Bonk

[mental python]

agreed?

[tulip violet]

maybe

wait

yes

i got this

[mental python]

you got what?

[tulip violet]

nothing

[mental python]

(are you sending a photo?)

[tulip violet]

no

🥲

[mental python]

ohhh, you understand the equation

okay

[tulip violet]

yes

yay

maybe

[mental python]

you really dont know how to solve 2x^2=36?

what could be a possible first step

[tulip violet]

our teachers arent that good

um

x2?

times the equation by 2

???

[mental python]

times?

[tulip violet]

divide..?

[mental python]

i think dividing left and right hand side by 2 is more preferable

[heady cloak]

pythagoras?

I can help

[tulip violet]

ok ok

i got this

lemme

write it down

[mental python]

you wanna take it over?

[tulip violet]

i got

x^2=18

[mental python]

yup

now how do you get rid of the ^2?